symplectic structure and non degenerate form

Question

I have a bivector field defined on R4 as

pi = dx1 ^ dy1 + exp(x2). dx2 ^ dy2

i need to prove that this is a symplectic structure and to find its explicit expression and then check that it is a closed non degenerate 2-form.

any help please?

Answer 1

To check that the form is closed, compute $d\left(dx_{1}\wedge dy_{1}+\exp(x_{2})dx_{2}\wedge dy_{2}\right)$, which equals $$d(dx_{1}\wedge dy_{1})+d\exp(x_{2})\wedge dx_{2}\wedge dy_{2}+\exp(x_{2})d(dx_{2}\wedge dy_{2}).$$ Here $$d(dx_{1}\wedge dy_{1})=d(dx_{1})\wedge dy_{1}-dx_{1}\wedge d(dy_{1})=0,$$ since $d^{2}=0$. Similarly, $d(dx_{2}\wedge dy_{2})=0$. As for the middle term, $$d\exp(x_{2})\wedge dx_{2}\wedge dy_{2}=\frac{d\exp(x_{2})}{dx_{2}}dx_{2}\wedge dx_{2}\wedge dy_{2}=0$$ due to skew-symmetry of $\wedge$ on one-forms.

For non-degeneracy, you may compute the matrix of the differential form with respect to the frame $\left\{\partial_{x_{1}},\partial_{y_{1}},\partial_{x_{2}},\partial_{y_{2}}\right\}$ and check that this matrix is always invertible. The matrix is $$\begin{pmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0& 0\\ 0 & 0 & 0 & \exp(x_{2}) \\ 0 & 0 & -\exp(x_{2}) & 0 \end{pmatrix},$$ which is clearly invertible for all values of $x_{2}$.

Answer 2

Hint:

You may want to look up the tautological 1-form $\theta$ and check whether the canonical symplectic structure $\omega = -d \theta$ defines a symplectic structure.

https://en.wikipedia.org/wiki/Tautological_one-form