# 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).

• The most straightforward method is to add a Hamiltonian function and find the equations of motion by means of a variational principle. This is just the generalization of how $\int pdq + H(p, q)dt$ variationally leads to $\dot{q} := \{H, q\}= \frac{\partial H}{\partial p}$. Then, you can just choose your Hamiltonian to be $H(p, q) = p$, to obtain the canonical Poisson brackets. – David Bar Moshe Nov 4 '19 at 9:50

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 $$a$$.

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
$$\left. d^2z\right|_{\rm Ref. 1}~=~2d^2z, \qquad \left.\delta^2(z)\right|_{\rm Ref. 1}~=~\frac{1}{2}\delta^2(z).$$