Standard Symplectic Form

Question

I am stuck at how to make the matrix of standard symplectic form.

The given conditions are

Definition Let $V$ be a vector space over $\mathbb{R}$. Then $\omega\in \Lambda^2(V)$ is called symplectic form on $V$ if $\omega' : V\rightarrow V^*$ defined by $\omega'(X)=\omega(X,\cdot)$ is an isomorphism.

Thus, if $\{\epsilon^i\}$ is a dual basis of $V$, then we can write $$\omega = \sum_{i<j} \omega_{ij} \epsilon^i\wedge \epsilon^j . $$

Thus, every symplectic form can be represented by matrix $[\omega_{ij}].$

The thing that I am stuck at is the matrix representation of the symplectic form $$\omega = \sum_{i=1}^n \epsilon^i\wedge \delta ^i$$

where $\{\epsilon^i,\delta^i\}_{i=1}^{n}$ (of course $dimV=2n$. ) is a standard dual basis of $V$.

The matrix representation of it should be form of

$$\begin{bmatrix} 0 &1&0&0& \cdots &0&0\\ -1 &0 &0 &0 &\cdots &0 &0 \\0&0&0&1 &\cdots &0&0 \\ 0&0&-1&0&\cdots &0&0\\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots &\vdots\end{bmatrix}.$$

But I don't know how to get it.

I was calculating $\omega(e^i,d^i)$. But $\omega(d^1,e^1)=0$ but it should it be $-1$ according to the matrix.

I hope I can get help :) Thank you in advance.

Answer 1

The order of the basis is, by convention, $(e^1,d^1,e^2,d^2,\ldots,e^n,d^n)$.

(And just as a reminder: for a bilinear form $\omega$, its matrix representation $A$ is given by $a_{ij} = \omega(f_i,f_j)$ where $(f_1,\ldots,f_m)$ is the ordered basis.)