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  Supertrace of holonomy of commutator

+ 3 like - 0 dislike

On page 47 of Surface operators in four-dimensional topological gauge theory and Langlands duality by Kapustin et al., the following expression is given \begin{equation} \delta\mathcal{N}=d(\omega_\bar{i}\eta^{\bar{i}}+T)+[\mathcal{N},\omega_\bar{i}\eta^{\bar{i}}+T]. \end{equation} It is then claimed that the supertrace of the holonomy of this expression vanishes, i.e., that \begin{equation} STr\textrm{ e}^{\oint\delta\mathcal{N}}=0. \end{equation} My question is, how does on show this?

Using Stoke's theorem, one can show that \begin{equation} STr\textrm{ e}^{\oint\delta\mathcal{N}}=STr\textrm{ e}^{\oint [\mathcal{N},\omega_\bar{i}\eta^{\bar{i}}+T]}. \end{equation} However, I have no idea as to how this expression should vanish.

This post imported from StackExchange Physics at 2016-08-14 09:23 (UTC), posted by SE-user Mtheorist

asked Mar 4, 2016 in Theoretical Physics by Mtheorist (100 points) [ revision history ]
edited Aug 14, 2016 by Dilaton

1 Answer

+ 1 like - 0 dislike

On the same page of the paper, $\mathcal{N}$ is defined as a connection on a bundle $\sigma^* E$. The claim that its BRST variation is $$\delta_{\text{BRST}} \,\mathcal{N} = \mathrm{d} \alpha + [\mathcal{N},\alpha],~~~ (\alpha = \omega \eta + T),$$ means that $\delta_\text{BRST} \mathcal{N}$ is just a gauge transformation of $\mathcal{N}$, $\delta_\alpha \mathcal{N} = \mathrm{d}_\mathcal{N}\, \alpha$, with $\mathrm{d}_\mathcal{N} = \mathrm{d} + [\mathcal{N}, \cdot]$ the gauge-covariant derivative. The trace of the holonomy around a curve $C$ is just a Wilson loop, $$W_C(\mathcal{N}) = \mathrm{str} \,\mathrm{P}\, e^{ \oint_C \mathcal{N}},$$ which is of course a gauge invariant operator. It is therefore also BRST invariant.

This post imported from StackExchange Physics at 2016-08-14 09:23 (UTC), posted by SE-user user81003
answered Mar 4, 2016 by user81003 (130 points) [ no revision ]

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