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I have some affine varieties $X$ defined over $\mathbb Z$, and associated integers $c(X)$, with the property that $\# X_{\mathbb Z/p} \equiv c(X) \bmod p$ for all $p$. (In particular $c(X)$ is usually $1$ in my examples, but not always.)

Is this property related to any other interesting properties of $X$, e.g. "ordinary", "unirational", etc.?

I am happy to entertain conjectures, necessary conditions, sufficient conditions, anything. I looked at the 33 occurrences of the word "constant" in Serre's Lectures on $N_X(p)$, but haven't delved deeper into that book.

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  • $\begingroup$ Reminds me of work of Dmitry Doryn on "The dual graph polynomials and a 4-face formula". He uses the coefficients of $q^2$. Maybe also look at Helene Esnault's articles ? $\endgroup$
    – F. C.
    Jun 25, 2015 at 20:06
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    $\begingroup$ Esnault proves that for a proper, regular variety $X_R$ over a finitely generated $\mathbb{Z}$-module $R$, if the geometric generic fiber has "coniveau 1", e.g., if it is rationally connected, then for every maximal ideal $\mathfrak{m}$ of $R$ with finite residue field $R/\mathfrak{m} \cong \mathbb{F}_q$, the number $X_{R/\mathfrak{m}}(\mathbb{F}_q)$ is congruent to $1$ modulo $q$. She may have considered affine varieties, but I am not aware of that. "Similar" results for affine varieties $Y$ usually have that $Y(\mathbb{F}_q)$ is congruent to $0$, not $1$ (e.g., affine spaces). $\endgroup$ Jun 25, 2015 at 20:33
  • $\begingroup$ @JasonStarr, is $\mathbf F_q$ a potentially general finite field and not one of prime order? If so, "modulo $q$" is not what you mean (e.g., $q = 8$). $\endgroup$
    – KConrad
    Jun 25, 2015 at 20:35
  • $\begingroup$ Can you work out a complete formula for $\#X({\mathbf F}_p)$ as $p$ varies, e.g., compute the zeta-function of $X_{/{\mathbf F}_p}$? Perhaps the counting formula for $\#X({\mathbf F}_p)$ is a universal polynomial in $p$ (like Hall polynomials). $\endgroup$
    – KConrad
    Jun 25, 2015 at 20:36
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    $\begingroup$ @KConrad. "If so, 'modulo $q$' is not what you mean (e.g., $q=8$)." I believe that this is what Esnault works with: a congruence for the number of points modulo the cardinality of the finite field. I do realize that the number of points changes when we base change from $\mathbb{F}_q$ to $\mathbb{F}_{q^r}$, but the number is congruent to $1$ modulo $q$, resp. $q^r$. $\endgroup$ Jun 26, 2015 at 0:24

4 Answers 4

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The Grothendieck-Lefschetz trace formula tells you that $$ \#X(\mathbf F_q) = \sum (-1)^i \mathrm{Tr}(\mathrm{Frob}_q \mid H^i_c(X)),$$ where all cohomology is with coefficients in $\mathbf Q_\ell$. Suppose all the cohomology of $X$ is of Tate type. For example, maybe there is a compactification $X \subset \overline X$ with a stratification such that the closure of each stratum has cohomology spanned by algebraic cycles. In this case, $$ \mathrm{Tr}(\mathrm{Frob}_q \mid H^i_c(X)) \equiv \dim W_0 H^i_c(X) \pmod q,$$ where $W$ denotes the weight filtration. Indeed, $\mathrm{Frob}_q$ acts as the identity on $W_0$ and as multiplication by $q^d$ on $\mathrm{Gr}^W_{2d}$. So in this case, your equation is satisfied, with $c(X) = \sum (-1)^i \dim W_0 H^i_c(X)$; equivalently, $c(X)$ is the Hodge-Deligne polynomial $E(X,u,v)$ specialized at $u=v=0$.

I know that Carel Faber (unpublished) used this to prove the existence of cohomology that is neither of Tate type, nor associated to Siegel modular forms of genus $\leq 3$, in the moduli space $\mathcal M_{3,n}$ for some $n$ that I can not recall. He computed the number of points mod $p$ for various $p$ and found that this was not a constant independent of $p$, nor was it a linear combination of traces on Galois representations attached to Siegel cusp forms (for the full modular group) of low enough weight and genus.

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  • $\begingroup$ Yes; varieties whose motives are mixed Tate give one of the "easy pieces of $K_0(Mot_{num})$". $\endgroup$ Jun 26, 2015 at 12:32
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Do you know that $\#X(\mathbb F_q)$ is a multiple of $p$?

If you do, then for all sufficiently large primes $p$, we have

$$\zeta(X) \equiv (1-t)^{-c(X)} \mod p$$

Hence each eigenvalue of Frobenius that appears with nonzero signed multiplicity is either $1$ or is a multiple of $p$.

In terms of Galois representations, you get that each Galois representation appearing in the cohomology of $X$ with nonzero signed multiplicity is either the trivial Galois representation or has its eigenvalues multiples of $p$ for almost all primes. (Well there's small issues if eigenvalues cancel each other but let's ignore those).

This is actually related not to ordinariness but to supersingularity. For instance, a supersingular elliptic curve always has $p+1$ points. However, we believe all Galois representations are ordinary most of the time. Moreover, even if this is false, it should be very unlikely that you can find a counterexample by some fixed explicit method, so I think we can assume it is true in your case.

So I guess all Galois representations of weight $i$ that appear with nonzero signed multiplicity should have Hodge numbers in the interval $[1, i-1]$.

This is a very weak geometric condition because it only concerns cohomology, as others have pointed out. But my understanding is that varieties like fake projective planes with simple cohomology but complicated geometry are hard to construct by explicit equations.

As others suggest, rational connectedness might provide a good geometric explanation for this. If the variety is also Fano, proving that might be easier than explicitly finding rational curves - one example of varieties where there is an elementary proof of this type of congruence is varieties given by low-degree equations, which are also always Fano.

Also, note that these statements should apply not just rationally but also to the boundary divisor and to any singularities, if they occur. More precisely, it would be interesting to see if you can explicitly write your variety as a class in the Grothendieck group of smooth projective varieties. In that case possibly all varieties that occur should have that nice condition.

One interesting thing - the simplest kind of affine variety, a smooth projective variety minus a smooth divisor, can only satisfy this kind of congruence with $c(X)=0$. (I see Jason Starr has already pointed this out.) In fact it's easy to see from the $\zeta$ function that if you write your variety as a sum and difference of smooth projective connected varieties, the total number of smooth projective connected varieties must be exactly $1$. That's sort of interesting, and maybe you can see that from another angle.

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Nobody seems to have mentioned Fulton's (?) trace formula. It says that the number of points mod p equals the alternating trace of Frobenius on coherent cohomology $H^i(X, \mathcal{O})$. So - and this is probably exactly the same as Sawin's point- the easiest reason that the number of points is congruent to $1$ modulo $p$ is that all the higher cohomology of the structure sheaf vanishes.

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  • $\begingroup$ There is, indeed, Fulton's trace formula. There is also the earlier work of Nick Katz giving a second proof of Ax's theorem (now the Ax-Katz theorem), again by relating congruences to Hodge level. A key point of Esnault's work is that, sometimes there are cycle-theoretic reasons that the trace of Frobenius on $H^i(X,\mathcal{O}_X)$ must vanish, even when the corresponding Dolbeault groups $H^i(X,\Omega^j_{X/k})$ are nonzero. Since those cycles may have coefficients that are divisible by $p$, the $p$-torsion Dolbeault groups are not the best cohomology theory to use . . . $\endgroup$ Jun 26, 2015 at 17:10
  • $\begingroup$ Correction: "trace of Frobenius on ... must vanish" --> "trace of Frobenius on ... must be divisible by $p$". $\endgroup$ Jun 26, 2015 at 18:02
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    $\begingroup$ A remark is that Fulton's trace formula only applies to compact $X$, whereas Allen is specifically considering affine varieties (whose coherent cohomology will be a bit uninformative). $\endgroup$ Jun 26, 2015 at 18:03
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Firstly I would like to note that there seems to be two reasonable versions of your question:

  1. For which $X$ is $\#X(\mathbb{F}_p)$ is divisible by $p$ for any prime $p$?

  2. For which $X$ is $\#X(\mathbb{F}_p)$ is divisible by $p$ for any sufficiently large prime $p$?

Besides, in both of these questions one can demand $\#X(\mathbb{F}_q)$ to be divisible by $q$ for $q$ being any power of a prime. I will denote the corresponding versions of the questions by 1' and 2', respectively.

Both of these questions are quite interesting; yet Question 1 (and 1') is more "arithmetic" and so harder to answer. Possibly I will treat it in a paper some day (thank you for asking it!).:) So, in this answer I will mostly treat Questions 2 and 2' (that are also quite hard).

Both of the questions are "motivic" since (if we ignore a single prime $l$) they depend on $Rx_!\mathbb{Q}_{l,X}$ (considered as a mixed complex of $\mathbb{Q}_{l}$-etale sheaves over $Spec\, \mathbb{Z}[1/l]$; here $x:X\to Spec\, \mathbb{Z}$ is the structure morphism) only. Besides, they are of "Euler characteristic type". So, ("the prime-to-$l$-part of") Question 1' and Question 2' can be easily reformulated in terms of the class of $Rx_!\mathbb{Q}_{l,X}$ in the Grothendieck ring of mixed Galois representations of $Gal(\mathbb{Q})$. Are you interested in an answer of this type

Now I will proceed to motives (and avoid fixing $l$). The class mentioned only depends on the class of the motif of $X$ (with compact support) with rational coefficients in the corresponding Grothendiek group of motives (that is also a ring). To get a sufficient condition for the first question one may consider the class of $X$ in a certain "complicated motivic Grothendieck" group over the integers. For the second question it suffices to consider the motif with compact support of $X$ over $\mathbb{Q}$. This motif (and so, $X$ itself) has a well-defined class in the Grothendiek group of Chow motives (this is a seminal result of Gillet and Soule that answers a question of Serre; you may also have a look at my results on weight complexes and $K_0$(motives)). Thus $X$ also has a well-defined class in $K_0(Mot_{num})$. $X$ satisfies the conditions of Question 2' whenever this class is congruent to the one of a point modulo the Lefschetz motif (i.e., if $[X]-[pt]$ is the twist of an effective class).

At this stage one can ask two more natural questions:

  1. It this $K_0$-condition a necessary one (also for Question 2)?

  2. What geometric information does this condition contain?

I suspect that one can deduce the positive answer to Question 3 from certain "standard motivic" conjectures (certainly including the Tate one); yet this doesn't seem to be easy (though the Fontaine-Mazur condition could help here).

Lastly, I would say that I do not expect any nice answers to Question 4. Certainly, the motivic conjectures predict that one can lift from $K_0(Mot_{num})$ to $K_0(Chow)$. Yet I see no way to rise from $K_0$ to motives themselves. Yet possibly I miss something here; then you may be interested in my results on Chow-weight homology: https://www.google.ru/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB0QFjAA&url=http%3A%2F%2Farxiv.org%2Fabs%2F1411.6354&ei=dxWNVZbXO8XgyQOR8YCQCQ&usg=AFQjCNGZOey1IoXyWM5On9S2AUDnUTTmFA&sig2=nRumNYSvux2R2QblD1jcqQ&bvm=bv.96782255,d.bGQ&cad=rjt Anyway, the motivic assertions that one can obtain this way do not seems to imply anything like rational connectedness of varieties.

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  • $\begingroup$ Yet my "general" pessimism does not mean that one cannot hope to obtain certain information in special cases (i.e., if $X belongs to a "simple" class of varieties; smooth and affine does not seem to be sufficient here). $\endgroup$ Jun 26, 2015 at 9:27
  • $\begingroup$ Your #4 sounds like "what does it imply?" rather than "what does it suggest?" (what would imply it, that one should be on the lookout for?). I have a certain class of varieties; I don't know what properties they have, but now I'm going to try to prove rational connectivity directly for them (not actually imply it from the point-counting fact). $\endgroup$ Jun 26, 2015 at 12:33
  • $\begingroup$ Yes; this seems to be a very reasonable idea! Yet I have an impressing that testing rational connectivity may be rather hard. So I would be glad to read any your further questions in this direction.:) $\endgroup$ Jun 26, 2015 at 12:38
  • $\begingroup$ Yet my impression is that the notions rationall connectivity/unirationality/etc. are mostly adapted to smooth projective varieties. Still you certainly may look at "nice compactifications" of your $X$ (and ask whether there exists a compactification with prescribed properties). $\endgroup$ Jun 26, 2015 at 12:54
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    $\begingroup$ @MikhailBondarko. Regarding the impression that testing rational connectedness is hard, that depends very much on the variety. There is a conjecture of Mumford that, if true, implies that testing rational connectedness is "computable" (if not particularly efficient). In many cases, it really is easy to test rational connectedness: for smooth projective varieties in characteristic $0$, it is a matter of finding a single morphism from $\mathbb{P}^1$ that pulls back the tangent bundle to be ample. $\endgroup$ Jun 26, 2015 at 14:06

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