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?


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.

| cite | improve this answer | |
  • $\begingroup$ 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! $\endgroup$ – BLI Nov 22 '11 at 5:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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