# Composite Operators in Topological/Conformal Field Theory

In virtually any reference on conformal or cohomological field theory, you'll eventually see a formula like: $$\frac{\delta}{\delta h^{\alpha \beta}}\langle\mathcal{O}_1(x_1)...\mathcal{O}_n(x_n)\rangle = \frac{1}{4\pi}\int d^2x \sqrt{h} \langle T_{\alpha \beta}(x)\mathcal{O}_1(x_1)...\mathcal{O}_n(x_n)\rangle$$

Here, $h_{\alpha \beta}$ is the worldsheet metric, $T_{\alpha \beta}$ is the energy-momentum tensor, and $\mathcal{O}_i$ are some local physical operators. Naively, we can derive this by just expressing the correlation function as a path integral over fields and taking the variational derivative directly. If you've ever worked with any theories explicitly, to get a good $T_{\alpha \beta}$ (e.g. in the CFT case, one which has the right commutators with itself and primary fields) you have to specify some choice of ordering to avoid composite operator singularities. How does the path integral argument take this into account? I am tempted to just hand-wave it away and say that the correct definition of the path integral measure does contain dependence on $h_{\alpha \beta}$ and this ensures that $T_{\alpha \beta}$ comes out correctly, but this seems far too sketchy. In the CFT case, this isn't so important since the path integral takes a backseat, but in TQFTs these sort of formal arguments are instrumental in proving that topological invariants are actually invariant, so I am curious as to how to connect the formal generalities with the more hands-on considerations like ordering.

More generally, when we deal with other important objects like supercharges which may be composite operators, how does the path integral picture take into account the proper ordering? It seems easy enough to just take the correctly ordered operators as our definition, but how are we sure that the path integral agrees with this definition?

For example, in the case of a 2d free field theory (such as superstrings in flat backgrounds), you can undo the normal ordering of $T_{ab}$ (or, conversely, you can implement it) by adding a counterterm to the 2d cosmological constant. As I'm addressing a possibly wider audience I'll mention also that although the latter breaks conformal invariance classically, it is required in any case to preserve conformal invariance at the quantum level. To be precise, this implies that a 2d cosmological constant is generated by the path integral measure (by Polchinski's principle of ultralocality), so for this reason your hand-waving argument is also correct.