-1
$\begingroup$

I have asked this on mse, but I did not get any responses even after a bounty.

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 are some:

Assumptions and notations: Let $V$ be a $\mathbb C$-vector space. 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 $V=(V_{\mathbb R},i^{\sharp})$.

  • Let $W$ be an $\mathbb R$-vector space. 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 $J(v,w):=(-w,v)$. Let $\chi: W^2 \to W^2$, $\chi(v,w):=(v,-w)$

  • For any map $f: V_{\mathbb R} \to V_{\mathbb R}$ and for any almost complex structure $I$ on $V_{\mathbb R}$, denote by $f^I$ as the unique map $f^I: (V_{\mathbb R}, I) \to (V_{\mathbb R}, I)$ such that $(f^I)_{\mathbb R} = f$. With this notation, the conditions '$f$ is $\mathbb C$-linear with respect to $I$' and '$f$ is $\mathbb C$-anti-linear with respect to $I$' are shortened to, respectively, '$f^I$ is $\mathbb C$-linear' and '$f^I$ is $\mathbb C$-anti-linear'.

  • The complexification, under $J$, of any $g \in End_{\mathbb R}W$ is $g^{\mathbb C} := (g \oplus g)^J$, i.e. the unique $\mathbb C$-linear map on $W^{\mathbb C}$ such that $(g^{\mathbb C})_{\mathbb R} = g \oplus g$

  • Let $\sigma: V_{\mathbb R}^2 \to V_{\mathbb R}^2$, $\gamma: W^2 \to W^2$ and $\eta: V_{\mathbb R} \to V_{\mathbb R}$ be any maps such that $\sigma^J$, $\gamma^J$ and $\eta^{i^{\sharp}}$ are conjugations. (The $J$'s are of course different, but they have the same formula.)

Questions:

  1. For $\sigma$, does there exist an almost complex structure $I$ on $V_{\mathbb R}^2$ such that $\sigma^I$ is $\mathbb C$-linear, and why/why not?

  2. Whenever we have such an $I$, is $I$ necessarily $I=k \oplus h$ for some almost complex structures $k$ and $h$?

  3. For $\gamma$, does there exist an almost complex structure $K$ on $W^2$ such that $\gamma^K$ is $\mathbb C$-linear, and why/why not?

    • Note: I think the answer to Question 3 is no if the answer to Question 1 is no. However, I think Question 3 is answered affirmatively and with explanation if the answer to Question is 1 is yes and the answer to Questions 2 is no.
  4. For $\eta$, does there exist an almost complex structure $H$ on $V_{\mathbb R}$ such that $\gamma^K$ is $\mathbb C$-linear, and why/why not?

    • Note: I think the answer to Question 4 is no if the answer to Question 3 is no.

Observations that led to above questions:

  1. $\chi^J$ is a conjugation, on $(V_{\mathbb R})^{\mathbb C}$, called the standard conjugation on $(V_{\mathbb R})^{\mathbb C}$.

  2. Let $\hat i: V_{\mathbb R}^2 \to V_{\mathbb R}^2$, $\hat i := i^{\sharp} \oplus i^{\sharp}$. $\hat i$ is an almost complex structure on $V_{\mathbb R}^2$.

  3. While $\chi^J$ and $\chi^{-J}$ are $\mathbb C$-anti-linear, we have that $\chi^{\hat i}$ is $\mathbb C$-linear.

  4. $k$ and $h$ are almost complex structures on $V_{\mathbb R}$ if and only if $k \oplus h$ is an almost complex structure on $V_{\mathbb R}^2$

  5. Actually, I think $\chi^{k \oplus h}$ is $\mathbb C$-linear, for any almost complex structures $k$ and $h$ on $V_{\mathbb R}$, not just $k=h=i^{\sharp}$.

$\endgroup$
1
$\begingroup$

You change complex linearity to conjugate linearity, and vice versa, by replacing $I$ by $-I$, but only on the domain or the range independently. If you want to change them both, as the same vector space with the same complex structure, it is trickier.

For question 3, a complex linear map, when realified, can only have an even number of -1 eigenvalues, so a conjugation can't be complex linear on $\mathbb{R}^2$, for example, for any complex structure. On the other hand, in any real dimension which is a multiple of 4 you can clearly have such a complex structure, and there are many. If you have a real linear map with simple eigenvalue, to become complex linear, the necessary and sufficient condition is that the real eigenvalues have even multiplicities. You can pick any complex structure on each (even dimensional) real eigenspace, and then pick out any complex eigenvalues in conjugate pairs, making one of them into an $\sqrt{-1}$ eigenspace, and the other into a $-\sqrt{-1}$ eigenspace. For generalized eigenvalues, it is more complicated.

It is easier to work in a complex linear coframing; see my lecture notes (https://arxiv.org/abs/1706.09697) where I compute some examples. You avoid this $J$ thing almost entirely.

$\endgroup$
1
  • $\begingroup$ Thanks Ben McKay. I have yet to analyse your answer. $\endgroup$ – John Smith Kyon Jun 24 '20 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.