Consider multivariate integer Laurent polynomials $f\in \mathbb{Z}[x_1, x_1^{-1}, \ldots, x_n, x_n^{-1}]$ satisfying the following condition:

A. For every odd prime $p$, the coefficient of $x_1^0\ldots x_n^0$ in $f^{p-1}$ is congruent to $1$ modulo $p$.

Here are some easy examples:

  1. $n=1$, $f = x_1 - x_1^{-1}$, due to ${{p-1} \choose {(p-1)/2}}\equiv 1$ mod $p$,
  2. $f$ such that $(0, \ldots, 0)$ is a vertex of the convex hull of the support of $f$, with constant term $\pm 1$ ,
  3. $n=2$, $f = x_1^{-1} (x_1-1)^2 - x_2^{-1} (x_2-1)^2$, due to the identity $\sum_{i+j=p-1} {2i\choose i}{2j\choose j} = 4^{p-1} \equiv 1$ mod $p$.

The question is: can you give some characterization of all $f$ satisfying A? Or at least some less trivial examples?

Actually, the following weaker condition is also interesting for me:

B. For every odd prime $p$, the coefficient of $x_1^0\ldots x_n^0$ in $f^{p-1}$ is not divisible by $p$.

Question related to Frobenius splitting. If it reminds you of the study of supersingularity of reductions of elliptic curves (e.g. Hartshorne IV, 4.21-4.23.6, p. 332-335) this is because of the fact that an elliptic curve Frobenius split if and only if it is not supersingular. In particular, the result of Elkies shows that for a certain family of polynomials $f$ (equations of elliptic curves) condition B is never satisfied.

  • $\begingroup$ See my question mathoverflow.net/questions/8003/… for examples of what we don't know. $\endgroup$ – David E Speyer Feb 9 '11 at 12:49
  • $\begingroup$ Thanks, David. You're question is much deeper and difficult than mine (I hope ;)). $\endgroup$ – Piotr Achinger Feb 9 '11 at 15:57
  • $\begingroup$ Doesn't case 2 generalize to when the constant term is $\pm 2^k$ for any natural number $k$? $\endgroup$ – ARupinski Feb 9 '11 at 19:19

I should explain how to think about these things in terms of Frobenius splitting. This perspective will let you write down lots more examples like the ones you already have. For example, if I haven't made any silly typos, $x - 2 + 3x^{-1} - x^{-1} y - x^{-1} y^{-1}$ should be an example.

$\def\Tr{\mathrm{Tr}}$ Here is the strategy. Let $\Tr$ be the following map from Laurent polynomials to themselves given by $$\Tr\left( \sum g_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n} \right) = \sum \sum g^{1/p}_{(p i_1) (p i_2) \cdots (p i_n)} x_1^{i_1} \cdots x_n^{i_n}.$$ Given any Laurent polynomial $f$, we define an almost splitting by $\phi_f(g) =\Tr(f^{p-1} g)$. This is a splitting if and only if $\Tr(f^{p-1}) =1$, meaning that the constant term of $f^{p-1}$ is $1$ and there are no other monomials in $f^{p-1}$ whose exponents are divisble by $p$.

Consider the following:

Condition 1: Let $N(f)$ be the Newton polytope of $f$, i.e., the convex hull of the monomials occurring in $f$. Then $(0,0,\ldots, 0)$ is the only lattice point in the interior of $f$.

In the presence of Condition 1, the only monomial in $f^{p-1}$ whose exponents are all divisible by $p$ is the constant term. So, in the presence of Condition 1, what you want to know is whether $\phi_f$ is a splitting. From now on, we assume Condition 1.

Now, there is a very powerful tool to force $\phi_f$ to be a splitting. If there is an $\mathbb{F}_p$ point $z$ of $f=0$, where the hypersurface $f=0$ has normal crossings, then $\phi_f$ will be a splitting. (This is in, for example, Brion and Kumar's book.) We can also weaken this condition in various ways. It is enough to have residual normal crossings (see V. Lakshmibai, V. B. Mehta, and A. J. Parameswaran, Frobenius splittings and blow-ups, J. Algebra 208 (1998), no. 1, 101–128. MR 1643983 (99i:14059)). Even more generally than this, let $y_1$, ..., $y_n$ be local coordinates at $z$; it is enough that there is some term order (with $1< y_1$, ..., $y_n$) such that $y_1 y_2 \cdots y_n$ is the minimal term of $f$ near $z$. Moreover, if the point $z$ only exists over an extension $\mathbb{F}_q$, or if the coordinates in which $f=0$ becomes normal crossings are only defined over $\mathbb{F}_q$, there are modifications of these results available, although I don't know quite what they say off the top of my head.

I realize that I am not including as many details as I perhaps should here, but it is a large topic. If you give me some clues as to how far you've gotten, I can adjust accordingly.

Now, I want to convince you that all your examples where the constant term is inside the Newton polygon are of this sort. Namely: $x-x^{-1}$ has two roots, $1$ and $-1$, both defined over $\mathbb{F}_p$ for any odd $p$. So $f=0$, near either of these points, is a zero dimensional "normal crossings" variety. If you instead work with $x+x^{-1}$, then the roots are only defined over $\mathbb{F}_p$ when $p \equiv 1 \mod 4$, which is why you got peculiar behavior depending on what $p$ was mod $4$.

Your third example factors as $(x_1-x_2)(1-x_1^{-1} x_2^{-1})$, the intersection of a line and a hyperbola. This has simple normal crossings at $(1,1)$ and $(-1, -1)$.

ARupinski's example is $n$ hypersurfaces which all meet transversely at $(-1,1,-1,1,\cdots,-1,1)$, if I didn't screw up any signs.

The example I gave at the beginning is a nodal cubic, with node at $(1,1)$, if I didn't screw it up.

What would be interesting is if anyone knew an example of this behavior which wasn't explained by a singularity of the hypersurface.

  • $\begingroup$ Considering I don't know much about singularities and absolutely nothing about Frobenius splittings, this is a rather interesting explanation, especially since it explains how my examples fit into the framework of the problem. +1 $\endgroup$ – ARupinski Feb 10 '11 at 3:56

I don't know if this qualifies as less trivial than your examples since it relies on the same identity that your example (1) does, but here goes:

Define $f_k(x_1,\ldots,x_k) = \frac{(1+x_1)(x_{k-1}+x_k^2)}{\prod_{i=1}^kx_i}\prod_{j=2}^{k-1}(x_{j-1}+x_j)$.

Then the constant term of $f_k^{p-1}$ is $\binom{p-1}{\frac{p-1}{2}}^k \equiv \pm 1 \mod p$ (note that your congruence in (1) should be $\pm 1$ as well). Furthermore, if one exchanges any number of the plus signs with minus signs, the constant term remains the same up to possibly a factor of -1; this does not affect the congruence $\mod p$ however. In light of this, your example (1) is the special case where the plus sign in $f_1$ is replaced by a minus sign.

The reason this particular sequence of $f_k$'s came to mind is that relative to certain coordinates on a maximal torus, via the Weyl Character Formula $f_k$ represents the character of the spinor representation of $Spin(2k+1)$. Extracting the constant term of $f_k^n$ is then equivalent to counting the multiplicity of the zero weight in the $n^{th}$ tensor power of the spinor representation.

I have been toying around with the Laurent polynomials corresponding to other representations of compact groups, but have not yet found any others that (provably) satisfy your criterion.

  • $\begingroup$ As a clarification on the definition of $f_k$ for $k=1,2$: $f_1(x_1) =(1+x_2^1)/x_1$ and $f_2(x_1,x_2) = \frac{(1+x_1)(x_1+x_2^2)}{x_1x_2}$ $\endgroup$ – ARupinski Feb 9 '11 at 23:20
  • $\begingroup$ Actually, your answer can be quite useful since my motivations also come from representation theory. Can you give some reference for these special coordinates on the torus? By the way, in your comment above, you probably meant $f_1(x_1) = (1+x_1^2)/x_1)$. $\endgroup$ – Piotr Achinger Feb 10 '11 at 9:14
  • $\begingroup$ What is the idea behind the fact that the multiplicity of the zero weight in $V^{\otimes (p-1)}$ gives $1$ modulo $p$ for a certain representation $V$ and all odd $p$? $\endgroup$ – Piotr Achinger Feb 10 '11 at 11:05
  • $\begingroup$ You are correct that was a minor typo on $f_1$. As far as the coordinates are concerned, they are "standard" coordinates: $x_i = \exp(\omega_i)$ for $\omega_i$ a fundamental weight of the corresponding Spin group (I put "standard" in quotes because to my recollection there are other indexing conventions used by physicists and perhaps by some mathematicians too). My numbering on the fundamental weights coincides with the Bourbaki numbering, so the highest weight of the spinor representation of $B_n$ is $\omega_n$. $\endgroup$ – ARupinski Feb 10 '11 at 17:56
  • $\begingroup$ So for example the 4-dimensional representation of Spin(5) has weights $\omega_2$, $\omega_1-\omega_2$, $-\omega_1+\omega_2$, and $-\omega_2$, leading to a character with monomial terms $x_2$, $x_1x_2^{-1}$, $x_1^{-1}x_2$, and $x_2^{-1}$; adding these up and factoring in the Laurent Ring $\mathbb{Z}[x_1,x_1^{-1},x_2,x_2^{-1}]$ gives $f_2$. $\endgroup$ – ARupinski Feb 10 '11 at 18:00

Here is a surprising connection between such questions and mirror symmetry for Fanos.

I know next to nothing about mirror symmetry, but I recently came across the preprint by Ilten and Przyjalkowski. The authors consider the mirrors to three-dimensional Fano varieties $X$, that is, their Landau-Ginzburg models. These are spaces $Y$ together with a potential $f\in \mathcal{O}(Y)$. In the cited paper, $Y = (\mathbb{C}^\ast)^3$, so $f$ is just a Laurent polynomial in three variables. We can take $X$ to be an $n$-dimensional Fano and consider Landau-Ginzburg models on $(\mathbb{C}^\ast)^n$.

On the Landau-Ginzburg side, we associate with $f$ the generating function $\Phi_Y(z) = \sum a_n z^n$ where $a_n$ is the constant term of $f^n$. On the Fano side, we first compute some Gromov-Witten invariants of $X$ and using them we form a differential operator $L_X \in \mathbb{C}[z, \frac{\partial }{\partial z}]$. We then define $I_X(z)$ to be the unique analytic solution to $L_X I_X = 0$. It is called "the fundamental term of the regularized $I$-series of $X$" (this tells me nothing unfortunately). All these definitions are on page 5 of the cited preprint.

As you may have already guessed, mirror symmetry is expected to yield $\Phi_Y = I_X$.

This of course is incredibly far from answering my question (it has nothing to do with prime numbers and divisibility, but at least it mentions the constant terms of powers of $f$ altogether). Maybe it could help us get our hands on the function $\Phi_Y$ aka $I_X$. We have already seen that it is not a rational function in general ($f = x + x^{-1}$ gives $\Phi_Y(z) = 1/\sqrt{1 - 4x^2}$), but maybe we can, I don't know, prove some "Weil type conjectures" for them?


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