Suppose $Y$ is a hypersurface of a smooth projective $X$. For which degree k values defined below, do we always get isomorphism when studying the cohomology of $Y$? What can be said about the general version of this? If $Y$ is a subvariety of codimension say $i$, to which codimension $i$ and to which $k$ do we get isomorphism between $H^{n - k}(X) \cong H^{n + k}(X)$?
$\begingroup$
$\endgroup$
6
-
2$\begingroup$ 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? $\endgroup$– Will SawinCommented Sep 6, 2022 at 19:12
-
5$\begingroup$ 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.) $\endgroup$– Will SawinCommented Sep 6, 2022 at 19:17
-
$\begingroup$ @WillSawin Just because the $Y$ comes from hyperplane section on $X$ so there is more control over the singularities. $\endgroup$– user490795Commented Sep 6, 2022 at 19:21
-
$\begingroup$ @WillSawin We know that 𝑋 is smooth and 𝑋−𝑌 is smooth is it possible to deduce $H^{n - k}(Y) \cong H^{n + k}(Y)$? $\endgroup$– user490795Commented Sep 6, 2022 at 19:51
-
1$\begingroup$ 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 $X$) $\endgroup$– Will SawinCommented Sep 6, 2022 at 19:59
|
Show 1 more comment