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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:

I understand for a finite dimensional $\mathbb R-$vector space $V=(V,\text{Add}_V: V^2 \to V,s_V: \mathbb R \times V \to V)$, the following are equivalent

  1. $\dim V$ even
  2. $V$ has an almost complex structure $J: V \to V$
  3. $V$ has a complex structure $s_V^{\#}: \mathbb C \times V \to V$ that agrees with its real structure: $s_V^{\#} (r,v)=s_V(r,v)$, for any $r \in \mathbb R$ and $v \in V$
  4. if and only if $V \cong \mathbb R^{2n} \cong (\mathbb R^{n})^2$ for some positive integer $n$ (that turns out to be half of $\dim V$) if and only if $V \cong$ (maybe even $=$) $W^2=W \bigoplus W$ for some $\mathbb R-$vector space $W$.

The last condition makes me think that the property 'even-dimensional' for finite-dimensional $V$ is generalised by the property '$V \cong W^2$ for some $\mathbb R-$vector space $W$' for finite or infinite dimensional $V$.

Question: For $V$ finite or infinite dimensional $\mathbb R-$vector space, are the following equivalent?

  1. $V$ has an almost complex structure $J: V \to V$

  2. Externally, $V \cong$ (maybe even $=$) $W^2=W \bigoplus W$ for some $\mathbb R-$ vector space $W$

  3. Internally, $V=S \bigoplus U$ for some $\mathbb R-$ vector subspaces $S$ and $U$ of $V$ with $S \cong U$ (and $S \cap U = \{0_V\}$)

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GreginGre's solution is, of course, perfectly lovely, but if we're just killing this with choice, I guess you can also prove it as follows:

Let $V$ be infinite dimensional and, using Zorn's Lemma, let $\{e_i\}_{i\in I}$ be a basis for $V$. Using choice again, there exists $I_1$ and $I_2$ such that both $I_1\cap I_2=\emptyset,$ $I_1\cup I_2=I$ and there exists a bijection $\varphi: I_1\to I_2$. Thus, let $S=\textrm{span}\{e_i\}_{i\in I_1}$ and $U=\textrm{span}\{e_i\}_{i\in I_2}$. Then, $V=S\oplus U$ and $A:S\to U$ given by $e_i\mapsto e_{\varphi(i)}$ is a linear isomorphism of the two. This just proves that any infinite dimensional vector space admits such a decomposition, so there is only something to prove in the finite dimensional case.

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    $\begingroup$ So it would seem (conditional on choice). $\endgroup$ – WoolierThanThou Jan 23 at 6:43
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    $\begingroup$ How is that literal and not up to isomorphism? You've identified those spans with $\mathbb{R}^n$. They weren't $\mathbb{R}^n$ to begin with. The same problem will arise whenever $V$ is not defined as a direct sum. $\endgroup$ – WoolierThanThou Feb 3 at 6:47
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    $\begingroup$ Let $V$ be the abstract $\mathbb{R}$ vector space with basis $\{ :),XD, :3,T_T\}$ which allows a complex structure. However, how is it a direct sum of two identical subspaces? The symbol representing a given vector is completely distinct. There's no reason to think $ :)+XD$ is the same as $:3+T_T$. $\endgroup$ – WoolierThanThou Feb 3 at 7:05
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    $\begingroup$ Wrong is a strong word. I'd say your proof involves a choice of $K$, which, to me, is as good as just saying that everything is up to isomorphism. For instance, I don't think there's a natural transformation between either of these functors and the identity. $\endgroup$ – WoolierThanThou Feb 3 at 13:08
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    $\begingroup$ "Are the addition and scalar multiplication structures the same." That's exactly the same as asking about isomorphism class, up to the names that you have given elements (and remember: "That which we call a rose by any other name would smell as sweet"). Question: I have two groups $\{a,b\}$ and $\{c,d\}$ with neutral elements $a$ and $c$ respectively. Are these groups different? $\endgroup$ – WoolierThanThou Feb 3 at 13:36
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Yes, they are. Note that 6. and 7. are clearly equivalent (if we have 6. take for $S$ and $U$ the images of $W\times \{0\}$ and $\{0\}\times W$ under an isomorphism $W^2\overset{\sim}{\to} V$. If we have 7., then $V\simeq S\times U\simeq S\times S$, so take $W=S$.)

Assume that we have $7.$ Since $S$ and $U$ are isomorphic , their bases have same cardinality (countable or not). Pick $(s_i)_{i\in I}$ a basis of $S$, and $(u_i)_{i\in I}$ a basis of $U$ (we can index the two bases by the same set, thanks to the previous remark).

Setting $J(e_i)=u_i$ and $J(u_i)=-e_i$ for all $i\in I$ yields an endomorphism $J$ satisyfing $J^2=-Id_V$.

Conversely, assume that we have an endomorphism $J$ of $V$ satisfying $J^2=-Id_V$.

The map $\mathbb{C}\times V\to {V}$ sending $(a+bi,v)$ to $av+ bJ(v)$ endows $V$ with the structure of a complex vector space which agrees on $\mathbb{R}\times V$ to its real structure.

Now pick a complex basis $(s_i)_{i\in I}$ of $V$, and set $u_i=i\cdot s_i=J(s_i)$. Then, gluing $(s_i)_{i\in I}$ and $(u_i)_{i\in I}$, we obtain a real basis of $V$. The real subspaces $S=Span_\mathbb{R}(s_i)$ and $U=Span_\mathbb{R}(u_i)$ then satisfy the conditions of 7.

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  • $\begingroup$ Is this what you're doing? For $\mathbb C$-basis $E$ of $(V,J)$, we get that the union $E \cup iE$ is an $\mathbb R$-basis of both $V$ and $(V,J)$ and then you choose $S=\mathbb R-\text{span}(E)$ and $U= \mathbb R-\text{span}(iE)$ ? $\endgroup$ – John Smith Kyon Jan 22 at 10:28
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    $\begingroup$ No. $E\cup iE$ is a basis of the $\mathbb{R}$-vector space $V$, period. Your interpretation of $S$ and $U$ are correct. I just prefer to pformulate it the way i did, making the complex structure diseappear. But really, I copied the standard proof in the finite dimensional case... $\endgroup$ – GreginGre Jan 22 at 16:52
  • $\begingroup$ Thanks. $E \cup iE$ is an $\mathbb R$-basis of $(V,J)$, but you didn't use this fact? $\endgroup$ – John Smith Kyon Jan 23 at 1:55
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    $\begingroup$ I don' t know what you mean by a basis of $(V,J)$. If this is a way to say that it is a basis of the complex vector space $V$ viewed as a real vector space, there is no need to do so, because it IS the real vector space we started with. $\endgroup$ – GreginGre Jan 23 at 8:16
  • $\begingroup$ Yeah, I was just checking. Thanks. $\endgroup$ – John Smith Kyon Jan 24 at 3:34
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As a supplement to the other answers, I'm going to prove (6 or) 7 implies 5 without axiom of choice. This is based on Joppy's answer and WoolierThanThou's comment:

Given an isomorphism $\theta: S \to U$, define $J: V \to V$ on the direct sum $V = S \bigoplus U$ by setting $J(s \oplus u) := - \theta^{-1}(u) \oplus \theta(s)$.

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