Symplectic structures from Lagrangians?

Question

In Witten's paper 'Quantization of Chern-Simons Gauge Theory with Complex Gauge Group,' he makes the statement (p. 35) that the symplectic form on the moduli space of flat $G_\mathbb{C}$ connections on a surface can be derived from the Chern-Simons Lagrangian and depends upon the coupling chosen.
Now I am familiar with the bracket on the moduli space of connections obtained via Goldman's construction, but what is this method for getting a symplectic form from a Lagrangian and where could I learn more about it? It sounds like this is a general construction, too, that could apply any time a phase space is derived from a configuration space, not just something that applies to this particular case?

Answer 1

The reference is Deligne-Freed, Classical Field Theory, chapter 2. I will follow their notation.

Let $M$ be a spacetime manifold (for simplicity assume oriented) and $\mathcal{F}$ the space of fields. $d$ is the de Rham differential along $M$ and $\delta$ the differential along $\mathcal{F}$.

If the action $S$ is local, in the sense that $S=\int_M L$ for $L\in\Omega^{0,n}(\mathcal{F}\times M)$, then the procedure is the following. Find a form (variational one-form) $\gamma\in\Omega^{1,n-1}(\mathcal{F}\times M)$, such that $\alpha=\delta L+d\gamma\in\Omega^{1,n}(\mathcal{F}\times M)$ is linear over functions. What this means is that $\alpha(f\xi)=f\alpha(\xi)$ for every vector field $\xi\in T\mathcal{F}$ and a function $f\in\mathcal{O}(M)$. Then the symplectic form is defined to be $\omega=\int_H\delta\gamma\in\Omega^2(\mathcal{F})$ for $H$ a hypersurface in $M$. It is closed on the space of classical solutions, but may be degenerate.

In the case of Chern-Simons, this is literally true if $\mathcal{F}$ is the affine space of connections before modding out by the gauge transformations. I will consider the compact group Chern-Simons, the complex group version is similar. Let $M=\Sigma\times\mathbf{R}$ and $H=\Sigma\times\{0\}$. $$S=\int_M Tr(A\wedge dA+\frac{2}{3}A\wedge A\wedge A).$$ Then $$\delta L=Tr(\delta A\wedge dA-A\wedge \delta dA+2\delta A\wedge A\wedge A).$$ In this formula only the second term is not linear over functions. It can be killed off if one takes $$\gamma=Tr(A\wedge \delta A).$$ Its derivative is $$d\gamma = Tr(dA\wedge \delta A-A\wedge d\delta A),$$ so the nonlinear term in $\delta L+d\gamma$ disappears, since $d\delta +\delta d=0$.

Here the choice of $\gamma$ is unique if one assumes it itself is linear over functions. In the end one gets the standard symplectic form on the space of flat connections $$\omega=\int_\Sigma Tr(\delta A\wedge \delta A).$$

One should note that the action is not local in the naive sense on the space of connections mod gauge: the Lagrangian changes by a closed form under gauge transformations. However, the symplectic form does descend to the space of classical solutions mod gauge.

Answer 2

I will give a much simpler answer than the others. Given a Lagrangian, there is a corresponding Hamiltonian obtained via Legendre transformation. The symplectic structure comes along with this.

Answer 3

Additional information at the nlab: covariant phase space

Answer 4

You can read about this in the famous paper of Atiyah and Bott, Yang-Mills Equations over Riemann Surfaces, pg587.

Answer 5

See section 20.6.7 in

[1] V.P. Nair. Quantum Field Theory: A modern perspective. Springer, 2005.

You might want to check chapter 3 first.