# Gauge and global symmetries in Chern-Simons/WZW correspondence

I am trying to understand how bulk gauge symmetry in 3d Chern-Simons theory becomes a global symmetry in the boundary 2d WZW theory. In particular, I am trying to understand the papers by Elitzur et al. and Moore and Seiberg. In these references, 3d nonabelian Chern-Simons theory with the classical action $$S=\frac{k}{4\pi}\int_{Y}\textrm{Tr}(A\wedge dA+\frac{2}{3}A\wedge A\wedge A)$$

is studied on a manifold $$Y$$ with boundary. In reference 2 (last paragraph of first page), Moore and Seiberg explain that to ensure the equations of motion are free from boundary terms, one needs to set one of the boundary components of the gauge field to be zero, since $$\delta S= \frac{k}{4\pi}\int_{\partial Y} \textrm{Tr}(\delta A\wedge A)+\frac{k}{2\pi}\int_Y\textrm{Tr}(\delta A\wedge F).$$ They then claim that "With these boundary conditions the functional integral is invariant only under gauge transformations which are one at the boundary." My first question is, why is this the case?

As a first step to getting WZW theory on the boundary, both references split $$Y$$ as $$\Sigma \times \mathbb{R}$$, and pick the boundary condition $$A_0=0$$, i.e., the component of $$A$$ along the time direction $$\mathbb{R}$$ vanishes.

Reference 1 then gives more detail (on page 110); they explain that "The symmetry of the theory IS the group of gauge transformations which do not change the boundary conditions. These are gauge transformations which are independent of time on the boundary. Only a subgroup of this group should be viewed as a gauge symmetry. This is the set of transformations which are one at the boundary. Time independent transformations on the boundary should be viewed as a global symmetry. My second question is, why should time independent transformations be viewed as a global symmetry?

Added note: If the time independent $$g$$ was independent of other coordinates on the boundary, it would be global, but this is not necessarily the case. Moreover, in the references, they do not require that the transformation be fixed to be considered global. For $$\Sigma=D$$ the boundary chiral WZW theory with group valued fields $$U$$ has a symmetry $$U\rightarrow \widetilde{V}(\varphi)UV(t)$$, where $$\widetilde{V}(\varphi)$$ is referred to as a global symmetry (see e.g. reference 1, equation 2.7). The reason given is that $$\widetilde{V}$$ does not go to one in the past and future, which I also am unclear about.

Attempted solution: Let us choose the simple case of $$\Sigma=D$$, the disk, with polar coordinates $$r$$ and $$\varphi$$. A variation of $$S$$ under a large gauge transformation generated by $$g\in SU(N)$$, i.e., $$A\rightarrow g A g^{-1}-dg g ^{-1}$$ gives $$\begin{equation} S\rightarrow S+ \frac{k}{4\pi}\int d^3x\bigg(\epsilon^{ijk}\partial_j\textrm{Tr}(\partial_igg^{-1}A_k)+\frac{1}{3}\epsilon^{ijk}\textrm{Tr}(g^{-1}\partial_{i}gg^{-1}\partial_{j}gg^{-1}\partial_{k}g)\bigg). \end{equation}$$ Now, using the boundary condition $$A_0=0$$ and restricting $$g$$ such that $$\partial_0g=0$$ at the boundary, the first term vanishes via Stoke's theorem. The second term does not vanish in general, but I believe if we choose the gauge transformations to approach the same value at the boundary of $$D\times \mathbb{R}$$ (which makes it equivalent to the $$Y=S^3$$ case), then since $$\pi_3(SU(N))=\mathbb{Z}$$, the second term integrates to $$2\pi k n$$ for $$n\in \mathbb{Z}$$, which leads to a trivial phase in the path integral. (I've referred to page 182 of Tong's notes extensively here.)

However, I do not see why we must choose the gauge transformations to approach one at the boundary, but not some other fixed value. Furthermore, I do not see why general time independent transformations should be viewed as a global symmetry.

• If the only allowed transformations are some fixed time independent $g$ on the boundary, isn't that basically the definition of a global symmetry (on the boundary)? – octonion Mar 30 at 21:02
• If the time independent $g$ was independent of other coordinates, then yes, it would be global, but this is not necessarily the case. In the references, they do not require that the transformation be fixed to be considered global. For $\Sigma=D$ the boundary chiral WZW theory with group valued fields $U$ has a symmetry $U\rightarrow \widetilde{V}(\varphi)UV(t)$, where $\widetilde{V}(\varphi)$ is referred to as a global symmetry. – Mtheorist Mar 31 at 4:05
• Might ocnsider this paper. – Cosmas Zachos Apr 2 at 19:03