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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 $d$?

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Can anyone please help me fixing the dieresis on "o"? (There is nothing worse than quoting names badly, sorry for that.) –  Brenin Jan 9 at 10:57
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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 . –  Vivek Shende Jan 10 at 4:01
    
@VivekShende: thanks, this is a very nice reference to know about! –  Brenin Jan 10 at 15:34

1 Answer 1

up vote 1 down vote accepted

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 $p_d$.

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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.) –  Brenin Jan 10 at 15:32
    
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 $n$. –  Will Sawin Jan 10 at 15:45
    
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! –  Brenin Jan 10 at 17:11
    
No the fiber is the disjoint union over all partitions of $n$ into $k$ numbers $a_1,\dots,a_k$ of the product of $Hilb^{a_i}(X)_x = Hilb^{a_i}(\mathbb C^d)_0$. –  Will Sawin Jan 10 at 17:13

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