# How does one "invert" such infinite-dimensional sympletic form?

1. Using the deWitt condensed notation, we formally invert components of the 2-form $$\Omega~=~\frac{1}{2} \mathrm{d}x^I \Omega_{IJ} \wedge \mathrm{d}x^J \tag{2.6.1}$$ into the Poisson bracket $$\frac{[\hat{A},\hat{B}]}{i\hbar}~\longleftrightarrow~\{A,B\}_{PB}~=~(\partial_IA) (\Omega^{-1})^{IJ} (\partial_JB).\tag{2.6.2}$$ [Ref. 1 puts Planck's constant $$\hbar=1$$ equal to one.]

2. Now let's apply this on E&M. Given the 2-form$$^1$$ \begin{align}\Omega ~=~&\frac{4}{e^2}\int \!dud^2z~ \partial_u\delta\hat{A}_{z}\wedge \delta\hat{A}_{\bar{z}} -4\int\! d^2z~\partial_z\delta \phi\wedge \partial_{\bar{z}}\delta N \tag{2.6.8}\cr ~=~&-\frac{4}{e^2}\int \!dud^2z~du^{\prime}d^2z^{\prime}~\delta\hat{A}_{z}(u,z,\bar{z})~ \partial_u\delta(u\!-\!u^{\prime})~\delta^2(z\!-\!z^{\prime}) \cr & \qquad\qquad\qquad \wedge \delta\hat{A}_{\bar{z}^{\prime}}(u^{\prime},z^{\prime},\bar{z}^{\prime})\cr\cr &+\int\! d^2z~d^2z^{\prime}~\delta \phi(z,\bar{z})~\Delta_z\delta^2(z\!-\!z^{\prime}) \wedge \delta N (z^{\prime},\bar{z}^{\prime}), \qquad \Delta_{z}~\equiv~4\partial_{z}\partial_{\bar{z}}, \end{align} the non-vanishing Poisson brackets become $$\left\{\hat{A}_{\bar{z}^{\prime}}(u^{\prime},z^{\prime}),\hat{A}_{z^{\prime\prime}}(u^{\prime\prime},z^{\prime\prime})\right\}_{PB} ~=~\color{red}{+}\frac{e^2}{8}{\rm sgn}(u^{\prime}\!-\!u^{\prime\prime})~\delta^2(z^{\prime}\!-\!z^{\prime\prime}), \tag{2.6.10}$$ $$\left\{N(z^{\prime}),\phi(z^{\prime\prime})\right\}_{PB} ~=~G(z^{\prime}\!-\!z^{\prime\prime})~\equiv~ \frac{1}{4\pi} \ln|z^{\prime}\!-\!z^{\prime\prime}|^2, \tag{2.6.12}$$ where $$\Delta_{z^{\prime}}G(z^{\prime}\!-\!z^{\prime\prime})~=~\delta^2(z^{\prime}\!-\!z^{\prime\prime}).$$ The "$$\color{red}{+}$$" sign marks a deviation from the results in Ref. 1. [Ref.1 has a "$$\color{red}{-}$$" sign in eq. (2.6.10). Disclaimer: We have not checked if there are any typos in the starting eq. (2.6.8) itself.]

References:

1. A. Strominger, Lectures on the Infrared Structure of Gravity and Gauge Theory, arXiv:1703.05448.

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$$^1$$ Conventions. We use the convention $$d^2z~=~d{\rm Re}z~d{\rm Im}z, \qquad \delta^2(z)~=~\delta({\rm Re}z)\delta({\rm Im}z).$$ From eq. (2.6.14) it is clear that Ref. 1 uses the convention
$$\left. d^2z\right|_{\rm Ref. 1}~=~2d^2z, \qquad \left.\delta^2(z)\right|_{\rm Ref. 1}~=~\frac{1}{2}\delta^2(z).$$