Natural isomorphism between the complexification of $V$ and $V^2$, when $V$ is already a complex vector space?

Question

I'm considering the question of:

What happens if you complexify a vector space that is already complex?

I basically believe the following. I take complexification of a complex vector space to mean the complexification of the underlying real vector space. Let $V$ be the space, an let $V^\Bbb C = V\otimes_\Bbb R \Bbb C$ be the complexification of $V$. If $V$ has $n$ complex dimensions, then it is $2n$-dimensional as a real space, and then $V^\Bbb C$ is $2n$-dimensional as a complex space. Thus, I think that $$\tag{*}V^\Bbb C \cong V^2$$ as complex vector spaces.

Main questions:

1. Is the above isomorphism true? (Hopefully yes, unless my dimension-counting is wrong).
2. Is it natural? (I felt like it should be, but...).
3. Is it natural in the restricted category of complex vector spaces endowed with a real structure (equivalently, a conjugation, meaning an antilinear involution), and with morphisms that preserve the real structure?

I think the answers are yes, no(?), yes. Here are my thoughts on it. We can decompose $V$ as $$ V = V_\Bbb R \oplus iV_\Bbb R, $$ where $V_\Bbb R$ is the subspace left invariant under the conjugation. In particular, we can decompose vector in $z\in V$ according to this direct sum, and I'll denote the components the real and imaginary parts like so: $$ z=\frac{z+\bar z}2 + i\cdot\frac{z-\bar z}{2i}=: \Re z + i\Im z. $$ Note in particular that $\Re$ and $\Im$ are real-linear maps $V\to V_\Bbb R$.

I then propose the following map for $(*)$, where I write $|z,w|$ for the general element $z\otimes 1 + w\otimes i$ of $V^\Bbb C$, and ``$(z,w)$ for elements of $V^2$. I define $$ |z,w| = \left|\Re z+i\Im z, \Re w+i\Im w\right| \mapsto \left(\Re z+i\Re w, -\Im w +i \Im z\right) \in V^2. $$ If my algebra was right, this map should be a (complex-)linear isomorphism. In particular, the complex scalar product on either side should agree, most notably $$ i\left|z,w\right| = \left|-w,z\right| \mapsto (-\Re w+i\Re z, -\Im z -i \Im w) = i\left(\Re z+i\Re w, -\Im w +i \Im z\right). $$

However:

• The map seems to require a choice for the real structure on $V$, so I don't think it's natural in the full category of linear spaces.
• That said, I believe I calculated it to indeed be natural over the maps that preserve the real structure.

Soooo, am I on the right track here? Is there anything more to say?

Answer 1

You are correct. The natural isomorphism is that if $V$ is a complex vector space, not necessarily finite-dimensional, then the complexification $V \otimes_{\mathbb{R}} \mathbb{C}$ of the underlying real vector space of $V$ can be written

$$V \otimes_{\mathbb{R}} \mathbb{C} \cong V \otimes_{\mathbb{C}} \left( \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \right)$$

where to understand the complex structure here we need to understand how $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ decomposes as a $(\mathbb{C}, \mathbb{C})$-bimodule. The answer (and this is a nice exercise) is that it decomposes into a direct sum of $\mathbb{C}$ (with the usual bimodule structure) and $\overline{\mathbb{C}}$ (where one of the multiplications is twisted by complex conjugation). Explicitly, $\mathbb{C}$ is spanned by $1 \otimes i + i \otimes 1$, and $\overline{\mathbb{C}}$ is spanned by $1 \otimes i - i \otimes 1$. This gives a natural isomorphism

$$V \otimes_{\mathbb{R}} \mathbb{C} \cong V \oplus \overline{V}$$

where $\overline{V}$ is the conjugate of $V$, namely $V$ but with scalar multiplication twisted by complex conjugation. If we are working with bare vector spaces then $V$ is always isomorphic to $\overline{V}$ but not naturally, although if $V$ has a real structure then we have a natural isomorphism. But we can work with, for example, complex representations of a group or complex vector bundles or complex Lie algebras, and in that generality $V$ can actually fail to be isomorphic to $\overline{V}$ and therefore can fail to have a real structure. (In fact a real structure can equivalently be defined as a sufficiently well-behaved isomorphism $V \cong \overline{V}$.)

This generalizes to a finite Galois extension $L/K$ with Galois group $G$ as follows. $L \otimes_K L$ decomposes into a direct sum of bimodules $L^g$, which is $L$ with left and right multiplication but where one of the multiplications has been twisted by $g \in G$, and this gives that if $V$ is an $L$-vector space then

$$V \otimes_K L \cong V \otimes_L (L \otimes_K L) \cong \bigoplus_{g \in G} V^g$$

where $V^g$ is $V$ but with scalar multiplication twisted by $g \in G$.