# Is it actually true that Chern-Simons theory is topological?

Chern-Simons theory has action $$\tag{1} S = \frac{k}{4\pi}\int_X tr(A\wedge dA + \frac{2}{3}A\wedge A\wedge A).$$

Here, $$X$$ is some compact 3-manifold, perhaps with boundary, and $$A$$ is a connection 1-form on a principal $$G$$-bundle for some compact Lie group $$G$$.

At first glance, the theory seems to be topological, for the simple reason that no background metric appears in the action. But I have two doubts about this (which may well be related, hence my including them both in the same question):

1. It is standard (e.g. see p3 of Elitzur et al) to split $$A$$ into "spatial" components $$A_i$$ and a "timelike" component $$A_0$$. Then the equation of motion for $$A_0$$ leads to the constraint $$F_{ij}=0$$ -- that is, the spatial components of the curvature vanish. But the $$F_{0i}$$ components are unconstrained. So there's an asymmetry between the "space" and "time" directions, which should be identical as far as topology is concerned.

2. Chern-Simons on a manifold-with-boundary is supposed to be dual to a chiral WZW theory on the boundary. The WZW theory is a CFT: it is defined on some background conformal structure. But if Chern-Simons is truly topological, where does the necessary conformal structure on the boundary come from?

So, is Chern-Simons really a topological theory? And if not, where does the extra geometric structure come from?

• It is topological on a manifold with no boundary. On a manifold with boundary, there are edge modes which have non-trivial dynamics described by WZW. Commented Apr 12, 2023 at 18:27
• To get a CFT you need to add by hand an additional structure on the boundary: i.e. something that in Lorentz signature determines the speed of light. Commented Apr 12, 2023 at 21:32
• @Prahar The action (1) is well-defined and manifestly topological regardless of whether there's a boundary. This is just because there's no metric anywhere to be seen. So my question still stands: where does the conformal structure on the boundary come from? Commented Apr 12, 2023 at 22:17

1. This is not necessary. It is just convenient. It is an easy way to quantise a theory when you have some way of choosing a "time direction". E.g. when $$X=\mathbb{S}^1\times\Sigma$$. Choosing $$\mathbb{S}^1$$ as (Euclidean) time is just a natural choice. On the other hand, when $$X$$ is e.g. an $$\mathbb{S}^3$$, this splitting is not amenable. In both cases you can explicitly perform the path integral, when $$X$$ is closed (see below for the case with a boundary), without having to appeal to Elitzur et al, and you get simply the Reshetikhin-Turaev invariant (normalising $$Z(\mathbb{S}^3)$$ to 1).
2. When the manifold has a boundary you can (and have to if you want to get a conformal theory on the boundary) provide boundary dynamics. The dynamics that are usually chosen are of the form (for simplicity I'm laying out the abelian case, but the non-abelian case has the same explanation) $$S_\text{bdy} \sim \int_{\partial X} a\wedge\star_{\scriptscriptstyle \partial X}\; a,$$ where $$a$$ is the restriction on $$\partial X$$ of $$A$$, or some linear combination of its components, that is compatible with the variational principle. It is this $$\star_{\scriptscriptstyle \partial X}$$ that generates some dependence on the boundary metric, and hence allows for a conformal structure.