Why are subfactors interesting? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:52:00Z http://mathoverflow.net/feeds/question/544 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/544/why-are-subfactors-interesting Why are subfactors interesting? Kate Liesinger 2009-10-15T02:09:30Z 2011-04-25T08:51:36Z <p>I get asked this question a lot, and am not very happy with any of the answers. </p> <p>Vaguely I think of subfactor theory as a generalization of representation theory of groups. That is, if you have a group/subgroup subfactor, and look at the fusion category, you get a category that has all of the induction and restriction data for the subgroup (I think). So maybe that is my first question, what group theoretic information can you extract from the fusion category of a subgroup subfactor? You can do a similar thing with representations of quantum groups, and you also get "sporadic" subfactors, which don't come from groups or quantum groups. This is interesting because it looks a little like the classification of finite groups, you have a bunch of families and then the fun unexpected sporadic groups. I would love to hear someone else's opinion, if anyone can make the subfactors/group theory analogy more formal. </p> <p>The other explanation as to the interest in subfactors that I hear is, "they have a lot of surprising connections to other topics and show up all over the place." Can someone tell me about such an instance? What are some other topics where subfactors unexpectedly showed up? I am vaguely aware of something to do with random matrices ...</p> <p>Of course, the easy answer is that they are really beautiful and cool in their own right, and no one has to convince me of that. </p> http://mathoverflow.net/questions/544/why-are-subfactors-interesting/558#558 Answer by Noah Snyder for Why are subfactors interesting? Noah Snyder 2009-10-15T04:05:49Z 2009-10-15T04:05:49Z <p>If you want to understand some collection of objects, naturally you should also want to understand all maps between them. Since factors have no two-sided ideals, every map between factors is an inclusion. So to understand maps between factors is the same thing as understand subfactors. (This is a way in which factors are noncommutative analogues of fields, and hence subfactors are a noncommutative version of Galois theory.)</p> <p>Also since all vN algebras are direct integrals of factors, understanding subfactors is the only interesting part of understanding all maps between all vN algebras.</p> http://mathoverflow.net/questions/544/why-are-subfactors-interesting/560#560 Answer by Noah Snyder for Why are subfactors interesting? Noah Snyder 2009-10-15T04:07:40Z 2009-10-15T05:41:02Z <p>Here was my explanation for why I'm interested in Subfactors, which has a different flavor than my other answer for why "people" are interested in them:</p> <p>Subfactor planar algebras are just unitary versions of something very natural. For example, they're unitary versions of 2-categories with 2 objects (tensor categories are 2-categories with 1 object, so this is a natural next step). Also they're just the unitary versions of a simple algebra object in a tensor category, so if you like simple algebras (and who doesn't really) then you'd like subfactors. The unitarity assumption is convenient for computations and thus is convenient to assume when you're trying to find new examples. Also the subfactor literature has lots of results which have applications to tensor categories, but written in a totally different language, so it's worth learning what subfactor people are talking about.</p> http://mathoverflow.net/questions/544/why-are-subfactors-interesting/573#573 Answer by Dave Penneys for Why are subfactors interesting? Dave Penneys 2009-10-15T05:57:00Z 2009-10-15T05:57:00Z <p>Here are some partial answers to some of the questions posed:</p> <p>Any finite group has an outer action on the hyperfinite $II_1$-factor $R$. This means from an inclusion of finite groups $H\leq G$, we can use the crossed product construction to form the subgroup subfactor $N=R\rtimes H\subseteq R\rtimes G=M$ whose index is $[G\colon H]$. The principal graph of the subgroup subfactor indeed encodes the induction/restriction data. In particular, if we take $H$ to be the trivial group, we can read off the dimensions of the irreducible representations of $G$. </p> <p>Just as there are sporadic simple groups, there are 5 exceptional complex simple Lie algebras: $E_6$, $E_7$, $E_8$, $F_4$, and $G_2$. By looking at graph norms, we see that the only possibilities for principal graphs for subfactors of index less than 4 are $A_n$, $D_n$, $E_6$, $E_7$, and $E_8$. It turns out that only $A_n$, $D_{2n}$, $E_6$, and $E_8$ occur as principal graphs, which raises one of the driving questions in subfactor theory: which finite bipartite graphs appear as principal graphs of finite depth, finite index subfactors? A complete classification is known up to $3+\sqrt{3}$.</p> <p>A completely different reason subfactors are interesting is their connection to knot theory. The Jones polynomial was first defined using a representation of the braid group in the $II_1$-factor generated by the Jones projections coming from the iterated basic construction of a finite index subfactor $N\subset M$. Kauffman then gave a skein theoretic definition of the Jones polynomial which also gives us the pictorial representation of the Temperley-Lieb algebras (also generated by the Jones projections). We use these planar diagrams extensively in planar algebras, which are "equivalent" to subfactors when we add some more adjectives (finite index, extremal subfactors and spherical, positive C^*-planar algebras, also known as subfactor planar algebras).</p> http://mathoverflow.net/questions/544/why-are-subfactors-interesting/2481#2481 Answer by Emily Peters for Why are subfactors interesting? Emily Peters 2009-10-25T15:17:35Z 2009-10-25T15:17:35Z <p>I think the original interest in subfactors probably came about from the search for invariants of factors, which in turn was partly motivated by the free group problem. The free group problem is one of those questions that drives mathematicians crazy; it's straightforward to state, and hard to do anything with. It's described below. In general, von Neumann algebras and factors are slippery creatures; they're hard to get your hands on. This makes people want to find invariants of factors. In group theory, the number and size of subgroups is an invariant of a group; analogously, for factors, the number and indices of subfactors is also an invariant.</p> <p><strong>Free group problem:</strong></p> <p>Open question: is the von Neumann algebra of F(n) (free group on n generators) isomorphic to the von Neumann algebra of F(m) if n, m aren't equal? </p> <p>Now let me describe the constructions at work here. Starting with a discrete group, one can build a von Neumann algebra: If G is a group, then l^2(G) is a Hilbert space. The left action of G on l^2(G) gives us a copy of G inside B(l^2(G)), bounded operators on l^2(G) (for g in G, define u(g) in B(l^2(G)) by u(g)(f)(h)=f(g^-1 h) ). Take the closure of this copy of G; you now have a von Neumann algebra!</p> <p>If the group G is i.c.c. (all conjugacy classes (except the identity's) are infinite), the resulting von Neumann algebra is a type II_1 factor. So, it's natural to ask how much information about the original group does this construction preserve? Well, that turns out to be a very difficult question ...</p> http://mathoverflow.net/questions/544/why-are-subfactors-interesting/62900#62900 Answer by Thomas Riepe for Why are subfactors interesting? Thomas Riepe 2011-04-25T08:51:36Z 2011-04-25T08:51:36Z <p>The title of <a href="http://www.diffusion.ens.fr/index.php?res=conf&amp;idconf=1707" rel="nofollow" title="link">this talk</a> suggests that there is a kind of refinement of Galois theory associated with subfactors, but I can't watch it here. Do you know where one can read about that? </p>