acrobatics with $2$-form in $\mathbb{R}^{2n}$

Question

In the space $V = \mathbb{R}^{2n}$ with coordinates $(x_1, \dots, x_n, y_1, \dots, y_n)$ consider the $2$-form $\omega = \sum_{i=1}^n x_i \wedge y_i$.

1. Let $A$ be a $n \times n$ matrix. Consider a map $L_A: \mathbb{R}^n \to \mathbb{R}^{2n}$, given by the formula $L_A(x) = (x, Ax)$, where we view the vector $x \in \mathbb{R}^n$ as a column-matrix. Compute $(L_A)^*\omega$ and show that $(L_A)^*\omega = 0$ if and only if the matrix $A$ is symmetric.
2. Consider the map $\mathcal{C}_\omega$ of the space $V = \mathbb{R}^{2n}$ to the dual space $V^*$ given by the formula$$\mathcal{C}_\omega(v)(Z) = \omega(v, Z),\text{ }v,Z\in V.$$Find the matrix $C$ of the map $\mathcal{C}_\omega$ with respect to the standard basis in $V = \mathbb{R}^{2n}$ and its dual basis in $V^*$.

Answer 1

Let $e_1, \dots, e_n, f_1, \dots, f_n$ be the standard basis on $\mathbb{R}^{2n}$ with $x_1, \dots, x_n, y_1, \dots, y_n$ the dual basis. Let $A: \mathbb{R}^n \to \mathbb{R}^n$ be given by $Ae_i = \sum_{j=1}^n A_{ji}e_j$.

1.

We have$$(L_A)^*\omega = \sum_{1 \le i < j \le n}(L_A)^*\omega(e_i, e_j)\,x_i \wedge x_j.$$Now,$$(L_A)^*\omega(e_i, e_j) = \omega(L_Ae_i, L_Ae_j) = \sum_{l=1}^n(x_l \otimes y_l - y_l \otimes x_l)\left(\left(e_i + \sum_{k=1}^n A_{ki}f_k\right), \left(e_j + \sum_{k=1}^n A_{kj}f_k\right)\right)$$$$=\sum_{l=1}^n\left[x_l\left(e_i + \sum_{k=1}^n A_{ki}f_k\right)y_l\left(e_k + \sum_{k=1}^n A_{kj}f_k\right) - y_i\left(e_i + \sum_{k=1}^n A_{ki}f_k\right)x_l\left(e_j + \sum_{k=1}^n A_{kj}f_k\right)\right]$$$$=y_i\left(e_j + \sum_{k=1}^n A_{kj}f_k\right) - y_j\left(e_i + \sum_{k=1}^n A_{ki}f_k\right) = A_{ij} - A_{ji}.$$So $$(L_A)^*\omega = \sum_{1 \le i < j \le n}(A_{ij} - A_{ji})\,x_i \wedge x_j$$ and $(L_A)^*\omega = 0$ if and only if $A$ is symmetric.

2.

Write $$\mathcal{C}_\omega(e_k) = \sum_j(\alpha_jx_j + \beta_jy_j).$$ We have that$$\alpha_j = \mathcal{C}_\omega(e_k)(e_j) = \omega(e_k,e_j) = \sum_j x_i \wedge y_i(e_k, e_j) = 0,$$and$$\beta_j = \mathcal{C}_\omega(e_k)(f_j) = \omega(e_k, f_j) = \sum_j x_i \wedge y_i(e_k, f_j) = \delta_{k,j}.$$Similarly, write $$C_\omega(f_k) = \sum_j(\alpha_jx_j + \beta_jy_j).$$We then have $$\alpha_j = \mathcal{C}_\omega(f_k)(e_j) \omega(f_k, e_j) = \sum_j x_i \wedge y_i(f_k, e_j) = -\delta_{k,j},\text{ }\beta_j = 0,$$ similar to before.

The vector $(\alpha_1, \dots, \alpha_n, \beta_1, \dots, \beta_n)$ corresponding to $(e_k, f_k)$ is the $k$th column of the required matrix. Thus $$C = \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix},$$where $0$ denotes the $n \times n$ zero matrix, and $I$ denotes the $n \times n$ identity matrix.