# Does this geometric rigidity condition forces the map to be the identity?


Let $B \in \GL(\bigwedge^2V)$, and suppose that

$$B(v_1) \wedge B(v_2) \wedge B(v_3)=v_1 \wedge v_2 \wedge v_3 \in \bigwedge^6V \tag{*}$$

for every multi-vectors $v_1,v_2,v_3 \in \bigwedge^2V$.

Is it true that $B=\text{Id}_V$?

If it helps, we can also assume that $B$ maps decomposable elements to decomposable elements.

It is important that $\dim V>6$. If $\dim V=6$, one can take any $B=\bigwedge^2 A$, where $A \in \GL(V)$ has determinant one.

Edit: This seems to be true for diagonal maps $B$. Thus it also holds for diagonalizable maps (by multiplicativity).

Indeed, suppose $v_i$ is a basis for $\bigwedge^2V$, and that $Bv_i=\lambda_i v_i$. Then condition $(*)$ implies $\lambda_{i_1} \lambda_{i_2} \lambda_{i_3}=1$ for every distinct triplet $1\le i_1,i_2,i_3 \le \binom{d}{2}=\dim(\bigwedge^2V )$.

Now we think of the diagonal matrix $A=\text{diag}(\lambda_i)$ as a linear map $W \to W$, where $W=\bigwedge^2V$ is a $\binom{d}{2}$-dimensional vector space. Then the $3$-th exterior power of $A$, $\bigwedge^3 A=\text{Id}_W$. Since $3 < \dim(W)$, this implies $A=\text{Id}$, so $B=\text{Id}$ as required.

We might try to generalize this claim for non-diagonalizable maps, perhaps via some density argument (over $\mathbb{C}$), but at the moment I don't see how to do this.

• Note that it is quite possible for $v_i$ to be a basis of $\Lambda^2(V)$ with $v_i \wedge v_j = 0$ for some $i \neq j$ so I'm not sure your argument works even for diagonal maps. – levap Jul 24 '18 at 10:45
• Thanks, your comment seems correct. I was careless. – Asaf Shachar Jul 24 '18 at 11:24