In Witten's paper 'Quantization of ChernSimons 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 ChernSimons 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?



The reference is DeligneFreed, 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 oneform) $\gamma\in\Omega^{1,n1}(\mathcal{F}\times M)$, 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 ChernSimons, 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 ChernSimons, 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 $$\delta L=Tr(\delta A\wedge dAA\wedge \delta dA+2\delta A\wedge A\wedge A).$$ 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 $$d\gamma = Tr(dA\wedge \delta AA\wedge d\delta A),$$ 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. 


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. 


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


Additional information at the nlab: covariant phase space 

