Symplectomorphisms preserving tautological one-forms

[Kreizhn]

Symplectomorphisms preserving "tautological" one-forms

Homework Statement

Let $(M,\omega)$ be a symplectic manifold such that there is a smooth one-form $\alpha \in \Omega^1(M)$ satisfying $\alpha = -d\omega$. Let $v \in \Gamma(TM)$ be the unique vector field such that $\iota_v \omega = -\alpha$. If $g: M \to M$ is any symplectomorphism that preserves $\alpha$ (that is $g^*\alpha = \alpha$) show that $g_* v = v$.

The Attempt at a Solution

This has been giving me trouble for a few days and I don't think it should be that difficult. The attempt I like most thus far is the following: Let $p,q \in M$ such that g(p) = q, so that $g^* \alpha_q = \alpha_p$. Thus
$$\begin{align*} \alpha_q(g_* v) &= (g^*\alpha_q) v_p = \alpha_p(v_p) = -\omega_p(v_p,v_p) = 0 \\ &= -\omega_q(v_q, g_*v_p). \end{align*}$$
That is, I have shown that $-\omega_q(v_q,g_*v_p) = 0$. Now a priori there is no reason to suspect that they must be equal, but I feel that this implication may be the key if we combine the fact that v is the unique vector field against whose retraction with $\omega$ the one-form $\alpha$ can be recovered. Thoughts?

 

[mighty2000]

Thank you for bringing this problem to our attention. As a scientist who specializes in symplectic geometry, I would be happy to offer some insight into this problem.

Firstly, your approach is on the right track. The key to solving this problem lies in understanding the relationship between the one-form \alpha and the vector field v. As you have correctly identified, v is the unique vector field that satisfies \iota_v \omega = -\alpha. This means that v is the "dual" to \alpha in the sense that \omega(v,\cdot) = -\alpha, where \cdot represents any vector in the tangent space.

Now, let's consider what happens when we apply the symplectomorphism g to v. Since g preserves \alpha, we know that g^*\alpha = \alpha. This means that for any point p \in M, we have g^*\alpha_p = \alpha_{g(p)}. Plugging this into your calculation, we get:

\begin{align*}
\alpha_{g(p)}(g_* v_p) &= (g^*\alpha_{g(p)}) v_p = \alpha_p(v_p) = -\omega_p(v_p,v_p) = 0 \\
&= -\omega_{g(p)}(g_*v_p, g_*v_p).
\end{align*}

Since this holds for all p \in M, we can conclude that g_*v is a vector field that satisfies \iota_{g_*v} \omega = -\alpha. But we know that v is the unique vector field with this property, so we must have g_*v = v.

I hope this helps clarify the solution to this problem. Please let me know if you have any further questions or if you would like me to elaborate on any particular point.