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I'm reading Polyakov's book, Gauge fields and strings. There is this formula (9.247) which I do not really understand how to get. The formula states that in two dimensions, taking $z$ as my holomorphic and $\bar{z}$ as my antiholomorphic variable, the following relation holds

$$\partial_{\bar{z}}\frac{1}{z-w}=-\pi \delta(z-w)$$

How does one find this?

bnado
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2 Answers2

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Multiplying the LHS for an holomorphic test function $f(z)$, which then satisfies $\partial_\bar{z}f(z) = 0$, and integrating over the unit disc $D$, we get $$\begin{align} \iint_D f(z)\color{blue}{\partial_\bar{z}\left(\frac{1}{z-w}\right)}\frac{\mathrm dz\wedge \mathrm d\bar z}{-2i}&=\iint_D \partial_\bar{z}\left(\frac{f(z)}{z-w}\right)\frac{\mathrm dz\wedge \mathrm d\bar z}{-2i}=\int_D \frac{f(z)}{z-w}\frac{\mathrm dz}{-2i}=-\pi f(w)\\ &=\iint_D \color{blue}{-\pi\delta(z-w)}f(z)\frac{\mathrm dz\wedge \mathrm d\bar z}{-2i} \end{align} $$where in the second and third equalities we used respectively Stokes' theorem and Cauchy's equality.

Thus $$ \partial_\bar{z}\left(\frac{1}{z-w}\right)=-\pi\delta(z-w) $$

alexjo
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  • How is stokes' theorem applied exactly? An extra step would be nice I think. Stokes would work with a full derivative not a partial one. The resulting integral afterwards should be over $\partial D$ thus it would be clearer to write $d\theta$ (integrating from 0 to $2\pi$) instead of dz as the meaning of the current z integral isn't clear. Finally, shouldn't the delta function be 2 dimensional, since it kills of a two dimensional integral? – Kvothe Nov 23 '17 at 11:29
  • @alexjo Thank you for posting this. The order of your equalities is confusing, however, well, at least to me. – AmorFati Apr 22 '19 at 23:43
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I am confused by the above answer, and also do not see where Stokes' theorem is being applied. The following is hopefully more clear.

Work on a sufficiently small disk $\Delta$ centred at $w$, and let $f$ be a holomorphic function on $\Delta$. The distributional derivative of $1/(z-w)$ with respect to $\overline{z}$ is the distribution $g$ such that \begin{eqnarray*} \int_{\Delta} f(z) \frac{\partial}{\partial \overline{z}} \left( \frac{1}{z-w} \right) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}} &=& \int_{\Delta} g(z) \frac{1}{z-w} \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}}. \end{eqnarray*}

Note that since $f$ is holomorphic, $\dfrac{\partial f}{\partial \overline{z}} \equiv 0$, and therefore \begin{eqnarray*} \frac{\partial}{\partial \overline{z}} \left( \frac{f(z)}{z-w} \right) &=& \frac{1}{(z-w)^2} \left( (z-w) \frac{\partial f}{\partial \overline{z}} - f(z) \frac{\partial}{\partial \overline{z}} (z-w) \right) \\ &=& - \frac{f(z)}{(z-w)^2} \frac{\partial}{\partial\overline{z}} (z-w) \\ &=& f(z) \frac{\partial}{\partial \overline{z}} \left( \frac{1}{z-w} \right). \end{eqnarray*}

Therefore, \begin{eqnarray*} \int_{\Delta} f(z) \frac{\partial}{\partial \overline{z}} \left( \frac{1}{z-w} \right) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}} &=& \int_{\Delta} \frac{\partial}{\partial\overline{z}} \left( \frac{f(z)}{z-w} \right) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}} \\ &=& \frac{1}{-2\sqrt{-1}} \int_{\Delta} \frac{f(z)}{z-w} dz \\ &=& -\pi f(w), \end{eqnarray*}

by the Cauchy integral formula. Observe, however, that $f(w)$ can be expressed as the distribution $\delta(z-w)$ acting on $f$. Hence, \begin{eqnarray*} -\pi f(w) &=& \int_{\Delta} -\pi \delta(z-w) f(z) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}}. \end{eqnarray*}

From this, we see that $g = -\pi \delta(z-w)$, as required.

Note also, that $dz \wedge d\overline{z} = -2\sqrt{-1} dx \wedge dy$, hence the reason for the $-2\sqrt{-1}$ term hanging around.

AmorFati
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