If $\omega \in \bigwedge^k\mathbb{R}^d$ is decomposable over $\mathbb C$, is it decomposable over $\mathbb R$?

Question

Let $k,d$ be positive integers, and let $\omega \in \bigwedge^k\mathbb{R}^d$ be decomposable in $ \bigwedge^k\mathbb{C}^d$.

Is $\omega$ decomposable in $\bigwedge^k\mathbb{R}^d$?

Edit:

Let me be more careful about the formulation of this question, as $\bigwedge^k\mathbb{C}^d$ can have two different non-isomorphic interpretation:

1. $\bigwedge^k_{\mathbb{R}}\mathbb{C}^d$: Here we take exterior power over $\mathbb R$. In particular, we think of $\mathbb{C}^d$ as a real $2d$-dimensional vector space. In that case $\mathbb{R}^d$ is a vector subspace (over $\mathbb{R}$), and so we have the general claim that “being decomposable” is a property which remains invariant under passing to a subspace. So, in that case, the answer is positive.

Sasha's answer refers to this interpretation of the question.

2. $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$: Here we take exterior power over $\mathbb C$, and we think of $\mathbb{C}^d$ as a complex vector space. In this case $\mathbb{R}^d$ is not a (complex) vector subspace of $\mathbb{C}^d$, but we can still view $\bigwedge^k\mathbb{R}^d$ as a subspace of $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$, via complexification. This because complexification commutes with exterior powers, so $(\bigwedge^k\mathbb{R}^d)^{\mathbb C}=\bigwedge^k_{\mathbb{C}}((\mathbb{R}^d)^{\mathbb C})=\bigwedge^k_{\mathbb{C}}\mathbb{C}^d.$ Any real vector space $V$ can be viewed as a subspace of its complexification $V^{\mathbb C}=V \otimes_{\mathbb R}C$ via the map $v \to v \otimes 1$. Thus, we can consider in this way $\bigwedge^k\mathbb{R}^d$ as a subspace of $(\bigwedge^k\mathbb{R}^d)^{\mathbb C}=\bigwedge^k_{\mathbb{C}}\mathbb{C}^d.$ Now we are given an element $\omega \in \bigwedge^k\mathbb{R}^d$, which is decomposable as an element in $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$, and we ask whether or not it is decomposable as an element in $ \bigwedge^k\mathbb{R}^d$.

What is the answer for this second variant of the question?

Comment:

Perhaps there is a way to view $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$ as a subspace of $\bigwedge^k_{\mathbb{R}}\mathbb{C}^d$ in a way which preserves decomposability, and the "real copy" $\bigwedge^k_{\mathbb{R}}\mathbb{R}^d$, thus reducing the second problem to the first one. I asked about the possible existence of such an embedding here.

So far, I only know that the answer is positive for $k=2$:

Let $\omega \in \bigwedge^2\mathbb{R}^d$ be decomposable in $\bigwedge^2\mathbb{C}^d$. $\omega$ can be written as

$$\omega=(u_1+iv_1) \wedge (u_2+iv_2), \tag{1}$$ where $u_1,u_2,v_1,v_2 \in \mathbb R^d$. Since

$$ \omega=(u_1 \wedge u_2 - v_1 \wedge v_2)+i (v_1 \wedge u_2+u_1 \wedge v_2), $$

$\omega \in \bigwedge^2\mathbb{R}^d$ if and only if $$v_1 \wedge u_2=-u_1 \wedge v_2, \tag{2}$$

where this is an equality in $\bigwedge^2\mathbb{R}^d$.

Suppose that $\omega \in \bigwedge^2\mathbb{R}^d$. If $\dim(\text{span}_{\mathbb R}(u_1,u_2) \cap \text{span}_{\mathbb R}(v_1,v_2)) \ge 1$ then $\omega=u_1 \wedge u_2 - v_1 \wedge v_2$ is decomposable in $\bigwedge^2\mathbb{R}^d$. Otherwise, $u_1,u_2,v_1,v_2$ are linearly independent over $\mathbb R$, which violates equation $(2)$.

I don't see an immediate generalization of this proof for $k \ge 3$. By expanding $$\omega=(u_1+iv_1) \wedge (u_2+iv_2) \wedge \dots \wedge (u_k+iv_k)$$ we get more than two summands in the real part.

Answer 1

The question can be reformulated as follows: is $$ Gr_{\mathbb{C}}(k,d) \cap \mathbb{P}_{\mathbb{R}}(\wedge^k \mathbb{R}^d) = Gr_{\mathbb{R}}(k,d)? $$ The answer is positive, because Plücker equations of the Grassmannian have real (in fact even integer) coefficients.

Alternatively, this can be shown as follows. Decomposibility of $\omega$ is equivalent to the fact that the dimension of the space of vectors annihilating $\omega$ (by wedge products) is equal to $k$. This is equivalent to a rank condition for a system of linear equations with real coefficients, and its enough to note that its rank over $\mathbb{R}$ is the same as its rank over $\mathbb{C}$.