# Infinite dimensional vector space has almost complex structure if and only if it is 'even-dimensional'?

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 $$S \cap U = \{0_V\}$$)

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.

• So it would seem (conditional on choice). – WoolierThanThou Jan 23 at 6:43
• 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. – WoolierThanThou Feb 3 at 6:47
• 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 $:3+T_T$. – WoolierThanThou Feb 3 at 7:05
• 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. – WoolierThanThou Feb 3 at 13:08
• "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? – WoolierThanThou Feb 3 at 13:36

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.

• 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 $U= \mathbb R-\text{span}(iE)$ ? – John Smith Kyon Jan 22 at 10:28
• 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... – GreginGre Jan 22 at 16:52
• Thanks. $E \cup iE$ is an $\mathbb R$-basis of $(V,J)$, but you didn't use this fact? – John Smith Kyon Jan 23 at 1:55
• 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. – GreginGre Jan 23 at 8:16
• Yeah, I was just checking. Thanks. – John Smith Kyon Jan 24 at 3:34

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 $$J(s \oplus u) := - \theta^{-1}(u) \oplus \theta(s)$$.