Timeline for Complex manifolds with the same cohomology
Current License: CC BY-SA 3.0
14 events
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| Dec 3, 2016 at 15:18 | comment | added | Nguyen lan Lee | @JasonStarr Thank you for the explanation, I think I start to understand now what is going on. | |
| Dec 3, 2016 at 14:18 | comment | added | Jason Starr | 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. | |
| Dec 3, 2016 at 13:40 | history | edited | Nguyen lan Lee | CC BY-SA 3.0 |
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| Dec 3, 2016 at 13:23 | comment | added | Will Sawin | 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 $\mathbb C$. | |
| Dec 3, 2016 at 11:36 | comment | added | Enrico | @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 $\mathbb{P}^1$. | |
| Dec 3, 2016 at 10:55 | comment | added | Jason Starr | 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/… | |
| Dec 3, 2016 at 9:46 | comment | added | Nguyen lan Lee | Professor @JasonStarr I just noticed your comment! Could you please give more details about $Y$ in your example. It will be nice if there is such example in "low" dimension. | |
| Dec 3, 2016 at 1:03 | comment | added | T. Amdeberhan | Here is a copy of a paper mathunion.org/ICM/ICM2010.1/Main/icm2010.1.0476.0503.pdf See if it might be helpful to you in some way; in particular page 496. | |
| Dec 3, 2016 at 0:36 | comment | added | Jason Starr | Isn't that already true for $X$ equal to $\mathbb{P}^3$ and $Y$ equal to a smooth quadric hypersurface in $\mathbb{P}^4$? | |
| Dec 3, 2016 at 0:10 | history | edited | YCor | CC BY-SA 3.0 |
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| Dec 2, 2016 at 23:55 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Dec 2, 2016 at 23:53 | history | edited | Nguyen lan Lee | CC BY-SA 3.0 |
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| Dec 2, 2016 at 23:52 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Dec 2, 2016 at 23:50 | history | asked | Nguyen lan Lee | CC BY-SA 3.0 |