mm-2009 === Subject: Diff equation: g(x^2 , y)= d/dy g(x ,2y) , please a method ! Epigone-thread: flyswoigerm Is this kind of differential functional equation known? g:R*R->R . Does anyone know clearly built solving methods for such equations. Alain. === Subject: Re: Diff equation: g(x^2 , y)= d/dy g(x ,2y) , please a method ! >Is this kind of differential functional equation known? >g:R*R->R . >Does anyone know clearly built solving methods for such >equations. Of the usual suspects, one is a separation-of-variables solution g(x,y) = f(x) h(y), which leads to f(x^2) h(y) = 2 f(x) h'(2 y) and thus f(x^2)/f(x) = 2 h'(2 y)/h(y) = c for some constant c. The case c = 0 is rather special, leading to f(x) = 0 for x >= 0 (but arbitrary for x < 0) and h(y) = constant. Otherwise, we'll suppose c <> 0. Note that f(-x) = f(x) = f(x^2)/c, so it suffices to find f(x) for x >= 0. Since x^2 = x for x = 0 or 1, we have either c = 1 or f(0) = f(1) = 0. I assume you want f to be continuous on all of R. Note that f(x^(2^n)) = c^n f(x) for all integers n As n -> +infinity we have x^(2^n) -> 0 if |x| < 1, so either (i) f(x) = f(0) for |x| <= 1, and c = 1; (ii) f(x) = 0 for |x| <= 1, and |c| > 1; or (iii) |c| < 1. As n -> -infinity we have x^(2^n) -> 1 for 0 < x < infinity. Thus in case (i) we have f(x) = f(0) for all x, and in case (iii) we have f(x) = 0 for all x. The only remaining nontrivial case is (ii), where for x > 1 we can have f(x) = (ln (x))^k P(ln(ln(x))) where k = ln(|c|)/ln(2), and P(t + ln(2)) = sgn(c) P(t). For h'(2y) = (c/2) h(y), or h'(t) = c/2 h(t/2), we have series solutions h(t) = sum_{j=0}^infinity a_j t^j with a_{j+1} = 2^(-j-1) c/(j+1) a_j, and thus a_j = c^j a_0/(j! 2^(j(j+1)/2)). The series converges rapidly so h(t) is an entire function: I don't know a closed form, but its Laplace transform is related to Jacobi theta functions. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Paper published by Algebraic and Geometric Topology The following paper has been published: Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-46.abs.html Title: Span of the Jones polynomial of an alternating virtual link Author(s): Naoko Kamada Abstract: For an oriented virtual link, L.H. Kauffman defined the f-polynomial (Jones polynomial). The supporting genus of a virtual link diagram is the minimal genus of a surface in which the diagram can be embedded. In this paper we show that the span of the f-polynomial of an alternating virtual link L is determined by the number of crossings of any alternating diagram of L and the supporting genus of the diagram. It is a generalization of Kauffman-Murasugi-Thistlethwaite's theorem. We also prove a similar result for a virtual link diagram that is obtained from an alternating virtual link diagram by virtualizing one real crossing. As a consequence, such a diagram is not equivalent to a classical link diagram. Secondary: 57M27 Keywords: Virtual knot theory, knot theory Author(s) address(es): Department of Mathematics, Osaka City University, Sugimoto, Sumiyoshi-ku Osaka, 558-8585, Japan Email: naoko@sci.osaka-cu.ac.jp === Subject: Open problem Epigone-thread: chongprorglen Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let B(H) be the C*-algebra of all bounded linear operators on the complex Hilbert space H and let C_{1}(H) denote the class trace of B(H). For A, B in B(H), it is true that the supremum of 2abs(tr(AX)tr(BX))>=N(A)N(B) over the unit sphere of C_{1}(H)? where abs(a) denotes the absolute value of the complex number a and N(A) denotes the usual norm of A. === Subject: Q-forms of p-adic representations Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let g be a semisimple (finite dimensional) Lie algebra over Q_p. Is it true that g has a Q-form, and that, more generally, every representation of g over Q_p has a Q-form? -- Yves de Cornulier === Subject: Why exp(-st) in the Laplace Transform? Does anyone have an explaination why the kernel function exp(-st) was used in the definition of the Laplace transform? Is there a physical meaning to the use of this function? I know s is a complex frequency, how can we visualize what the Laplace transform is doing? --Eric Erpelding === Subject: Re: Why exp(-st) in the Laplace Transform? > Does anyone have an explaination why the kernel function exp(-st) was > used in the definition of the Laplace transform? > Is there a physical meaning to the use of this function? Here is a group-theoretic answer. The existence (and utility) of both the Laplace and Fourier transforms come from the symmetry of the real line under translations (t -> t + a, where a is some constant shift). This is a one-dimensional Lie group, so it is Abelian (commutative), and its unitary irreducible representations are therefore one-dimensional. These are the Fourier transform kernels, exp(-i w t), for w any real number. Under the translation operation (t -> t + a), these get multiplied by the (unitary) number exp(-i w a). In other words, when you take the Fourier transform of a well-behaved complex function, you are splitting it into functions which transform in a simple way under translations! This splits the infinite- dimensional space of functions over the real line (t) into an infinite number of one-dimensional subspaces (the Fourier kernels) which are dense (in nice functions), and which transform trivially (multiplication by a constant) under translations. If you draw the Fourier kernels as their graph in the complex plane versus t, they are constant-pitch helices. Moving a constant-pitch helix by a constant is equivalent to rotating it! If you take the Laplace transform, you are splitting a well-behaved function into its *real* irreducible representations: exponentials! Shifting an exponential (exp(-st)) by a constant (t -> t + a) is equivalent to multiplying it by a real number (exp(-sa)). Taking the Laplace transform is splitting the infinite-dimensional space of well-behaved functions into an infinite number of one-dimensional subspaces which are dense in well-behaved functions, and which each behave nicely under translations. So, both the Laplace and Fourier transforms are sort of splitting functions into the normal modes of the group of translations of the real line. === Subject: Re: Why exp(-st) in the Laplace Transform? Paul, I am not sure that I comprehend all of your commentary, however I do = note one of your statements If you take the Laplace transform, ... Shifting an exponential (exp(-st)) by a constant (t -> t + a) is equivalent to multiplying it by a real number (exp(-sa)). Your statement is not quite valid, in that, for the case of a Laplace = transform, the transform variable s has both real and imaginary = components, thus exp(-sa) has both real and imaginary components as = well. As in the Fourier transform, you would integrate over the = imaginary component. It is the real component, however, which = distinguishes the Laplace Transform from the Fourier Transform. That = is, the real component of our multiplier will tend to attenuate the = function. An example of the use of a Laplace transform would be to = transform the unit step function U(t). Note that it's Laplace Transform = is determined to be 1/s, under the restriction re(s) > 0. The Fourier = Transform of U(t) does not exist. Ed -- Edward Hyman EdwardH@email.uophx.edu Other EMail: e.hyman@worldnet.att.net > Does anyone have an explaination why the kernel function exp(-st) = was > used in the definition of the Laplace transform? > Is there a physical meaning to the use of this function? Here is a group-theoretic answer. The existence (and utility) of both the Laplace and Fourier transforms come from the symmetry of the real line under translations (t -> t + a, where a is some constant shift). This is a one-dimensional Lie group, so it is Abelian (commutative), and its unitary irreducible representations are therefore one-dimensional. These are the Fourier transform kernels, exp(-i w t), for w any real number. Under the translation operation (t -> t + a), these get multiplied by the (unitary) number exp(-i w a). In other words, when you take the Fourier transform of a well-behaved complex function, you are splitting it into functions which transform in a simple way under translations! This splits the infinite- dimensional space of functions over the real line (t) into an infinite number of one-dimensional subspaces (the Fourier kernels) which are dense (in nice functions), and which transform trivially (multiplication by a constant) under translations. If you draw the Fourier kernels as their graph in the complex plane versus t, they are constant-pitch helices. Moving a constant-pitch helix by a constant is equivalent to rotating it! If you take the Laplace transform, you are splitting a well-behaved function into its *real* irreducible representations: exponentials! Shifting an exponential (exp(-st)) by a constant (t -> t + a) is equivalent to multiplying it by a real number (exp(-sa)). Taking the Laplace transform is splitting the infinite-dimensional space of well-behaved functions into an infinite number of one-dimensional subspaces which are dense in well-behaved functions, and which each behave nicely under translations. So, both the Laplace and Fourier transforms are sort of splitting functions into the normal modes of the group of translations of the real line. === Subject: Re: Why exp(-st) in the Laplace Transform? <7c3wd.1100621$Gx4.435402@bgtnsc04-news.ops.worldnet.att.net Paul, > I am not sure that I comprehend all of your commentary, however I do > note one of your statements > If you take the Laplace transform, ... > Shifting an exponential (exp(-st)) by a constant (t -> t + a) is > equivalent to multiplying it by a real number (exp(-sa)). > Your statement is not quite valid, in that, for the case of a Laplace > transform, the transform variable s has both real and imaginary > components, thus exp(-sa) has both real and imaginary components as > well. Terminology issue. In many fields a Laplace transform means exclusively real-valued exponent. If it's complex than it's a complex Laplace transform or even generalized Laplace transform, though that's also used for other extensions, such as adding discrete portions. > As in the Fourier transform, you would integrate over the > imaginary component. It is the real component, however, which > distinguishes the Laplace Transform from the Fourier Transform. That > is, the real component of our multiplier will tend to attenuate the > function. An example of the use of a Laplace transform would be to > transform the unit step function U(t). Note that it's Laplace Transform > is determined to be 1/s, under the restriction re(s) > 0. The Fourier > Transform of U(t) does not exist. Well, that depends on what you mean by exist. It sure does, if you allow distributions, rather than just functions as Fourier transforms. Unsurprisingly, the non-singular portion is -i/k, or what you get for 1/s, when you replace s with (ik). The singular portion is just a delta distribution, caused by the (1/2) offset that is the even portion of the unit step. One way to do this is to take a series of function that converge to the unit step but whose Fourier transforms are functions -- e.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving the delta distribution noticed above. Write signum(x) as the limit a -> 0 of f_a(t) = e^{-at} for t > 0, -e^{at} for t < 0. The fourier transform of these will be -1/(a - ik) + 1/(a + ik) = (-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a -> 0 = 2/ki, but we need to divide by 2 to get the step function. Physicists often just use it via formal manipulation in integrals, which while not normally made rigorous, does give workable results, along the lines of: The unit step is just the integral of the delta function. Integrating a function in the real domain can be done by dividing the function in the Fourier domain by ik. Again this gives the non-singular portion as 1/ik. I have left out constant multiplicative factors, as they'll depend on one's convention. -- Aaron Denney -><- === Subject: Re: Why exp(-st) in the Laplace Transform? <7c3wd.1100621$Gx4.435402@bgtnsc04-news.ops.worldnet.att.net> Aaron, Your point is well made. Indeed, the Laplace transform of U(t) is determined by the real part of s, the transform variable corresponding to t. So, we would have Laplace transform of U(t) is 1/s for re(s) > 0 Fourier transform of U(t) is pi*delta(k) + 1/ik for the case re(s) = 0. s, in this case, being equal to ik. As you point out, each piece of the signum function which you define is Fourier transformable. Note that your selection of exp(-at) for the U(t) portion is, in essence, adding the positive real component a to the Fourier transform variable ik. That is, lim a->0 of Fourier transform of exp(-at)U(t) is identical to the limit re(s)->0 of Laplace transform of U(t), having re(s) > 0. A similar argument is made for the U(-t) portion. One could therefore argue that the Fourier transform of U(t) exists, since its Laplace transform exists in the limiting sense as re(s)->0. Ed -- Edward Hyman EdwardH@email.uophx.edu Other EMail: e.hyman@worldnet.att.net > Paul, > I am not sure that I comprehend all of your commentary, however I do > note one of your statements > If you take the Laplace transform, ... > Shifting an exponential (exp(-st)) by a constant (t -> t + a) is > equivalent to multiplying it by a real number (exp(-sa)). > Your statement is not quite valid, in that, for the case of a Laplace > transform, the transform variable s has both real and imaginary > components, thus exp(-sa) has both real and imaginary components as > well. Terminology issue. In many fields a Laplace transform means exclusively real-valued exponent. If it's complex than it's a complex Laplace transform or even generalized Laplace transform, though that's also used for other extensions, such as adding discrete portions. > As in the Fourier transform, you would integrate over the > imaginary component. It is the real component, however, which > distinguishes the Laplace Transform from the Fourier Transform. That > is, the real component of our multiplier will tend to attenuate the > function. An example of the use of a Laplace transform would be to > transform the unit step function U(t). Note that it's Laplace Transform > is determined to be 1/s, under the restriction re(s) > 0. The Fourier > Transform of U(t) does not exist. Well, that depends on what you mean by exist. It sure does, if you allow distributions, rather than just functions as Fourier transforms. Unsurprisingly, the non-singular portion is -i/k, or what you get for 1/s, when you replace s with (ik). The singular portion is just a delta distribution, caused by the (1/2) offset that is the even portion of the unit step. One way to do this is to take a series of function that converge to the unit step but whose Fourier transforms are functions -- e.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving the delta distribution noticed above. Write signum(x) as the limit a -> 0 of f_a(t) = e^{-at} for t > 0, -e^{at} for t < 0. The fourier transform of these will be -1/(a - ik) + 1/(a + ik) (-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a -> 0 = 2/ki, but we need to divide by 2 to get the step function. Physicists often just use it via formal manipulation in integrals, which while not normally made rigorous, does give workable results, along the lines of: The unit step is just the integral of the delta function. Integrating a function in the real domain can be done by dividing the function in the Fourier domain by ik. Again this gives the non-singular portion as 1/ik. I have left out constant multiplicative factors, as they'll depend on one's convention. -- Aaron Denney -><- === Subject: Re: Why exp(-st) in the Laplace Transform? B. Girod, R. Rabenstein, A. Stenger. Signals and Systems. Chichester: Wiley 2001 first introduce and visualize LT, then FT as a special case of it. Decide yourself whether or not this somewhat demanding approach is the pedagogically best one. Pierre-Simon, marquis de Laplace (1749-1825) first of all embodies determinism. The famous integral transform carries his name, but seems to have been used first by Denis Poisson (1781-1840) in 1815. Heaviside (1850-1925) was not aware of it when he nearly reinvented a slightly modified calculus with the operator pt =3D d/dt instead of st. Laplace di= d not trust in the work by Jean Baptiste Fourier (1768-1830) and rejected his famous paper which was nonetheless published with several years delay. Determinism is the wrong belief that anything can be predicted in advance by calculation. Correspondingly, one-sided LT assumes that anything starts at t=3D0. Pertaining signals are misleadingly called causal signals. The name deterministic signals would be more correct. One-sidedness of LT and the option to consider initial values make LT attractive for the first glance. In order to find a fan of LT you might Google for Kastrup. As far as I can judge, FT is much more in use than LT, mainly because of difficulties with inversion. In reality, all physical quantities are one-sided because future merely exists in imagination. LT is, however, not suited for performing a frequency analysis of such quantities only existing within past, not within future. Real-valued cosine transform is, in principle, adequate for that task. In other words: LT belongs to deterministic prediction of processes to come. CT belongs to causal analysis of the past. Causality is hidden in influences from what already happened. The laws of physics are symmetrical with respect to time. the caveat =91in principle=92 refer= s to time t in the kernel of LT. In case of CT in IR+, time has to be replaced by elapsed time. LT generally assumes an event-related time scale. CT makes sense with an observer-bound time-scale permanently sliding relatively to the event-related one. Having dealt with LT, Ernst Terhardt of MMK Munich copied the trick of making each finite function to converge, as does any real process. However, he finds fault with the widespread practice to start integration a little bit left from zero. In this case, definition of LT is not to blame for the fallacy. Instead, notion of real numbers should be revised. What about the minus sign of the imaginary part of s or omega in LT or FT, respectively, this is an arbitrarily chosen option, enforced by transformation into the inevitably redundant complex representation. (The only way out of that dilemma would be use of the observer-bound time-scale that has a natural zero to rely on.) Physicists still prefer the opposite sign of imaginary part. This discrepancy in sign does not matter because inverse transform returns the correct input. In order to bridge the gap between imaginable reality and visualization of complex representations like LT or FT, one might get aware of their half-reality in the sense the other half of reality is just economized. They are exhibiting just one out of two phasors, one of which rotates clockwise, the other one anticlockwise. Omission of the half picture paradoxically results in redundant doubling of data. Being a complex transform, at least the bilateral LT should possibly lead to ambiguous and even non-causal results, as does FT. Eckard Blumschein === Subject: Re: Why exp(-st) in the Laplace Transform? > Does anyone have an explaination why the kernel function exp(-st) was > used in the definition of the Laplace transform? > Is there a physical meaning to the use of this function? > I know s is a complex frequency, how can we visualize what the Laplace > transform is doing? > --Eric Erpelding Here's an economic interpretation of the LT. Let T be a positive random variable with density function f(). Then E[exp(-sT)] is the LT of f(). Suppose you will receive a dollar at time T. The expected present value with (real)continuous interest rate s is the expection as shown. This interpretation is useful in operations research. Dan Heyman === Subject: Re: Why exp(-st) in the Laplace Transform? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Does anyone have an explaination why the kernel function exp(-st) was >used in the definition of the Laplace transform? >Is there a physical meaning to the use of this function? >I know s is a complex frequency, how can we visualize what the Laplace >transform is doing? >--Eric Erpelding For all time derivative and time integral operations and all analytic functions of t, there is a transformed equivalent function LT of s that is self-consistent and algebraic. Derivatives are represented by s and integrals by 1/s, the unit step function 1/s and impulse by 1.For example if the transform is 1/(s + a) then 1/s(1/s+a) is the integral. The original functions become disguised, but after the desired algebraic operations, the whole thing can be unrolled by the inverse LT. The Laplace transform makes it possible to show block diagrams containing s factories able to cascade operations by simple multiplication. Accompanying this work should be a sturdy table of transforms and their inverses. Formal study of LT can be hairy, but for engineering work you can usually get by with the rudiments. The Laplace transform actually maps the function onto the s plane = sigma + j omega. If you limit yourself to sigma = 0, you stay on the frequency axis and get the Fourier transform. Is that any help? Mr. Dual Space If you have something to say, write an equation. If you have nothing to say, write an essay === Subject: Re: Why exp(-st) in the Laplace Transform? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Does anyone have an explaination why the kernel function exp(-st) was > used in the definition of the Laplace transform? > Is there a physical meaning to the use of this function? > I know s is a complex frequency, how can we visualize what the Laplace > transform is doing? The way I think of it, is that the Laplace transform is really a Fourier transform of a signal f(t) that has been secretly multiplied by an exponential exp(-at)) before it was Fourier transformed. This secrtet exponetial is related to s and w by s = a + iw By doing this, you can tame an exponentially growing signal before you apply Fourier to it. Gerard === Subject: Re: Why exp(-st) in the Laplace Transform? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >> Does anyone have an explaination why the kernel function exp(-st) was >> used in the definition of the Laplace transform? Instead of trying to find reasons applicable today, we should be looking at what was the situation at the time of Laplace, the late 18th and early 19th century. Laplace found it a useful tool in handling functions on the positive reals, and in solving differential equations, and other processes. At that time, fiddling around to find ways to get answers to what are now routine problems was very common, and much was discovered in this manner. >> Is there a physical meaning to the use of this function? It can be given one; the discrete version is that of a generating function. For positive s, which is the way it was used most of the time, it converts functions on the positive reals to smooth functions which can sometimes be better analyzed. Its derivates can be used to get moments, and using the Gamma function, its integrals can be used to obtain negative moments. >> I know s is a complex frequency, how can we visualize what the Laplace >> transform is doing? I suggest that this is not the way to look at it. Laplace precedes Fourier in this matter. There are NOW many other uses for the Laplace and bilateral Laplace transforms, and that one corresponds to imaginary values of the other can be used both ways. In explaining its numerical uses, I find the Laplace version easier, using complex values of s. It is quite useful in this way. >The way I think of it, is that the Laplace transform is really >a Fourier transform of a signal f(t) that has been secretly >multiplied by an exponential exp(-at)) before it was Fourier >transformed. >This secrtet exponetial is related to s and w by > s = a + iw >By doing this, you can tame an exponentially growing signal >before you apply Fourier to it. For probability distributions, this approach is very useful, and is used for large deviation results, reducing them to ordinary deviation results with an exponential factor, and another term which is somewhat tricky. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Why exp(-st) in the Laplace Transform? > Does anyone have an explaination why the kernel function exp(-st) was > used in the definition of the Laplace transform? > Is there a physical meaning to the use of this function? > I know s is a complex frequency, how can we visualize what the Laplace > transform is doing? Be s = sigma + j omega . With sigma > 0 , the factor exp(-sigma t) makes each finite function of time converge. Thus, you can integrate up to each time value. You get a quite exact image (and the exact origin at back- transformation). Ulrich Bruchholz === Subject: Re: Why exp(-st) in the Laplace Transform? > Does anyone have an explaination why the kernel function exp(-st) was > used in the definition of the Laplace transform? > Is there a physical meaning to the use of this function? > I know s is a complex frequency, how can we visualize what the Laplace > transform is doing? > --Eric Erpelding Write s=u+iv, then you have Fourier transform (dual variable v) of the damped function exp(-ut)f(t), depending on the real parameter u. It is assumed that f(t) vanishes for all t<0. And the physical, or signal-processing meaning of Fourier transform is well-known. Hope it helps, ZVK(Slavek). === Subject: Re: Why exp(-st) in the Laplace Transform? > Does anyone have an explaination why the kernel function exp(-st) was > used in the definition of the Laplace transform? > Is there a physical meaning to the use of this function? In a nutshell, the exp(-st) comes from the fact that convolution integrals can be written as integrals over an exp(t.d/dx) operator. But you should read the whole story (= heuristics): Especially read the subsection Laplace and Statistics. Hope it helps in providing some background understanding. > I know s is a complex frequency, how can we visualize what the Laplace > transform is doing? If s is complex - but it doesn't have to be - then what you probably want is a visualization of the complex Fourier transform. I don't have it at hand, though. Han de Bruijn === Subject: Maximal number of distinct factors Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Given a number n, is there a tight upper bound on the number of its distinct factors (need NOT be prime), d(n)? Specifically, is d(n) = O(logn)? abey Notes: 1. The average number of distinct prime factors, omega(n), is known to be ~loglogn [see http://mathworld.wolfram.com/DistinctPrimeFactors.html ]. So average d(n) is probably theta(logn). 2. When n is power of a prime number, then d(n) is trivally theta(logn). 3. When n is square-free i.e. it is a product of distinct primes, average omega(n) is known to be ~ logn/(loglogn) [see url above]. If this is also an upper bound, then maximal d(n) is ~ n^(1/loglogn). 4. I expect that maximal d(n) would occur for a distribution of prime factors of n that is a combination of the cases 2 and 3 above. === Subject: Re: Maximal number of distinct factors Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Given a number n, is there a tight upper bound on the number of its > distinct factors (need NOT be prime), d(n)? > Specifically, is d(n) = O(logn)? No. See Hardy & Wright. In the 5-th edition: Theorem 314 states that d(n) is not O(log(n)^k) for any k. Theorem 315 states that d(n) is O(n^k) for any k >0. Theorem 317 states that limsup (log d(n) log log n/log n) = log 2, and so for k > 0 d(n) < 2^{(1+k)log n/log log n} for all large n, and d(n) > 2^{(1-k)log n/log log n} infinitely many n. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Recursive relation Epigone-thread: skanslaslin Originator: bergv@math.uiuc.edu (Maarten Bergvelt) To all who read my former reply, I apologize. My reading of the === Subject: Re: Recursive relation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I appreciate your time and effort. Regretfully though, both solutions seem to be incorrect. Incorrectness of the second solution can be demonstrated by substitution of initial conditions, namely, C_{00} = D_{00} = 1 and all parameters with negative subindex equals 0. Such substitution gives u=v=1, which means that C_{n,k} = D_{n,k}, which is clearly incorrect. Discrete mathematics is not my field. The problem formulation seems to be so simple and so classical, that it seems to me that someone should have encountered a similar problem before. Any ideas or references? Alex === Subject: This Week's Finds in Mathematical Physics (Week 209) Originator: baez@math-cl-n03.math.ucr.edu (John Baez) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Also available at http://math.ucr.edu/home/baez/week209.html This Week's Finds in Mathematical Physics - Week 209 John Baez Time flies! This June, Peter May and I organized a workshop on n-categories at the Institute for Mathematics and its Applications: 1) n-Categories: Foundations and Applications, http://www.ima.umn.edu/categories/ I've been meaning to write about it ever since, but I keep putting it off because it would be so much work. The meeting lasted almost two weeks. It was an intense, exhausting affair packed with talks, conversations, and Russian-style seminars where the audience interrupted the speakers with lots of questions. I took about 50 pages of notes. How am I supposed to describe all that?! Oh well... I'll just dive in. I'll quickly list all the official talks in this conference. I won't describe the many interesting impromptu talks, some of which you can see on the above webpage. Nor will I explain what n-categories are, or what they're good for! If you want to learn what they're good for, you should go back to week73 and read The Tale of n-Categories. And if you want to know what they *are*, try this brand-new book: 2) Eugenia Cheng and Aaron Lauda, Higher-Dimensional Categories: an Illustrated Guide Book, available free online at: http://www.dpmms.cam.ac.uk/~elgc2/guidebook/ packed with pictures and it's lots of fun. I'm just going to list the talks.... Throwing etiquette to the winds, I kicked off the conference myself with two talks explaining some reasons why n-categories are interesting and what they should be like: 3) John Baez, Why n-Categories? and What n-categories should be like. Notes available at http://www.ima.umn.edu/categories/#mon If you're a long-time reader of This Week's Finds you'll know what I said: n-categories give a new world of math in which equations are always replaced by isomorphisms, and this world is incredibly rich in structure. The n-categories called n-groupoids magically know everything there is to know about homotopy theory, while those called n-categories with duals know everything there is to know about the topology of manifolds. There are, unfortunately, some details that still need to be worked out! After my talks there was a reception. Later, over dinner, Tom Leinster gave a Russian style seminar outlining the different approaches to n-categories: 4) Tom Leinster, Survey and Taxonomy. Talk based on chapter 10 of his book Higher Operads, Higher Categories, Cambridge U. Press, You'll notice these young n-category people are smart: they force their publishers to keep their books available for free online! All scientists should do this, since the only ones who make serious money from scientific monographs are the publishers. What scientists get from writing technical books is not money but attention. As George Franck said, Attention is a mode of payment... reputation is the asset into which the attention received from colleagues crystallizes. The next morning began with a triple-header talk on weak categories: 5) Andre Joyal, Peter May and Timothy Porter, Weak categories. Notes available at http://www.ima.umn.edu/categories/#tues Here a weak category means a category where the usual laws hold only up to homotopy, where the homotopies satisfy laws of their own up to homotopy, ad infinitum. If you know what weak infinity-categories are, you can define a weak category to be one of these where all the j-morphisms are equivalences for j > 1. But, the nice thing is that there are ways to define weak categories without the full machinery of infinity-categories! People have come up with different approaches: categories enriched over simplicial sets, Segal categories, A_infinity categories and also Joyal's quasicategories. The talk was a nice introduction to all these approaches. Then Michael Batanin explained his definition of infinity-categories. This was a blackboard talk, so there are no notes on the web, but you can try his original paper: 6) Michael Batanin, Monoidal globular categories as natural environment for the theory of weak n-categories, Adv. Math. 136 (1998), 39-103, also available at http://www.ics.mq.edu.au/~mbatanin/papers.html and when you get stuck, try the books by Cheng-Lauda and Leinster. Over dinner, Eugenia Cheng and Tom Leinster explained the concepts of operad and multicategory which play such an important role in so much work on n-categories. Again there are no notes, so try their books. I forget when it happened, but sometime around the second or third day of the conference people decided it was too much of a nuisance listening to math lectures while eating dinner - mainly because there wasn't enough room in the dining hall to take notes, and the blackboards weren't big enough. So at that point, we switched to having lectures *after* dinner. As I said, this workshop was not for wimps! The morning of the third day began with a no-holds-barred minicourse on model categories by Peter May: 7) Peter May, Model categories. Notes available at http://www.ima.umn.edu/categories/#wed Model categories are a wonderful framework for relating different approaches to homotopy theory, and a bunch of people hope they can also be used to relate different approaches to n-categories. Then Clemens Berger explained Andre Joyal's approach to weak n-categories: 8) Clemens Berger, Cellular definitions. Notes available at http://www.ima.umn.edu/categories/#wed Then, either during or after dinner, Eugenia Cheng explained various opetopic approaches to weak n-categories. Again, the best way to learn Leinster. On the morning of the fourth day, Andre Joyal explained his work on quasicategories - an approach to weak categories in which they are simplicial sets satisfying a restricted version of the Kan condition. They've been around a long time, but Joyal is redoing all of category theory in this context! He's been writing a book about this, which deserves to be called Quasicategories for the Working Mathematician. Since Joyal is a perfectionist, this will take forever to finish. However, we're hoping to extract a preliminary version from him for the proceedings of this conference. For now, you can read a bit about quasicategories in Tim Porter's notes mentioned in item 5) above. Then Tom Leinster and Nick Gurski spoke about Ross Street's definition to weak infinity-categories, where they are simplicial sets satisfying an even more subtly restricted version of the Kan condition. 9) Nick Gurski and Tom Leinster, Simplicial definition. Notes available at http://www.ima.umn.edu/categories/#thur Street's definition is tough to understand at first, but it should eventually include Joyal's quasicategories as a special case, which is nice. For Street's own discussion, see: 10) Ross Street, Weak omega-categories, in Diagrammatic Morphisms and Applications, eds. David Radford, Fernando Souza, and David Yetter, Also available as www.maths.mq.edu.au/~street/Womcats.pdf It relies on some work by Dominic Verity which has finally been written up after many years of unpublished limbo: 11) Dominic Verity, Complicial sets, available as math.CT/0410412. After dinner we took a turn towards applications, and Larry Breen explained his work on n-stacks and n-gerbes. An n-stack is like a sheaf that has an (n-1)-category of sections, while an n-gerbe has an (n-1)-groupoid of sections. Such things show up a lot in algebraic geometry, and more recently in mathematical physics inspired by string theory. Alas, the audience was rather tired this evening, so Larry only got to 1-stacks and 1-gerbes! But he gave an impromptu talk later where he reached n = 2, and the notes for both talks are available in combined form here: 12) Larry Breen, n-Stacks and n-gerbes: homotopy theory. Notes available at http://www.ima.umn.edu/categories/#thur You've heard about David Corfield's quest for a philosophy of real mathematics in week198. He's one of the few philosophers who understands enough math to realize how cool n-categories are - which may explain why he's having trouble getting a job. On the morning of the fourth day, he gave a talk on the impact n-categories could have in philosophy: 13) David Corfield, n-Category theory as a catalyst for change in philosophy. Notes available at http://www.ima.umn.edu/categories/#fri Later that day, Bertrand Toen explained Segal categories, which are another popular approach to weak categories: 14) Bertrand Toen, Segal categories. Notes by Joachim Kock available at http://www.ima.umn.edu/categories/#fri After dinner, he spoke about n-stacks and n-gerbes: 15) Bertrand Toen, n-Stacks and n-gerbes: algebraic geometry. Notes by Joachim Kock available at http://www.ima.umn.edu/categories/#fri Everyone slept all weekend long. Then on Monday of the second week, the homotopy theorist Zbigniew Fiedorowicz spoke about his work on a kind of n-fold monoidal category that has an n-fold loop space as its nerve. He has some good papers on the web about this, too: 16) Zbigniew Fiedorowicz, n-Fold categories. Notes available at http://www.ima.umn.edu/categories/#mon2 C. Balteanu, Z. Fiedorowicz, R. Schwaenzl and R. Vogt, Iterated monoidal categories, available at math.AT/9808082. Z. Fiedorowicz, Constructions of E_n operads, available at math.AT/9808089. Stefan Forcey continued this theme by discussing enrichment over n-fold monoidal categories. He also has a number of papers about this on the arXiv, of which I'll just mention one: 17) Stefan Forcey, Higher enrichment: n-fold Operads and enriched n-categories, delooping and weakening. Notes available at http://www.ima.umn.edu/categories/#mon2 Stefan Forcey, Enrichment over iterated monoidal categories, at http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-7.abs.html Also available as math.CT/0403152. After dinner we discussed how to relate different definitions of weak n-category. On Tuesday of the second week, the logician Michael Makkai presented his astounding project of redoing logic in a way that completely eliminates the concept of equality. This *forces* you to do all of mathematics using weak infinity-categories. I thought this stuff was great, in part because I finally understood it, and in part because it leads naturally to the opetopic definition of n-categories that James Dolan and I introduced. The idea of eliminating equality was very much on our mind in inventing this definition, but we didn't create a system of logic that systematizes this idea. There are no notes for Makkai's talk online, but you can get a lot of good stuff from his website, including: 18) Michael Makkai, On comparing definitions of weak n-category, available at http://www.math.mcgill.ca/makkai/ and this more technical paper which works out the details of his vision: 19) Michael Makkai, The multitopic omega-category of all multitopic omega-categories, available at http://www.math.mcgill.ca/makkai/ After Makkai's talk, Mark Weber spoke on n-categorical generalizations of the concept of monad, which is a nice way of describing mathematical gadgets. There are no notes for this talk, but his work on higher operads is at least morally related: 20) Mark Weber, Operads within monoidal pseudo algebras, available as math.CT/0410230. Again, after dinner we talked about how to relate different definitions of weak n-category. On Wednesday of the second week, Michael Batanin spoke about his recent work relating n-categories to n-fold loop spaces. Again no notes, but you can read these papers: 21) Michael Batanin, The Eckmann-Hilton argument, higher operads and E_n-spaces, available at http://www.ics.mq.edu.au/~mbatanin/papers.html Michael Batanin, The combinatorics of iterated loop spaces, available at http://www.ics.mq.edu.au/~mbatanin/papers.html Then Joachim Kock laid the ground for a discussion of n-categories and topological quantum field theories, or TQFTs, by explaining the definition of a TQFT and the classification of 2d TQFTs: 22) Joachim Kock, Topological quantum field theory primer. Notes available at http://www.ima.umn.edu/categories/#wed2 In the evening, Marco Mackaay and I said more about the relation between TQFTs and n-categories: 23) Marco Mackaay, Topological quantum field theories. Notes available at http://www.ima.umn.edu/categories/#wed2 24) John Baez, Space and state, spacetime and process. Notes available at http://www.ima.umn.edu/categories/#wed2 On Thursday, Ross Street started the day in a pleasantly different way - he gave a historical account of work on categories and n-categories in Australia! Australia is home to much of the best work on these subjects, so if you can understand his history you'll wind up understanding these subjects pretty well: 25) Ross Street, An Australian conspectus of higher category theory. Notes available at http://www.ima.umn.edu/categories/#thur2 As a younger exponent of the Australian tradition, it was then nicely appropriate for Steve Lack to speak about ways of building a model category of 2-categories: 26) Steve Lack, Higher model categories. Notes available at http://www.ima.umn.edu/categories/#thur2 In the afternoon we had a blast of computer science. First John Power gave a hilarious talk phrased in terms of how one should convince computer theorists to embrace categories, then 2-categories, and then maybe higher categories: 27) John Power, Why tricategories? Notes available at http://www.ima.umn.edu/categories/#thur2 I spoke about Power's paper with this title back in week53; now you can get it online! Then Phillipe Gaucher, Lisbeth Falstrup and Eric Goubault spoke about higher-dimensional automata and directed homotopy theory: 28) Phillipe Gaucher, Towards a homotopy theory of higher dimensional automata. Notes available at http://www.ima.umn.edu/categories/#thur2 Lisbeth Falstrup, More on directed topology and concurrency, Notes available at http://www.ima.umn.edu/categories/#thur2 Eric Goubault, Directed homotopy theory and higher-dimensional automata, Notes available at http://www.ima.umn.edu/categories/#thur2 On Friday, Martin Hyland and Tony Elmendorf gave a double-header talk on higher-dimensional linear algebra and how some concepts in this subject can be simplified using symmetric multicategories. There are, alas, no notes for this talk. You just had to be there. Finally, my student Alissa Crans gave a talk on higher-dimensional linear algebra, with an emphasis on categorified Lie algebras: 29) Alissa Crans, Higher linear algebra. Notes available at Notes available at http://www.ima.umn.edu/categories/#fri2 Hers was the last talk in the workshop! I would like to say more about it, but I'm exhausted... and her talk fits naturally into a discussion of higher gauge theory, which deserves a Week of its own. By the way, you can see pictures of this workshop here: 30) John Baez, IMA, http://math.ucr.edu/home/baez/IMA/ If you want to see what these crazy n-category people look like, you can see most of them here. Hmm. If you wanted me to actually *explain* something this week, I'm afraid you'll be rather disappointed - so far everything has just been pointers to other material. of some material on Picard groups and Brauer groups. There's a Spanish school of higher-dimensional algebra, centered in Granada, and this spring Aurora del Rio Cabeza came from Granada to visit UCR. She and James Dolan spent a lot of time talking about categorical groups (also known as 2-groups) and cohomology theory. I was, alas, too busy to keep up with their conversations, but I learned a little from listening in... and here's my writeup! Higher categories show up quite naturally in the study of commutative rings and associative algebras over commutative rings. I'd heard of things called Brauer groups and Picard groups of rings, and something called Morita equivalence, but I only understood how these fit together when I learned they were part of a marvelous thing: a weak 3-groupoid! Here's how it goes. You don't need to know much about higher categories for this to make some sense... at least, I hope not. Starting with a commutative ring R, we can form a weak 2-category Alg(R) where: an object A is an associative algebra over R a 1-morphism M: A -> B is an (A,B)-bimodule a 2-morphism f: M -> N is a homomorphism between (A,B)-bimodules. This has all the structure you need to get a 2-category. In particular, we can compose an (A,B)-bimodule and a (B,C)-bimodule by tensoring them over B, getting an (A,C) bimodule. But since tensor products are only associative up to isomorphism, we only get a *weak* 2-category, not a strict one. This weak 2-category has a tensor product, since we can tensor two associative algebras over R and get another one. All the stuff listed above gets along with this process! When an n-category has a well-behaved tensor product we call it monoidal, so Alg(R) is a weak monoidal 2-category. But using a standard trick we can reinterpret this as a weak 3-category with one object, as follows: there's only one object, R a 1-morphism A: R -> R is an associative algebra over R a 2-morphism M: A -> B is an (A,B)-bimodule a 3-morphism f: M -> N is a homomorphism between (A,B)-bimodules. Note how all the morphisms have shifted up a notch. What used to be called objects, the associative algebras over R, are now called 1-morphisms. We compose them by tensoring them over R. Next, recall a bit of n-category theory from week35. In an n-category we define a j-morphism to be an equivalence iff it's invertible... up to equivalence! This definition may sound circular, but really just recursive. To start it off we just need to add that an n-morphism is an equivalence iff it's invertible. What does equivalence amount to in the 3-category Alg(R)? It's easiest to figure this out from the top down: A 3-morphism f: M -> N is an equivalence iff it's invertible, so it's an isomorphism between (A,B)-bimodules. A 2-morphism M: A -> B is an equivalence iff it's invertible up to isomorphism, meaning there exists N: B -> A such that: M tensor_B N is isomorphic to A as an (A,A)-bimodule, N tensor_A M is isomorphic to B as a (B,B)-bimodule. In this situation people say M is a Morita equivalence from A to B. A 1-morphism A: R -> R is an equivalence iff it's invertible up to Morita equivalence, meaning there exists a 1-morphism B: x -> x such that: A tensor_R B is Morita equivalent to R as an associative algebra over R, B tensor_R A is Morita equivalent to R as an associative algebra over R. In this situation people say A is an Azumaya algebra. Here's a nice example of how Morita equivalence works. Over any commutative ring R there's an algebra R[n] consisting of n x n matrices with entries in R. R[n] isn't usually isomorphic to R[m], but they're always Morita equivalent! To see this, suppose M: R[n] -> R[m] is the space of n x m matrices with entries in R, N: R[m] -> R[n] is the space of m x n matrices with entries in R. These become bimodules in an obvious way via matrix multiplication, and a little calculation shows that they're inverses up to isomorphism! So, all the algebras R[n] are Morita equivalent. In particular this means that they're all Morita equivalent to R, so they are Azumaya algebras of a rather trivial sort. If we take R to be real numbers there is also a more interesting Azumaya algebra over R, namely the quaternions H. This follows from the fact that H tensor_R H = R[4] This says H tensor_R H is Morita equivalent to R as an associative algebra over R, which implies (by the definition above) that H is an Azumaya algebra. Morita equivalence is really important in the theory of C*-algebras, Clifford algebras, and things like that. Someday I want to explain how it's connected to Bott periodicity. Oh, there's so much I want to explain.... But right now I want to take our 3-category Alg(R), massage it a bit, and turn it into a topological space! Then I'll look at the homotopy groups of this space and see what they have to say about our ring R. To do this, we need a bit more n-category theory. A weak n-category where all the 1-morphisms, 2-morphisms and so on are equivalences is called a n-groupoid. For example, given any weak n-category, we can form a weak n-groupoid called its core by throwing out all the morphisms that aren't equivalences. So, let's take the core of Alg(R) and get a weak 3-groupoid. Here's what it's like: there's one object, R the 1-morphisms A: x -> x are Azumaya algebras over R the 2-morphisms M: A -> B are Morita equivalences the 3-morphisms f: M -> N are bimodule isomorphisms. Next, given a weak n-groupoid with one object, it's very nice to compute its homotopy groups. These are easy to define in general, but I'll just do it for the core of Alg(R) and let you guess the general pattern. First, notice that: the identity 1-morphism 1_R: R -> R is just R, regarded as an associative algebra over itself in the obvious way the identity 2-morphism 1_{1_R}: 1_R -> 1_R is just R, regarded as an (R,R)-bimodule in the obvious way the identity 3-morphism 1_{1_{1_R}}: 1_{1_R} -> 1_{1_R} is just the identity function on R, regarded as an isomorphism of (R,R)-bimodules. At this point we let out a cackle of n-categorical glee. Then, we define the homotopy groups of the core of Alg(R) as follows: the 1st homotopy group consists of equivalence classes of 1-morphisms from R to itself the 2nd homotopy group consists of equivalence classes of 2-morphisms from 1_R to itself the 3rd homotopy group consists of equivalence classes of 3-morphisms from 1_{1_R} to itself Here we say two morphisms in an n-category are equivalent if there is an equivalence from one to the other (or if they're equal, in the case of n-morphisms). I hope the pattern in this definition of homotopy groups is obvious. In fact, n-groupoids are secretly the same - in a subtle sense I'd rather not explain - as spaces whose homotopy groups vanish above dimension n. Using this, the homotopy groups as defined above turn out to be same as the homotopy groups of a certain space associated with the ring R! So, we're doing something very funny: we're using algebraic topology to study algebra. But, we don't need to know this to figure out what these homotopy groups are like. Unraveling the definitions a bit, one sees they amount to this: The 1st homotopy group consists of Morita equivalence classes of Azumaya algebras over R. This is also called the BRAUER GROUP of R. The 2nd homotopy group consists of isomorphism classes of Morita equivalences from R to R. This is also called the PICARD GROUP of R. The 3rd homotopy group consists of invertible elements of R. This is also called the UNIT GROUP of R. People had been quite happily studying these groups for a long time without knowing they were the homotopy groups of the core of a weak 3-category associated to the commutative ring R! But, the relationships between these groups are easier to explain if you use the n-categorical picture. It's a great example of how n-categories unify mathematics. For example, everything we've done is functorial. So, if you have a homomorphism between commuative rings, say f: R -> S then you get a weak 3-functor Alg(f): Alg(R) -> Alg(S) This gives a weak 3-functor from the core of Alg(R) to the core of Alg(S), and thus a map between spaces... which in turn gives a long exact sequence of homotopy groups! So, we get interesting maps going from the unit, Picard and groups of R to those of S - and these fit into an interesting long exact sequence. For more, try the following papers. The first paper is actually about a generalization of Azumaya algebras called Azumaya categories, but it starts with a nice quick review of Azumaya algebras and Brauer groups: 31) Francis Borceux and Enrico Vitale, Azumaya categories, available at http://www.math.ucl.ac.be/AGEL/Azumaya_categories.pdf Category theorists will enjoy the generalization: since algebras are just one-object categories enriched over Vect, the concept of Azumaya algebra really *wants* to generalize to that of an Azumaya category. I'm sure most of the Brauer-Picard-Morita stuff generalizes too, but I haven't checked that out yet. This second paper makes the connection between Picard and Brauer groups explicit using categorical groups: 32) Enrico Vitale, A Picard-Brauer exact sequence of categorical groups, Journal of Pure and Applied Algebra 175 (2002) 383-408. Also available as http://www.math.ucl.ac.be/membres/vitale/cat-gruppi2.pdf ----------------------------------------------------------------------- mathematics and physics, as well as some of my research papers, can be obtained at http://math.ucr.edu/home/baez/ For a table of contents of all the issues of This Week's Finds, try http://math.ucr.edu/home/baez/twf.html A simple jumping-off point to the old issues is available at http://math.ucr.edu/home/baez/twfshort.html If you just want the latest issue, go to http://math.ucr.edu/home/baez/this.week.html === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Content-Length: 2443 Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Higher categories show up quite naturally in the study of > commutative rings and associative algebras over commutative rings. > I'd heard of things called Brauer groups and Picard groups > of rings, and something called Morita equivalence, but I only > understood how these fit together when I learned they were part > of a marvelous thing: a weak 3-groupoid! Be careful. You're secretly doing some string theory here. > Here's how it goes. You don't need to know much about higher > categories for this to make some sense... at least, I hope not. > Starting with a commutative ring R, we can form a weak 2-category > Alg(R) where: > an object A is an associative algebra over R Or a string boundary state. > a 1-morphism M: A -> B is an (A,B)-bimodule An open string stretching between two boundaries. > a 2-morphism f: M -> N is a homomorphism between (A,B)-bimodules. Beats me on this one. > This has all the structure you need to get a 2-category. In particular, > we can compose an (A,B)-bimodule and a (B,C)-bimodule by tensoring them > over B, getting an (A,C) bimodule. But since tensor products are only > associative up to isomorphism, we only get a *weak* 2-category, not a > strict one. Sounds like an A_oo type structure to me, just like open strings. Anyways, the connection here is this. Noncommutative geometry is a decoupling limit of open string theory. So, let's let our string boundary state be a noncommutative torus, certainly an example of an associative algebra. A simple example is the noncommutative T^2 with parameter theta. We can ask when two noncommutative tori are Morita equivalent. For T^2, the answer is when theta' = (a theta + b) / (c theta + d) ad - bc = 1 This is just T-duality in string theory. The duality groups work out to be the same for higher dimensional torii, too. Azumaya algebras also show up in strings when you have a torsion three form floating around. (Fun thing I learned recently: the A_oo structure of the Yoneda pairing of Ext groups -- which I don't really understand as yet -- actually encodes topological string amplitudes of a sort. It should be thought of as the cohomology of the A_oo structure (with nontrivial m_1) associated with open strings. This isn't new -- I was just happy to understand it somewhat.) Aaron === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Originator: baez@math-cl-n03.math.ucr.edu (John Baez) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Anyways, the connection here is this. Noncommutative geometry is a >decoupling limit of open string theory. Let's see if I can understand that. I think a conformal field theory should assign associative algebras A to 0-manifolds (A,B)-bimodules to 1-manifolds with boundary (A,B)-bimodule homomorphisms to 2-manifolds with corners where our 2-manifolds should also be equipped with complex structure. The whole setup should be a (projective) symmetric monoidal 2-functor. In particular, disjoint unions should get mapped to tensor products. In particular, the algebra assigned to be a bunch of points should be a tensor product of copies of some given algebra A. This is the algebra of string boundary states: >So, let's let our string >boundary state be a noncommutative torus, certainly an example of an >associative algebra. I don't get the role the decoupling limit plays here. Maybe you're saying that in some limit the algebra becomes commutative, and we're *almost* but not quite taking that limit, so we're getting a noncommutative deformation of some commutative algebra, as folks do in deformation quantization? >A simple example is the noncommutative T^2 with >parameter theta. We can ask when two noncommutative tori are Morita >equivalent. For T^2, the answer is when >theta' = (a theta + b) / (c theta + d) ad - bc = 1 Cool! >This is just T-duality in string theory. The duality groups work out to >be the same for higher dimensional torii, too. Here duality groups means Brauer groups, i.e. groups whose elements are Azumaya algebras mod Morita equivalence? >(Fun thing I learned recently: the A_oo structure of the Yoneda pairing >of Ext groups -- which I don't really understand as yet -- actually >encodes topological string amplitudes of a sort. It should be thought of >as the cohomology of the A_oo structure (with nontrivial m_1) associated >with open strings. This isn't new -- I was just happy to understand it >somewhat.) I'd be happy to understand it more. Sounds like there's some homological algebra I haven't gotten around to yet: the A_infinity structure of the Yoneda pairing of Ext groups! I know what an Ext group is, but not their Yoneda pairings... A_infinity stuff is no problem for me but I don't know what an A_infinity structure on a pairing would be.... === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Anyways, the connection here is this. Noncommutative geometry is a >decoupling limit of open string theory. > Let's see if I can understand that. I think a conformal field > theory should assign > associative algebras A to 0-manifolds > (A,B)-bimodules to 1-manifolds with boundary > (A,B)-bimodule homomorphisms to 2-manifolds with corners > where our 2-manifolds should also be equipped with complex structure. > The whole setup should be a (projective) symmetric monoidal 2-functor. > In particular, disjoint unions should get mapped to tensor products. > In particular, the algebra assigned to be a bunch of points should > be a tensor product of copies of some given algebra A. This is the > algebra of string boundary states: Hm, this is like identifying the CFT with a functor from some 2-category of paths and bigons to the above 2-category of (A,B). This is at least not what I thought Aaron was talking about, and I am also not sure why it should be true. To me, it is not a given slice of a given worldsheet of the open string wich gives the (A,B)-bimodule, but instead the mere existence of any open string stretching between a brane configuration described by A and one described by B. No? That's why I thought the (A,B)-homomorphisms must map different such suggests that the homomorphism are associated with patches of worldsheet. === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >So, let's let our string >boundary state be a noncommutative torus, certainly an example of an >associative algebra. > I don't get the role the decoupling limit plays here. Maybe > you're saying that in some limit the algebra becomes commutative, > and we're *almost* but not quite taking that limit, so we're getting > a noncommutative deformation of some commutative algebra, as folks > do in deformation quantization? Well, we can look at the string worldsheet action: int G_ab + a X^*(B) + ... where < , > is the pairing with the worldsheet metric, D is the derivative on the worldsheet, X is the worldsheet field mapping into spacetime and B is a two form on spacetime. Now, take a limit of this action such that the metric term drops out. We're left with just the pullback of the B field. Assume B is a symplectic form and this is exactly the Cattaneo-Felder sigma model for deformation quantization in the symplectic case. So, if you take the correct limit -- the Seiberg-Witten limit -- you get, living on the D-brane as these are open strings, a deformation quantization. >A simple example is the noncommutative T^2 with >parameter theta. We can ask when two noncommutative tori are Morita >equivalent. For T^2, the answer is when >theta' = (a theta + b) / (c theta + d) ad - bc = 1 > Cool! >This is just T-duality in string theory. The duality groups work out to >be the same for higher dimensional torii, too. > Here duality groups means Brauer groups, i.e. groups whose elements > are Azumaya algebras mod Morita equivalence? When I hear Azumaya algebras, I think, take a torsion three form, define a SU(N)/Z_n bundle associated with that form and take the algebra of sections. As for what I mean by duality group, for the purposes of math, I mean whatever's in this: math.QA/9803057. >(Fun thing I learned recently: the A_oo structure of the Yoneda pairing >of Ext groups -- which I don't really understand as yet -- actually >encodes topological string amplitudes of a sort. It should be thought of >as the cohomology of the A_oo structure (with nontrivial m_1) associated >with open strings. This isn't new -- I was just happy to understand it >somewhat.) > I'd be happy to understand it more. Sounds like there's some homological > algebra I haven't gotten around to yet: the A_infinity structure of the > Yoneda pairing of Ext groups! I know what an Ext group is, but not their > Yoneda pairings... A_infinity stuff is no problem for me but I don't know > what an A_infinity structure on a pairing would be.... The Yoneda pairing is defined in Griffiths and Harris. All I know about the A_oo structure is that m_1 vanishes so the pairing is associative. I was trying to read a few other things that purported to give more detail, but I decided there were more important things I needed to understand first. Here's a quote from an intro by B. Keller: > Yoneda algebras. Let B be a unital associative algebra, M a (right) > Bmodule and P -> === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Content-Length: 1308 Originator: rusin@vesuvius Aaron Bergman schrieb im Newsbeitrag > The Yoneda pairing is defined in Griffiths and Harris. All I know about > the A_oo structure is that m_1 vanishes so the pairing is associative. Since the associator is [m_2, m_2] = [m_1,m_3] + [m_4,m_0] (in terms of Gerstenhaber brackets, maybe up to a sign here and there) it seems that the A_oo structure that you are talking about is either 2-term, so that m_i vanishes for all i > 3, or standard, so that [m_i,m_0] vanishes? > Sounds like an A_oo type structure to me, just like open strings. necessarily imply that the A_oo algebra associated with it is related to that enocuntered in string field theory. If the (A,B)-bimodules that we are talking about describe brane configurations (as I think it should) there might be some relation, but its a little vague so far. I find it interesting that in a somewhat similar context, where the non-associativity is that in weak 2-groups, the A_oo algebra (or L_oo rather) that appears *seems* to describe non-associative concatenations of membranes with boundary. (But don't ask for details. This is either wrong or top-secret ;-) === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Content-Length: 1186 Originator: rusin@vesuvius > Aaron Bergman schrieb im Newsbeitrag > The Yoneda pairing is defined in Griffiths and Harris. All I know about > the A_oo structure is that m_1 vanishes so the pairing is associative. > Since the associator is > [m_2, m_2] = [m_1,m_3] + [m_4,m_0] I don't know what definition of an A_oo algebra you're using, but mine doesn't include an m_0. In particular, the structure includes a set of mappings m_i : A^i -> A where A^i denotes a tensor power. The associator is m_2(1 x m_2 - m_2 x 1) = m_1 m_3 + m_3(m_1 x 1 x 1 + cycles) which vanishes when m_1 does. ('x' denotes tensor product.) Written differently, this is (A,(B,C)) - ((A,B),C) = Q(A,B,C) + (QA,B,C) + (A,QB,C) + (A,B,QC) up to some signs I won't bother trying to figure out. These are all from Keller's Introduction to A-Infinity Algebras and Modules. This sort of thing seems to happen when you take the homology of a different A_oo algebra (like the open string algebra) with respect to m_1 (the BRST operator). Aaron === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Content-Length: 3123 Originator: rusin@vesuvius Aaron Bergman schrieb im Newsbeitrag >> Since the associator is >> [m_2, m_2] = [m_1,m_3] + [m_4,m_0] > I don't know what definition of an A_oo algebra you're using, but mine > doesn't include an m_0. Ok, in some texts it is not included. But it easily can and is necessary for some applications. My current reference to A_oo algebras is Ezra Getzler & John D. S. Jones: A_oo-Algebras and the Cyclic Bar Complex Ill. J. Math, 34(2), 256-283 (1990) , where m_0 is included. For the applications that I am concerned with (which is not, at least not directly (see below), related to open string field theory) the m_0 is a necessary ingredient. It specifies the (possibly nonabelian) 2-form field when you do nonabelian gauge theory over path space. In this context m_1(a) = da + A(a), where A(a) is the (adjoint) action of the 1-form A on the element B and m_2 is the wedge product times some product on the Lie algebra. This is more or less what Hofmann describes in his hep-th/0207017, though in the context of 2-groups there are slight modifications to what he does. > The associator is > m_2(1 x m_2 - m_2 x 1) = m_1 m_3 + m_3(m_1 x 1 x 1 + cycles) down. Given a n1-ary m1 and an n2-ary m2 we have the Gerstenhabe bracket [m1,m2] = sum m1(1 x ...x 1 x m2 ( .... ) x ... x 1) - sum m2(1 x ...x 1 x m1 ( .... ) x ... x 1) up to signs which are determined by the natural grading of the m1, m2. This reduces to the formula you give above when m2 = m_2 and m1 = m_1 . > which vanishes when m_1 does. ('x' denotes tensor product.) Yes, I understand that. I was asking because if you include m_0 there is also the term [m_0,m_4], which you did not have. Now I undertand that you assumed m_0 = 0. > These are all from Keller's Introduction to A-Infinity Algebras and > Modules. > This sort of thing seems to happen when you take the homology > of a different A_oo algebra (like the open string algebra) with respect > to m_1 (the BRST operator). Yes. As I mentioned above, in other contexts m_1 is an exterior differential coming from the exterior differential on path space. But in fact I would expect there to be some relation. That path space gauge theory is really living in the context of string boundary states. On those the BRST cohomology is characterized by reparameterization invariance plus its supersymmetric square root, G + ibar G . Rep-invariance is really a constraint very easily solved by suitable constructions, so that the BRST operator here reduces to that polar sum of worldsheet supercharged G + i bar G . This is essentially the exterior derivative on path space that I was talking about. Conjugating it by a nonabelian Wilson line gives path space connection with the above mentioned m_0, m_1 and m_2 operators appearing. === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) o Secret Service: Vile Persecution of Ed Cummings o Secret Service: Harassment of Steve Jackson Games * The New York Times, CyberTimes, June 20, 1997 * * Panel Chief Says Computer Attacks Are Sure to Come * * By THE ASSOCIATED PRESS * * WASHINGTON -- It is only a matter of time before critical U.S. computer * systems face major attack, the head of a White House panel on the nation's * infrastructure systems warned. * * Robert Marsh is the head of the President's Commission on Critical * Infrastructure Protection. Whatever should we do about those nasty hackers? **************************************************************************** ** Secret Service: Harassment of 2600 ------ ------- ---------- -- ---- A group of above-ground hackers associated with 2600 were having a lawful peaceful public meeting at the Pentagon City Mall on November 6, 1992. The meeting was busted up by mall police for no apparent reason. Identification was demanded from everyone. Bags were searched. It's the 1990s now. The harassment was publicized by 2600, and a reporter talked to the head of the mall's security: he let slip that the Secret Service ordered them to harass 2600's lawful peaceful public meeting. That was definitely news. The mall security manager then denied what he said about Secret Service ordering the harassment: luckily the reporter recorded his conversation. CPSR [Computer Professionals for Social Responsibility] and Marc Rotenberg of EPIC [Electronic Privacy Information Center] began FOIA [U.S. Freedom of Information Act] proceedings to find out about this incident. The case raises significant issues of freedom of speech and assembly, privacy and government accountability. In response to an FOIA asking why this happened, the Secret Service responded: We are sure no one knows why we had the meeting disrupted. They have made a === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Originator: baez@math-cl-n03.math.ucr.edu (John Baez) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >> Higher categories show up quite naturally in the study of >> commutative rings and associative algebras over commutative rings. >> I'd heard of things called Brauer groups and Picard groups >> of rings, and something called Morita equivalence, but I only >> understood how these fit together when I learned they were part >> of a marvelous thing: a weak 3-groupoid! >Be careful. You're secretly doing some string theory here. Eeek! You're right... but I don't mind. I like the math of string theory; it's the physics that gets kinda problematic. >> Starting with a commutative ring R, we can form a weak 2-category >> Alg(R) where: >> an object A is an associative algebra over R >Or a string boundary state. Right! There are some Moore-Segal axioms governing topological open/closed string theories, which talk about this sort of thing, but apparently trying to downplay the n-category theory. >> a 1-morphism M: A -> B is an (A,B)-bimodule >An open string stretching between two boundaries. Right! Or more precisely, I guess, the space of states of such a thing. >> a 2-morphism f: M -> N is a homomorphism between (A,B)-bimodules. >Beats me on this one. A time evolution operator for a string, no? String worldsheets are 2-dimensional so we naturally expect a 2-category X where: objects are finite sets of points morphisms are 1-dimensional manifolds with boundary 2-morphisms are 2-dimensional manifolds with corners going between these where the last guys are equipped with conformal structure if we want a full-fledged open/closed string theory, but not if we just want a topological one... and then dynamics is described by a symmetric monoidal 2-functor Z: X -> Y where Y is a 2-category of roughly the sort I just described. Here I'm not including anomalies, which make Z into a mere projective functor. >> This has all the structure you need to get a 2-category. In particular, >> we can compose an (A,B)-bimodule and a (B,C)-bimodule by tensoring them >> over B, getting an (A,C) bimodule. But since tensor products are only >> associative up to isomorphism, we only get a *weak* 2-category, not a >> strict one. >Sounds like an A_oo type structure to me, just like open strings. A_infinity categories are related to weak infinity-categories, which have weak 2-categories as a special case, so probably yeah. Much more to say, but dinner-time. === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Content-Length: 544 Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Right! There are some Moore-Segal axioms governing topological > open/closed string theories, which talk about this sort of thing, > but apparently trying to downplay the n-category theory. There's a lot of fun stuff along these lines at Caldararu has apparently proven some interesting things regarding this and the derived category of coherent sheaves but I don't claim to understand it. Aaron === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Aaron Bergman schrieb im Newsbeitrag >> Higher categories show up quite naturally in the study of >> commutative rings and associative algebras over commutative rings. >> I'd heard of things called Brauer groups and Picard groups >> of rings, and something called Morita equivalence, but I only >> understood how these fit together when I learned they were part >> of a marvelous thing: a weak 3-groupoid! > Be careful. You're secretly doing some string theory here. Probably lots categorization has more or less to do with going from points the config space of a (super)string. Categorizing susy QM on that space yields 2D SCFTs, roughly. >> Here's how it goes. You don't need to know much about higher >> categories for this to make some sense... at least, I hope not. >> Starting with a commutative ring R, we can form a weak 2-category >> Alg(R) where: >> an object A is an associative algebra over R > Or a string boundary state. >> a 1-morphism M: A -> B is an (A,B)-bimodule > An open string stretching between two boundaries. >> a 2-morphism f: M -> N is a homomorphism between (A,B)-bimodules. > Beats me on this one. At least one can say that if the homomorphism is an isomorphism this should be some kind of duality/gauge equivalence of (A,B) strings with (A',B') strings. Otherwise the (A,B) string is either embedded in some larger structure or projected on a smaller structure. === Subject: Invariant symmetric bilinear form on Lie algebras Hi all, Let g be a finite-dimensional real Lie algebra. For each @ in the dual of g, let g_@ be the isotropic subalgebra associated with @; in other words, g_@ = { X in g | Y in g => @([X,Y]) = 0 }. Consider the following property: (P) For every @, there is some invariant symmetric bilinear form B_@ on g x g such that the restriction of B_@ to g_@ x g_@ is non-degenerate. It is clear that if g is semisimple of compact type, then (P) holds; simply take B_@ = Killing form, for each @. My question is: is the property (P) *equivalent* to the fact that g is semisimple of compact type? Jose Carlos Santos === Subject: approximate entropy Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I would appreciate any answers to me directly (george@netvision.net.il), george I am a math and science writer for the Swiss newspaper Neue Entropy measure. (Pincus, PNAS Vol. 88. 1991. See for example Ivar Peterson's (http://www.maa.org/mathland/mathtrek_10_11_04.html ) In the course of my research I have been hearing some criticism about ApEn. I would be very happy to receive anybody's comments. 1) ApEn seems like a reasonable measure of randomness or complexity. But one of the first tests of its applicability turned out to be counter-intuitive: the ApEn's of transcendental and algebraic numbers are quite intermingled. (Pincus and Kalmann, PNAS Vol. 94., 1997) At that point one option would have been to say that ApEn produces unreasonable results and should be thrown out. Rather, an artificially created puzzle remained in the literature. 2) Following that, many authors published papers which compute ApEn of various physiological variables (heart rhythm, EEG, hormone levels). The papers seem to be purely descriptive and non-refutable. The authors write up the results and leave it to future research to figure out why ApEn is high in certain instances and low in others. When computing a measure - any measure - one will always get some numerical results. 3) Recently, ApEn was applied to financial time-series (Pincus and Kalmann, be totally anecdotal: (A) two one-week periods of the Dow Jones index are picked out and their ApEn's are declared significantly different (without any indication of how significant). A more comprehensive comparison of all low ApEn periods versus high ApEn periods would have maybe given an indication of what is going on, but that was not done. (B) The evidence on the Hang-Seng index describes one episode other periods with high ApEn can be gleaned with no crash following, and vice versa. So? Furthermore, the evidence ends just after the crash in (C) The section on the Black and Scholes model seems to be in error since it claims that the B-S model implies Brownian motion of the underlying stock prices. But Brownian motion is an assumption of the B-S model, not an implication. -- -------- George Szpiro Neue Z.9frcher Zeitung (Switzerland) POB 6278 Jerusalem 91060 Israel === Subject: Re: Galois group of a given quartic equation Epigone-thread: twomprendlex Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I'm kicking myself for not having noticed this earlier, but the condition for getting (I) x^4 - 2c * x^3 + (c^2 - a^2 - b^2) * x^2 + 2 * a^2 * c * x - a^2 * c^2 with Galois group Z4 can be simplified considerably: in order that (I) have cyclic Galois group Z4 (a, b, c integers as described previously) it is necessary that the discriminant be twice a square. For, it was found that, if a zero of (I) defines a Z4 extension K of Q, and p is an *odd* prime totally ramified in K/Q, then p must satisfy the conditions p divides c, p divides a^2 + b^2 and, assuming gcd(a, b, c) = 1, then p does not divide ab. Assume K/Q is cyclic of degree 4, and p is totally ramified in K/Q. Let R be the ring of algebraic integers in K, with P the prime ideal of R containing p, pR = P^4. Suppose r1, r2, r3 and r4 are the four zeroes of (I); then r1 * r2 * r3 * r4 = -a^2 * c^2. Since the r's are conjugate, and P is stable under conjugation, the factorization of the ideals r1*R, r2*R, r3*R, and r4*R all contain the exact same prime-power factor P^m. That is, the P-adic values (exponents) v_P(r1), v_P(r2), v_P(r3), and v_P(r4) are all equal to the same integer, say m. Since v_P(-a^2 * c^2) = 2*V_P(c), we have v_P(C) = 2m. Since p divides c, m is positive. (In fact m must be even since v_P(c) is divisible by 4, but that isn't needed in what follows). Looking at the coefficient of x, we have on the one hand v_P(2 * a^2 * c) = v_P(c) = 2m since p is odd and does not divide a. On the other hand, expressing the coefficient as an elementary symmetric function of the r's, 2 * a^2 * c = -(r1*r2*r3 + r1*r2*r4 + r1*r3*r4 + r2*r3*r4), so v_P(2 * a^2 * c) = v_P(r1*r2*r3 + r1*r2*r4 + r1*r3*r4 + r2*r3*r4), whence v_P(-2 * a^2 * c) >= 3m, because every term in the sum has P-adic value 3m. But m is positive, so 2m >= 3m is false. The assumption that K/Q is cyclic of degree 4 and p is totally ramified in K/Q must therefore also be false. Thus, it is not possible for (I) give a Z4 extension K/Q in which any odd prime is totally ramified. The only remaining possibility is that 2 is the only prime totally ramified in K/Q; and in this case, the discriminant of K/Q, hence that of the quartic (I), must be twice a square. The small examples satisfying this condition that turned up in my search gave four irreducible quartics with Galois group S4, a = 27 b = 77 c = 148 a = 32 b = 155 c = 247 a = 1 b = 352 c = 395 a = 19 b = 315 c = 404 and one quartic with a linear factor and irreducible cubic cofactor with Galois group S3: a = 63 b = 64 c = 195 With so simple a condition as the discriminant being twice a square, further numerical searching is obviously feasible. It might also be possible to look for families of examples in which the discriminant is twice a square (as was done for square discriminants). In the small examples found so far, the Galois group had a 3-cycle (resolvent cubic irreducible). === Subject: Presentation of a category Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Could someone introduce what is a presentation of a category (if it exists)? I guess a presentation should be a set S of objects and arrows (generators), plus a relation R on arrows. A category would then be the quotient by R of the free category over S. Well... what is this free category over S? Does every category have a presentation? I guess a presentation is not unique but is there a canonical one? In this context, what is the equivalent of the word problem for groups. David. === Subject: Re: Presentation of a category Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Could someone introduce what is a presentation of a category (if it >exists)? >I guess a presentation should be a set S of objects and arrows >(generators), plus a relation R on arrows. A category would then be the >quotient by R of the free category over S. Well... what is this free >category over S? You can see these things explained in Mac Lane's Categories for the working mathematician but here goes a a summing up of subject. The situation is (almost) the same as in groups, algebras, etc. If A is a category of algebraic objects there is a forgetful functor U: A->Set and a left adjoint free functor F:Set -> A that to a set X associates the free algebra in A F(A). With categories it is the same thing except that we replace Set by Graph, the category of graphs. To each category A associate the graph U(A) - just forget about composition and identities. The left adjoint is the free path category on a graph G: roughly F(G) has as objects the vertices of G and for arrows v->w finite sequences of edges (e1, ..., en) such that dom(e1) = v cod(en) = w cod(ei) = dom(ei+1) where cod is codomain and dom domain of the edge. Composition is concatenation and the identities are the empty sequences (). A presentation of a category A is then an isomorphism (1) A -> F(G)/R where G is some graph and R a relation on F(G). >Does every category have a presentation? I guess a presentation is not >unique but is there a canonical one? Of course, just do the same thing as you do with algebras: take the underlying graph U(A) and then form the free category F(U(A)). There is an obvious functor F(U(A))->A (the counit of the adjunction) that gives the needed relation R to get (1). G. Rodrigues === Subject: Re: Presentation of a category designed to benefit individuals, but in practice they usually serve as methods for inducing individuals to think and behave as the system requires. (There is no contradiction here; an individual whose attitudes or behavior bring him into conflict with the system is up against a force that is too powerful for him to conquer or escape from, hence he is likely to suffer from stress, frustration, defeat. His path will be much easier if he thinks and behaves as the system requires. In that sense the system is acting for the benefit of the individual when it brainwashes him into conformity.) Child abuse in its gross and obvious forms is disapproved in most if not all cultures. Tormenting a child for a trivial reason or no reason at all is something that appalls almost everyone. But many psychologists interpret the concept of abuse much more broadly. Is spanking, when used as part of a rational and consistent system of discipline, a form of abuse? The question will ultimately be decided by whether or not s === Subject: Re: Presentation of a category Content-Length: 956 Originator: rusin@vesuvius > You can see these things explained in Mac Lane's Categories for the > working mathematician but here goes a a summing up of subject. http://www.cwru.edu/artsci/math/wells/pub/pdf/sketch.pdf At the end of Section 2 it is written that 'a sketch is a presentation of a category in the usual sense of presentation'. This seems more complicate than what you explained. Indeed in a sketch there is not only a graph and a relation, but there are also cones and cocones. So what is really a presentation of a category? A sketch or your definition? By the way, you refer to Mac Lane's book (I guess you refer to Chapter II, Section 8 on Quotient categories). But Mac Lane does not use the word presentation. Maybe it is on purpose. David. === Subject: Re: Presentation of a category before technical necessity there would be economic problems, unemployment, shortages or worse. The concept of mental health in our society is defined largely by the extent to which an individual behaves in accord with the needs of the system and does so without showing signs of stress. 120. Efforts to make room for a sense of purpose and for autonomy within the system are no better than a joke. For example, one company, instead of having each of its employees assemble only one section of a catalogue, had each assemble a whole catalogue, and this was supposed to give them a sense of purpose and achievement. Some companies have tried to give their employees more autonomy in their work, but for practical reasons this usually can be done only to a very limited extent, and in any case employees are never given autonomy as to ultimate goals -- their autonomous efforts can never be directed toward goals that they select personally, but only toward their employer's goals, such as the survival and growth of the company. Any company would soon go out of business if it permitted its employees to act otherwise. Similarly, in any enterprise within a socialist system, workers must direct their efforts toward the goals of the === Subject: Re: Presentation of a category Content-Length: 1586 Originator: rusin@vesuvius >> You can see these things explained in Mac Lane's Categories for the >> working mathematician but here goes a a summing up of subject. > http://www.cwru.edu/artsci/math/wells/pub/pdf/sketch.pdf >At the end of Section 2 it is written that 'a sketch is a presentation of >a category in the usual sense of presentation'. This seems more >complicate than what you explained. Indeed in a sketch there is not only >a graph and a relation, but there are also cones and cocones. So what is >really a presentation of a category? A sketch or your definition? Sketches are something more -- they are graph, a relation on it *plus* a set of cones and a set of cocones. You can view them as presentations of a category with *prescribed limits and colimits*. As for your question what is really a presentation of a category the best answer is: (if you haven't done it yet) study the theory of monads -- this is in Mac Lane. The Barr-Wells book (available online) also deals at length with monads (there under the name of triples). And jumping ahead: yes, categories are monadic over graphs (I think this is in Barr-Wells but it's a good exercise) so, IMHO, the above notion of a presentation is a good one. G. Rodrigues P.S: Sorry for the short answer, tight on time. === Subject: Re: Presentation of a category Content-Length: 478 Originator: rusin@vesuvius > And jumping ahead: yes, categories are monadic over graphs (I think > this is in Barr-Wells but it's a good exercise) so, IMHO, the above > notion of a presentation is a good one. You mean the forgetful functor from Cat to Graph is monadic. I understand that it means that Cat is equivalent to its category of Eilenberg-Moore algebras. But why is it interesting? How does it relate to presentation of category? === Subject: Re: Presentation of a category Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Could someone introduce what is a presentation of a category (if it >exists)? >I guess a presentation should be a set S of objects and arrows >(generators), plus a relation R on arrows. A category would then be the >quotient by R of the free category over S. Well... what is this free >category over S? >Does every category have a presentation? Surely yes; just as a group G has the presentation in which the generators are (in bijection with) the elements of G, and the relations are (in appropriate bijection with) the entries of the multiplication table of G (and it is understood that the presentation is a group presentation, and not, say, a semigroup presentation or a loop presentation). Presumably, you mean to ask whether there are always category presentations that are somewhat more interesting, or smaller, or..., than the tautologous presentation. (Note that I don't claim to know what a category presentation *is*, I merely claim that any construction worthy of the name must have tautologous instances analogous to those I described in the case of groups.) Lee Rudolph === Subject: Functor as a category Originator: bergv@math.uiuc.edu (Maarten Bergvelt) pairs (X, FX) where X is an object in C, and morphims are pairs (f, Ff) where f is a morphism in C. What can be said about this category |F|? Are there properties of C and/or D which also hold in this category? If F:C-->D and G:C'-->D' are functor, a functor from |F| to |G|. Is something more general than a natural transformation. Do such functors have interesting properties? David. === Subject: Re: Functor as a category legislation across California. * * Because the San Jose City Attorney's office brought action against the * defendants under a civil procedure, the defendants were not guaranteed * the standard protections of criminal law. * * * The defendants may not engage in any form of public association: standing, * sitting, walking, driving, gathering or appearing anywhere in public view * in the neighborhood, or face 6 months in jail. * * They further may not: climb trees or fences, make loud noises, possess * wire cutters or marbles [What???], wear particular clothes, make certain * hand signs, or carry marking pens or pagers. The worst-case version of a National ID Card is a biometric-based one. Biometrics means that card is numbered with your fingerprint, or retina scan, or other unique physical characteristic. You would be enumerated. Numbered for all time. * CARROLLTON, Texas--(BUSINESS WIRE)--May 16, 1997--Sandia Imaging, a * majority-owned subsidiary of Lasertechnics, Inc. (Nasdaq: LASX) and a * leader in secure card technologies, will premiere the world's first * printer === Subject: Re: Functor as a category over, he said for crossing lanes while going through an intersection. ABC's cameras then showed cars doing that constantly at the same intersection. They said they counted hundreds the same night. The police chief then tacitly admitted they were pulling over black people on purpose. ---- [ Yes, I am aware of the cocaine/crack sentencing discrimination. ] You monitor any group real close, you'll get many arrests. The implications of heavy monitoring are serious. * * Blacks make up 51 percent of the 1.1 million inmates in state and * Federal prisons, the Sentencing Project study said, though * blacks are only 14 percent of the nation's population. * * Of a total voting age population of 10.4 million black men nationwide, * an estimated 1.46 million have lost the right to vote [as a result]. Wow. This highly focused monitoring of blacks should be way illegal. Blacks make up 14% of our U.S. population. Blacks make up 51% of our prison population. Never forget what it means to be heavily monitored: there is no place to hide. Noone is an angel. Are you? What's in store next for black Americans? # This Modern World, by Tom Tomorrow [political cartoon, in NYT] # # Biff: You know why we should eliminate welfare, Wanda? # It's been A COMPLETE FAILURE! # After all -- t === Subject: Re: Functor as a category # # The Police even pointed a machine gun at the # head of Mrs. Kuriatnk's six-year-old daughter. Actually, I think the government says 'because organized crime would use cryptography'; but criminals with guns is the generalization. If you were a criminal, would you select cryptography that is 'Key Recovery' (GAK Government Access to Keys) compliant? I don't think so. But we'll catch stupid criminals using GAK crypto! ---Scott Charney, Computer Crime Unit, Department of Justice, at 5/22/97 NYC cryptography conference And equally stupid Congress members will support that logic. The subject of law enforcement v. guns brings up a subtopic: Burn Baby Burn ---- ---- ---- I was watching MSNBC's Time and Again with Jane Pauley, and they went over the Patty Hearst kidnapping by the Symbionese Liberation Army. The terrorists kept Ms. Hearst locked in a closet so small she couldn't move. They would play games with her head, like suddenly dragging her out of the closet, hold a gun to her head, and sometimes pull the trigger (on an empty chamber). Then they started screwing her. Again and again. They took complete control of a sweet young college kid. The government knows how that works. Yet the government charged her with bank robbery even though she was under their mental control. She had renounced her rich parents and taken a new name. Happens all the time. The FBI called her a 'common criminal'. Then put her on their Top Ten Wanted List. Maybe they thought she had taken Terrorism 101 at school and ambitiously decided to start robbing banks of her own accord. Anyway, there was an episo === Subject: Re: Functor as a category of a styrofoam cup. **************************************************************************** ** National ID Card -------- -- ---- * C-SPAN Congressional Television: outside coming down the Senate building's * steps, Senator Biden with Senator Simpson in tow proclaims: What's wrong * with a National ID Card? It's the same tired old arguments against it. As if sane people shouldn't be paranoid about a National ID Card. * New Rules Mean Job-Hunters Need Proof of Identity, The New York Times * * Passports, driver's licenses, Social Security cards or birth certificates * will be allowed to serve as identity papers. * * A 1982 proposal to catch illegal aliens by giving American workers * counterfeit-proof identity cards was hooted off the boards as a * threat to individual liberty. How bad would a National ID Card be? Bad. Real bad. You would be required to carry it at all times. It's all about surveillance and control. This section is about the National ID Card, plus deployment of a mix of surveillance and control techniques for tracking people. In California a few years back the police kept hassling a black man who liked to walk around at night to think. Unfortunately, he wasn't white, but liked to walk around white neighborhoods. [Anyone with detailed info email me.] He didn't carry id with him: the polic === Subject: Re: Functor as a category o Other firms' IUO (Internal Use Only) inbound o Our detailed systems inventory o Determined what PGP (encrypted) traffic was occurring. Among others, we had constant small traffic back-and-forth with Military contractor Rockwell. o Salomon's Official Restricted List being repeatedly transmitted outbound (list of securities Salomon can't purchase without a conflict of interest) o Unreleased Financing Summaries and unreleased IPO's: SEC violations o Internal Use Only documents o Trade confirmations o JobTalk hits concerning internal budget details by an SOO. o JobTalk hit of a resume of a risk management person who wanted to explain how it works here o Hundreds of router (security) configurations o 42,000 lines of OASYS data o router and bridge passwords o Hostname/username/password for unmonitored outbound ISDN access from Salomon o RadioMail: spotted that all the big cheeses who use it have all their highly sensitive email going out over the unprotected Internet, because we were too cheap to buy a transmitter, and so are forwarding all the email over the Internet to RadioMail's transmitter!!! o The key to one's financial life: Social Security numbers of Salomon retirees transmitted in/out the Internet. Names, birth dates, sex, life insurance amount, date of spouse's birth... o caught our proprietary infrastructure code running at JP Morgan ********** end excerpt from 'Corruption at Salomon Brothers' ********** So...how have I done, to indicate how powerful keyword monitoring is? NSA employees would go to jail for ten years for describing the effectiveness of DIC === Subject: Re: Functor as a category kidding about having a baggie of pot. Students and parents are stunned. The teacher said she believed her students had better sense than that, and since she inspected it and it smelled like oregano she was sure they were kidding her. Students and their parents protest, the school board asks her back, but she says no, she is too disgusted at her treatment. Zero Tolerance victims, falling into the abyss. State troopers really know their business: : Robert Fitches, a 22 year-old said in his Federal lawsuit that he was : humiliated when state troopers ordered him to drop his pants during a : drug search along Interstate 15 in Davis County. : Source: Salt Lake City Tribune 7/8/95 Maybe this is an accurate analogy of why dragnet-monitoring is wrong: : The Sheraton Boston Hotel was discovered videotaping employees changing : clothes in locker rooms. The 1991 surveillance caught employees using : drugs, Sheraton said. Source: Senate Labor Committee on Employment, 6/93 If you strip us naked you will detect more crime, but also, you strip individuals naked without specific individuals being suspected of a crime. Dragnet monitoring should not be the American way. Unrestricted cryptography must be made legal now, so we are no longer naked to ECHELON monitoring. It will be a beginning. : Privacy === Subject: Re: Functor as a category And so it is up to all of you. To arm yourselves. With writing implements. COMPLAIN LIKE HELL! Write to all your Congressional representatives. Send them a copy of any/all of this manifesto with a cover letter stating the specific questions you demand be answered. Write to your local papers, radio stations, state supreme courts (make them aware of fingerprinting drivers is a violation of the 1974 Privacy Act). Write to all your state representatives. Take copies of this manifesto and go to your neighbors and ask they consider doing the same. Contact all your friends. : The New York Times, 2/10/87 : Is This America?, by Anthony Lewis : : When we speak out, she said, that's our protection. : : She still believes in America. Network. Creep back at the bastards who are destroying America! Be persistent. It is almost too late. ---guy@panix.com It must always be remembered that crime statistics are highly inflammatory---an explosive fuel that powers the nation's debate over a large number of important social issues---and that FBI Director Louis Freeh today is the leading official shoveling the fuel into the blazing firebox. ---David Burnham Indeed, the Scary Man has === Subject: Re: Functor as a category Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > What can be said about this category |F|? Are there properties of C and/or > D which also hold in this category? I think this category is isomorphic to C, so all the properties of C hold. Have I misunderstood your definition? Boris === Subject: Re: Functor as a category Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > pairs (X, FX) where X is an object in C, and morphims are pairs (f, Ff) > where f is a morphism in C. > What can be said about this category |F|? Are there properties of C and/or > D which also hold in this category? The category |F| is equivalent to C. Indeed, consider the functor H:C-->|F| defined by X|-->(X,FX) on objects and by f|-->(f,Ff) on morphisms. Then |F|((X,FX),(Y,|FY|)=C(X,Y) so H:C-->|F| is full and faithful. And for any object Z of |F|, there exists T object of C such that (T,FT)=Z and by definition H(T)=Z. So H is essentially surjective. > If F:C-->D and G:C'-->D' are functor, a functor from |F| to |G|. Is > something more general than a natural transformation. Do such functors > have interesting properties? > David. A functor from |F| to |G| is nothing else but a functor from C to C'. If you want something close to a natural transformation, try the notion of comma category. If F:C-->D and G:C'-->D are functors, the comma category Fdownarrow G is the category whose objects are the arrows FX-->GX' and whose morphisms are the commutative squares FX --> GX' ^ ^ | | FY --> GY' I dont know what you are looking for but I think that 'comma category' is a good keyword. Some references are Borceux's Handbook of categorical algebra (3 volumes : I dont remember which one and the Math library is closed during the week-end) and Maclane's book 'categories for the working mathematician' p46. Try also : http://en.wikipedia.org/wiki/Comma_category pg. === Subject: Question on harmonic analysis on non LCA groups Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi everyone, I know there doesn't exist a general theory for harmonic analysis on non-locally compact groups. What useful applications would a theory for harmonic analysis on non-locally compact groups have? Are there real world practical applications? Does anyone have an idea as to what has been done already in this area, what people are currently working on, and how people think that this problem can be solved? Any references? Isaac === Subject: Re: Question on harmonic analysis on non LCA groups methodologically distinguishable from economics. The process whereby you * analyzed, managed, and controlled an economy was not essentially different * from the way you managed a war, except that one was an economy of produc- * tion and the other was an economy of force. The principal underlying both * was the doctrine of efficiency: maximizing the benefits received from the * efforts and expenditures---a cost benefit analysis. Cyberneticians hope to use their capabilities for the betterment of the human race, of which they are a part. They are not naive when it comes to the government and politics, either. * The Rise of the Computer State, David Burnham, 1984 * * Norbert Wiener, the MIT professor who is generally credited with being * one of the principal minds behind the development of the computer, * refused to take research money from the Pentagon because he was * convinced it would corrupt his research and undermine his independence. When Stafford Beer monitored factories and banks to give the government the necessary tools to govern the economy effectively, he chose to monitor national infrastructure of the industrial variety. However, even he knows what can happen with cybernetic control in the wrong hands: * Brain of the Firm, Stafford Beer, 1986, ISBN 0 471 27687 1 * * If Project Cybersyn were altered, and the tools used are not the * tools we made, they === Subject: Re: Question on harmonic analysis on non LCA groups patriot's nightmare. Someday, the Holy Bible * prophesies, that planetary dictator will emerge on the scene, lusting * for blood... * * There can be no doubt about it. * * The REAL Chief Executive Officer of the NSA is not a human being. * * The CEO MUST be Lucifer himself. Amen. ---- It is technology driving the capabilities, it is our government using them ruthlessly: without letting us vote on it. Never before could someone walk up to you and number you by scanning your fingerprints. A number that is yours and yours alone. You have been numbered for all time. No ID card needed once portable fingerprint scanners are deployed all over! If the government suddenly ordered all citizens to be numbered with an indelible invisible ink on their arms so they were permanently numbered; so law enforcement could scan them at will: there would be a revolt. Yet that is what is happening. Fingerprints, scanned into a computer, are a number. The number is inescapably yours. Modern technology means they don't have to put the number on you, t === Subject: Re: Question on harmonic analysis on non LCA groups Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Hi everyone, > I know there doesn't exist a general theory for harmonic analysis on > non-locally compact groups. What useful applications would a theory for > harmonic analysis on non-locally compact groups have? Are there real > world practical applications? I don't really know anything about this stuff, but I've heard that some diffeomorphism groups are not locally compact. A theory of unitary representations of these groups would certainly be useful for quantum mechanics of diffeomorphism invariant theories. > Does anyone have an idea as to what has been done already in this area, > what people are > currently working on, and how people think that this problem can be > solved? Any references? I wish I knew any. Igor === Subject: primitve polynomial generation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) hi all, Are there algorithms for generating primitive polynomials over GF(2^x). I need to be able to generate ALL the primitive polynomials over GF(2^x) for x = 1 to 64. Even an randamoized (but efficient) algortithm would do. I could not find references that might help me. So references to those algorithms will be helpful to me. chax. --- Any time things appear to be going better, you have overlooked something === Subject: Re: primitve polynomial generation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > hi all, > Are there algorithms for generating primitive polynomials > over GF(2^x). I need to be able to generate ALL the primitive > polynomials over GF(2^x) for x = 1 to 64. > Even an randamoized (but efficient) algortithm would do. > I could not find references that might help me. So references > to those algorithms will be helpful to me. Text: http://www.jjj.de/fxt/#fxtbook Code: http://www.jjj.de/fxt/ (the FXT package) Note FXT's binary polynomials are machine words (i.e. use a 64-bit machine for degrees >32) With the text you should have no difficulty to roll your own code. If that fails I can give you a (addmittedly cryptic) pari/gp script that I used to create several lists ( http://www.jjj.de/mathdata/ ) of irreducible and primitive polynomials with additional properties. all the best, jj -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die. === Subject: Re: primitve polynomial generation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > hi all, Are there algorithms for generating primitive polynomials > over GF(2^x). I need to be able to generate ALL the primitive > polynomials over GF(2^x) for x = 1 to 64. > Even an randamoized (but efficient) algortithm would do. > I could not find references that might help me. So references > to those algorithms will be helpful to me. > Text: http://www.jjj.de/fxt/#fxtbook > Code: http://www.jjj.de/fxt/ (the FXT package) > Note FXT's binary polynomials are machine words > (i.e. use a 64-bit machine for degrees >32) > With the text you should have no difficulty to roll your own code. > If that fails I can give you a (addmittedly cryptic) pari/gp script that > I used to create several lists ( http://www.jjj.de/mathdata/ ) of > irreducible and primitive polynomials with additional properties. > all the best, > jj Thanx a lot !! I partially got what I wanted. I suppose the code generates primitive polynomials of varying degrees belonging to the ring GF(2)[x] for degree = 1 ... 64. But are there algorithms to generate the primitive polynomials over GF(4), GF(8), or any other GF(2^x) for that matter. x = 1.....8 atleast. something like - (01) + (11)x + (11)x^2 being primitive in GF(4^2) i.e GF(16) (please note that I'm not using GF(16) as GF(2^4) but as GF(4^2)) even if it is difficult. can you send me some references? thanx in advance. chax -- heisenberg .. could be .. here or there.... === Subject: Re: primitve polynomial generation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > [...] > > Thanx a lot !! I partially got what I wanted. > I suppose the code generates primitive polynomials > of varying degrees belonging to the ring GF(2)[x] for > degree = 1 ... 64. > But are there algorithms to generate the primitive polynomials > over GF(4), GF(8), or any other GF(2^x) for that matter. > x = 1.....8 atleast. > something like - (01) + (11)x + (11)x^2 being primitive in > GF(4^2) i.e GF(16) (please note that I'm not using GF(16) > as GF(2^4) but as GF(4^2)) > even if it is difficult. can you send me some references? > thanx in advance. Check section 21.2 (irred. & primitive polynomials) and especially 21.2.3 (testing for primitivity). Use pari's polisirreducible() to test for irreducibility then compute the order with the given code. === Subject: Re: primitve polynomial generation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Check section 21.2 (irred. & primitive polynomials) > and especially 21.2.3 (testing for primitivity). > Use pari's polisirreducible() to test for irreducibility then compute > the order with the given code. May be I'm not making myself clear. If we take GF(16) ... its elements can be represented as degree-3 polynomials over GF(2). 1001 - x^3 + 1 1011 - x^3 + x + 1 where a possible primitive polynomial(degree 4) picked up for the sake of GF(16) multiplication could be x^4 + x + 1 (lets call it p(x)) a * b in GF(16) is a(x) * b(x) mod (p(x)) where a(x) and b(x) are polynomial representations of elements 'a' and 'b' repsectively. ................ OR ..................... The same GF(16) elements can be represented as degree-1 polynomials over GF(4) 1001 - 10x + 01 1011 - 10x + 11 where a possible primitive polynomial (degree-2) this time over GF(4) say something like (01) + (11) x + (10) x^2 I'm looking for primitive polynomials of this type. I need to generate ALL primitive polynomials OVER GF(4), GF(8) -- (coefficients are 3 bits now) ... GF(256) of degrees 16, 8, 4 .. respectively.. chax. === Subject: properties of ergodic matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) could anyone suggest a good reference listing the properties of ergodic matrices? thank you === Subject: One problem on linear algebra Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi all, I am unable to solve the following problem. Any help is welcome. PROBLEM: Let V be an n dimensional vector space over R. Does there exists an (n-1)-dimensional subspace W of E^2(V) such that for all b in V{0} dimension S_b=1 where S_b={w in W| w#b = 0}. NOTATIONS: Notation: We work in the exterior powers of an n-dimensional vector space V over R. E^k(V) will be the k-th exterior power, and # will denote exterior product. --------------------------- for n=3, i know there does not exist any. for arbitrary n i dont know how to proceed. akash === Subject: Linear homogeneous functions Epigone-thread: strirgharperd Originator: bergv@math.uiuc.edu (Maarten Bergvelt) It is well known (see, e.g., Aczel J., Dhombres J. Functional Equations in Several Variables. 1989.) that general continuous solution of a functional equation for linear homogeneous function f(a*x[1],...,a*x[n]) = a * f(x[1],...,x[n]), for all strictly positive a,x[1],...,x[n], is: f(x[1],...,x[n]) = x[1] * g(x[2]/x[1],...,x[n]/x[1]), where g(.) is an arbitrary function of strictly positive variables. The question: what can be said about general continuous solution of the same functional equation f(a*x[1],...,a*x[n]) = a * f(x[1],...,x[n]), when a > 0, but x[1],...,x[n] are non-negative? Tnanks in any advance, Mikhail. === Subject: Re: Linear homogeneous functions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >It is well known (see, e.g., Aczel J., Dhombres J. Functional >Equations in Several Variables. 1989.) that general continuous >solution of a functional equation for linear homogeneous function >f(a*x[1],...,a*x[n]) = a * f(x[1],...,x[n]), >for all strictly positive a,x[1],...,x[n], is: >f(x[1],...,x[n]) = x[1] * g(x[2]/x[1],...,x[n]/x[1]), >where g(.) is an arbitrary function of strictly positive variables. If you want f to be continuous, g must be continuous. >The question: what can be said about general continuous solution of >the same functional equation >f(a*x[1],...,a*x[n]) = a * f(x[1],...,x[n]), >when a > 0, but x[1],...,x[n] are non-negative? You could write it as f(x) = S(x) g(x/S(x)) for S(x) > 0, f(0) = 0 where S(x) = x[1]+...+x[n], and g is an arbitrary continuous function on the (n-1)-simplex {x in R^n: all x_j >= 0, S(x) = 1}. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Linear homogeneous functions Epigone-thread: strirgharperd Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >>The question: what can be said about general continuous solution of >>the same functional equation >>f(a*x[1],...,a*x[n]) = a * f(x[1],...,x[n]), >>when a > 0, but x[1],...,x[n] are non-negative? >You could write it as >f(x) = S(x) g(x/S(x)) for S(x) > 0, >f(0) = 0 >where S(x) = x[1]+...+x[n], and g is an arbitrary continuous function >on the (n-1)-simplex {x in R^n: all x_j >= 0, S(x) = 1}. May be you know, is there a way to solve the same functional equation for the case when a > 0, but x[1],...,x[n] are reals? It seems me more difficult... === Subject: Re: Linear homogeneous functions Epigone-thread: strirgharperd Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >>The question: what can be said about general continuous solution of >>the same functional equation >>f(a*x[1],...,a*x[n]) = a * f(x[1],...,x[n]), >>when a > 0, but x[1],...,x[n] are non-negative? >You could write it as >f(x) = S(x) g(x/S(x)) for S(x) > 0, >f(0) = 0 >where S(x) = x[1]+...+x[n], and g is an arbitrary continuous function >on the (n-1)-simplex {x in R^n: all x_j >= 0, S(x) = 1}. So, for the case a > 0, x[1],...,x[n] are reals we can choose, say, any metrics M in R^n, then f(x) = M(x)*g(x/M(x)), where g is an arbitrary continuous function on a unit ball: M(x) < 1 without zero point. But the case of f(0) still confuses me (as we seek only continouos solutions)... === Subject: Re: Linear homogeneous functions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >The question: what can be said about general continuous solution of >the same functional equation >f(a*x[1],...,a*x[n]) = a * f(x[1],...,x[n]), >when a > 0, but x[1],...,x[n] are non-negative? >>You could write it as >>f(x) = S(x) g(x/S(x)) for S(x) > 0, >>f(0) = 0 >>where S(x) = x[1]+...+x[n], and g is an arbitrary continuous function >>on the (n-1)-simplex {x in R^n: all x_j >= 0, S(x) = 1}. >So, for the case a > 0, x[1],...,x[n] are reals we can choose, say, >any metrics M in R^n, then A norm will do, or more generally any continuous function satisfying M(ax) = a M(x) for a > 0, and M(x) > 0 for x <> 0. >f(x) = M(x)*g(x/M(x)), >where g is an arbitrary continuous function on a unit ball: M(x) < 1 >without zero point. No, g is a continuous function on the unit sphere {x: M(x) = 1}. >But the case of f(0) still confuses me (as we seek only continouos >solutions)... Since the unit sphere is compact, g is bounded, and as x -> 0 we have M(x) -> 0, so f(x) -> 0, and the solution is continuous. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Three papers published by Algebraic and Geometric Topology Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following three papers have been published: (1) An indecomposable PD_3-complex : II by Jonathan A. Hillman URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-47.abs.html (2) Alexander polynomial, finite type invariants and volume of hyperbolic knots by Efstratia Kalfagianni URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-48.abs.html (3) Transverse contact structures on Seifert 3-manifolds by Paolo Lisca Gordana Matic URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-49.abs.html Details follow: (1) Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-47.abs.html Title: An indecomposable PD_3-complex : II Author(s): Jonathan A. Hillman Abstract: We show that there are two homotopy types of PD_3-complexes with fundamental group S_3*_{Z/2Z}S_3, and give explicit constructions for each, which differ only in the attachment of the top cell. Secondary: 55M05 Keywords: Indecomposable, Poincare duality, PD_3-complex Author(s) address(es): School of Mathematics and Statistics F07 University of Sydney, NSW 2006, Australia Email: jonh@maths.usyd.edu.au (2) Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-48.abs.html Title: Alexander polynomial, finite type invariants and volume of hyperbolic knots Author(s): Efstratia Kalfagianni Abstract: We show that given n>0, there exists a hyperbolic knot K with trivial Alexander polynomial, trivial finite type invariants of order <=n, and such that the volume of the complement of K is larger than n. This contrasts with the known statement that the volume of the complement of a hyperbolic alternating knot is bounded above by a linear function of the coefficients of the Alexander polynomial of the knot. As a corollary to our main result we obtain that, for every m>0, there exists a sequence of hyperbolic knots with trivial finite type invariants of order <=m but arbitrarily large volume. We discuss how our results fit within the framework of relations between the finite type invariants and the volume of hyperbolic knots, predicted by Kashaev's hyperbolic volume conjecture. Secondary: 57M27, 57N16 Keywords: Alexander polynomial, finite type invariants, hyperbolic knot, hyperbolic Dehn filling, volume. Author(s) address(es): Department of Mathematics, Michigan State University E. Lansing, MI 48824, USA or School of Mathematics, Institute for Advanced Study Princeton, NJ 08540, USA Email: kalfagia@math.msu.edu, kalfagia@math.ias.edu URL: http://www.math.msu.edu/~kalfagia (3) Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-49.abs.html Title: Transverse contact structures on Seifert 3-manifolds Author(s): Paolo Lisca Gordana Matic Abstract: We characterize the oriented Seifert-fibered three-manifolds which admit positive, transverse contact structures. Keywords: Transverse contact structures, Seifert three-manifolds Author(s) address(es): Dipartimento di Matematica L. Tonelli Universita di Pisa I-56127 Pisa, ITALY and Department of Mathematics University of Georgia Athens, GA 30602, USA Email: lisca@dm.unipi.it, gordana@math.uga.edu === Subject: Gray codes through Lyndon words Originator: bergv@math.uiuc.edu (Maarten Bergvelt) In section 6.5 (Gray code for Lyndon words) of the fxtbook ( http://www.jjj.de/fxt/#fxtbook ) an algorithm is given that (by optimized graph search) finds Gray codes (in fact cycles) for binary Lyndon words of length n for all prime n<=37. Is an algorithm known that finds Gray codes without using memory proportional to the size (2^n-1)/n of the result? ( The graph search used needs 0.25 bits times (2^n-1). ) 1 week of single CPU time (possibly more time with NUMA systems). === Subject: Jobs in London Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Imperial College London has just advertised (a) 2 post-doc positions (b) a pure maths lectureship (c) a pure maths chair See http://www.ma.ic.ac.uk/ for details (the links under Appointments:). Imperial College has one of the strongest pure maths departments in the UK; in the last research assessment exercise it got the (top) rating of 5*, one of only 4 departments in the country to do so. Note also that the starting salary for the lectureship is 35,000 pounds (Imperial pays significantly more than the standard wages for a young lecturer at a UK university). The post-docs are open to both pure and applied mathematicians but the other jobs are specifically pure; in fact they are specifically geometry---on the other hand this word is broadly interpreted (so e.g. it definitely includes algebraic and arithmetic geometry, for example). The geometry group at Imperial is very strong (headed up by Simon Donaldson and Richard Thomas) but there are also very active groups in algebraic number theory, algebra, analysis, PDEs, dynamical systems... See http://www.ma.ic.ac.uk/ for more details. Kevin Buzzard === Subject: Re: Linear homogeneous functions Epigone-thread: strirgharperd Originator: bergv@math.uiuc.edu (Maarten Bergvelt) We can solve a more general case. f(g(x1),g(x2),...g(xn))=g(f(x1,x2,...xn) 1) If phi(x)is Abel 'counting' function of g(u): phi(g(u))=phi(u)+1 and g(u)=phi^-1(phi(x)+1) A direct solution is: f(x1,x2,x3,...xn)=phi^-1( sum(ai.phi(xi))+k) 2) with {sum(ai) i=1 to n} = 1 ai,k ,c constants. For your case g(x)=a*x and phi(x)=ln(x)/ln(2)+ c. You've got an homogenous case in sci.math.research The book you spoke about is basic,rigorous but doesn't show nice and elaborate equations... Friendly,Alain === Subject: Re: Re: Linear homogeneous functions crooks. ..which is also apparently their view of the American public. WE ARE NOW AT AN HISTORICAL CROSSROAD ON THE ENCRYPTION ISSUE. BE ENHANCED FOR DECADES TO COME. [1984 Newspeak:] BUT IF NARROW INTERESTS PREVAIL, LAW ENFORCEMENT WILL BE UNABLE TO PROVIDE THE LEVEL OF PROTECTION THAT PEOPLE IN A DEMOCRACY PROPERLY EXPECT AND DESERVE. ANY SOLUTION THAT IGNORES THE PUBLIC SAFETY AND NATIONAL SECURITY CONCERNS RISK GRAVE HARM TO BOTH. And what was a critical public safety and national security item the FBI insisted on in the first version of CALEA? They wanted all cellular phones to continually monitor the location of the owner, EVEN WHEN NOT IN USE. Every cellular phone would become a location tracking monitor for the government. And why would this be a critical public safety and national security item? Because: The NSA/FBI are raving rabid frothing-at-the-mouth lying looneys. I hope you understand that by now. * Above the Law, by David Burnham, ISBN 0-684-80699-1, 1996 * * A few months after his appointment as the new director of the Federal * Bureau of Investigation, Louis J. Freeh delivered a speech at the National * Press Club in Washington. * * More than two hundred Washington-based reporters, congressional staffers * and interested lobbyists had come, and because the speech was carried by * C-Span, National Public Radio and the Global Internet Computer Network, * States, Freeh was also delivering his message to a much larger national * and international audience. * * The people of this country are fed up with crime, Freeh declared. The * media report it, the statistics support it, the polls prove it. * * To drive home his point and authenticate the national menace, Freeh said, * the rate of === Subject: Re: Re: Linear homogeneous functions victims from California were the fastest and easiest to identify, * the Medical Examiner, Dr. Charles V. Wetli, said, because the fingerprint * of their right thumb was on their driver's license. * * It was a nightmare for the other families to wait for identification. ---- Just how much does the government want to track us, by issuing tracking devices? Metrocard is a re-writeable magnetic card. It's new to us New Yorkers. They are individually serial-numbered. * Metrocards to Replace School Transit Passes * By John Sullivan, The New York Times, 8/26/1996 * * About 500,000 students will now have their bus and subway usage tracked by * Metrocards, in an effort to save money. Unlike current passes, which * students can use anytime between 6 A.M. and 7 P.M. on weekdays, the * Metrocard pass can be programmed to restrict the students to a set number * of trips a day. * * Ms. Gonzalez-Light, a spokeswoman for the Board of Education, said they * would work with the Transit Authority to individualize the number of * trips per student to adjust for extra-curricular activities. Then you could track each individual student? Decide if they might be truant including if they didn't use it, or went the wrong way? * Last Clink for Token-Only Turnstiles * By Garry Pierre-Pierre, The New York Times, May 14 1997 * * The last token-only turnstile was ripped out today. * * Officials have spent $700 million over a four-year period to === Subject: Re: Re: Linear homogeneous functions admissions, coverups, and more denials by Australian political leaders. * * is the world surveillance headquarters, and (2) Australia has it's own * secret computer center, linked with the NSA via satellite, which * illegally watches over Australia's citizenry. Article snippets... capitalization by the original authors... * On a fateful fall day in America, on November 4th, 1952, a new United * States government agency quietly was brought into existence through * presidential decree. * * The birth of the National Security Agency on that day so long ago * heralded the beginning of the world's most sophisticated and all * encompassing surveillance system, and the beginnings of the greatest * threat to individual liberty and freedom not only in Australia, but * the entire planet will ever see. * * The NSA grew out of the post war Signals Intelligence section of the * U.S. War Department. It is unique amongst government organizations in * America, and indeed most other countries, in that there are NO specified * or defined limits to its powers. * * The NSA can (and does) do just about whatever it wants, whenever, and * wherever it wants. Although little known in both the U.S. and elsewhere, * the NSA is quite literally the most powerful organization in the world. * * Not limited by any law, and answerable only to the U.S. National Security * Council through COMSEC, the NSA now controls an information and * surveillance network around the globe that even Orwell, in his novel * 1984, could not have imagined. * * Most people believe that the current computer age grew out of either * the space program or the nuclear weapons race; it did not. * * ALL significant advances in computer technology over the last thirty * years, from the very beginnings of IBM, through to the super computers * o === Subject: principal driection please help Originator: bergv@math.uiuc.edu (Maarten Bergvelt) For a point on a surface S(x, y), it has two principal curvatures k1, k2. What are the correspondent principal directions? I am using the the following notations(Sx mean derivative with S to x, Sxx means second order derivate...:) E = 1 + Sx * Sx; F = Sx * Sy; G = 1 + Sy * Sy; e = Sxx / sqrt(E * G - F * F); f = Sxy / sqrt(E * G - F * F); g = Syy / sqrt(E * G - F * F); a = E * G - F * F; b = - E * g - G * e + 2 * F * f; c = e * g - f * f; k1 and k2 are the solution of the equation: a * k * k + b * k + c = 0; I would like to find the the angle between the principal direction and the X axis. qq === Subject: Re: principal driection please help and seized his diary and address book. # # The FBI admitted to interviewing more than 100 people who visited # Nicaragua, but said they were acting under Presidential Executive Order. # # Two women have come forward to complain the IRS audited them IMMEDIATELY # AFTER RETURNING FROM NICARAGUA. # # The IRS denied it had anything to do with political views: One woman has # never earned more than 12,000 a year, and we found that suspicious. FBI director Sessions ended up apologizing BIG TIME on C-SPAN, saying that sort of thing would NEVER happen again. We have put procedures in place so that that will NEVER happen again. But, after having been granted the special powers of the court by Congress, noone was arrested and tried for this MASSIVE abuse of power, which was granted by Congress in the good faith that the government would not trade off the Bill of Rights in order to pursue political objectives. It was a worst-case disaster. Even after investigating, Congress basically yawned: The CISPES case was an aberration, it was lower-level FBI employees who got carried away by their national security mandate. It was not politically motivated --- The Senate Select Intelligence Committee. * Above the Law, by David Burnham, ISBN 0-684-80699-1, 1996 * * ...something much more sinister was at work. In his carefully documented * analysis of the CISPES matter, 'Break-ins, Death Threats and the FBI: the * Covert War Against the Central American Movement', Boston writer Ross * Gelbspan argues that a much more extensive conspiracy may have been at * work. Far from being a low-level operation, Gelbspan reports, hundreds * of documents in the CISPES file had been initialed by Oliver Buck * Revell, then the number two official in the FBI. [Further evidence * implicates the CIA] Congress is unable to investigate t === Subject: Re: principal driection please help by the government. Indeed, it seems to be accomplished in the quietest way possible, giving citizens the least amount of opportunity to choose their fate. Odd, since tax-payer paid-for government services is what gives them the power. But elected representatives will do, you say? Did you hear any of them mention it during campaigning? Did Alabama elected officials even mention it with their press release of a new driver's license, despite that being the plan? No. What does that tell you? We need a cabinet-level Privacy Commission, with the power to intervene nationwide. Power to protect us little people from fanatical personal data collection. We are losing it piece by piece. Who would have thought the United States would collect fingerprints from all citizens? Collect biometric information from everyone... law enforcement's Evil Holy Grail. * U.S. Has Plan to Broaden Availability tests of DNA Testing * By Fox Butterfield, The New York Times, undated but 1996 implied. * * In a little known provision of the Clinton Administration's 1994 Crime * Control Act was a call for the establishment of a nationwide DNA data * bank like the current national system for fingerprints, run by the FBI. * * In the two years since then, 42 states have passed laws requiring prison * inmates give blood or saliva samples for a DNA fingerprint. * * In a report today, the Justice Department said it is stepping === Subject: Re: principal driection please help Anyone can be made a criminal in the monitoring net. Or seduced into a crime, like Qubilah Shabazz. Without you realizing it, the person you met was taking advantage of knowing all your most passionate likes and dislikes. It is INSANE to design our systems for government monitoring. CONGRESS WAKE UP NOW FOR CHRIS'SAKES!!! * Dispute Arises Over Proposal for Wiretaps * By John Markoff, The New York Times, February 15 1997 * * The telephone companies, after meeting with the FBI, said they wanted to * be able to monitor tens of thousands of conversations simultaneously in * metropolitan areas, much more than their stated intention of simply * trying to transfer its current surveillance capabilities into the * digital era. * * And the Cellular Telephone Industry Association said the FBI wanted to * monitor 103,190 cellular calls simultaneously nationwide. * * Lawyers for AT&T Wireless Services said, The numbers alone are astounding. * * This is kind of scary, said Tom Wheeler, CTIA president. What does * the FBI know about our future that we don't? ---- You cannot assign people one-to-one to control everyone in a society. But you can control society in a HIGHLY effective way using cybernetics, and do so COST EFFECTIVELY. That's one of the things CALEA is about, cost effectiveness of maintaining the spying infrastructure when there are so many companies, new technologies, so many different data formats. Recording), which is a computer that monitors phone call logs and attaches to a PBX within a company and can generate long-distance expense reports by department, person, etc. We had to write a different program interface for every damn PBX manufacturer. The data format was different for each. NSA's spying operations are so massive and all encompassing, and the maintenance burden for interfacing t [CapitalEth].87.99.8fnewsdbm06.news.prodigy.com!newsdst01.n ews.prodigy.com!newsmst01a.news.prodigy.com!prodigy.com!news.glorb.com!news. k s.uiuc.edu!not-for-mail ¨.98.95.92Annales de Toulouse .91.8c.85.97.8d.98.95[O Tilde][Eth].97sci.math.research .93.98.8d.87.94.8e.9c[ AAcute].99.8e.95.94Annales de la facult,bC)(B des sciences de Toulouse .92.8e.94.8c.97.bc .b2.b2 .8d[Eth][Eth].98.95.9a.8c[ADo ubleDot]Maarten Bergvelt , moderator for sci.math.research .90.8c.97.97.87.8d.8c[Hyph en].83Ä .91.91.95[CapitalEth][Hyphen][CapitalEth ].95.97.99.8e.94.8d[Hyphen][CapitalEG rave].95.97.99.bc .9b.b2Ǭ.92.87.99.8f[Re gisteredTrademark].9b.8e.9b.8b¬ .8c.8a .9bí.95.94.99.8c.94.99[ Hyphen].95.98.87.94.97.be.8c.98 [Hyphen].81.94.8b.95.8a.8e[ IHat].8d8bit ø[Hyphen]í.95.92[Eth][IGrav e].87.8e.94.99.97[Hyphen].95[IDoubleDo t].bc .94.8c.85.97.82.87.89.9b[O Acute].8cè.91.97¬[O Tilde].8e.9b.8b¬.8c.8a .9b ø[Hyphen]í.95.8a.8e .94.8d[Hyphen].94.9d.97.99.8c .92.bc .8e.97.95[Hyphen].9f.9f[Micro].b9[Hyphen][PlusMi nus][Micro][Hyphen].9b.94.8eÀ .93.98.8e.8d.8e.94.87[ OHat].95.98bergv@math.uiuc.edu (Maarten Bergvelt) ø.98.8c.benewsmst01a.news.prodigy.com sci.math.research:24271 .8f.87appeared. It is vol. 1/2 of the proceedings of a conference in the .8f.95.94.95of Jean Ecalle, which was held in Nov. 2002. í.95.94.99.8c.94.99.97 .bc.be.8c.87Ecalle, Bruno Vallet .83.94.99.8c.98.99.85.8e[IH at].8cmappings í.8f.98.8e.97.99.8e[AAcute ]Even Ä.8c.8d.98å.89(Bs de libert,bi(B des moyennes de convolution pr,bi(Bservant la r,bi(Balit,bi(B ¨.98å.89(Bd,bi(Bric Menous .8d.94 .8cÀ.87.92[Eth].93of nonlinear q-difference equation .90.87.98.8e.9b.97 .85.87Der Put .82.87.93.95.8e.97 [Eth].98.95[Eth].8c.98.99.8e.8c[ OAcute] .95.be .93.8e.94.8c.87.98 .8a.8e.be.be.8c.98.8c.94.99[EAcu te].87.93 .8c.96.9b.87.99.8e.95.94 .95.8fabstracts and some full texts can be downloaded at: .8f.99.99[Eth].bc¿¿[Eth].8e.8b.87[ OGrave].8a¬.9b[Eth].97[Hyphen][ OHat].93.97.8c¬.be.98¿[Thor n].87.94.94.87.93.8c.97 Â.be.98.8c.94.8bversion) .8f.99.99[Eth].bc¿¿[Eth].8e.8b.87[ OGrave].8a¬.9b[Eth].97[Hyphen][ OHat].93.97.8c¬.be.98¿[Thor n].87.94.94.87.93.8c.97¿.8e[IH at].8a.8cÀ¤.8c.94¬[ EGrave].99.92.93 Â.8c.94.8d.93.8e.97.8f .9a.8c.98.97.8e.95.94[Copyrig ht] [CapitalEth].87.99.8fnewsdbm06.news.prodigy.com!newsdst01.n ews.prodigy.com!newsmst01a.news.prodigy.com!prodigy.com!news.maxwell.syr.edu ! newsfeed.yul.equant.net!news-santiago!news1.nivel5.cl!chi1.usenetserver.com! n ews.usenetserver.com!newsfeed-east.nntpserver.com!nntpserver.com!news.glorb. c om!news.ks.uiuc.edu!not-for-mail ¨.98.95.92Annales de Toulouse .91.8c.85.97.8d.98.95[O Tilde][Eth].97sci.math.research ¨.95.93.93.95.85.9b[ Eth][Hyphen].95.95.bc .94.8c.85.97¬.87.8a .92.8e.94¬.94.8c.99[Hyphen ].87.89.9b.97.8c¬.8c[IA cute].87.8e.93 .93.98.8d.87.94.8e.9c[ AAcute].99.8e.95.94Annales de la facult,bC)(B des sciences de Toulouse .92.8e.94.8c.97.bc .a6± .8d[Eth][Eth].98.95.9a.8c[ADo ubleDot]Annales de Toulouse .90.8c.97.97.87.8d.8c[Hyph en].83Ä .96.8c.be.8c.98.8c.94.8b[ARin g].97 .91.91.95[CapitalEth][Hyphen][CapitalEth ].95.97.99.8e.94.8d[Hyphen][CapitalEG rave].95.97.99.bc .b2ÁÁ¬.9f.b3¬26 2Á.b9¬±.b9[Micro] ø[Hyphen]í.95.92[Eth][IGrav e].87.8e.94.99.97[Hyphen].95[IDoubleDo t].bc .94.8c.85.97è.94.8c[CapitalODouble Dot][OA cute]±¬.94.8e.9a.8c .93[Micro]¬.8b.93 ø[Hyphen].90.87.8e.93.8c .98Mozilla 4.61 [en] (WinNT; I) ø.98.8c.benewsmst01a.news.prodigy.com sci.math.research:24313 .94.95have the power to seize your .89.9b.97.8e.94.8c.97.97home, bank account, records and personal property, .87.93without indictment, hearing, or trial. Everything you have can be taken away at the whim of one or two .be.8c.8a.8c.98.87or state officials operating in secret The loss of basic American constitutional guarantees: due process, .99.8fpresumption of innocence, and the right to own and enjoy [Eth].98.8e.9a.87.99property .83.92.87.8d.8e.94all that happened on one day. [Times].8f.87.99 .8a.95 .9d.95think would have happened next? í.8e.9a.8ewar would have broken out. [Times].8c .94.95 .93.95.94.8d.8c.98 .93.8e.9a.8c .8e.94 .99.8fhome of the brave, land of the free. [Times].8c .87.98controlled by the hand of the Freeh, beating the Drum of Fear. .83.99 .8f.87[Eth][Eth].8c.94.8c.8a .97.93.95.85.93.9d .95.9a.8c.98 .8a.8c.8b.87.8a.8c.97å steady .8a.98.9b.92[Hyphen].89.8c.87of destruction of the American Way. .94.95.93.8c.93.9d .be.95the benefit of those in power. .91.95for the people. .83.99Û.97 .97.9b[Eth][Eth].95.97.8c.8a .99.95 .89.8c Û.8d.95.9a.8c.98.94[IAcu te].8c.94*for* the people, by the people'. [Times].8c .8f.87.9a.8c .97.93.95.85.93.9d .98.8c.87.8b.8f.8ca state of McCarthyism against any elected .95.be.be.8e.8b.8e.87.93 .85.8fshows ANY SIGNS OF SOFTNESS in the War against Crime. .95.8fconstant state of War against imaginary enemies must end. .8c.9d .8e.92.87.8d.8e.94.87.98[UGr ave]å mean crime was going down the whole War time. .8d.93we are saying, is give peace a chance. [Hyphen][Hyphen][Hyphen][Hyphen] repeat: Civil war would have broken out. [Hyphen][Hyphen][Hyphen][Hyphen] Ä.8e.98.8c .97.9b.97[Eth].8c.94.97.8e[IDoubleDo t].94 .95.be í.95.94.97.99.8e.99.9b .99.8e.95.94.87.93 [Eth].98.95.99.8c.8b.99.8e[IDouble Dot].94happens during War: .8d.89.98.87.8f.87Lincoln ordered thousands of people detained without hearings, .87.94opposition newspapers shut down during the Civil War. During [Times].95.98.93War II: the president orders Japanese and such to be held in .8e.94.99.8c.98.94.92.8c.94[OHat ] .8b.87.92[Eth].97¬ .94.95 .85.8fdo we have all these loss of freedoms during peacetime? ¨.98.95.92Annales de Toulouse .91.8c.85.97.8d.98.95[O Tilde][Eth].97sci.math.research ¨.95.93.93.95.85.9b[ Eth][Hyphen].95.95.bc .94.8c.85.97¬.87.8a .92.8e.94¬.94.8c.99[Hyphen ].87.89.9b.97.8c¬.8c[IA cute].87.8e.93 .93.98.8d.87.94.8e.9c[ AAcute].99.8e.95.94Annales de la facult,bC)(B des sciences de Toulouse .92.8e.94.8c.97.bc .b3[Micro] .8d[Eth][Eth].98.95.9a.8c[ADo ubleDot]Annales de Toulouse .90.8c.97.97.87.8d.8c[Hyph en].83Ä .96.8c.be.8c.98.8c.94.8b[ARin g].97 .91.91.95[CapitalEth][Hyphen][CapitalEth ].95.97.99.8e.94.8d[Hyphen][CapitalEG rave].95.97.99.bc .b2ÁÁ¬.9f.b3¬26 2Á.b9¬±.b9[Micro] ø[Hyphen]í.95.92[Eth][IGrav e].87.8e.94.99.97[Hyphen].95[IDoubleDo t].bc .94.8c.85.97è.94.8c[CapitalODouble Dot][OA cute]±¬.94.8e.9a.8c .93[Micro]¬.8b.93 ê.97.8c.98[Hyphen].8d[CCedill a].8c.94.99Turnpike/6.00-Beta-3-U (<9IQbmnKYDsWNGxL7c08tE>) [CapitalEth].87.99.8fnewsdbm06.news.prodigy.com!newsdst01.n ews.prodigy.com!newsmst01a.news.prodigy.com!prodigy.com!newsfeed.cwix.com!ma u le!news.psinet.cl!news-santiago!news1.nivel5.cl!newsfeed.yul.equant.net!c03. a tl99.usenetserver.com!chi1.usenetserver.com!news.usenetserver.com!newsfeed-e a st.nntpserver.com!nntpserver.com!news.glorb.com!news.ks.uiuc.edu!not-for-mai l ø.98.8c.benewsmst01a.news.prodigy.com sci.math.research:24295 .97.8f.95.85.94¬ .91.95.85å .8c.9a.8c.98.9d.95.94.8c .85.8e.93.93 .87.8d.98.8c .8cÈ .99.8f.87.99 .94.95 .95.94should attempt to drive a train while high on marijuana. But a man's voice says that anyone who tells you 'marijuana is harmless' is lying, because his wife was killed in the train accident. This contradicts the direct sworn testimony of the engineer responsible for that disaster; that the accident was not caused by marijuana. It deliberately ignores his admissions of drinking alcohol, snacking, watching TV, generally failing to pay adequate attention to his job, accident. .97.97.94.8e[Eth][CapitalYAcute] ÈÈ .8d.94in yet another ad, the lies finally got them in trouble. The ad showed two brain wave charts which it said showed the brain waves of a 14-year- old on marijuana. Outraged, researcher Dr. Donald Blum from the UCLA neurological studies center told KABC TV (Los Angelos) news November 2, 1989, that the chart actually shows the brain waves of someone in a deep sleep --- or a coma. He said he had previously complained directly to the Partnership for a Drug Free America and they ignored him. They finally pulled the ads. .97.97.94.8e[Eth][CapitalYAcute] ÈÈ .95.8fHeath/Tulane Study of 1974 has been widely sited nationally as evidence that marijuana is harmful. One set of Rhesus monkeys began to atrophy and die after 90 days of pot smoking. When Playboy and NORMAL finally received the methodology of the study in 1980 after six years of trying, they were stunned. The Rhesus monkeys had b [CapitalEth].87.99.8fnewsdbm06.news.prodigy.com!newsdst01.n ews.prodigy.com!newsmst01a.news.prodigy.com!prodigy.com!atl-c02.usenetserver . com!feed5.newsreader.com!newsreader.com!border2.nntp.dca.giganews.com!nntp.g i ganews.com!wn13feed!worldnet.att.net!204.71.34.3!newsfeed.cwix.com!maule!hel c araxe.dcc.uchile.cl!news-santiago!news1.nivel5.cl!news.gsl.net!gip.net!newsf e ed.yul.equant.net!c03.atl99.usenetserver.com!chi1.usenetserver.com!news.usen e tserver.com!newsfeed-east.nntpserver.com!nntpserver.com!news.glorb.com!news. k s.uiuc.edu!not-for-mail ¨.98.95.92Annales de Toulouse .91.8c.85.97.8d.98.95[O Tilde][Eth].97sci.math.research ¨.95.93.93.95.85.9b[ Eth][Hyphen].95.95.bc .94.8c.85.97¬.87.8a .92.8e.94¬.94.8c.99[Hyphen ].87.89.9b.97.8c¬.8c[IA cute].87.8e.93 .93.98.8d.87.94.8e.9c[ AAcute].99.8e.95.94Annales de la facult,bC)(B des sciences de Toulouse .92.8e.94.8c.97.bc Ç.b2 .8d[Eth][Eth].98.95.9a.8c[ADo ubleDot]Annales de Toulouse .90.8c.97.97.87.8d.8c[Hyph en].83Ä .96.8c.be.8c.98.8c.94.8b[ARin g].97 .91.91.95[CapitalEth][Hyphen][CapitalEth ].95.97.99.8e.94.8d[Hyphen][CapitalEG rave].95.97.99.bc .b2ÁÁ¬.9f.b3¬26 2Á.b9¬±.b9[Micro] ø[Hyphen]í.95.92[Eth][IGrav e].87.8e.94.99.97[Hyphen].95[IDoubleDo t].bc .94.8c.85.97è.94.8c[CapitalODouble Dot][OA cute]±¬.94.8e.9a.8c .93[Micro]¬.8b.93 ø[Hyphen].91.8c.85.97.98[ ARing].87.8a.8c.98IBM NewsReader/2 v1.2.5 ø.98.8c.benewsmst01a.news.prodigy.com sci.math.research:24276 .93.98 .85.95.98.97.8c¬ Above the Law, by David Burnham, ISBN 0-684-80699-1, 1996 At 4:00 A.M. on December 4, 1969, for example, a special fourteen-man squad of Chicago police officers raided a house used by the Black Panther Party. During the shoot-first-ask-questions-later raid, police fired at least ninety-eight rounds into the apartment. Illinois chairman Fred Hampton and Peoria chairman Mark Clark were killed. An FBI informant gave the bureau specific information about where Hampton was probably sleeping, and a detailed floor plan of the house which the special squad used during its raid. Thirteen years later, in November 1982, District Court Judge John F. Grady determined that there was sufficient evidence of an FBI-led conspiracy to deprive the Panthers of their civil rights, and awarded the plaintiffs $1.85 million in damages. 5/30/97 MSNBC After more than a quarter of a century in prison, a Black Panther activist has won the right to a new trial. A judge ruled there had been prosecutorial misconduct. The judge overturned the conviction when it was disclosed the government prosecutors withheld critical evidence: o They never said the informer was working with and paid by the FBI. o A former FBI agent also agrees with his alibi: that he was in the Black Panther HQ at the time of the murder. That the FBI knew this because they were monitoring the HQ. o And the jury never knew the eyewitness, who has since died, had misidentified people in other cases. === Subject: job in Pure Mathematics at the University of Adelaide Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The School of Mathematical Sciences is advertising a tenurable lectureship. See http://www.adelaide.edu.au/jobs/10878.html Details are on that web site or contact me directly. Professor Michael Murray Head of School michael.murray@adelaide.edu.au === Subject: Re: job in Pure Mathematics at the University of Adelaide for their * Electronic Funds Transfer System (EFTS). ALL financial transactions * for the banks pass through there via subsidiary company, Funds * Transfers Services Pty Ltd. (FTS) * The New York Times INTERNATIONAL Wednesday, May 21, 1997 * by Clyde H. Farnsworth, Woomera, Australia. * * As Darth Vader's Death Star is blown to bits in the newly remastered Star * Wars at the local theater, the audience of Australian and American Air * Force personnel, and squadrons of their children, lets out a whoop. As the * lights go on, everyone is beaming. * * Pine Gap employs nearly 1,000 people, mainly from the CIA and the U.S. * National Reconnaissance Office. * * It is the ground station for a U.S. satellite network that intercepts * telephone, radio, data links and other communications around the world. Worldwide telephone interception. **************************************************************************** ** New Zealand: Unhappy Campers --- ------- ------- ------- You'd be unhappy too if French terrorists - er - French intelligence agency operatives sank a ship (GreenPeace!) on the shores of your country, and the USA controlled ECHELON system failed to warn you. Here comes the usual - GROAN: the usual - PLUS a description of the basic mechanisms. Spy tools, come 'n' get yer spy tools...a comprehensive look at ECHELON DICTIONARY. A look at the Beast in your phone. This is the big one. In the section after this, 'On Monitoring', I give detailed examples of the capability of ECHELON DICTIONARY to seek out information from noise. To pick out conversations from a massive dragnet. I even give the keyword monitoring logic for spotting conversations of people actively searching to leave their current job for another employer. First read this section. *** Secret Power by Nicky Hage === Subject: Re: job in Pure Mathematics at the University of Adelaide George Orwell, 1949, ISBN 0-679-41739-7 * * Hardly a week passed in which the Times did not carry a paragraph * describing how some eavesdropping little sneak --- 'child hero' was * the phrase generally used --- had overheard some compromising remark * and denounced his parents to the Thought Police. # The Emperor Wears No Clothes, by Jack Herer, 1992, ISBN 1-878125-00-1 # # The Police-taught DARE program encourages students to turn in # friends and family by becoming a police informant. : Real life: a child in school answers the friendly and inquiring police : officer teaching about drug dangers that yes their parents have some : of the displayed paraphernalia. : : A search warrant is issued, the parents are arrested, and : the child is put into custody of Child Welfare workers. # The Feds Under Our Beds, By James Bovard, The New York Times, 9/6/1995 # # The Justice Department confiscated the home of an elderly Cuban-American # couple in Miami after the couple was arrested for playing host to a weekly # poker game for family and friends. * Nynex Mistake Brings Scholarship Offer, NYT, 4/26/1995 * * Walter Ray Hill, 18, was arrested and jailed for two days based solely on * his phone number being used for a hoax bomb threat. * * Nynex eventually realized one of its employees transp === Subject: Paper published by Geometry and Topology Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following paper has been published: URL: http://www.maths.warwick.ac.uk/gt/GTVol8/paper39.abs.html Title: Limit groups and groups acting freely on R^n-trees Author(s): Vincent Guirardel Abstract: We give a simple proof of the finite presentation of Sela's limit groups by using free actions on R^n-trees. We first prove that Sela's limit groups do have a free action on an R^n-tree. We then prove that a finitely generated group having a free action on an R^n-tree can be obtained from free abelian groups and surface groups by a finite sequence of free products and amalgamations over cyclic groups. As a corollary, such a group is finitely presented, has a finite classifying space, its abelian subgroups are finitely generated and contains only finitely many conjugacy classes of non-cyclic maximal abelian subgroups. Secondary: 20E26 Keywords: R^n-tree, limit group, finite presentation Proposed: Martin Bridson Seconded: Benson Farb, Walter Neumann Author(s) address(es): Laboratoire E. Picard, UMR 5580, Bat 1R2 Universite Paul Sabatier, 118 rte de Narbonne 31062 Toulouse cedex 4, France Email: guirardel@picard.ups-tlse.fr === Subject: Approx triangularity by permutation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) In my research I've encountered a problem which seems likely to have been solved or at least considered by mathematicians already. However, I haven't been able to find much on the net. Given a square matrix A, the task is to permute the rows and the columns to yield a matrix which is as lower-diagonal as possible. The cost function could be the sum-of-squares of elements above the diagonal, or sum-of-absolute-values, or something similar. The elements of A will seldom be exactly zero, so number-of-nonzero-elements would not be very useful. In equations, given a matrix A, I'm looking to minimize something like C(P,Q) = sum_{i I am interested to know whether there is a > way to obtain all othogonal antisymmetric matrices > over the reals. Another obvious reformulation is that you look for orthogonal matrices A such that A^2 = -1. > Other then by generating all orthogonal > matrices O and then contructing O^T J O where J is the > block diagonal matrix with 2x2 blocks [0,-1;1,0]. What does it mean? Do you want a discription which is 1-to-1? This beast is called [*] isotropic Grassmannian. It is as complicatedv as Grassmannians go: Schubert cells, cohomology, and all this bruhaha. Each Schubert cell is easily 1-to-1-parameterized, but there are many of them: they should correspond to certain elements of the Weyl group of the orthogonal group. This group may be identified with the group of permutations of a certain form (one particular case was already noted on this thread). [Actually, I can't remember now any canonical text describing this; but googling for `Schubert cell OR cells isotropic Grassmannian` should get some hits...] The problem is equivalent to discription of n-dimensional planes on 2n-dimentional (complex) hyperboloid (i.e., a quadric in 2n+1-dimensional projective space; here your initial matrix is 2n+2 x 2n+2.) This is quite easy in small dimensions: if n=0, the quadric consists of 2 points, each of them is the plane in question. If n=1, the hyperboloid in 3-space has 2 families of lines: alpha-lines, and beta-lines. Each alpha-line intersects each beta-line in one point; through each point there passes one line of each family; thus the set of alpha-lines is in 1-to-1 correspondence with the set of points on a fixed beta-line. Thus the Grassmannian in question consists of 2 disjoint projective lines. The usual way to treat these Grassmannians is by induction: given a quadric Q of dimension 2n, take a point P on it. The quadric intersects its tangent plane at P over a quadratic cone (of dimension 2n-1). The base of this cone is a quadric Q' of dimension 2n-2 in a projective space. An n-plane in Q passing through P induces an n-1-plane in Q'. This is a step of induction; e.g., for n=2 it says that though any point of hyperboloid there passes 2 lines. In general, the Grassmannian always has two connected components (alpha and beta). Two alpha-planes intersect each other over a plane of even codimension. Any alpha-plane intersects any beta-plane over a plane of odd codimension. (Here empty set has dimension -1...) The largest Schubert cell is especially easy to describe: fix any alpha-plane S. Take a complementary isotropic plane S' (it will be alpha or beta depending on n mod 2). An isotropic plane close to S (automatically alpha) is a graph of a B: mapping S --> S'; the mappings which appear here are all antisymmetric mappings (this notion is canonically defined, but for the sake of simple discussion, consider dual bases in S and S'). Actually, not only alpha-planes close to S are described this way, but all alpha-planes except several families with smaller number of parameters (other, smaller, Schubert cells). For example, in a vector space with orthogonal basis X1,Y1,...,Xn,Yn let W be the subspace spanned by Xk, let T be a linear mapping exchanging X's with Y's. Then two subspaces S = {w+iTw | w in W}, S' = {w-iTw | w in W} are complementary isotropic; given a skewsymmetric n x n matrix B (considered as a mapping W --> W), an isotropic subspace I close to W is {w+iTw+Bw-iTBw | w in W}. The complex conjugate space I' is {w-iTw+B'w+iTB'w | w in W}; here B' is the complex conjugate matrix. The corresponding matrix A is -2 Im Pr; here Pr is the projector on I along I'. Calculation of Pr it simple, but tedious; I did its lower-left block-corner: is it inverse to (1+B)/(1-B)-i(1+B')/(1-B') (maybe this may be simplified more?). Hope this helps, Ilya [*] The correspondence associates to a matrix A its eigenspace of eigenvalue i. This eigenspace is obviously isotropic. In the other direction, given an isotropic subspace S, it cannot contain any non-0 real vector. Thus it does not intersect its complex conjugate S' (if v and its complex conjugate vector v' are both in S, then consider v+v' and i(v-v')). Consider a mapping A with S and S' eigenspaces with eigenvalues i and -i; since A is invariant w.r.t. complex conjugation, it is real. Taking arbitrary basis Zk in S, and the dual basis Zk' in S', consider the basis Zk+Zk', Zk-Zk'; it is othogonal, and normalized by 2. In this basis the matrix of A is orthogonal, thus the mapping A is orthogonal. That A^2 = -1 is obvious. === Subject: Re: orthogonal antisymmetric real matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > I am interested to know whether there is a >way to obtain all othogonal antisymmetric matrices >over the reals. Other then by generating all orthogonal >matrices O and then contructing O^T J O where J is the >block diagonal matrix with 2x2 blocks [0,-1;1,0]. Perhaps >some background would help. >These are precisely the antisymmetric matrices B such that >B^2 + I = 0. >Due to nonsingularity, all are of even order. >Given two or more such matrices, we can arrange >them along the diagonal to make a new matrix. Thus >we have a form of summation to get new solutions from old. >The examples of order 4 are sums aA+bB+cC with >a2+b2+c2 = 1 and the matrices A,B,C corresponding to >directed matchings on 4 points. There are three such matchings, and >the directions have to be chosen properly. >A = [0,1,0,0;-1,0,0,0;0,0,0,1;0,0,-1,0] >B = [0,0,1,0;0,0,0,-1;-1,0,0,0;0,1,0,0] >C = [0,0,0,1;0,0,1,0;0,-1,0,0;-1,0,0,0] This is not the only possibility. For example, you could also have [0,a,b,c; -a,0,-c,b; -b,c,0,-a; -c,-b,a,0] >One definite questions is: is there an example of order 6 which is >not the sum of solutions of order 2 and 4? Yes. I'm pretty sure the manifold of possible B's has dimension 6, with tangent space at [0,-1,0,0,0,0;1,0,0,0,0,0;0,0,0,-1,0,0;0,0,1,0,0,0;0,0,0,0,0,-1;0,0,0,0,1,0] given by matrices of the form [0,0,a,b,c,d;0,0,b,-a,d,-c;-a,-b,0,0,-f,-e;-b,a,0,0,-e,f;-c,-d,f,e,0,0; -d,c,e,-f,0,0] For an explicit example, you might e.g. take B = 625^(-1) [0, -500,-300,-180,-108, -81] [500, 0,-225, 240, 144, 108] [300, 225, 0,-400,-240,-180] [180,-240, 400, 0,-225, 300] [108,-144, 240, 225, 0,-500] [81, -108, 180,-300, 500, 0] Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: orthogonal antisymmetric real matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >> I am interested to know whether there is a >>way to obtain all othogonal antisymmetric matrices >>over the reals. Other then by generating all orthogonal >>matrices O and then contructing O^T J O where J is the >>block diagonal matrix with 2x2 blocks [0,-1;1,0]. Perhaps >>some background would help. >>The examples of order 4 are sums aA+bB+cC with >>a2+b2+c2 = 1 and the matrices A,B,C corresponding to >>directed matchings on 4 points. There are three such matchings, and >>the directions have to be chosen properly. >>A = [0,1,0,0;-1,0,0,0;0,0,0,1;0,0,-1,0] >>B = [0,0,1,0;0,0,0,-1;-1,0,0,0;0,1,0,0] >>C = [0,0,0,1;0,0,1,0;0,-1,0,0;-1,0,0,0] > This is not the only possibility. For example, you could also have > [0,a,b,c; -a,0,-c,b; -b,c,0,-a; -c,-b,a,0] Yes, you are quite right. However this (and other possibilities) are all permutation similar - essentially there is only one class of solutions here. >>One definite questions is: is there an example of order 6 which is >>not the sum of solutions of order 2 and 4? > Yes. I'm pretty sure the manifold of possible B's has dimension 6, > with tangent space at > [0,-1,0,0,0,0;1,0,0,0,0,0;0,0,0,-1,0,0;0,0,1,0,0,0;0,0,0,0,0,-1;0,0,0,0,1,0] > given by matrices of the form > [0,0,a,b,c,d;0,0,b,-a,d,-c;-a,-b,0,0,-f,-e;-b,a,0,0,-e,f;-c,-d,f,e,0,0; > -d,c,e,-f,0,0] This is interesting. You start with the matrix with three 2x2 blocks on the diagonal, then it looks like you act on them to move them off a bit. This would correspond to conjugating the block diagonal matrix by orthogonal matrices. I can explain this to myself as follows: the dimension of the Orthogonal group (or should I say the Lie algebra?) is 15. We lose 6 dimensions corresponding to the subgroup that moves the three blocks around (this is the symmetric group on 3 elements) and then three more for the 2x2 orth matrices that fix the blocks. So we are left with a 15 - 6 - 3 = 6D space of solutions. However this argument doesn't carry on. If we are looking at examples of dimension 2n, then the orthogonal group dimension is n(2n-1), while the symmetric group shifting blocks around has dimension n!, which increases too quickly. So I think I am using something wrongly here. It doesn't even work for n= 2 or 4. > For an explicit example, you might e.g. take > B = 625^(-1) [0, -500,-300,-180,-108, -81] > [500, 0,-225, 240, 144, 108] > [300, 225, 0,-400,-240,-180] > [180,-240, 400, 0,-225, 300] > [108,-144, 240, 225, 0,-500] > [81, -108, 180,-300, 500, 0] I would be interested to know how you constructed this example. Best, Tim === Subject: Re: orthogonal antisymmetric real matrices Content-Length: 3377 Originator: rusin@vesuvius >> I am interested to know whether there is a >>way to obtain all othogonal antisymmetric matrices >>over the reals. Other then by generating all orthogonal >>matrices O and then contructing O^T J O where J is the >>block diagonal matrix with 2x2 blocks [0,-1;1,0]. >>One definite questions is: is there an example of order 6 which is >>not the sum of solutions of order 2 and 4? > Yes. I'm pretty sure the manifold of possible B's has dimension 6, > with tangent space at > [0,-1,0,0,0,0;1,0,0,0,0,0;0,0,0,-1,0,0;0,0,1,0,0,0;0,0,0,0,0,-1;0,0,0,0,1,0] > given by matrices of the form > [0,0,a,b,c,d;0,0,b,-a,d,-c;-a,-b,0,0,-f,-e;-b,a,0,0,-e,f;-c,-d,f,e,0,0; > -d,c,e,-f,0,0] > This is interesting. You start with the matrix with three > 2x2 blocks on the diagonal, then it looks like you act on > them to move them off a bit. This would correspond to conjugating > the block diagonal matrix by orthogonal matrices. That's right. I looked at all conjugations by infinitesimal rotations in planes given by two coordinate directions. > I can explain this to myself as follows: the > dimension of the Orthogonal group (or should I say the Lie > algebra?) is 15. We lose 6 dimensions corresponding to the > subgroup that moves the three blocks around (this is the symmetric > group on 3 elements) and then three more for the 2x2 orth matrices > that fix the blocks. So we are left with a 15 - 6 - 3 = 6D > space of solutions. > However this argument doesn't carry on. If we are > looking at examples of dimension 2n, then the orthogonal group > dimension is n(2n-1), while the symmetric group shifting > blocks around has dimension n!, which increases too quickly. > So I think I am using something wrongly here. It doesn't even > work for n= 2 or 4. Just look at infinitesimal transformations. If B is an antisymmetic (2n)x(2n) matrix, JB - BJ is antisymmetric with (i,j) entry (+/-) B_{i',j} (+/-) B_{i,j'} where i' and j' are the indices paired with i and j respectively in the blocks of J. In particular if i and j are in the same block, this is 0. We get two elements of a basis of these matrices for each pair of distinct blocks (e.g. for the blocks {1,2} and {3,4} there's one matrix with its (1,3) and (4,2) entries 1 and its (2,4) and (3,1) entries -1, all else 0, and another with its (1,4) and (2,3) entries 1 and its (3,2) and (4,1) entries -1, all else 0). That should make the dimension n(n-1). > For an explicit example, you might e.g. take > B = 625^(-1) [0, -500,-300,-180,-108, -81] > [500, 0,-225, 240, 144, 108] > [300, 225, 0,-400,-240,-180] > [180,-240, 400, 0,-225, 300] > [108,-144, 240, 225, 0,-500] > [81, -108, 180,-300, 500, 0] > I would be interested to know how you constructed this example. Conjugating with rotations given by a 2 x 2 block, I think it was [ 4/5 3/5 ] [ -3/5 4/5 ] in various planes. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: properties of ergodic matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) It seems that no-one knows what the properties of an ergodic matrix are. I am still searching and nothing comes up. If anybody can help, i'd appreciate it a lot thank you === Subject: Re: properties of ergodic matrices of accidents? Dr. Teller saw only one side of the question. Clearly his emotional involvement with nuclear power arose not from a desire to benefit humanity but from a personal fulfillment he got from his work and from seeing it put to practical use. 89. The same is true of scientists generally. With possible rare exceptions, their motive is neither curiosity nor a desire to benefit humanity but the need to go through the power process: to have a goal (a scientific problem to solve), to make an effort (research) and to attain the goal (solution of the problem.) Science is a surrogate activity because scientists work mainly for the fulfillment they get out of the work itself. 90. Of course, it's not that simple. Other motives do play a role for many scientists. Money and status for example. Some scientists may be persons of the type who have an insatiable drive for status (see paragraph 79) and this may provide much of the motivation for their work. No doubt the majority of scientists, like the majority of the general population, are more or less susceptible to advertising and marketing techniques and need money to satisfy their craving for goods and services. Thus science is not a PURE surrogate activity. But it is in large part a surrogate activity. 91. Also, science and technology constitute a mass power movement, and many scientists gratify their need for power through identification with this mass movement (see paragraph 83). 92. Thus science marches on blindly, without regard to the real welfare of the human race or to any other standard, obedient only to the psychological needs of the scientists and of the government officials and corporation executives who provide the funds for research. THE NATURE OF FREEDOM 93. We are going to argue that industrial-technological society cannot be reformed in such a way === Subject: Re: properties of ergodic matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >It seems that no-one knows what the properties of an ergodic matrix are. >I am still searching and nothing comes up. >If anybody can help, i'd appreciate it a lot I think you may be referring to an irreducible aperiodic stochastic matrix. In this case, you might look at F.R. Gantmakher, Theory of Matrices, Chelsea 1959. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: properties of ergodic matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > It seems that no-one knows what the properties of an ergodic matrix are. > I am still searching and nothing comes up. > If anybody can help, i'd appreciate it a lot > thank you Google search on ergodic matrix shows 72 hits. It looks like http://wwwhome.math.utwente.nl/~doornea/papers/peisdev.pdf gives some information in the introduction -- provided they talk about the same thing as the one you inquire about. === Subject: The space of smoothly embedded n-disks in R^n Epigone-thread: khaisnemblen Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Consider the set of all *images* of smoothly embedded n-disks in R^n, and put a reasonable topology on it (e.g., coming from a metric on the set of their boundary S^(n-1)'s given, say, by the length of the (in some sense) shortest isotopy of R^n carrying either one onto the other). Call the resulting topological space EmbIm(n). ----------------------------------------------------- QUESTION: What is known about the topology of EmbIm(n)? (My guess would be that each EmbIm(n) is contractible.) Can anyone point me to, e.g., literature on this subject? (It would be especially nice if an explicit construction were given for a contraction of EmbIm(n) onto the standard unit disk.) Dan Asimov === Subject: About one mathematical problem. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Given one divergence-free vector field u is known in bounded open domain Omega of the space, say 2-D, with one continuous boundary gamma, does there exist one scalar function rho defined in Omega + gamma which can make vector field rho*u be divergence-curl-constant, that is: div( rho * u) = c1 curl(rho * u) = c2 c1 and c2 are arbitrary constants? Here the 2-D curl operator is scalar: curl(u) := -d_y u_x + d_x u_y. J. Yuan === Subject: Re: About one mathematical problem. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Given one divergence-free vector field u is known in bounded open >domain Omega of >the space, say 2-D, with one continuous boundary gamma, does there >exist one scalar function rho defined in Omega + gamma which can make >vector field rho*u be divergence-curl-constant, that is: >div( rho * u) = c1 >curl(rho * u) = c2 >c1 and c2 are arbitrary constants? Here the 2-D curl operator is >scalar: >curl(u) := -d_y u_x + d_x u_y. Try e.g. u = exp(y)*x * i + j in an arbitrary domain. I get rho_x * exp(y) + rho_y = c1 - rho_x + rho_y * exp(y) + exp(y) * rho = c2 The general solutions to these are rho = F(exp(y)+x) - c1*y and rho = exp(-y)*G(exp(-y)+x) + c2*y*exp(-y) and it's clear that these are incompatible unless rho = c1 = c2 = 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: About one mathematical problem. mean those aspects of the functioning of the human individual that are not subject to regulation by organized society but are products of chance, or free will, or God (depending on your religious or philosophical opinions). 184. Nature makes a perfect counter-ideal to technology for several reasons. Nature (that which is outside the power of the system) is the opposite of technology (which seeks to expand indefinitely the power of the system). Most people will agree that nature is beautiful; certainly it has tremendous popular appeal. The radical environmentalists ALREADY hold an ideology that exalts nature and opposes technology. [30] It is not necessary for the sake of nature to set up some chimerical utopia or any new kind of social order. Nature takes care of itself: It was a spontaneous creation that existed long before any human society, and for countless centuries many different kinds of human societies coexisted with nature without doing it an excessive amount of damag === Subject: Re: About one mathematical problem. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Given one divergence-free vector field u is known in bounded open >domain Omega of >the space, say 2-D, with one continuous boundary gamma, does there >exist one scalar function rho defined in Omega + gamma which can make >vector field rho*u be divergence-curl-constant, that is: >div( rho * u) = c1 >curl(rho * u) = c2 >c1 and c2 are arbitrary constants? Here the 2-D curl operator is >scalar: >curl(u) := -d_y u_x + d_x u_y. > Try e.g. u = exp(y)*x * i + j in an arbitrary domain. I get > rho_x * exp(y) + rho_y = c1 > - rho_x + rho_y * exp(y) + exp(y) * rho = c2 > The general solutions to these are > rho = F(exp(y)+x) - c1*y > and > rho = exp(-y)*G(exp(-y)+x) + c2*y*exp(-y) > and it's clear that these are incompatible unless > rho = c1 = c2 = 0. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Hi Robert, The original vector field u must be divergence-free. This means: div(u) = 0 or u can be expressed as the gradient of one potential field g: u = grad (g). Jing Yuan === Subject: Re: About one mathematical problem. Content-Length: 310 Originator: rusin@vesuvius Sorry. I made one error in the previous message. The divergence-free vector field can be also expressed as the curl of another vector field g: u = curl x g . In 2-D, u can be expressed the perp-gradient of one potential field g: u = grad^{T} (g) . Here grad^{T} (g) = -d_y g i + d_x g j . Jing Yuan === Subject: Glimm schemes Originator: bergv@math.uiuc.edu (Maarten Bergvelt) In 1965 J. Glimm introduced a technique for approximations of hyperbolic nonlinear systems that has been later known as the Glimm scheme. What is the state of the art today with respect to this technique? Is is still considered as a good method or are there better methods today? === Subject: Re: Glimm schemes Content-Length: 1402 Originator: rusin@vesuvius + ingehap@start.no (Inge H. A. Pettersen): | In 1965 J. Glimm introduced a technique for approximations of | hyperbolic nonlinear systems that has been later known as the Glimm | scheme. What is the state of the art today with respect to this | technique? Is is still considered as a good method or are there | better methods today? It was never considered a very efficient method for numerical computation. Glimm and his group were themselves pursuing other numerical methods, termed front tracking methods explicitly tracking the fronts and using higher order methods for the smooth parts of the solutions. Another method was pioneered in Oslo, somewhat confusingly also called front tracking. Then there is are lots of finite volume methods. Some random references: Helge Holden, Nils Henrik Risebro: Front tracking for hyperbolic conservation laws. Randall J. LeVeque: Numerical methods for conservation laws. Randall J. LeVeque: Finite Volume Methods for Hyperbolic Problems. -- * Harald Hanche-Olsen - Debating gives most of us much more psychological satisfaction than thinking does: but it deprives us of whatever chance there is of getting closer to the truth. -- C.P. Snow [I posted a somewhat longer followup before, but it must have gotten lost on its way to the moderators, or something. Stupidly I did not keep a copy around.] === Subject: Re: Glimm schemes Content-Length: 1710 Originator: rusin@vesuvius various techniques for hyperbolic nonlinear system with respect to numeric efficiency and with respect to analytical properties like convergence and stability? > + ingehap@start.no (Inge H. A. Pettersen): > | In 1965 J. Glimm introduced a technique for approximations of > | hyperbolic nonlinear systems that has been later known as the Glimm > | scheme. What is the state of the art today with respect to this > | technique? Is is still considered as a good method or are there > | better methods today? > It was never considered a very efficient method for numerical > computation. Glimm and his group were themselves pursuing other > numerical methods, termed front tracking methods explicitly tracking > the fronts and using higher order methods for the smooth parts of the > solutions. Another method was pioneered in Oslo, somewhat confusingly > also called front tracking. Then there is are lots of finite volume > methods. > Some random references: > Helge Holden, Nils Henrik Risebro: Front tracking for hyperbolic > conservation laws. > Randall J. LeVeque: Numerical methods for conservation laws. > Randall J. LeVeque: Finite Volume Methods for Hyperbolic Problems. > -- > * Harald Hanche-Olsen than thinking does: but it deprives us of whatever chance there is > of getting closer to the truth. -- C.P. Snow > [I posted a somewhat longer followup before, but it must have gotten > lost on its way to the moderators, or something. Stupidly I did not > keep a copy around.] === Subject: A continuous iteration of f(x,y) r>0 ; f(x,y)^[r] Epigone-thread: ghamsahslimp Originator: bergv@math.uiuc.edu (Maarten Bergvelt) f:R*R->R .Let us start with a simple case: g(x,g(x,y))= f(x,y) f given (1) or f(x,y)^[1/2]=g(x,y) Example: f(x,y)=y/(1+x*y) , g(x,y)=2*y/(2+x*y) verifies (1). By the same way will have:f(x,y)^[1/p]=p*y/(p+x*y)=g(x,y) p integer verifying g(x,g(x....)))n times =f(x,y)=y/(1+x*y). We may generalize: if f(x,y)=phi^-1(phi(y)+n(x)) or m^[n(x)](y) , phi(m(y))=phi(y)+1 ; then f(x,y)^[r]=phi^-1(phi(y)+r.n(x)) or m^[r.n(x)](y) , here r is a positive real number. Please your comments and ideas,alain. === Subject: Re: A continuous iteration of f(x,y) r>0 ; f(x,y)^[r] Originator: israel@math.ubc.ca (Robert Israel) > f:R*R->R .Let us start with a simple case: > g(x,g(x,y))= f(x,y) f given (1) > or f(x,y)^[1/2]=g(x,y) > Example: f(x,y)=y/(1+x*y) , g(x,y)=2*y/(2+x*y) verifies (1). > By the same way will have:f(x,y)^[1/p]=p*y/(p+x*y)=g(x,y) p integer > verifying g(x,g(x....)))n times =f(x,y)=y/(1+x*y). > We may generalize: > if f(x,y)=phi^-1(phi(y)+n(x)) or m^[n(x)](y) , > phi(m(y))=phi(y)+1 ; > then f(x,y)^[r]=phi^-1(phi(y)+r.n(x)) or m^[r.n(x)](y) , > here r is a positive real number. > Please your comments and ideas,alain. Alain, my tetration.org web site is dedicated to continuous iteration. See http://www.tetration.org/Dynamics/index.html for resources on continuous iteration and http://www.tetration.org/Combinatorics/index.html for an outline of my combinatorial approach to the subject. For ideas consider Stephen Wolfram's question: > How can one extend recursive function definitions to continuous > numbers? What is the continuous analog of the Ackermann function? The > symbolic forms of the Ackermann function with a fixed first argument > seem to have obvious interpretations for arbitrary real or complex > values of the second argument. But is there a general way to extend > these kinds of recursive definitions to continuous cases? Given a way > to do this, how does it apply to recursive definitions like those on > page 130? What happens to all the irregularities when one is between > integer values? Or is it only possible to find simple continuous > generalizations to functions that show fundamentally simple behavior? > Can this be used as a characterization of when the behavior is simple? For my non-peer-reviewed response to Wolfram's question on the NKS Forum see http://forum.wolframscience.com/showthread.php?s=&threadid=488 Here's why I'm interested in the subject. A theory of continuously iterated smooth matrix functions would probably encompass all of dynamics in physics. A theory of continuously iterated smooth complex functions would be powerful enough to extend the Ackermann function to complex numbers. Daniel Geisler === Subject: Re: Galois group of a given quartic equation Epigone-thread: twomprendlex Originator: bergv@math.uiuc.edu (Maarten Bergvelt) This continues a line of argument using the primes that are totally ramified in cyclic quartic extensions of Q. The conclusion is that if a, b, and c are positive integers, f(x) = x^4 - 2c*x^3 + (c^2 - a^2 - b^2)*x^2 + 2c*a^2 * x - a^2 * c^2 can never have Galois group Z4. It would be interesting if a more elementary argument could be found. As before, we assume that gcd(a, b, c) = 1. of Z4 as Galois group of f(x) to cases where the discriminant D = (4*a*b*c)^2 * ((c^2 - a^2 - b^2)^3 - 27*(abc)^2) and therefore the factor D1 = (c^2 - a^2 - b^2)^3 - 27*(abc)^2, is twice a square. The examples a = 32 b = 155 c = 247, D1 = 2 * 288560^2 a = 187 b = 224 c = 617, D1 = 2 * 62368976^2 give f(x) having discriminants that are twice squares, and Galois groups S4 and D4, respectively. If K is the splitting field of f(x), and K/Q is cyclic of degree 4, from previous arguments we know 2 must be the unique prime which is totally ramified in K/Q. For further consideration of the possibility of K/Q being cyclic of degree 4 when D is twice a square, we can use the criterion Mr. Holing mentioned in to distinguish the cases Z4 and D4, because we now know the field E = Q(sqrt(D)) is Q(sqrt(2)). If f(x) has Galois group Z4, it must factor over E. Clearly the factors are irreducible quadratics over E, and are algebraically conjugate. And since here f(x) is monic, the factors are monic, and the other coefficients are algebraic integers. So in our situation the factorization is of the form f(x) = (x^2 + (u + s*v)*x + u' + s*v')(x^2 + (u - s*v)*x + (u' - s*v') where s = sqrt(2) and u, v, u' and v' are rational integers. Equating coefficients, coeff of x^3: 2u = -2c coeff of x^2: 2u' + u^2 - 2v^2 = c^2 - a^2 - b^2 coeff of x: 2uu' - 4vv' = 2a^2 * c constant term: (u')^2 - 2(v')^2 = -a^2 * c^2 Substituting -c for u in the expression for the coefficient of x^2 gives a^2 + b^2 = 2v^2 - 2u', so if K/Q is cyclic of degree 4, a^2 + b^2 must be even (in the D4 example above, this sum is odd). Now since K/Q is cyclic of degree 4, and 2 is totally ramified in K/Q, we have from previous considerations f(x) == x^4 or (x+1)^4 (mod 2). This requires the coefficient c^2 - a^2 - b^2 of x^2 to be even. Since we are assuming gcd(a, b, c) = 1 the only possibility is that a and b are both odd, and c is even. Since any odd square is congruent to 1 modulo 8, we have c^2 - a^2 - b^2 == 2 (mod 4), so (c^2 - a^2 - b^2)^3 == 8 (mod 16) If c == 2 (mod 4) then 27*(abc)^2 == 4 (mod 8), which would force D1 == 4 (mod 8) which is inconsistent with being twice a square. Therefore, c must be divisible by 4. A similar argument shows that if 8|c then D1/8 == -1 (mod 8) which is incinsistent with D1 being twice a square. Therefore, c is divisible by 4 but not by 8. So, -a^2 * c^2 is divisible by 16 but not by 32. Now, using r1*r2*r3*r4 = -a^2 * c^2 where r1,r2, r3 and r4 are the zeroes of f(x), we have as in a previous argument, that if P is the unique prime of K containing 2, v_P(r1) = v_P(r2) = v_P(r3) = v_P(r4) = v_P(2). Thus, r1/2, r2/2, r3/2 and r4/2 are algebraic integers (but not in P). This causes no obvious difficulty with the coefficients of x or x^3. But with the coefficient of x^2, we have r1*r2 + r1*r3 + r1*r4 + r2*r3 + r2*r4 + r3*r4 = c^2 - a^2 - b^2. Each term on the left side is 4 times an algebraic integer, hence so is the sum. But as previously noted, c^2 - a^2 - b^2 == 2 (mod 4) so the coefficient of x^2 in f(x) is not 4 times an algebraic integer. This (finally!) eliminates the last possibility for f(x) to have Galois group Z4. === Subject: Re: Galois group of a given quartic equation Content-Length: 1159 Originator: rusin@vesuvius >This continues a line of argument using the primes that are totally >ramified in cyclic quartic extensions of Q. The conclusion is that if >a, b, and c are positive integers, >f(x) = x^4 - 2c*x^3 + (c^2 - a^2 - b^2)*x^2 + 2c*a^2 * x - a^2 * c^2 >can never have Galois group Z4. It would be interesting if a more >elementary argument could be found. I observe that f(x) = 0 is equivalent to a^2/x^2 + b^2/(x - c)^2 - 1 = 0 . If a/x = sin(theta) and b/(x - c) = cos(theta), then x = a*csc(theta) and x - c = b*sec(theta), so we are looking for solutions to a*csc(theta) - b*sec(theta) = c. Change our variable to t = tan(theta/2): a*(1 + t^2)/(2*t) - b*(1 + t^2)/(1 - t^2) = c . Clearing fractions gives a*t^4 + (2*b - 2*c)*t^3 + (2*b + 2*c)*t - a = 0 . This polynomial is not monic, but may be easier to analyze than the original. The two have identical Galois groups. -- If Scott Peterson had not been convicted, he would be Scott free. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Interesting new problems in combinatorial geometry Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I thought of these problems yesterday, 11/29/04. 1. There are n labeled points in the plane in general position, no 3 on a line. Draw a directed line from the lower numbered point Pi to the higher numbered point Pj, j>i. Some of the other points lie strictly to the left of the line, some strictly to its right. Let Eij be the # of points on the left minus the # of points on the right. Let E = Sum(n >= j > i >= 1)Eij. How should the points be arranged and labeled to maximize or minimize E (or make it exactly zero, if possible)? Can E=0 for any odd n? (No for n=3.) There are papers on the number of lines which cut a point set in half but this is different. 2. There are n points in the plane in general position: no 3 on a line, no 4 on a circle. There are n(n-1)(n-2)/6 circles through all triplets of points. Each circle Ci, 1 <= i <= n(n-1)(n-2)/6, strictly excludes Ti points and strictly encloses n-3-Ti. How should the n points be arranged to maximize or minimize Sum(i) of Ti, and what are the maximum and minimum? For about 6 quick trials with 5 points, I get exclusion totals over all 10 circles of only 10, 11, and 12. Is it always true that the total number of excluded points is at least equal to the number of circles? 3. Put n points are on a sphere and disallow any two being polar opposites, and define the exteriors of all circles as larger than their interiors, and ask the same question. 4. Are there arrangements where some points are always enclosed or excluded? What is the distribution of points enclosed or excluded, over all circles, and what property of the arrangement does this depend on? 5. The same question can be asked of plane squares and ellipses through the appropriate number of points. Conjectures? Proofs? Steve Gray === Subject: Re: Interesting new problems in combinatorial geometry know his lessons and patting him on the head when he does know them. It is becoming a scientific technique for controlling the child's development. Sylvan Learning Centers, for example, have had great success in motivating children to study, and psychological techniques are also used with more or less success in many conventional schools. Parenting techniques that are taught to parents are designed to make children accept fundamental values of the system and behave in ways that the system finds desirable. Mental health programs, intervention techniques, psychotherapy and so forth are ostensibly designed to benefit individuals, but in practice they usually serve as methods for inducing individuals to think and behave as the system requires. (There is no contradiction here; an individual whose attitudes or behavior bring him into conflict with the system is up against a force that is too powerful for him to conquer or escape from, hence he is likely to suffer from stress, frustration, defeat. His path will be mu === Subject: Re: Interesting new problems in combinatorial geometry Content-Length: 2300 Originator: rusin@vesuvius > I thought of these problems yesterday, 11/29/04. > 1. There are n labeled points in the plane in general position, no 3 on a line. Draw a > directed line from the lower numbered point Pi to the higher numbered point Pj, j>i. Some of the > other points lie strictly to the left of the line, some strictly to its right. Let Eij be the # of > points on the left minus the # of points on the right. Let E = Sum(n >= j > i >= 1)Eij. How should > the points be arranged and labeled to maximize or minimize E (or make it exactly zero, if possible)? > Can E=0 for any odd n? (No for n=3.) There are papers on the number of lines which cut a point set > in half but this is different. Writing Eij as sum_{k neq i,j} {1, if Pk is left of Pi->Pj; else, -1}, we can rewrite E as sum_{iPj; else -1}. In this formulation, each 3-element subset {a,b,c} of {1..n}, with a < b < c, contributes exactly three times to this sum: i = a, j = b, k = c; i = a, j = c, k = b; and i = b, j = c, k = a. If Pc is left of Pa->Pb, then this triplet contributes +1 to the sum for E, getting +1 contributions when i = a, j = b, k = c, and i = b, j = c, k = a, and a -1 for i = a, j = c, k = b. Call such a triplet {a,b,c} positively oriented. Similarly if Pc is right of Pa->Pb, the contribution of this triplet is -1 to the sum for E, and we can call such a triplet negatively oriented. With this in hand, we can write E as E = sum_{{a,b,c} in S(n,3)} {+1, if {a,b,c} is positively oriented {-1, if {a,b,c} is negatively oriented Thus it is clear that the maximum possible value for E is n choose 3, which occurs when all triplets {a,b,c} are positively oriented, and similarly that the minimum is the negative of n choose 3. These values are both realizable... evenly distribute the n points along the semicircle x^2+y^2=1, x>=0. Labelling the points 1 through n, increasing in a counterclockwise fashion from the bottom yields a labelled point-set where all triplets are positively oriented, and reversing this labelling produces the opposite. As far as E=0, since E is a sum of n choose 3 +/- 1's, it can only yield a zero sum when n choose 3 is even, which requires n neq 3 (mod 4). Don't know about the other cases. Good luck with your investigations!