Return to Answer

The (pre)-sympletic form ${\pmb \Omega}$ can be constructed from $\mathbf{L}$ in a precise way very well explained in the reference I provided. It turns out that it is one integral over such Cauchy slice $\Sigma$, which of course does not depend on which slice you choose. The reason for calling it pre-sympletic form instead of sympletic form is at the heart of your question and I'll explain in a moment. Before that, let me just state what I said more precisely. There exists one object, called pre-sympletic density ${\pmb \omega}$ which is a $(d-1)$-form in spacetime and a two-form in $\Gamma$, from which we obtain ${\pmb \Omega}$ as follows:

The (pre)-sympletic form ${\pmb \Omega}$ can be constructed from $\mathbf{L}$ in a precise way which is very well explained in the references I provided. It turns out that it is one integral over such Cauchy slice $\Sigma$, which of course does not depend on which slice you choose. The reason for calling it pre-sympletic form instead of sympletic form is at the heart of your question and I'll explain in a moment. Before that, let me just state what I said more precisely. There exists one object, called pre-sympletic density ${\pmb \omega}$ which is a $(d-1)$-form in spacetime and a two-form in $\Gamma$, from which we obtain ${\pmb \Omega}$ as follows:

In particular, a gauge transformation with parameter $\varepsilon$ corresponds to a vector field $\mathbf{X}_\varepsilon$ on $\Gamma$ and one may study ${\pmb \Omega}(\mathbf{X}_\varepsilon,\mathbf{Y})$ to study such symmetry and understand its charge. The key thing is that when one writes ${\pmb \Omega}$ as in (1) choosing a slice $\Sigma$, it turns out one has ${\pmb \Omega}(\mathbf{X}_\varepsilon,\mathbf{Y})$ as one integral over $\partial \Sigma$.

In particular, a gauge transformation with parameter $\varepsilon$ corresponds to a vector field $\mathbf{X}_\varepsilon$ on $\Gamma$ and one may study ${\pmb \Omega}(\mathbf{X}_\varepsilon,\mathbf{Y})$ to study such symmetry and understand its charge. The key thing is that when one writes ${\pmb \Omega}$ as in (1) choosing a slice $\Sigma$, it turns out one has ${\pmb \Omega}(\mathbf{X}_\varepsilon,\mathbf{Y})$ as one integral over $\partial \Sigma$. This is a consequence of Noether's second theorem, which applies to local symmetries.

where ${\pmb \omega}[\Phi,\mathbf{X}_\Phi,\mathbf{Y}_\Phi]$ means the two-form in $\Gamma$ at the point $\Phi\in\Gamma$ evaluated at the vectors $\mathbf{X}_\Phi,\mathbf{Y}_\Phi$ at $T_\Phi\Gamma$. The result is a $(d-1)$-form in spacetime which can be integrated over a $(d-1)$-dimensional slice which we take to be $\Sigma$. This gives the two-form ${\pmb \omega}$ in $\Gamma$.

where ${\pmb \omega}[\Phi,\mathbf{X}_\Phi,\mathbf{Y}_\Phi]$ means the two-form in $\Gamma$ at the point $\Phi\in\Gamma$ evaluated at the vectors $\mathbf{X}_\Phi,\mathbf{Y}_\Phi$ at $T_\Phi\Gamma$. The result is a $(d-1)$-form in spacetime which can be integrated over a $(d-1)$-dimensional slice which we take to be $\Sigma$. This gives the two-form ${\pmb \Omega}$ in $\Gamma$.