2
$\begingroup$

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.

  1. $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$: 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.

$\endgroup$
6
$\begingroup$

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 $\mathbb{C}$.

$\endgroup$
  • $\begingroup$ Thank you. I think that I was confused about some particular aspects of this problem, which are related to the the fact that we should distinguish whether we are taking the exterior (or tensor) power over $\mathbb R$ or over $\mathbb C$. I will try to clarify my confusion soon... (Anyway I think your answer is definitely relevant to whatever formulation I had in mind...). $\endgroup$ – Asaf Shachar Feb 3 at 11:56
  • $\begingroup$ (Note that TeX codes for diacritical marks don't work in math.SE) $\endgroup$ – Pedro Tamaroff Feb 4 at 13:06
  • $\begingroup$ Indeed, I have now clarified things. The question can be viewed via two different interpretations, depending on which field we are tensoring over. The less trivial case, which interests me more is when we tensor over $\mathbb C$. I am quite sure that your answer is only applicable to the case where we tensor over $\mathbb R$. I have updated the question thoroughly to include a discussion of the different cases, if you are interested. Thanks again for your help. $\endgroup$ – Asaf Shachar Feb 4 at 13:10

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.