Alternating forms and isometries of the underlying vector space Does a linear isometry of $\mathbb{R}^n$ yields a linear isometry of
$\mathbb{R}^n \wedge \mathbb{R}^n$ ?
If yes, why ? How to prove it elegantly, without explicit coordinate
computations ? Using, for instance, well-known properties of the determinant
and of the scalar product.
 A: Let $V$ be a (real, say) vector space and let $g(-,-):V\times V\to V$ be an inner product on $V.$ Then to say that $A:V\to V$ being an isometry means that $g(Au,Av)=g(u,v)$ for all $u,v\in V.$ If we consider $V\wedge V$ then this gets an induced metric $\widetilde g$ as follows: we define $$\widetilde g(v_1\wedge v_2,u_1\wedge u_2):=\det\begin{pmatrix}g(v_1,u_1) & g(v_2,u_1)\\ g(v_1,u_2) & g(v_2,u_2)\end{pmatrix}.$$ We then extend to all of $V\wedge V$ by linearity. Also, $A:V\to V$ induces a map $\widetilde A:V\wedge V\to V\wedge V$ by defining $\widetilde A(u\wedge v)=Au\wedge Av$ and then extending by linearity. Well, putting all this together we see that \begin{align*}\widetilde g(\widetilde A(v_1\wedge v_2),\widetilde A(u_1\wedge u_2)) &= g(Av_1\wedge Av_2,Au_1\wedge Au_2) = \det\begin{pmatrix}g(Av_1,Au_1) & g(Av_2,Au_1)\\ g(Av_1,Au_2) & g(Av_2,Au_2)\end{pmatrix}\\ &= \det\begin{pmatrix}g(v_1,u_1) & g(v_2,u_1)\\ g(v_1,u_2) & g(v_2,u_2)\end{pmatrix} = \widetilde g(v_1\wedge v_2,u_1\wedge u_2).\end{align*}
We've checked the isometry condition on a spanning set for $V\wedge V,$ which then extends to the whole of $V\wedge V$ by linearity. Thus $\widetilde A$ is an isometry if $A$ is.
So really this has nothing to do with the determinant. It only matters how $A$ and $g$ interact.
