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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.

Thank you all for your very nice comments, it is helpful. I want to add a question with slightly different hypothesis.

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 ?

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    $\begingroup$ Isn't that already true for $X$ equal to $\mathbb{P}^3$ and $Y$ equal to a smooth quadric hypersurface in $\mathbb{P}^4$? $\endgroup$
    – Jason Starr
    Commented Dec 3, 2016 at 0:36
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    $\begingroup$ 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/… $\endgroup$
    – Jason Starr
    Commented Dec 3, 2016 at 10:55
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    $\begingroup$ @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 $\mathbb{P}^1$. $\endgroup$
    – Enrico
    Commented Dec 3, 2016 at 11:36
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    $\begingroup$ 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 $\mathbb C$. $\endgroup$
    – Will Sawin
    Commented Dec 3, 2016 at 13:23
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    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented Dec 3, 2016 at 14:18

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