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Assume we have a Hamiltonian system on $(\mathbb{R}^{2n},\omega)$ with Hamiltonian $H = H(q,p)$. In a paper I read, it says, without clarification, that the natural Liouville measure $\mu$ obtained by the volume form $\Omega = \omega \wedge \cdots \wedge \omega $ restricts on the energy level surfaces $N = \left\{(q,p) \in \mathbb{R}^{2n}: H(q,p) = c\right\}$ to


where F(q,p) is a function that satisfies $\omega(X_{H},F(q,p)) = dH(F(q,p)) = 1$ and $\iota(\cdot)$ is the interior product. A possible choice would be $F(q,p) = \frac{grad H}{\|grad H\|}$.

Why is this so?

Thanks in advance!

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It's often helpful to look at the original source; could you give a reference to the paper? – user53153 Feb 11 '13 at 1:21

1 Answer 1

Given a hypersurface $S$ in a manifold $M$ with volume form $\omega$ you can construct the volume element on $S$ induced by $\omega$ as follows: Take a field of unit normal vectors $N$ along $S$ and contract $\omega$ by $N$ that is define the differential form of 1 less rank as

$$\omega_S(\cdot,...)= \omega(N,\cdot,...)$$

The resulting element is called the volume (or sometimes area) element on $S$ induced by $\omega$.

In this setting we want however a volume form on $S$ which is invariant under the flow. So the usual volume element may not work for us. Now first to construct the volume element on an energy surface, we note that $\frac{\nabla H}{|\nabla H|}$ is unit vector field normal to $S$ (since $S$ is a constant level set of the function $H$). Therefore the volume element on $S$ is given as

$$\omega_S(\cdot,...) = \omega(\frac{\nabla H}{|\nabla H|},\cdot,...)$$

This means that we can write

$$\omega = \omega_s \wedge \frac{dH}{|\nabla H|}=\frac{\omega_S}{|\nabla H|}\wedge dH$$

Now under your flow $\phi$, the forms $\omega$ and $dH$ are invariant that is $\phi^*\omega= \omega$ and $\phi^*dH =dH$. Then

$$\omega= \phi^* \omega = \phi^*\frac{\omega_S}{|\nabla H|} \wedge \phi^*dH$$

So we get that

$$ \frac{\omega_S}{|\nabla H|}\wedge dH = \phi^*\frac{\omega_S}{|\nabla H|} \wedge dH$$

Now take any collection of basis vectors $v_i$ from the tangent space of $TS$ and contract both sides. Since $dH(v_i)=0$ what we get is that

$$ \frac{\omega_S}{|\nabla H|}(v_1,...,v_k) dH = \phi^*\frac{\omega_S}{|\nabla H|} (v_1,...,v_k)dH$$

for all vector $v_i$ which implies that

$$\frac{\omega_S}{|\nabla H|} = \phi^*\frac{\omega_S}{|\nabla H|}$$

so $\frac{\omega_S}{|\nabla H|}$ is invariant.

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