Let $1<k<d$ be an integer. Let $A \in \text{End}(\bigwedge^k \mathbb{R}^d)$, and suppose that $A=\bigwedge^k B$ for some complex $B \in \text{End}(\mathbb{C}^d)$.

Does there exist $M \in \text{End}(\mathbb{R}^d)$ such that $A=\bigwedge^k M$ or $A=-\bigwedge^k M$?

More formally, I mean that we have an element $A \in \text{End}(\bigwedge^k \mathbb{C}^d)$, such that $A(\bigwedge^k \mathbb{R}^d) \subseteq \bigwedge^k\mathbb{R}^d$ (so in this sense $A$ is real), and $A$ has a complex "root". The question is whether $A$ must be a real power up to a sign.

The minus option can occur: Take $A = -\operatorname{Id}_{\bigwedge^2\mathbb{C}^3}$; then $A|_{\bigwedge^2\mathbb{R}^3}=-\operatorname{Id}_{\bigwedge^2\mathbb{R}^3}$, and $A=\bigwedge^2 (i\operatorname{Id}_{\mathbb{C}^3})$. $A$ does not admit a "real source".

Here is a possible approach for the invertible case $A \in \text{GL}$:

(In that case, if a real source exist, then it is unique up to a sign).

Since $A=\bigwedge^k B$, and $A(\bigwedge^k \mathbb{R}^d) \subseteq \bigwedge^k\mathbb{R}^d$, $Bv_1 \wedge \ldots \wedge Bv_k \in \bigwedge^k\mathbb{R}^d$ for every $v_1,\ldots,v_k \in \mathbb{R}^d$.

In other words, $Bv_1 \wedge \ldots \wedge Bv_k $ is decomposable in $ \bigwedge^k\mathbb{C}^d$, and belongs to $\bigwedge^k\mathbb{R}^d$.

If this implies that it is also decomposable in $\bigwedge^k\mathbb{R}^d$, then $B$ is a complex matrix which maps real $k$-dimensional subspaces (over $\mathbb{C}$) to real subspaces (over $\mathbb{C}$).

We need to be careful here:

$Bv_1 \wedge \ldots \wedge Bv_k $ is decomposable in $\bigwedge^k\mathbb{R}^d$, means that there exist $w_1,\ldots,w_k \in \mathbb{R}^d$ such that $$ Bv_1 \wedge \ldots \wedge Bv_k =w_1 \wedge \ldots \wedge w_k.$$

Now, thinking on the last equality as an equality of elements in $\bigwedge^k\mathbb{C}^d$, we deduce that $\text{span}_{\mathbb{C}}(Bv_1,\ldots,Bv_k)=\text{span}_{\mathbb{C}}(w_1,\ldots,w_k)$.

The stronger statement $\text{span}_{\mathbb{R}}(Bv_1,\ldots,Bv_k)=\text{span}_{\mathbb{R}}(w_1,\ldots,w_k)$ is false in general! Indeed, take $B=i\operatorname{Id}_{\mathbb{C}^3}$ from the example above.

Maybe this fact forces $B$ to be a (complex) scalar multiple of a real matrix.

However, I am not sure that that every "real" element which is "complex-decomposable" is also "real-decomposable".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.