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:

Let $V$ be $\mathbb R$-vector space, possibly infinite-dimensional.

**Complexification of space definition**: Its complexification can be defined as $V^{\mathbb C} := (V^2,J)$ where $J$ is the almost complex structure $J: V^2 \to V^2, J(v,w):=(-w,v)$ which corresponds to the complex structure $s_{(J,V^2)}: \mathbb C \times V^2 \to V^2,$$ s_{(J,V^2)}(a+bi,(v,w))$$:=s_{V^2}(a,(v,w))+s_{V^2}(b,J(v,w))$$=a(v,w)+bJ(v,w)$ where $s_{V^2}$ is the real scalar multiplication on $V^2$ extended to $s_{(J,V^2)}$. In particular, $i(v,w)=(-w,v)$.

**Complexification of map definition**: See a question I posted previously.

**Proposition** (Conrad, Bell): Let $f \in End_{\mathbb C}(V^{\mathbb C})$. We have that $f$ is the complexification of a map if and only if $f$ commutes with the standard conjugation map $\chi$ on $V^{\mathbb C}$, $\chi(v,w):=(v,-w)$. In symbols:

If $f \circ J = J \circ f$, then the following are equivalent:

Condition 1. $f=g^{\mathbb C}$ for some $g \in End_{\mathbb R}(V)$

Condition 2. $f \circ \chi = \chi \circ f$

I think Bell would rewrite Condition 2 as $f = \chi \circ f \circ \chi$ and say $f$ 'equals its own conjugate'.

**Questions:** Considering the half of the above proposition that says '$f$ commutes with both $J$ and $\chi$ implies $f$ is complexification of a map', what do we get if we instead have the following?

commutes with $J$ and anti-commutes with $\chi$ ($f \circ \chi = - \chi \circ f$)

anti-commutes with $J$ ($f \circ J = - J \circ f$, i.e. $f$ is $\mathbb C$-anti-linear) and commutes with $\chi$

anti-commutes with $J$ and anti-commutes with $\chi$

**Motivation**: $f=J$ satisfies the case in Question 1, and $f=\chi$ satisfies the case in Question 2.

**Guess** (for Question 2):

Similar to this (the $K=-J$ part), I kind of had the idea to define something like anti-complexification of a map: for $g \in End_{\mathbb R}(V)$, $g^{anti-\mathbb C}$ is any $\mathbb C$-anti-linear map such that $g^{anti-\mathbb C} \circ cpx = cpx \circ g$, where $cpx: V \to V^{\mathbb C}$ is the complexification map, as Roman (Chapter 1) calls it, or the standard embedding, as Conrad calls it. I think $g^{anti-\mathbb C}$ turns out to always exist uniquely as $g^{anti-\mathbb C}(v,w)=(g(v),-g(w))$.

Then, I think the answer for Question 2 is that $f$ is the anti-complexification of a map. We can strengthen the result to: Let $f$ be $\mathbb C$-anti-linear on $V^{\mathbb C}$, i.e. $f$ anti-commutes with $J$. We have that $f$ is the anti-complexification of a map $g \in End_{\mathbb R}V$, i.e. $f=g^{anti-\mathbb C}$ if and only if $f$ commutes with the standard conjugation map $\chi$, i.e. $f \circ \chi = \chi \circ f$.

In the case of $f=\chi$ for Question 2, $f=\chi = g^{anti-\mathbb C}$ for $g=id_{V}$, the identity map on $V$, which by the way gives us $(id_{V})^{\mathbb C} = id_{V^{\mathbb C}}$