# Complex manifolds with the same cohomology

Is there an example of two non-homeomorphic projective smooth complex varieties $X$ and $Y$ such that there exists an isomorphism $H^{\ast}(X,\mathbb{C})\rightarrow H^{\ast}(Y,\mathbb{C})$ of commutative-graded algebras compatible with Hodge filtration.

second question: Suppose we have a holomorphic map $f:Y\rightarrow X$ between two smooth projective complex varieties such that the induced map $H^{\ast}(X;\mathbb{Q})\rightarrow H^{\ast}(Y;\mathbb{Q})$ is an isomorphism. Is it true that $f$ is an isomorphism ?
• Isn't that already true for $X$ equal to $\mathbb{P}^3$ and $Y$ equal to a smooth quadric hypersurface in $\mathbb{P}^4$? – Jason Starr Dec 3 '16 at 0:36
• Please refer to the following MathOverflow answer by Francesco Polizzi to see that for $n$ odd, both linear and quadratic hypersurfaces in $\mathbb{P}^{n+1}$ have equal Betti numbers. Moreover, the Hodge filtration on each is trivial. Finally, since you are working with $\mathbb{C}$-coefficients rather than something smaller, the cup product algebra structures are also trivial. MO link: mathoverflow.net/questions/42038/… – Jason Starr Dec 3 '16 at 10:55
• @NguyenlanLee moreover you cannot extend this example in lower dimension. If $Q$ is a smooth quadric in $\mathbb{P}^3$ then $H^2(Q) \cong \mathbb{C}^2$, where a smooth plane quadric is of course isomorphic to $\mathbb{P}^1$. – Enrico Dec 3 '16 at 11:36
• In dimension $2$, one can take $\mathbb P^1 \times \mathbb P^1$ and $\mathbb P^2$ blown up at a point. They are not homeomorphic because the cup product pairing on $H^2$ is even on the first but not even on the second, but they have the same cohomology algebras because those pairings become isomorphic after tensoring with $\mathbb C$. – Will Sawin Dec 3 '16 at 13:23
• Regarding your second question, let $Y$ be a smooth quadric hypersurface in $\mathbb{CP}^4$, let $X$ be $\mathbb{CP}^3$, and let $f:Y\to X$ be a linear projection of degree $2$. The pullback ring homomorphism $f^*:H^*(X^{\text{an}};\mathbb{Z})\to H^*(Y^{\text{an}};\mathbb{Z})$ is not an isomorphism, but the pullback ring homomorphism $f^*:H^*(X^{\text{an}};\mathbb{Q})\to H^*(Y^{\text{an}};\mathbb{Q})$ is an isomorphism. – Jason Starr Dec 3 '16 at 14:18