# What is the difference between projective space and weighted projective space?

I know their definitions but I don't really know what the differences are conceptually. I know that $\mathbb{C}P^n$ can be thought of as the complex lines through the origin of a complex Euclidean space but I don't know how that differs from weighted projective space.

Theorem 2.1 states without proof that they have the same cohomology. Can anyone provide intuition for this? I have only a loose understanding of cohomology.

Here is the first step of the proof, taken from the paper by T. Kawasaki, "Cohomology of Twisted Projective Spaces and Lens Complexes".

The idea is to use the covering $p : \Bbb P^n \to \Bbb P(a_0, \dots, a_n)$, $[x_0: \dots : x_n] \mapsto [x_0^{a_0}: \dots :x_n^{a_n}]$. If $G$ is the deck group of the covering (the product of $\mu_{a_i}$, where $\mu_k$ is the group of $k$-th roots of unity), then we get an homeomorphism $\Bbb P^n/G \cong \Bbb P(a_0, \dots, a_n)$. Now there is a theorem by Borel saying that for a finite group $G$ acting on a space $X$ the pullback $$\pi^* : H^{\bullet}(X/G, S) \to H^{\bullet}(X,S)^G$$

is an isomorphism as long as $|G|^{-1} \in S$. So it doesn't work for integer coefficients but at least it gives you the result for $\Bbb Q$, or even the localisation $\Bbb Z_a$ where $a = \prod a_i$. This is getting technical after this so I'll stop here.

On the other hand, it is always instructive to see example by hand. So let $X = \Bbb P(1,1,n)$ be the projective space. If we look at space of functions of degree $n$, by definition it is $x^n,x^{n-1}y, \dots, y^n, z$. (A degree $n$ monomial $M$ means that $t \cdot M = t^nM$). We get an embedding $$f : \Bbb P(1,1,n) \to \Bbb P^n, (x,y,z) \mapsto (x^n:\dots:y^n:z)$$

This is exactly the (projective) cone over $(0:0: \dots :1)$ of the Veronese curve $(x^n:\dots:y^n:0)$. If it was an affine cone it would be contractible, but projective cones are not and in fact they have an interesting cohomology. Let $Y$ be this cone, $X$ be our Veronese curve curve and $p = (0:0: \dots : 0:1)$. Then, we use Mayer-Vietoris in the following way : $U$ is the complement of the $p$ and $V$ is the intersection of a a small ball around $p$ with $Y$, notice $V$ is contractible and $U$ retracts on $X \cong \Bbb P^1$. The Mayer-Vietoris sequence gives

$$H^0(Y) \to H^0(U) \oplus H^0(V) \to H^0(U \cap V) \to H^1(Y) \to H^1(U) \to H^1(U \cap V) \to H^2(Y) \to H^2(U) \to H^2(U \cap V) \to H^3(Y) \to H^3(U) \to H^3(U \cap V) \to H^4(Y) \to 0$$

Since $H^1(U) = H^3(U) = 0$ the interesting part of the sequence becomes

$$H^1(U \cap V) \to H^2(Y) \to H^2(U) \to H^2(U \cap V) \to H^3(Y)$$

Now I'll skip details since this is already a bit long : to understand $U \cap V$ we can image that we get a $D^*$-bundle over $X$, homotopically equivalent to a $S^1$-bundle with positive Euler class. A very close look at the Gysin sequence gives : $H^3(U \cap V) = \Bbb Z$, $H^1(U \cap V) = 0$ and that the map restriction map $H^2(U) \to H^2(U \cap V)$ is zero. If you believe it (or better check it) it follows that $\Bbb P (1,1,n)$ has the additive cohomology of $\Bbb P^2$, despite its conical singularity ! I should say this is pretty surprising.

Finally let me give you a quick application of weighted projective space. Say you have a curve $f(x,y,z) = 0$ in $\Bbb P^2$, maybe you want to look at the cyclic cover $w^n = f(x,y,z)$. This equation does not make sense until you go in a weighted projective space.