I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier
I have several questions on the concepts of almost complex structures and complexification. Here is one:
Based on my previous question, it appears that (at least under axiom of choice), every $\mathbb C$-vector space $W$ both has conjugations and is the internal complexification of several of its $\mathbb R$-subspaces. However, it seems that $W$ need not be the external complexification of some $\mathbb R$-vector space $U$.
- For example: $\mathbb C$ is the external complexification of $\mathbb R$ and the internal complexification of $\mathbb R+0i$, $0+\mathbb Ri$ and all its other 1-dimensional $\mathbb R$-subspaces. However, while $\overline{\mathbb C}$ is the internal complexification of all its 1-dimensional $\mathbb R$-subspaces, $\overline{\mathbb C}$ is never the external complexification of $\mathbb R$ (at least under the standard definition of complexification; we might say $\overline{\mathbb C}$ is the external anti-complexification of $\mathbb R$).
Question: What are sufficient or necessary conditions for a complex vector space to be the external complexification of a real vector space?
All I got so far: In Roman (Exercise 1.26), we get that for any real $\mathbb R$-vector space $V$, a $\mathbb C$-subspace $A$ of $V$'s complexification $V^{\mathbb C}$ is (literally equal to, not merely isomorphic to) the (external) complexification of some $\mathbb R$-subspace $S$ of $V$ if and only if $A$ is a subset of image of the complexification map $cpx: V \to V^{\mathbb C}$, $cpx(v)=(v,0_V)$ if and only if $A$ is a subset of the fixed points of the standard conjugation $\chi: V^{\mathbb C} \to V^{\mathbb C}$, $\chi(v,w) := (v,-w)$ (Actually, $image(cpx)=$ fixed points of $\chi$. )
I guess, then, $W$ is the external complexification of some $\mathbb R$-vector space $U$ if there exists $\mathbb R$-vector space $V$ such that $W$ is a $\mathbb C$-subspace of some $V^{\mathbb C}$ and then $\chi(W) \subseteq W$. In which case, we pick $U=A$, which I think is $U=A=cpx^{-1}(W)$.
For $\overline{\mathbb C}$, maybe we could argue that for any $\mathbb R$-vector space $V$, $\overline{\mathbb C}$ is never a $\mathbb C$-subspace of $V^{\mathbb C}$.
Best Answer
The way I see these constructions is as follows.
complex structures
Given a $\mathbb C$-vector space $W$, we can regard it as an $\mathbb R$-vector space (via restriction of scalars), in which case multiplication by $i$ yields an $\mathbb R$-endomorphism $I$ satisfying $I^2=-\mathrm{id}_W$.
Conversely, for an $\mathbb R$-vector space $W$, a complex structure on $W$ is precisely such an endomorphism $I$ satisfying $I^2=-\mathrm{id}_W$. Given such an $I$, we can then define an action of $\mathbb C$ on $W$ via $(a+bi)\cdot w:=aw+bI(w)$, and in this way $W$ becomes a $\mathbb C$-vector space.
Note that complex structures are not unique (and if we have two such complex structures $I$ and $J$ satisfying $IJ=-JI$, then we get a quaternionic structure...).
real forms
Going in the other direction, starting with an $\mathbb R$-vector space $V$, we can complexify (extend scalars) to obtain the $\mathbb C$-vector space $\mathbb C\otimes_{\mathbb R}V$. One may identify this with the (external) direct sum $V\times V$, together with the complex structure given by $I(u,v):=(-v,u)$.
Conversely, a real form for a $\mathbb C$-vector space $W$ is an isomorphism $\mathbb C\otimes_{\mathbb R}V\cong W$ for some $\mathbb R$-vector space $V$.
main result
Every $\mathbb C$-vector space $W$ has a real form (or as you might say, is isomorphic to the external complexification of some $V$).
In fact, $W$ is isomorphic to the complexification of some $\mathbb R$-subspace $V$. For, take any $\mathbb C$-basis of $W$, and let $V$ be the $\mathbb R$-span of these basis vectors. Then scalar multiplication induces an isomorphism $\mathbb C\otimes_{\mathbb R}V\to W$.
Note:
real forms of subspaces
Suppose we have fixed a real form $\mathbb C\otimes_{\mathbb R}V\cong W$. Then an $\mathbb R$-subspace $V'\leq V$ yields the $\mathbb C$-subspace $\mathbb C\otimes_{\mathbb R}V'$ of $W$. Conversely, given a $\mathbb C$-subspace $W'\leq W$, set $V':=W'\cap V$. This is an $\mathbb R$-subspace of $V$, and $\mathbb C\otimes_{\mathbb R}V'\leq W'$. Thus a $\mathbb C$-subspace $W'\leq W$ is induced from an $\mathbb R$-subspace of $V$ if and only if $\dim_{\mathbb R}(W'\cap V)=\dim_{\mathbb C}W'$.
Note however that, given $W'\leq W$, we can always choose a basis for $W'$ and extend to a basis for $W$. The real span of this basis gives $V'\leq V$ such that $\mathbb C\otimes_{\mathbb R}V'\cong W'$, so every $W'$ is compatible with some real form of $W$.
Finally, I'm not sure it makes sense to ask when a $\mathbb C$-vector space $W$ is equal to the complexification of some $\mathbb R$-vector space (unless that structure is already given). Your question is related to: when is a vector space equal to an external direct sum, so given $W$, when do we have $W=V\times V'$. The problem is that elements of $V\times V'$ are pairs, which is not a structure we can see on $W$. Alternatively, if we are given $W$ as an internal direct sum $W=V\oplus V'$, we have a natural isomorphism $V\times V'\to W$ but this is not equality.