for $\pi= dx_1 \wedge dy_1 + x_2^3\, dx_2 \wedge dy_2$ a bivector field on $\mathbb R^4.$
check that $dy_2$ is a poisson vector field, is it a hamiltonian vector field?
What should i do to prove it is a poisson vector field? or hamiltonian? What is the procedure?
Thank you
I guess you meant to write $$\pi=\partial_{x_{1}}\wedge\partial_{y_{1}}+x_{2}^{3}\partial_{x_{2}}\wedge\partial_{y_{2}}.$$ To see that $\partial_{y_{2}}$ is a Poisson vector field, you should check that the Lie derivative $\mathcal{L}_{\partial_{y_{2}}}\pi$ is zero. Do this by using the derivation property $$\mathcal{L}_{X}(V_{1}\wedge V_{2})=\mathcal{L}_{X}(V_{1})\wedge V_{2} + V_{1}\wedge\mathcal{L}_{X}(V_{2})$$ for vector fields $X,V_{1},V_{2}$.
One way to see that $\partial_{y_{2}}$ is not Hamiltonian, is the following. Hamiltonian vector fields are tangent to all symplectic leaves. Given a point $p\in\mathbb{R}^{4}$, the tangent space to the leaf through $p$ is $$\text{Span}\{\left.\partial_{x_{1}}\right|_{p},\left.\partial_{y_{1}}\right|_{p},x_{2}^{3}(p)\left.\partial_{x_{2}}\right|_{p},x_{2}^{3}(p)\left.\partial_{y_{2}}\right|_{p}\}$$ So at any point $p$ of the form $p=(x_{1},y_{1},0,y_{2})$, the vector field $\partial_{y_{2}}$ is not tangent to the leaf through $p$. Therefore, $\partial_{y_{2}}$ is not Hamiltonian.