Incompatible charts on a complex manifold

Question

Consider the following subset of $\mathbb{R}^2$. $$U=\{(x,y):x,y \in \mathbb{R}, x>0\}$$ We can make this into a 1-dimensional complex manifold by giving it a global chart on to the open set of complex numbers with positive real part. $$V=\{z \in \mathbb{C}:Re(z)>0\}$$ Consider the following two identifications, $\phi,\phi^{\prime}$: $$1)\phi(x,y)=x+iy$$ $$2)\phi^{\prime}(x,y)=x+i\frac{y}{a}$$ where $a$ is any non-zero number. In general these two charts aren't holomorphically compatible, so they define two different complex manifolds. I have a hard time wrapping my head around this, what exactly is the difference between the two complex manifolds. I am thinking that they should be very similar since the charts are compatible when you consider them as smooth manifolds. Is there any good example where two incompatible charts give bizarrely different complex manifolds?

Answer 1

I don't have an example of two "bizarrely different complex manifolds", but I can answer your basic question: Let's call the underlying topological space of your manifolds $V$, and the manifolds themselves $M$ and $M'$. The difference between the two manifolds is how you define angles on their underlying topological space. Obviously, there is a standard notion of angles on subspaces of $\mathbb R^n$ via the standard inner product, but when making a complex manifold out of $V$, you're throwing away this notion and instead use the charts to define angles. Take any two smooth, regular (i.e. the tangents are non-zero) curves $\gamma$ and $\eta$ in $M$ and assume they intersect. Then $\tilde\gamma:=\phi\circ\gamma$ and $\tilde\eta:=\phi\circ\eta$ are smooth, regular curves in $\mathbb C$, and they also intersect. Here we can consider the angle between the curves at their intersection, and define: The angle between $\gamma$ and $\eta$ is the angle between $\tilde\gamma$ and $\tilde\eta$. We're essentially reading the angle out of their charts. And another chart $\phi'$ is compatible with $\phi$ only if defining angles via this second chart gives the same angles. Essentially, compatible charts always agree about angles. The reason is that this is one of the main features of conformal maps: conformal maps (which injective holomorphic maps are) are exactly the differentiable maps which preserve angles, and compatibility requires the change of coordinates to be conformal (or injective and holomorphic, which is the same).

Now getting back to your example, you can see that the two charts do not agree about angles. Take two lines: one is the $x$-axis, the other is the line through $(1,1)$ and the origin. The chart $\phi$ tells us that the angle between them is $\frac{\pi}{4}$, or $45^\circ$. But the chart $\phi'$ tells us that the angle is $\operatorname{atan2}(\frac{1}{a},1)$, which is not necessarily $\frac{\pi}{4}$. See here for a description of $\operatorname{atan2}$. It's essentially a function which takes the coordinates of a point and returns its polar angle. One can say that $M'$ is essentially a squished version of $M$.

However, the two are still conformally equivalent. It's just that the conformal map establishing this equivalence will not be the identity on $V$, but the "reverse squishing". Basically, a conformal map $\varphi:M\longrightarrow M'$ must send $(x,y)$ to $(x,ay)$. Because then if we have two intersecting, smooth, regular curves $\gamma$ and $\eta$ on $M$, the curves $\varphi\circ\gamma$ and $\varphi\circ\eta$ will have the same angle between them according to the complex structure of $M'$ as $\gamma$ and $\eta$ have according to the complex structure of $M$.