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Let $k$ be a field and let $R$ be a commutative (standard) graded $k$-algebra, that is, $R=\bigoplus_{n=0}^\infty R_n$ with $R_0=k$ (and $R=k[R_1]$). The Hilbert function $h_R:\mathbb{N}\rightarrow \mathbb{N}$ is given by $h_R(n)=\dim_k R_n$. When $n\gg 0$, it is known that $h_R(n)$ agrees with a polnyomial $p_R(n)$ in $n$ with rational coefficients.

In many theorems in combinatorics, polynomials which count objects have combinatorially sensible meaning when evaluated at negative numbers. For example, the chromatic polynomial $\chi_G(x)$ of a graph, which counts the number of proper colorings of $G$ with $x$ colors, has the delightful property that $\chi_G(-1)$ is the number of acyclic orientations of $G$.

One other example, intimately related to the Hilbert function, is that of Ehrhart polynomials. The Ehrhart polynomial $E_P(n)$ of a convex integral polytope $P$ inside $\mathbb{R}^m$ is the number of lattice points of $\mathbb{Z}^m$ inside the $n$th dilate $nP$. It is also known that $E_P(-n)$ is the number of interior lattice points (up to sign) of $nP$. For some (all?) polytopes, the Ehrhart polynomial agrees with the Hilbert polynomial of an associated affine semigroup ring.

In general, or in the specific context of affine semigroup rings $R$, is there some sensible (algebraic or combinatorial) meaning to $p_R(-n)$, or even $p_R(-1)$?

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    $\begingroup$ This is a cross-post from Math StackExchange, where it has received no answers or comments. math.stackexchange.com/questions/4670444/… $\endgroup$
    – walkar
    Commented Apr 14, 2023 at 14:04
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    $\begingroup$ You can also do this with nonstandardly graded commutative $k$-algebras, and get interesting answers. The Hilbert function is then not polynomial, but "polynomial on residue classes" - in other words, there exists a modulus $m$, and a set of $m$ polynomials $p_i$ such that if $n$ is congruent to $i$ modulo $m$, you apply $p_i$ to get $h_R(n)$. As an example, if $G$ is a compact Lie group, and $R=H^*(BG,k)$ (in any characteristic) then $h_r(n)$ is zero for $n$ between $-\dim G$ and $-1$. This follows from a calculation of Castelnuovo-Mumford regularity, contained in a paper of Peter Symonds. $\endgroup$ Commented Apr 14, 2023 at 14:32
  • $\begingroup$ @DaveBenson Thank you for the reference! I believe this is called "quasi-polynomial" behavior, isn't that right? It is very interesting to know that there are some negative zeroes of $h_R(n)$. Is there any, say topological, interpretation about the Lie group that comes about from knowing the zeroes of the Hilbert polynomial of these rings? $\endgroup$
    – walkar
    Commented Apr 14, 2023 at 16:14
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    $\begingroup$ It goes by various names, including quasi-polynomial. Graham Higman called it PORC, for "polynomial on residue classes". The functions I was referring to are only eventually quasi-polynomial, which Higman called "almost PORC". The zeroes in the case of a compact Lie group are related to a sort of derived Gorenstein property with a shift equal to $-\dim G$. $\endgroup$ Commented Apr 14, 2023 at 16:20
  • $\begingroup$ @DaveBenson Yes, I was thinking of it asymptotically, so "almost" is good enough for me. $\endgroup$
    – walkar
    Commented Apr 14, 2023 at 16:51

2 Answers 2

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This answer is essentially the same as that of Phil Tosteson, written before I saw that post. I also mention a non-Cohen-Macaulay example at the end.

If $R$ is Cohen-Macaulay (but not necessarily generated in degree one), then $R$ has associated with it a canonical module $\Omega(R)$ which can be graded so its Hilbert function agrees with $(-1)^d p_R(-n)$ for $n$ sufficiently large, where $d$ is the Krull dimension of $R$ or $\Omega(R)$. If $R$ has $n$ generators then it can be regarded as a module over the polynomial ring $A=k[x_1,\dots,x_n]$. One can then define $\Omega(R) = \mathrm{Ext}^{n-d}_A(R,A)$. If $R$ is not Cohen-Macaulay, then there are "correction terms" to the formula $p_R(-n) = (-1)^d\mathrm{HQ}(\Omega(R))$, where $\mathrm{HQ}$ denotes Hilbert quasipolynomial. Namely, $$ p_R(-n) = (-1)^d \sum_{i=0}^d (-1)^i\mathrm{HQ}\left( \mathrm{Ext}^{n-d+i}_A(R,A)\right). $$ (If $R$ is Cohen-Macaulay, then only the term indexed by $i=0$ doesn't vanish.) I don't know where this result is stated in precisely this form, but it is equivalent to Theorem 6.4 of my book Combinatorics and Commutative Algebra, second ed. Theorem 8.2 gives an example, stated in terms of Hilbert series rather than Hilbert quasipolynomials.

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  • $\begingroup$ Thank you! I had looked at your book and several others when thinking about Ehrhart polynomials recently. In fact, your paper about combinatorial reciprocity inspired my initial thoughts about this question! $\endgroup$
    – walkar
    Commented Apr 14, 2023 at 17:46
  • $\begingroup$ Although it is quite obvious, $A = k[x_1,\ldots,x_n]$ in the above, yes? ($A$ is not defined). $\endgroup$
    – Pedro
    Commented Apr 14, 2023 at 17:54
  • $\begingroup$ @Pedro: yes $A=k[x_1,\dots,x_n]$. $\endgroup$ Commented Apr 14, 2023 at 18:55
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For $n \gg 0$, we have that ${\rm dim~} R_n = {\rm dim~} H^0(X, \mathcal O(n))$, where $X = {\rm Proj}(R) \subseteq \mathbb P(R_1^{\vee})$ is the projective scheme associated to $R$ and $\mathcal O(n)$ is the degree $n$ line bundle.

In fact, we have that for all integer values that $$p_R(n) = \chi(H^*(X,\mathcal O(n))) := \sum_{i} (-1)^i {\rm dim} ~ H^i(X, \mathcal O(n)).$$

This gives an interpretation of $p_R(n)$ for negative values, but it is somewhat abstract in general. However in special cases (e.g. when $X$ is Cohen-Macaulay) there are more concrete interpretations of these Euler characteristics.

Let $K$ be a dualizing complex for $X$. Then ${\rm Ext}^*_X(\mathcal O,\mathcal O(-n))$ is graded dual to $Ext^*(\mathcal O(-n),K) = H^*(X, K(n))$. Thus in particular, if $X$ is Cohen-Macaulay of dimension $d$, so that $K = \omega_X[d]$ for some dualizing sheaf $\omega_X$, then we get that for all $n \gg 0$, that $$p_R(-n) = (-1)^d \chi(H^*(X, \omega_X(n)) = (-1)^d {\rm dim~} H^0(X, \omega_X(n)).$$ (We used that $n \gg 0$ in the last equality). So in this case, the negative values of $p_R(n)$ can be interpreted as the graded dimensions of the dualizing module of $R$.

As an example, you can take $R = k[x_0, \dots, x_d]$ so that $X = \mathbb P^d$ and $\omega_X = \mathcal O(-d-1)$ and one recovers the combinatorial reciprocity $${-n + d \choose d} = (-1)^d {n -1 \choose d}.$$

Also this story is closely related to Ehrhart reciprocity, because Ehrhart functions are Hilbert functions on toric varieties.

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  • $\begingroup$ Very nice! Thank you for the answer. I suppose I should have expected a canonical module would show up. I am most interested in non-Cohen-Macaulay examples, but this gives me somewhere to start. $\endgroup$
    – walkar
    Commented Apr 14, 2023 at 16:55
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    $\begingroup$ (I liked both answers, but accepted Richard Stanley's because I was looking for references and he included a reference.) $\endgroup$
    – walkar
    Commented Apr 15, 2023 at 15:46

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