Prove that the $2$ form defines a symplectic structure

Question

Prove that the $2$ form

$$\omega = -2[(1+x_2^2)dx_1 \wedge dx_2 + dx_1 \wedge dx_3 + dx_3 \wedge dx_4]$$

defines a symplectic structure on $\mathbb{R}_x^4$.

My definition of as symplectic structure is that a $2$ form $\omega \in \Omega^2(U)$ where $U$ is an open subset is a symplectic structure if it is of the form

$$\omega = 2 \sum_{i = 1}^{n} (dq_i \wedge dp_i) = 2\{(dq_1 \wedge dp_1) + \cdots + (dq_n \wedge dp_n)\}$$

and it satisfies

• $\omega$ is closed, i.e $d \omega = 0$
• That matrix of $\omega$ is a bilinear function which is invertible.

I checked the determinant of $\omega$ and got that to be not $0$ (correctly) but I can't seem to get $d \omega = 0$. When I do it, I get:

$$d \omega = d\{-2[(1+x_2^2)dx_1 \wedge dx_2 + dx_1 \wedge dx_3 + dx_3 \wedge dx_4] \}$$ $$ = -2 \{2x_2 dx_2 \wedge dx_1 \wedge dx_2 + dx_1 \wedge dx_3 + dx_3 \wedge dx_4\}$$ $$= -2\{dx_1 \wedge dx_3 + dx_3 \wedge dx_4\} \neq 0.$$

Where have I gone wrong?

Answer 1

You forgot to take the exterior derivative of the $dx_1 \wedge dx_3$ and $dx_3 \wedge dx_4$ pieces of $\omega$.