Dwork Congruences and reflexive Polytopes

TL;DR

This paper demonstrates that the coefficients of the principal period's power series for Laurent polynomials obey significant congruence relations, aiding in p-adic analytic continuation of the unit root.

Contribution

It establishes strong congruence properties for these coefficients, linking Dwork congruences with reflexive polytopes.

Findings

Coefficients satisfy strong congruences.

Facilitates p-adic analytic continuation.

Connects Dwork congruences with reflexive polytopes.

Abstract

We show that the coefficients of the power series expansion of the principal period of a Laurent polynomial satisfy strong congruence properties. These congruences play a key role in the explicit p-adic analytic continuation of the unit root.

Equations

$$\frac{a(n)}{a(\lfloor n/p\rfloor )}\in {\mathbb{Z}}_p$$

$$\frac{a(n+m{p}^{s+1})}{a(\lfloor n/p\rfloor +m{p}^s)}\equiv \frac{a(n)}{a(\lfloor n/p\rfloor )}\mathrm{mod}{p}^{s+1}$$

$$a(n+m{p}^{s+1})a(\lfloor \frac{n}{p}\rfloor )\equiv a(n)a(\lfloor \frac{n}{p}\rfloor +m{p}^s)\mathrm{mod}{p}^{s+1}.$$

$$a(n_0+mp)\equiv a(n_0)a(m)\mathrm{mod}p$$

$$n=n_0+p{n}_1+\dots +n_r{p}^r,0\le n_i\le p-1,$$

$$a(n)\equiv a(n_0)a(n_1)\dots a(n_r)\mathrm{mod}p$$

$$a(n_0+...+n_{s+1}{p}^{s+1})a(n_1+...+n_s{p}^{s-1})\equiv$$

$$a(n_0+...+n_s{p}^s)a(n_1+...+n_{s+1}{p}^s)\mathrm{mod}{p}^{s+1}.$$

$$n\mapsto a(n)/a(\lfloor n/p\rfloor )$$

$$F(t)=\sum _{n=0}^{\infty }a(n)t^n$$

$$F^s(t)=\sum _{n=0}^{p^s-1}a(n)t^n.$$

$$\mathfrak{D}:=\{x\in {\mathbb{Z}}_p,|F^1(x)|=1\}.$$

$$f(x)=\underset{s\to \infty }{\lim }{F}^{s+1}(x)/F^s(x^p).$$

$$X^{\mathbf{a}}=X_1^{a_1}{X}_2^{a_2}\dots X_n^{a_n},\mathbf{a}=(a_1,a_2,\dots ,a_n)\in {\mathbb{Z}}^n$$

$$f=\sum _{\mathbf{a}}{c}_{\mathbf{a}}{X}^{\mathbf{a}}\in \mathbb{Z}[X_1,X_1^{-1},X_2,X_2^{-1},\dots ,X_n,X_n^{-1}].$$

$$\text{supp}(f):=\{\mathbf{a}\in {\mathbb{Z}}^n|c_{\mathbf{a}}\ne 0\}$$

$$\Delta (f):=\text{convex}(\text{supp}(f))$$

$$\mathcal{A}=\left({\bf a}_{1},{\bf a}_{2},\ldots,{\bf a}_{m}\right)=\left(\begin{array}[]{cccc}a_{1,1}&a_{1,2}&\ldots&a_{1,m}\\ \vdots&&&\vdots\\ a_{n,1}&a_{1,2}&\ldots&a_{n,m}\\ \end{array}\right)$$

$$f=\sum _{j=1}^m{c}_j{X}^{{\mathbf{a}}_j}=\sum _{j=1}^m{c}_j\prod _{i=1}^n{X}^{a_{i,j}}$$

$${\mathbb{R}}^m\stackrel{\mathcal{A}}{⟶}{\mathbb{R}}^n$$

$$\underset{i=1,...,n}{\gcd }(\sum _{j=1}^m{a}_{i,j}{\ell }_j)\le \sum _{j=1}^m{\ell }_j$$

$$g:=\underset{i=1,...,n}{\gcd }(\sum _{j=1}^m{a}_{i,j}{\ell }_j)>\sum _{j=1}^m{\ell }_j.$$

$${\bf a}_{1}\ell_{1}+...+{\bf a}_{m}\ell_{m}=\mathcal{A}\left(\begin{array}[]{c}\ell_{1}\\ \vdots\\ \ell_{m}\end{array}\right)=\left(\begin{array}[]{c}\sum_{j=1}^{m}a_{1,j}\ell_{j}\\ \vdots\\ \sum_{j=1}^{m}a_{n,j}\ell_{j}\end{array}\right).$$

$$v:=\frac{{\ell }_1}{g}{\mathbf{a}}_1+...+\frac{{\ell }_m}{g}{\mathbf{a}}_m\in {\mathbb{Z}}^n$$

$$\sum _j\frac{{\ell }_j}{g}<1$$

$${\alpha }_1{\mathbf{a}}_1+...+{\alpha }_m{\mathbf{a}}_m$$

$$\Phi (t):=\sum _{k=0}^{\infty }a(k)t^k,a(k):=[f^k{]}_0$$

$$\begin{array}[]{rcl}\Phi(t)&=&\sum_{k=0}^{\infty}[f^{k}]_{0}t^{k}\\[5.69054pt] &=&\sum_{k=0}^{\infty}\frac{1}{(2\pi i)^{n}}\int_{T}f^{k}t^{k}\Omega\\[5.69054pt] &=&\frac{1}{(2\pi i)^{n}}\int_{T}\sum_{k=0}^{\infty}f^{k}t^{k}\Omega\\[5.69054pt] &=&\frac{1}{(2\pi i)^{n}}\int_{T}\frac{1}{1-tf}\Omega\\[5.69054pt] &=&\int_{\gamma_{t}}\omega_{t}\\ \end{array}$$

$${\omega }_t=Re{s}_{X_t}(\frac{1}{1-tf}\Omega )$$

Full text

This work is entirely supported by DFG Sonderforschungsbereich/Transregio 45

1. Dwork congruences

Definition 1.1**.**

Let $(a(n))_{n\in{\mathbb{N}}_{0}}$ be a sequence of integers with $a(0)=1$ and let $p$ be a prime number. We say that $(a(n))_{n}$ satisfies the Dwork congruences if for all $s,m,n\in{\mathbb{N}}_{0}$ one has

1. D1

[TABLE] 2. D2

[TABLE]

In fact, the validity of these congruences is implied by those for which $n<p^{s+1}$, as one sees by writing $n=n^{\prime}+mp^{s+1}$ with $n^{\prime}<p^{s+1}$. By cross-multiplication, $D2$ becomes

1. D3

[TABLE]

The congruences for $s=0$ say that for $0\leq n_{0}\leq p-1$ one has

[TABLE]

So if we write $n$ in base $p$

[TABLE]

we find by repeated application that

[TABLE]

In fact, this is easily seen to be equivalent to $D3$ for $s=0$.

Similarly, for higher $s$ the congruences $D3$ are equivalent to

[TABLE]

[TABLE]

The congruences express a strong $p$-adic analyticity property of the function

[TABLE]

and play a key role in the $p$-adic analytic continuation of the series

[TABLE]

to points on the closed $p$-adic unit disc. More precisely, one has the following theorem (see [Dw], Theorem 3.)

Theorem 1.2**.**

Let $(a(n))_{n}$ be a $\mathbb{Z}_{p}-$valued sequence satisfying the Dwork congruences D1 and D2 . Let

[TABLE]

and

[TABLE]

Let ${\mathfrak{D}}$ be the region in $\mathbb{Z}_{p}$

[TABLE]

Then $F(t)/F(t^{p})$ is the restriction to $p\mathbb{Z}_{p}$ of an analytic element $f$ of support ${\mathfrak{D}}$:

[TABLE]

The congruences were used in [SvS] to determine Frobenius polynomials associated to Calabi-Yau motives coming from fourth order operators of Calabi-Yau type from the list [AESZ]. Although there are many examples of sequences that satisfy these congruences, the true cohomological meaning remains obscure at present. For a recent interpretation in terms of formal groups, see [Yu]. In this paper we will give a completely elementary proof of the congruences $D3$ for sequences $(a(n))_{n}$ that arise as constant term of the powers of a fixed Laurent polynomial with integral coefficients and whose Newton polyhedron contains a unique interior point. These include the series that come from reflexive polytopes.

2. Laurent polynomials

We will use the familiar multi-index notation for monomials and exponents

[TABLE]

to write a general Laurent-polynomial as

[TABLE]

The support of $f$ is the set of exponents $\bf a$ occuring in $f$, i.e.

[TABLE]

The Newton polyhedron $\Delta(f)\subset{\mathbb{R}}^{n}$ of $f$ is defined as the convex hull of its support

[TABLE]

When the support of $f$ consists of $m$ monomials, we can put the information of the polyhedron $\Delta:=\Delta(f)$ in an $n\times m$ matrix $\mathcal{A}\in Mat(m\times n,{\mathbb{Z}})$, whose columns ${\bf a}_{j}$, $j=1,2,\ldots,m$ are the exponents of $f$;

[TABLE]

so that we can write

[TABLE]

The polyhedron $\Delta$ is the image of the standard simplex $\Delta_{m}$ under the map

[TABLE]

The following theorem will play a key role in the sequel.

Theorem 2.1**.**

Let $\Delta$ be an integral polyhedron with [math] as unique interior point. Then for all non-negative integral vectors $(\ell_{1},\ell_{2},\ldots,\ell_{m})\in{\mathbb{Z}}^{m}$ such that $\sum_{i=1}^{m}a_{i,j}\ell_{j}\not=0$ for some $1\leq i\leq n$, one has

[TABLE]

Proof.

Assume that there exists a non-negative integral vector $\ell=(\ell_{1},...,\ell_{m})\in{\mathbb{Z}}^{m}$ such that $\sum_{i=1}^{m}a_{i,j}\ell_{j}\not=0$ for some $1\leq i\leq n$ and

[TABLE]

We have

[TABLE]

The components of the vector at the right hand side are all divisible by $g$, so that after division by $g$ we obtain a non-zero lattice point

[TABLE]

of $\Delta$ with

[TABLE]

The interior points of $\Delta$ (i.e. the points that do not lie on the boundary) consist of the combinations

[TABLE]

of the columns of $\mathcal{A}$ with $\sum_{j=1}^{m}\alpha_{j}<1$. As [math] was assumed to be the only interior lattice point of $\Delta$ we arrive at a contradiction. ∎

We remark that the above statement applies in particular to reflexive polyhedra.

3. The fundamental period

Notation 3.1**.**

For a Laurent-polynomial we denote by $[f]_{0}$ the constant term, that is, the coefficient of the monomial $X^{0}$.

Definition 3.2**.**

The fundamental period of $f$ is the series

[TABLE]

Note that the function $\Phi(t)$ can be interpreted as the period of a holomorphic differential form on the hypersurface $X_{t}:=\{t.f=1\}\subset({\mathbb{C}}^{*})^{n}$, as one has

[TABLE]

Here $\Omega:=\frac{dX_{1}}{X_{1}}\frac{dX_{2}}{X_{2}}\ldots\frac{dX_{n}}{X_{n}}$, $T$ is the cycle given by $|X_{i}|=\epsilon_{i}$ and homologous to the Leray coboundary of $\gamma_{t}\in H_{n-1}(X_{t})$ and

[TABLE]

In particular, $\Phi(t)$ is a solution of a Picard-Fuchs equation; the coefficients $a(k)$ satisfy a linear recursion relation.

Theorem 3.3**.**

*Let $f\in{\mathbb{Z}}[X_{1},X_{1}^{-1},\ldots,X_{n},X_{n}^{-1}]$ with integral coefficients. Assume that the Newton polyhedron $\Delta(f)$ has [math] as its unique interior lattice point.

Then the coefficients $a(n)=[f^{n}]_{0}$ of the fundamental period satisfy for each prime number $p$ and $s\in{\mathbb{N}}$ the congruence*

[TABLE]

[TABLE]

where $0\leq n_{i}\leq p-1$ for $0\leq i\leq s-1$.

We remark that already for the simplest cases where the the Newton polyhedron contains more than one lattice point, like $f=X^{2}+X^{-1}$, the coefficients $a(n)$ do not satisfy such simple congruences.

4. Proof for the congruence $\mod p$

For $s=1$ we have to show that for all $n_{0}\leq p-1$

[TABLE]

The proof we will give is completely elementary; the key ingredient is theorem 2.1, which states that for all non-negative integral $\ell=(\ell_{1},...,\ell_{m})$ one has,

[TABLE]

Proposition 4.1**.**

Let $f$ be a Laurent polynomial as above and $n_{0}<p$. Then

[TABLE]

Proof.

As $f$ has integral coefficients, we have $f^{n_{1}p}(X)\equiv f^{n_{1}}(X^{p})\mod p$. So the congruence is implied by the equality

[TABLE]

which means: the product of a monomial from $f^{n_{0}}(X)$ and a monomial from $f^{n_{1}}(X^{p})$ can never be constant, unless the two monomials are constant themselves. It is this statement that we will prove now.

For the product of a non-constant monomial from $f^{n_{0}}(X)$ and a non-constant monomial from $f^{n_{1}}(X^{p})$ to be constant, the monomial coming from $f^{n_{0}}(X)$ has to be a monomial in $X_{1}^{p},...,X_{n}^{p}$, since all monomials in $f^{n_{1}}(X^{p})$ are monomials in $X_{1}^{p},...,X_{n}^{p}$.

A monomial

[TABLE]

appearing in $f^{n_{0}}(X)$ corresponds to a partition

[TABLE]

of $n_{0}$ in non-negative integers $\ell_{i}$. If $M$ were a monomial in $X_{1}^{p},...,X_{n}^{p}$, then we would have the divisibility

[TABLE]

and hence

[TABLE]

On the other hand, by 2.1 we have

[TABLE]

So we conclude that $\sum_{i=1}^{m}a_{i,j}\ell_{j}=0$ for $1\leq j\leq n$ and that the monomial $M$ is the constant monomial $X^{0}$. Hence it follows that

[TABLE]

and since

[TABLE]

the proposition follows. ∎

We remark that the congruence has the following interpretation. By a result of [DvK] (Theorem 4.) one can compactify the map $f:({\mathbb{C}}^{*})^{n}\longrightarrow{\mathbb{C}}$ given by the Laurent polynomial to a map $\phi:{{\mathcal{X}}}\longrightarrow{\mathbb{P}}^{1}$ such that the differential form $\Omega$ extends to a form in $\Omega^{n}(({{\mathcal{X}}}\setminus\phi^{-1}(\{\infty\})))$. In the case $\Delta(f)$ is reflexive one has

[TABLE]

see [DK], $(8.3)$. On the other hand, from this and under an additional condition $(R)$, it follows from [Yu] corollary 3.7 that the$\mod p$ Dwork-congruences hold.

5. Strategy for higher $s$

The idea for the higher congruences is basically the same as for $s=1$, but is combinatorically more involved. Surprisingly, one does not need any statements stronger than 2.1. To prove the congruence 3.1, we have to show that

[TABLE]

To do this, we will use the following expansion of $f^{np^{s}}(X)$:

Proposition 5.1**.**

We can write

[TABLE]

where $g_{n,k}$ is a polynomial of degree $np^{k}$ in the monomials of $f$, independent of $s$, defined inductively by $g_{n,0}(X)=f^{n}(X)$ and

[TABLE]

Proof.

We have to prove that the right-hand side of equation 5.2 is divisible by $p^{k}$. This is proved by induction on $k$ and an application of the congruence

[TABLE]

For $k=1$, the divisibility follows directly by (5.3). Assume that the statement is true for $m\leq k-1$. Write $f(X)^{np^{k-1}}=\sum_{j=0}^{k-1}p^{j}g_{n,j}(X^{p^{k-1-j}}).$ Then, $\sum_{j=0}^{k-1}p^{j}g_{n,j}(X^{p^{k-j}})=f(X^{p})^{np^{k-1}}\equiv f(X)^{np^{k}}\mod p^{n},$ and thus $f(X)^{np^{k}}-\sum_{j=0}^{k-1}p^{j}g_{n,j}(X^{p^{k-j}})\equiv 0\mod p^{n}$.∎

The congruences involve constant term expressions of the form

[TABLE]

Thus, equation (5.1) translates into

[TABLE]

Since this congruence is supposed to hold modulo $p^{s}$, on the left-hand side, only the summands with $\sum_{k=0}^{s}i_{k}+\sum_{k=1}^{s-1}l_{k}\leq s-1$ contribute, and on the right-hand side, only those with $\sum_{k=0}^{s-1}i_{k}+\sum_{k=1}^{s}l_{k}\leq s-1$ play a role.

Now, we proceed by comparing these summands on both sides of equation 5.1. We will prove that each summand on the right-hand side is equal to exactly one summand on the left-hand side and vice versa.

6. Splitting positions

So we are led to study for $a\leq b$ expressions of the type

[TABLE]

where the $0\leq n_{k}\leq p-1$ are fixed for $a\leq k\leq b$ and $I:=(i_{a},...,i_{b})$ is a sequence with $0\leq i_{k}\leq k$.

Definition 6.1**.**

We say that $G(a,b;I)$ splits at $\ell$ if

[TABLE]

The number of entries of $I$ is determined implicitly by $a$ and $b$, so that by $G(a,\ell-1;I)$ we mean the expression corresponding to the sequence $(i_{a},...,i_{\ell-1})$, while by $G(\ell,b;I)$, we mean the expression corresponding to $(i_{\ell},...,i_{b})$. Note that $\ell=a$ represents a trivial splitting, but splitting at $\ell=b$ is a non-trivial property.

Proposition 6.2**.**

If $k-i_{k}\geq\ell$ for all $k\geq\ell$, then $G(a,b;I)$ splits at $\ell$.

Proof.

A monomial $\prod_{j=1}^{m}(X^{p^{k-i_{k}}})^{{\bf a}_{j}\beta_{j,k}}$ occuring in $g_{n_{k},i_{k}}(X^{p^{k-i_{k}}})$ corresponds to a partition

[TABLE]

of the number $p^{i_{k}}n_{k}$ in non-negative integers $\beta_{1,k},...,\beta_{m,k}$. So we have

[TABLE]

It follows from the assumptions that the product $G(\ell,b;I)=\prod_{k=\ell}^{b}g_{n_{k},i_{k}}(X^{p^{k-i_{k}}})$ is a Laurent-polynomial in $X^{p}$. As a consequence, the product of a monomial in $G(a,\ell-1;I)=\prod_{k=a}^{\ell-1}g_{n_{k},i_{k}}(X^{p^{k-i_{k}}})$ and a monomial of $G(\ell,b;I)$ can be constant only if the sum

[TABLE]

is divisible by $p^{\ell}$ for $1\leq i\leq n$.

Set

[TABLE]

so that

[TABLE]

It follows that

[TABLE]

Hence, it follows that

[TABLE]

where the first inequality follows from Theorem 2.1. This implies $\sum_{j=1}^{m}a_{i,j}\gamma_{j}=0$ for $1\leq i\leq n$. But this means that the monomial in $\prod_{k=t}^{s-1}g_{n_{k},i_{k}}(X^{p^{k-i_{k}}})$ is itself constant. ∎

Now that we know that we can split up expressions $G(a,b;I)$ satisfying the condition given in Proposition 6.2, we proceed by proving that all the summands on both sides of equation 5.5 that do not have a coefficient divisible by $p^{s}$ satisfy this splitting condition.

7. Three combinatorical Lemmas

In this section, we prove three simple combinatorical lemmas which will be applied to split up expressions $G(0,s;I)G(1,s-1;J+1)$ that occur in the congruence (5.1).

Definition 7.1**.**

Let $a\leq b$ and $I=(i_{a},i_{a+1},\ldots,i_{b})$ a sequence with $0\leq i_{k}\leq k$ for all $k$ with $a\leq k\leq b$. We say that $\ell$ is a splitting index for $I$ if $\ell>a$ and for $k\geq\ell$ one has

[TABLE]

Remark that for a splitting index $\ell$ one can apply 6.2 and that $i_{\ell}=0$.

Lemma 7.2**.**

Let $I$ as above and assume that

[TABLE]

Then there exists at least one splitting index for $I$.

Proof.

Let ${\mathcal{N}}:=\{k\;\;|i_{k}=0\}$ be the set of all indices $k$ such that the corresponding $i_{k}$ is zero. Since the sum has $b-a+1$ summands $i_{k}$, the set ${\mathcal{N}}$ has at least two elements. So there exists at least one index $k\not=a$ such that $i_{k}=0$.

We will show by contradiction that one of these zero-indices is a splitting index.

We say that $\nu>k$ is a violating index with respect to $k\in{\mathcal{N}}$ if $i_{\nu}>\nu-k$. Assume now that all $k\in{\mathcal{N}}$ posses a violating index.

It follows directly that for each violating index $\nu$, $i_{\nu}\geq 2$. Furthermore, if $\nu$ is a violating index for $m$ different zero-indices $k_{1}<...<k_{m}$, it follows that $i_{\nu}\geq m+1$.

Now assume that we have $\mu$ different violating indices $\nu_{1},...,\nu_{\mu}$ and that $\nu_{j}$ is a violating index for all $j\in{\mathcal{N}}_{j}$, where we partition ${\mathcal{N}}$ into disjoint subsets

[TABLE]

Then $\sum_{j=1}^{\mu}i_{\nu_{j}}\geq\sum_{j=1}^{\mu}(\#{\mathcal{N}}_{j}+1)=\#{\mathcal{N}}+\mu$, and

[TABLE]

a contradiction.∎

We can sharpen lemma 7.2 to:

Lemma 7.3**.**

Let $I$ be as above and assume that

[TABLE]

*Then there exist at least $m$ different splitting indices for $I$ *

Proof.

We proceed by induction on $m$. The case $m=1$ is just Lemma 7.2. Assume that for all $n\leq m$, we have proven the statement.

Now assume $\sum_{k=a}^{b}i_{k}=b-a-(m+1)$. Since $m+1>1$, there exists a splitting index $\nu$. We can split up the set of indices $\{i_{a},...,i_{b}\}=\{i_{a},...,i_{\nu-1}\}\cup\{i_{\nu},...,i_{b}\}$ in position $\nu$ such that $\sum_{k=a}^{\nu-1}i_{k}=N_{\nu}$ and $\sum_{k=\nu}^{b}i_{k}=b-a-m-1-N_{\nu}$. Depending on $N_{\nu}$, we have to distinguish between the following cases:

1. (1)

$N_{\nu}>(\nu-1)-a-1$. It follows that $b-a-m-1-N_{\nu}<b-a-m-((\nu-1)-a-1)=b-m-(\nu-1)$, and thus $\sum_{k=\nu}^{b}i_{k}\leq b-\nu-m$. By induction, there exists at least $m$ splitting indices in $(i_{\nu},...,i_{b})$, and thus for the whole $(i_{a},...,i_{b})$, there exist at least $m+1$ such indices. 2. (2)

The case $N_{\nu}\leq(\nu-1)-a-1$ splits up in two cases:

1. (a)

$N_{\nu}\leq(\nu-1)-a-m$. By induction, $(i_{a},...,i_{\nu-1})$ has at least $m$ splitting indices, and the whole $(i_{a},...,i_{b})$ has at least $m+1$ such indices. 2. (b)

$N_{\nu}=(\nu-1)-a-n$, where $1\leq n\leq m$. Since $\sum_{k=a}^{\nu-1}i_{k}=(\nu-1)-a-n$, by induction for $(i_{a},...,i_{\nu-1})$ exist at least $n$ splitting indices. Since $\sum_{k=\nu}^{b}i_{k}=b-\nu-(m-n)$, for $(i_{\nu},...,i_{b})$, there exist at least $m-n$ splitting indices. Thus, for the whole $(i_{a},...,i_{b})$ exist at least $n+(m-n)+1=m+1$ splitting indices.

∎

Lemma 7.4**.**

1. (1)

Let $I=(i_{0},...,i_{s})$ and $J=(j_{1},...,j_{s-1})$ with

[TABLE]

Let $S_{I}$ be the set of splitting indices of $I$ and $S_{J}$ be the set of splitting indices of $J$. Then,

[TABLE] 2. (2)

Let $I=\{i_{0},...,i_{s-1}\}$ and $J=(j_{1},...,j_{s})$ with

[TABLE]

Let $S_{I}$ be the set of splitting indices of $I$ and $S_{J}$ be the set of splitting indices of $J$. Then,

[TABLE]

Proof.

1. (1)

Note that since $S_{I}\cup S_{J}\cup\{1,s\}\subset\{1,2,...,s\}$, it follows that $\#(S_{I}\cup S_{J}\cup\{1,s\})\leq s$. Note that $\sum_{k=0}^{s}i_{k}\geq s-\#S_{I}$ by Lemma 7.3. This implies that $\sum_{k=1}^{s-1}j_{k}\leq s-2-(s-(\#S_{I}+1))$, and hence that $\#S_{J}\geq s-(\#S_{I}+1)$ by Lemma 7.3. But $\#S_{I}+\#S_{J}+2=\#S_{I}+s-(\#S_{I}+1)+2=s+1>s$, which implies $\#(S_{I}\cap(S_{J}\cup\{1,s\}))\geq 1$, and thus the statement follows. 2. (2)

Note that since $(S_{I}\cup\{s\})\cup(S_{J}\cup\{1\})\subset\{1,...,s\}$, it follows that $\#(S_{I}\cup\{s\})\cup(S_{J}\cup\{1\})\leq s$. Now $\sum_{k=0}^{s-1}i_{k}\geq s-1-\#S_{I}$, which implies $\sum_{k=1}^{s}j_{k}\leq s-1-(s-\#S_{I}-1)$, and $\#S_{J}\geq s-\#S_{I}-1$. But $\#S_{I}+1+\#S_{J}+1\geq\#S_{I}+1+s-\#S_{I}=s+1>s$, which implies that $\#((S_{I}\cup\{s\})\cap(S_{J}\cup\{1\}))\geq 1$, and the statement follows.

∎

8. Proof for higher $s$

We will use the combinatorical lemmas on splitting indices from the last section to prove the congruence (5.1) modulo $p^{s}$.

For a sequence $I=(i_{a},...,i_{b})$, we write

[TABLE]

For a sequence $J=(j_{a},...,j_{b})$, we define $J+1:=(j_{a}+1,...,j_{b}+1)$.

Note that if $k-j_{k}>0$ for $a\leq k\leq b$, then we have

[TABLE]

since the constant term of a Laurent-polynomial $f(X)$ is the same as the constant term of the Laurent-polynomial $f(X^{p})$.

Let

[TABLE]

be a summand on the left-hand side of (5.5) defined by the tuple $(I,J)$ with $\sum_{k=0}^{s}i_{k}+\sum_{k=1}^{s-1}j_{k}\leq s-1$, and let $1\leq\nu\leq s$ be such that $G(0,s;I)$ splits in position $\nu$ and either $G(1,s-1;J+1)$ splits in position $\nu$ or $\nu\in\{1,s\}$. Such a $\nu$ exists by Lemma (7.4). Define $I^{\prime}=(i_{0}^{\prime},...,i_{s-1}^{\prime})$ and $J^{\prime}=(j_{1}^{\prime},...,j_{s}^{\prime})$ by

[TABLE]

To show that $p^{I^{\prime}+J^{\prime}}G(0,s-1;I^{\prime})G(1,s;J^{\prime}+1)$ is in fact a summand on the right-hand side of (5.5), we have to explain why $i_{k}^{\prime}\leq k$ and $j_{k}^{\prime}\leq k-1$. Note that $j_{k}\leq k-1$ for $1\leq k\leq s-1$ and $i_{k}\leq k$ for $0\leq k\leq s$. Furthermore, we have $i_{k}\leq k-1$ for $k\geq\nu$ since $i_{\nu}=0$ and $G(0,s;I)$ splits in position $\nu$, which means that $k-i_{k}\geq\nu\geq 1$ for $k\geq\nu$.

By definition of $j_{k}^{\prime}$ and $i_{k}^{\prime}$, it now follows that $j_{k}^{\prime}\leq k-1$ for $1\leq k\leq s$, and $i_{k}^{\prime}\leq k$ for $0\leq k\leq s-1$.

Now that we know that $p^{I^{\prime}+J^{\prime}}G(0,s-1;I^{\prime})G(1,s;J^{\prime}+1)$ is in fact a summand on the right-hand side of congruence (5.5), we prove the following Proposition. Remark that obviously, we have $p^{I+J}=p^{I^{\prime}+J^{\prime}}$.

Proposition 8.1**.**

Let $I,J,I^{\prime}$ and $J^{\prime}$ be defined as above. Then,

[TABLE]

Thus, we can identify each summand on the left-hand side of (5.5) with a summand on the right-hand side.

Proof.

By a direct computation:

[TABLE]

the statement follows. Note that the last equality follows since by definition of $I^{\prime}$ and $J^{\prime}$, $i_{\nu}^{\prime}=j_{\nu}^{\prime}=0$, $k-i_{k}^{\prime}\geq\nu$ and $k-j_{k}^{\prime}\geq\nu$ for $k>\nu$. Thus, $G(0,s-1;I^{\prime})$ and $G(1,s;J^{\prime}+1)$ both split at $\nu$.∎

Since by Proposition 8.1, we can identify every summand on the left-hand side of equation (5.5) satisfying $I+J\leq s-1$ with a summand on the right-hand side, both sides are equal modulo $p^{s}$ and the proof of Theorem 3.3 is complete.

Remark: The above arguments to prove the congruence $D3$ can be slightly simplified, as was shown to us by A. Mellit.

9. An Example

Let $f$ be the Laurent-polynomial

[TABLE]

It is No. 24 in the list of Batyrev and Kreuzer [BK], so $\Delta(f)$ is a reflexive polytope and our theorem 3.3 applies: the coefficients $a(n):=[f^{n}]_{0}$

[TABLE]

satisfy the congruence $D3$ modulo $p^{s}$ for arbitrary $s$.

The power series $\Phi(t)=\sum_{n=0}^{\infty}a(n)t^{n}$ is solution to a fourth order linear differential equation $PF=0$, where the differential operator $P$ is of Calabi-Yau type

[TABLE]

(where $\theta:=t\partial/\partial t$) that was determined in [PM].

10. Behaviour under Covering

Let $f$ be a Laurent-polynomial corresponding to a reflexive polyhedron, let $\mathcal{A}$ be the exponent matrix corresponding to $f$, and consider the vectors with integral entries in the kernel of $\mathcal{A}$. If there exists a positive integer $k$ such that

[TABLE]

then it follows that

[TABLE]

since for $l\in\mathbb{N}$,

[TABLE]

where

[TABLE]

We are interested in the congruences

[TABLE]

which we will prove in general for $s=1$, and which we will prove for one example by proving that the following condition is satisfied:

Condition 1**.**

For a tuple $(\ell_{1},...,\ell_{m})$ with

[TABLE]

it follows that

[TABLE]

Note that the proof is simliar for many other examples which we will not treat in here.

First of all, before we come to the example, we give a general proof of (10) for $s=1$.

Proposition 10.1**.**

Let $a(n),n\in\mathbb{N}$ be an integral sequence satisfying

[TABLE]

for $0\leq n_{0}\leq p-1$ and $a(n)\not=0$ iff $k|n$. Then

[TABLE]

Proof.

If $kn_{0}<p$, then the proposition follows directly. Hence assume that $kn_{0}=n_{0}^{\prime}+n_{0}^{\prime\prime}p>p-1$. Then

[TABLE]

Since $k\not|n_{0}^{\prime}$ and $a(n_{0}^{\prime})=0$ by assumption, it follows that

[TABLE]

On the other hand, $a(kn_{0})=a(n_{0}^{\prime}+n_{0}^{\prime\prime}p)\equiv a(n_{0}^{\prime})a(n_{0}^{\prime\prime})\mod p$ where $a(n_{0}^{\prime})=0$, and thus $a(kn_{0})\equiv 0\mod p$ and

[TABLE]

and the proposition follows. ∎

10.1. An Example

Let $f$ be the Laurent-polynomial No. 62 in the list of Batyrev and Kreuzer [BK], which is given by

[TABLE]

Then, the coefficients $a(n)$ are given by $a(n)=0$ if $n\not=0\mod 3$ and

[TABLE]

The Newton polyhedron $\Delta(f)$ is reflexive (see [BK]), and hence by Theorem 3.3, the coefficients $a(n)$ satisfy the congruence $D3$ modulo $p^{s}$ for arbitrary $s$.

The power series $\Phi(t)=\sum_{n=0}^{\infty}a(3n)t^{n}$ is solution to a fourth order linear differential equation $PF=0$, where the differential operator $P$ is of Calabi-Yau type and is given by

[TABLE]

In this example, the exponent matrix is

[TABLE]

A basis of $\ker(\mathcal{A})$ is given by

[TABLE]

and thus it follows that $[f^{n}]_{0}\not=0\Rightarrow 3|n$ and $k=3$. We prove that Condition 1 is satisfied in this example. Assume that $p\not=3$ and that

[TABLE]

This means that there exist $x_{1},x_{2},x_{3},x_{4}\in\mathbb{Z}$ such that

[TABLE]

which implies

[TABLE]

Thus, it follows that $(x_{1}+...+x_{4})=3z$ for some $z\in\mathbb{Z}$ and that

[TABLE]

Since $\ell_{5},...,\ell_{8}$ are nonnegative integers, it follows directly that $z\leq 0$. Now, consider the following cases:

1. (1)

$z=0$: Then,

[TABLE]

Assume that $x_{i}<0$, i.e. $x_{i}\leq-1$ for some $1\leq i\leq 4$. Since $\ell_{1},...,\ell_{4}$ are nonnegative integers, it follows that either $\ell_{5}+\ell_{6}+\ell_{7}+2\ell_{8}\geq p$ or $\ell_{j}+\ell_{8}\geq p$ for some $5\leq j\leq 7$, a contradiction to (10.2). Thus, since $x_{1}+x_{2}+x_{3}+x_{4}=0$, it follows that $x_{1}=x_{2}=x_{3}=x_{4}=0$ and that

[TABLE]

in this example. 2. (2)

$z<0$: Assume that $\ell_{5}+\ell_{6}+\ell_{7}+2\ell_{8}<(-z+1)p$. Since $\ell_{1}\geq 0$, it follows that $x_{1}>z-1$, and since $x_{1}$ is integral, that $x_{1}\geq z$. Since $x_{1}+x_{2}+x_{3}+x_{4}=3z$, it follows that $x_{2}+x_{3}+x_{4}\leq 2z$. Now assume that $x_{i}\geq z$ for $2\leq i\leq 4$. Then $x_{2}+x_{3}+x_{4}\geq 3z$,a contradiction. Hence there exists an index $i$ such that $x_{i}<z$, and hence $x_{i}\leq z-1$. Since $\ell_{i}\geq 0$, it follows that $\ell_{i+2}+\ell_{8}\geq(-z+1)p$, a contradiction since $\ell_{i+2}+\ell_{8}\leq\ell_{5}+\ell_{6}+\ell_{7}+2\ell_{8}<(-z+1)p$ by assumption. Thus, we have $\ell_{5}+\ell_{6}+\ell_{7}+2\ell_{8}\geq(-z+1)p$, which implies $p\leq\ell_{5}+\ell_{6}+\ell_{7}+2\ell_{8}+zp\leq p-1$, a contradiction.

Thus, it follows that the only possible case is $z=0$, and $x_{1}=x_{2}=x_{3}=x_{4}=0$, which proves that Condition 1 is satisfied in this example.

11. The statement D1

For the proof of congruence (3.1), the coefficients $c_{\bf a}$ of

[TABLE]

did not play a role. This is different if one is interested in the proof of part D1 of the Dwork congruences. Let $n\in\mathbb{N}$, and write $n=n_{0}+pn_{1}$, where $n_{0}\leq p-1$. Then, to prove D1 for the sequence $a(n):=[f^{n}]_{0}$ means that one has to prove that

[TABLE]

Sticking to the notation of the previous sections, we write

[TABLE]

Assume that $p^{k}|[f^{n_{1}}]_{0}$. To prove (11.1), one has to prove that $p^{k}|[f^{n_{0}+n_{1}p}]_{0}$. By (11.2), this is equivalent to proving that $p^{k-1}|[f^{n_{0}}g_{n_{1},1}(X)]_{0}$. Thus, the proof of part D1 of the Dwork congruences requires an investigation in the $p-$adic orders of the constant terms of $f^{n_{1}}$ and $g_{n_{1},1}$ for arbitrary $n_{1}$, and requires methods that are completely different from the methods we applied to prove the congruence $D3$.

Acknowledgement. We thank A. Mellit for his comments. The work of the first author was funded by the SFB Transregio 45.