Subspace of $\Bbb C^2 $ as Riemann Surface

Question

I am reading on the construction of Riemann Surface. From the myriad of theorems and lemmas, I am unable to understand a clear 'method' of identifying Riemann surfaces.

For example, consider $$X = \{ (z, w) \in \mathbb{C}^2 \mid zw = 0 \}.$$ I have a hunch that this is not a Riemann Surface. One reason could be this:

Set $P(z, w) = zw$.

Then $ \frac {\partial P}{ \partial w} = z $. This will be $0$ when $z = 0$. But given $z = 0$ , $P(z,w) = zw = 0$ for infinitely many values of $w$.

However, for some reason which is not clear to me, this creates a problem.

Can someone explain to me clearly

1. how to identify Riemann surfaces from equations such as above?
2. is my reasoning (mentioned above) is in the right track
3. how to find the atlas once I have identified a subspace (of $ \mathbb{C}^2 $ ) as a Riemann surface? For example $ X = \{ (z, w) \in \mathbb{C}^2 \mid 3z - 14w^2 = 0 \} $ has what atlas? Is it $ \phi _1 = \sqrt {\frac{3}{14} z}, \phi _2 = -\sqrt {\frac{3}{14} z} $ ?

Answer 1

For the particular surface $$X := \{zw = 0\} ,$$ we can see that the Jacobian of $P(z, w) = zw$, of which $X$ is a level set, is $$\pmatrix{\frac{\partial P}{\partial z} & \frac{\partial P}{\partial w}} = \pmatrix{w&z}$$ and so vanishes only at $(0, 0)$, which is contained in $X$. Thus, if there is a problem, it will occur at that point. Put more precisely, $X - \{(0, 0)\}$ is a Riemann surface, as it is the level set at a regular value of the holomorphic function $P\vert_{\Bbb C^2 - \{(0, 0)\}}$.

Now, can you show that $X$ isn't a Riemann surface by considering the topology of $X$ near $(0, 0)$? (Note that on the other hand, $X$ is the union of the two Riemann surfaces $\{z = 0\}$ and $\{w = 0\}$.)

In general, finding an explicit atlas for a Riemann surface can be difficult, but note that the particular surface $$Y := \{3 z - 14 w^2 = 0\}$$ in (3) is the graph of a function $f(w)$, so $Y$ admits a preferred global chart, $$\Bbb C \to Y, \qquad w \mapsto (w, f(w)) .$$

One can use graph charts like the $\phi_1, \phi_2$ in your solution, but one must be a good deal more careful: The two functions $\phi_i$ are not smooth at $0$, and simply by writing down $\sqrt{\cdot}$ we must (at least implicitly) make a choice of branch cut. So, $\phi_1$ and $\phi_2$ do not together cover $Y$.

Answer 2

There's a Riemann surface for $z = w^2$ (or $w = \sqrt{z}$); for this case $$ P(z, w) = z - w^2 $$ and $\frac{\partial P} {\partial w} = 0$ for $w = 0$, so having a partial derivative that's zero isn't a surface-killer. So I guess that means that the answer to your question 2 is "no".

As a kind of general rule, think about a curve in the plane, like, say, the unit circle. At almost every point (except $(\pm 1, 0)$) the map

$$ (x, y) \mapsto x $$

is a good coordinate function. That's because the tangent space to the curve projects to the $x$-axis nicely (i.e., projects to the whole axis rather than to a single point, as it does where there are vertical tangents). In those two "bad" places, we can tilt our heads and use the $y$-axis and the map $$ (x, y) \mapsto y $$ to map a neighborhood of each point in a nice way to the $y$-axis (which we regard as $\Bbb R$).

This generally works: take a nice curve, and for most point $P$ on the curve, a neighborhood of $P$ projects to the $x$-axis nicely, and when it doesn't, some neighborhood projects to the $y$-axis nicely.

There are exceptions, however: The curve defined by

$$ t \mapsto \begin{cases} (-t^2, 0) & x \le 0 \\ (0, t^2) & x > 0\end{cases}, $$ for instance, projects badly (near $t =0$, i.e., $x = y = 0$, onto both axes.

As Travis notes, the critical thing is for your defining polynomial to have a rank-1 jacobian at every point (the same thing's true for curves in the plane!).

And essentially the same argument applies: at almost every point, projection $(z, w) \mapsto z$ will be a chart; at points where it's not, projection $(z, w) \mapsto w$ will be chart. The transition function between these charts may, however, be complex.