## Implicit orientation of a holomorphic manifold

When I first started learning analytic geometry (the complex analogue to algebraic geometry), I recall asking my advisor why the local intersection number is defined with no regards for orientation. At the time, he simply said that *complex manifolds are assumed to be positively oriented*, which I didn't pay much attention to since I was too busy trying to wrap my head around more elementary concepts, such as the Zariski topology of the spectre of a Noetherian ring.

Fast forward to a few days ago. A friend was telling us about Chern classes, and he mentioned once again orientation and holomorphic manifolds. However, he was more specific on the question:

...and we can assume that [a complex manifold] always has a well-defined orientation since “the complex numbers have an implicit orientation”, or something like that [sic].

I recently happened to have attended a mini-course about Floer homology, which involves endowing real manifolds of even dimension with a complex structure on the tangent spaces, so I realised that you could use the latter to explain the former. For our intents and purposes, a complex structure is a $2\times 2$ matrix $J$ such that $J^2$ is the identity matrix with a minus sign.

Any complex number $x+iy$ can be represented in the real plane in two ways: as a vector $(x,y)$ and as a matrix

$$J(x+iy)=\begin{pmatrix}x&-y \\ y&x\end{pmatrix}.$$

In particular, $J(i)$ is a complex structure (as per our definition)! Not only that, but we also have $J(a+ib)=a J(1)+ b J(i)$, so complex multiplication can be recovered using this $J$ matrix. Note that $J$ (seen as a mapping $\mathbb{R}^2 \to \mathbb{R}^2$) describes an counter-clockwise rotation, which corresponds to the positive orientation on the plane.

The transpose of $a+ib$ can be described in terms of the $J$ matrix as $aJ(1)+bJ(-i)$. However, $J(-i)$ corresponds to a clockwise rotation, which is the *negative* orientation on the plane.

Since the mapping $z \mapsto \overline z$ is not holomorphic, this means that only one of the two orientations on the plane is compatible with the complex derivative!