Riemann surface from $x^2 + y^2 = 1$ for $x,y \in \mathbb{C}$

Question

I am reading Edward Frenkel's book Love and Math. In Chapter 9, it is talked about the one-to-one correspondence of solution of algebraic function of complex numbers and Riemann surfaces. can anyone show me what is the corresponding Riemann surface for $x^2 + y^2 = 1$ for $x,y \in \mathbb{C}$ and why?

Answer 1

The most simple algebraic object attached to an algebraic equation $f(x,y) = 0$ like $f(x,y)= 1 - (x^2 + y^2)$ is the zero set of $f$:

$$V(f):=\{(x,y) \in \mathbb C^2: f(x,y) = 0\}.$$

The zero set $V(f)$ is an example of an affine algebraic variety.

Things become easier when taking the projective closure of $V(f)$, which is analogous to compactification. The projective closure is obtained by replacing $\mathbb C^2$ by the compley projective plane $\mathbb P^2$. After introducing an additional variable $z$, the polynomial $f$ extends to the homogenous polynomial $F(x,y,z) := z^2 - (x^2 + y^2)$. Its zero set is the projective algebraic set

$$V(F):=\{(x:y:z) \in \mathbb P^2: F(x,y,z) = 0\}.$$

$V(F)$ is an example of a compact Riemannian surface when equipped with the Euclidean topology as a subset of $\mathbb P^2$.