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

BCLC
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3 Answers3

5

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.

GreginGre
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  • 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 $U= \mathbb R-\text{span}(iE)$ ? – BCLC Jan 22 '20 at 10:28
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    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 '20 at 16:52
  • Thanks. $E \cup iE$ is an $\mathbb R$-basis of $(V,J)$, but you didn't use this fact? – BCLC Jan 23 '20 at 01:55
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    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 '20 at 08:16
  • Yeah, I was just checking. Thanks. – BCLC Jan 24 '20 at 03:34
  • Follow-up question about the 'maybe even' in my post (it was not in original version of my post when you answered): Okay so $V$ has an almost complex structure if and only if $V \cong W^2$ for some $W$, but does $V$ have an almost complex structure if and only if $V = W^2$ (as in literally and not just in isomorphism) for some $W$? For example, for $V=(\mathbb C^n){\mathbb R}$ and $V=(\overline{\mathbb C^n}){\mathbb R}$ (and I guess$V=(\overline{\mathbb C}^n)_{\mathbb R}$), we get $W = \mathbb R^n$. (FYI: Related to this) – BCLC Feb 03 '20 at 06:36
  • Update: WoolierThanThou and I have a discussion in comments in WoolierThanThou's answer. – BCLC Feb 03 '20 at 13:29
  • GreginGre, is it that we use axiom of choice for proving that 5 implies (6 or) 7, but we actually don't need axiom of choice for proving that 6 (or 7) implies 5? Based on this, I think having both $S$ and $U$ is equivalent to both $J$ and a $\sigma$. Therefore, $J$ alone doesn't give us $S$ and $U$. However, $S$ and $U$ alone give us $J$ (and $\sigma$). – BCLC Feb 28 '20 at 11:36
5

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.

WoolierThanThou
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  • Wait so $\mathbb R$-vector space $V$ has an almost complex structure if and only if $V \cong W^2$ if and only if $V=S \bigoplus U$ if and only if $V$ is infinite dimensional or finite but even-dimensional? – BCLC Jan 23 '20 at 01:59
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    So it would seem (conditional on choice). – WoolierThanThou Jan 23 '20 at 06:43
  • Thanks! I'll pick your answer since you proved something stronger. – BCLC Jan 24 '20 at 03:36
  • Follow-up question about the 'maybe even' in my post (it was not in original version of my post when you answered): Okay so $V$ has an almost complex structure if and only if $V \cong W^2$ for some $W$, but does $V$ have an almost complex structure if and only if $V = W^2$ (as in literally and not just in isomorphism) for some $W$? For example, for $V=(\mathbb C^n){\mathbb R}$ and $V=(\overline{\mathbb C^n}){\mathbb R}$ (and I guess$V=(\overline{\mathbb C}^n)_{\mathbb R}$), we get $W = \mathbb R^n$. (FYI: Related to this) – BCLC Feb 03 '20 at 06:33
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    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 03 '20 at 06:47
  • Well maybe it's a matter of convention. I view $\mathbb C_{\mathbb R}=\mathbb R^2$ because I view $\mathbb C:=\mathbb R^{\mathbb C}:=(\mathbb R^2,J(v,w))$. Anyway, that's just for our familiar $\mathbb C$ and $\mathbb R$. For arbitrary $V$ and $W$... – BCLC Feb 03 '20 at 06:53
  • ...right so I wanted to see counterexamples - please - where we don't have $V=W^2$ (but we still assume axiom of choice). I mean, we already have $V$ as isomorphic to an external direct sum $W^2$ and then $V$ literally equal to an internal direct sum of isomorphic $S$ and $U$. Just wondering if we might have $S=T \times 0$ and $U=0 \times T$ for some $T$. – BCLC Feb 03 '20 at 06:54
  • Okay so yeah I'm wrong. Please just give me some counterexamples then. – BCLC Feb 03 '20 at 06:56
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    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 03 '20 at 07:05
  • Thanks WoolierThanThou. Well we might say that $\text{:) + XD} = u \times 0$ and $\text{3 +} T_T = 0 \times u$ (Oh I just figure out that $T_T$ refers to $\text{T_T}$. Clever.) for some...but I guess those equalities are actually isomorphisms. Anyway I really thought about your comment in terms of free modules and stuff, but also in terms of almost complex structures: – BCLC Feb 03 '20 at 11:35
  • If we have '$V \cong W^2$ for some $W$ if and only if $V$ has an almost complex structure $K$', then what is $V = W^2$ equivalent to? My guess: 1. The equivalent condition is that '$V$ has an almost complex structure $K$ such that $(V,K)$ is the external complexification of some $\mathbb R$-vector space $T$' 2. In proving this equivalence in either direction we get $T=W$ – BCLC Feb 03 '20 at 11:35
  • The point is that exact set equalities don't actually mean anything as vector spaces - you caring about the differences between isomorphic objects necessarily comes from you caring about your objects as something else. If I give you the isomorphic vector spaces $V$ and $V'$ where one of them, I promise, is $W^2$ and the other arises from somewhere different, you will never be able to tell which one is which, because as vector spaces, they are exactly identical. – WoolierThanThou Feb 03 '20 at 12:11
  • Wait so is 'my guess' wrong? I now understand how existence of almost complex structure is not equivalent to literal equality with some $W^2$ and now am just wondering about the equivalent strengthening of 'existence of almost complex structure' for the strengthening of $V \cong W^2$ to $V = W^2$. – BCLC Feb 03 '20 at 12:53
  • There's nothing in the data of $V$ as a vector space that would ever tell you that it's an external complexification, right? Because the vector space data can only ever tell you its isomorphism type. – WoolierThanThou Feb 03 '20 at 12:56
  • Is it wrong to say '$V=W^2$ (as in literal $=$ and not $\cong$) for some real $W$ if and only if $(V,K)=T^{\mathbb C}$ for some $T$ and $K$' ? My proof: For 'only if', choose $T=W$ and $K(w_1,w_2):=(-w_2,w_1)$. For 'if', $V=(V,K)_{\mathbb R}=T^2$. Choose $W=T$. QED – BCLC Feb 03 '20 at 13:04
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    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 03 '20 at 13:08
  • WoolierThanThou, is this the part in the legal drama/comedy where the defense attorney says 'Objection: Doesn't answer question' and then the witness says 'No, but--' and then the defense attorney interrupts with '--Thank you. You answered the question.' ? – BCLC Feb 03 '20 at 13:12
  • Specifically, what does it mean that 'choice of $K$' means 'up to isomorphism' ? Look, it's like this. $V \cong W^2$ if and only if $V$ has an almost complex structure. Now that it's been established that '$V = W^2$' is strictly stronger than '$V \cong W^2$', what possible strengthening is there of '$V$ has an almost complex structure' to be equivalent to '$V = W^2$' ? Ok, say I'm (insert some less strong word than 'wrong') about saying $V$ is the external complexification of $W$ (or of some $T$ that turns out to be $W$). Your answer then is that there is no such strengthening because... – BCLC Feb 03 '20 at 13:15
  • ...'exact set equalities don't actually mean anything as vector spaces' ? Actually, in this case, I'm trying to make sense about exact set equalities. Otherwise, it's pointless to talk about external complexifications. Okay, it's pointless to talk about external complexifications, but whenever we do, we are talking about exact set equalities. Therefore, it's meaningful to ask about $V=W^2$ in the same way that it's meaningful to ask about external complexifications. Am I wrong? – BCLC Feb 03 '20 at 13:17
  • I simply disagree that $V=W^2$ is a stronger statement in $V\simeq W^2$ in anything but semantics. Say $V=W^2$, but I simply give you a basis such that you can't recognise this fact. What, as a vector space, have you really lost? What's so important about you thinking of $V$ as $W^2$ that you would refuse to consider certain bases? – WoolierThanThou Feb 03 '20 at 13:24
  • Well in terms of isomorphisms, nothing really (assuming I understand your question and what you mean by 'semantics'), but I'm really talking about the case of 'semantics'/literal equalities/set equalities. (Oh actually, I'm talking about literal equalities in that not only are the underlying sets equal but also are the addition and scalar multiplication structures the same. – BCLC Feb 03 '20 at 13:28
  • I assume this, rather than 'underlying set equalities', is what you mean by 'set equalities'). In fact the very question about external complexifications I had is about 'semantics', right? I mean, the meaning in discussing $V=W^2$ vs $V \cong W^2$ is the same thing as the meaning in discussing when a $\mathbb C$-vector space is the external complexification of something in re external vs internal complexifications right? – BCLC Feb 03 '20 at 13:28
  • Anyway, re 'Wrong is a strong word.', um: Thank you. You answered the (that) question...I guess – BCLC Feb 03 '20 at 13:32
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    "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 03 '20 at 13:36
  • WoolierThanThou thanks but new question now: Is it that we use axiom of choice for proving that 5 implies (6 or) 7, but we actually don't need axiom of choice for proving that 6 (or 7) implies 5? Based on this, I think having both $S$ and $U$ is equivalent to both $J$ and a $\sigma$. Therefore, $J$ alone doesn't give us $S$ and $U$. However, $S$ and $U$ alone give us $J$ (and $\sigma$). – BCLC Feb 28 '20 at 11:36
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    $6$ or $7$ implies $5$, yes, whether we have choice or not. The axiom of choice gives us that any infinite dimensional vector space satisfies $7$. Thus, $5$ and $7$ become equivalent in the world of infinite dimensional vector spaces. $5$ alone doesn't seem to give you a good way of "cutting the space in half" without some basis (where this amounts to cutting the index set in half). – WoolierThanThou Feb 28 '20 at 12:24
  • WoolierThanThou THANK YOU – BCLC Mar 02 '20 at 03:48
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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 $J(s \oplus u) := - \theta^{-1}(u) \oplus \theta(s)$.

BCLC
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