Why is the matrix representation of an almost complex structure like this? Let $M$ be a Riemann surface, $J$ be an almost complex structure (i.e. a 1-1 tensor such that $J^2=-I$ and for any $x \in M,v,w \in T_xM, \{v,Jv\}$ is oriented). Consider a conformal coordinate at a point $x \in M$. Then the text (page 24) that I am reading says that in conformal coordinate, $J=\begin{bmatrix} 0 & -1\\
1 & 0
\end{bmatrix}$. 
This doesn't make any sense to me for the following reason: I know that there can be more than one almost complex structure, and the matrix representation uniquely determines a linear operator. As we just start with an arbitrary $J$, it is not possible for us to obtain a matrix representation that takes the special form I mentioned above, unless there is only one almost complex structure on $M$.
My thought is that $J$ only takes a form $S\begin{bmatrix} 0 & -1\\
1 & 0
\end{bmatrix}S^{-1}$, but the drawback is that I have no idea how to prove the theorem in the text if $J$ takes this form.
Thank you very much for answering this question!
 A: In the context of Riemann surfaces, $J$ is not an arbitrary complex structure but rather a particular complex structure that, depending on your definition of Riemann surface, is either essentially built therein or is fast consequence thereof.
In the latter case, if we define, e.g., a Riemann surface to be a (connected) oriented real smooth manifold $M$ of dimension $2$ equipped with a conformal structure $\bf c$, we can define $J$ as follows: Fix any representative metric $g \in {\bf c}$. Then, for $p \in M$ and $X \in T_p M$, define $JX \in T_p M$ to be the unique vector for which


*

*The oriented angle between $X$ and $JX$ is $\frac{\pi}{2}$ (equivalently, $g(X, JX) = 0$ and $(X, JX)$ is an oriented basis of $T_p M$), and

*$X$ and $JX$ have the same length ($g(X, X) = g(JX, JX)$).


Both statements are invariant under a change of metric $g \in {\bf c}$, so $J$ is well-defined (a priori rough) bundle map $TM \to TM$. From this definition, we see that $J$ is fiberwise linear and that for any $X \in T_p M$ the matrix representation of $J_p = J \vert_{T_p M}$ with respect to the basis $(X, JX)$ is the familiar
$$\phantom{(\ast)}\qquad [J_p] = \pmatrix{0&-1\\1&0} . \qquad (\ast)$$
In any (oriented, smooth) conformal coordinates $(x, y)$, the conformal class contains $dx^2 + dy^2$, in the induced basis $(\partial_x, \partial_y)$ at any point $p$ applying the above definition shows that $J \partial_x = \partial_y)$, so the matrix representation $[J_p]$ of the coordinate basis is just $(\ast)$ as claimed. Since the components of $[J_p]$ are smooth functions, $J$ is a smooth bundle endomorphism of $TM$.
Remark A similar procedure lets us recover ${\bf c}$ and the orientation on $M$ from $J$. This equivalence of the conformal structure (plus an orientation) and the (almost) complex structure is a consequence of the isomorphism between the oriented conformal group $CSO(2)$ and the complex general linear group $GL(1, \Bbb C) \cong \Bbb C^*$ special to this dimension.
