mm-1539 I have been trying to determine the asymptotic behaviour of the following integral int |r-A|^2m |r+A|^2m Exp[-|r-B|^2] dr managed to do this so any answers or suggestions would be gratefully received. Darragh Originator: bergv@math.uiuc.edu (Maarten Bergvelt) No one has mentioned this yet, but you mean 2 - delta where you have written 2 + delta. And of course, one only needs this locally. That is, delta (which clearly depends on epsilon) can also depend on y). Dan -- Dan Luecking Department of Mathematical Sciences University of Arkansas Fayetteville, Arkansas 72701 To reply by email, change Look-In-Sig to luecking Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Well T(B) = T(-B), so surely (T(B) intersect S) contains (Q union -Q). In the complex case, it would also contain e^{itheta}Q (but that's as far as it goes). It is quite possible (in the complex case) for Q to be non-Borel and contained in a one-dimensional subspace, so the range of T is a hyperplane and T(B) is the unit ball of that hyperplane. So you have to start with a Q that is nonBorel and closed under Dan -- Dan Luecking Department of Mathematical Sciences University of Arkansas Fayetteville, Arkansas 72701 To reply by email, change Look-In-Sig to luecking Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following paper has been published: Algebraic and Geometric Topology Volume 5 (2005), paper no. 11, pages 207--217 URL: http://www.maths.warwick.ac.uk/agt/AGTVol5/agt-5-11.abs.html Title: All roots of unity are detected by the A--polynomial Author(s): Eric Chesebro Abstract: For an arbitrary positive integer n, we construct infinitely many one-cusped hyperbolic 3-manifolds where each manifold's A-polynomial detects every n-th root of unity. This answers a question of Cooper, Culler, Gillet, Long, and Shalen as to which roots of unity arise in this manner. Secondary: 57M50 Keywords: character variety, ideal point, A-polynomial Received: 23 February 2005 Accepted: 6 March 2005 Published: 28 March 2005 Author(s) address(es): Department of Mathematics, The University of Texas at Austin Austin, TX 78712-0257, USA Email: chesebro@math.utexas.edu URL: http://www.ma.utexas.edu/users/chesebro/ Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I have collected together some older and new results about extending x^^y to the Reals. Using the latest extension, one could conceivably define pentation, You are welcome to take a look at: http://users.forthnet.gr/ath/jgal/math/ExtensionsPaper.html The link is a redirection link which points to a .pdf paper. -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable Originator: bergv@math.uiuc.edu (Maarten Bergvelt) So in the book Representation Theory by Fulton and Harris, they're classifying Lie algebras of dimension 3 in one of their beginning chapters. It made me want to ask some questions about Lie groups which are not matrix Lie groups, i.e. have no faithful finite dimensional representations. Let me briefly describe the stuff in the text. They get to the real Lie algebra [X,Y]=[X,Z]=0, [Y,Z]=X, and goes on to exponentiate some representation of the algebra to get the group G of upper triangular 3x3 matrices with 1s along the diagonal. This group has center R, matrices with the only nonzero entry in the upper corner. To get other Lie groups with some algebra, they take the quotient of G/N with N any discrete normal subgroup, which must therefore be central. Z being a discrete normal subgroup, G/Z is the only other group with this algebra, presumably G/nZ will be isomorphic. Then he proves that this group has no faithful irreps: if it did, the center, S^1, would be proportional to the identity in that irrep, and we would have [Y,Z]~1, but tr[Y,Z]=0. My first question is about this proof. I can't understand why this proof doesn't apply to the covering group G (which clearly does have finite dimensional faithful reps). I think its center R should also be proportional to the identity in an irrep. The difference between the two cases is that S^1 is compact, while R is not. But so what? Did I miss the theorem in the book where he showed that a compact central subgroup must be proportional to the identity? Does not Schur's lemma apply to anything that commutes with the whole group? He also talks about how the tower of covers of SL(2) are not matrix groups. And I seem to recall once hearing that the universal cover of GL(n) is not a matrix group. These Lie groups that are not matrix groups seem a bit mysterious. Is there a general way to find groups who are not matrix groups? Or turn one that is not into one that is? What are some others? I think I've heard that the universal cover of GL(n) is not. Is that right? What about Spin(n)? The Spin(n)s that I've met have been, but I've never seen a general case. zig Originator: bergv@math.uiuc.edu (Maarten Bergvelt) [snipped setup - sorry] You are correct in that compactness of the center is the key. I dunno if it was covered elsewhere in F&H, but the starting point of their argument was that a compact abelian group must be diagonalizable. to prove this, but if you can prove that all the matrices in a compact abelian subgroup of GL_n(R) are semisimple, then you are basically done, because then the elements of the said group are individually diagonalizable, and as they commute, they are simultaneously diagonalizable. I think that the argument used the fact that the semisimple and unipotent parts of a non-singular linear mapping T commute with everything that commutes with T. I don't remember how compactness implies that the unipotent parts aren't there, sorry:) Anyway, the argument in F&H depended heavily on the fact that the center must act via diagonal matrices. The upshot is that you can't deduce this, when the center isn't compact! Your argument could be viewed as a proof for the fact that the said group (when nothing was moded out of the center) has no such finite dimensional representations, where the (non-compact!) center is mapped into diagonal matrices. While we are at it: I'm not entirely happy with the F&H argument in that why it is sufficient to consider irreducible reps only? The 3-D rep by upper triangular matrices certainly isn't irreducible! This Lie algebra is solvable, so all the irreducible reps are 1-D? Jyrki Lahtonen, Turku, Finland Originator: bergv@math.uiuc.edu (Maarten Bergvelt) F&H also describe this group as a group of unipotent 3x3 matrices. Does unipotent mean that X^n=1 for some n? I can't see how that description applies to my group. For example the matrix [1 b 0] [0 1 0] [0 0 1] does not satisfy that for any n. So what does unipotent mean? This is one of the things that bothered me. The center of this group is the set of matrices like [1 0 a] [0 1 0] [0 0 1] which certainly does not act diagonally. Not in any basis, I don't think. So why isn't this a violation of Schur's lemma? I was thinking today that maybe this rep isn't irreducible, a prerequisite of Schur. Does the noncompactness of the center imply the reducibility of this rep? Hmm... you're right. That's one way to define these guys, right? They preserve a chain of vector subspaces (a flag?), each is an invariant subspace. So Schur's lemma cannot apply to the center of this rep. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I see now that the definition of unipotent is simply upper triangular matrices with 1s on the diagonal. OK, good. thinking Yes, it's all clear now. If a Lie algebra is semisimple, then we have complete reducibility. But in the nilpotent case, we have no such guarantee. So not only is this rep not irreducible, it won't even decompose into irreps, and Schur's lemma does not apply. But the same applies to the compact center, so how does compactness get us to diagonalizablity? Well I found the answer to that too today in F&H: if a group is compact, when we can use the unitary trick. Get an invariant inner product, and then the orthogonal complement of an invariant space is also invariant, and the rep is completely reducible. But I'm still having trouble filling in the rest: so in our hypothetical rep of the Heisenberg group carries a rep of the center S^1, which should be completely reducible, since its compact. S^1 should reduce to a multiple of the identity on each of these irreps, but that doesn't quite imply that S^1 is the identity on the full rep of H. So you're complaining because it's unreasonable to expect this group to have an irreducible representation, while it may very well have a reducible, non-fully-reducible, rep. Is that it? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) This is the Heisenberg algebra. Its standard representation is in terms of vector fields (first-order differential operators) acting on functions of a single variable t. Set e.g. Y = d/dt, Z = t, X = 1. IOW, the Lie algebra of G has 0s in the diagonal. Hence tr X = tr Y = tr Z = 0, and the rep is not faithful. It has not. Recall how you build reps of su(2). In a Cartan-Weyl basis, the generators are J+, J-, J0, with brackets [J0, J+] = J+, [J0, J-] = -J-, [J+, J-] = 2J0. j is an integer of half-integer, the state with n = 2j+1 has norm zero, and we can proclaim that it is annihilated by J+. The rep has dimension 2j+1. However, if j is not a (half-)integer, we just keep generating new states with J+, and the rep is infinite-dimensional (and not unitary, but that is beside the point). Since a matrix rep is finite-dimensional, this is not a matrix rep. The reps of other Lie algebras are built in a similar fashion. They all have su(2) subalgebras, and the construction above applies to these subalgebras. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) [snipped] Given that the covering group was defined to be the group of upper triangular unipotent real 3x3 matrices, it is somewhat obvious that the group *HAS* a 3-dimensional faithful representation. I think that you are missing out on something? Jyrki Originator: bergv@math.uiuc.edu (Maarten Bergvelt) This realization is not in terms of vector fields, of course. It is easy to write down a realization as vector fields in two variables (s,t), though: Y = d/ds - s d/dt, Z = d/ds + s d/dt, X = 2 d/dt. Of course it is faithful. It is also reducible, though not decomposable. The statement is that in any *irreducible* rep, a central element is proportional to the unit matrix. The rep above contains the subrep where X is replaced by the zero matrix. This subrep is irreducible but evidently not faithful. That the Heisenberg algebra has no reps which are both faithful, irreducible, and finite-dimensional should at least be true. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) irreducible, Yes, I think reducibility was the criterion that I was failing to appreciate. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) real terms functions Hmm... I probably agree that Y is a first-order differential operator, but Z and X do not appear to be. triangular tr Z = 0, But I think the rep *is* faithful. The Lie algebra is 3-dimensional, and the space of upper triangular 3x3 matrices is 3-dimensional, after all. A homomorphism from a 3-dimensional Lie algebra to a 3-dimensional Lie algebra ought to be invertible, no? of groups is? of that all have subalgebras. I've heard this statement about simple Lie algebras, but not about generic Lie algebras. The Heisenberg algebra, it's nilpotent, and therefore not simple. Does it have an su(2) subalgebra? It does not appear to. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Can anything interesting be said about a field K such that, for any polynomial f over K which is split (all its root are in K), f' is also split? Does this property have a name? If K is real-closed, it is easy to see that it satisfies the property. I'm not sure I can come up with any other example, in fact. What about the smallest subfield of the algebraic closure of the rationals which satisfies the property in question (it is clear that it exists)? What does it look like? -- David A. Madore (david.madore@ens.fr, http://www.dma.ens.fr/~madore/ ) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Ballantine, C.(1-DTM); Roberts, J.(1-BOWD) A simple proof of Rolle's theorem for finite fields. Amer. Math. Monthly 109 (2002), no. 1, 72--74. 12E20 (11T06) Using terminology drawn from a consequence of Rolle's theorem for the real field, the authors say that an arbitrary field $F$ obeys Rolle's property if the formal derivative of any polynomial in $F[x]$ that splits completely over $F$ also splits completely over $F$. Kaplansky asked for a classification of such fields, and T. Craven and G. Csordas ref[Illinois J. Math. 21 (1977), no. 4, 801--817; MR0568321 (58 #27921)] and Craven ref[Proc. Amer. Math. Soc. 125 (1997), no. 11, 3147--3153; MR1401731 (97m:12010)] carried out general investigations that brought the particular conclusion that the finite fields that obeyed Rolle's property were precisely those with 2 or 4 elements. Claiming the approach of these papers to be technical, the authors now provide an elementary proof of the finite field result. {Reviewer's remarks: It seems to the reviewer that an equally simple proof is inherent in Craven's paper, especially Theorems 2.2 and 2.5. For finite $F$ of odd characteristic $p$, consider $x^{p-1}(x+1)(x+a)$, where $a$ is a non-square in $F$. For $p=2$, the crucial fact that, if $F$ has Rolle's property, then the set ${gamma ^2 + gamma colon gamma in F }$ is closed under multiplication (and so a subfield for finite $F$), is already in Craven. Also, the paper's title is somewhat misleading in that its main thrust is that even the weak version of Rolle's theorem discussed is invalid except in two small fields.} Originator: bergv@math.uiuc.edu (Maarten Bergvelt) An old Monthly problem is relevant. Let F be an ordered field. [...] (b) If Rolle's Theorem holds in F, does it follow that F is real-closed? Here Rolle's Theorem means: If a polynomial f has zeros a