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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.

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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 $\mathbb P^n$.

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