# How the canonical symplectic form acts

I've read that the canonical symplectic form $\omega$ on $\mathbb R^{2n}$ is given by $$\omega=\sum_{i=1}^n dp_i\wedge dq_i,$$ where $(p_1,\dots,p_n,q_1,\dots,q_n)$ are the coordinates on $\mathbb R^{2n}$.

If I've understood correctly, this means that $\omega$ is a differential 2-form on $\mathbb R^{2n}$, i.e. $$\forall p\in \mathbb R^{2n} \quad\omega_p\in \Lambda^2(T_p\mathbb R^{2n})^*,$$ thus $\omega_p$ is a skew 2-form, so $$\omega_p:T_p\mathbb R^{2n} \times T_p\mathbb R^{2n}\to \mathbb R$$ is bilinear and alternating.

Now, we can put in 1-1 correspondence $T_p\mathbb R^{2n}$ with $\mathbb R^{2n}$, because to every derivation of $T_p\mathbb R^{2n}$ corresponds its direction, which is a vector of $\mathbb R^{2n}$. So, given $u,v\in \mathbb R^{2n}$, how should I compute $\omega_p(u,v)$?

With block matrix notation, where $0_n$ is the $n \times n$ zero matrix and $I_n$ is the $n \times n$ identity matrix, it's simply the matrix product $$\omega_p(u,v)= (u_1,\dots,u_{2n}) \begin{pmatrix} 0_n & I_n \\ -I_n & 0_n \end{pmatrix} \begin{pmatrix} v_1 \\ \vdots \\ v_{2n} \end{pmatrix} .$$ To see this, call the basis vectors $$(\partial/\partial p_1,\dots,\partial/\partial p_n,\partial/\partial q_1,\dots,\partial/\partial q_n) ,$$ and then use $$dp_i(\partial/\partial q_j)=0 ,\qquad dp_i(\partial/\partial p_j)=\delta_{ij}$$ (and the similar formulas for $dq_i$) together with the definition of the action of a wedge product: $$(\alpha \wedge \beta)(u,v) = \begin{vmatrix} \alpha(u) & \alpha(v) \\ \beta(u) & \beta(v) \end{vmatrix} .$$
• No, I mean, why is that the matrix associated to the bilinear form $\omega_p$? I know that its coefficients are given by $\omega_p(e_i,e_j)$. right? So how can I compute $\omega_p(e_i,e_j)$? Oct 7 '15 at 16:56