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As I understand it, there is a subfactor whose principal graph is the affine Dynkin diagram $\tilde{A}_n$. Since every vertex has two neighbors, does that mean the space of 1-boxes is two dimensional? Is that allowed?

flag	asked Mar 16 2011 at 16:18 Stephen Bigelow 235●6

Kevin Walker · Accepted Answer · 2011-03-16 18:05:07Z UTC

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One concrete realization of this planar algebra goes as follows.

Start with the (generalized) PA freely generated by an oriented strand.
Impose "$Z$-homology" relations: (a) oriented saddle moves, and (b) erase small loops (loop value = 1).
Now introduce an unoriented strand which is the formal direct sum of an upward pointing strand and a downward pointing strand.
If we now draw the principal graph of this PA from the point of view of the newly introduced unoriented strand type, we get the $\tilde{A}_\infty$ graph.
I just noticed that you want $\tilde{A}_n$ , not $\tilde{A}_\infty$. For this case, start with $Z/n$ homology instead of $Z$ homology. If you want more details let me know.

answered Mar 16 2011 at 18:05

Kevin Walker
6,147●1●10●32

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Kevin, can you tell me more about what $Z$-homology and $Z/n$ homology mean here? – Emily Peters May 24 2012 at 19:09

Z/n homology is a bit of an abbreviation, perhaps... If you're working on a closed surface, Z/n homology just means you can take n parallel (considering orientations, too) circles and remove then (even if they are essential on the surface). If you want to work on open surfaces (e.g. disks, to define a planar algebra), I think you need to add two n-valent vertices, one with all edges oriented in, the other with all oriented out. Now you can replace n parallel edges and replace them with a pair of these. ... – Scott Morrison♦ May 25 at 5:46

If you do the dimension count, each disk space is 1 or 0 dimensional, depending on whether the signed count of the boundary points is 0 mod n or not. It seems this should give $\widehat{A_n}$ graphs per Kevin's prescription above (although maybe not with exactly the same n?). – Scott Morrison♦ May 25 at 5:50

I suspect one can also freely chose the rotational eigenvalue of the two n-valent vertices, and these give the different non-isomorphic subfactor planar algebras with principal graph $\widehat{A_n}$. – Scott Morrison♦ May 25 at 5:50

Finally (to record all the thoughts from talking to Dave Penneys this afternoon), you ought to get the affine D subfactors via the automorphism in Kevin's picture which reverses all orientations. – Scott Morrison♦ May 25 at 5:51

Dave Penneys · Answer 2 · 2011-03-16 17:28:50Z UTC

Yes, it is two dimensional, and this is allowed. It just means the planar algebra is not irreducible. I don't know of anyone that has thought about a presentation by generators and relations of this planar algebra yet.

One issue here is that since $d=[M\colon N]^{1/2}=2$ is not generic ($>2$), one has to be careful about the annular multiplicities of the subfactor (arXiv:math/0105071). So I don't know if the planar algebra qualifies as "annular multiplicities $*10$" like (extended) Haagerup.

What is the subfactor planar algebra of type $\tilde{A}_n$, of index 4?

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