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:

The questions (asked towards the end of this post) are related to these questions:

Assumptions, definitions and notations: Let $V$ be an $\mathbb R$-vector space. Define $K \in Aut_{\mathbb R} (V^2)$ as anti-involutive if $K^2 = -id_{V^2}$. Observe that $K$ is anti-involutive on $V^2$ if and only if $K$ is an almost complex structure on $V^2$. Let $\Gamma(V^2)$ be the $\mathbb R$-subspaces of $V^2$ that are isomorphic to $V$. Let $AI(V^2)$ and $I(V^2)$ be, respectively, the anti-involutive and involutive maps on $V^2$.


  1. Let $J: V^2 \to V^2$, $J(v,w):=(-w,v)$ be the canonical almost complex structure on $V^2$. It appears $\chi: V^2 \to V^2$, $\chi(v,w):=(v,-w)$ is the unique involutive $\sigma \in Aut_{\mathbb R} (V^2)$ on $V^2$ such that $\sigma$ anti-commutes with $J$ (i.e. $\sigma$ is $\mathbb C$-anti-linear with respect to $J$) and the set of fixed points of $\sigma$ is equal to $V \times 0$.

In other words: For any $\sigma \in Aut_{\mathbb R} (V^2)$, we actually have that $\sigma = \chi$ if and only if $\sigma$ satisfies

  • 1.1. $\sigma \circ J = - J \circ \sigma$,

  • 1.2. $\sigma \circ \sigma = id_{V^2}$

  • 1.3. The set of fixed points of $\sigma$ is equal to $V \times 0$,

  1. I believe Conrad's Theorem 4.11 without reference to complex numbers can be restated as:

Let $V$ be $\mathbb R$-vector space. Let $J(v,w):=(-w,v)$. There exists a bijection between $\Gamma(V^2)$ and involutive $\mathbb R$-linear maps that anti-commute with $J$. $\tag{2A}$


Question 1. Can we generalise $(2A)$, as follows, to arbitrary $\mathbb R$-linear map anti-involutive map $K$?

Let $V$ be an $\mathbb R$-vector space. Let $K \in AI(V^2)$. There exists a bijection between $\Gamma(V^2)$ and involutive $\mathbb R$-linear maps $\sigma$ that anti-commute with $K$.

Question 2. If no to Question 1: what's so special about $K=J$ that works as opposed to some other $K$ that doesn't necessarily work? If yes to Question 1: I believe half of the bijection allows us to define a map $\hat \sigma: \Gamma(V^2) \times AI(V^2) \to I(V^2)$, $\hat \sigma(A,K) =: \sigma_{A,K}$, the unique element of $I(V^2)$ that anti-commutes with $K$ and has $A$ equal to the set of its fixed points. What's the formula for $\sigma_{A,K}$?


The following answer is based on Joppy's answer here.

Answer to Question 1. Yes, assuming axiom of choice.

  • A1. Given a $\sigma$, we get $A_{\sigma}$ as follows: Actually, any $\sigma \in I(V^2)$, whether or not $\sigma$ anti-commutes with $K$ is such that $fixed(\sigma) \bigoplus fixed(-\sigma) = V^2$, where $fixed(\cdot)$ denotes the set of the fixed points (see here). Choose $A_{\sigma} = fixed(\sigma)$.

  • A2. Given an $A$, we get a $\sigma_A$ as follows: See answer to Question 2.

  • A3. We must show that for $\gamma(A)=\sigma_A$ and $\delta(\sigma)=A_{\sigma}$, we have that $\gamma \circ \delta(\sigma)=\sigma_{A_{\sigma}}=\sigma$ and $\delta \circ \gamma(A)=A_{\sigma_{A}}=A$.

  • A3.1. For $\delta \circ \gamma(A)=A$: $A_{\sigma_{A}} := fixed(\sigma_{A})$ and then by definition of $\sigma_{A}$, $fixed(\sigma_{A})=A$.

  • A3.2. For $\gamma \circ \delta(\sigma)=\sigma$: $\sigma_{A_{\sigma}}$ is the unique element $\eta \in End_{\mathbb R}(V^2)$ such that $\eta = id_{A_{\sigma}}$ on $A_{\sigma}$ and such that $\eta=-id_{K(A_{\sigma})}$ on $K(A_{\sigma})$. Let's show that $\sigma \in End_{\mathbb R}(V^2)$ satisfies this property: Let $v \in A_{\sigma} = fixed(\sigma)$.

  • A3.2.1. $\sigma = id_{A_{\sigma}}$ on $A_{\sigma}$: $\sigma(v)=v=id_{A_{\sigma}}(v)$

  • A3.2.1. $\sigma=-id_{K(A_{\sigma})}$ on $K(A_{\sigma})$: (I am using the fact that $K$ is injective) $\sigma(K(v)) = -K(\sigma(v))=-K(v)$

Answer to Question 2. For any such $A$,

  • Step 1. First, note that axiom of choice gives us $A \bigoplus K(A) = V^2$ (see here; I actually can't think of a way to prove this without axiom of choice and without deducing some $\sigma$ from $A$ and $K$, the latter of which is circular).

  • Step 2. By Step 1, it makes sense to say that there exists unique $\eta \in End_{\mathbb R}(V^2)$ such that $\eta = id_A$ on $A$ and such that $\eta=-id_{K(A)}$ on $K(A)$. This $\eta$ is uniquely given by the formula $\eta(a \oplus K(b))=a \oplus K(-b)$

  • Step 3. Choose $\sigma = \sigma_{A,K} := \eta$: We can see that $\sigma$ anti-commutes with $K$, is involutive and has $A$ as its fixed points.


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