Say we have a manifold $M$ with local coordinates $q_i$. Then, $T^{\star}M$ has elements of the form $\sum_i p_i \mathrm{d}q_i$, for $p_i \in \mathbb{R}$; so $T^{\star} M$ has local coordinates $(q_i, p_i)$.
Now, we define the $1$-form $\lambda = \sum_i p_i \mathrm{d} q_i$ on $T^{\star} M$ (which means that $\lambda$ is a section of $T^{\star} T^{\star} M$), and $\mathrm{d}\lambda$ turns out to be symplectic.
Now, nlab contains a coordinate-free definition, stating that $\lambda$ as above, is determined uniquely by the fact that for every $1$-form $\sigma$ on $X$, we have:
\begin{equation*}
\sigma^{*}\lambda = j\sigma
\end{equation*}
Where $j$ is the natural isomorphism $j : \Gamma(T^{\star} M) \overset{\sim}{\longrightarrow} \Omega^1(M)$ (where $\Gamma$ denotes the space of sections, and $\Omega^1(M)$ the space of $1$-forms on $M$)). I am trying to start from the coordinate expression of $\lambda$; to show that this holds (or the other way around, really. I'm trying to understand why these two definitions are equivalent).
My working:
Take $\sigma$ an arbitrary $1$-form on $M$. So $\sigma : M \to T^{\star}M$; and hence induces $\sigma^{\star} : \Omega^1(T^\star M) \to \Omega^1(M)$.
Now, an element of $\Omega^1(T^\star M)$ is a section of $T^\star T^\star M$, hence a map $\alpha : \big(T^\star M \ni y \mapsto (\alpha_y : T_y T^\star M \to \mathbb{R})\big)$.
The natural way of pulling it back through $\sigma$, should give us something like:
$\sigma^{*} \alpha : M \to T^\star M : x \mapsto \alpha\big(\sigma(x)\big)$
Now, this is where I am getting a bit lost; I think I am getting confused between the different cotangent bundles. Ultimately, I want to show that this pullback above coincides with $j\alpha$ iff $\alpha = \lambda$; but I'm unsure how to proceed.
I am mostly confused as to what the isomorphism $j : \Gamma(T^\star M) \to \Omega^1(M)$ looks like. In my head, I have always identified $\Gamma(T^\star M)$ and $\Omega^1(M)$ as being the same space; so unless $j$ is the identity (lol), I'm not really sure what it represents.
Are $\Gamma(T^\star M)$ and $\Omega^1(M)$ "presented differently", even though they correspond to the same concept, which is why we need $j$ to bridge the gap between them?
Best Answer
I would regard $\Gamma(T^*M)$ and $\Omega^1(M)$ as the same thing - I guess the reason a distinction is being made here is that in the first case, we regard $\sigma:M\to T^*M$, and so can perform the pullback as with any smooth function, while in the second you think of $j\sigma$ as a 1-form. But its a bit pedantic IMO. So taking $\lambda\in\Omega^1(T^*M)$ and $\sigma:M\to T^*M$, we have that $\sigma^*\lambda\in\Omega^1(M)$. The characterisation of $\lambda$ is that this element equals $\sigma$ again (this time considered as a 1-form on $M$). Using the property that for $x\in M$ $$ (\sigma^*\lambda)_x = \lambda_{\sigma_x}\circ T_x\sigma $$ where $T_x\sigma$ denotes the differential of $\sigma$ at $x$, this says that for any $\sigma\in\Omega^1(M)$ $$ \lambda_{\sigma_x}\circ T_x\sigma = \sigma_x. $$
Now write $\lambda$ in local coordinates as $\lambda(q,p) = \alpha_i(q,p) dq^i + \beta^j(q,p) dp_j$. Also, writing $\sigma(q) = \sigma_j(q)dq^j = (q_i,\sigma_j(q))$, we have that $$T_x\sigma(\frac{\partial}{\partial q^i}) = (\frac{\partial}{\partial q^i}, \frac{\partial \sigma_j}{\partial q^i}\frac{\partial}{\partial p_j})$$
and so
$$ (\sigma^*\lambda)(q)(\frac{\partial}{\partial q^i}) = (\alpha_i(q,\sigma(q))dq^i+\beta^j(q,\sigma(q))dp_j)(\frac{\partial}{\partial q^i}, \frac{\partial \sigma_j}{\partial q^i}\frac{\partial}{\partial p_j}) \\ = \alpha_i(q,\sigma(q))+\beta^j(q,\sigma(q))\frac{\partial \sigma_j}{\partial q^i}$$.
For this to equal $\sigma(\frac{\partial}{\partial q^i}) = \sigma_i(q)$ for all $i$ requires that $\alpha_i(q,\sigma(q)) = \sigma_i(q)$ and $\beta^j(q,\sigma(q)) = 0$, i.e. $$ \alpha_i(q,p) = p_i, \quad \beta^j(q,p) = 0. $$ Hence $\lambda = p_idq^i$.
I should say, though, I prefer to use the definition of $\lambda$ as $$ \lambda_{\sigma_x} := \sigma_x\circ T_{\sigma_x}\pi $$ where $\pi:T^*M\to M$ is the canonical projection - it's a bit more direct, and a bit easier to derive the coordinate expression from.