# Fixed a symplectic form, any differential of a regular function is a contraction of the symplectic form

Let $$M$$ be a smooth manifold over $$\mathbb{R}$$ and $$\omega$$ a symplectic form on $$M$$, i.e. a regular, non-degenerate, closed 2-form. Given a vector field on $$M$$, $$X \in \mathcal{V}(M)$$, we define the contraction of $$\omega$$ with respect to $$X$$ as $$i_X\omega:=\omega(X,-).$$ The problem I am addressing is the following one

For any $$f \in \mathcal{O}(M)$$ there exists a vector field $$X$$ such that $$i_X\omega = df$$

I have two ideas to address this problem. First, fixed $$f$$ I write explicitly the $$1$$-forms $$df$$ and $$i_X\omega$$ locally and then I set the equality. This gives me sufficient local conditions to determine $$X$$.

The second idea (which, if correct, is equivalent to the first one) is the following. Let me write the symplectic form as a skew-symmetric non-degenerate bilinear map $$\omega : \mathcal{V}(M) \times \mathcal{V}(M) \rightarrow \mathcal{O}(M).$$ Then, if we restrict ourselves in a local chart, we have a basis for $$\mathcal{V}(M)$$ and there exists a matrix $$M \in M_{ n \times n}(\mathcal{O}(M))$$ such that it expresses $$\omega$$ with respect to the fixed basis of $$\mathcal{V}(M)$$. For the property of $$\omega$$ I would say that $$M$$ has an inverse matrix in $$M_n(\mathcal{O}(M))$$. This is sufficient to conclude. In fact, in coordinates, by the identity $$i_X\omega = df$$, we would have that $$\begin{bmatrix} X_1, & \dots & X_n \end{bmatrix}=\begin{bmatrix} \partial_{x_1}f, & \dots & ,\partial_{x_n}f \end{bmatrix} \cdot M^{-1}$$ which define $$X$$ locally. Now remain to prove that these "local definitions" can be put together to define a global vector fields. I would prove this last step using just the definition (if is there any shortcut I will be happy to know that).

A third way to prove it could be the following one.

Since $$\omega$$ is a symplectic form, it defines a smooth bundle isomorphism $$TM \rightarrow T^*M, \ T_pM \ni (p,\nu) \mapsto \omega_p(\nu,-) \in T_p^*M$$ (This can be seen as a consequence of the Bundle homomorphism characterization Lemma) So, the thesis is straightforward.

I would like to know if this is a correct way to proceed.

Thank you

• The third approach works. The fact is a pointwise one, so you can focus on only the fiber over one point. The tangent space $T_p$ is a vector space with a symplectic form, which is nondegenerate. This, by definition, means that the linear map $v \in T_p \mapsto \omega_p(v,\cdot) \in T_p^*$ is an isomorphism. It therefore has an inverse map, which is what you want. Aug 5 at 16:49

Your approaches are not quite correct since you seem to assume that you are given $$i_Xw=df$$. The whole point is to prove that such a vector field $$X$$ exists. This is why you can't just assume that you are given $$i_Xw=df$$ and solve for $$X$$, you are not guaranteed a priori that there exists an object $$X$$ that satisfies this identity.

Your second approach is close to the correct one (although, again, at the end you assume the identity). the approach is as follows. As you observed, at every point $$p \in M$$, $$w$$ has an invertible matrix, hence define the matrix $$P^{ij}$$ such that $$P^{ij}w_{jk}=\delta^{i}_{k}$$

Then define the rank-2 contravariant tensor (the so called Poisson-bivector) to be:

$$P=P^{ij} \dfrac{\partial}{\partial x^i} \wedge \dfrac{\partial}{\partial x^j}$$ You can prove this is a legitimate tensor by checking how $$P^{ij}$$ transform under a change of coordinates using its definition; it's a straightforward exercise but requires a bit computation so I'll leave that to you. Now define the two maps:

$$F:\Omega(M) \to V(M)$$ by $$F(\alpha) = P(\alpha, -)$$ and a second map $$G: V(M) \to \Omega(M)$$ $$G(X)=w(X,-)$$

Given how we defined $$P$$ it is easy to show that $$G \circ F = Id : \Omega(M) \to \Omega(M)$$ is in fact the identity map.

Hence for each $$f \in C^{\infty}(M)$$ define $$X=P(df,-)$$

Note that $$i_Xw(-)=w(X,-)= G \circ F (df) = Id(df)=df$$

• Thank you very much. However, I don't understand why my approaches are not correct. In fact, I set the identity to find conditions for $X$ and then I state that these conditions determine a unique vector field. So, "if $X$ exists then it satisfies that condition. If there exists a (rough) vector field satisfying those conditions, it is pointwise uniquely determined. Then we have that this rough vector field is smooth."
– wood
Aug 5 at 16:47
• @wood: Perhaps I'm being overly fastidious but the problem with your statement is that it begins with "if X exists", what if X doesn't exist at all? After all, your proof requires that you proof the existence of an object, so you can't start by just assuming it does. What you can do is the following: You can guess how this vector field looks like on scratch (by doing what you did), then you start the actual proof by claiming that this aformentioned vector field (which you have now explicit form for, and hence exists) satisfies your identity. It's a minor, but imo important, distinction. Aug 5 at 18:31
• Yes, I understand. But what I wanted to point out was the method I had decided to use to guess the "aspect" of the vector fields. Thank you for your answer
– wood
Aug 5 at 21:47
• Furthermore, if $X$ had not be existed then I would not have found $X$ but an absurd.
– wood
Aug 5 at 22:34

I decided to post as an answer a clear and easy way to develop the third way to prove the statement.

By the previous point we have that the symplectic form $$\omega$$ define a smooth bundle isomorphism $$\Phi_\omega:TM \rightarrow T^*M, \ T_pM \ni (p,v) \mapsto \omega_p(v,-) \in T_p^*M$$

Now, since $$\Omega^1(M)=\{M \xrightarrow{\lambda} T^*M, \ p\mapsto \lambda_p \in T^*_pM \ \text{such that it is smooth}\}$$ $$\mathcal(V)(M) = \{M \xrightarrow{X} TM, \ p \mapsto X_p \in T_pM \ \text{such that it is smooth}\}$$ We have that given $$f \in \mathcal{O}(M)$$ we can define $$X=\Phi_\omega^{-1} \circ df$$ that defines $$X : M \rightarrow T^*M$$ which is clearly a (smooth since it is composition of smooth maps) vector field satisfying the requested property.