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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:

Assumptions, notations and what I understand so far: Let $V$ be a $\mathbb C$-vector space. Let $W$ be an $\mathbb R$-vector space.

  1. Let $V_{\mathbb R}$ be the realification of $V$. For any almost complex structure $I$ on $V_{\mathbb R}$, denote by $(V_{\mathbb R},I)$ as the unique $\mathbb C$-vector space whose complex structure is given $(a+bi) \cdot v := av + bI(v)$. Let $i^{\sharp}$ be the unique almost complex structure on $V_{\mathbb R}$ such that $V=(V_{\mathbb R},i^{\sharp})$.

  2. Let $W^{\mathbb C}$ denote the complexification of $W$ given by $W^{\mathbb C} := (W^2,J)$, where $J$ is the 'canonical' almost complex structure on $W^2$ given by $J(v,w):=(-w,v)$. The map $\chi: W^2 \to W^2$, $\chi(v,w):=(v,-w)$ is such that $\chi^J: W^{\mathbb C} \to W^{\mathbb C}$, which is $\chi$ now viewed as a map on $W^{\mathbb C}$ instead of $W^2$, is the 'canonical' conjugation/real structure. Here, 'canonical' is meant in the sense that we would use $J$ and $\chi$ to define complexifications of $W$ and of elements of $End_{\mathbb R}(W)$. (See here.)

  3. Then the complex conjugate of $V$ is defined $\overline V := (V_{\mathbb R},-i^{\sharp})$.

Question: What exactly is the relationship between the concept of $\overline V$, the conjugation of $V$ and the concept of conjugations/real structures on $V$?

BCLC
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    The second coordinate being the imaginary axis is based solely on your initial identification of $\mathbb R^2$ with $\mathbb C$, and doesn't match with the notion of $i$ you obtain from choosing a complex structure compatible with $-\chi$ as conjugation. For $-\chi$ the imaginary coordinate is the first one and the real coordinate is second. – Ben Feb 06 '20 at 10:11
  • @Ben Thanks. I actually don't quite really get you unless: is your comment perhaps related to this (or this or this)? – BCLC Feb 25 '20 at 08:41
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    The complex conjugate space of $(V, J)$ is $(V, -J)$. Maps that were previously complex-linear become complex-antilinear. – Joppy Feb 27 '20 at 07:05
  • @Joppy Thanks. Actually my question was not so clear. I edited it. – BCLC Feb 28 '20 at 11:30
  • @Ben Actually, I realise my question was not so clear. I edited it. – BCLC Feb 28 '20 at 11:30
  • @Joppy Oh I forgot to mention I added bounty. ah well. – BCLC Nov 20 '20 at 16:19
  • @Ben Oh I forgot to mention I added bounty. ah well. – BCLC Nov 20 '20 at 16:19

1 Answers1

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A common misconception I've run across is to conflate the conjugate space $\overline V$ with $V$ in too strong a fashion. I will attempt to describe what I mean by this below. (Disclaimer: I am not a complex geometer. Far from it. This answer is just sharing what I have understood from attempts to resolve my previous and lasting confusions. ;-;)

a) For a complex vector space $(V,+,\cdot)$, the conjugate space $\big(\overline V,+,\star\big)$ is defined to have the same underlying set and additive structure, but scalar multiplication is conjugated: $\lambda \star v=\bar\lambda\cdot v.$ Since $V$ and $\overline V$ have the same dimension, there is a $\mathbb C$-linear isomorphism $V\cong \overline V,$ but it is not canonical. The $\mathbb C$-linear isomorphisms $V\cong \overline V$ are in natural bijection with $\mathbb C$-antilinear bijections $V\rightarrow V$ (which does not include $\text{Id}_V$).

b) Now what is a real structure on $V$? There are several ways to look at it, but one is as a $\mathbb C$-antilinear involution $\varphi:V\rightarrow V.$ That means that it has the extra condition that $\varphi^2=\text{Id}_V$, which is stricter than just having some $\mathbb C$-antilinear bijection $V\rightarrow V$. Thus identifying the space $V$ with its conjugation $\overline V$ is a strictly weaker notion than choosing a real structure on $V$.

To see this in action, check out this question that I had a while back. Constructions tend to only work on bundles if you can do them canonically (perhaps up to homotopy), because otherwise things may twist up and clash globally. The answer to the linked question gives a complex vector bundle that is isomorphic to its conjugate bundle, but is not a complexification of a real vector bundle. In other words, you can do (a) globally, but you can't do (b) globally.

This "weakness" isn't to say that conjugate vector spaces aren't useful; in fact, they will probably occur far more than real structures in your reading of Huybrechts. For example, everything that you do in terms of $(p,q)$-forms is based on a decomposition $V\otimes_{\mathbb R}\mathbb C\cong V\oplus \overline V$.

Update: Here's another "false resemblance." A real structure on a $\mathbb C$-vector space $(V,J)$ is a way of identifying $V$ as the complexification of an $\mathbb R$-vector space. But what if we complexify something that already has a complex structure? By this, I mean that we form $(V_\mathbb{R})^\mathbb{C}=V\otimes_\mathbb{R}\mathbb C$. We can write this as $V\otimes_\mathbb{R}\mathbb C=V\oplus iV$ or $V\otimes_\mathbb{R}\mathbb C=V\oplus \overline V$. This seems to imply that we can identify $iV$ and $\overline V$, but this is not the case. If we are careful, we actually see that the two copies of $V$ are different:

  • In the decomposition $V\otimes_\mathbb{R}\mathbb C=V\oplus iV$, we embed $V\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ and $iV\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ by the maps $v\mapsto v\otimes 1$ and $v\mapsto v\otimes i$, respectively. This "remembers" how $V\otimes_\mathbb{R}\mathbb C$ was formed as a complexification, but it doesn't make $J$ and $i$ work in a particularly compatible way. Note that $V$ and $iV$ are both closed under $J$, but they are interchanged when we multiply by $i$. Thus, we have two complex structures, but these are only complex subspaces with respect to one.

  • In the decomposition $V\otimes_\mathbb{R}\mathbb C=V\oplus \overline V$, we embed $V\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ and $\overline V\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ by the maps $v\mapsto v\otimes 1-Jv\otimes i$ and $v\mapsto v\otimes 1+Jv\otimes i$, respectively. This is different from the above, although perhaps not super enlightening. More useful is the description $$V=\{w\in V\otimes_\mathbb{R}\mathbb C:Jv=iv\}\quad\text{and}\quad \overline V=\{w\in V\otimes_\mathbb{R}\mathbb C:Jv=-iv\}.$$ This shows us that $V$ and $\overline V$ are complex subspaces under both complex structures. These two structures are equal on $V$ and opposite on $\overline V,$ which allows us to talk about "holomorphic" and "anti-holomorphic" things on a complex (or almost complex) manifold.

Nikhil Sahoo
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    Great answer! For me, the most concrete description of a real structure on $V$ is a half-dimensional real subspace $V_{re} \subseteq V$ such that $V = V_{re} \oplus i V_{re}$ as real vector spaces. It is then clear that one can obtain $V_{re}$ directly from $\varphi$ as the $1$-eigenspace, and it sheds some light on why an isomorphism $V \cong \overline{V}$ is not the same thing: how would you get the half-dimensional subspace? – Joppy Nov 17 '20 at 00:36
  • Thanks! :) I agree that the half-dimensional subspace is more concrete. I chose to phrase it this way because, in clarifying my own understanding, one thing that really helped (somewhat paradoxically) was phrasing the definitions in (a) and (b) as closely as possible, so that I could precisely identify their difference. – Nikhil Sahoo Nov 17 '20 at 00:44
  • To be more specific, before we got the concrete counterexample in the question I linked, my classmate and I sought to see the difference by showing that the space of complex-antilinear automorphisms of $V$ does not deformation-retract onto the subspace of complex-antilinear involutions. To see the difference in this fashion, it was easiest to view one notion as a "subspace" of the other. – Nikhil Sahoo Nov 17 '20 at 05:11
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    Thanks Nikhil Sahoo! I have yet to analyse your answer – BCLC Nov 20 '20 at 16:20
  • in re 'is a strictly weaker notion', did I make a mistake here ? – BCLC Nov 22 '20 at 08:56