Infinite dimensional vector space has almost complex structure if and only if it is ‘even-dimensional’

abstract-algebraalmost-complexcomplex-analysiscomplex-geometrylinear algebra

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

Best Answer

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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