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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.

I've been reading about weighted projective space here: https://arxiv.org/pdf/1108.1938.pdf

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.

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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.

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