The zero set of $z_0^2+z_1^2-1$ in $\mathbb{C}^2$.

Recently, I read the notes "Vector bundles on Riemann surfaces" by Sabin Cautis (http://www-bcf.usc.edu/~cautis/classes/notes-bundles.pdf). On the sixth page of these notes, there is a statement without any explanation to it: "For example, $z_0^2+z_1^2-1 = 0$ is isomorphic (as a complex manifold) to $\mathbb{C}$." We are dealing with the zero set of this polynomial in $\mathbb{C}^2$.

Since I didn't manage to find a proof of this statement in the time available, I discussed it with some colleagues over a few beers, one evening. Even though approaches were attempted, we didn't quite get anywhere. The approach that looked most promising is to prove the given zero set is simply connected and then invoke the uniformisation theorem. By determining the automorphism group, we would then be able to find out that it's indeed $\mathbb{C}$. Would this approach be able to solve the question? (i.e. can anyone fill in the details?) And if not, how could the statement be proven differently?

Remark. This zero-set of course looks very sphere-like at first sight, yet an absence of $|$ prevents this.

(I wouldn't be surprised if this question is around on this site already, in which case I couldn't find it and would appreciate a link t)

• Maybe I'm missing something, but isn't that simply wrong? $z_0^2+z_1^2-1=0$ can be transformed linearly into the hyperbola $z_0z_1-1=0$, and the hyperbola is isomorphic to $\Bbb{C} \setminus \{0\}$, which itself is clearly not isomorphic to $\Bbb{C}$. Apr 22 '13 at 11:54
• I would most certainly be willing to believe it isn't even true, since that was my first thought as well, though for completely different reasons: by standard projection on the first coordinate it is possible to consider it as a double cover of $\mathbb{C}$ with exactly two ramification points, i.e. as two planes pinched together in two points. My head fails to see how this would be contractible. Could you just present your suggestion slightly extended (with some more details) as an answer?
– HSN
Apr 22 '13 at 16:00

The statement is simply false. As requested, here is an extension of my above comment.

I denote by $Z(z_0^2+z_1^2-1) \subset \Bbb{C}^2$ the vanishing locus of the polynomial $z_0^2+z_1^2-1$.

Consider the linear change of coordinates $$z_0 \mapsto \frac{1}{2} \left(\frac{w_0}{1+i}+w_1 \right)$$ $$z_1 \mapsto \frac{-i}{2}\left(\frac{w_0}{1+i}-w_1 \right)$$ This gives an isomorphism of the complex variety $Z(z_0^2+z_1^2-1)$ onto the hyperbola $Z(w_0w_1-1)$. Now we clearly have $$Z(w_0w_1-1)=\{(t,t^{-1}) \ \; \vert \; t \in \Bbb{C}^*\}$$ Projection onto the first factor gives the desired isomorphism of $Z(w_0w_1-1)$ with $\Bbb{C}^*$, and $\Bbb{C}^*$ is clearly not isomorphic to $\Bbb{C}$, e.g. $\Bbb{C}$ is simply connected, while $\Bbb{C}^*$ isn't. Hence we can conclude that $Z(z_0^2+z_1^2-1)$ is not isomorphic to $\Bbb{C}$ as well.

• Thanks a lot for extending the answer!
– HSN
Apr 22 '13 at 19:34
• You're welcome. Apr 22 '13 at 23:00

The variety $z^2+w^2=1$ is a smooth conic which by the degree-genus formula is of genus zero. Any such affine curve is isomorphic to $\mathbb{C}$.

• Every smooth, projective conic is isomorphic to $\Bbb{P}^1$, but not every smooth, affine conic is isomorphic to $\Bbb{C}$, e.g. the hyperbola $zw-1=0$ is isomorphic to $\Bbb{C} \setminus \{0\}$, and the latter is not isomorphic to $\Bbb{C}$. Apr 22 '13 at 15:27
• There is a problem there indeed. The Riemann sphere $\hat{\mathbb{C}}$ and $\mathbb{C}$ both have genus 0, but aren't isomorphic.
– HSN
Apr 22 '13 at 15:55
• Thanks for pointing that out. Apr 24 '13 at 2:07