Timeline for Does the hard Lefschetz theorem hold for hyperplane sections for singular subvarieties of smooth projective varieties?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2022 at 10:43 | history | edited | Todd Trimble | CC BY-SA 4.0 |
edits from a reposting of intended question
|
| Sep 7, 2022 at 0:22 | comment | added | Will Sawin | I think one can probably prove the map $H^{n-k} \to H^{n+k}$ is an isomorphism for $k$ greater than the dimension of the singular locus of $X$ plus one. | |
| S Sep 6, 2022 at 21:16 | history | suggested | Chef- | CC BY-SA 4.0 |
I made a new account can't access old account.
|
| Sep 6, 2022 at 20:23 | review | Suggested edits | |||
| S Sep 6, 2022 at 21:16 | |||||
| Sep 6, 2022 at 19:59 | comment | added | Will Sawin | No, this is not sufficient. The cone on a smooth projective hypersurface of lower dimension is a good example - it is itself a singular projective hypersurface, but has complicated cohomology in the "wrong" degree and thus doesn't satisfy hard Lefschetz. But please edit your question to include all the relevant information (for example that $Y$ is a hypersurface in $X$) | |
| Sep 6, 2022 at 19:51 | comment | added | user490795 | @WillSawin We know that 𝑋 is smooth and 𝑋−𝑌 is smooth is it possible to deduce $H^{n - k}(Y) \cong H^{n + k}(Y)$? | |
| Sep 6, 2022 at 19:41 | history | edited | LSpice | CC BY-SA 4.0 |
Tidying
|
| Sep 6, 2022 at 19:21 | comment | added | user490795 | @WillSawin Just because the $Y$ comes from hyperplane section on $X$ so there is more control over the singularities. | |
| Sep 6, 2022 at 19:17 | comment | added | Will Sawin | If everything takes place in $Y$, how is $X$ relevant? Ever singular projective variety is a subvariety of a smooth projective variety (e.g. $\mathbb P^n$ for $n$ sufficiently large.) | |
| Sep 6, 2022 at 19:12 | comment | added | Will Sawin | Do you mean the Lefschetz hyperplane theorem, which covers a hyperplane section of a variety, and not the hard Lefschetz theorem, which involves only the cohomology of a single variety? | |
| Sep 6, 2022 at 19:11 | history | edited | user490795 | CC BY-SA 4.0 |
added 37 characters in body; edited title
|
| S Sep 6, 2022 at 19:06 | review | First questions | |||
| Sep 6, 2022 at 19:48 | |||||
| S Sep 6, 2022 at 19:06 | history | asked | user490795 | CC BY-SA 4.0 |