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I am studying CFT, where I encounter Stokes' theorem in complex coordinates: $$ \int_R (\partial_zv^z + \partial_{\bar{z}}v^{\bar{z}})dzd\bar{z} = i \int_{\partial R}(v^{z}d\bar{z} - v^{\bar{z}}dz). $$ I am trying to prove this by starting from the form of Stokes'/Greens theorem: $$ \int_R(\partial_xF^y - \partial_yF^x)dxdy = \int_{\partial R}(F^xdx + F^ydy $$ and transforming to complex coordinates. The reason I ask this here and not on Math exchange is that in CFT we have the distinction between indices up and indices down: $$ v^z = v^{\tau}+iv^{\sigma} \\ v^{\bar{z}} = v^{\tau} - iv^{\sigma} \\ v_z = v^{\tau}-iv^{\sigma} \\ v_{\bar{z}} = v^{\tau} + iv^{\sigma}$$, with $z = \tau +i\sigma$ and $\bar{z} = \tau - i\sigma$. The substitution is kind of straight forward, but I get: $$ \int_R(\partial_zv^{z} - \partial_{\bar{z}}v^{\bar{z}})dzd\bar{z} = i\int_{\partial R}v^zd\bar{z}+v^{\bar{z}}dz, $$ so I get two relative minus signs wrong. As the calculation is not so difficult, I have the feeling I am missing something crucial here. Does anyone have an idea?

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    $\begingroup$ Your expression of standard Stokes theorem is false : index are not coherent and not correct. The correct expression is : $\int_R(\partial_xF_y - \partial_yF_x)dxdy = \int_{\partial R}(F_xdx + F_ydy)$ $\endgroup$
    – Trimok
    Jun 29, 2013 at 11:22
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    $\begingroup$ I would say that the complex form of the theorem is simpler so you are carrying coal to Newscastle if you're converting it to the higher-dimensional real case. $\endgroup$ Jun 29, 2013 at 11:34
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    $\begingroup$ And, for your final formula, if you make the transformation $w^z = v^z$, $w^{\bar z} = -v^{\bar z}$, it seems that you get the correct formula. $\endgroup$
    – Trimok
    Jun 29, 2013 at 11:46
  • $\begingroup$ Hi @Erik, which references are you using? $\endgroup$
    – Qmechanic
    Jun 29, 2013 at 13:22
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    $\begingroup$ @Qmechanic David Tong's chapter 4, damtp.cam.ac.uk/user/tong/string.html and Polchinski's String Theory (Volume 1) $\endgroup$
    – Funzies
    Jun 29, 2013 at 14:28

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I think I can help. Pick up the expression you started with. Convert to coordinates $\tau$ and $\sigma$ as you did. Use the Green's theorem in these coordinates and then convert back to the complex coordinates again. I am not good with word by this transliterates in the following expression

$$2∫∫dσ dτ (∂_{σ}v^{σ} + ∂_{τ}v^{τ})$$ = (now use Green’s theorem and the next integral is a contour counterclockwise bounding the area defined by the previous double integral $$= 2∫ [v^{σ} dτ - v^{τ} dσ] .$$ Now convert back to $v^{z}$ and $v^{\bar z}$ and to the differentials ($dz$, $d\bar z$) and you get to the correct result.

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    $\begingroup$ Hello, and welcome to Physics Stack Exchange. This site supports mathjax for writing mathematical symbols. Please use it instead of unicode. $\endgroup$
    – DanielSank
    Nov 25, 2015 at 18:20

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