Corners in the worldsheet and Ricci scalar term in the Polyakov action

Question

I have a question about a passage in Polchinski's textbook [1], regarding the topological term in the Polyakov action.

In the Polyakov action for a closed manifold, we can add a term proportional to the Euler characteristic, which we can write as a local functional using the Gauss-Bonnet theorem: $$\Delta S_{\text{closed}} = \lambda\chi = \frac{\lambda}{4\pi}\int_\Sigma |g|^{1/2}R\text d^2 \sigma.$$

This is invariant under diffeomorphisms and Weyl transformations. For a manifold with boundary, we have to include a boundary term in order to preserve Weyl invariance. This is also the formula for the Euler characteristic given by the Gauss-Bonnet theorem: $$\Delta S_{\text{open}} = \lambda \chi = \frac{\lambda}{4\pi}\int_\Sigma |g|^{1/2}R\text d^2 \sigma + \frac{\lambda}{2\pi}\oint_{\partial M}k\text ds.$$

Now, if there are corners on the boundary, with inner angles $\alpha_i$ (Polchinski considers only $\alpha_i = \pi/2$), the Gauss-Bonnet theorem becomes: $$\chi = \underset{\tilde \chi}{\underbrace{\frac{1}{4\pi}\int_\Sigma |g|^{1/2}R\text d^2 \sigma + \frac{1}{2\pi}\oint_{\partial M}k\text ds }}+ \frac{1}{2\pi}\sum_i (\pi -\alpha_i).$$

Now, it is not clear (to me) what should be the term in the action : the topological term $\lambda\chi$ or only the contribution from the integrals $\lambda\tilde\chi$, disregarding the contributions from the corners, as both are Weyl-invariant.

In Polchinski's textbook [1,p.84], the author claims that the latter is correct. He says:

This follows from unitarity: the $\lambda$-dependence must be unchanged if we cut through a world-sheet, leaving fewer internal lines and more external sources. This will be the case for the weight $\exp(-\lambda \tilde \chi)$ because the surface integrals cancels on the cut edges.

Why does unitarity imply that the "$\lambda$-dependence" be unchanged under this cutting procedure?

[1] String Theory: Volume 1, An introduction to the bosonic string. J. Polchinski; Cambridge Monographs on Mathematical Physics.

Answer 1

You can (in this context) interpret unitarity as meaning factorisation, namely the ability to cut and glue path integrals across different cycles of your Riemann surface without changing the physics. In particular, by cutting a (say, non-separating) internal line and inserting a complete set of states in an appropriate gauge slice for the integral over moduli space, you can see that you could have instead computed a higher point amplitude at one genus lower and obtained the same thing. So then the Euler characteristics must also match up in the two viewpoints.