1
$\begingroup$

Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.

Are $(\bigwedge^k V)^{\mathbb{C}}$ and $\bigwedge^k (V^{\mathbb{C}})$ naturally isomorphic?

They both have the same complex dimension. I guess that one possible map $\bigwedge^k (V^{\mathbb{C}}) \to (\bigwedge^k V)^{\mathbb{C}}$ is given by $$ v_1+i\tilde v_1 \wedge \dots \wedge v_k+i\tilde v_k \to (v_1 \wedge \dots \wedge v_k)+i\,\,( \tilde v_1 \wedge \dots \wedge \tilde v_k).$$

Or maybe the correct one is given by "expanding explicitly" $v_1+i\tilde v_1 \wedge \dots \wedge v_k+i\tilde v_k$ in the usual way?

I think I might be confusing here with the way that exterior power commutes with direct sum, since we can think on $V^{\mathbb{C}}$ as $V\oplus iV$, right? (though then we need to "remember" the way we multiply $v+iw$ by complex numbers...

Edit: The map proposed above cannot be the correct one, since it is not injective. I guess we need to define our map by using a basis (and the basis it induces on the exterior algebra or complexification) and then prove that the map we built is independent of the original basis chosen.

However, maybe there is a basis-free approach? (e.g. using some universal properties etc).

$\endgroup$
1
$\begingroup$

It depends what field you take the exterior power over.

$\Lambda_{\mathbb R}(V)\otimes\mathbb C = \Lambda_{\mathbb C}(V\otimes \mathbb C)$ is true and the map is just by rewriting eg $(v\otimes i)\wedge(w\otimes 1) = (v\wedge w)\otimes i$.

On the other hand if you use the real exterior power on both sides the dimensions won’t be the same. In degree $k$ you get ${2n \choose k} \neq 2{n\choose k}$. (Not to mention the thing on the right wont be a complex vector space...)

$\endgroup$
  • $\begingroup$ Thanks. Indeed, I meant to take powers over $\mathbb{C}$. $\endgroup$ – Asaf Shachar Jan 28 at 15:07
  • $\begingroup$ Sure, no problem! $\endgroup$ – Ben Jan 28 at 15:23
0
$\begingroup$

Here is an attempt to write explicitly the isomorphism $(\bigwedge^k V)^{\mathbb{C}} \to \bigwedge^k (V^\mathbb{C})$. First, we fix some $z \in \mathbb{C}$. Then we look at the map

$$ (v_1,\ldots,v_k) \to z\cdot(v_1 \otimes 1) \wedge \ldots \wedge (v_k \otimes 1).$$

This map is multilinear and alternating, hence it lifts to an $\mathbb{R}$-linear map $\phi_z:\bigwedge^k V \to \bigwedge^k (V^\mathbb{C})$, given by

$$ v_1 \wedge \ldots \wedge v_k \to z \cdot(v_1 \otimes 1) \wedge \ldots \wedge (v_k \otimes 1).$$

Now, we define an $\mathbb{R}$-bilinear map $\bigwedge^k V \times \mathbb{C} \to \bigwedge^k (V^\mathbb{C})$ given by $(\omega,z) \to \phi_z(\omega)$.

This lifts to an $\mathbb{R}$-linear map $T:\bigwedge^k V \otimes_{\mathbb{R}} \mathbb{C} = (\bigwedge^k V)^{\mathbb{C}} \to \bigwedge^k (V^\mathbb{C})$, $T(\omega \otimes z)= \phi_z(\omega)$.

So, to conclude

$$ T\big( (v_1 \wedge \ldots \wedge v_k) \otimes z \big)=z \cdot(v_1 \otimes 1) \wedge \ldots \wedge (v_k \otimes 1).$$

It is easy to that $T$ is complex-linear isomorphism.

$\endgroup$

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.