mm-317 === Subject: : University of Sheffield: Lectureship and Chair in Pure MathematicsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)The University of Sheffield is seeking to appoint a Lecturer in PureMathematics and a new Professor of Pure Mathematics. Details are summarizedbelow, and are also on the University's Jobs and Recruitment Informationwebsite at http://www.shef.ac.uk/jobs/ (click on 'Academic') under ReferenceNumbers R3186 and R3188 respectively.------------------------------------------------- ---------------------------LECTURER IN PURE MATHEMATICSIt is expec that the appointee will have a proven research record(possibly with some postdoctoral experience), and will contribute to thehigh-quality research and teaching of the department. Preference may begiven to candidates whose research interests would place them in one of theexisting research clusters in the department: these are in algebraictopology, differential geometry, algebraic number theory, non-commutativering theory and commutative algebra. Pure Mathematics at the University ofSheffield was awarded a grade 5 in the 2001 Research Assessment Exercise.Informal enquiries may be made to Professor Rodney Sharp (0114 2223746,email R.Y.Sharp@sheffield.ac.uk) or Professor John Greenlees (0114 2223786,email J.Greenlees@sheffield.ac.uk). More information about the department'sresearch strengths is on the department's website athttp://www.shef.ac.uk/~puremath/.A scale (22,191 - 25,451 per annum), but it should be no that this postattracts a special HEFCE-funded 'Golden Hello' to the value of 9,000 paidin 3 instalments over 3 years, subject to individuals' satisfying theeligibility criteria.----------------------------------------------------- -----------------------PROFESSOR OF PURE MATHEMATICSApplications are invi from outstanding candidates for a new Professorshipin the Department of Pure Mathematics. Applicants should be able todemonstrate an excellent research record and future plans for the pursuit ofhigh-quality, internationally-leading research in areas of pure mathematicsthat link with some of those of the department's established researchclusters in algebraic topology, differential geometry, algebraic numbertheory, non-commutative ring theory and commutative algebra. PureMathematics at the University of Sheffield was awarded a grade 5 in the 2001mutually agreeable date. Informal enquiries may be made to Professor RodneySharp (0114 2223746, email R.Y.Sharp@sheffield.ac.uk) or Professor JohnGreenlees (0114 2223786, email J.Greenlees@sheffield.ac.uk). Moreinformation about the department's research strengths is on the department'swebsite at http://www.shef.ac.uk/~puremath/.for non-clinical professorial appointments.------------------------------------------------- -------------------------=== === Subject: : Paper published by Algebraic and Geometric TopologyOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)The following paper has been published:Algebraic and Geometric TopologyURL:http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-2 .abs.htmlTitle:L_delta groups are almost convex and have a sub-cubic Dehn functionAuthor(s):Murray ElderAbstract:We prove that if the Cayley graph of a finitely genera group enjoysthe property L_delta then the group is almost convex and has asub-cubic isoperimetric function.Secondary: 20F67Keywords:Almost convex, isoperimetric function, property L_deltaAuthor(s) address(es):School of Mathematics and Statistics University of St. Andrews North Haugh, St. Andrews Fife, KY16 9SS, ScotlandEmail: murray@mcs.st-and.ac.ukURL: http://turnbull.mcs.st-and.ac.uk/~murray/=== === Subject: : groupEpigone-thread: pramthashirOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)I have a few question: Let G a sub-group of the group of homeomorphisms of a metric space E,Homeo(E), such that each orbit of G is finite (the cardinality oforbits is not uniformly limi i.e for all n in IN there exists xin E such that card(G(x)) > n (G(x) is the orbit of x by G).A point x in E is regular if there exists an open neighboorhod U_x andan integer k_x such that for all yin U_x we have card(G(y)) < k_x.I want to study the set of irregular points (not regular).Is it true that if x is irregular then for all gin G g(x) = x.?Let m be the measure of Lebesgue on E and let S be the set ofirregular points.Is it true that if G is equicontinuous than m(S) = 0.?Thanks in advance.=== === Subject: : Re: groupOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)No, the answer to both questions is negative and this may be shown byone and the same example. Let x_n be a dense subset of S^2, take for any n a set A(n)consisting of 2^n points which are situa at distance < 1/n to x_n.Consider the set M=S^2U {A(n)U(-A(n)), n=1,...}. Now we may define in M a map f which is the antipodal map on S^2 andis permuting cyclically the points of A(n)U(-A(n)). Defining G={f^n,n=1,...}, then obviously the set of irregular points is S^2 and m(S^2)=1, where m is the 2-dimensional measure on M. Moreover, it iseasy to see that G is equicontinuous. If you are looking for a connec example, you may take the coneover M with a map F acting as f on each fiber, except the vertex. It would be interesting to find a group action as above in somecompact X with ALL points being irregular. Simeon > I have a few question:> Let G a sub-group of the group of homeomorphisms of a metric space E,> Homeo(E), such that each orbit of G is finite (the cardinality of> orbits is not uniformly limi i.e for all n in IN there exists xin> E such that card(G(x)) > n (G(x) is the orbit of x by G).> A point x in E is regular if there exists an open neighboorhod U_x and> an integer k_x such that for all yin U_x we have card(G(y)) < k_x.> I want to study the set of irregular points (not regular).> Is it true that if x is irregular then for all gin G g(x) = x.?> Let m be the measure of Lebesgue on E and let S be the set of> irregular points.> Is it true that if G is equicontinuous than m(S) = 0.?> Thanks in advance.=== === Subject: : Re: groupOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)No, the answer to both questions is negative and this may be shown byone and the same example. Let x_n be a dense subset of S^2, take for any n a set A(n)consisting of 2^n points which are situa at distance < 1/n to x_n.Consider the set M=S^2U {A(n)U(-A(n)), n=1,...}. Now we may define in M a map f which is the antipodal map on S^2 andis permuting cyclically the points of A(n)U(-A(n)). Defining G={f^n,n=1,...}, then obviously the set of irregular points is S^2 and m(S^2)=1, where m is the 2-dimensional measure on M. Moreover, it iseasy to see that G is equicontinuous. If you are looking for a connec example, you may take the coneover M with a map F acting as f on each fiber, except the vertex. It would be interesting to find a group action as above in somecompact X with ALL points being irregular. Simeon > I have a few question:> Let G a sub-group of the group of homeomorphisms of a metric space E,> Homeo(E), such that each orbit of G is finite (the cardinality of> orbits is not uniformly limi i.e for all n in IN there exists xin> E such that card(G(x)) > n (G(x) is the orbit of x by G).> A point x in E is regular if there exists an open neighboorhod U_x and> an integer k_x such that for all yin U_x we have card(G(y)) < k_x.> I want to study the set of irregular points (not regular).> Is it true that if x is irregular then for all gin G g(x) = x.?> Let m be the measure of Lebesgue on E and let S be the set of> irregular points.> Is it true that if G is equicontinuous than m(S) = 0.?> Thanks in advance.=== === Subject: : Paper published by Geometry and TopologyOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)The following paper has been published:URL:http://www.maths.warwick.ac.uk/gt/GTVol8/paper1. abs.htmlTitle:Modular circle quotients and PL limit setsAuthor(s):Richard Evan Schwartz Abstract:We say that a collection Gamma of geodesics in the hyperbolic planeH^2 is a modular pattern if Gamma is invariant under the modular groupPSL_2(Z), if there are only finitely many PSL_2(Z)-equivalence classesof geodesics in Gamma, and if each geodesic in Gamma is stabilized byan infinite order subgroup of PSL_2(Z). For instance, any finite unionof closed geodesics on the modular orbifold H^2/PSL_2(Z) lifts to amodular pattern. Let S^1 be the ideal boundary of H^2. Given twopoints p,q in S^1 we write p~q if p and q are the endpoints of ageodesic in Gamma. (In particular p~p.) We show that ~ is anequivalence relation. We let Q_Gamma=S^1/~ be the quotient space. Wecall Q_Gamma a modular circle quotient. In this paper we will give asense of what modular circle quotients `look like' by realizing themas limit sets of piecewise-linear group actionsSecondary: 54E99, 51M15Keywords:Modular group, geodesic patterns, limit sets, representationsProposed: David GabaiSeconded: Martin Bridson, Walter NeumannAuthor(s) address(es):Department of Mathematics, University of Maryland College Park, MD 20742, USAEmail: res@math.umd.edu=== === Subject: : Integer Linear Problem and Dynamic ProgrammingOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)i have the follwoing integer linear problem. sum(Min(4a11+6a21,1),Min(4a12+6a22+3a31,4),Min(6a23+5a41,10), Min(5a42,4))S.T:a11+ a12+....=1 //we can just choose one to be 1 at the same timea21+ a22+....=1a31+ a32+....=1......an1+ an2+....=1for all i,j: aij={0,1} //the solution must be 1/0 I know how to solve it in (n^2)logn when using dynamic pogramming. (nis the number of a's)Does someone know a better way to solve this kind of problem? (lowercomlexity)Michal=== === Subject: : Re: An Inequality with Dirichlet SeriesOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> Hi. Is it true that for a real valued function f(x),> sum_{m=1}^infty m^{-r}(sum_{d|m} f(d))^2 <=> zeta^3(r)sum_{m=1}^infty f(m)/m^r?[snip]No.Take f a constant function, say a. Then your LHS isa^2 * sum_m m^{-r}*d(m)^2where d(m)=number of divisors of m, hence d(m)>1 when m>1therefore LHS > a^2 * sum_m m^{-r} = a^2 zeta(r).On the other hand RHS = a*zeta(r)^4 < LHS.The last inequality is neither for all a nor for all r, but it is easy tochoose such a and r that it holds. E.g. r>1, a>=1.All the best,Teijo=== === Subject: : Re: An Inequality with Dirichlet SeriesEpigone-thread: salmerljongOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)Oops. Typo...The f should be squared on the LHS in> sum_{m=1}^infty m^{-r}(sum_{d|m} f(d))^2 <=> zeta^3(r)sum_{m=1}^infty f(m)/m^r.That is, the question should read, is it always true thatsum_{m=1}^infty m^{-r}(sum_{d|m} f(d))^2 <=zeta^3(r)sum_{m=1}^infty f(m)^2/m^r?In this case, the inequality is true for constant functions f.Apologies for the misleading typo,Brad=== === Subject: : Assistance with combination problemOriginator: bergv@math.uiuc.edu (Maarten Bergvelt) I would appreciate is anybody could help me with the following question: lets A be such that : 0 <= A <= 4294967296 (i.e. 2^32) what is the formula giving me the number of 4 integers combinations (N1, N2, N3, N4) (order does not matter) 0 <= N1,N2,N3,N4 <= 4294967296 that comply to the following requirement: N1 + N2 + N3 + N4 = Ae.g. A = 53 (53,0,0,0) (24,13,7,9) (50,2,1,0) ......Thanks for the help,Dan=== === Subject: : Re: Assistance with combination problemOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> I would appreciate is anybody could help me with the following question:> lets A be such that : 0 <= A <= 4294967296 (i.e. 2^32)> what is the formula giving me the number of 4 integers combinations> (N1, N2, N3, N4) (order does not matter)> 0 <= N1,N2,N3,N4 <= 4294967296> that comply to the following requirement: N1 + N2 + N3 + N4 = A> e.g. > A = 53> (53,0,0,0) (24,13,7,9) (50,2,1,0) ......> Thanks for the help,> DanDan,Here's an easy-to-remember recursion, reminiscent of Pascal'striangle. Since another post in this thread has used p(n,k) for thenumber of partitions of n into at most k parts, I'll use q(n,k) forthe number of partitions of n into exactly k parts.Each such partition has a smallest part, say x. If x > 1, we cansubtract 1 from each member and get a partition of n-k into k parts.The remaining partitions (x = 1) have smallest part 1, so deletingthat (delete only one if there are several) part gives a partition ofn-1 into k-1 parts.So q(n,k) = q(n-k,k) + q(n-1,k-1). FarrDPRA, Inc.=== === Subject: : Re: Assistance with combination problemOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> I would appreciate is anybody could help me with the followingquestion:> lets A be such that : 0 <= A <= 4294967296 (i.e. 2^32)> what is the formula giving me the number of 4 integers combinations> (N1, N2, N3, N4) (order does not matter)> 0 <= N1,N2,N3,N4 <= 4294967296> that comply to the following requirement: N1 + N2 + N3 + N4 = A..Since order does not matter and you allow 0, what you want is the number ofpartitions of A into at most 4 parts.There is no such thing as the formula. There are many formulas.For A = 4294967296, I obtained the value 550195574937260409780728264using the formula (in Maple):1/144 *n^3 +5/48 *n^2 +15/32 *n +175/288 +Sum(1/16 *_alpha^(-n),_alpha =RootOf(_Z^2 +1)) +Sum(1/27 *(-1 +_alpha) *_alpha^(-1-n),_alpha = RootOf(_Z^2+_Z +1)) +1/32 *(-1)^(-n) *n +5/32 *(-1)^(-n)given at http://algo.inria.fr/bin/encyclopedia?Search=ECSnb&argsearch= 353More generally p(n,k) is the number of partitions of n into at most k parts.And the formula above is for k = 4. I found a couple of other formulasonline which turned out to be incorrect for small values of n. So it pays tocheck any other formulas you come across.In addition to the above you might check out http://www.research.att.com/projects/OEIS?Anum=A001400and the references give there. Clark=== === Subject: : Re: Assistance with combination problemOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> I would appreciate is anybody could help me with thefollowing question:> lets A be such that : 0 <= A <= 4294967296 (i.e.2^32)> what is the formula giving me the number of 4 integerscombinations> (N1, N2, N3, N4) (order does not matter)> 0 <= N1,N2,N3,N4 <= 4294967296> that comply to the following requirement: N1 + N2 +N3 + N4 = A> e.g.> A = 53> (53,0,0,0) (24,13,7,9) (50,2,1,0) ......******************************************************** ******You can think of the problem this way. Suppose A=8. Wewill represent a decomposition this way: **|***|*|** for (2, 3, 1, 2).All decompositions can be made with eight *'s and three |'s.We have 11 positions in the line to fill and we want toselect three positions for the |'s. In how many ways can wedo this? (Note that | can appear anywhere in the line. Forexample, |||******** would represent (0, 0, 0, 8).) Nowgeneralize.________________________________________________ _________Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics and StatisticsYoungstown State UniversityOne University PlazaYoungstown, OH 44555-0001330-941-1817=== === Subject: : Re: Assistance with combination problemOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>> lets A be such that : 0 <= A <= 4294967296 (i.e.>2^32)>> what is the formula giving me the number of 4 integers>combinations>> (N1, N2, N3, N4) (order does not matter)>> 0 <= N1,N2,N3,N4 <= 4294967296>> that comply to the following requirement: N1 + N2 +>N3 + N4 = A>You can think of the problem this way. Suppose A=8. We>will represent a decomposition this way:> **|***|*|** for (2, 3, 1, 2).>All decompositions can be made with eight *'s and three |'s.>We have 11 positions in the line to fill and we want to>select three positions for the |'s. In how many ways can we>do this? (Note that | can appear anywhere in the line. For>example, |||******** would represent (0, 0, 0, 8).) Now>generalize.The OP said order doesn't matter, but it would matter in this solution, e.g. (0,0,0,8) and (0,0,8,0) would be coun separately.If P_j(n) is the number of ways to write n as the sum of j nonnegativeintegers (where order doesn't count), we have the recurrenceP_j(n) = sum_{k=0}^{floor(n/j)} P_{j-1}(n - j k)with P_1(n) = 1. Then I getP_2(n) = n/2 + 3/4 + (-1)^n/4P_3(n) = n^2/12 + n/2 + 47/72 + 2/9*cos(2*Pi*n/3) + (-1)^n/8;P_4(n) = n^3/144 + 5*n^2/48 + (15 + (-1)^n)*n/32 + 175/288 + 5*(-1)^n/32 + 1/8*cos(Pi*n/2) + 1/9*cos(2*Pi*n/3) + sqrt(3)/27*sin(2*Pi*n/3)This is sequence A001400 at the Online Encyclopedia of Integer Sequences,Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: mesh network questionOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> X> o--+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+> Rv Rv Rv | | Rv Rv> +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+> Rv Rv Rv | | Rv Rv> . . . .(m rows) .--Rh--+--Rh--+> Rv Rv Rv | | Rv Rv> +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+> Rv Rv Rv | | Rv Rv> +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+--o YThe resistance can be found by dividing the number of spanning treesinto the number of spanning trees there would be if the ends wereshor together. The number of spanning trees is (mn) divided intothe product of the non-zero eigenvalues of the Laplacian matrix.For each extra resistor connec directly across from X to Y, thenumber of spanning trees increases by the number of spanning treesthere would be if X and Y were shor together. The derivative ofthis function can be found by adding an infinitesimal increment ofconductance directly from X to Y and calculating the spanning treefunction increase using a perturbation method that is commonly usedfor quantum calculations.The mesh eigenvalues are Gh(2-2cos(Pi i/n)) + Gv(2-2cos(Pi j/m)).The eigenvectors: sqrt(c/(mn)) cos(Pi(k+1/2) i/n) cos(Pi(h+1/2) j/m),with c = 1, 2 or 4 if both, one or neither of i and j is zero.The eigenvector matrix transforms the coordinates of the Laplacianmatrix L producing the diagonalized matrix U*LU with the eigenvalueson the diagonal.To perturb, add p to L(X,X) and L(Y,Y), and -p to L(X,Y) and L(Y,X).For X, k=0 and h=0. For Y, k=n-1 and h=m-1.Since for k=n-1, cos(Pi(k+1/2) i/n) = (-1)^i cos(Pi/2 i/n),the remapped perturbation adds the following to each eigenvalue:q(cos(Pi/2 i/n)cos(Pi/2 j/m)-(-1)^(i+j)cos(Pi/2 i/n)cos(Pi/2 j/m))^2equaling: (pc)/(mn) (2 cos(Pi/2 i/n)cos(Pi/2 j/m))^2 for (i+j) odd,and zero otherwise. The off-diagonal elements of the remappedperturbation don't contribute to the first order result.Using (E1 E2 ...)' = (E1 E2 ...)(E1'/E1 + E2'/E2 + ...),and then dividing by (E1 E2 ...) which is (mn) times the tree count,where the Es are eigenvalues and E' means dE/dp,the resistance formula comes out:1/(2mn) Sum_{i=0 to n-1 and j=0 to m-1, and i+j is odd}c(1+cos(Pi i/n))(1+cos(Pi j/m))/(Gh(1-cos(Pi i/n))+Gv(1-cos(Pi j/m)))where c depends on i and j as above.=== === Subject: : International Journal of Mathematics - Vol 14 No 10Originator: bergv@math.uiuc.edu (Maarten Bergvelt)International Journal of MathematicsView table-of-contents and abstracts athttp://www.worldscinet.com/ijm.htmlContents:On Weakly Hyperbolic Spaces And A Convergence-Extension Theorem InWeakly Hyperbolic SpacesPham Viet DucOn Branched Coverings Of Pn By Products Of DiscsA. Muhammed UludagCoordinates for the Teichm.9fller Space of Planar Surface NEC GroupsB. Estrada and E. MartinezRationality Properties Of Manifolds Containing Quasi-LinesPaltin Ionescu And Daniel NaieReflexivity Of The Translation-Dilation Algebras On L2(R)R. H. Levene and S. C. PowerA Remark On A Question Of LempertHenkinViet Anh NguyenStable Rank-2 Bundles On CalabiYau ManifoldsWei-Ping Li And Zhenbo QinFor more information, go to http://www.worldscinet.com/ijm.html=== === Subject: : Re: Large cardinalsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>I'm not acquain with serious set theory but I've had to look for>some info on real valued measurable cardinals needed for a proof.>Fortunately there are many info sources quickly and freely available>in the web and I mostly found what I was searching for, but there is>an interesting bit for which I don't have a reference.>I've read that the existence of a measurable cardinal (a large>cardinal indeed, I believe) cannot be proven within ZFC because the>existence of large cardinals entails the consistency of ZFC.>Question:>Is that an effect of there existing a model of ZFC where large>cardinals do not exist? Or does the statement refer rather to the>unprovability by means of ZFC of the existence of large cardinal in>every model where large cardinals do actually exist?It means that we cannot prove in ZFC that there is amodel even of ZF in which large cardinals exist. Wecan prove that there is a model in which no largecardinals exist, as restricting elements to thosewhose rank is less than the smallest large cardinalmust form a model. This does not do anything to settle the question of whether the existence oflarge cardinals is consistent.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: Large cardinalsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> I've read that the existence of a measurable cardinal (a large> cardinal indeed, I believe) cannot be proven within ZFC because the> existence of large cardinals entails the consistency of ZFC.> Question:> Is that an effect of there existing a model of ZFC where large> cardinals do not exist? If large cardinals exis in all models, this would be provable in ZFC by the completeness theorem. And since ZFC does not prove its consistency, if they exis in all models, they could not imply the consistency of ZFC. But not everything that has this property - not holding in all models -, entails consistency. For example, the statement ~Cons(ZFC) asserting the inconsistency of ZFC is independent of ZFC, and is thus false in some models, but this doesn't mean from ~Cons(ZFC) Cons(ZFC) follow. Thus the consistency implication is not an effect of there existing a model of ZFC where large cardinals do not exist, although there existing a model of ZFC where large cardinals do not exist is a necessary condition for large cardinal axioms to imply the consistency of ZFC. > Or does the statement refer rather to the> unprovability by means of ZFC of the existence of large cardinal in> every model where large cardinals do actually exist?Theories do not prove things in a model. What is a theorem of a theory is true in all models of the theory. I can't think of any sensible way to parse unprovability by means of ZFC of the existence of large cardinal in every model where large cardinals do actually exist? that doesn't render it completely uninteresting. Perhaps you can rephrase your question?> Anyway, it would be great if someone provided a reference and the> exact meaning of that statement.Large cardinal axioms prove the cosistency of ZFC because the weakest of them, which asserts the existence of a strongly inaccessible cardinal does so. The larger large cardinal axioms imply the existence of a strongly inaccessible cardinal, and thus also the consistency of ZFC. Basicly a strongly inaccessible cardinal kappa is such that the powerset operation and replacement can't reach from below, or formally cf(kappa)=kappa and for all lambda < kappa 2^lambda < kappa. kappa. The reason it implies consistency of ZFC is that the cumulative hierarchy defined by V_0 = emptyset V_alpha+1 = P(V_alpha) V_lambda = Union over alphaThat's interesting, because intuitively one would have a structure and>good reason to affirm that some statements about what happens inside>it are true, then try to deduce useful consequences from those basic>statements. Those statements and consequences would be expressed in>some language used to describe the structure.There's nothing incompatible between this view and the other view that youdescribe below. The point is that once you create a formal language tocapture some of the properties of your structure, it can take a life of itsown; it may turn out that other structures satisfy the same axioms, forexample.ZFC is potentially a confusing case because it has the reputation ofcapturing all of mathematics and this sometimes causes people tosubconsciously assume that there is just one model of ZFC (the classof all sets---which by the way isn't even a model in the usual sensebecause it's a proper class, not a set) and to identify ZFC with thatmodel. Naturally this subconscious bias leads to trouble when one istrying to understand other models of ZFC. Psychologically, I've foundthat the best way to root out this subconcious bias is (initially atleast) to treat ZFC as a formal language that people have developed butthat might not have any models at all.>a language and structures are in some sense subordina to it, in the>sense that once a set of axioms is fixed, the set of its consequences>does not depend on the properties of the structure which supports the>axioms.This is right, but it doesn't mean that you have to abandon the structureyou star with---just that you shouldn't *identify* it too closely withthe formal language.>That seems to assume that every property of the structure can be>expressed in the language in which axioms are expressed.No, the structures might have other properties that are not expressible inthe formal language under study. If you're thinking that a formal languagecreates structures and that a crea object cannot be greater than itscreator, that's not the right intuition. Don't think of formal languages ascreating anything.>However, no obvious reason occurs to me as to why all relevant>properties of the structure should be able to be expressed in the>language.Indeed, there is no reason for this because it's not true.>Does that prove that large cardinals are consistent with ZFC, or not?>And, if it doesn't, are they known to be?It's more accurate to say that large cardinals are not known to beinconsistent with ZFC.The familiar word known unfortunately becomes slippery in the context oflarge cardinals. Known presumably means proved, but proved in whatsystem? One can arbitrarily choose ZFC as one's system. In that case,it is not known that ZFC itself is consistent, let alone that strongersystems are consistent. We can get around this by backing off somewhatand saying that we know that something is consistent if it is consistent*relative* to ZFC, e.g., we know that the continuum hypothesis (CH) isconsistent with ZFC because it is provable in ZFC that if ZFC is consistentthen so is ZFC+CH. The trouble with large cardinals is that theirconsistency strength is higher than that of ZFC, i.e., even if we assumethat weaker systems such as ZFC or ZFC+ZFC is consistent are consistent,we cannot deduce the consistency of the stronger systems. (Cannot hereis again slightly subtle---you might ask, do we know that we cannot dothis? We do not know this in an absolute sense because in principle ZFCcould be inconsistent and everything could be deducible from it. However,this thread.)Therefore in some sense large cardinals are not known to be consistentwith ZFC, but you should not conclude from this that there is aninteresting open problem to be tackled here, because the situation isunderstood pretty well, to the extent that there is a sense in which it isknown that you can't prove a theorem that would settle the consistencyof large cardinals. (Exception: You could prove that a large cardinalaxiom is *inconsistent*. But if it's actually consistent, then you can'treally do much more than has already been done, at least with the morecommon types of large cardinals. That is, the best you can do is to rankthem in order of consistency strength, which has been done.)-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences=== === Subject: : Re: Large cardinalsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>> I've read that the existence of a measurable cardinal (a large>> cardinal indeed, I believe) cannot be proven within ZFC because the>> existence of large cardinals entails the consistency of ZFC.> Or does the statement refer rather to the>> unprovability by means of ZFC of the existence of large cardinal in>> every model where large cardinals do actually exist?> Theories do not prove things in a model. [...]> Perhaps you can rephrase your question?I think what he means is the following:Assume that ZFC is consistent, i.e. there is a model of it.One could think that then one might use this model somehow to constructa model of ZFC + I (using some tricky techniques like forcing or whatever).The question is: Could this be true (although currently nobody knows how)or is it known that this is impossible?In other words: Is it known that the implication ZFC is consistent => ZFC + I is consistentis unprovable (within ZFC)?=== === Subject: : Re: Large cardinalsOriginator: tchow@maclaurin.mit.edu.mit.edu (Timothy Chow)Originator: bergv@math.uiuc.edu (Maarten Bergvelt)>In other words: Is it known that the implication> ZFC is consistent => ZFC + I is consistent>is unprovable (within ZFC)?If there exists an inaccessible cardinal, then ZFC is consistent. That is,ZFC is consistent is a theorem of ZFC + I. So if ZFC + I is consistent,then so is ZFC + ZFC is consistent. This argument is easily formalized,i.e., it is a theorem of ZFC that ZFC + I is consistent => ZFC + `ZFC is consistent' is consistent.So, suppose your proposed implication were in fact a theorem of ZFC. Thencombining it with the theorem of the previous paragraph, we find that ZFC is consistent => ZFC + `ZFC is consistent' is consistentis a theorem of ZFC. It follows that ZFC + `ZFC is consistent' is consistentis a theorem of ZFC + ZFC is consistent. By Goedel's 2nd incompletenesstheorem, this implies that ZFC + ZFC is consistent is inconsistent.Summary: Your proposed implication is unprovable in ZFC, provided thatZFC + ZFC is consistent is consistent.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences=== === Subject: : Re: Large cardinalsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>> I've read that the existence of a measurable cardinal (a large>> cardinal indeed, I believe) cannot be proven within ZFC because the>> existence of large cardinals entails the consistency of ZFC.> Or does the statement refer rather to the>> unprovability by means of ZFC of the existence of large cardinal in>> every model where large cardinals do actually exist?> Theories do not prove things in a model. [...]> Perhaps you can rephrase your question?I think what he means is the following:Assume that ZFC is consistent, i.e. there is a model of it.One could think that then one might use this model somehow to constructa model of ZFC + I (using some tricky techniques like forcing or whatever).The question is: Could this be true (although currently nobody knows how)or is it known that this is impossible?In other words: Is it known that the implication ZFC is consistent => ZFC + I is consistentis unprovable (within ZFC)?=== === Subject: : One more follow up quaestionEpigone-thread: derdcrongkirOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)All this proves that if we have a Pythagorean triangle with arms oflength a and b, 1 + ab cannot divide a^2 + b^2 if the triangle isprimitive (i.e. gcd(a,b) = 1).Can anybody prove this if the triangle is non-primitive (i.e. gcd(a,b)= d > 1)?Kent Holing === === Subject: : Where c^c^c^... divergesOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)About 2 years ago when I finished the proofs of itera exponentials, Iintentionally left out the case where the exponential tower diverges.Turns out this case is pretty simple, too, so I've added a short excerptthat shows that complex infinite towers c^c^c^... diverge whenever c doesnot belong to the image of h(t)=e^{t*e^(-t)}, |t|<=1.The short proof can be found at the end of my second page on InfiniteExponentials, right after the 5-cycle graphic:Of course, this only proves divergence for 1-cycles. There exist cycles ofarbitrary length in there, but proving this is still an open problem.--------------------------------------------Eventually , _everything_ is understandable=== === Subject: : Looking for a binary sequence with specific propertiesX-mailer: epigoneEpigone-thread: hurrursexOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)I have this problem for which I have a not-pretty-at-all solution;I look for a pretty one (but I don't have much experience inthe field).The problem:Given an integer q, build a 0/1-sequence S such that the followingproperties hold simultaneously:- S has exactly q characters 1 - the length of S is polynomial with respect to q- when one alignes S with a copy of S shif k positions withrespect to the beginning of S, there is at most one character 1 in S which corresponds to a character 1 in the copy of S, and this for every k>0.For instance, when q=5, S=101001000100001 is not a good sequencesince the alignment S: 101001000100001copy of S: 101001000100001 (k=9) x x allows to have two pairs of 1's.A pretty solution would have an easy definition of the way tobuild S, and a short proof that S satisfies the required properties.Do you have any idea?Thank you,Irena.=== === Subject: : Re: Looking for a binary sequence with specific propertiesOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> I have this problem for which I have a not-pretty-at-all solution;> I look for a pretty one (but I don't have much experience in> the field).> The problem:> Given an integer q, build a 0/1-sequence S such that the following> properties hold simultaneously:> - S has exactly q characters 1 > - the length of S is polynomial with respect to q> - when one alignes S with a copy of S shif k positions with> respect to the beginning of S, there is at most one character > 1 in S which corresponds to a character 1 in the copy of S, > and this for every k>0.Well, I have a short and simple solution, but I don't knowwhether you'd consider it pretty:The requirement is that no distance-between-1s occursmore than once. Here's a simple construction. Suppose(by induction) that we've done the job for q, and we wantto do it for q+1. Then there are O(q^2) distances thatwe need to avoid, and q things to avoid being any of thosedistances away from; so there are O(q^3) positions to avoid,so in particular the earliest good position is O(q^3),and we're done.-- Gareth McCaughan..sig under construc=== === Subject: : Re: Looking for a binary sequence with specific propertiesOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> For instance, when q=5, S=101001000100001 is not a good sequence> since the alignment> S: 101001000100001> copy of S: 101001000100001 (k=9)> x x > allows to have two pairs of 1's.Your string above has 1's at indicies: 101001000100001index 1 3 6 10 15And the problem is that 15-10 = 6-1. You are looking for perfect difference sets which are rela to Golomb rulers (rulers where all the marks define distinct differences between them.) Most of the efforts in constructing these is put into the construction of optimal such structures... I believe perfect difference sets exist in quadratic size of your q, but there's no simple way to construct them (I might be incorrect about that.) But nothing says constructing something that is cubic in the size of q would be difficult. Trying a cubically sized string: instead of separating the 1's with 1,2,3,4,5, etc 0's, if we separate the 1's with 1,4,9,16,25, etc. 10100001000000000100000000000000001index 1 3 8 18 35we have 35-18 = 18-1 :( Have you considered the greedy method for placing 1's? This creates the Mian-Chowla sequence... I don't know how fast it grows, however. I know itgrows at least quadratically (since the sum of the reciprocals of the indices is less than 3) but I'm not sure if there are any upper bounds on its growth rate out there.J=== === Subject: : Definitive Mathematical Functions Site Now OnlineOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)87,160 formulas and 10,828 graphics about mathematical functionsare now available free at The Wolfram Functions Site, animportant new resource for mathematicians, scientists, engineers,and students at:http://functions.wolfram.comSeveral widely used handbooks of mathematical functions have beenpublished, the largest of which contained about 15,000 formulas,meticulously compiled from thousands of technical papers.Traditional handbooks have also included only handfuls ofgraphics illustrating the properties of functions. WithMathematica, a huge number of new visualizations of functionshave become possible. The Wolfram Functions Site assembles over10,000 of these, with many more being planned.A major tour de force of reference website construction, TheWolfram Functions Site contains over 30 gigabytes of data.Material in The Wolfram Functions Site can be downloaded inseveral standard formats, including Mathematica InputForm andStandardForm, MathML, and PDF. Formulas can be copied from thesite and immediately used as input to a computer system. For easeof citation, each formula has been assigned a unique permanent ID.While having already far surpassed previous knowledge bases formathematical functions, continued growth is projec for TheWolfram Functions Site, with new searching capabilities, externalcontributions, and new classes of graphics and information. Visitthe site at:http://functions.wolfram.comMichael Trott=== === Subject: : Functional equations for zeta functionsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)I was looking a Hecke'sLectures on Dirichlet series, modular functions and quadratic forms (ISBN 3-525-40727-0),and wondering how general was his specification of Dirichlet seriessatisfying a functional equation?He considers Dirichlet series f(s) (convergent somewhere)such that F(s) = g(s)f(s) extends to a meromorphic function on Cwith a simple pole at s = k,satisfying the functional equation F(k-s) = +/- F(s),whereg(s) = (2pi/l)^s Gamma(s)^a Gamma(s/2)^b Gamma((s+1)/2)^c.Later he assumes g(s) takes the simpler form(a,b,c) = (1,0,0), (0,1,0), (0,0,1) or (0,1,1).My question is --and it may be an ignorant one as I am not too familiar with this area --whether in practice whenever a zeta- or L-functionsatisfies a functional equationthe equation takes this form(either with the general g(s) or the more specialized one)?-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland=== === Subject: : Paper published by Geometry and TopologyOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)The following paper has been published:URL:http://www.maths.warwick.ac.uk/gt/GTVol8/paper2. abs.htmlTitle:Rohlin's invariant and gauge theory II. Mapping toriAuthor(s):Daniel Ruberman, Nikolai SavelievAbstract:This is the second in a series of papers studying the relationshipbetween Rohlin's theorem and gauge theory. We discuss an invariant ofa homology S^1 cross S^3 defined by Furuta and Ohta as an analogue ofCasson's invariant for homology 3-spheres. Our main result is acalculation of the Furuta-Ohta invariant for the mapping torus of afinite-order diffeomorphism of a homology sphere. The answer is theequivariant Casson invariant (Collin-Saveliev 2001) if the action hasfixed points, and a version of the Boyer-Nicas (1990) invariant if theaction is free. We deduce, for finite-order mapping tori, theconjecture of Furuta and Ohta that their invariant reduces mod 2 tothe Rohlin invariant of a manifold carrying a generator of the thirdhomology group. Under some transversality assumptions, we show thatthe Furuta-Ohta invariant coincides with the Lefschetz number of theaction on Floer homology. Comparing our two answers yields an exampleof a diffeomorphism acting trivially on the representation variety butnon-trivially on Floer homology.Secondary: 57R58Keywords:Casson invariant, Rohlin invariant, Floer homologyProposed: Ronald SternSeconded: Robion Kirby, Tomasz MrowkaAuthor(s) address(es):Department of Mathematics, MS 050, Brandeis University Waltham, MA 02454, USA and Department of Mathematics, University of Miami PO Box 249085, Coral Gables, FL 33124, USAEmail: ruberman@brandeis.edu, saveliev@math.miami.edu===Originator: bergv@math.uiuc.edu (Maarten Bergvelt)CALL FOR PAPERS AND WORKSHOPS~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~ 12th International Symposium on the Foundations of Software Engineering=================================================== ============================ Hyatt Newporter, Newport Beach, California, USA http://www.isr.uci.edu/FSE-12 Sponsored by ACM SIGSOFT In cooperation with ACM SIGPLAN======================================================= ======================== Call for Papers ---------------and industry to exchange new results rela to both traditional and emergingfields of software engineering. We invite submission of technical papers whichreport results of theoretical, empirical, and experimental work, as well asexperience with technology transition.Topics of InterestWe encourage submissions in any field of software engineering, including, butnot limi to:* Component-Based Software Engineering* Distribu, Web-Based, and Internet-Scale Software Engineering* Empirical Studies of Software Tools and Methods* Feature Interaction and Crosscutting Concerns* Generic Programming and Software Reuse* Requirements Engineering* Software Analysis and Model Checking* Software Architectures* Software Configuration Management* Software Engineering and Security* Software Engineering Tools and Environments* Software Information Management* Software Metrics* Software Performance Engineering* Software Process and Workflow* Software Reengineering* Software Reliability Engineering* Software Safety* Software Testing* Specification and Verification* User InterfacesSubmission GuidelinesPaper submissions will be made electronically via the Web, at the conferenceweb site. Both an abstract and the full paper must be submit by their respective submission deadlines. Papers must not exceed ten (10) 8.5x11 single spaced pages using the ACM SIG proceedings formatting instructions, which are accessible from the conference web site. Accep papers must be accompanied by a signed ACM copyright release form.Paper submissions will be reviewed by the program committee for originality,significance, soundness, and quality of presentation. Papers must clearlypresent an original contribution to the state-of-the-art. Experience papers arewelcome, but they must clearly present general lessons learned that would be ofinterest and benefit to a broad audience of both researchers and practitioners.Papers must describe work that has not been submit to or presen atanother forum. Any double submissions will be rejec without review, and theother forum will be informed of the situation.March 24 --- Abstract submission dueMarch 29 --- Technical paper submissions dueJune 4 --- Author notificationAugust 15 -- Camera ready dueGeneral Chair-------------Richard N. Taylor, University of California, Irvine, USA; taylor at uci.eduProgram Chair-------------Matthew B. Dwyer, Kansas State University, USA; dwyer at csi.ksu.eduProgram Committee-----------------Ken Anderson, University of Colorado, Boulder, USAAnnie Ant.97n, North Carolina State University, USAJoanne Atlee, University of Waterloo, CanadaPrem Devanbu, University of California, Davis, USAMary Jean Harrold, Georgia Institute of Technology, USAJohn Hatcliff, Kansas State University, USAJim Herbsleb, Carnegie Mellon University, USAAndr.8e van der Hoek, University of California, Irvine, USAPaola Inverardi, University of L'Aquila, ItalyGregor Kiczales, University of British Columbia, CanadaJeff Kramer, Imperial College London, UKShriram Krishnamurthi, Brown University, USAShinji Kusumoto, Osaka University, JapanAxel van Lamsweerde, Universit.8e Catholique de Louvain, BelgiumBashar Nuseibeh, The Open University, UKMauro Pezz.8e, Universita' di Milano-Bicocca, ItalyGian Pietro Picco, Politecnico di Milano, ItalyDavid Rosenblum, University College London, UKGregg Rothermel, Oregon State University, USAWilhelm Sch.8afer, Universit.8at Paderborn, Germanylas C. 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