Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider subschemes $F$ of the Grassmannian $\mathbb{G}(k,n)$ satisfying the condition that each point of $\mathbb{P}^n$ is contained in only finitely many of the $k$-planes in $F$. Does this give us some sort of partial map of Chow groups $A^d(\mathbb{G}(k,n))\to A^{something}(\mathbb{P}^n)?$

(Here I'm pretty sure $something=n-((k+1)(n-k)-d+k).$)

Can this condition be dropped by correctly picking representatives of $A^d(\mathbb{G}(k,n))$?

The motivation behind this is the following. The class in $A^3(\mathbb{G}(1,3))$ of lines contained in a quadric surface $X$ is $[F_1(X)]=4\sigma_{2,1}.$ These lines trace out $X$ with multiplicity 2, i.e. each line is contained in exactly two lines of $F_1(X).$ So the map of Chow groups that I want to exist would be $[F_1(x)]=4\sigma_{2,1}\mapsto 2[X]=4[H],$ where $[H]$ is the class of a plane. The fours showing up in both places seems potentially not a coincidence.

Here's a very concrete question I (and seemingly Google) don't know the answer to. If I calculated everything correctly, the class of lines on a general quartic surface in $\mathbb{P}^4$ is $544\sigma_{3,2}.$ Does anyone happen to know if it's true that through each point there are 544/4=136 lines? (This would correspond to the map $544\sigma_{3,2}\mapsto544[H]=136[Q].$)

I apologize if this question is too basic, but I didn't get any answers at stackexchange and couldn't find any answers on Google. My stackexchange post is here and is more detailed, if a little bit less coherent.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

There is a natural $\mathbb P^k$-bundle on the Grassmanian, with a natural map to $\mathbb P^n$. Pull back your cycle to the bundle, then push forward to $\mathbb P^n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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