Diffeomorphism invariance of scattering amplitude in bosonic string theory

Question

It is mentioned in Polchinski's book (vol 1) that the diffeomorphism invariance of the scattering amplitude (see Polchinski, vol 1, eq 5.3.9) follows from the equation of motion of $b_{ab}$ (see Polchinski, vol 1, eq 5.4.4). It is also mentioned that the corresponding contact terms due to $c^a$ insertions precisely cancel the effect of the diffeomorphism on the fixed vertex operators. Can someone explain how I can show this?

More concretely, the variation in $b_{ab}$ insertion term is of the form (eq 5.4.4)

$$ \delta(b,\partial_k \hat g) = -2(b,P_1 \partial_k \xi) = -2(P_1^Tb, \partial_k \xi)=0, $$

where $\xi^a(\sigma;t)$ is an infinitesimal diffeomorphism, and the last equality follows from equation of motion of $b_{ab}$, i.e., $(P_1^T b)_a=0$. I expect the contact terms to be of the form

$$[\partial_k \xi^1(\hat \sigma;t)c^2(\hat \sigma) + c^1(\hat \sigma)\partial_k \xi^2(\hat \sigma;t)] \sqrt{\hat g(\hat\sigma;t)} V(\hat\sigma) $$

for each fixed coordinate $\hat \sigma$ [basically, $\partial_k \xi^a(\hat \sigma;t)$ replaces $c^a(\hat \sigma)$ for $a=1,2$]. There is a derivative with respect to moduli, $\partial_k\equiv\partial/\partial{t_k}$, here. I don't see how such a term can cancel the variation of $c^1(\hat \sigma)c^2(\hat \sigma)\sqrt{\hat g(\hat\sigma;t)} V(\hat\sigma)$ under the diffeomorphism. Can someone explain how the cancellation happens?

A related question: how can I show that, leaving everything else unchanged, the scattering amplitude (eq 5.3.9) is independent of the choice of the fixed coordinates $\hat\sigma$? I can see that this is true in specific cases like 3-tachyon amplitude, Veneziano amplitude, etc., but is there a way to show this directly from eq 5.3.9?

EDIT: Here are the relevant equations,

The scattering amplitude is (eq 5.3.9)

$$ S_{j_1\ldots j_n}=\underset{\text{topologies}}{\sum_\text{compact}} \int_F d^\mu t \int [DX Db Dc] \exp\left(-S_{X}-S_{g}-\lambda\chi\right) \prod_{k=1}^{\mu}\frac{1}{4\pi}\left(b,\partial_{k}\hat{g}\right)\\ \times \prod_{i=1}^{\kappa}\left(\prod_{a=1,2}c^{a}\left(\hat{\sigma}_{i}\right)\right)\sqrt{\hat{g}\left(\hat{\sigma}_{i};t\right)}V_{j_{i}}\left(\hat{\sigma}_{i}\right)\prod_{i=\kappa+1}^{n}\int d^{2}\sigma_{i}\sqrt{\hat{g}\left(\sigma_{i};t\right)}V_{j_{i}}\left(\sigma_{i}\right), $$

where

• $F$ is moduli space, and $\mu$ its dimension,
• $t_k$ are coordinates on moduli space, $k=1,\ldots,\mu$,
• $\hat g_{ab}(\sigma;t)$ is the fiduciary metric,
• $S_X$ is the Polyakov action for the string,
• $S_g=\frac{1}{2\pi}(b,P_1 c)$ is the ghost action,
• $(A,B)=\int d^2 \sigma\sqrt{\hat g}\ A^{ab}B_{ab}$ for any two rank-2 tensors $A$ and $B$ on the world sheet,
• $(P_1 c)^{ab} = \frac{1}{2}(\hat{\nabla}^a c^b + \hat{\nabla}^b c^a -\hat{g}^{ab} \hat{\nabla}\cdot c)$, and $P_1^T$ is defined using integration by parts as $(P_1^T b,c)=(b,P_1 c)$,
• $\chi$ is the Euler number of the world sheet,
• $\lambda$ is a constant [related to dilaton zero mode, $\Phi_0$],
• $\kappa$ is the number of conformal Killing vectors (CKVs),
• $\hat \sigma_i$, for $i=1,\ldots,\kappa$, are the fixed coordinates.

I am guessing that the sum over topologies and hence the term $\lambda\chi$ are not really relevant to my question but I included them anyway to stick to Polchinski's equation.

Answer 1

I finally understand how this works out and I strongly feel that the explanation given in Polchinski's book is not the correct one. As I mentioned in the question, the variation of any $b$-insertion vanishes by the equation of motion of $b_{ab}$ but there are contact terms to be taken care of.

However, one crucial thing to remember about the scattering amplitude is that it makes sense only when the number of $b$-insertions is $\mu$ and the number of $c$-insertions is $\kappa$. Else, the path integral over $b$ and $c$ zero-modes (there are $\mu$ and $\kappa$ of them respectively) vanishes because $b$ and $c$ are Grassmannian fields. This is discussed around equation (5.3.17) in Polchinski's vol 1.

When we use the equation of motion of $b_{ab}$ and get contact terms, there is one less $b$-insertion and one less $c$-insertion in each of the contact terms. For example, we know that

$$ \int [DbDc] \frac{\delta}{\delta c^a(\sigma)}\big[\exp(-S_g)c^b(\sigma')\big] = 0 $$ because the integrand is a total derivative. By Leibniz rule, we get

$$ \big\langle -\frac{\delta S_g}{\delta c^a(\sigma)} c^b(\sigma')\big\rangle + \langle \delta^b_a \delta^2(\sigma-\sigma') \rangle = 0 \\ \implies \frac{1}{2\pi}\langle (P_1^Tb)_a(\sigma) c^b(\sigma')\rangle=\delta^b_a \delta^2(\sigma-\sigma'). $$

Clearly, the numbers of $b$ and $c$-insertions reduce by one in the contact terms. In our case, the contact terms, therefore, have $(\mu-1)$ $b$-insertions and $(\kappa-1)$ $c$-insertions, which are not enough to satisfy the Grassmannian integrals over $b$ and $c$ zero-modes. Hence, these terms vanish identically.

We still have to confirm the diff-invariance of the fixed vertex operators but this is easy enough to see even in the case of a finite coordinate transformation.

Given the diff-Weyl invariance of the scattering amplitude, we can now answer my question about the independence of scattering amplitude on the fixed positions. Recall that conformal transformations are those diff-Weyl transformations which leave the fiducial metric $\hat{g}_{ab}$ invariant. Using such transformations, we can move the fixed positions $\hat{\sigma}_i$ anywhere we want leaving everything else the same, including $\hat g_{ab}$, and the scattering amplitude remains invariant.