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Let $\omega = dp_i \wedge dq^i$ be the standard symplectic form on $\mathbb{R}^{2n}(q^i, p_i)$. (We use the Einstein summation convention throughout). Assume $dq^i = h^{ij} dp_j$ on a Lagrangian near $0$ for some function $h^{ij} = h^{ij}(p)$. Then, in particular, $h^{ij} = h^{ji}$. I want to show that there exists a function $f$ such that $$ h^{ij} = \frac{\partial f}{\partial p_i \partial p_j} $$ and $\frac{\partial f}{\partial p_i}(0) = 0$.

I am stuck: Clearly, $0 = d^2 q^i = dh^{ij} \wedge dp_j$ and hence, $$ \frac{\partial h^{ij}}{\partial p_k} dp_k \wedge dp_j = 0. $$ How to conclude from here?

Edit: We see that $h^{ij} = \frac{\partial q^i}{\partial p_j}$. Is it always true that in the case where $h^{ij} = h^{ji}$, we have $h^{ij}$ is a second partial derivative of a function (due to Schwarz)?

warzasch
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    Let $\eta=q^i dp_i$ and note that $d\eta=-\omega=0$ along the Lagrangian. This tells that $\eta$ is closed, hence is exact on every simply connected open subset. Now let $U$ be an open neighborhood of $p=0$ in the Lagrangian. Then this observation tells that $\eta=df$ for some function $f$ on $U$, which we may assume satisfies $f(0)=0$. Now check that $$q^i=\frac{\partial f}{\partial p_i}$$ and hence the desired conclusion follows. – Sangchul Lee Jan 17 '24 at 16:18
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    Right. So, just use Poincaré lemma on the ball around 0. – warzasch Jan 17 '24 at 16:29

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