5
$\begingroup$

Let $d>1$ be an integer. If $n\geq 0$ is an integer we have a notion of $d$-dimensional partitions of $n$; the number of these, denoted $p_d(n)$, is the number of ways we can stack $n$ ($d$-dimensional) boxes in a corner of a $d$-dimensional "room". No closed formula is known for $p_d$, for any $d>1$. As far as I know, the generating function $\mathcal P_d$ for $p_d$ is known for $d=2,3$, but for no higher $d$'s: \begin{align} \mathcal P_2=\sum_{n\geq 0}p_2(n)t^n&=\prod_{k\geq 1}(1-t^k)^{-1},\notag\\ \mathcal P_3=\sum_{n\geq 0}p_3(n)t^n&=\prod_{k\geq 1}(1-t^k)^{-k}.\notag \end{align}

However, it seems to me that to find $p_d(n)$ is to find the number of "higher dimensional Young Tableaux", and these correspond to monomial ideals in $\mathbb C^d$. So it should be true that $$p_d(n)=\chi(\textrm{Hilb}^n(\mathbb C^d)_0),$$ the topological Euler characteristic of the punctual Hilbert scheme. It is also true that, if $S$ is a smooth projective surface and $Y$ is a smooth projective threefold, then \begin{align} \sum_{n\geq 0}\chi(\textrm{Hilb}^nS)t^n&=\mathcal P_2^{\chi(S)}\,\,\,\,\,\,\,\textrm{(Göttsche's formula)}\notag\\ \sum_{n\geq 0}\chi(\textrm{Hilb}^nY)t^n&=\mathcal P_3^{\chi(Y)} \,\,\,\,\,\,\,\textrm{(Cheah's formula)}\notag \end{align}

Question: do we have such formulas for any $d$? in other words, do we have $$\sum_{n\geq 0}\chi(\textrm{Hilb}^nX)t^n=\mathcal P_d^{\chi(X)}$$ for any smooth projective $X$ of dimension $d$?

$\endgroup$
3
  • $\begingroup$ Can anyone please help me fixing the dieresis on "o"? (There is nothing worse than quoting names badly, sorry for that.) $\endgroup$
    – Brenin
    Jan 9, 2014 at 10:57
  • 1
    $\begingroup$ In fact, a version of that equality holds even for 'universal' Euler characteristics, i.e. in the Grothendieck ring of varieties. See arxiv.org/pdf/math/0407204v1.pdf . $\endgroup$ Jan 10, 2014 at 4:01
  • $\begingroup$ @VivekShende: thanks, this is a very nice reference to know about! $\endgroup$
    – Brenin
    Jan 10, 2014 at 15:34

1 Answer 1

2
$\begingroup$

Yes.

Write $\mathcal P_d= 1 + p_d$, so $\mathcal P_d^{\chi(X)}= \sum_{k=0}^{\infty} \left( \begin{array}{c} \chi( X) \\ k \end{array}\right) p_d^k$.

I will show that $\left( \begin{array}{c} \chi( X) \\ k \end{array}\right) p_d^k$ is the generating function for the stratum of $Hilb^n X$ consisting of subschemas that are supported on $k$ distinct points.

This subscheme is a fibration over the variety $\left( \begin{array}{c} X \\ k \end{array}\right)$, the variety of all sets of $k$ distinct points in $X$. We can easily check that the Euler characteristic of $\left( \begin{array}{c} X \\ k \end{array}\right)$ is $\left( \begin{array}{c} \chi(X) \\ k \end{array}\right)$. The Euler characteristic of a fibration is the Euler characteristic of the base times the Euler characteristic of the fiber. So we must show that the Euler characteristic of the fiber is $p_d^k$. But this is clear - it's just the Hilbert scheme of subschemes supported exactly at $k$ distinct fixed points, which is just a $k$-fold product of the hilbert scheme of nonempty subschemes supported at a single point, which is $p_d$.

$\endgroup$
4
  • $\begingroup$ Thanks for your answer! Following your proof, I find instead $\chi(\textrm{Hilb}^n_kX)=\binom{\chi(X)}{k}p_d(n)^k$ (small $p$), so that the generating series $\sum_n\chi(\textrm{Hilb}^n_kX)t^n=\binom{\chi(X)}{k}\sum_np_d(n)^kt^n\neq \binom{\chi(X)}{k}P_d^k$. (the latter is big $P$.) Where is my mistake? (Here $\textrm{Hilb}^n_kX$ is the Hilb of subschemes supported on $k$ distinct points.) $\endgroup$
    – Brenin
    Jan 10, 2014 at 15:32
  • $\begingroup$ It's not $p_d(n)^k$, because the whole thing having degree $n$ does not mean the points have degree $n$. Instead, the degrees sum to $n$. $\endgroup$
    – Will Sawin
    Jan 10, 2014 at 15:45
  • $\begingroup$ Sorry, I do not get it. According to what $\mathcal P_d$ is, we have $p_d=\sum_{n\geq 1}p_d(n)t^n$. How can $p_d^k$ equal $\chi(\textrm{fiber})$? If I got your last comment, the fiber is $\prod_{1\leq i\leq k}\textrm{Hilb}^i(X)_{x}$, so its $\chi$ is $\prod_{1\leq i\leq k}p_d(i)$. Sorry to bother you! $\endgroup$
    – Brenin
    Jan 10, 2014 at 17:11
  • $\begingroup$ No the fiber is the disjoint union over all partitions of $n$ into $k$ numbers $a_1,\dots,a_k$ of the product of $Hilb^{a_i}(X)_x = Hilb^{a_i}(\mathbb C^d)_0$. $\endgroup$
    – Will Sawin
    Jan 10, 2014 at 17:13

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.