# A complex vector subspace is stable for complex conjugation if and only if it is the complexification of a real vector subspace

A conjugation on a complex vector space $$U$$ is a map $$c\colon U\to U$$ that satisfies the following three properties:

• $$c(u + u^\prime) = c(u) + c(u^\prime)$$ for every $$u,u^\prime\in U$$;
• $$c(\alpha u) = \bar a c(u)$$ for every $$u\in U$$ and $$\alpha\in \mathbb C$$;
• $$c(c(u)) = u$$ for every $$u\in U$$.

We write $$U_c$$ for the set of all $$u\in U$$ such that $$c(u) = u$$.

I'm trying to show that if $$V$$ is a (complex) subspace of a complex vector space $$U$$ equipped with a conjugation $$c$$ such that $$c(v) = V$$, then $$V$$ is (isomorphic to) the complexification of the real vector subspace $$V\cap U_c$$.

Recall that if V is a vector space over the real numbers then it's complexification is a complex vector space $$V_{\mathbb C}$$ equipped with an $$\mathbb R$$-linear map $$\iota\colon V\hookrightarrow V_{\mathbb C}$$ such that the following universal property holds: for every complex vector space $$F$$ and for every $$\mathbb R$$-linear map $$\phi\colon V\to F$$, there exists a unique $$\mathbb C$$-linear map $$\tilde\phi\colon V_{\mathbb C}\to F$$ such that the diagram $$\require{AMScd} \begin{CD} V @>\iota>> V_{\mathbb C}\\ @| @VV\tilde\phi V \\ V @>>\phi> F \end{CD}$$ commutes.

In my case, I want to show that $$V$$ equipped with the obvious injection $$V\cap U_c\hookrightarrow V$$ satisfies this universal property. Given some $$\phi\colon V\cap U_c\to F$$ I tried to define $$\tilde\phi$$ by noticing that a vector $$v\in V$$ can be written as $$v = \frac{v + c(v)}{2} + \frac{v - c(v)}{2}$$ where $$\frac{v + c(v)}2\in V\cap U_c$$, and by putting $$\tilde\phi(v) = \phi\Bigl(\frac{v + c(v)}{2}\Bigr)\text{,}\qquad \tilde\phi(v) = \phi\Bigl(\frac{v + c(v)}{2}\Bigr) +i\phi\Bigl(\frac{v - c(v)}{2}\Bigr)\text{,}\qquad \cdots$$ but none of this choices work.

How can I show my statement working only with the "abstract" characterization of $$V_{\mathbb C}$$ I gave? (i know that $$\mathbb V_{\mathbb C}\cong \mathbb C\otimes V$$, but I want to be as implementation independent as possible).

You need to take $$\tilde\phi(v)=\phi\left(\frac{v + c(v)}{2}\right) + i\phi\left(\frac{v - c(v)}{2i}\right)$$ because $$c\left(\frac{v-c(v)}{2}\right)=\frac{c(v)-c(c(v))}{2}=-\frac{v-c(v)}{2}$$ which is not in $$V_c=V\cap U_c$$. On the other hand $$c\left(\frac{v-c(v)}{2i}\right)=\frac{c(v)-c(c(v))}{2\bar{i}}=-\frac{c(v)-v}{2i} = \frac{v-c(v)}{2i}$$ so $$(v-c(v))/2i\in V_c$$ and you can apply $$\phi$$ to it and get what you need.
I don't understand why you start with $$U$$ and restrict to $$V$$. You may as well start with $$V$$ (i.e. assume $$V$$ is all $$U$$.)
• Whops, I forgot a bunch of $i$s... Commented Dec 11, 2023 at 2:35