5
$\begingroup$

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!

$\endgroup$
6
$\begingroup$

I may be missing some subtlety in your question, but: Yes, Chern-Simons theory is a conformal QFT. It does not depend on the metric, so it is invariant under changes of the metric, which is what conformal transformations are. Conformal transformations are symmetries, because they act trivially. This is true of any topological field theory. It's not especially helpful, however, because it's a degenerate case of conformality. Knowing a QFT is conformal is useful because you can use the representation theory of the conformal group to decompose the fields and states into understandable pieces. If the conformal group acts trivially, you don't gain anything.

$\endgroup$
  • $\begingroup$ Thank you for the useful notes, and sorry for the delayed reply. So all TQFTs are also conformally invariant -- in particular for the case of the Chern-Simons theory, all interesting quantities (such as $Z(Y)$) transform trivially under the conformal group and hence we get nothing new. On the other hand, introducing Wilson loops (or Wilson lines), we obtain static charges on a cross section of $Y$. Consequently, interesting quantities no longer transform trivially under the conformal group, and it is now useful to talk about the corresponding 2D CFT. I hope I'm getting this right. $\endgroup$ – KSP Jan 7 '18 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.