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)

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


  • $F$ 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.


1 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.


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