User noah snyder - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:25:05Z http://mathoverflow.net/feeds/user/22 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/132878/are-there-small-examples-of-non-pivotal-finite-tensor-categories Are there small examples of non-pivotal finite tensor categories? Noah Snyder 2013-06-05T21:30:46Z 2013-06-11T20:25:59Z <p>I'm looking for small concrete examples of non-pivotal finite tensor categories to do some calculations with.</p> <p>Here a finite tensor category is, according to <a href="http://arxiv.org/abs/math/0301027" rel="nofollow">Etingof-Ostrik</a>, a rigid monoidal category whose underlying category is equivalent to the category of finite dimensional modules over a finite dimensional algebra. A tensor category is pivotal if there's an isomorphism of tensor functors between the identity functor and the double dual functor.</p> <p>The example I was able to find in the literature is the category of representations of a 72-dimensional Hopf algebra in Remark 2.11 of <a href="http://arxiv.org/abs/1204.5807" rel="nofollow">Andruskiewitsch-Angiono-Iglesias-Torrecillas-Vay</a>. (In fact, they give three such examples.) But I was hoping for a smaller example.</p> <p>Bonus question: What I would really like is an example of a category where not only is the double dual functor nontrivial, but there's no invertible object X such that the double dual is isomorphic as a tensor functor to conjugation by X. (Note that the 72 dimensional Hopf algebras mentioned above do not give counterexamples, because their duals are pivotal.)</p> <p>Easier question: I'd also love to hear about <em>any</em> other examples of non-pivotal finite tensor categories beyond the three from AAITV, even if they're not smaller.</p> http://mathoverflow.net/questions/132881/is-there-an-infinite-depth-irreducible-finite-index-maximal-subfactor-other-than/132882#132882 Answer by Noah Snyder for Is there an infinite depth irreducible finite index maximal subfactor (other than Temperley Lieb) ? Noah Snyder 2013-06-05T21:46:25Z 2013-06-05T21:46:25Z <p>The infinite depth subfactor coming from SU(3) at any index above 9 gives an example. Here the Q-system is $V_{(1,0)} \otimes V_{(0,1)} \cong V_{(1,1)} \oplus V_{(0,0)}$ so the only possible sub-objects are the whole thing or the trivial, so it's certainly maximal.</p> http://mathoverflow.net/questions/41210/does-every-frobenius-algebra-in-a-monoidal-category-give-a-q-system Does every Frobenius algebra in a monoidal *-category give a Q-system? Noah Snyder 2010-10-05T21:51:50Z 2013-05-23T02:27:01Z <p>Suppose that C is a fusion C*-cateogry and that A is an irreducible Frobenius algebra object in C, is there always a Frobenius algebra A' isomorphic to A such that A' is a Longo Q-system (that is the coproduct on A' is an isometry)? In other words, wlog can one assume that the coproduct is the * of the product?</p> <p>The motivation of this question is to understand whether the theory of irreducible Frobenius algebra objects in monoidal *-categories actually agrees on the nose with the theory of irreducible subfactors or whether there's a small loophole.</p> http://mathoverflow.net/questions/71681/are-there-interesting-semisimple-algebras-in-non-semisimple-categories Are there interesting semisimple algebras in non-semisimple categories? Noah Snyder 2011-07-30T20:03:31Z 2013-05-23T00:40:24Z <p>Are there any interesting examples of semisimple algebras in nonsemisimple categories which don't "come from" a semisimple algebra in a semisimple category? That is, if you want to study semisimple algebra objects can you assume wlog that the underlying category is semisimple?</p> <p>Here's one way of trying to make this question precise. Suppose that C is a finite tensor category, and A is an algebra object in C. Further suppose that A-mod (the category of left A-module objects in C) is a semisimple category. Does there always exist: a <strong>semisimple</strong> tensor category C', a monoidal functor F: C'->C, and an algebra object A' in C' such that F(A') = A? (Perhaps I should also require some relationship between A-mod and A'-mod, at the very least A'-mod should be semisimple.)</p> <p>I would also accept an answer explaining why "A-mod is semisimple" doesn't correctly capture the notion of a semisimple algebra if C is non-semisimple.</p> http://mathoverflow.net/questions/16735/is-a-non-analytic-proof-of-dirichlets-theorem-on-primes-known-or-possible/28216#28216 Answer by Noah Snyder for Is a "non-analytic" proof of Dirichlet's theorem on primes known or possible? Noah Snyder 2010-06-15T04:48:54Z 2013-05-08T15:34:47Z <p>For more on Murty's result that the usual elementary approach is doomed to failure look at Paul Pollack's paper <a href="http://www.math.uga.edu/~pollack/impossibility.pdf" rel="nofollow">Hypothesis H and an impossibility theorem of Ram Murty</a> There he shows that a commonly believed conjecture implies a generalization of Murty's result to a broader type of Euclidean proof.</p> http://mathoverflow.net/questions/3309/are-there-two-groups-which-are-categorically-morita-equivalent-but-only-one-of-wh Are there two groups which are categorically Morita equivalent but only one of which is simple Noah Snyder 2009-10-29T18:27:03Z 2013-05-03T05:24:53Z <p>Can you find two finite groups G and H such that their representation categories are Morita equivalent (which is to say that there's an invertible bimodule category over these two monoidal categories) but where G is simple and H is not. The standard reference for module categories and related notions is <a href="http://arxiv.org/PS%5Fcache/math/pdf/0111/0111139v1.pdf" rel="nofollow">this paper</a> of Ostrik's</p> <p>This is a much stronger condition than saying that C[G] and C[H] are Morita equivalent as rings (where C[A_7] and C[Z/9Z] gives an example, since they both have 9 matrix factors). It is weaker than asking whether a simple group can be isocategorical (i.e. have representation categories which are equivalent as tensor categories) with a non-simple group, which was shown to be impossible by <a href="http://arxiv.org/abs/math/0007196" rel="nofollow">Etingof and Gelaki</a>.</p> <p>Matt Emerton asked me this question when I was trying to explain to him why I was unhappy with any notion of "simple" for fusion categories. It's of interest to the study of fusion categories where the dual even Haagerup fusion category appears to be "simple" while the principal even Haagerup fusion category appears to be "not simple" yet the two are categorically Morita equivalent.</p> http://mathoverflow.net/questions/1420/whats-the-best-proof-of-quadratic-reciprocity/1472#1472 Answer by Noah Snyder for What's the "best" proof of quadratic reciprocity? Noah Snyder 2009-10-20T19:31:56Z 2013-04-20T22:01:01Z <p>I think by far the simplest easiest to remember elementary proof of QR is due to <a href="http://tinyurl.com/3z62fdw" rel="nofollow">Rousseau</a>. All it uses is the Chinese remainder theorem and Euler's formula $a^{(p-1)/2}\equiv (\frac{a}{p}) \mod p$. The mathscinet review does a very good job of outlining the proof. I'll try to explain how I remember it here (but the lack of formatting is really rough for this argument).</p> <p>Here's the outline. Consider $(\mathbb{Z}/p)^\times \times (\mathbb{Z}/q)^\times = (\mathbb{Z}/pq)^\times$. We want to split that group in "half", that is consider a subset such that exactly one of x and -x is in it. There are three obvious ways to do that. For each of these we take the product of all the elements in that "half." The resulting three numbers are equal up to an overall sign. Calculating that sign on the $(\mathbb{Z}/p)^\times$ part and the $(\mathbb{Z}/q)^\times$ part give you the two sides of QR.</p> <p>In more detail. First let me describe the three "obvious" halves:</p> <ol> <li>Take the first half of $(\mathbb{Z}/p)^\times$ and all of the other factor</li> <li>Take all of the first factor and the first half of $(\mathbb{Z}/q)^\times$</li> <li>Take the first half of $(\mathbb{Z}/pq)^\times$</li> </ol> <p>The three products are then (letting P = (p-1)/2 and Q=(q-1)/2):</p> <ol> <li>$(P!^{q-1}, (q-1)!^P)$</li> <li>$((p-1)!^Q, Q!^{q-1})$</li> <li>$(\frac{(p-1)!^Q P!}{q^P P!},\frac{(q-1)!^P Q!}{p^Q Q!})$</li> </ol> <p>All of these are equal to each other up to overall signs. Looking at the second component it's clear that the sign relating 1 and 3 is $\left(\frac{p}{q}\right)$. Similarly, the sign relating 2 and 3 is $\left(\frac{q}{p}\right)$. So the sign relating 1 and 2 is $\left(\frac{p}{q}\right) \left(\frac{q}{p}\right)$. But to get from 1 to 2 we just changed the signs of $\frac{p-1}{2} \frac{q-1}{2}$ elements. QED</p> http://mathoverflow.net/questions/118536/delignes-1996-note-on-exceptional-lie-groups/118550#118550 Answer by Noah Snyder for Deligne's 1996 note on exceptional Lie groups Noah Snyder 2013-01-10T16:25:09Z 2013-01-10T16:25:09Z <p>The key papers to read about this are <a href="http://www.math.jussieu.fr/~vogel/" rel="nofollow">Vogel's</a> who conjectures an even better 2-dimensional family which includes all simple Lie algebras, not just the exceptional ones. Deligne's family would correspond to a certain line in this plane. Personally, I find Vogel's papers difficult to read, but you have to read them if you really want to understand this stuff. Some good expositional sources are Cvitanovic's <a href="http://birdtracks.eu/" rel="nofollow">Bird Tracks</a> and <a href="http://www.pdmi.ras.ru/~duzhin/papers/cdbook/" rel="nofollow">Introduction to Vassiliev knot invariants</a> by Chmutov-Duzhin-Mostovoy. In particular, Cvitanovic developed some of the key ideas independently.</p> <p>The basic idea is that you write down diagrammatically the universal metric Lie algebra object in a symmetric tensor category, using the category of Jacobi diagrams. This is very closely related to Vassiliev's finite type invariants by work of Bar-Natan. The resulting Lie algebra object won't be simple, but you can add the assumption that it is simple by assuming that certain Hom spaces are 1-dimensional. Roughly, conjecturally this added assumption gets you down to a 2-dimensional space of possible simple metric Lie algebra objects.</p> <p>To get from the universal Lie algebra in the last paragraph to Deligne's exceptional family, you need to impose one more relation which is satisfied by the exceptional Lie algebras. Namely there is a certain invariant vector in the 4th power of the adjoint representation of any exceptional Lie algebra, and this gives an extra relation. This relation is written down explicitly on the first page of Dylan's paper mentioned in comments.</p> <p>So where are we stuck? Well there's two problems:</p> <ul> <li>Do the relations that we can write down suffice to calculate the value of all closed diagrams?</li> <li>Are the relations consistent for all values of the parameters? That is, if you evaluate the same closed diagram in two different ways do you get the same answer?</li> </ul> <p>The first of these questions should be easier than the second. But both are wide open.</p> <p>What Dylan did in the paper mentioned in comments was look at some analogous situations and calculated directly that it appeared the answers to the first question in those circumstances was yes, but the answer to the second question in those situations was no, and in fact the relations imposed a polynomial relation on the parameter forcing you back to finitely many possibilities. But for Deligne's question, the calculation would require looking at very large examples which were beyond the computer's power.</p> http://mathoverflow.net/questions/118324/what-is-known-about-arbitrary-subfactors-of-integer-index/118330#118330 Answer by Noah Snyder for What is known about arbitrary subfactors of integer index? Noah Snyder 2013-01-08T03:39:52Z 2013-01-08T03:39:52Z <p>If you also assume finite depth, then there's a hope (it's too vague to call it a conjecture) that all integer index subfactors can be classified "using only finite group theory." That is, if you had a black box which could answer all questions about finite groups and their cohomology you'd be able to understand all finite depth integer index subfactors. The key words to look for to get the flavor of this program are "weakly integral" and "weakly group theoretical" fusion categories.</p> <p>There are some results in this direction for small integer global index (I'll try to locate them when I'm at a real computer), but if you just assume small index then we only know the answer for index up to and including 5. The integrality assumption doesn't help in the approach we've been using. So index 6 is way out of reach even assuming finite depth.</p> <p>One could be very optimistic and hope that in infinite depth all integer index subfactors come from (possibly infinite) discrete groups in some suitably general fashion. For example, Bisch-Haagerup subfactors come from certain triples of discrete groups.But I have no idea how one would even try to make "suitably general fashion" precise.</p> <p>On the negative side, Bisch-Nicoara-Popa have shown that at integral index 6 the classification becomes somewhat wild.</p> http://mathoverflow.net/questions/117933/submit-the-second-part-of-a-paper/118240#118240 Answer by Noah Snyder for submit the second part of a paper Noah Snyder 2013-01-07T01:46:47Z 2013-01-07T01:46:47Z <p>One option to consider, which we did with the small index subfactor papers, is to submit to the same editor but at <em>different</em> journals. This retains the logistical advantage of having the same referees, but doesn't require as much space in one journal. Furthermore you can put the better half in a better journal.</p> http://mathoverflow.net/questions/107556/generalizing-the-reshitikhin-turaev-construction-possible/107558#107558 Answer by Noah Snyder for Generalizing the Reshitikhin-Turaev construction possible? Noah Snyder 2012-09-19T12:46:28Z 2012-09-19T12:46:28Z <p>Indeed any ribbon tensor category gives knot invariants. There's no reason why the ribbon tensor category has to come from a Lie group. However, we just don't know many other examples of ribbon tensor categories.</p> <p>It's certainly plausible that you can get knot invariants from quantum versions of other points on the Vogel plane. This is closely related to the theory of finite type invariants and to Bar Natan's program to generalize Etingof-Kazhdan quantization. However, showing that such a braided tensor category exists is <em>harder</em> than just showing that the corresponding Lie algebra object exists (which is itself completely open and hard).</p> <p>It is an interesting question to try to work out what knot invariants you would get from a quantized version of the E7 series. I don't know of any known results in that direction.</p> http://mathoverflow.net/questions/107098/can-reidemeister-3-be-weakened/107187#107187 Answer by Noah Snyder for Can Reidemeister 3 be weakened? Noah Snyder 2012-09-14T14:43:51Z 2012-09-14T14:43:51Z <p>I am very very skeptical that you can recover R3 from your weak R3. Instead here's what I expect is happening in your examples. Let V be the vector space spanned by diagrams with 6 boundary points. There's a pairing on V given by connect up all the boundary points and apply your invariants. You might as well treat as 0 anything in V that's in that's in the radical of this form, it won't change your invariants. So what we want to know is whether the difference of the two sides of R3 (which I'll call v) is in the radical of this form for your particular examples. When you take inner products of v with various other vectors in V you do the calculation by applying whatever skein relations you have. It's quite possible that <em>in the presence of your other specific skein relations</em> your weak R3 is all you need in order to show that v has inner product 0 with everything. But this would be because you've imposed enough other special skein relations.</p> http://mathoverflow.net/questions/105208/are-there-better-upper-bounds-on-the-rank-of-the-commutant-of-a-fusion-module-tha/105243#105243 Answer by Noah Snyder for Are there better upper bounds on the rank of the commutant of a fusion module than the global dimension? Noah Snyder 2012-08-22T14:50:15Z 2012-08-22T14:50:15Z <p>In the special case of Vec(G) and Rep(G) there's a lot of results in the literature. Usually the phrasing is in terms of conjugacy classes (e.g. the strongest proved bound as far as I know is Keller's <a href="http://arxiv.org/abs/0812.2590" rel="nofollow">"Finite groups have even more conjugacy classes"</a>). I learned about this from Pavel Etingof when Eric Rowell asked him about a similar question for ranks of centers.</p> <p>Do I understand correctly though that you're happy to have conditions that involve more than just the rank of C? E.g. if I say wanted to have the full list of dimensions of objects in C as input to the bound would that be a problem?</p> http://mathoverflow.net/questions/105151/mini-fusion-categories-via-6j-symbols/105186#105186 Answer by Noah Snyder for "Mini" fusion categories via 6j symbols Noah Snyder 2012-08-21T18:57:16Z 2012-08-21T21:43:22Z <p>The number of non-isomorphic simple objects is called the <em>rank</em> of the fusion category.</p> <p>You've made an error somewhere, as in rank 2 you should also get an example where the nontrivial simple object squares to $1$. (In your notation $2 \otimes 2 \cong 1$.) </p> <p>For rank 2, Ostrik gave a complete classification (http://arxiv.org/abs/math/0203255). Ostrik's argument is somewhat indirect (via the Drinfel'd center). As far as I know, no one has given a direct classification in rank 2 via 6j symbols. The hard part is figuring out why there's no fusion categories with the fusion rule $X^2 = 1 + n X$ for $n>1$.</p> <p>For rank 3 if you only look at fusion categories that give knot invariants, then Ostrik has a complete classification (http://arxiv.org/abs/math/0503564).</p> <p>A great source of all information about "minis" that give knot invariants is Section 5.3 of Rowell-Strong-Wang (http://arxiv.org/abs/0712.1377). "Modular" means that you get knot invariants plus you have a certain "non-degeneracy" condition that's a bit hard to explain.</p> http://mathoverflow.net/questions/101725/what-is-the-ray-in-ray-class-group/101737#101737 Answer by Noah Snyder for What is the "ray" in ray class group? Noah Snyder 2012-07-09T04:30:45Z 2012-07-09T04:30:45Z <p>A "ray class group" is constructed from a "ray" $K_{m,1}$, so the question is why $K_{m,1}$ is called a ray. I don't have direct evidence for this, but it seems pretty clear to me that the reason for this terminology is that the condition coming from a real primes in m is that $x \in K_{m,1}$ lies in the ray (in the high school geometry sense) of positive real numbers (under the appropriate real embedding).</p> http://mathoverflow.net/questions/98735/luries-virtual-fundamental-classes-and-geometric-derived-stacks/98901#98901 Answer by Noah Snyder for Lurie's "Virtual fundamental classes" and "Geometric derived stacks" Noah Snyder 2012-06-05T18:59:38Z 2012-06-05T18:59:38Z <p>It's worth noting that Lurie's thesis is not on his webpage, and is not regularly updated like the papers which are on is webpage. I would take this to mean that you're supposed to be reading the current versions of this material in the DAG series, and not reading the 8 year old outline.</p> http://mathoverflow.net/questions/94624/jonesunlinkphi/98202#98202 Answer by Noah Snyder for Jones(unlink)=phi Noah Snyder 2012-05-28T16:20:33Z 2012-05-28T16:20:33Z <p>In some sense this is the smallest possible quantum group, so it's perhaps not surprising that it comes up often. In fact, if you have only 2 objects, then there are very few possibilities, see Ostrik's paper <a href="http://arxiv.org/abs/math/0203255" rel="nofollow">http://arxiv.org/abs/math/0203255</a>.</p> http://mathoverflow.net/questions/96069/status-of-pl-topology/96083#96083 Answer by Noah Snyder for Status of PL topology Noah Snyder 2012-05-05T20:32:42Z 2012-05-05T20:32:42Z <p>PL topology is popular in quantum topology where some invariants (e.g Turaev-Viro) are defined by fixing a triangulation and the checking invariance under some standard moves.</p> http://mathoverflow.net/questions/88828/module-categories-over-symmetric-braided-monoidal-categories/88845#88845 Answer by Noah Snyder for Module categories over symmetric/braided monoidal categories Noah Snyder 2012-02-18T17:18:10Z 2012-02-18T17:18:10Z <p>Consider C = Rep(G) for a finite group G. This is a symmetric tensor category. Any subgroup $H \subset G$ yields a module category Rep(H) with the action given by restricting and then tensoring. This module category is simple in any appropriate sense, and in no interesting sense is it "1-dimensional."</p> http://mathoverflow.net/questions/88463/role-of-fiber-functor-monoidal-structure-in-tannakian-bialgebra-reconstruction/88838#88838 Answer by Noah Snyder for Role of fiber functor monoidal structure in Tannakian bialgebra reconstruction Noah Snyder 2012-02-18T16:23:33Z 2012-02-18T16:23:33Z <p>To give a different answer to your meta-question, you can certainly have a non-monoidal fiber functor from C -> Vec where C is not the category of representations of any bialgebra. Just take the category of twisted G-graded vector space Vec(G,w) where w is a nontrivial 3-cocycle. This is not the category of representations of any bialgebra, but it has an obvious non-monoidal fiber functor to vector spaces. (It is, however, the category of representations of a quasi-bialgebra.)</p> http://mathoverflow.net/questions/88828/module-categories-over-symmetric-braided-monoidal-categories/88835#88835 Answer by Noah Snyder for Module categories over symmetric/braided monoidal categories Noah Snyder 2012-02-18T15:25:01Z 2012-02-18T15:25:01Z <p>If A is a commutative Noetherian A algebra in Vec, then A-mod is not typically equivalent to Vec. First, if it's not a semisimple algebra then there are interesting non-simple modules, and even in the semisimple case you don't just have <em>one</em> 1-dimensional A-module. So I don't think your question makes sense as currently written.</p> <p>At any rate, you can have a commutative semisimple finite dimensional algebra A in a braided tensor category with simple A-modules that aren't one-dimensional. For example, your tensor category is U_q(sl_2) at an appropriate root of unity, your commutative algebra is the sum of the two 1-dimensional objects out at the ends, and the category of modules is quite interesting and goes by the name D_2n (the easiest place for you to see more about this is probably <a href="http://arxiv.org/abs/math/0101219" rel="nofollow">http://arxiv.org/abs/math/0101219</a>). I'm not totally sure about the symmetric case.</p> http://mathoverflow.net/questions/84161/hopf-algebras-examples/84171#84171 Answer by Noah Snyder for Hopf algebras examples Noah Snyder 2011-12-23T17:02:51Z 2011-12-24T04:37:32Z <p>There's a (unique) semisimple noncocommutative Hopf algebra of dimension 8 that makes a nice example. (Unfortunately I don't remember where to find information on it at the moment, I learned about it in a survey paper of Susan Montgomery's.)</p> http://mathoverflow.net/questions/32432/does-the-quantum-subgroup-of-quantum-su-2-called-e-8-have-anything-at-all-to-do-w Does the quantum subgroup of quantum su_2 called E_8 have anything at all to do with the Lie algebra E_8? Noah Snyder 2010-07-19T01:06:57Z 2011-12-13T15:51:20Z <p>The ordinary McKay correspondence relates the subgroups of SU(2) to the affine ADE Dynkin diagrams. The correspondence is that the vertices correspond to irreducible representations of the subgroup, and edges correspond to tensor product with the 2-dimensional representation (which is given by the inclusion of G in SU(2)).</p> <p>The quantum McKay correspondence does the same thing for "quantum subgroups" of the quantum group U_q(su(2)) for q a root of unity and the ordinary ADE Dynkin diagrams. One way to make this precise is that "quantum subgroups" are module categories over semisimplified categories of modules for U_q(su(2)), so the simple objects in the module category are the vertices, and the edges come from the tensor product with the defining rep of U_q(su(2)). In this language the result is due to <a href="http://arxiv.org/abs/math/0101219" rel="nofollow">Kirillov-Ostrik</a>, but it goes back further in both the subfactor literature (due to Ocneanu and others) and the conformal field theory literature (which I don't know as well). These quantum subgroups come in two sorts, type 1 and type 2, and the type 1 ones can also be realized as fusion categories where the module structure comes from "restrict and tensor." The A_n, the D_2n, E_6 and E_8 are Type 1.</p> <p>Ok, so now for the question, does the E_8 "quantum subgroup of quantum su(2)" have anything whatesoever to do with Lie algebra E_8?</p> <p>I think the answer should be "no." The main reason I'm guessing this is that although the D_2n fusion categories are related to certain SO quantum groups under level-rank duality, the level-rank dual fusion category is <em>not</em> related to the quantum group for the Lie algebra D_2n it's instead related to the Lie algebra D_n-1 (<a href="http://arxiv.org/abs/1003.0022" rel="nofollow">see section 4.3 of my paper with Scott and Emily</a> and the references therein).</p> <p>The reason that I ask is in reference to <a href="http://mathoverflow.net/questions/32315/has-the-lie-group-e8-really-been-detected-experimentally" rel="nofollow">Borcherds's question</a> about detecting E_8 experimentally. I don't totally understand Will Orrick's answer but it sure sounds to me like he's saying that they've detected *the quantum subgroup E_8*, not the Lie algebra E_8. In particular its the quantum subgroup E_8 that has 8 objects ("particle types" in physics speak) one of which (all the way at the far end) has dimension the golden ratio (and so the golden ratio should come up in comparing it with the trivial object in some experiment).</p> http://mathoverflow.net/questions/79044/what-items-must-appear-on-a-mathematicians-cv/79161#79161 Answer by Noah Snyder for What items MUST appear on a mathematician's CV? Noah Snyder 2011-10-26T13:29:11Z 2011-10-26T13:29:11Z <p>Your PhD advisor's name (listed under education).</p> http://mathoverflow.net/questions/79065/example-of-a-commutative-algebra-object-in-a-braded-monoidal-category-c/79076#79076 Answer by Noah Snyder for Example of a commutative algebra object in a braded monoidal category C Noah Snyder 2011-10-25T13:19:07Z 2011-10-25T13:19:07Z <p>The standard example here is where the braided tensor category is the Drinfeld center Z(C) and the algebra object is the induction of the trivial object from C to Z(C). If C is semsimple over an algebraically closed field then this can be written explicitly as $\sum_x x \otimes x^*$ with half braiding given by Theorem 2.3 of <a href="http://arxiv.org/abs/1004.1533" rel="nofollow">Kirillov-Balsam</a>.</p> <p>There are plenty of trivial examples when the category is allowed to be symmetric (which presumably you don't want), for example any ordinary commutative algebra is an algebra object in the symmetric (and hence braided) tensor category Vec.</p> http://mathoverflow.net/questions/75519/oriented-vs-unoriented-knot-polynomials/75530#75530 Answer by Noah Snyder for Oriented vs. unoriented knot polynomials Noah Snyder 2011-09-15T15:33:12Z 2011-09-15T15:33:12Z <p>If I'm remembering correctly, the situation here is a bit subtle. On the one hand, morally the invariants do depend on orientation. In particular, my recollection is that for links they actually depend on orientation. But, on <em>knots</em> they can't detect orientation.</p> <p>So you should expect your individual formulas which you use to compute the knot invariant to depend on orientation, but in the end the <em>answers</em> for <em>knots</em> won't depend on the orientation.</p> http://mathoverflow.net/questions/75232/to-what-extent-can-one-prescribe-degrees-of-irreducible-representations-of-a-grou/75254#75254 Answer by Noah Snyder for To what extent can one prescribe degrees of irreducible representations of a group? Noah Snyder 2011-09-12T19:57:34Z 2011-09-12T19:57:34Z <p>There are lots of results like this, but they're pretty piecemeal. For example, the number of 1-dimensional representations divides the size of the group. There are some results about how large the largest dimensions can be (see Durfee &amp; Jensen (J. Alg, 2011) for the most recent results). There are also some results about the small end of the distribution (e.g. if you have a 2-dimensional representation then you can use the classification of subgroups of SU(2) to get info).</p> <p>In general if you write down a sequence that passes all the obvious tests then no one has any idea whether a group exists which realizes that.</p> http://mathoverflow.net/questions/74095/maiden-names-vs-married-names/74166#74166 Answer by Noah Snyder for Maiden Names vs. Married Names Noah Snyder 2011-08-31T14:51:47Z 2011-08-31T15:00:12Z <p>Like all questions involving names and marriage, there is no set convention (at least in the US). I know a male mathematician who publishes under his wife's last name which he took at marriage and I know people who have started publishing under a new name before they took it legally. As Ben says, there's also no rules for names that don't involve marriage: not only do people pick whether they use their full first name or a nickname, some people use initials, and some people use nicknames which are not related to their legal first name.</p> <p>Although there are no set rules or conventions, most people seem to agree that early on in your career it's unwise to change the name that you are publishing under. Your name is your brand and diluting it is likely to hurt you professionally. Thus there's a strong tendency for people to publish under a fixed name. Nonetheless this is not a fixed rule, a particularly striking example is a <em>theorem</em> that's changed names: Nichols-Richmond nee Nichols-Zoeller.</p> http://mathoverflow.net/questions/73922/homotopy-fixed-points-of-so2-on-fully-dualizable-algebras/73965#73965 Answer by Noah Snyder for Homotopy Fixed Points of SO(2) on Fully Dualizable Algebras Noah Snyder 2011-08-29T12:38:40Z 2011-08-29T12:38:40Z <p>I'm not comfortable enough with the $\infty$ setting to give a full answer, but for ordinary algebras the condition you're looking for here should be that of a <em>symmetric</em> Frobenius algebra (i.e. the Frobenius bilinear form is symmetric). So you're just looking for the $\infty$ analogue of the symmetric condition on Frobenius algebras.</p> <p>There's a lot of great information about oriented 2d extended TFTs in <a href="http://sites.google.com/site/chrisschommerpriesmath/" rel="nofollow">Chris Schommer-Pries's thesis</a>.</p> http://mathoverflow.net/questions/39273/is-the-following-map-from-zg-x-h3g-c-h2g-c-ever-nontrivial Is the following map from Z(G) x H^3(G, C*) --> H^2(G, C*) ever nontrivial? Noah Snyder 2010-09-19T04:18:17Z 2011-08-25T11:14:18Z <p>Suppose that G is a finite group, then we have the following map f which takes an element z in the center of G and a 3-cohomology class w and returns a 2-cohomology class f(z,w) (for concreteness let's take coefficients to be C* everywhere).</p> <p>$f(z,w)(x,y) = \frac{w(z, x, y) w(x, y, z)}{w(x, z, y)}.$</p> <p>Is this map ever nontrivial? That is can you find a group G, a central element z, and an element of H^3 such that f(z,w) is not the trivial element of H^2?</p> <p>The motivation for this question is that it should give an example where Z(G) did not lift to a subcategory of the Drinfel'd center Z(Vec(G,w)).</p> http://mathoverflow.net/questions/133816/problem-5-15-i-martin-isaacs-character-theory Comment by Noah Snyder Noah Snyder 2013-06-15T16:14:21Z 2013-06-15T16:14:21Z As the person who said Yemon's post may be on topic for MO, I agree with Fernando that just copying the question isn't acceptable. Furthermore, I think there's a big difference between a mathematician needing to know the answer to an exercise for research and is genuinely stuck, and a new anonymous user who is probably asking about homework. http://mathoverflow.net/questions/131546/unitary-structures-on-fusion-categories Comment by Noah Snyder Noah Snyder 2013-06-13T04:37:42Z 2013-06-13T04:37:42Z But isn't that just polar decomposition? That is to say, if you can conjugate one unitary by $X$ to get another, then you can also get between them by conjugating by the unitary matrix $X \sqrt{X X^\dagger}^{-1}$? http://mathoverflow.net/questions/72651/if-c-is-a-fusion-category-over-a-field-of-nonzero-characteristic-and-dim-c-0-i Comment by Noah Snyder Noah Snyder 2013-06-08T04:23:41Z 2013-06-08T04:23:41Z I think we've worked out that Z(C) fusion implies $\mathrm{dim} C \neq 0$. I will post an answer soon when the proof is posted. http://mathoverflow.net/questions/133118/how-to-compute-the-abelianization-of-the-representation-theory-of-a-hopf-algebra Comment by Noah Snyder Noah Snyder 2013-06-08T04:18:46Z 2013-06-08T04:18:46Z Another name is Dim(C), since it's the value of the (1-framed) circle in a 1-dimensional TFT. Or Tr(C) to avoid confusion with global dim. In 3 dimensions (which is the right dimension for tensor cats) there are two 3-framed circles: one is Hochschild cohomology (aka Z(C)) and the other is Hochschild homology (your $C_{ab}$). In our upcoming paper (should be this month?) with CD and CSP we give the above mentioned formula (all the techniques are already in ENO's papers but you have to be careful to not assume pivotally). If you want an unfinished draft earlier than that send me an email. http://mathoverflow.net/questions/133118/how-to-compute-the-abelianization-of-the-representation-theory-of-a-hopf-algebra Comment by Noah Snyder Noah Snyder 2013-06-08T03:45:57Z 2013-06-08T03:45:57Z A priori, you have to be a little careful about whether those are left duals or right duals and whether you're twisting on the left or on the right, but once you know Radford's theorem you can canonically identify all the variations (nonetheless one of them is the right one to be Hochschild homology). That description works for any finite tensor category (in the sense of Etingof-Ostrik) not just Rep(H). But I don't actually know where to put in the duals to modify the definition of the Drinfel'd double which is why I hadn't written a full answer. http://mathoverflow.net/questions/133118/how-to-compute-the-abelianization-of-the-representation-theory-of-a-hopf-algebra Comment by Noah Snyder Noah Snyder 2013-06-08T01:30:39Z 2013-06-08T01:30:39Z Note that Z(C) is $\mathrm{Fun}_{C\text{-mod-}C}(C,C)$ while your H(C) is $\mathrm{Fun}_{C\text{-mod-}C}(C,S)$ where S is C where one of the actions is twisted by the double dual. (I can send you details about that calculation if you need them.) So I'd guess that you could somehow twist the Drinfel'd double of the Hopf algebra by inserting some double duals into the definition to get the algebra that you're looking for. http://mathoverflow.net/questions/132881/is-there-an-infinite-depth-irreducible-finite-index-maximal-subfactor-other-than/132882#132882 Comment by Noah Snyder Noah Snyder 2013-06-05T22:47:01Z 2013-06-05T22:47:01Z In general it'd probably be a bit tricky to work out whether the subfactor is maximal. But for SU(n) the same argument should work. http://mathoverflow.net/questions/131546/unitary-structures-on-fusion-categories Comment by Noah Snyder Noah Snyder 2013-05-30T21:54:16Z 2013-05-30T21:54:16Z I'm not totally sure, but I think that if a unitary structure exists then it's unique. This is because unitarity should be equivalent to saying that the F-matrices can be gauged to become unitary matrices, and that's a condition not a structure. http://mathoverflow.net/questions/16735/is-a-non-analytic-proof-of-dirichlets-theorem-on-primes-known-or-possible/28216#28216 Comment by Noah Snyder Noah Snyder 2013-05-08T15:35:32Z 2013-05-08T15:35:32Z @Chandrasekhar: Fixed. http://mathoverflow.net/questions/119219/4d-tqft-from-a-modular-tensor-category/119223#119223 Comment by Noah Snyder Noah Snyder 2013-01-18T14:55:13Z 2013-01-18T14:55:13Z When you say it's less info is that because you're talking about the 34 theory not the 234 theory? The oriented 234 theory is exactly the same info as the anomalous 23 theory, right? http://mathoverflow.net/questions/118536/delignes-1996-note-on-exceptional-lie-groups/118550#118550 Comment by Noah Snyder Noah Snyder 2013-01-10T22:26:20Z 2013-01-10T22:26:20Z Excellent! You can't drink MO reputation points. http://mathoverflow.net/questions/118536/delignes-1996-note-on-exceptional-lie-groups/118550#118550 Comment by Noah Snyder Noah Snyder 2013-01-10T16:39:36Z 2013-01-10T16:39:36Z Another source I found helpful was the papers of Kneissler though they are somewhat technical. <a href="http://arxiv.org/find/math/1/au:+Kneissler_J/0/1/0/all/0/1" rel="nofollow">arxiv.org/find/math/1/au:+Kneissler_J/0/1/0/all/0/&hellip;</a> http://mathoverflow.net/questions/118536/delignes-1996-note-on-exceptional-lie-groups/118547#118547 Comment by Noah Snyder Noah Snyder 2013-01-10T16:32:00Z 2013-01-10T16:32:00Z Isn't the subexceptional (E7) series also still open? http://mathoverflow.net/questions/117785/not-so-symmetric-monoidal-categories Comment by Noah Snyder Noah Snyder 2013-01-02T12:26:41Z 2013-01-02T12:26:41Z Oh and for Sp(4) at the smallest interesting root of unity Morrison-Peters have a result along these lines without assuming braided (but assuming unitary). http://mathoverflow.net/questions/117785/not-so-symmetric-monoidal-categories Comment by Noah Snyder Noah Snyder 2013-01-02T12:22:31Z 2013-01-02T12:22:31Z Kazhdan-Wenzl I think is on google books, it does all type A quantum groups. Tuba-Wenzl assumes braided. Without assuming braided Pinhas and I proved a result like this for SO(3) (that is, the sub tensor category of integer spin reps). While assuming braided Hans has done the Spin case, and I know how to do G2 roughly. The remaining exceptionals look too hard with current techniques.