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

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?

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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 $\Delta(\sigma, \sigma')$

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Thank you once more for your help! I have two questions (1) $V_0$ in the first line of (3.6.1) has no renormalization symbol $[]_r$, your Eq. (1) starts from $\delta_W [e^{i k \cdot X}]_r$, where does the $[]_r$ arise? (2) $e^{i k \cdot X}$ has no explicit Weyl dependence, is that because $k \cdot X = k^{\mu} X^{\nu} \eta_{\mu\nu}$, the metric is for the embeded space, not the 1+1 dimension one $\gamma_{ab} (1.2.21)$? –  user26143 Aug 5 '13 at 14:56
(1) The renormalization operation is a generalization of the normal ordering (see text p. 102 just before (3.6.5) The equation (3.6.1) is written with normal ordering$:e^{ik.X(\sigma)}:$, and we generalize it with renormalized operators $:[e^{ik.X(\sigma)}]_r$. Equation (1) is just equation $(3.6.7)$ with $\mathcal{F} = e^{i k.X(\sigma)}$, so you have a renormalized indice $_r$. –  Trimok Aug 5 '13 at 15:50
(2) You are perfectly right. There is no dependence on the worldsheet metrics for $e^{ik.X(\sigma)}$ –  Trimok Aug 5 '13 at 15:51
Thank you very much! –  user26143 Aug 5 '13 at 15:53