Homological dimension of a curve

Question

I'm studying Huybrechts' "Fourier-Mukai transforms in algebraic geometry". In the $3^{rd}$ chapter, when he shows that any object in $D^b(C)$, with $C$ a smooth projective curve, is $\oplus \mathcal{E}_i[i]$ (up to isomorphism), he says: "The homological dimension of a curve is one". What does it mean? I know the definition of homological dimension for a category (the maximum $p$ such that there exists two object $X$ e $Y$ with $Ext^p(X,Y) \neq 0$), but not for a curve.

Answer 1

For $X$ a smooth projective variety, the cohomological dimension of $X$ is the cohomological dimension of the Abelian category $\operatorname{Coh}(X)$. We might hope that this agrees with the ordinary notion of dimension, and in this case it does.

Indeed, if $X$ is a smooth projective variety and $\mathcal{F},\mathcal{G}$ are coherent sheaves, then the cohomological dimension of $X$ is $n=\dim X$. If $i>n$ then by Serre Duality, $$ \operatorname{Ext}^i(\mathcal{F},\mathcal{G})\cong \operatorname{Ext}^{n-i}(\mathcal{G},\mathcal{F}\otimes\omega_X)^*=0 $$ because $n-i<0$. So, the cohomological dimension is $\le n$. On the other hand, $$0\ne H^0(X,\mathcal{O}_X)=\operatorname{Ext}^0(\mathcal{O}_X,\mathcal{O}_X)\cong \operatorname{Ext}^n(\mathcal{O}_X,\omega_X)^*$$ so that the cohomological dimension is $\ge n$. Hence, $\operatorname{cd}\operatorname{Coh}(X)=\dim X.$