I have a function $f$ defined as $\frac{\overline{z}-\overline{w}}{(z-w)(1 + z\overline{z})(1+w\overline{w})}$.

I want to prove that $$\partial_{\overline{z}}^2 f + \frac{2z}{1+z\overline{z}}\partial_{\overline{z}}f = \pi \frac{2}{(1+ z\overline{z})^2}\delta(z-w).$$

This is basically a double covariant derivative wrt to $\overline{z}$ taking the metric in unit sphere in stereographic coordinates.

I found that $\partial_{\overline{z}}f = \delta(z-w)\frac{(1+z\overline{z})(\overline{z}-\overline{w})}{1+w\overline{w}} + \frac{1}{z-w}\frac{(1+2z\overline{z}-z\overline{w})}{1+w\overline{w}}$.

If I differentiate again I am encountering a derievative of a delta function which I do not know how to do.

But since I know the result I calculated back what the derivative of delta function should be a found out to be $\frac{2\delta(z-w)}{\overline{z}-\overline{w}}\frac{w\overline{w}-z\overline{z}}{1+z\overline{z}}$.

Basically during the calculation I used the product rule and took $\frac{1}{z-w}$ as one function and rest as another and did the diffrentiation.I also used the fact that $\partial_{\overline{z}}\frac{1}{z-w}=\delta(z-w)$.

I would like to know whether the procedure I have done is correct.Is the derivative of the delta function I have mentioned is correct or not.Pls help me.

This problem is basically involved during computing the BMS charge

  • 1
    $\begingroup$ -1. This is a purely mathematical question which belongs on Math SE. The context (computing BMS charge) has no impact on the question. $\endgroup$ Apr 30, 2017 at 16:39
  • $\begingroup$ Notice that the $\delta$-function is being multiplied by ${\bar z} - {\bar w}$ so it is zero. $\endgroup$
    – Prahar
    Apr 30, 2017 at 18:47

1 Answer 1



$$ D_{\bar z}^2f = \gamma_{z\bar z} \partial_{\bar z}( \gamma^{z\bar z} \partial_{\bar z} f ) $$ We are solving the equation \begin{align} D_{\bar z}^2f &= \pi \gamma_{z\bar z} \delta^2(z-w)\\ \gamma_{z\bar z} \partial_{\bar z}( \gamma^{z\bar z} \partial_{\bar z} f ) &= \pi \gamma_{z\bar z} \delta^2(z-w) \\ \partial_{\bar z}( \gamma^{z\bar z} \partial_{\bar z} f ) &= \frac{1}{2} \partial_{\bar z} \frac{1}{z-w} \\ \gamma^{z\bar z} \partial_{\bar z} f &= \frac{1}{2} \frac{1}{z-w} + g ( z,w,\bar w)\\ \partial_{\bar z} f &= \frac{1}{2} \frac{\gamma_{z\bar z} }{z-w} + \gamma_{z\bar z} g( z,w,\bar w) \end{align}

Note that we have

$$ f = \frac{ \bar z - \bar w}{ (z-w)(1+z\bar z)(1+w\bar w)} $$

so that

$$ \partial_{\bar z} f = \frac{ 2\pi \delta^2 ( z - w )( \bar z - \bar w )}{ (1+z\bar z)(1+w\bar w)} + \frac{1}{2} \frac{\gamma_{z\bar z} }{z-w} + \frac{1}{2} \gamma_{z\bar z} \frac{ \bar w }{ ( 1 + w \bar w ) } $$ The first term is zero. The rest of this is precisely the of the required form.

  • $\begingroup$ thank you prahar.This was exactly what I was looking for.i was confused when the dereivative of the $\delta$ came.But I didnt notice the fact that it is with factor of $\overline{z-w}$ so it will be zero.I guess the reason why this should be zero is when you are acting that on a test function you you get the integrand as zero right ? $\endgroup$
    – budi
    May 1, 2017 at 2:39
  • $\begingroup$ @AnupamAh yep, that's right. $\endgroup$
    – Prahar
    May 1, 2017 at 2:40
  • $\begingroup$ I have another doubt .How do you get the last piece from $f$.How does the $\frac{\overline{w}}{(1+w\overline{w})^2}$ $\endgroup$
    – budi
    May 1, 2017 at 14:41
  • $\begingroup$ @AnupamAh - There was a typo. I fixed it. $\endgroup$
    – Prahar
    May 1, 2017 at 17:30
  • $\begingroup$ @Prahar-Suppose I multiply $f$ by $\gamma^{z\bar z}$ will I get rid of the $\gamma_{z\bar z}$ in the double covariant derivative?What I thought was since the covariant derivative of metric is zero this will act as a constant and i can take it outside to cancel the $\gamma_{z\bar z}$ and get just the delta function.Am I thinking right ? $\endgroup$
    – budi
    May 15, 2017 at 13:37

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