Finding all alternating bilinear $T$ that preserve a certain group of isometries of $\mathbb{R}^{n+1}$ Let $$G=\left\{\begin{pmatrix} H & 0 \\ 0 & 1\end{pmatrix} \ | \ H\in O(n), HJ=JH \right\}\subset \mathrm{Lin}(\mathbb{R}^{n+1},\mathbb{R}^{n+1}) $$ where: $n=2m$, $J$ is the standard complex structure on $\mathbb{R}^{2m}$ $\left(\text{that is}, J=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}\right)$ and $O(n)$ is the orthogonal group.
I need help with this:

Find all antisymmetric bilinear $\mathbb{R}^{n+1}$-valued functions T on $\mathbb{R}^{n+1}\times \mathbb{R}^{n+1}$ such that $$g\circ T(u,v)=T(g(u),g(v))$$ for all $g\in G$, $u,v\in\mathbb{R}^{n+1}$.

I tried a coordinate approach but it got too messy and I arrived at expressions that say nothing to me. I suspect that (for instance, if $m=1$ ($n=3$)), $T$ acts as a wedge product on pairs of vectors of the $xy$-plane, but that's all I've been able to tell.
I'll appreciate any help.
 A: Messy or not, the coordinate approach should be a viable approach. Here is a sketch of solution. Let $T_1,\ldots,T_{n+1}\in M_{n+1}(\mathbb{R})$ be the $n+1$ skew symmetric matrices such that
$$
T(u,v)=\pmatrix{u^TT_1v\\ \vdots\\ u^TT_{n+1}v}
$$
for all $u,v\in\mathbb{R}^{n+1}$. Then the given requirement means that
$$
\mathbf{g}\pmatrix{u^TT_1v\\ \vdots\\ u^TT_{n+1}v}
=\pmatrix{u^T\mathbf{g}^TT_1\mathbf{g}v\\ \vdots\\ u^T\mathbf{g}^TT_{n+1}\mathbf{g}v}\tag{1}
$$
for all $\mathbf{g}\in G$ and $u,v\in\mathbb{R}^{n+1}$. Substitute $\mathbf{g}_1=\pmatrix{I_n\\ &1}$ and $\mathbf{g}_2=\pmatrix{-I_n\\ &1}$ into $(1)$ and add the two resulting equations together, we obtain $\mathbf{g}_1^TT_k\mathbf{g}_1+\mathbf{g}_2^TT_k\mathbf{g}_2 = 0$ for $k\le n$, and therefore $T_k=\pmatrix{0&x_k\\ -x_k^T&0}$ for some $x_k\in\mathbb{R}^n$. Define $X=(x_1,\ldots,x_n)\in M_n(\mathbb{R})$. Then $(1)$ implies that $XH^T=H^TX$ for all $H\in O(n)$ that commutes with $J$. Hence $X=\pmatrix{aI_m&bI_m\\ -bI_m&aI_m}$ for some real numbers $a$ and $b$.
Now, look at the last row on both sides. We see that $T_{n+1}$ is a block diagonal matrix that conforms to the partitioning of $\mathbf{g}$ and centralises $G$. Since $T_{n+1}$ is also antisymmetric, it must be of the form $\pmatrix{cJ\\ &0}$ for some real number $c$. Putting the results together, we get $T(u,v)=(w_1,\ldots,w_n)^T$, where
\begin{cases}
w_k &= (au_k-bu_{m+k}) v_{n+1} - (av_k-bv_{m+k}) u_{n+1}
\quad (k=1,2,\ldots,m),\\
w_{m+k} &= (bu_k+au_{m+k}) v_{n+1} - (bv_k+av_{m+k}) u_{n+1}
\quad (k=1,2,\ldots,m),\\
w_n &= c\sum_{j=1}^m(u_jv_{m+j}-u_{m+j}v_j).
\end{cases}
