# Cohomology ring of a projective variety

Let $X$ be a projective variety with ample sheaf $\mathcal{O}_X(1)$. Then $H^*(\oplus_n \mathcal{O}_X(n))$ is a graded algebra via the cup product: $H^i(\mathcal{O}(n)) \otimes H^j(\mathcal{O}(m)) \to H^{i+j}(\mathcal{O}(n) \otimes \mathcal{O}(m)) \cong H^{i+j}(\mathcal{O}(n+m))$. Remark that each individual homogeneous component is a graded abelian group (via $n$). Does this "bigraded" cohomology ring have a name, and is it studied in the literature?

If $X,Y$ are projective varieties such that the corresponding "bigraded" cohomology rings are isomorphic, do we then have $X \cong Y$? If this is false, what about the special case $Y=\mathbb{P}^d$ (here the cohomology ring is quite simple).

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I think you can have two nonisomorphic elliptic curves with the same bigraded cohomology ring, since you're essentially ignoring the $Pic^0$ component. – Parsa Apr 2 '12 at 6:23
I don't know a lot about sheaf cohomology of varieties, but I am under the impression that, for example, there are complex tori which are not algebraically isomorphic, but they're all 'just' tori(in the analytic topology) so they are homotopic and thus have identical singular cohomology rings. EDIT: and these would agree with sheaf cohomology, since these are secretly smooth manifolds. – John Stalfos Oct 25 '12 at 12:46
@MartinBrandenburg: you probably want the cohomological rings to be isomorphic as bigraded $k$-algebras (if $X, Y$ are projective varieties over $k$). What is the definition of an isomorphism of bigraded objects ? – user18119 Oct 25 '12 at 21:54
@QiL: Isn't it obvious how to define the category of bigraded algebras? – Martin Brandenburg Oct 26 '12 at 7:44
@MartinBrandenburg: If an isomorphism must preserve the bi-degrees, then the components of degrees $(0, *)$ already determine $X$. So I don't understand your initial question. – user18119 Oct 26 '12 at 15:18

If I understand your construction correctly, the bi-graded algebra is $$\oplus_{i,j} H^i(X, O_X(j))$$ and the component of degree $(i,j)$ is $H^i(X, O_X(j))$. So if an isomorphism $$\oplus_{i,j} H^i(X, O_X(j)) \simeq \oplus_{i,j} H^i(Y, O_Y(j))$$ of bi-graded algebras is required to preserve bi-degrees, hence induces linear isomorphisms $$H^0(X, O_X(j))\simeq H^0(Y, O_Y(j))$$ then $$X\simeq \mathrm{Proj}\left(\oplus_{j\ge 0} H^0(X, O_X(j))\right) \simeq \mathrm{Proj}\left(\oplus_{j\ge 0} H^0(Y, O_Y(j))\right) \simeq Y.$$