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We are on a manifold $M$, and we have a 2 form $\omega: T M \times T M \rightarrow \mathbb R$. We can also think of $\omega$ as $\omega: TM \rightarrow (TM \rightarrow \mathbb R$) where $\omega$ is now a linear function from the tangent bundle $TM$ to the space of linear functions $TM \rightarrow \mathbb R$. But this is just the space of linear functionals, so we can write it as $\omega: TM \rightarrow T^*M$.

Now, on being given a function $H: M \rightarrow \mathbb R$, we know that $dH \in T^* M$. Hence, we can try to find a vector field $X_H : M \rightarrow TM$ such that $\omega (X_H) = dH$. This is the question posed by symplectic geometry, where $\omega$ is the symplectic data, and $H$ is the hamiltonian.

But I can also write this as $X_H = \omega^{-1} dH$. Question: when is $\omega$ invertible?

$\omega$ is a linear function from $TM$ to $T^*M$. So $\omega$ is invertible when it gives us a bijection from $TM$ to $T^* M$. Over $\mathbb R^n$ this is easy to check: pick a basis write a matrix, compute determinant. What about in the abstract on a manifold? I am hoping that the condition works out to be $d \omega = 0$, thereby justifying this condition on a symplectic form.

Arctic Char
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Siddharth Bhat
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    Hm, so you can always calculate the determinant of the arising matrices that come from writing the form in local coordinates, and check if all those are nonequal to zero. But I don't think you're going to get anything simpler than this. Closedness will, in any case, not be what you're looking for: for example, the zero-form $\omega = 0$ is obviously non-invertible, but closed. If you have any specific goal in mind, perhaps it is better to ask in what kind of data you'd rather want to express the invertibility question? –  Aug 15 '20 at 13:59
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    this page does a good job of explaining why $d\omega=0$. – Steve D Aug 15 '20 at 14:06

2 Answers2

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Condition is that $\omega$ is non-degenerate. By definition, a two form is non-degenerate when the mapping $TM \to T^*M$ is invertible. On a symplectic manifold $(M, \omega)$, $\omega$ is assumed to be non-degenerate. It has nothing to do with the closed-ness of $\omega$.

In a local coordinate, write $$ \omega = \omega_{ij} dx^i \wedge dx^j, $$ then the mapping $TM \to TM^*$ you defined send $\frac{\partial }{\partial x^i}$ to $$ \sum_j \omega_{ij} dx^j.$$

So the matrix representation of the mapping $TM\to T^*M$ with respect to the basis $$\left\{ \frac{\partial }{\partial x^1} , \cdots, \frac{\partial }{\partial x^n}\right\}, \ \ \ \{ dx^1, \cdots, dx^n\}$$ is given by $(\omega_{ij})$. So the mapping is non-degenerate if and only if the matrix $(\omega_{ij})$ is invertible.

Arctic Char
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Maybe this answer is a little besides the point, since Arctic Char's answer correctly addresses that there's not really much you can do except go to local coordinates. There is one fancy equivalence that says if $M$ is $2n$-dimensional, then a 2-form $\omega$ is non-degenerate if and only if $\omega \wedge \dots \wedge \omega \neq 0$ for the $n$-fold wedge product of the form with itself (see here for the proof of this on vector spaces, from which the statement for differential forms follows straightforwardly by doing this at every point).

However, since you conclude your question by saying that you'd like to justify this condition on symplectic manifolds, let me make two remarks: The reason why one requires closedness of $\omega$ on a symplectic manifold $M$ is because that way, it induces a Poisson bracket on the space of smooth functions $C^\infty(M)$. This is then important for doing all kinds of physics-shenanigans related to Hamiltonian mechanics, which is, from my experience, the main motivation for considering symplectic manifolds at all.

Additionally, perhaps it's interesting to you that there is a very similar concept where one does not require any non-degeneracy of anything, namely Poisson manifolds. They are a little different in the sense that you don't require the existence of a closed 2-form, but you just require the existence of a Poisson bracket on the space of smooth functions $C^\infty(M)$. This comes down to the existence of a Poisson bivector $\pi \in \Gamma(\Lambda^2 TM)$ which fulfils an analogue of the closedness condition of symplectic forms, necessary for the Leibniz rule of the Poisson bracket. So, rather than a 2-form, you need to work with 2-vectors, and it turns out that this generalization is very often a whole lot more complicated. Turns out having an isomorphism $TM \to T^*M$ is an extremely handy tool!

Hence, from this perspective, I'd see the non-degeneracy in the symplectic setting as a kind gift you're allowed to use that makes a lot of questions simpler, but which is not philosophically required if you just want to do emulate Hamiltonian mechanics on your manifold.

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    Nice, thanks a lot. So it is possible to write classical mechanics on top of a Poisson manifold as well? Without needing a symplectic form? – Siddharth Bhat Aug 15 '20 at 14:19
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    Yes, I'd say so, since all you really need to do mechanics is a notion of functions $C^\infty(M)$ (for your observables and Hamiltonians) and a Poisson bracket ${\cdot,\cdot}$ on $C^\infty(M)$. In standard examples, you will a posteriori notice that you are in the symplectic setting anyway ($e.g. M = \mathbb{R}^{2n}$ or $M$ a cotangent bundle), but once you start doing more freaky things, like quotienting out symmetries, you may very well end up in a situation where you don't have a symplectic form anymore, but still want to do mechanics. Then Poisson manifolds come into play. –  Aug 15 '20 at 14:23