A question about the Weyl transformation for the vertex operator of the closed-string tachyon

Question

I met a problem of derving the Weyl transformation on the closed-string tachyon, Eq. (3.6.8) in Polchinski's string theory, vol 1, p 103.

Given the vertax operator of the closed-string tachyon

$$V_0 = 2g_c \int d^2 \sigma g^{1/2} e^{ik \cdot X} \tag{3.6.1}$$

It is said the Weyl transformation of Eq. (3.6.1) is

$$\delta_W V_0 = 2g_c \int d^2 \sigma g^{1/2} \left( 2 \delta \omega( \sigma) - \frac{k^2}{2} \delta_W \Delta (\sigma, \sigma) \right) [ e^{ik \cdot X(\sigma)} ]_r \tag{3.6.8}$$

here renormalized operator $[]_r$ and $\Delta (\sigma, \sigma)$ are defined as

$$ [ \mathcal{F}]_r = \exp \left( \frac{1}{2} \int d^2 \sigma d^2 \sigma' \Delta(\sigma, \sigma') \frac{ \delta}{\delta X^{\mu}(\sigma)} \frac{ \delta}{\delta X_{\mu}(\sigma')} \right) \mathcal{F} \tag{3.6.5} $$ $$ \Delta(\sigma,\sigma') = \frac{ \alpha'}{2} \ln d^2(\sigma,\sigma') \tag{3.6.6} $$ where $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$.

I can get the first term in the big parathesis $()$ of Eq. (3.6.8), namely consider $$ \delta g^{1/2} = \frac{1}{2} \frac{1}{g^{1/2}} \delta g= \frac{1}{2} \frac{1}{g^{1/2}} g g^{ab} \delta g_{ab} = \frac{1}{2} \frac{1}{g^{1/2}} g g^{ab} 2 \delta \omega g_{ab} = 2 g^{1/2} \delta \omega $$ I try to get the second term in the big parathesis of Eq. (3.6.8) $$ \delta e^{i k \cdot X} = e^{i k \cdot X} i k^{ab} X^{ab} \delta g_{ab} = e^{i k \cdot X} i k^{ab} X^{ab} 2 \delta \omega g_{ab} = 2 i k \cdot X e^{ik \cdot X} \delta \omega,$$

but where do the $\Delta(\sigma,\sigma)$ and renormalized opeator $ [ e^{ik \cdot X(\sigma)} ]_r $ in (3.6.8) come from?

Answer 1

I think it works like this : Looking at the Weyl dependence of renormalized operators:

$$\delta_W[ \mathcal{F}]_r = [ \delta_W \mathcal{F}]_r + \frac{1}{2} \int d^2 \sigma ~~d^2 \sigma' \delta_W \Delta(\sigma, \sigma') \frac{ \delta}{\delta X^{\mu}(\sigma)} \frac{ \delta}{\delta X_{\mu}(\sigma')} [ \mathcal{F}]_r \tag{3.6.7}$$

We apply this formula to the operator $ \mathcal{F} = e^{i k.X(\sigma'')}$.

There is no explicit Weyl dependence of $e^{i k.X(\sigma'')}$, so the first term $ [ \delta_W e^{i k.X(\sigma'')}]_r$ is zero.

The second term is, using the fact that we keep only the $\sigma = \sigma' =\sigma''$ terms because $e^{i k.X(\sigma'')}$ depends only of $\sigma''$:

$$\frac{1}{2} \delta_W \Delta(\sigma'', \sigma'') (-ik^\mu)(-ik_\mu) [ e^{i k.X(\sigma'')}]_r$$

So, finally :

$$\delta_W[ e^{i k.X(\sigma'')}]_r = -\frac{ k^2}{2} \delta_W \Delta(\sigma'', \sigma'') [ e^{i k.X(\sigma'')}]_r \tag {1}$$

Now, we have :

$$V_0 = 2g_c \int d^2 \sigma g^{1/2} [e^{ik \cdot X(\sigma)}]_r \tag{2}$$

You already get the the Weyl dependence of $V_0$ relatively to $g^{\frac{1}{2}}$.

The Weyl dependence of $V_0$ relatively to $[e^{ik \cdot X(\sigma)}]_r$ is :

$$\delta_W V_0 = 2g_c \int d^2 \sigma g^{1/2} \delta_W[e^{ik \cdot X(\sigma)}]_r\tag{3}$$

So, finally, this dependence is, using $(1)$ :

$$\delta_W V_0 = 2g_c \int d^2 \sigma g^{1/2} (-\frac{ k^2}{2} \delta_W \Delta(\sigma, \sigma) [ e^{i k.X(\sigma)}]_r)\tag{4}$$

Remark : The formula $(3.6.7)$ comes directly from the formula $(3.6.5)$ and could be interpreted like this: you have 2 terms, the first corresponds to an explicit Weyl dependence of the operator, and the second term to a Weyl dependence via $\Delta(\sigma, \sigma')$