on the matrix representation of a canonical linear map on the vector space of $4\times 4$ skew-symmetric matrices

Question

For $t_1,t_2,...,t_6\in \mathbb R$, let $P_{(t_1,t_2,...,t_6)}=\begin{pmatrix} 0&t_1&t_2&t_3\\ -t_1&0&t_4&t_5\\ -t_2&-t_4 &0&t_6\\-t_3&-t_5&-t_6&0\end{pmatrix}$.

Let $V=\{P_{(t_1,t_2,...,t_6)} : t_1,...,t_6\in \mathbb R\}$ i.e. $V$ is the vector space of all $4\times 4$ skew-symmetric matrices with real entries .

Then $V$ is a $6$-dimensional vector space with a basis given by

$\beta=\{P_{(1,0,...,0)}, P_{(0,1,0,...,0)}, P_{(0,0,1,0,...,0)},...,P_{(0,...,0,1)}\}$ .

Now given $A\in M(4,\mathbb R)$, consider the map $\phi_A:V \to V$ given by $\phi_A(P)=APA^t,\forall P\in V$. Let $\widehat A$ be the matrix of $\phi_A$ w.r.t. the basis $\beta $ of $V$ (so $\widehat A \in M(6,\mathbb R)$ ). It is easy to see that if $A \in GL(4,\mathbb R)$ then $\phi_A$ is an isomorphism, hence $A \in GL(4,\mathbb R)$ implies $\widehat A\in GL(6,\mathbb R) $. Now consider the map $\psi :GL(4,\mathbb R) \to GL(6,\mathbb R)$ given by $\psi(A)=\widehat A$ . Since $\phi_{AB}=\phi_A \circ \phi_B$, hence $\widehat {AB}=\widehat A . \widehat B ,\forall A,B \in GL(4,\mathbb R)$, thus $\psi$ is a group homomorphism.

My question is: How can we describe the kernel of $\psi$ i.e. how can we describe the set $\{A \in GL(4,\mathbb R) : \widehat A=Id\}$ ?

My try: $\widehat A=Id $ if and only if $\phi_A$ is the identity map if and only if $APA^t=P$ for every $4\times 4$ skew-symmetric matrix $P$. But I am unable to simplify this further.

Please help.

Answer 1

By my comment above, $A = (\begin{smallmatrix}B & 0\\0 & E\end{smallmatrix})$, where $\det B = \det E = 1$. Now, let $X$ be any $2\times 2$ matrix and set $P = (\begin{smallmatrix}0 & X\\-X^T & 0\end{smallmatrix})$. Then $P$ is skew-symmetric and \begin{align*} \begin{pmatrix}0 & X\\-X^T & 0\end{pmatrix} &= P = APA^T = \begin{pmatrix}B & 0\\0 & E\end{pmatrix}\begin{pmatrix}0 & X\\-X^T & 0\end{pmatrix}\begin{pmatrix}B^T & 0\\0 & E^T\end{pmatrix}\\ &= \begin{pmatrix}0 & BXE^T\\-EX^TB^T & 0\end{pmatrix}. \end{align*} Hence, $BXE^T = X$ for all $X\in\Bbb R^{2\times 2}$. Setting $X = I_2$ gives $E^T = B^{-1}$ and thus $BX = XB$ for all $X\in\Bbb R^{2\times 2}$. This implies $B = bI_2$ with $b\in\Bbb R$. But $\det B = 1$, so $b^2 = 1$, i.e., $B = \pm I$. Finally, $$ A = \begin{pmatrix}\pm I_2 & 0\\0 & \pm I_2\end{pmatrix} = \pm I_4. $$