Are Projective Varieties Secretly Affine Varieties?

Last time I recoiled from my statement that projective spaces are manifolds, but let’s explore this a little. We don’t want manifolds, but an idea comes up in showing projective space is one that we can use. The fact that we have natural local “coordinate patches” on projective space. i.e. $\mathbb{P}_k^n=U_0\cup \cdots \cup U_n$, where $U_i=\{(x_0 : \ldots : x_n)\in \mathbb{P}_k^n : x_i\neq 0$. For a manifold, we would show these are diffeomorphic to copies of $\mathbb{R}^n$, but this isn’t the right topology (and has far too restricting of structure), so we really want to show that each is homeomorphic to affine space $p_i: U_i\to \mathbb{A}_k^n$ by $(x_0 : \ldots : x_i : \ldots : x_n)\mapsto \left(\frac{x_0}{x_i}, \ldots , \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \ldots , \frac{x_n}{x_i}\right)$.

So to reiterate, we have the Zariski topology on projective space by closed sets defined as zero sets of homogeneous polynomials, and the Zariski topology on affine space by zero sets of polynomials. What we want to show is that we can cover projective space with affine spaces, so that locally projective space looks affine.

Let’s just work with $\pi_0$, since there is nothing special about any of the coverings. As far as sets are concerned, $\pi_0$ is certainly bijective, so we just need to check that $\pi_0$ and $\pi_0^{-1}$ are continuous. So we’ll check that they are closed maps.

The polynomials that we are concerned with defining closed sets on $U_0$ are from the set $S=\{f\in k[x_0, \ldots , x_n] : f ~ homogeneous\}$ and the polynomials defining closed sets on $\mathbb{A}_k^n$ are from $k[x_1, \ldots , x_n]$.

So now let $\alpha : S\to A$ by $f\mapsto f(1, x_1, \ldots , x_n)$, and let $\beta: A\to S$ by $g\mapsto x_0^{deg(g)}g(\frac{x_1}{x_0}, \ldots , \frac{x_n}{x_0})$.

It is easy to check that $\alpha\circ \beta=Id$.

Now let $X$ be any closed set of $U_0$. Then $X=E\cap U_0$ where $E$ is a closed set in $\mathbb{P}_k^n$. i.e. $E$ is a projective variety, so there is a set $T\subset S$ such that $E=V(T)$ and $\pi_0(X)=V(\alpha(T))$.

Any closed set of $\mathbb{A}_k^n$ has the form $W=V(T')$ where $T'\subset A$ and $\pi_0^{-1}(W)=V(\beta(T'))\cap U_0$.

But now the image of closed sets under $\pi_0$ is precisely $\pi_0(V(T)\cap U_0)=V(\alpha(T))$. So $\pi_0$ is a closed map. Also, the image of a closed set $\pi_0^{-1}(V(T'))=V(\beta(T')))\cap U_0$, so both are closed and hence continuous, so $\pi_0$ is a homeomorphism.

But this proof actually gives us more. Namely that irreducible projective varieties $X\subset \mathbb{P}_k^n$ such that $X\cap \{(x_0: \ldots : x_n) : x_0=0\}=\emptyset$ is in 1-1 correspondence with irreducible affine varieties $X_0\subset \mathbb{A}_k^n$.