Lately I have been trying to understand the chapter in Abraham and Marsden's Foundations of Mechanics on infinite-dimensional Hamiltonian systems. Now that I've finally got a feeling for the canonical 1-form on an infinite-dimensional cotangent bundle, I can't get how to arrive at the expression of its exterior derivative. I have looked through Chernoff and Marsden's Properties of infinite-dimensional Hamiltonian systems and finally also at Abraham, Marsden and Ratiu's Manifolds, Tensor Analysis, and Applications book to get more exposure to some details, backgrounds and conventions.
My precise question is how exactly to arrive from the expression of
$\theta (x,\alpha) \cdot (e,\beta)= \alpha(e)$
-where $(x,\alpha)\in T^{\star}M$ is the footpoint and $(e,\beta)\in T_{(x,\alpha)}T^{\star}M$ describes the inserted tangent vector-
to the given expression for the canonical symplectic form $\omega= -d\theta$ which is
$\omega (e,\alpha) \cdot ((e_{1},\alpha_{1}),(e_{2},\alpha_{2}))= \alpha_{2}(e_{1})-\alpha_{1}(e_{2}) $.
In Abraham and Marsden's chapter they are talking about an induced/pulled back version of this on the tangent bundle, but my question is more about the original idea on the cotangent bundle as briefly presented in Chernoff and Marsden. They supposedly use the local formula for exterior derivative in terms of Lie Derivatives, but I just don't understand how to apply or rather utilize that here, to arrive at this result for $\omega$, so any comment or help with this step would be much appreciated!
Thank you for taking the time and interest :)
I am not completely sure which formula they are referring to, but I would use the following: for a $1$-form $\eta$ on $M$, a point $p\in M$, and two tangent vectors $X_1,X_2\in T_pM$, one has $$\mathrm{d}\eta_p(X_1,X_2)=X_1\big(\eta(X_2)\big)-X_1\big(\eta(X_2)\big)-\eta_p\big([X_1,X_2]\big).$$ Of course to compute this expression one needs to first extend $X_1$ and $X_2$ to vector fields in a neighbourhood of $p$.
In the finite-dimensional setting, one can then obtain the expression you are looking for by extending $X_1$ and $X_2$ to constant vector fields in a neighbourhood of $p$. A way to do that is to fix a local coordinate system, write $X_i(p)=\sum_aX_i^a\partial_a$ for some numbers $X_i^a$ and then declare that for any $x$ in the coordinate neighbourhood, $X_i(x)=\sum_aX_i^a\partial_a$.
In an infinite-dimensional setting this becomes slightly more complicated, you have to be a bit careful about what you mean by "vector field", and most importantly what are the local coordinate neighbourhoods of the manifold. But it can usually be done in the same way without worrying too much, once you are in a convenient setting. I would refer you to the aptly-named The Convenient Setting of Global Analysis by Kriegl and Michor, especially section $33$.
Assuming it is possible to (locally) extend the tangent vectors to constant vector fields, here is how you could prove the formula for the canonical $2$-form of $T^*M$: following your notation, fix a point $(x,\alpha)\in T^*M$ and two tangent vectors $(e_1,\beta_1), (e_2,\beta_2)\in T_{(x,\alpha)}T^*M$. We extend these to constant vector fields $(E_1,B_1)$ and $(E_2,B_2)$ near $(x,\alpha)$, so we can use the above formula to obtain $$\theta_{(x,\alpha)}\left((e_1,\beta_1),(e_2,\beta_2)\right)=-\mathrm{d}\tau_{(x,\alpha)}\left((e_1,\beta_1),(e_2,\beta_2)\right)=(e_2,\beta_2)\left(\tau(E_1,B_1)\right)-(e_1,\beta_1)\left(\tau(E_2,B_2)\right)$$ as the fields commute, being locally constant. By definition of $\tau$ now we find $$(e_2,\beta_2)\left(\tau(E_1,B_1)\right)=\partial_t\Bigr|_{t=0}\left[\tau_{(x+t e_2,\alpha+t\beta_2)}(E_1,B_1)\right]=\partial_t\Bigr|_{t=0}\left[(\alpha+t\beta_2)(E_1)\right]=\beta_2(e_1).$$ A similar expression holds for the other piece, and putting them together you find the desired formula. Let me remark that this computation is the one you would normally carry out in the finite-dimensional setting.