Let $\newcommand{\bbh}{\mathbb H}\bbh$ denote the set of all quaternions. It is easy to see that $\bbh$ can be regarded as a $2$-dimensional complex vector space with basis $\{1,j\}$ by writing $$q=q_0+q_1{ i}+q_2{ j}+q_3{ k}=(q_0+q_1{ i})+(q_2+q_3{ i})j.$$ $\newcommand{\bbr}{\mathbb{R}}\newcommand{\bbc}{\mathbb{C}}$ In this arXiv paper(section $2.3$), the author describes an almost complex structure on $\bbh$(I'm attaching a picture).
At first glance, I thought this is same as the one I described above. Then I found out the following definition on Huybrechts' book Complex Geometry: An Introduction:
Let $V$ be a real vector space. An almost complex structure on $V$ is a map $J : V \to V$ such that $J^2 = -\mathrm{Id}_V$. An almost complex structure gives $V$ the structure of a complex vector space by defining $(a+bi)v = av + bJ(v)$.
Then, the construction of $L_q$ satisfying $L_q^2=-\mathrm{Id}$ made more sense. I know that the complexification of a real vector space $V_{\bbc}=V^{(1,0)}\oplus V^{(0,1)}$, where $V^{(1,0)}=\{v\in V_{\bbc}\mid J(v)=iv\}$ and $V^{(0,1)}=\{v\in V_{\bbc}\mid J(v)=-iv\}$(here, $J$ denotes the $\bbc$-linear extension of $J$!). And, given a $v\in V_{\bbc}$, we can split $v=\frac12(v-iJ(v))\oplus\frac12(v+iJ(v))$.
So I expected the $(1,0)$ coordinates of $v$ to be $\frac12(v-iJ(v))$(I don't know the exact definition). But, when I tried to find out the $(1,0)$ coordinates for the quaternion case described in the paper, it turned out to be different. And, I cannot understand the last part of the picture given above. After solving $L_{\mathbf i}q=iq$(note that there are two i here; the quaternion $\mathbf i$ and the complex $i$), how do they conclude $a=x_0+ix_1$ and $b=x_2+ix_3$ are the holomorphic coordinates?? Why is it different from $\frac12(v-iJ(v))$?
Thank you.
The natural map $\pi: V \to V^{(1,0)}$ which embeds $V\hookrightarrow V_\mathbb C$ and then projects to the $i$ eigenspace of $L_\mathbf i$ via $v\mapsto \frac12(v-iJv)$ is given by: \begin{align} 2(x + y\,\mathbf i + z\,\mathbf j + w\,\mathbf k)\ \ \ &\mapsto\ \ \ (x+iy) + (y-ix)\mathbf i + (z+iw)\mathbf j + (w-iz)\mathbf k\\ \\ &=\ \ \ \ (x+iy)(1-i\mathbf i) + (z+iw)(\mathbf j - i \mathbf k) \end{align}
It is important that in the last step we write the coordinates in terms of a eigenbasis, instead of in the basis $\{1,\mathbf i, \mathbf j, \mathbf k\}$ (which are not even in the eigenspace).
The pair of $\frac12(1-i\mathbf i)$ and $\frac12(1-i\mathbf i)\cdot \mathbf j$ is a natural choice of basis given that they are the projections of $1$ and $\mathbf j$, but you might do the decomposition with respect to another eigenbasis, for instance replacing $$(x+iy)(1-i\mathbf i) = (y-ix)(i+\mathbf i),$$ thus establishing a different isomorphism $\mathbb H \cong \mathbb C\oplus \mathbb C$. But this amounts to only modifying the original isomorphism by an element of $GL_2\mathbb C$, so we get the same $\mathbb C$ vector space structure, and an equally valid coordinate chart as a complex manifold.