How does one "invert" such infinite-dimensional sympletic form? In the book "Lectures on the IR structure of gravity and gauge theories" by Strominger the author considers the sympletic form for free electrodynamics: $$\Omega_\Sigma[A;\delta_1 A,\delta_2A]=-\frac{1}{e^2}\int_{\Sigma}\delta(\star F)\wedge \delta A\tag{2.6.3}$$
In the above the $\wedge$ on the right encodes a wedge-product on the infinite dimensional phase space so that we have
$$\delta(\star F)\wedge \delta A=\delta_1(\star F)\wedge \delta_2A-\delta_2(\star F)\wedge \delta_1A$$ 
and where now on the RHS the $\wedge$ means for each term a usual wedge-product on spacetime. 
Here one may interpret $\Omega_\Sigma$ as a two-form on the infinite-dimensional covariant phase space defined to be the space of solutions to the classical equations of motion.
Next, Strominger argues that taking $\Sigma=\mathcal{I}^+$ and properly considering the soft photon mode one gets $$\Omega_{\mathcal{I}^+}[A;\delta_1 A,\delta_2A]=\frac{2}{e^2}\int dud^2z \partial_u\delta \hat{A}_{z}\wedge \delta\hat{A}_{\bar{z}}-2\int d^2z\partial_z\delta \phi\wedge \partial_{\bar{z}}\delta N\tag{2.6.8}$$
Finally he says:

The next thing to discuss is the commutators. As we see from equation (2.6.2), to derive the commutators one must invert the symplectic form. Since it is a sum of terms, one of which only involves $\hat{A}$ and the other which only involves the boundary fields $N$ and $\phi$, we can invert them separately. Consider the first one. It implies the commutator $$[\partial_u\hat{A}_z(u,z,\bar{z}),\hat{A}_\bar{w}(u',w,\bar{w})]=-\frac{ie^2}{2}\delta(u-u')\delta^2(z-w)\tag{2.6.9}$$

The mentioned equation (2.6.2) is just the equation defining the commutator out of $\Omega$ to be $$[A,B]=i \Omega^{IJ}\partial_I A \partial_J B.\tag{2.6.2}$$
Now, this $\Omega_\Sigma$ is a two-form on one infinite dimensional space. How does one invert such a thing? Because I can't see it as something obvious how does one go from Eq. (2.6.8) to Eq. (2.6.9).
 A: *

*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.]

*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:


*

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


--
$^1$ 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
$$ \left. d^2z\right|_{\rm Ref. 1}~=~2d^2z, \qquad  \left.\delta^2(z)\right|_{\rm Ref. 1}~=~\frac{1}{2}\delta^2(z).$$
A: The way I inverted the symplectic form was to write it out in coordinate notation. Not keeping track of overall constants (homework exercise!) we can write
$$
\Omega \sim \int d^2 z du \partial_u \delta A_z \wedge \delta A_{\bar z} \sim \int du du' d^2 z d^2 z' \partial_u \delta(u-u') \delta^2(z-z') \delta A_z(u) \wedge \delta A_{{\bar z}'} (u')
$$
In matrix form, we then have
$$
\Omega_{(u,z,{\bar z})|(u',z',{\bar z}')} \sim \partial_u \delta(u-u') \delta^2(z-z') .
$$
The inverse matrix $\Omega^{(u,z,{\bar z})|(u',z',{\bar z}')}$ is one that satisfies
$$
\int du'' d^2 z'' \Omega^{(u,z,{\bar z})|(u'',z'',{\bar z}'')}\Omega_{(u'',z'',{\bar z}'')|(u',z',{\bar z}')} = \delta(u-u') \delta^2(z-z') \qquad \qquad (1)
$$
We can now make an guess for the inverse matrix
$$
\Omega^{(u,z,{\bar z})|(u',z',{\bar z}')} = a \Theta(u-u') \delta^2(z-z')
$$
for some constant $a$. Note that the matrix has to antisymmetric which is the case since $\Theta(u)$ is the sign function. We can plug this ansatz into equation (1) and then work out the coefficient $a$. 
