Let $\Delta$ be a finite simplicial complex on $n$ vertices. Let $S = \mathbb{k}[\mathbf{x}]$ be a polynomial ring over a field $\mathbb{k}$ in $n$ variables and $I$ be the Stanley-Reisner ideal of $\Delta$ in $S$. It is well-known that the Betti numbers (graded or otherwise) appearing in the free resolution of $I$ may depend on the ground field $\mathbb{k}$. A classic example of this is the minimal triangulation of the real projective plane with Stanley-Reisner ideal $$I = \langle abc, abe, acf, ade, adf, bcd, bdf, bef, cde, cef \rangle.$$ When the characteristic of $\mathbb{k}$ is $2$ then the projective dimension of $I$ (as an $S$-module) is 3 while in all other cases it is 2. This example and others can be found in Dalili and Kummini's paper Dependence Of Betti Numbers On Characteristic and the references there.

In all of the examples I've seen of complexes whose Betti numbers are characteristic-dependent, the offending characteristic has always been 2, and only 2. That is, the Betti numbers are stable in all cases except when the characteristic of $\mathbb{k}$ is 2. My intuition tells me there must be ideals with other offending characteristics. As I have no idea how to construct one and my computer experiments (using RandomIdeals) have come up empty my questions are the following.

Question 1: Are there known examples of square-free monomial ideals with characteristic-dependent Betti numbers for characteristics other than two?

Question 2: Is there one whose only offending characteristic is 3?

  • 1
    $\begingroup$ According to the paper you linked, the dependence on the characteristic in your example seems to be related to the torsion in the homology group $H_1(\mathbb{PR}_2,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$. My guess would thus be that Betti numbers differ in characteristic 3 for a simplicial complex having $\mathbb{Z}/3\mathbb{Z}$ as its first homology group. Have you already tried this? $\endgroup$ – Antoine Labelle Nov 8 '20 at 22:03
  • $\begingroup$ @AntoineLabelle I don't know how to construct such a thing so I'm looking into Moore spaces and the like. If you have an example at hand I'm happy to run the calculation to see if it's characteristic dependent. $\endgroup$ – Aaron Dall Nov 8 '20 at 22:14
  • $\begingroup$ I looked at Lutz' 12 vertex triangulation of the lens space L(3,1). Neither the Stanley-Reisner ideal nor the facet ideal of this complex has the desired property. $\endgroup$ – Aaron Dall Nov 9 '20 at 0:30
  • 1
    $\begingroup$ For a 2 dimensional example, you might try yo look at the presentation complex (en.wikipedia.org/wiki/Presentation_complex) of the presentation $\mathbb{Z}/3\mathbb{Z}=\langle x | x^3=1 \rangle$. It has a triangulation with 9 vertices and facets $abd,acd,abf,acf,abe,ace,def,bcg,ceg,deg,bdg,bch,cfh,efh,beh,bci,cdi,dfi,bfi$. $\endgroup$ – Antoine Labelle Nov 9 '20 at 1:29
  • $\begingroup$ That works! The projective dimension of the Stanley-Reisner ideal of that complex is 6 in characteristic 3 and is 5 in other characteristics. If you want to post it as an answer I'd happily accept it (and add the Betti tables if necessary). Thanks! $\endgroup$ – Aaron Dall Nov 9 '20 at 8:01

Consider the presentation complex of $\mathbb{Z}/3\mathbb{Z}=\langle x|x^3=1 \rangle$, consisting of a full triangle with all three edges identified in the same orientation. It has a triangulation with 9 vertices and facets $abd,acd,abf,acf,abe,ace,def,bcg,ceg,deg,bdg,bch,cfh,efh,beh,bci,cdi,dfi,bfi$.

The projective dimension of the corresponding Stanley-Reisner ideal is 6 in characteristic 3 and 5 in all others characteristics, which answers question 2 positively.


Question 1: There is a 3-dimensional simplicial complex $\Delta$ with $f$-vector $(1, 9, 36, 84, 42)$ whose Stanley-Reisner ideal has characteristic-dependent Betti numbers. Here is some Macaulay2 code to construct such an example.

-- set up some polynomial rings
X = vars(0..8);
char2 = ZZ/2[X];
char3 = ZZ/3[X];
char5 = ZZ/5[X];
char0 = QQ[X];

-- the square-free ideal
I = ideal "abcd,abce,abde,acde,bcde,abcf,abdf,acdf,bcdf,abef,acef,bcef,abcg,abdg,bcdg,aceg,bceg,adeg,bdeg,cdeg,bcfg,cdfg,befg,defg,abch,abdh,acdh,aceh,cdeh,acfh,bcfh,adfh,bdfh,cdfh,aefh,defh,abgh,acgh,bcgh,adgh,bdgh,cdgh,aegh,degh,bfgh,dfgh,efgh,abci,abdi,abei,acei,bdei,abfi,bcfi,adfi,aefi,befi,cefi,defi,abgi,acgi,bcgi,adgi,aegi,begi,cegi,degi,cfgi,dfgi,efgi,abhi,achi,bchi,bdhi,cehi,dehi,afhi,bfhi,cfhi,dfhi,efhi,dghi,eghi,fghi"

-- make the betti computations
betti res module sub(I, char0)
betti res module sub(I, char2)
betti res module sub(I, char3)
betti res module sub(I, char5)

This settles Question 1 but doesn't quite answer Question 2 because I don't have a proof that the only offending characteristic is 3 but it's a start in that direction.

Edit: Using the SimplicialComplexes package in Macaulay2 one can show that $\Delta$ has one non-trivial homology group over $\mathbb{Z}$: $\tilde{H}_2(\Delta; \mathbb{Z})= \mathbb{Z}^{14}\oplus\mathbb{Z}/3\mathbb{Z}$. So by Remark 2.2 in the paper cited in the OP its Betti numbers are only characteristic dependent when the characteristic is three.

The ideal $I$, like Reisner's example in the OP, has a linear resolution in most characteristics.


This is not a direct answer to either of your questions, but I don’t have enough reputation to comment. Perhaps the recent preprint of Booms, Erman and Yang might be of interest to you: https://arxiv.org/pdf/2007.13914.pdf

The title references Veronese syzygies (Edit: looks like the title changed in a recent update and no longer references Veronese syzygies), but the heuristic developed is based on syzygies of random Stanley-Reisner ideals.

  • $\begingroup$ Thanks! That's a great reference. $\endgroup$ – Aaron Dall Nov 9 '20 at 8:02

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.