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 for question: Let $W = (W,\text{Add}_W: W^2 \to W,s_W: \mathbb R \times W \to W)$ be $\mathbb R$-vector space, which may be infinite-dimensional. Suppose $W$ has an almost complex structure $H: W \to W$ uniquely corresponding to the $\mathbb C$-vector space $(W,H)$ where scalar multiplication is given by the complex structure $s_W^{H}: \mathbb C \times W \to W$, $s_W^{H}(a+ib,v) := s_W(a,v) + s_W(b,H(v))$ that agrees with the original real scalar multiplication $s_W$. I understand that (at least for finite-dimensional $W$) $K: W \to W$ is another almost complex structure on $W$ if and only if $K=S \circ H \circ S^{-1}$ for some $S \in Aut_{\mathbb R}W$ based on Moore (Section 9.1).
Question: Based on Gauthier (specifically Chapter 14.3 which is for finite-dimensional), it seems $(W,K)$ and $(W,H)$ are $\mathbb C$-isomorphic, but not necessarily by the identity map (at least for finite-dimensional $W$). Actually, are $(W,K)$ and $(W,H)$ $\mathbb C$-isomorphic by the identity map if and only if $K=H$ (whether finite-dimensional or infinite-dimensional)?
Best Answer
Yes because $id: (W,K) \to (W,H)$ is indeed $\mathbb R$-linear and thus is $\mathbb C$-linear if and only if $id$ commutes with scalar multiplication by i, i.e. for all $w \in W$,
$$id(i \cdot w) = i \cdot id(w) \tag{A}$$
For the left hand side is $i \cdot w := K(w)$ and for the right hand side, $i \cdot id(w) := H(id(w))$. Therefore, condition $(A)$ is equivalent to $(id \circ K) (w) = (H \circ id)(w)$, which is equivalent to $K(w)=H(w)$