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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 $V^{\mathbb C}$.

BCLC
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2 Answers2

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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:

  • this construction is not canonical (functorial), and every choice of $\mathbb C$-basis yields such an isomorphism.
  • given the $\mathbb R$-subspace $V$ of $W$ we have $W=V\oplus iV$ (internal direct sum).
  • given any isomorphism $\phi\colon\mathbb C\otimes_{\mathbb R}V\to W$ we have that $\phi(V)$ is an $\mathbb R$-subspace of $W$ and $W=\phi(V)\oplus i\phi(V)$.

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.

Andrew Hubery
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  • Thanks Andrew Hubery. Your last paragraph seems to suggest there might be a misunderstanding. Both $\mathbb C$ and its conjugate have $\mathbb R+0i$ as internal complexifications. Both $\mathbb C$ and its conjugate have $\mathbb 0+Ri$ as internal complexifications. But only $\mathbb C$ has $\mathbb R$ as its external complexification. For external: I view $\mathbb R^{external-\mathbb C} = \mathbb C$ as not merely isomorphism but literal equality in the sense that $\mathbb C = (\mathbb R^2,J)$, $J(v,w)=(-w,v)$ as literal equality. – BCLC Jan 30 '20 at 14:02
  • For internal: I view $(\mathbb R+0i)^{internal-\mathbb C} = \mathbb C = (0+\mathbb Ri)^{internal-\mathbb C}$ as literal equality (see the definition of internal vs external complexifications in the linked previous question please) since $\mathbb R+0i \bigoplus i(\mathbb R+0i) = 0+\mathbb Ri \bigoplus i(0+\mathbb Ri) = \mathbb C$ and $\mathbb R+0i \cap i(\mathbb R+0i) = 0$ and $i(\mathbb R+0i) = \mathbb R+0i$ – BCLC Jan 30 '20 at 14:05
  • You said 'Every $\mathbb C$-vector space $W$ has a real form (or as you might say, is isomorphic to the external complexification of some $V$).' Here's what I understand: Every $W$ is the literal internal complexification of some real subspace $V$ and is isomorphic to the external complexification of $V$ and is isomorphic to the external complexification of some $U$ real vector space that isn't a subset of $W$ (pick your favourite $U$ isomorphic to $V$). My question is when or only when is $W$ literally equal to and not merely isomorphic to the external complexification of some $U$. – BCLC Jan 30 '20 at 14:10
  • ' 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)' Well, $\mathbb C$ is the external complexification of $\mathbb R$ and then $\mathbb C^2$ is for $\mathbb R^2$ (Here I view $\mathbb C^2 = (\mathbb R^2 \times \mathbb R^2,J \times J)$ such that $(\mathbb C^2){\mathbb R} = \mathbb R^2 \times \mathbb R^2$ rather than $(\mathbb C^2){\mathbb R} = \mathbb R^4$. Of course, $(\mathbb C^2)_{\mathbb R} = \mathbb R^2 \times \mathbb R^2 \cong \mathbb R^4$) – BCLC Jan 30 '20 at 14:12
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    Let $V$ be a real vector space. Then $W=(V\times V,I)$ becomes a complex vector space, where $I(u,v):=(-v,u)$, and the 'conjugate' is $\bar W=(V\times V,-I)$. In this case both $W$ and $\bar W$ are already given as external direct sums, and as the 'external complexification' of $V$ is precisely $(V\times V,I)$, only $W$ can be equal to this. (The external complexification of the second copy of $V$ gives $\bar W$ after swapping the order of the components.) – Andrew Hubery Jan 30 '20 at 14:24
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    My point is therefore that the question only makes sense once you have already written $W$ as an external product of spaces, and then there is only one way it can be equal to the external complexification, which is when we have $W=(V\times V,I)$. An arbitrary complex vector space is not an external product of two real spaces, and so the question does not make sense. – Andrew Hubery Jan 30 '20 at 14:27
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    Explicitly, for you favourite one dimension examples, am I at liberty to write $\mathbb C=\mathbb R\times\mathbb R$ such that $a+bi=(b,a)$? If so, then the external complexification is $\bar{\mathbb C}$. – Andrew Hubery Jan 30 '20 at 14:33
  • Andrew Hubery, thanks. I was thinking about your statement 'An arbitrary complex vector space is not an external product of two real spaces, and so the question does not make sense.' and so went to an earlier question: Based on the comments here, can we say that 1. a complex vector space $V$ is the external complexification of some real vector space $W$ if and only if $V_{\mathbb R} = T^2$ for some real vector space $T$ 2. if (1) is true, then the way to prove (1) is to choose $T=W$ for both the 'if' part and the 'only if' part? – BCLC Feb 03 '20 at 11:38
  • As for your comment that the external complexification of $\mathbb R$ being $\overline{\mathbb C}$, well, that's if we do a different definition of external complexification...though I think you can do away with this example and focus on your stronger argument that 'An arbitrary complex vector space is not an external product of two real spaces, and so the question does not make sense.' – BCLC Feb 03 '20 at 11:40
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Based on other answer of Andrew Hubery, the comments on it and the comments in answer here of WoolierThanThou, I guess that for a real vector space $V$,

$V \cong W^2$ for some real vector space $W$ if and only if $V$ has an almost complex structure $K$

and then

$V = W^2$ for some real vector space $W$ if and only if $V$ has an almost complex structure $K$ such that $(V,K)$ is the external complexification of some real vector space $T$.

It can be shown that we can or will get $T=W$ in proving either directions.

Anyway, there's not really a bigger answer than this since apparently for complex vector space $E$, $E_{\mathbb R}$ is not necessarily equal to the Cartesian square of some $W$.

BCLC
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