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Questions on Yuji Tachikawa's talk Recent advances in Supersymmetry

+ 2 like - 0 dislike
334 views

I've been watching Tachikawa's talk.

Recent advances in Supersymmetry

around 38 minutes he talks about objects in \(H^3(M,\mathbb{Z}_N)\) with the property that

\(\int_C a \in \mathbb{Z}_N\)

He says due to self duality in 6D dimensions two intersecting cycles C and C' with \(C \cap C' \neq 0\) are mutually non local.

It seems that mutually non-local means \(\int_M a \wedge b \neq 0\)

But why is this true, what happens when a theory is self dual?

What happens when you add supersymmetry?

He talks about the self dual 3 form what about other fields in the theory?

Is there a fermionic(or supersymmetric) generalization to \(H^3(M,\mathbb{Z}_N)\)?

What is the recipe to construct the partition vector?

Is there any detailed exposition of the subject ? What books discuss the object \(H^3(M,\mathbb{Z}_N)\)? I am familiar with the basic ideas of co-homology for U(1) gauge groups, but I have never really studied its generalization to other gauge groups. 

Please excuse me if my questions are naive or too simple.

Edit: I have attached the related slides

asked May 15, 2015 in Theoretical Physics by Prathyush (695 points) [ revision history ]
edited May 19, 2015 by Prathyush

What does mutually non-local mean?

@RyanThorngren From what I understood form the talk mutually non local means

\(\int a\wedge b \neq 0 \)

While I did not understand exactly what he meant I will tell you the argument he makes.

He says one wants to fix \(\int_C a \in H^3(M,\mathbb{Z})\) so that \(\int_C a\in \mathbb{Z}_N \) is the magnetic flux through C.

However he goes on to say one cannot do that because of the self duality of 3 form tensor when you have 2 intersecting cycles C and C' the fluxes will be mutually non local.

Instead He says we have to split \(H^2(M,\mathbb{Z}) = A \oplus B\) where A and B are of equal sizes.

Such that the fluxes in A are mutually local to each other and fluxes in b are mutually local to each other.

i.e
\(\int a\wedge a' = 0\) and \(\int b\wedge b' = 0\)

1 Answer

+ 2 like - 0 dislike

This is something that happens for the 2d CFT on the boundary of 3d Chern-Simons theory. The Chern-Simons gauge field is its own canonical conjugate (a self-duality), so you can only specify some of the holonomies (like how the wavefunction depends on position, say, but not momentum). This is why the Hilbert space grows like $K^g$ rather than $K^{2g}$ on a genus $g$ surface.

There is a 7d TQFT that carries the 6d CFT conformal blocks in the same way.

answered May 18, 2015 by Ryan Thorngren (1,605 points) [ no revision ]

@RyanThorngren

Interesting I did not there is a 7d TQFT related to 6d theory.

Can you give me a reference that discusses the issue in 3d Chern Simons.

Ed Witten discusses the 7d TQFT here: http://arxiv.org/abs/hep-th/9610234 . There are many references for the 3d story, but Ed Witten also discusses it in ``quantum field theory and the Jones polynomial".

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