Is this a Lie algebra homomorphism?

Question

Let $g\in \mathrm{SL}(4, \mathbb{C})$ and $v, w\in \mathfrak{so}(4, \mathbb{C})$, i.e. $v^{T}+v =0$ and $w^{T}+w=0$. We can define an action of $\mathrm{SL}(4, \mathbb{C})$ on $\mathfrak{so}(4, \mathbb{C})$ as $g.v = gvg^{T}$, and we have a corresponding linear map $\phi_{g}:\mathfrak{so}(4, \mathbb{C})\to \mathfrak{so}(4, \mathbb{C})$. Is it true that $\phi_{g}$ is a Lie algebra homomorphism, even if $g\not\in \mathrm{SO}(4, \mathbb{C})$? So I want to check whether $$ vw-wv = vg^{T}gw-wg^{T}gv $$ is true for any $g\in \mathrm{SL}(4, \mathbb{C})$ and $v, w\in \mathfrak{so}(4, \mathbb{C})$. This seems false, but actually I hope this to be true. Thanks in advance.

Answer 1

This is indeed false. For instance, consider $$g=\begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix},\qquad v=\begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix},\qquad w=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$

Then by my calculations, $$vw-wv=\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ whereas $$vg^Tgw-wg^Tgv=\begin{pmatrix} 0 & 0 & 2 & 0 \\ 0 & 0 & -1 & 0 \\ -2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$