# A characterisation of Hamiltonian vector fields.

Let $$(M,\omega)$$ be a symplectic manifold and let $$X\colon M\rightarrow TM$$ be a vector field on $$M$$.

Theorem. The vector field $$X$$ is Hamiltonian if and only if the following two conditions are fulfilled:

• If $$(\phi_t)_t$$ is the flow of $$X$$, one has $${\phi_t}^*\omega=\omega$$,

• For all loop $$c\colon]-1,1[\rightarrow M$$, one has $$\displaystyle\int_ci_X\omega=0$$.

Proof. Let us proceed by double implication.

• Assume $$X$$ is Hamiltonian, then there exists a smooth map $$H\colon M\rightarrow\mathbb{R}$$ such that: $$i_X\omega=\mathrm{d}H.$$ Therefore, the differential form $$i_x\omega$$ of degree $$1$$ is exact and its integral vanish on all loops drawn on $$M$$. Furthermore, using Cartan's formula for the Lie derivative and since $$\omega$$ is closed, one gets: $$\frac{\mathrm{d}}{\mathrm{d}t}({\phi_t}^*\omega)_{\vert t=0}=0.$$ I do not know how to deduce that $${\phi_t}^*\omega=\omega$$. If I use directly the definition of the pullback, I get: $$({\phi_t}^*\omega)_x(v,w)=\omega_{\phi_t(x)}(\mathrm{d}_x\phi_t\cdot v,\mathrm{d}_x\phi_t\cdot w),$$ which does not look to promising.

• I will assume that $$M$$ is connected (is it necessary?). Assume that $$X$$ fulfils the two given conditions, let $$m\in M$$, for all $$x\in M$$, I would like to define: $$H(x):=\int_{c_x}i_X\omega,$$ where $$c_x$$ is any path from $$m$$ to $$x$$. According to the second condition, the result will not depend on $$c_x$$. Therefore, the only difficulty is to choose it so that $$H$$ is smooth and I don't know how to proceed.

Whence the result. $$\Box$$

Any enlightenment will be greatly appreciated.

## 1 Answer

$$\frac{\mathrm{d}}{\mathrm{d}t}({\phi_t}^*\omega)_{\vert t=t_0}=\phi_{t_0}^*(\frac{\mathrm{d}}{\mathrm{d}t}({\phi_t}^*\omega)_{\vert t=0}).$$ implies that $\phi_t^*(\omega)$ is constant and is equal to $\phi_{0}^*(\omega)=\omega$ since $\phi_0^*$ is the identity.