# Bijection for involutive maps and $\mathbb R$-subspaces given almost complex structure (anti-involutive)? Formula for conjugation?

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 $$V^2$$.

Observations:

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 $$J$$. $$\tag{2A}$$

Questions:

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 $$\sigma_{A,K}$$?

## 1 Answer

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.