most recent 30 from http://mathoverflow.net 2013-05-18T19:44:00Z http://mathoverflow.net/feeds/user/401 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110855/product-of-conjugacy-classes-is-there-an-analog-of-tanaka-krein-reconstruction/125060#125060 Bruce Bartlett 2013-03-20T12:02:17Z 2013-03-20T12:02:17Z

<p>One nice fact, I think, is that the formula of Burnside that Geoff Robinson gives above,</p> <p>$$ N^{C_z}_{C_x C_y} = \frac{|G|}{|C_G(x)| |C_G(y)|} \frac{\chi(x) \chi(y) \chi(z^{-1})}{\chi(1)} $$</p> <p>can be understood nicely from a geometric / topological quantum field theory perspective. I think it is precisely the "Verlinde formula" for the modular category Rep(/\G), the representation category of the Drinfeld double of C[G]. The Verlinde formula says that in a general modular category, we have</p> <p>$$ N^k_{ij} = \sum_r \frac{s_{ir} s_{jr} s_{k^* r}}{s_{0 r}} $$</p> <p>where $s_{ij}$ is the S-matrix. </p> <p>More concretely, this is to say that it has a natural interpretation in terms of G-bundles on the torus. This perspective also comes out in the appendix of Zagier to "Graphs on surfaces and their applications" by Lando and Zvonkin. </p>

http://mathoverflow.net/questions/108476/does-the-following-symmetric-2nd-cohomology-group-of-a-finite-group-with-coeffi/109519#109519 Bruce Bartlett 2012-10-13T08:32:46Z 2012-10-13T08:32:46Z

<p>I'm pretty sure now that $H^2_{sym} (\mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2) = \mathbb{Z}_2 \times \mathbb{Z}_2$, so it can be nonzero. For comparison, in ordinary group cohomology $H^2 (\mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2) = (\mathbb{Z}_2)^3$.</p>

http://mathoverflow.net/questions/108476/does-the-following-symmetric-2nd-cohomology-group-of-a-finite-group-with-coeffi Bruce Bartlett 2012-09-30T15:05:17Z 2012-10-13T08:32:46Z

<p>Let $G$ be a finite group. Usually, a 2-cocycle on $G$ with values in $\mathbb{Z}_2 = \{+1, -1\}$ is a collection of signs $\epsilon_{g,h} \in \{+1, -1\}$, $g,h \in G$, satisfying the cocycle equation (written multiplicatively) $$ \epsilon_{g,h} \epsilon_{gh, k} = \epsilon_{h,k} \epsilon_{g,hk} $$ And a 2-coboundary is a 2-cocycle with $\epsilon_{g,h} = \frac{t_g t_g}{t_{gh}}$ for all $g,h \in G$, with $t : G \rightarrow \{+1, -1\}$ an arbitrary map. Then the second cohomology group is $H^2(G, \mathbb{Z}_2)$ = {2-cocycles} / {2-coboundaries}. </p> <p>But suppose we demand that our 2-cocyles satisfy the 2-cocycle equation above together with the "conjugate-cyclic" symmetry $$ \epsilon_{g,h} = \epsilon_{h^{-1} g^{-1}, g} $$ as well as the "conjugate symmetric" symmetry, $$ \epsilon_{g,h} = \epsilon_{h^{-1}, g^{-1}}. $$ These symmetries make sense from a TQFT perspective if you draw the 2-cocycle as a bunch of trivalent vertices, when the first symmetry corresponds to counterclockwise rotation of the diagram and the second to a kind of vertical flip.</p> <p>And suppose now that the coboundaries given by $\{t_g\}$ satisfy $t_1 = 1$ and $t_g t_{g^{-1}} = 1$. This ensures that the "symmetric" 2nd cohomology group $H^2_{sym} (G) := ${ "symmetric" 2-cocycles} / {"symmetric" 2-coboundaries} makes sense. </p> <p>Question: Does this symmetric 2nd cohomology group always vanish?</p> <p>I've only checked one example, namely $G = \mathbb{Z}_2 \times \mathbb{Z}_2 = \langle a,b : a^2 = b^2 =(ab)^2 = 1 \rangle$. In normal cohomology, we have $H^2(G, \mathbb{Z}_2)$ = $(\mathbb{Z}_2)^3$ but according to my calculations, none of the non-trivial 2-cocycles respect the above symmetries, so $$ H^2_{sym} (G, \mathbb{Z}_2) = 0. $$ Is this perhaps always true for arbitrary $G$? Or perhaps I have made a silly mistake. </p>

http://mathoverflow.net/questions/42439/how-can-one-express-the-dedekind-eta-function-as-a-sum-over-the-lattice Bruce Bartlett 2010-10-16T23:01:40Z 2010-10-18T05:01:17Z

<p>The Dedekind eta function $\eta(\tau)$ can be regarded as a formula which assigns a number to a lattice $L \subset \mathbb{C}$. The algorithm is: rotate the lattice so that one of its basis vectors lies along the real axis, then pick another basis vector $\tau$ in the upper half plane, then compute the usual formula $$ \eta(\tau) = q^\frac{1}{24} \prod_{n=1}^\infty (1-q^n) $$ where $q=e^{2 \pi i \tau}$. It would be nice if there were a more "canonical" way to compute it directly from the lattice $L$ (i.e. without this rotate-and-pick-a-basis-vector story which breaks symmetry). I'm looking for a formula similar to that of the Eisenstein series where one sums over all points of the lattice: $$ G_n (L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^n} $$ We have the theorem of Jacobi that the 24th power of $\eta$ computes as the discriminant of the lattice, $$ (2\pi)^{12} \eta(\tau)^{24} = 20G_4(\mathbb{Z} + \tau \mathbb{Z})^3 -49 G_6 (\mathbb{Z} + \tau \mathbb{Z})^2, $$ which is great, since it shows that the 24th power of $\eta$ can be defined canonically via a sum over the lattice points... but how about $\eta$ itself? </p>

http://mathoverflow.net/questions/42480/how-can-one-see-the-hopf-fibration-in-the-space-of-lattices-in-the-plane Bruce Bartlett 2010-10-17T09:57:52Z 2010-10-17T22:46:59Z

<p>This question is inspired from Etienne Ghys's talk on Knots and Dynamics from ICM 2006. </p> <p>The map $L \mapsto (G_4(L), G_6(L))$ gives a bijection between all lattices $L\subset \mathbb{C}$ (including the degenerate ones) and $\mathbb{C}^2 - {0}$. Here $G_4$ and $G_6$ are the Eisenstein series of a lattice,</p> <p>$$ G_n (L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^n}. $$</p> <p>By scaling the lattice $L \mapsto t L$ we can arrange that these two numbers satisfy $|z_1|^2 + |z_2|^2 = 1$; in other words we have that </p> <p>$$ { \mbox{lattices in } \mathbb{C} \mbox{ up to rescaling} } \cong S^3. $$</p> <p>Now $S^3$ carries a nice action of $S^1$ given by sending $(z_1, z_2) \mapsto (e^{i \theta}z_1, e^{i \theta}z_2)$, the quotient being $S^2$; this is the Hopf fibration $S^3 \rightarrow S^2$.</p> <p>Since the collection of lattices up to rescaling identifies with $S^3$ in such a nice way, it is tempting to try and see this action of $S^1$ at the level of lattices. It would be nice if it were to correspond to rotation of the lattice! But alas, it does not. Firstly, it can't, because the action of $S^1$ on $S^3$ is free, while rotating a lattice might `click' it back into itself before one has rotated a full rotation. In fact we see that if we rotate the lattice via $L\mapsto e^{i\theta} L$, we find that the invariants change as $$ (G_4(L), G_6(L)) \mapsto (e^{-4i\theta}G_4(L), e^{-6i\theta}G_6(L)) $$ which is not the behaviour we are looking for. </p> <p>So what <em>does</em> the action of $S^1$ on $S^3$ correspond to in the space of lattices up to rescaling? In other words, what is $L'$ in terms of $L$ if </p> <p>$$ (G_4(L'), G_6(L')) = (e^{i \theta} G_4(L), e^{i \theta} G_6(L))? $$</p>

http://mathoverflow.net/questions/31897/does-the-function-which-sends-a-right-angled-triangle-to-its-area-produce-infinit Bruce Bartlett 2010-07-14T19:50:06Z 2010-07-18T20:19:47Z

<p>Let $T$ be the set of pythagorean triples, that is, triples of integers (a,b,c) satisfying a² + b² = c². We think of $T$ as the set of right angles triangles with integer lengths. And let $f : T \rightarrow \mathbb{Z}$ be the function $(a,b,c) \mapsto \frac{ab}{12}$ which computes the area of a triangle (divided by 6, which seems to always be a factor for some reason). </p> <p>I was wondering: what are the number theoretic propertires of $f$? It seems to produce numbers with few prime factors. What is the reason for this? For instance, $f(3,4,5) = 1$, $f(36,77,85) = 3 * 11 * 7$, and $f(65,72,97)=39*5*2$. Can we put a bound on the number of prime factors in the numbers that $f$ spits out? Or at least, can we give a 'generic' statement such as 'The output of $f$ almost always spits out numbers with less than 8 factors' or something?</p>

http://mathoverflow.net/questions/15466/what-do-the-numbers-g-4-and-g-6-of-a-lattice-actually-measure Bruce Bartlett 2010-02-16T15:40:03Z 2010-02-17T02:38:09Z

<p>If you have a lattice $L \subset \mathbb{C}$, you can compute the following numbers:</p> <p>$ G_4(L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^4}, \quad G_6(L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^6}. $</p> <p>By a 'lattice', I just mean a closed discrete additive non-trivial subgroup of $\mathbb{C}$ (so I'm allowing degenerate lattices like $\mathbb{Z}$). Anyhow, these numbers are important invariants of the lattice, because they set up a bijection</p> <p>{${ \mbox{Lattices in }\mathbb{C}}$} $\rightarrow$ {$\mathbb{C}^2 - 0$}</p> <p>$L \mapsto (G_4(L), G_6(L))$.</p> <p>But what do these numbers actually measure about the lattice, geometrically? Some kind of combination of angles? Some area? I'm confused. We have these numbers that get used over and over, but what do they actually <i>measure</i>?</p> <p>I guess one possible answer is: consider the Riemann surface (torus) $\mathbb{C} / \Lambda$. Then $G_4$ and $G_6$ can be recovered as certain period integrals along the fundamental cycles of the elliptic curve. Is that right? Is there a more direct geometric understanding of $G_4$ and $G_6$?</p>

http://mathoverflow.net/questions/5711/is-the-fell-doran-problem-trivial-in-a-topological-setting Bruce Bartlett 2009-11-16T16:27:53Z 2009-11-16T20:05:42Z

<p>The Fell-Doran problem is a problem in functional analysis. It goes as follows: Let $A$ be a complex unital algebra, $X$ a locally convex space, and $L(X)$ the algebra of all continuous endomorphisms of $X$. Suppose that we have a representation of $A$ on $X$, by which we simply mean an algebra homomorphism $$ T : A \rightarrow L(X) $$ which is irreducible (no proper closed invariant subspace) and has trivial commutant (any bounded operator commuting with all the $T_a$ must be a multiple of the identity). The Fell-Doran problem is: Is $T(A)$ dense in $L(X)$ in the strong operator topology? </p> <p>My question is: Is this a problem having to do with the fact that we didn't require a topology on our algebra? In other words, what can be said about the case when $A$ is actually a 'topological algebra' and the map $T$ is required to be continuous in some sense? Does that make the problem trivial, i.e. is the answer then automatically yes?</p> <p>By the way, I have heard that so far there is almost no progress on the Fell-Doran problem in general; not even for Hilbert spaces! The only thing that is known is that there exists a certain concrete space where the answer is affirmative.</p>

http://mathoverflow.net/questions/245/are-abelian-nondegenerate-tensor-categories-semisimple/618#618 Bruce Bartlett 2009-10-15T17:28:01Z 2009-10-15T17:28:01Z

<p>To my understanding, the answer is "yes"; at least if everything is sufficiently linear over a decent field k. Isn't this proved in Ulrieke Tillmann's great paper, "S-structures for k-linear categories and the definition of a modular functor"? By the way, one needs to obtain the journal version of that paper; the arXiv version doesn't have the pictures :-(</p>

http://mathoverflow.net/questions/108476/does-the-following-symmetric-2nd-cohomology-group-of-a-finite-group-with-coeffi Bruce Bartlett 2012-10-01T08:10:25Z 2012-10-01T08:10:25Z

I have corrected the error Alexander Cherkov pointed out above. But my question still stands. Any takers?

http://mathoverflow.net/questions/42480/how-can-one-see-the-hopf-fibration-in-the-space-of-lattices-in-the-plane/42540#42540 Bruce Bartlett 2012-10-01T08:06:07Z 2012-10-01T08:06:07Z

By the way, earlier this year I got an AIMS essay student Ihechukwu Chinyere to give a computer animation of the fibers of this &quot;modular fibration&quot;, see <a href="http://www.youtube.com/watch?v=eqeqbjec97w" rel="nofollow">youtube.com/watch?v=eqeqbjec97w</a>.

http://mathoverflow.net/questions/108476/does-the-following-symmetric-2nd-cohomology-group-of-a-finite-group-with-coeffi Bruce Bartlett 2012-09-30T21:32:50Z 2012-09-30T21:32:50Z

Actually, my statement that the &quot;symmetric&quot; cohomology is zero in this example is still true - checked by a direct calculation. Also, of the 4 extensions of $Z_2 x Z_2$ by $Z_2$, only the product extension and $Z_2 \times Z_4$ satisfy the second symmetry above, so the quaternion and the dihedral group don't come into it. There are only 2 cocycles satisfying both symmetries above, and they are both coboundaries.

http://mathoverflow.net/questions/108476/does-the-following-symmetric-2nd-cohomology-group-of-a-finite-group-with-coeffi Bruce Bartlett 2012-09-30T18:16:00Z 2012-09-30T18:16:00Z

Thanks. Yes I'm afraid I had built my whole argument around the fact that the &quot;2nd cohomology&quot; was Z_2, which I got from a quick lookup, but in fact what I had looked up was that H^2(G, U(1)) = Z_2, whereas I needed H^2(G ,Z_2) = (Z_2)^3.

http://mathoverflow.net/questions/55898/topos-theory-reference-suitable-for-undergraduates/55923#55923 Bruce Bartlett 2011-03-04T10:38:33Z 2011-03-04T10:38:33Z

I discovered these Tom Leinster notes from this post via a Google search. They are fantastic, thanks. There is an even shorter (a few pages) &quot;informal introduction&quot; to topos theory in Tom's category theory course notes (see pg 110), which can be seen as an &quot;introduction to this introduction&quot;: <a href="http://www.maths.gla.ac.uk/~tl/msci/all.pdf" rel="nofollow">maths.gla.ac.uk/~tl/msci/all.pdf</a>

http://mathoverflow.net/questions/42480/how-can-one-see-the-hopf-fibration-in-the-space-of-lattices-in-the-plane/42540#42540 Bruce Bartlett 2010-10-18T19:05:10Z 2010-10-18T19:05:10Z

Yes, the $S^1$ action on the space of lattices given by rotating the lattice is an interesting action to think about. But it's not the one I'm interested in... I'm trying to figure out the $S^1$ action on the space of lattices that corresponds to the $S^1$ action on $S^3$. We know one exists... what is it? Tim Perutz above suggests that there simply won't be a &quot;nice&quot; lattice interpretation... but I'm still holding out hope.

http://mathoverflow.net/questions/42439/how-can-one-express-the-dedekind-eta-function-as-a-sum-over-the-lattice Bruce Bartlett 2010-10-18T08:25:02Z 2010-10-18T08:25:02Z

Right - fair enough.

http://mathoverflow.net/questions/42480/how-can-one-see-the-hopf-fibration-in-the-space-of-lattices-in-the-plane Bruce Bartlett 2010-10-17T19:16:27Z 2010-10-17T19:16:27Z

Thanks, yes I did, I changed it now.

http://mathoverflow.net/questions/42439/how-can-one-express-the-dedekind-eta-function-as-a-sum-over-the-lattice Bruce Bartlett 2010-10-17T10:05:28Z 2010-10-17T10:05:28Z

Thanks to those who answered my question about the eta function. I've grabbed the part of the question about trying to &quot;see&quot; the Hopf fibration in the space of lattices and put it into a [separate question](<a href="http://mathoverflow.net/questions/42480/how-can-one-see-the-hopf-fibration-in-the-space-of-lattices-in-the-plane" rel="nofollow" title="how can one see the hopf fibration in the space of lattices in the plane">mathoverflow.net/questions/42480/&hellip;</a>).

http://mathoverflow.net/questions/42439/how-can-one-express-the-dedekind-eta-function-as-a-sum-over-the-lattice/42444#42444 Bruce Bartlett 2010-10-17T09:19:19Z 2010-10-17T09:19:19Z

Thanks, I hadn't known this, this is a very useful point. I've clarified the nature of my question above by adding to the last paragraphs at the end.

http://mathoverflow.net/questions/42439/how-can-one-express-the-dedekind-eta-function-as-a-sum-over-the-lattice/42445#42445 Bruce Bartlett 2010-10-17T09:17:23Z 2010-10-17T09:17:23Z

Thanks, that's useful, I hadn't known about the Euler identity, and the reference looks relevant. I had indeed wanted an expression which calculated the eta function $\eta(L)$ as a lattice sum, the reason being that then it is easy to see how $\eta$ will transform when you rotate the lattice $\Lambda \mapsto e^{i\theta} \Lambda$. The single sum isn't useful for this since one first has to rotate into standard position,so you can't &quot;see&quot; rotations! I'd like to know if one can see the Hopf fibration in the space of lattices! I've explained this point now further at the end of my question above.

http://mathoverflow.net/questions/31897/does-the-function-which-sends-a-right-angled-triangle-to-its-area-produce-infinit/31939#31939 Bruce Bartlett 2010-07-16T11:42:44Z 2010-07-16T11:42:44Z

Ah, I see there is a proof of something like this in 'Equidistribution and Primes' by Peter Sarnak...

http://mathoverflow.net/questions/31897/does-the-function-which-sends-a-right-angled-triangle-to-its-area-produce-infinit Bruce Bartlett 2010-07-15T05:40:49Z 2010-07-15T05:40:49Z

Thanks to all commenters, very useful comments indeed. (And sorry for my mistake that 39 was prime, I typed this up way too late.)

http://mathoverflow.net/questions/31897/does-the-function-which-sends-a-right-angled-triangle-to-its-area-produce-infinit/31939#31939 Bruce Bartlett 2010-07-15T05:38:41Z 2010-07-15T05:38:41Z

Thanks Gerry, very handy!

http://mathoverflow.net/questions/31897/does-the-function-which-sends-a-right-angled-triangle-to-its-area-produce-infinit Bruce Bartlett 2010-07-14T22:35:00Z 2010-07-14T22:35:00Z

Thanks. I know that f can produce numbers with arbitrarily many prime factors. I'm interested if it can produce infinitely many numbers with, say, less than $n$ factors, and then to see how small we can make $n$.