# Polchinski book, doubt related to $X_{0}$

I'm reading polchinski's book and I was asking if I've made a wrong calculation or I didn't understand sth. Well, basically. $$Z[p]=\int [DX^{\mu}]e^{-S[X]}e^{ipX}$$ $$=(2\pi)^{d}\delta^{d}(p_{0})\det{}^{\prime}\left(\frac{-\nabla^{2}}{4\pi^{2}\alpha'}\right)^{-d/2}\exp(-\frac{1}{2}\int d^{2}\sigma_{1} d^{2}\sigma_{2}p_{\mu}(\sigma_{1})p^{\mu}(\sigma_{2})G'(\sigma_{1},\sigma_{2})).\tag{6.2.6}$$ Here, $$(2\pi)^{d}\delta^{d}(p_{0})=\prod_{i=1}^{d}\int dx^{i}_{0}\exp\left(ix^{i}_{0}\int d^{2}\sigma X_{0}(\sigma)p_{i}(\sigma)\right).$$ These equalities, You can find them on the chapter 6, page 170. So, I was thinking about the meaning of $X_{0}$, which it's for me the kernel of the laplacian operator. Indeed, I tried to make this.

I'm starting from this equation:($-\frac{\delta}{\delta X_{\mu}(\sigma)}\langle X^{\nu}(\sigma')\rangle=0$) $$\frac{1}{2\pi\alpha'}\nabla^{2}\langle X^{\mu}(\sigma)X^{\nu}(\sigma')\rangle=-\eta^{\mu\nu}g^{-1/2}\delta^{2}(\sigma-\sigma')\langle 1\rangle$$ $$-\frac{1}{2\pi\alpha'}\nabla^{2}\frac{\delta}{\delta p_{\mu}(\sigma)}\frac{\delta}{\delta p_{\nu}(\sigma')}Z[p]|_{p=0}=-\eta^{\mu\nu}g^{-1/2}\delta^{2}(\sigma-\sigma')Z[p]|_{p=0}$$ $$-\frac{1}{2\pi\alpha'}\nabla^{2}\frac{\delta}{\delta p_{\mu}(\sigma)}\frac{\delta}{\delta p_{\nu}(\sigma')}Z[p]|_{p=0}=\frac {1}{2\pi\alpha'}\nabla^{2}[...]$$ This is the whole expression of $[...]$ $$[...]=\det\left(\frac{-\nabla^{2}}{4\pi^{2}\alpha'}\right)\left[(2\pi)^{d}\delta^{d}(p_{0})\eta^{\mu\nu}G'(\sigma,\sigma')\exp\left(\frac{-1}{2}ppG'\right)\right.$$ $$-\frac{\delta}{\delta p_{\mu}(\sigma)}\left((2\pi)^{d}\delta^{d}(p_{0})\right)\left\lbrace-\int d^{2}\sigma_{1}p^{\nu}(\sigma_{1})G'(\sigma_{1},\sigma')\right\rbrace\exp\left(\frac{-1}{2}ppG'\right)+(\mu\leftrightarrow\nu,\sigma\leftrightarrow\sigma')$$ $$-(2\pi)^{d}\delta^{d}(p_{0})\left\lbrace\int d^{2}\sigma_{1}d^{2}\sigma_{2}p^{\nu}(\sigma_{1})p^{\mu}(\sigma_{2})G'(\sigma_{1},\sigma')G'(\sigma_{2},\sigma)\right\rbrace\exp\left(\frac{-1}{2}ppG'\right)$$ $$-\left.\frac{\delta}{\delta p_{\mu}(\sigma)}\frac{\delta}{\delta p_{\nu}(\sigma')}\left((2\pi)^{d}\delta^{d}(p_{0})\right)\exp \left(\frac{-1}{2}ppG'\right)\right]\vert_{p=0}$$

So,I need the next result: $$[..2..]=-\frac{\delta}{\delta p_{\mu}(\sigma)}\frac{\delta}{\delta p_{\nu}(\sigma')}\left[(2\pi)^{d}\delta^{d}(p_{0})\right]_{p=0}=-\frac{\delta}{\delta p_{\mu}(\sigma)}\frac{\delta}{\delta p_{\nu}(\sigma')}\left[\prod_{i=1}^{d}\int dx^{i}_{0}\exp\left(ix^{i}_{0}\int d^{2}\sigma p_{i}(\sigma)X_{0}(\sigma)\right)\right]_{p=0}$$

$$[..2..]=\left[\prod_{i=1,i\neq\nu\neq\mu}^{d}(2\pi)\delta(p_{i0})\int dx^{\nu}_{0}x^{\nu}_{0}X_{0}(\sigma')\exp\left(ix^{\nu}_{0}\int d^{2}\sigma_{1} p_{\nu}(\sigma_{1})X_{0}(\sigma_{1})\right)\times\right.$$ $$\int dx^{\mu}_{0}x^{\mu}_{0}X_{0}(\sigma)\exp\left(ix^{\mu}_{0}\int d^{2}\sigma_{1} p_{\mu}(\sigma_{1})X_{0}(\sigma_{1})\right)$$ $$+\left.\prod_{i=1,i\neq\nu=\mu}^{d}(2\pi)\delta(p_{i0})\int dx^{\mu}_{0}(x^{\mu}_{0})^{2}X_{0}(\sigma)X_{0}(\sigma')\exp\left(ix^{\mu}_{0}\int d^{2}\sigma_{1} p_{\mu}(\sigma_{1})X_{0}(\sigma_{1})\right)\right]$$ That is, $\frac{1}{2\pi\alpha'}\nabla^{2}[..2..]=0$ ($\nabla^{2}X_{0}=0$) , Am I right?.

Figuring every operation out: $$\nabla^{2}[...]=\det\left(\frac{-\nabla^{2}}{4\pi^{2}\alpha'}\right)\exp \left(\frac{-1}{2}ppG'\right)\left[(2\pi)^{d}\delta^{d}(p_{0})\eta^{\mu\nu}\nabla^{2}G'(\sigma,\sigma')\right.$$ $$-\frac{\delta}{\delta p_{\nu}(\sigma')}((2\pi)^{d}\delta^{d}(p_{0}))\left\lbrace -\int d^{2}\sigma_{1}p^{\mu}(\sigma_{1})\nabla^{2}G'(\sigma_{1},\sigma \right\rbrace$$ $$\left.-(2\pi)^{d}\delta^{d}(p_{0})\left\lbrace\int d^{2}\sigma_{1}d^{2}\sigma_{2}p^{\nu}(\sigma_{1})p^{\mu}(\sigma_{2})G'(\sigma_{1},\sigma')\nabla^{2}G'(\sigma_{2},\sigma)\right\rbrace\right]\vert_{p=0}$$

At the end I got the next result: $$-\frac{1}{2\pi\alpha'}\nabla^{2}G'(\sigma,\sigma')=g^{-1/2}\delta^{2}(\sigma-\sigma')$$ So, I'm assuming that I made a mistake but I don't have any idea where. Right solution: $$-\frac{1}{2\pi\alpha'}\nabla^{2}G'(\sigma,\sigma')=g^{-1/2}\delta^{2}(\sigma-\sigma')-X_{0}^{2}$$

NOTE 1: I'm trying to get that result is because of I would like to find any relation between $X_{0}$ and $\frac{\alpha'}{2}\ln d^{2}(\sigma,\sigma')$. As I understand this last one amount is the quantity which allows me to renormalize operators and work with OPEs. The first time I saw this amount was when I was trying to figure out $\langle X^{\mu}(\sigma)X^{\nu}(\sigma')\rangle$ and Polchinski defined the normal ordering.

The problem is that you are making a wrong assumption. The equation

$$\frac{1}{2\pi\alpha'}\nabla^{2}\langle X^{\mu}(\sigma)X^{\nu}(\sigma')\rangle _{S^2}=-\eta^{\mu\nu}g^{-1/2}\delta^{2}(\sigma-\sigma')\langle 1\rangle_{S_2}$$

is not right. This equation is only right if there is no zero mode $X_0$. In fact this equation does not have solutions in compact surfaces.

To understand this, just recall the first time this equation shows up in Polchinski's book, in $(2.1.15)$ and $(2.1.18)$. The main assumption that sustain this formulas is that

$$\int \left[dX\right]\frac{\delta}{\delta X_{\mu}(\sigma)}(...\exp(-S)))=0$$

This is only true if $\exp(-S)$ works as a convergent factor for all the $X(\sigma)$'s. Other way to say it is that $\exp(-S)(...)$ should vanish for $X(\sigma)\rightarrow \pm\infty$. In the presence of zero modes this is not true.

Let us assume that $S$ is Gaussian, after all, this is our case. The local operator $X(\sigma)$ can be decomposed into

$$X^{\mu}(\sigma)=\sum_{I}\textsf{X}(\sigma)\,x_{I}^{\mu}$$

and the measure can be writen as

$$\left[dX\right]=\prod_{I,\mu}dx_{I}^{\mu}$$

Now, the action $S(\{x_{I}^{\mu}\})$ will be a function of all the $x_I^{\mu}$, except the $x_0^{\mu}$. This is striking. The zero mode does not shows up in the action, so the $\exp(-S)$ does not act as a convergent factor for $x_0^{\mu}$, and in fact the path integral diverges even after renormalization. The only way to make it converges is by "putting $X$ in a box", i.e. by putting an upper and lower bound for $X$, or compactfying it.

This divergence is crucial in string theory. It is what make up the deltas of momentum conservation $\delta^d(k_1+...+k_n)$. Is the integral in $dx_0^{\mu}$ that will be producing it. And is precisely this integral that prohibit you to do

$$\int \left[dX\right]\frac{\delta}{\delta X_{\mu}(\sigma)}(...\exp(-S)))=0$$

What you can do is decouple $x_0^{\mu}$ from the $X^{\nu}$ and $\delta/\delta X_{\mu}$ in equation $(2.1.18)$ getting an extra term $\textsf{X}_0^2\eta^{\mu\nu}$ in your equation:

$$\frac{1}{2\pi\alpha'}\nabla^{2}\langle X^{\mu}(\sigma)X^{\nu}(\sigma')\rangle _{S^2}=-\eta^{\mu\nu}g^{-1/2}\delta^{2}(\sigma-\sigma')\langle 1\rangle_{S_2}+\textsf{X}_0^2\eta^{\mu\nu}\langle 1\rangle_{S_2}$$

• Ok, I've updated my question, Can you tell me where I made a mistake?
– 7919
Commented Jan 14, 2018 at 6:17
• I did the calculation , $[..2..]$ , but when you act on the result with $\nabla^{2}$, you got a 0.
– 7919
Commented Jan 14, 2018 at 18:34
• @7919, sorry for the delay, took me long time to realize what is going on here. Commented Feb 21, 2018 at 1:15
• @7919 I'v update the answer Commented Feb 21, 2018 at 2:39
• @7919 there is an excellent answer here in stackexchange about the existence of Green's functions on compact manifold: https://physics.stackexchange.com/questions/379769/electrostatic-field-on-compact-surfaces Commented Feb 21, 2018 at 16:24