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Let $V$ be a linear space, $V_1$, $V_2$ and $V_3$ be linear subspaces of $V$. Consider $\mathbb P(V/V_1)\times\mathbb P(V/V_2)\times\mathbb P(V/V_3)$ as a space parametrizing the triples of linear subspaces $(W_1, W_2,W_3)$, with $V_i\subset W_i\subset V$ and $\dim(W_i)-\dim(V_i)=1$ for $i=1,2,3$, and also $V_1\cap V_2\cap V_3=0$.

Now consider the closed subspace P consisting of the points $(W_1,W_2,W_3)\in \mathbb P(V/V_1)\times\mathbb P(V/V_2)\times\mathbb P(V/V_3)$, where $\bigcap_IW_i\supsetneqq \bigcap_IV_i$, with $I$ arbitrary subset of $\{1,2,3\}$.

Then the question is: now that P is a closed subscheme of $\mathbb P(V/V_1)\times\mathbb P(V/V_2)\times\mathbb P(V/V_3)$, how to compute trivariate Hilbert Polynomial of P without explicitly writing down the coordinates of P?

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Your question isn't correct. Such $W_1$ are parametrized by $\mathbb P (V\setminus V_1)$ not by $\mathbb P (V/V_1)$ as you wrote. And $\mathbb P (V\setminus V_1)$ is not a closed variety.

In any case, $W_1\cap W_2\cap W_3\ne 0$ means that there is vector $v\in W_1\cap W_2\cap W_3$. So, you can parametrize almost all such triples by vectors $v$. So, we found one component of $P$ (it is $\mathbb P(V) $). The other ones are when $W_1\cap W_2\subset V_3$ and the same triples with permuting indices (and if $V_1\cap V_2\cap V_3 = 0$ then these triples are parametrized by $\mathbb P(V_3)$). Etc. It is easy to compute Hilbert polynomial for linear subspaces. But if $V_1\cap V_2\cap V_3\ne 0$ then situation is more complicated, and< I think it is strange to expect a good answer.

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The question is correct. Such $W_1$ are parametrized by $\mathbb P(V/V_1)$. You are right in the sense that if I do not require $\V_1\cap V_2\cap V_3=0$, the problem is really complicated. So I add this condition after revising my problem! –  BLI Nov 22 '11 at 5:47
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