Question Regarding Proof of Hodge Index Theorem

Question

I am reading Voisin's proof of the Hodge Index Theorem on pp. 153-154 of her Hodge Theory and Complex Algebraic Geometry I. The proof is mostly clear except for one technical point. Let $n$ denote an even number, let $X$ denote a compact Kähler manifold, and let $h^{a,b}=h^{a,b}(X)=\dim H^{a,b}(X).$ Voisin claims that using Poincaré duality, we can get $$ 2\sum_{a+b=n-2r,r>0}(-1)^a h^{a,b}=\sum_{a+b\equiv n(2), a+b\ne n} (-1)^ah^{a,b}.$$ It's likely that I'm simply misunderstanding what is meant here, but I understand the left hand side as saying $$ 2\sum_{r=1}^{n/2}\sum_{a+b=n-2r}(-1)^ah^{a,b}=\sum_{a+b\equiv n(2), a+b\ne n} (-1)^ah^{a,b}.$$ However, given my interpretation, choosing $n=2$, the left hand side is $$ 2\sum_{a+b=0}(-1)^ah^{a,b}=2h^{0,0}$$ because $a+b\ge 0$ implicitly. The right hand side becomes a sum over $a,b$ so that $a+b\equiv 0 \pmod{2}$, and $a+b\ne 0$ subject to the same constraint of $a+b\ge 0$. So, it looks like the right hand side reads $h^{0,0}$. Then it looks like this implies that $$ 2h^{0,0}=h^{0,0}$$ so that $h^{0,0}=0$. But that implies $h^{0,0}(X)=h^0(X,\mathbb{C})=0$, which is false for instance in the case of $X=\mathbb{P}^2$.

Answer 1

A friend of mine pointed out that my example calculation is incorrect. Indeed, because we have an $n-$dimensional complex manifold, and we are manipulating real forms, we must have $0\le a+b\le 2n$ and $0\le a,b\le n$. In particular, the right hand side should really read: $$ h^{0,0}+h^{2,2}+h^{3,1}+h^{1,3}=h^{0,0}+h^{2,2}=2h^{0,0}$$ by Poincaré duality. Indeed, by Poincaré duality, we know that $h^{2,2}=h^{0,0}$. This pretty much explains the general case, too. On the right hand side, we have a sum over $0\le a+b\le 2n$, with $a+b$ even and $a+b\ne n$. It's then clear that $$\{(a,b): 0\le a+b< n\:\text{with $a+b$ even}\}\leftrightarrow\{(a,b):n< a+b\le 2n\:\text{with $a+b$ even}\}$$ with the bijection induced by the Poincaré duality map $H^{p,q}(X)\to H^{n-p,n-q}(X)$. Indeed, $n<a+b\le 2n$ implies $0\le n-p+n-q<n$ and conversely. Expanding the right hand sum $$ \sum_{a+b\equiv n(2), a+b\ne n} (-1)^ah^{a,b}, $$ we see that each summand of $$ \sum_{r=1}^{n/2}\sum_{a+b=n-2r} (-1)^ah^{a,b}$$ is double counted, and the result follows.