Let's consider a 3D manifold $Y$ for brevity, and a topological invariant $Z(Y)$ which does not depend on the geometry of $Y$.
A Schwarz-type topological quantum field theory (TQFT) has the defining feature that the action $S$ is independent of the geometry of $Y$. More explicitly, we may write \begin{equation} {\delta S\over \delta g_{\mu\nu}}=0, \end{equation} where $g_{\mu\nu}$ is a metric defined on $Y$. However, it is also well known that the expression on the left hand side is nothing more than the stress-energy tensor $T^{\mu\nu}={\delta S\over \delta g_{\mu\nu}}$. So equivalently, we may rewrite this as \begin{equation} T^{\mu\nu}=0. \end{equation}
This tells us a couple of things:
- We have a vanishing Hamiltonian $H=T^{00}=0$ -- there are no dynamics involved in such a TQFT. This fact also holds true for a Witten-type TQFT.
- Having vanishing diagonal terms also mean that we have $\text{Tr}\,T=T^\mu_{\phantom{\mu}\mu}=0$.
The first point is easy to understand. Intuitively a topological invariant should be unaffected when we stretch $Y$ without tearing or gluing. So if we identify one of these dimensions as being time-like (so that we really have a 2+1D manifold, rather than a 3D one), then translations in time should leave $Z(Y)$ unchanged.
The second point needs clarifying. The condition $\text{Tr}\,T=0$ also implies conformal invariance. Does this necessarily mean that the TQFT on $Y$ is also conformally invariant?
In Witten's work on the Jones polynomial for example, a 2D conformal field theory (CFT) is indeed defined on a cross-section of $Y$, upon choosing the Chern-Simons action. However, there is no explicit mention of a 3D CFT being defined on $Y$. Is the Chern-Simons theory also conformally invariant, and if so, does it hold true for higher dimensional Schwarz-type TQFTs?
I really appreciate any help that can point me in the right direction. Thanks in advance!