Characterization of contact vector fields

Question

I am trying to prove Theorem 22.33 in Lee's Introduction to Smooth Manifolds, whose proof was left as a Problem:

If $(M, H)$ is a contact manifold and $\theta$ is a contact form for $H$, then a smooth vector field on $M$ is a contact vector field if and only if it is a contact Hamiltonian vector field.

A contact vector field is defined as a vector field $X \in \mathfrak{X}(M)$ whose flow $\psi$ preserves the contact structure, in the sense that $$d(\psi_t)_p(H_p) = H_{\psi_t(p)}\text{ for all }(t, p)\text{ in the domain of }\psi$$

For any $f \in C^{\infty}(M)$, a contact Hamiltonian vector field $X_f \in \mathfrak{X}(M)$ is defined as the unique vector field such that $$\theta(X_f) = f\text{ and }\left(\iota_{X_f} d\theta\right)\big\vert_H = -df\big\vert_H$$

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What I tried:

I can prove the implication "contact vector field" $\implies$ "contact Hamiltonian vector field".

Suppose that $X$ is a contact vector field on $(M, H)$ with flow $\psi: \mathcal{D} \to M$.
Since $\theta$ is a contact form for $H$, it annihilates $H$ so $\theta_p(H_p) = 0$ for all $p \in M$. Then for all $(t, p) \in \mathcal{D}$ we have $$\left(\psi_t^*\theta\right)_p\left(H_p\right) = \theta_{\psi_t(p)}\left(d(\psi_t)_p\left(H_p\right)\right) = \theta_{\psi_t(p)}\left(H_{\psi_t(p)}\right) = 0$$

This shows that for all $p \in M$, $$\left(\mathcal{L}_X\theta\right)_p\left(H_p\right) = \frac{d}{dt}\bigg\vert_{t=0}\left(\psi_t^*\theta\right)_p\left(H_p\right) = 0$$

Therefore $\left(\mathcal{L}_X\theta\right)\big\vert_H = 0$, so using Cartan's magic formula we find $$\left(\iota_X d\theta\right)\big\vert_H + \left(d\left(\iota_X \theta\right)\right)\big\vert_H = \left(\iota_X d\theta + d\left(\iota_X\theta\right)\right)\big\vert_H = \left(\mathcal{L}_X\theta\right)\big\vert_H = 0$$

So we can define $f = \theta(X)$ which is in $C^\infty(M)$, then the previous equation also shows that $$\left(\iota_X d\theta\right)\big\vert_H = -\left(d\left(\theta(X)\right)\right)\big\vert_H = -df\big\vert_H$$

Therefore, $X = X_f$ for this choice of $f$ so $X$ is contact Hamiltonian.

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I can't prove the converse.

Suppose $X_f$ is a contact Hamiltonian vector field. Let $\psi: \mathcal{D} \to M$ be its flow, and let $\theta$ be the contact form for $(M, H)$.

We want to show that $d(\psi_t)_p(H_p) = H_{\psi_t(p)}$ for all $(t, p) \in \mathcal{D}$.

By definition of the contact form, $\theta$ annihilates $H$ so $H_p = \ker \theta_p$ for each $p \in M$.

So we want to show that $d(\psi_t)_p(H_p) = \ker \theta_{\psi_t(p)}$, or in other words that $\theta_{\psi_t(p)}\left(d(\psi_t)_p(H_p)\right) = 0$.

We recognize this as a pullback, so we want to prove that $\left(\psi_t^*\theta\right)_p(H_p) = 0$.

Now we know that $\psi_0 = \text{id}_M$ so this holds when $t = 0$ because $\theta_p(H_p) = 0$.

Differentiating with respect to $t$, we have $$\frac{d}{dt}\left(\psi_t^*\theta\right)_p = \left(\psi_t^*\left(\mathcal{L}_{X_f}\theta\right)\right)_p$$

So the problem is equivalent to showing that $$\left(\psi_t^*\left(\mathcal{L}_{X_f}\theta\right)\right)_p(H_p) = 0$$

By Cartan's Magic Formula and the property $\theta(X_f) = f$, we can write $$\mathcal{L}_{X_f}\theta = \iota_{X_f} d\theta + d\left(\iota_{X_f}\theta\right) = \iota_{X_f} d\theta + d\left(\theta(X_f)\right) = \iota_{X_f} d\theta + df$$

We also know that $\left(\iota_{X_f} d\theta\right)\big\vert_H = -df\big\vert_H$, and this shows that $\left(\mathcal{L}_{X_f}\theta\right)\big\vert_H = 0$, but it is not clear that $\mathcal{L}_{X_f}\theta = 0$ outside of $H$ so this doesn't prove that $\psi_t^*\left(\mathcal{L}_{X_f}\theta\right) = 0$.

Since we don't know that $d(\psi_t)_p(H_p) = H_{\psi_t(p)}$ yet, we also don't know that $$\left(\psi_t^*\left(\mathcal{L}_{X_f}\theta\right)\right)_p(H_p) = \left(\mathcal{L}_{X_f}\theta\right)_{\psi_t(p)}\left(d(\psi_t)_p(H_p)\right) \stackrel{?}{=} 0$$

So unfortunately I am stuck here.

Answer 1

You do not need to prove $\mathcal{L}_{X_f} \theta = 0$ because this is the statement that $X_f$ preserves the contact form $\theta$, which is a stricter condition than preserving the contact structure. Instead one needs to check that the flow of $X_f$ preserves $H = \ker(\theta)$, which in Lie derivatives can be expressed as $(\mathcal{L}_{X_f}\theta)|_H = 0$.

Eg observe for any $g: M \to \mathbb{R}_{>0}$ one has $g\theta$ is also a contact form for $H$. For instance if the flow $\psi_t$ of a vector field $X$ satisies $\psi_t^* \theta = \exp(t) \theta$, then $X$ is a contact vector field (since $\ker(\theta)$ is preserve by $\psi_t$) but in terms of Lie derivative we have $\mathcal{L}_{X} \theta = \theta$.

Your proof that $\mathcal{L}_{X_f}\theta|_H = 0$ suffices, but perhaps unwinding it strictly by the definitions are you getting stuck because you are missing an observation about the definition of the contact Hamiltonian vector field.

A choice of contact form $\theta$ actually gives a splitting: $$TM = \ker(\theta) \oplus \ker(d\theta) = H \oplus span(R_\theta)$$ where $R_\theta$ is the unique vector field such that $\iota_{R_\theta} d\theta = 0$ and $\theta(R_\theta) = 1$ (it is called the Reeb vector field associated to the contact form $\theta$). By $\ker(d\theta)$ we are abusing notation and talking about the kernel of the map $T_mM \to T_m^*M$ given by $X \mapsto \iota_Xd\theta$ and that this kernel is one dimensional follows from the equation $\theta \wedge (d\theta)^n \not= 0$.

From the above considerations it must be that the contact vector field $X_f$ satisfies $\theta(X_f) = f$ and $$ \iota_{X_f} d\theta = -df + df(R_\theta) \theta $$ because this is what is must be given $d\theta(X_f, R_\theta) = 0$ (remember $R_\theta$ is in the kernel). Hence your computation of the Lie derivative actually gives $$ \mathcal{L}_{X_f} \theta = d(\iota_{X_f}\theta) + \iota_{X_f}d\theta = df(R_\theta) \theta $$ and this is enough to prove that if $\theta(Y) = 0$ then $\theta(d\psi_t(Y)) = 0$ and hence $\psi_t$ preserves $H = \ker(\theta)$.

To bring this discussion full circle consider $(\mathbb{R}^3, H = \ker(\theta))$ where $\theta = dz + xdy$ so $H = span(\partial_x, x\partial_z - \partial_y)$. Consider the vector field $X = z \partial_z + x \partial_x$. You can directly check it preserves H by computing the Lie derivatives $\mathcal{L}_{X}(\partial_x)$ and $\mathcal{L}_X(x\partial_z+ \partial_y)$. Or you can verify that it is the contact vector field associated to $f(x,y,z) = z$.