4
$\begingroup$

This is probably a very simple question, but I can't seem to see the answer. Given a (Zariski) closed subset $Z$ in $\mathbb P^n$ of codimension $d$, we can always find a linear subspace $L \cong \mathbb P^{d-1}$ such that $L \cap Z = \varnothing $. I'm wondering if a similar statement can be said about Grassmannians.

Specifically, let $G= \mathbb G(1,n)$ be the Grassmannian of lines in $\mathbb P^n$ with $n \geq 3$. If $Z$ is a closed subset in $G$ of codimension $2$, can we always find a $\Sigma_{x,P} \cong \mathbb P^1$ in $G$ with $\Sigma_{x,P} \cap Z = \varnothing $, where $\Sigma_{x,P}$ corresponds to the locus of lines through $x$ lying in a plane $P= \mathbb P^2 \subset \mathbb P^n$ containing $x$ ?

Similarly, if $Z$ is of codimension $3$ in $G$, can we always find a $\Sigma_P \cong \mathbb P^2$ with $\Sigma_P \cap Z = \varnothing $, where $\Sigma_P$ corresponds to the locus of lines lying on a plane $P= \mathbb P^2 \subset \mathbb P^n$ ?

$\endgroup$
1

1 Answer 1

10
$\begingroup$

Indeed much more is true.

Suppose $X$ is an irreducible algebraic manifold admitting a transitive action of a linear algebraic group $G$. If $Y$ and $Z$ are irreducible subvarieties of $X$ then for a general $g \in G$ the intersection of $gY$ (the translate of $Y$ by $g$) and $Z$ is empty or equidimensional of dimension $\mathrm{dim}(Y) + \mathrm{dim}(Z) -\mathrm{dim}(X)$.

See Kleiman's "The transversality of a general translate".

$\endgroup$
2
  • $\begingroup$ Note that some of Kleiman's results only hold in characteristic 0, but my guess (without having it at hand) would be that this one is general. $\endgroup$ Mar 27, 2011 at 0:49
  • $\begingroup$ You are right the result above holds true in general, no need to restrict to char zero. If one further assume that $Y$ and $Z$ are smooth and want to conclude that the intersection is also smooth then one does need to restrict to char zero. $\endgroup$ Mar 27, 2011 at 2:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.