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25 votes
2 answers
5k views

How to understand Chern-Simons action

Hi all. The question I have should be a rather simple one, but I just can't think it through. So the Chern-Simons action reads \begin{equation} S = \int_M {\rm tr} (A\wedge dA + \frac{2}{3} A\wedge A ...
Lelouch's user avatar
  • 857
5 votes
1 answer
735 views

Gauge invariance of Chern-Simons functional integral for a 3-manifold with boundary

I am trying to understand how the functional integral for Chern-Simons theory for a possibly non-compact 3-manifold with boundary is made gauge invariant. For a compact 3-manifold, $M$, without ...
Mtheorist's user avatar
  • 1,155
5 votes
0 answers
240 views

Does Dijkgraaf-Witten theory have a time-reversal symmetry?

By having a time-reversal symmetry I mean that there is a local anti-unitary symmetry (representing the non-trivial element of $Z_2$) of the state-sum construction (or, if you want, of the associated ...
Andi Bauer's user avatar
  • 3,001
4 votes
0 answers
211 views

3d Chern-Simons TQFT of gauge group (E8)$_1$ = SO(16)$_1 \otimes$ a trivial spin TQFT = Cartan E$_8$ matrix

In this post, we like to relate the following 3 bosonic TQFTs that can be defined on generic non-spin manifold $M^3$. Given a non-abelian Chern-Simons (CS) TQFT of a gauge group $G$ and the $k$ named ...
wonderich's user avatar
  • 10.5k
4 votes
0 answers
153 views

tertiary characteristic class: integration of the Chern-Simons form

Let $P \to M$ be a trivial principal circle bundle with connection $A$ over a closed 3-manifold $M$. The Chern-Simons 3-form of the connection is defined by $\mathrm{CS}(A) = A \wedge dA$. Suppose ...
Tobias Diez's user avatar
  • 5,824