# Riemann surface of zeroes

I am trying to figure out why $X =\{(z,w) \in \mathbb {C}^2 | z^2 -w^3 = 0\}$ is not a Riemann Surface. I have a hunch that there is an issue at (0,0). It is possibly not related to connectedness. How can we show that no neighborhood of (0,0) in X is homeomorphic to open unit Disk in complex plane?

• $X$ is not smooth at $0$, since the gradient of the defining equation $F(z,w) = z^2 - w^3$ is zero there. Are you familiar with this language? – Lorenzo May 17 '17 at 0:58
• I understand that the gradient and the function P(z,w) = z^2-w^3 are both zero at (0,0). What do you mean by smoothness? – ardhajya May 17 '17 at 1:05
• A curve is the zero set of a single polynomial on $\mathbb{C}^2$. Smoothness at a point x on a curve means that the tangent space is one dimensional. The tangent space is the zero set of the gradient. For curves, smooth = Riemann surface. You can see any book on algebraic curves for much more on this. Your particular curve has a cuspidal singularity at the origin. – Lorenzo May 17 '17 at 1:29
• However ... I see now that you are asking about a homeomorphism of a neighborhood of $(0,0)$ with the disk. This is a slightly different question (I think, I'm not an expert in singularities), but I'll write up something about how you can compute the link (intersection with 3-sphere around the origin). – Lorenzo May 17 '17 at 1:30

What I show below (in "on the link"...) is that every intersection of $X$ with a small sphere around the origin is topologically distinct from the intersection of a plane with a small sphere around the origin.

This is not exactly what you are asking, and I'm convinced that you are not asking precisely what you mean to ask.... Topologically, $X$ is a sphere minus a point, and in particular there is a neighborhood of $0$ which is homeomorphic to a disk. This is not sufficient for it to be a Riemann surface - recall that you also need a complex structure, which involves compatible holomorphic patches. (A remark that uses some more algebraic geometry: The homeomorphism type of $X$ is easier to see if you work with the projective varieties -- then you get a map from $\mathbb{CP^1} \to V(uz^2 - w^3) \subset \mathbb{C}P^2$, the normalization, and this is a continuous bijection. In the Euclidean topology, the domain is compact, and the target is Hausdorff, and hence this is a homeomorphism).

However, $X$ is not a Riemann surface, because it fails to be "smooth" at the origin. Perhaps the most explicit way to understand this is to consider the limit of secant lines whose defining points approach $0$ - if the zero set was smooth at $0$, the limit of these lines would sit inside the one (complex) dimensional tangent space to $X$ at $0$.

By taking $a(t) = (t^3,t^2)$, the line from $0$ to $a(t)$ is cut out by $t^3w - t^2 z = 0$, or $tw - z = 0$. As $t \to 0$, this converge to the line $z = 0$.

We can produce a linearly independent direction as a limit of secant lines. For example, we let $a(t) = (t^3,t^2)$ and $b(t) = (-t^3,t^2)$. Then the secant line connecting these points is $w = t^2$, so the limit of the secant lines as $t \to 0$ is $w = 0$.

But $z$ and $w$ are independent coordinates over $\mathbb{C}$.

Let us define $B_{\epsilon}$ to be the ball of radius epsilon around the origin, and $S_{\epsilon} = \partial B_{\epsilon}$, its boundary sphere. This is a 3-sphere in $\mathbb{C}^2$.

In other words, $S_{\epsilon} = \{(z,w) \in \mathbb{C}^2 : |z|^2 + |w|^2 = \epsilon\}$.

Let will let $X$ be the zero set of a single polynomial $F(z,w)$. This is a "curve", since it has 1 complex dimension. Suppose that $X$ contains $(0,0)$ (equivalently, $F$ has no constant part).

Then, using $S_{\epsilon}$, you can define a topological invariant of the point $0 \in X$, called the link. (I'm not totally how "invariant" this is - under what transformations is it preserved?)

Consider $S_{\epsilon} \cap X$. For most $\epsilon > 0$, this will be a real 1 manifold in $S^3$ (since it still have the 2-real codimension inherited from X in $\mathbb{C}^2)$, so it will be a disjoint collection of circles. The isotopy type this collection of circles inside $S^3$ is an invariant of the singularity.

(We take the $\epsilon$ for which the intersection of the 3-sphere with $X$ is transverse, and not intersecting any singular points of $X$ -- this guarantees that the dimension is correct, and it guarantees that the intersection is a manifold. There are only finitely many singular points of $X$, so most of the time the second condition is satisfied. For the transversality, we rely on the generic nature of transverse ... this is discussed in Guillemin and Pollack, for instance.)

If $X$ was a Riemann surface at $0$, then near zero (up to diffeomorphism) it is a 2 dimensional plane passing through the origin. This intersects the 3-sphere in an (unknotted) circle.

Suppose now that $X$ was cut out by $z^2 = w^3$. We will compute the link and learn that it is a trefoil knot.

First, it's useful to know that the solutions of equation are parametrized by $(t^3,t^2)$ for $t \in \mathbb{C}$. (*)

If $(z,w) \in X \cap S_{\epsilon}$, then $|z|^2 + |w|^2 = \epsilon$ and $z^2 = w^3$, so $|w|^3 + |w|^2 = \epsilon$. For $c$ the unique positive real solution of $c^2 + c^3 = \epsilon$, the corresponding point on the intersection $(z,w)$ has $|z|^2 = |w|^3 = c^3$ and $|w|^3 = c^2$.

Returning to the parametrization (*), this means that $|t|^6 = c^2$, so $|t| = c^{1/2}$, so $t = c^{1/2} e^{i \theta}$ and so a paremetrization of the intersection is given by $(c e^{2 i \theta}, c^{3/2} e^{3 i \theta}) \subset S^1 \times S^1 \subset S_{\epsilon}$. (The first $S^1$ is the points of modulus $c$ in $\mathbb{C}$, similarly for the second $S^1$.)

This is a circle that travels twice around the first $S^1$ and 3 times around the second $S^1$ before returning to it's beginning. This is a torus knot, and non-trivial. (I think a proof of non-triviality can be found in Hatcher's algerbaic topology.)

• This is beautiful! I will study this explanation more closely. So basically you are showing that every neighborhood of 0 in X has non trivial fundamental group? – ardhajya May 19 '17 at 2:11
• @ardhajya It's not that. As I mentioned, this zero set is homeomorphic locally to discs (the cuspidal cubic in projective plane is homeomorphic to a sphere, unless I tricked myself). The problem is that it has a cusp. These knots have to do with the way the neighborhoods are embedded in $\mathbb{C}^2$. – Lorenzo May 19 '17 at 2:15