# Fuss-Catalan algebras and non-commutative Hilbert schemes

Hello, this is a question regarding Reineke's paper "Cohomology of non-commutative Hilbert schemes", http://arxiv.org/abs/math/0306185, and more precisely the formula on page 4 there (at $n=1$), namely: $$\chi(H_{d,1}^{(m)})=\frac{1}{(m-1)d+1}\binom{md}{d}$$

Here at left we have the cohomological Euler characteristic of the non-commutative Hilbert scheme $H_{d,1}^{(m)}$, and at right we have the Fuss-Catalan numbers.

The point now is that the Fuss-Catalan numbers appear as well in a key place in subfactor theory, namely as coefficients of the Poincaré series of the Fuss-Catalan algebra of Bisch and Jones (for a quick introduction to this algebra, see e.g. Bisch's survey paper http://arxiv.org/abs/math/0304340).

• Question: is there any serious relation between Reineke's formula and the Bisch-Jones formula?

(Note: regarding my own motivation, some 10 years ago I happened to attend a talk by Reineke, and, as a subfactor aficionado, when I saw that numbers on the blackboard I jumped on my seat! I asked him after his talk, he didn't know, and then for some time I kept asking people around, either in AG or in subfactors. But, so far, couldn't find anyone knowing well enough both subjects, in order to answer.)

-
Nice question! If it had just been the ordinary Catalan numbers, I'd be much less surprised but I've seen very few instances of the Fuss-Catalan numbers in my time (which many moons ago included some work on the Fuss-Catalan algebras...). –  Jan Grabowski Jan 11 '13 at 16:41
Catalan numbers count among other things triangulations of polygons (another example from my own interests, in cluster algebras). The OEIS page oeis.org/A000108 lists lots more! So $m=2$ is likely to be speculation and I'd concur that $m=3$ is the place to start. –  Jan Grabowski Jan 11 '13 at 17:15
@Teo: just to note that if no other answer is submitted before the deadline for the bounty, then the bounty is automatically given to the answer with the most votes... –  Yemon Choi Feb 3 '13 at 7:59
@Yemon Choi: the way things are now (and were when you commented) no answer would be automatically accepted. (The answer needs two votes!) The bounty would just expire (points still lost). –  quid Feb 4 '13 at 17:40
@Teo B: In case you did not notice, Vivek Shende actually answered something detailed but then got unhappy with it for some reason and deleted it. And, interesting philosophy. –  quid Feb 4 '13 at 19:58

Yet another comment, if you're interested in such connections, then.. better look at good old algebraic varieties! These are technically speaking at about the same abstraction level as the OA stuff that you mention, i.e. lots of combinatorics and fun. You can check for instance matrix varieties $M\subset M_n(\mathbb C)$, by doing inside some spin model statistics you might perhaps get into FC stuff?