# Computing Hamiltonian vector field

Let $$R=(R_1,R_2) \in \mathbb{R}^2$$ and consider the product of $$2$$-spheres $$S^2 \times S^2 \subset \mathbb{R}^3 \times \mathbb{R}^3$$ with cartesian coordinates $$(u_1,u_2)=((x_1,y_1,z_1),(x_2,y_2,z_2))$$ and symplectic form $$\omega=-(R_1\omega_{st} \oplus R_2\omega_{st})$$, where $$\omega_{st}$$ is the standard symplectic form on $$S^2$$, i.e. $$\omega(p)(u,v)=\langle p, u \wedge v \rangle$$ for all $$p \in S^2$$ and $$u,v \in T_pS^2$$. Here, $$\langle .,. \rangle$$ denotes the standard Euclidean product on $$\mathbb{R}^3$$. Define $$J: S^2 \times S^2 \to \mathbb{R}: (u_1,u_2) \mapsto R_1z_1 + R_2z_2$$

I have to check that the Hamiltonian vector field $$X^J$$ of $$J$$ is given by $$X^J(u_1,u_2)=(\gamma \wedge u_1, \gamma \wedge u_2)$$ where $$\gamma = (0,0,1) \in \mathbb{R}^3$$. I know that $$X^J$$ is defined by $$\omega(X^J, .)=-dJ(.)$$. It's not clear to me how one uses this to determine $$X^J$$. Plugging in a random vector doesn't give you an explicit form of $$X^J$$, right? On the other hand, writing $$\omega$$ in its matrix form and computing the inverse of $$\omega$$ will give you some nasty calculations. Does anynone care to explain?

• Maybe choose a coordinate system, and just compute $dJ$ and $\omega(X^J,-)$ in those coordinates, and check they are the same one-form.
– Nick
Nov 16 '18 at 17:47

You can do this by considering each sphere independently, and then taking linear combinations of the result for each sphere. So let's stick to one sphere for now. The tangent space $$T_uS^2$$ is spanned by vectors of the form $$a\wedge u$$ for $$a\in\mathbb{R}^3$$. For such a vector, we have $$(\omega_{st})_u((a\wedge u),\cdot) = \langle u, (a\wedge u)\wedge\cdot\rangle = \langle u\wedge(a\wedge u),\cdot\rangle = \langle a,\cdot\rangle = d_uH^a$$ where $$H^a:S^2\to\mathbb{R}$$ is the function $$H^a(u) = \langle a,u\rangle$$. In other words for $$a\in\mathbb{R}^3$$, $$u\mapsto a\wedge u$$ is a Hamiltonian vector field on $$S^2$$ with (using your conventions) Hamiltonian $$-H^a$$. Taking $$a=\gamma$$, the Hamiltonian becomes $$-H^\gamma(u) = -z$$.

Considering now the two spheres (and corresponding scaled symplectic forms), $$X(u_1,u_2) = (\gamma\wedge u_1,\gamma\wedge u_2)$$ is a Hamiltonian vector field corresponding to Hamiltonian $$R_1z_1 + R_2z_2$$.