I would like to know if what I'm doing is right.

Firstly, $$\frac{1}{2\pi\alpha'}\nabla^{2}\left[\frac{\alpha'}{2}\ln d^{2}(\sigma,\sigma') \right]=g^{-1/2}(\sigma)\delta^{2}(\sigma-\sigma')$$ Assuming this equality. Secondly, $$-\frac{1}{2\pi\alpha'}\nabla^{2}G'(\sigma,\sigma')=g^{-1/2}\delta^{2}(\sigma-\sigma')-X_{0}^{2}$$ Solving the equation, $$G'(\sigma,\sigma')=-\frac{\alpha'}{2}\ln d^{2}(\sigma,\sigma')+\frac{\alpha'X^{2}_{0}}{2}\int d^{2}\sigma'g^{1/2}(\sigma')\ln d^{2}(\sigma,\sigma')$$ Is this solution right? How can I get from this solution the next result in the case in which we are working over a sphere? $$G'(\sigma_{1},\sigma_{2})=-\frac{\alpha'}{2}\ln|z_{1}-z_{2}|^{2}+f(z_{1},\bar{z}_{1})+f(z_{2},\bar{z}_{2})$$

where: $$f(z,\bar{z})=\frac{\alpha'X_{0}^{2}}{4}\int d^{2}z'\exp(2\omega(z',\bar{z}'))\ln|z-z'|^{2}+k$$

I'm pretty sure the first solution has some of truth, but I'm confused how can I express $\sigma$ coordinates to z,using the fact $z=\sigma_{1}+i\sigma_{2}$


1 Answer 1


Here solution is written in conformal gauge $g_{ab}=e^{2 \omega} \eta_{ab}$. After gauge fixing then use the definition of $z= \sigma^1+i \sigma^2$ and conjugate for $\bar{z}$. So space is locally flat and coordinates are complex defined earlier.

  1. $ d^{2}(\sigma-\sigma')\rightarrow |z_1-z_2|^2$
  2. $g^{1/2}\rightarrow e^{2 \omega}$
  3. $d^2z\rightarrow 2 d\sigma^1 d\sigma^2$

After substituting this it is easy to check the given solution.


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