2.1 Mellit’s theorem

Let $\Lambda (t)$ be a Laurent polynomial in variables $t=(t_1,\ldots ,t_r)$ with coefficients in ${\mathbb Z}_p$ and constant term $\operatorname {CT}_t(\Lambda )$ . Assume that the Newton polytope of $\Lambda (t)$ contains only one interior point $\{0\}$ .

For a tuple $a=(a_0,a_1,\ldots ,a_{l-1})$ , denote by $l(a)=l$ its length. For two tuples a and b, the concatenation product $a*b$ is the tuple of length $l(a)+l(b)$ obtained by gluing a and b together. For $a=(a_0$ , …, $a_{l-1})$ of length l, denote by $a'=(a_1,\ldots ,a_{l-1})$ the ‘derivative’ tuple of length $l-1$ . If a is a tuple of numbers, denote $|a|=\sum _{i=0}^{l-1}a_i$ .

For a tuple $m=(m_0,\ldots ,m_{l-1})$ of integers from $\{1,\ldots ,p-1\}$ , denote by $\operatorname {CT}_t(\Lambda ^m)$ the constant term of the Laurent polynomial $\Lambda (t)^{m_0+m_1p+m_2p^2+\cdots +m_{l-1}p^{l-1}}$ .

Theorem 2.1 [Reference MellitMe].

Let $a,b,c$ be tuples of integers from $\{1,\ldots ,p-1\}$ , where $b,c,a'$ can be empty, that is, of length 0. Then,

$$ \begin{align*} \operatorname{CT}_t(\Lambda^{a*b}) \operatorname{CT}_t(\Lambda^{a'*c}) \equiv \operatorname{CT}_t(\Lambda^{a'*b}) \operatorname{CT}_t(\Lambda^{a*c}) \,(\mathrm{mod}\,{p^{l(a)}}). \end{align*} $$

We modify the statement and three-page Mellit’s proof of Theorem 2.1 and prove Theorem 2.9.

2.2 Convex polytopes

Given a positive integer r, we consider convex polytopes, which are convex hulls of finite subsets of ${\mathbb Z}^r\subset {\mathbb R}^r$ .

Definition 2.2. A tuple $(N_0, N_1,\ldots ,N_{l-1})$ of convex polytopes is called $\operatorname {admissible}$ if for any $0\leqslant i\leqslant j \leqslant l-1$ ,

$$ \begin{align*} (N_i + pN_{i+1}+ \cdots + p^{j-i}N_j)\cap p^{j-i+1}{\mathbb Z}^r = \{0\}. \end{align*} $$

2.3 Definition of ghosts

Let $\Lambda (t,z)$ be a Laurent polynomial in some variables $t=(t_1,\ldots ,t_r)$ , $z=(z_1,\ldots ,z_{r'})$ with coefficients in ${\mathbb Z}_p$ . We define the ghost terms $R_m(\Lambda )$ , $m\geqslant 0,$ as the unique sequence of Laurent polynomials in $t,z$ satisfying the following two properties.

1. (i) For each m,

   $$ \begin{align*} \Lambda(t,z)^{p^m} = R_{0}(\Lambda)(t^{p^{m}}, z^{p^{m}}) +R_{1}(\Lambda)(t^{p^{m-1}}, z^{p^{m-1}})+\cdots + R_{m}(\Lambda)(t, z). \end{align*} $$

2. (ii) For each m, the coefficients of $R_m(\Lambda )(t,z)$ are divisible by $p^m$ in ${\mathbb Z}_p$ .

Properties (i) and (ii) recursively determine the polynomials $R_{m}(\Lambda )(t, z)$ . Namely,

$$ \begin{align*} R_m(\Lambda)(t,z) = \Lambda(t,z)^{p^m} - \Lambda(t^p,z^p)^{p^{m-1}}, \quad R_0(\Lambda)(t,z) = \Lambda(t,z). \end{align*} $$

Let $F(t,z)$ be a Laurent polynomial in $t,z$ . Let $N(F)$ be the Newton polytope of $F(t,z)$ with respect to the t variables only. Clearly,

$$ \begin{align*} N(R_m(\Lambda))\subset p^mN(\Lambda). \end{align*} $$

2.4 Composed ghosts

Let $\lambda = (\Lambda _0(t,z), \ldots , \Lambda _{l-1}(t,z))$ be a tuple of Laurent polynomials with coefficients in ${\mathbb Z}_p$ . We decompose the product

$$ \begin{align*} \tilde \lambda(t,z) := \Lambda_0(t,z) (\Lambda_1(t,z))^p\cdots (\Lambda_{l-1}(t,z))^{p^{l-1}} \end{align*} $$

into the sum of ghost terms of $\Lambda _0, \ldots , \Lambda _{l-1}$ . As the result, we obtain that $\tilde \lambda $ is the sum of the products

$$ \begin{align*} R_{m,\lambda}(t,z) &:= R_{m_0}(\Lambda_0)(t,z)\cdot R_{m_1}(\Lambda_1)(t^{p^{1-m_1}},z^{p^{1-m_1}})\cdot R_{m_2}(\Lambda_2)(t^{p^{2-m_2}},z^{p^{2-m_2}})\cdots \\& \quad R_{m_{l-1}}(\Lambda_{l-1})(t^{p^{l-1-m_{l-1}}},z^{p^{l-1-m_{l-1}}}), \end{align*} $$

where $m=(m_0,\ldots ,m_{l-1})$ runs over the set of all l-tuples of integers satisfying ${0\leqslant m_i \leqslant i}$ . Clearly,

$$ \begin{align*} R_{m,\lambda}(t,z) \equiv 0\, (\mathrm{mod}\,{p^{|m|}}) \end{align*} $$

and

$$ \begin{align*} N(R_{m,\lambda}(t,z)) \subset N(\Lambda_0(t,z)) + pN(\Lambda_1(t,z)) + \cdots + p^{l-1} N(\Lambda_{l-1}(t,z)). \end{align*} $$

2.5 Indecomposable tuples

Denote by $S_k$ the set of all k-tuples $m=(m_0,\ldots ,m_{k-1})$ of integers such that $0\leqslant m_i\leqslant i$ . Put $S=\bigcup _{k=1}^\infty S_k$ . A tuple $m\in S$ is called indecomposable if it cannot be presented as $m'*m"$ for $m',m"\in S$ . Denote by $S_k^{\operatorname {ind}}$ the set of all indecomposable k-tuples and put $S^{\operatorname {ind}} = \bigcup _{k=1}^\infty S_k^{\operatorname {ind}}$ .

Proof. The proof is by induction on $l(m)$ . If $l(m)=1$ , then $m=(m_0)=(0)$ and m is indecomposable. Let us prove the induction step. Let

$$ \begin{align*} m=m^1*\cdots *m^r = n^1*\cdots *n^s \end{align*} $$

be two decompositions into indecomposable factors. We may assume that $l(n^s)\geqslant l(m^r)$ . If $l(n^s) = l(m^r)$ , then $n^s = m^r$ . In this case, we can conclude that $m^1*\cdots *m^{r-1} = n^1*\cdots *n^{s-1}$ , and the statement follows from the induction hypothesis. If $l(n^s)> l(m^r)$ , then the sequence $n^s$ contains the sequence $m^r$ as its last $l(m^r)$ -part. This contradicts to the indecomposability of $n^s$ . The lemma is proved.

2.6 Polynomials $I_\lambda $

For an l-tuple $\lambda =(\Lambda _0(t,z),\Lambda _1(t,z),\ldots ,\Lambda _{l-1}(t,z))$ of Laurent polynomials with coefficients in ${\mathbb Z}_p$ , define

$$ \begin{align*} I_\lambda(t,z) = \sum_{m\in S_{l}^{\operatorname{ind}}} R_{m,\lambda}(t,z). \end{align*} $$

We have

$$ \begin{align*} I_\lambda(x,z) \equiv 0\, (\mathrm{mod}\,{p^{l-1}}) \end{align*} $$

by Lemma 2.3 and

$$ \begin{align*} N(I_{\lambda}(t,z)) \subset N(\Lambda_0(t,z)) + pN(\Lambda_1(t,z)) + \cdots + p^{l-1} N(\Lambda_{l-1}(t,z)). \end{align*} $$

Lemma 2.5. We have

$$ \begin{align*} \tilde \lambda(t,z) =\sum_{\lambda=\lambda^1*\cdots *\lambda^s} I_{\lambda^1}(t,z) I_{\lambda^2}(t^{p^{l(\lambda^1)}},z^{p^{l(\lambda^1)}}) \cdots I_{\lambda^s}(t^{p^{l(\lambda^1)+\cdots +l(\lambda^{s-1})}},z^{p^{l(\lambda^1)+\cdots+l(\lambda^{s-1})}}), \end{align*} $$

where the sum is over the set of all possible decompositions of the tuple $\lambda $ into a product of tuples.

Proof. We have

$$ \begin{align*} \tilde \lambda(t,z) = \sum_{m\in S_{l}} R_{m,\lambda}(t,z). \end{align*} $$

For any $m\in S_{l}$ , let $m=m^1*\cdots *m^s$ be its unique indecomposition into indecomposable factors. Let $\lambda =\lambda ^1*\cdots *\lambda ^s$ be the corresponding factorization of the sequence $\lambda $ . Then,

(2-1) $$ \begin{align} R_{m,\lambda}(t,z) = R_{m^1,\lambda^1}(t,z) R_{m^2,\lambda^2}(t^{p^{l(\lambda^1)}},z^{p^{l(\lambda^1)}}) \cdots R_{m^s,\lambda^s}(t^{p^{l(\lambda^1)+\cdots+l(\lambda^{s-1})}},z^{p^{l(\lambda^1)+\cdots+l(\lambda^{s-1})}}). \end{align} $$

This product contributes to the expansion of the product

(2-2) $$ \begin{align} I_{\lambda^1}(t,z) I_{\lambda^2}(t^{p^{l(\lambda^1)}},z^{p^{l(\lambda^1)}}) \cdots I_{\lambda^s}(t^{p^{l(\lambda^1)+\cdots +l(\lambda^{s-1})}},z^{p^{l(\lambda^1)+\cdots+l(\lambda^{s-1})}}) \end{align} $$

into the sum, and conversely each summand in the expansion of Equation (2-2) comes from Equation (2-1) for a unique indecomposable factorization $m=m^1*\cdots *m^s$ .

2.7 Admissible tuples of Laurent polynomials

Denote by $\operatorname {CT}_t(\Lambda )(z)$ the constant term of the Laurent polynomial $\Lambda (t,z)$ with respect to the variables t. The constant term $\operatorname {CT}_t(\Lambda )(z)$ is a Laurent polynomial in z.

Lemma 2.7. Let $\lambda =(\Lambda _0(t,z),\Lambda _1(t,z),\ldots ,\Lambda _{l-1}(t,z))$ be an admissible tuple of Laurent polynomials with coefficients in ${\mathbb Z}_p$ and $\lambda =\lambda ^1*\cdots *\lambda ^s$ . Then,

$$ \begin{align*} \operatorname{CT}_t\!\bigg(\prod_{i=1}^s I_{\lambda^i}(t^{p^{l(\lambda^1)+\cdots +l(\lambda^{i-1})}},z^{p^{l(\lambda^1)+\cdots+l(\lambda^{i-1})}})\bigg)(z) =\,\prod_{i=1}^s \operatorname{CT}_t (I_{\lambda^i}(t,z))(z^{p^{l(\lambda^1)+\cdots+l(\lambda^{i-1})}}). \end{align*} $$

Proof. We have

$$ \begin{align*} N(I_{\lambda^1}(t,z)) \subset N(\Lambda_0(t,z)) + pN(\Lambda_1(t,z)) + \cdots + p^{l(\lambda^1)-1} N(\Lambda_{l(\lambda^1)-1}(t,z)). \end{align*} $$

Hence,

$$ \begin{align*} N(I_{\lambda^1}(t,z))\cap p^{l(\lambda^1)}{\mathbb Z}^r = \{0\} \end{align*} $$

and

$$ \begin{align*} &\operatorname{CT}_t\!\bigg(\prod_{i=1}^s I_{\lambda^i}(t^{p^{l(\lambda^1)+\cdots +l(\lambda^{i-1})}},z^{p^{l(\lambda^1)+\cdots+l(\lambda^{i-1})}})\bigg)(z) \\ &\quad = \operatorname{CT}_t(I_{\lambda^1}(t,z))(z) \operatorname{CT}_t\!\bigg( \prod_{i=2}^s I_{\lambda^i}(t^{p^{l(\lambda^2)+\cdots+l(\lambda^{i-1})}},z^{p^{l(\lambda^2)+\cdots+l(\lambda^{i-1})}})\bigg) (z^{p^{l(\lambda^1)}}). \end{align*} $$

Thus, by induction on s, we prove the statement.

Corollary 2.8. We have

$$ \begin{align*} \operatorname{CT}_t(\tilde \lambda)(z) =\sum_{\lambda=\lambda^1*\cdots *\lambda^s} \operatorname{CT}_t(I_{\lambda^1})(z)\cdot \operatorname{CT}_t(I_{\lambda^2})(z^{p^{l(\lambda^1)}}) \cdots \operatorname{CT}_t(I_{\lambda^s})(z^{p^{l(\lambda^1)+\cdots+l(\lambda^{s-1})}}), \end{align*} $$

where the sum is over the set of all decompositions of $\lambda $ into a product of tuples.

2.8 Dwork congruence for tuples of Laurent polynomials

Theorem 2.9. Let $a,b,c$ be tuples of Laurent polynomials in $t,z$ with coefficients in ${\mathbb Z}_p$ , where $b,c,a'$ can be empty. Assume that the tuples $a*b, a*c, a'*b, a'*c$ of Laurent polynomials are admissible. Then,

(2-3) $$ \begin{align} \operatorname{CT}_t(\widetilde{a*b})(z)\, \operatorname{CT}_t(\widetilde{a'*c})(z^p) \,\equiv\, \operatorname{CT}_t(\widetilde{a'*b})(z^p) \,\operatorname{CT}_t(\widetilde{a*c})(z) \,(\mathrm{mod}\,{p^{l(a)}}). \end{align} $$

Proof. The left-hand side and right-hand side of Equation (2-3) are

$$ \begin{align*} \sum_{{a*b=x^1*\cdots *x^q\atop a'*c=y^1*\cdots *y^s}} \prod_{i=1}^q \operatorname{CT}_t(I_{x^i})(z^{p^{l(x^1)+\cdots+l(x^{i-1})}}) \prod_{i=1}^s \operatorname{CT}_t(I_{y^j})(z^{p^{1+l(y^1)+\cdots+l(y^{j-1})}}) \end{align*} $$

and

(2-4) $$ \begin{align} \sum_{{a'*b=x^1*\cdots *x^q\atop a*c=y^1*\cdots *y^s}} \prod_{i=1}^q \operatorname{CT}_t(I_{x^i})(z^{p^{1+l(x^1)+\cdots+l(x^{i-1})}}) \prod_{i=1}^s \operatorname{CT}_t(I_{y^j})(z^{p^{l(y^1)+\cdots+l(y^{j-1})}}), \end{align} $$

respectively. Since we work modulo $p^{l(a)}$ , all the terms with

$$ \begin{align*} \sum_{i=1}^q l(x^i) + \sum_{j=1}^s l(y^j) -q-s\geqslant l(a) \end{align*} $$

may be dropped off from consideration. That inequality can be reformulated as $l(a)+l(b)+l(a)+l(c)-1 - q-s \geqslant l(a)$ , equivalently, as

(2-5) $$ \begin{align} l(a)+l(b)+l(c) \geqslant q+s+1. \end{align} $$

Let us prove that the remaining terms in both expressions are in a bijective correspondence such that the corresponding terms are equal.

Namely, take one of the remaining summands on the left-hand side:

(2-6) $$ \begin{align} \prod_{i=1}^q \operatorname{CT}_t(I_{x^i})(z^{p^{l(x^1)+\cdots+l(x^{i-1})}}) \prod_{i=1}^s \operatorname{CT}_t(I_{y^j})(z^{p^{1+l(y^1)+\cdots+l(y^{j-1})}}), \end{align} $$

the summand corresponding to the presentation $a\kern1.2pt{*}\kern1.2ptb\kern1.2pt{=}\kern1.2pt x^1\kern1.2pt{*}\kern1.2pt\cdots *x^q$ , $a'\kern1.2pt{*}\kern1.2ptc\kern1.2pt{=}\kern1.2pt y^1 \kern1.2pt{*}\cdots \kern1.2pt{*}\kern1.2pt y^s$ .

Lemma 2.10. There exist indices $i\geqslant 1$ and $j\geqslant 0$ such that

(2-7) $$ \begin{align} l(x^1)+\cdots+l(x^i) = l(y^1)+\cdots+l(y^j) + 1\leqslant l(a). \end{align} $$

Proof. If $l(x^1)=1$ , then $i=1$ and $j=0$ are the required indices.

Assume that $l(x^1)>1$ and the required i, j do not exist. Then each number in $\{2, \ldots ,l(a)\}$ cannot be represented simultaneously as $l(x^1)+\cdots +l(x^i)$ and $l(y^1)+\cdots +l(y^j) + 1$ . Therefore, the sum of the total number of $i\geqslant 1$ , such that $l(x^1)+\cdots +l(x^i)\leqslant l(a)$ , and the total number of $j\geqslant 1$ , such that $l(y^1)+\cdots +l(y^j) + 1\leqslant l(a)$ is at most $l(a)-1$ . The number of remaining i is at most $l(b)$ and the number of remaining j is at most $l(c)$ . Therefore, $q+s \leqslant l(a) - 1 + l(b) + l(c)$ , which is the same as Equation (2-5). Hence, the corresponding summand must have been dropped off. This establishes the existence of indices i and j required.

Now we return to the remaining summand in Equation (2-6). Choose the minimal indices $i\geqslant 1$ and $j\geqslant 0$ such that Equation (2-7) holds. Then it is easy to see that

(2-8) $$ \begin{align} \hspace{-12pt}a'*b=y^1*\cdots *y^j *x^{i+1}*\cdots *x^q, \quad a*c =x^1*\cdots *x^i * y^{j+1}*\cdots *x^s, \end{align} $$

and the summand in Equation (2-4) corresponding to the presentations in Equation (2-8) equals the product in Equation (2-6). This clearly gives the desired bijection.

4.1 Polynomials $P_s(x)$

The function

(4-1) $$ \begin{align} _2F_1\bigg(\frac12,\frac 12; 1; x\bigg) = \frac 1\pi\,\int_1^\infty t^{-1/2}(t-1)^{-1/2}(t-x)^{-1/2}\,dt =\sum_{k = 0}^\infty\binom{-1/2}{k}^2x^k \end{align} $$

satisfies the hypergeometric differential equation

(4-2) $$ \begin{align} x(1-x) I" +(1-2x)I'-\tfrac14I=0. \end{align} $$

Define the master polynomial

$$ \begin{align*} \Phi_{p^s}(t,x) = t^{(p^s-1)/2}(t-1)^{(p^s-1)/2}(t-x)^{(p^s-1)/2}. \end{align*} $$

The number $M=({p^s-1})/2=({p-1})/2+({p-1})/2 p+\cdots +({p-1})/2 p^{s-1}$ is the unique positive integer such that $1\leqslant M\leqslant p^s$ and $M\equiv -1/2 \ \,(\mathrm {mod}\,{p^s})$ . Define the $p^s$ -approximation polynomial $P_s(x)$ as the coefficient of $t^{p^s-1}$ in the master polynomial $\Phi _{p^s}(t,x)$ . Then,

(4-3) $$ \begin{align} P_s(x) = (-1)^{(p^s-1)/2} \sum_{k=0}^{(p^s-1)/2}\binom{(p^s-1)/2}{k}^2 x^k. \end{align} $$

Define $P_0(x)=1$ .

Recall the hypergeometric function ${}_2F_1(a,b;c;x)$ . Then,

(4-4) $$ \begin{align} P_s(x) = (-1)^{(p^s-1)/2} {}_2F_1\bigg(\frac{1-p^s}2, \frac{1-p^s}2; 1; x\bigg). \end{align} $$

The polynomial $P_s(x)$ is a solution of the hypergeometric equation in Equation (4-2) modulo $p^s$ . This follows from Theorem 3.1 or from Equation (4-4).

4.2 Baby congruences

Let $\varphi _s(x)=(x+1)^{(p^s-1)/2}$ . Then,

(4-5) $$ \begin{align} \varphi_{s+1}(x)\varphi_{s-1}(x^p)\equiv\varphi_s(x)\varphi_s(x^p)\,(\mathrm{mod}\,{p^s}). \end{align} $$

This follows from $(x+1)^{p^s}\equiv (x^p+1)^{p^{s-1}}\,(\mathrm {mod}\,{p^s})$ .

Lemma 4.1. The master polynomials $\Phi _{s}(t,x)$ satisfy the baby congruence

(4-6) $$ \begin{align} \Phi_{s+1}(t,x)\Phi_{s-1}(t^p,x^p) \equiv \Phi_{s}(t,x)\Phi_{s}(t^p,x^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

4.3 Congruences for $P_s(x)$

Theorem 4.2. The approximation polynomials $P_s(x)$ satisfy the congruence

(4-7) $$ \begin{align} P_{s+1}(x) P_{s-1}(x^p) \equiv P_{s}(x) P_{s}(x^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

This theorem follows from a more general Theorem 4.4.

Using Equation (4-4), we may rewrite Equation (4-7) as the congruence

(4-8) $$ \begin{align} &{}_2F_1\bigg(\dfrac{1-p^{s+1}}2, \dfrac{1-p^{s+1}}2; 1; x\bigg) {}_2F_1\bigg(\dfrac{1-p^{s-1}}2, \dfrac{1-p^{s-1}}2; 1; x^p\bigg) \notag \\ & \quad \equiv {}_2F_1\bigg(\dfrac{1-p^s}2, \dfrac{1-p^s}2; 1; x\bigg){}_2F_1\bigg(\dfrac{1-p^s}2, \dfrac{1-p^s}2; 1; x^p\bigg) \,(\mathrm{mod}\,{p^s}). \end{align} $$

Let $\alpha $ be a rational number which is a p-adic unit, $\alpha = \alpha _0 + \alpha _1 p+ \alpha _2p^2+\cdots $ . Denote by $[\alpha ]_s$ the sum of the first s summands. Then the congruence in Equation (4-8) takes the form:

$$ \begin{align*} &{}_2F_1\big([-\tfrac12]_{s+1}, [-\tfrac12]_{s+1};1;x\big)\, {}_2F_1\big([-\tfrac12]_{s-1}, [-\tfrac12]_{s-1};1;x^p\big) \notag \\ &\quad \equiv {}_2F_1\big([-\tfrac 12]_{s}, [-\tfrac12]_{s};1;x\big)\, {}_2F_1\big([-\tfrac12]_{s}, [-\tfrac12]_{s};1;x^p\big) \,(\mathrm{mod}\,{p^s}). \end{align*} $$

4.4 Coefficients of master polynomials

Consider

$$ \begin{align*} \hat\Phi_s(t,x) &:= t^{-(p^s-1)}\Phi_s(t,x) = t^{-(p^s-1)/2}((t-1)(t-x))^{(p^s-1)/2} \\ &= ((t-1)(1-x/t))^{(p^s-1)/2} =\sum_{j=-(p^s-1)/2}^{(p^s-1)/2} C_{s,j}(x)t^j, \end{align*} $$

where

$$ \begin{align*} C_{s,j}(x) = (-1)^{(({p^s-1})/2)-j}\sum_{m} \binom{\frac{p^s-1}2}{m+j} \binom{\frac{p^s-1}2}{m}\, x^m. \end{align*} $$

In particular, $C_{s,0}(x) = P_s(x)$ . Every coefficient $C_{s,j}(x)$ is a hypergeometric function:

$$ \begin{align*} C_{s,j}(x) = (-1)^{(({p^s-1})/2)-j} \binom{\frac{p^s-1}2}{j} {}_2F_1\bigg( \frac{1-p^s}2, \frac{1-p^s}2+j; j+1; x\bigg) \quad\text{for}\; j\geqslant0, \end{align*} $$

while a hypergeometric expression in the case $j<0$ comes out from the following simple fact.

Lemma 4.3. We have $\hat \Phi _s(x/t,x) = \hat \Phi _s(t,x)$ and hence

(4-9) $$ \begin{align} C_{s,-j}(x) = x^j C_{s,j}(x). \end{align} $$

We expand the congruence in Equation (4-6) into a congruence of polynomials $C_{s,j}(x)$ . The constant term in t gives us

(4-10) $$ \begin{align} \sum_{k} C_{s+1,kp}(x)C_{s-1,-k}(x^p) \equiv \sum_{k} C_{s,kp}(x)C_{s,-k}(x^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

The following Theorem 4.4 establishes the congruences of individual pairs of terms in Equation (4-10).

Theorem 4.4. For any k appearing in Equation (4-10),

(4-11) $$ \begin{align} C_{s+1,kp}(x)C_{s-1,-k}(x^p) \equiv C_{s,kp}(x)C_{s,-k}(x^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

In particular, for $k=0$ , we have the congruence in Equation (4-7).

Proof. Every index k appearing in Equation (4-10) can be written uniquely as

$$ \begin{align*} \phantom{aaa} k= k_0 + k_1p+\cdots + k_{s-2}p^{s-2}, \quad -(p-1)/2\leqslant k_i \leqslant (p-1)/2. \end{align*} $$

Using Equation (4-9), we reformulate Equation (4-11) as

(4-12) $$ \begin{align} C_{s+1,kp}(x)C_{s-1,k}(x^p) \equiv C_{s,kp}(x)C_{s,k}(x^p)\ (\mod {p^{3}}) \end{align} $$

and prove it. We have

$$ \begin{align*} C_{s+1,kp}(x) &= \operatorname{CT}_t\bigg[\hat\Phi_1(t,x) \bigg(\prod_{i=0}^{s-2} ( t^{-k_i}\hat\Phi_1(t,x))^{p^{i+1}}\bigg) \hat\Phi_1(t,x)^{p^s}\bigg], \\ C_{s-1,k}(x) &= \operatorname{CT}_t\bigg[\prod_{i=0}^{s-2} ( t^{-k_i}\hat\Phi_1(t,x))^{p^{i}}\bigg], \\ C_{s,kp}(x) &=\operatorname{CT}_t\bigg[\hat\Phi_1(t,x) \prod_{i=0}^{s-2} ( t^{-k_i}\hat\Phi_1(t,x))^{p^{i+1}}\bigg], \\ C_{s,k}(x) &= \operatorname{CT}_t\bigg[\bigg(\prod_{i=0}^{s-2} ( t^{-k_i}\hat\Phi_1(t,x))^{p^{i}}\bigg) \hat\Phi_1(t,x)^{p^{s-1}}\bigg]. \end{align*} $$

It is easy to see that the $(s+1)$ -tuple of Laurent polynomials

$$ \begin{align*} \hat\Phi_1(t,x),\, t^{-k_0}\hat\Phi_1(t,x), \,\ldots, \,t^{-k_{s-2}}\hat\Phi_1(t,x),\, \hat\Phi_1(t,x) \end{align*} $$

is admissible in the sense of Definition 2.6. Now the application of Theorem 2.9 gives the congruence in Equation (4-12) and hence the congruence in Equation (4-11).

Remark 4.5. Denote $A(n,x) :={}_2F_1(-n,-n;1;x) = \sum _k \binom {n}{k}^2 x^k$ . Let

$$ \begin{align*} n = n_0+n_1p+ \cdots + n_{s-1}p^{s-1}, \quad [n/p] = n_1 +\cdots + n_{s-1}p^{s-2}, \end{align*} $$

where $0\leqslant n_i<p$ . Then for any $m\in {\mathbb Z}_{\geqslant 0}$ ,

$$ \begin{align*} A(n +mp^s, x) A([n/p],x^p) \equiv A(n,x) A([n/p]+mp^{s-1},x^p) \,(\mathrm{mod}\,{p^s}). \end{align*} $$

The proof follows from Theorem 2.9 and the identity

$$ \begin{align*} A(n,x) = \operatorname{CT}_t[((t+1)(1+x/t))^n]. \end{align*} $$

4.5 Limits of $P_s(x)$

For $\alpha \in {\mathbb Z}_p$ , there exists a unique solution $\omega (\alpha )\in {\mathbb Z}_p$ of the equation $\omega (\alpha )^p=\omega (\alpha )$ that is congruent to x modulo p. The element $\omega (\alpha )$ is called the Teichmüller representative of $\alpha $ . For $\alpha \in {\mathbb F}_p$ , $r>0$ , define the disc

$$ \begin{align*} D_{\alpha,r} = \{ x\in {\mathbb Z}_p \mid |x-\omega(\alpha)|_p < r \}. \end{align*} $$

Denote

$$ \begin{align*} \bar P_s(x) := (-1)^{(p^s-1)/2} P_s(x) = {}_2F_1\big(\big[\!-\tfrac 12\big]_{s},\big[\!-\tfrac12\big]_{s};1;x\big); \end{align*} $$

see Equation (4-4). Denote

$$ \begin{align*} \mathfrak D = \{x\in {\mathbb Z}_p \mid |\bar P_1(x)|_p=1\}. \end{align*} $$

Proof. We have ${\mathbb Z}_p = \bigcup _{\alpha \in {\mathbb F}_p}D_{\alpha ,1}$ and also $\mathfrak D= \bigcup _{\alpha \in {\mathbb F}_p,\, |\bar P_1(\omega (\alpha ))|_p =1} D_{\alpha ,1}$ since $\bar P_1(x)$ has coefficients in ${\mathbb Z}_p$ . In particular, $D_{0,1}\subset \mathfrak D$ . We also have

$$ \begin{align*} \{x\in {\mathbb Z}_p \mid |\bar P_1(x^p)|_p=1\} = \bigcup_{\alpha\in{\mathbb F}_p,\, |\bar P_1(\omega(\alpha))|_p =1} D_{\alpha,1} = \mathfrak D \end{align*} $$

for the same reason.

We have $\bar P_s(x) \equiv \bar P_1(x) \bar P_1(x^p)\cdots \bar P_1(x^{p^{s-1}}) \ \,(\mathrm {mod}\,{p})$ . Indeed, the polynomial $ P_s(x)$ is the coefficient of $x^{p^s-1}$ in the master polynomials $\Phi _{p^s}(t,x)$ and

$$ \begin{align*} \Phi_{p^s}(t,x) \equiv \Phi_{p}(t,x)\Phi_{p}(t^{p},x^{p})\cdots \Phi_{p}(t^{p^{s-1}},x^{p^{s-1}}) \,(\mathrm{mod}\,{p}). \end{align*} $$

Hence, $|\bar P_s(x)|_p = |\bar P_s(x^p)|_p=1$ for $s\geqslant 1$ , $x\in \mathfrak D$ . Hence, the rational functions ${\bar P_{s+1}(x)}/{\bar P_s(x^p)}$ are regular on $\mathfrak D$ .

The congruence in Equation (4-7) implies that

$$ \begin{align*} \bigg|\frac{\bar P_{s+1}(x)}{\bar P_{s}(x^p)} - \frac{\bar P_{s}(x)}{\bar P_{s-1}(x^p)}\bigg|_p\leqslant p^{-s} \quad\operatorname{for}\ x\in\mathfrak D. \end{align*} $$

This shows the uniform convergence of our sequence of rational functions on the domain $\mathfrak D$ . For the limiting function $f(x)$ , we have $|f(x)|_p=1$ for $x\in \mathfrak D$ .

Clearly, for any fixed index k, the coefficient $\binom {(p^s-1)/2}{k}^2$ of $x^k$ in $\bar P_s(x)$ converges p-adically to the coefficient $\binom {-1/2}{k}^2$ of $x^k$ in $F(x)$ . Hence, the sequence $(\bar P_s(x))_{s\geqslant 1}$ converges to $F(x)$ on $D_{0,1}$ , so that $f(x) ={F(x)}/{F(x^p)}$ on $D_{0,1}$ . The theorem is proved.

Dwork gives in [Reference DworkDw] a different construction of analytic continuation of the ratio ${F(x)}/{F(x^p)}$ from $D_{0,1}$ to a larger domain. He considers the sequence of polynomials

$$ \begin{align*} F_s(x) = \sum_{k=0}^{p^s-1}\binom{-1/2}{k}^2 x^k, \end{align*} $$

which are truncations of the hypergeometric series $F(x)$ , and shows that the sequence of rational functions $({F_{s+1}(x)}/{F_s(x^p)})_{s\geqslant 1}$ uniformly converges on the domain $\mathfrak D^{\operatorname {Dw}} =\{x\in {\mathbb Z}_p\ |\ |g(x)|_p=1\}$ , where the polynomial

(4-13) $$ \begin{align} g(x) =\sum_{k=0}^{(p-1)/2} \binom{-1/2}{k}^2x^k \end{align} $$

is attributed by Dwork to Igusa [Reference IgusaIg]. Clearly, his limiting function $f^{\operatorname {Dw}}(x)$ equals the ratio ${F(x)}/{F(x^p)}$ on $D_{0,1}$ .

It is easy to see that the two sequences of rational functions $({\bar P_{s+1}(x)}/{\bar P_s(x^p)})_{s\geqslant 1}$ and $({F_{s+1}(x)}/{F_s(x^p)})_{s\geqslant 1}$ have the same limiting functions on the same domain. Indeed, $\bar P_1(x)\equiv g(x) \ \,(\mathrm {mod}\,{p})$ and hence $\mathfrak D=\mathfrak D^{\operatorname {Dw}}$ . Also, $f(x) = f^{\operatorname {Dw}}(x)$ on $D_{0,1}$ and hence on $\mathfrak D$ .

Dwork shows in [Reference DworkDw] interesting properties of the function $f(x)$ . For example, let $\alpha \in {\mathbb F}^\times _p-\{1\}$ be such that $\omega (\alpha ) \in {\mathfrak D}$ . Dwork shows that the zeta function of the elliptic curve defined over ${\mathbb F}_p$ by the equation $y^2=x(x-1)(x-\alpha )$ has two zeros, which are $1/((-1)^{(p-1)/2} f(\omega (\alpha )))$ and $(-1)^{(p-1)/2} f(\omega (\alpha ))/p$ . Clearly, $|f(\omega (\alpha ))|_p=1$ . The number $(-1)^{(p-1)/2} f(\omega (\alpha ))$ is called the unit root of that elliptic curve.

According to our discussion, this unit root can be calculated as the value at $x=\omega (\alpha )$ of the limit as $s\to \infty $ of the ratio ${\bar P_{s+1}(x)}/{\bar P_s(x^p)}$ of approximation polynomials multiplied by $(-1)^{(p-1)/2}$ .

5.1 Two hypergeometric integrals

The integral

$$ \begin{align*} I^{(C)}(x) =\int_C t^{-1/3}(t-1)^{-1/3} (t-x)^{-2/3}\,dt, \end{align*} $$

where $C\subset {\mathbb C}-\{0,1,x\}$ is a contour on which the integrand takes its initial value when t encircles C, satisfies the hypergeometric differential equation

(5-1) $$ \begin{align} x(1-x) I" +(1-2x)I'-\tfrac29I=0. \end{align} $$

For a suitable choice of C, the integral $I^{(C)}(x)$ presents the hypergeometric function

$$ \begin{align*} {}_2F_1\bigg(\frac23,\frac 13; 1,x\bigg) = \sum_{k = 0}^\infty\binom{-1/3}{k}\binom{-2/3}{k}\,x^k. \end{align*} $$

The integral

$$ \begin{align*} J^{(D)}(x) =\int_D t^{-2/3}(t-1)^{-2/3} (t-x)^{-1/3}\,dt, \end{align*} $$

where $D\subset {\mathbb C}-\{0,1,x\}$ is a contour on which the integrand takes its initial value when t encircles D, satisfies the same hypergeometric differential equation. For a suitable choice of D, the integral $J^{(D)}(x)$ presents the same hypergeometric function $_2F_1(2/3,\tfrac 13; 1,x)$ .

The differential form $t^{-1/3}(t-1)^{-1/3}(t-z)^{-2/3}\,dt$ is transformed to the differential form $-t^{-2/3}(t-1)^{-2/3}(t-z)^{-1/3}\,dt$ by the change of variable $t\mapsto (t-z)/(t-1)$ .

In this section, we discuss the $p^s$ -approximations of the integrals $I^{(C)}(x)$ and $J^{(D)}(x)$ .

5.2 The case $p=3\ell +1$

The master polynomial for $I^{(C)}(x)$ is given by the formula

$$ \begin{align*} \Phi_s(t,x) = t^{(p^s-1)/3}(t-1)^{(p^s-1)/3}(t-x)^{2(p^s-1)/3}. \end{align*} $$

The $p^s$ -approximation polynomial $Q_s(x)$ is defined as the coefficient of $t^{p^s-1}$ in $\Phi _s(t,x)$ ,

$$ \begin{align*} Q_s(x) = (-1)^{(p^s-1)/3}\sum_{k} \binom{2(p^s-1)/3}{k} \binom{(p^s-1)/3}{k}\, x^k. \end{align*} $$

Define $Q_0(x)=1$ . We have

(5-2) $$ \begin{align} Q_s(x) = {}_2F_1\bigg(\frac{2-2p^s}3, \frac{1-p^s}3; 1; x\bigg), \end{align} $$

since $(-1)^{(p^s-1)/3}=1$ .

The polynomial $Q_s(x)$ is a solution of the hypergeometric equation in Equation (5-1) modulo $p^s$ . This follows from Theorem 3.1 or from Equation (5-2).

The master polynomial for $J^{(D)}(x)$ is given by the formula

$$ \begin{align*} \Psi_s(t,x) = t^{2(p^s-1)/3}(t-1)^{2(p^s-1)/3}(t-x)^{(p^s-1)/3}. \end{align*} $$

The $p^s$ -approximation polynomial $R_s(x)$ is defined as the coefficient of $t^{p^s-1}$ in $\Psi _s(t,x)$ ,

$$ \begin{align*} R_s(x) = \sum_{k} \binom{2(p^s-1)/3}{k} \binom{(p^s-1)/3}{k}\, x^k. \end{align*} $$

Define $R_0(x)=1$ . We have

(5-3) $$ \begin{align} Q_s(x) = {}_2F_1\bigg(2\frac{1-p^s}3, \frac{1-p^s}3; 1; x\bigg). \end{align} $$

The master polynomials satisfy the baby congruences,

(5-4) $$ \begin{align} \begin{aligned} &\kern0.2pt\Phi_{s+1}(t,x) \Phi_{s-1}(t^p,x^p) \equiv \Phi_{s}(t,x) \Phi_{s}(t^p,x^p) \,(\mathrm{mod}\,{p^s}),\\ &\Psi_{s+1}(t,x) \Psi_{s-1}(t^p,x^p) \equiv \Psi_{s}(t,x) \Psi_{s}(t^p,x^p) \,(\mathrm{mod}\,{p^s}) , \end{aligned}\qquad \end{align} $$

by Equation (4-5).

Theorem 5.1. For $p=3\ell +1$ , the approximation polynomials $R_s(x)$ and $Q_s(x)$ satisfy the congruences

(5-5) $$ \begin{align} &Q_{s+1}(x) Q_{s-1}(x^p) \equiv Q_{s}(x) Q_{s}(x^p) \,(\mathrm{mod}\,{p^s}), \end{align} $$

(5-6) $$ \begin{align} &R_{s+1}(x) R_{s-1}(x^p) \equiv R_{s}(x) R_{s}(x^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

Equations (5-2) and (5-3) imply that for $p=3\ell +1$ , $s\geqslant 1$ ,

$$ \begin{align*} &{}_2F_1\bigg(\dfrac{2-2p^{s+1}}3, \dfrac{1-p^{s+1}}3; 1; x\bigg) {}_2F_1\bigg(\dfrac{2-2p^{s-1}}3, \dfrac{1-p^{s-1}}3; 1; x^p\bigg) \notag\\ &\quad \equiv {}_2F_1\bigg(\dfrac{2-2p^s}3, \dfrac{1-p^s}3; 1; x\bigg) {}_2F_1\bigg(\dfrac{2-2p^s}3, \dfrac{1-p^s}3; 1; x^p\bigg) \,(\mathrm{mod}\,{p^s}). \end{align*} $$

Using the expansions

$$ \begin{align*} -1/3 &= \ell + \ell p+\ell p^2+\cdots, &\quad -2/3 &= 2\ell + 2\ell p+2\ell p^2+\cdots,\\ (p^s-1)/3 &= \ell + \ell p+\ell p^2+\cdots + \ell p^{s-1}, &\ \ \, (p^s-2)/3 &= 2\ell + 2\ell p+2\ell p^2+\cdots + 2\ell p^{s-1}, \end{align*} $$

we conclude that for $p=3\ell +1$ and $s\geqslant 1$ ,

(5-7) $$ \begin{align} &{}_2F_1\big([-\tfrac23]_{s+1}, [-\tfrac13]_{s+1};1;x\big)\, {}_2F_1\big([-\tfrac23]_{s-1}, [-\tfrac13]_{s-1};1;x^p\big) \notag \\ &\quad \equiv {}_2F_1\big([-\tfrac 23]_{s}, [-\tfrac13]_{s};1;x\big)\, {}_2F_1\big([-\tfrac23]_{s}, [-\tfrac13]_{s};1;x^p\big) \,(\mathrm{mod}\,{p^s}). \end{align} $$

5.3 The case $p=3\ell +2>2$

The master polynomial for $I^{(C)}(x)$ is given by the formulas

$$ \begin{align*} \Phi_s(t,x) &= t^{(2p^s-1)/3}(t-1)^{(2p^s-1)/3}(t-x)^{(p^s-2)/3}, \quad \!\!\operatorname{odd}\,s,\\ \Phi_s(t,x) &= t^{(p^s-1)/3}(t-1)^{(p^s-1)/3}(t-x)^{2(p^s-1)/3}, \quad \operatorname{even}\,s. \end{align*} $$

The $p^s$ -approximation polynomial $Q_s(x)$ is defined as the coefficient of $t^{p^s-1}$ in $\Phi _s(t,x)$ ,

$$ \begin{align*} Q_s(x) &= (-1)^{(2p^s-1)/3}\sum_{k} \binom{(2p^s-1)/3}{k} \binom{(p^s-2)/3}{k}\, x^k, \quad \!\!\operatorname{odd}\,s,\\ Q_s(x) &= (-1)^{(p^s-1)/3}\sum_{k} \binom{2(p^s-1)/3}{k} \binom{(p^s-1)/3}{k}\, x^k, \quad \operatorname{even}\,s. \end{align*} $$

Define $Q_0(x)=1$ . We have

(5-8) $$ \begin{align} \begin{aligned} \kern1pt Q_s(x) &= -\, {}_2F_1\bigg(\frac{2-p^s}3, \frac{1-2p^s}3; 1; x\bigg), \quad\! \!\!\operatorname{odd}\,s,\\ Q_s(x) &= {}_2F_1\bigg(\frac{2-2p^s}3, \frac{1-p^s}3; 1; x\bigg),\ \quad \operatorname{even}\,s. \end{aligned}\qquad \end{align} $$

Here we use the fact that for $p=3\ell +2>2$ , we have $(-1)^{(2p^s-1)/3}=-1 $ for odd s and $(-1)^{(p^s-1)/3}=1$ for even s.

The polynomial $Q_s(x)$ is a solution of the hypergeometric equation in Equation (5-1) modulo $p^s$ . This follows from Theorem 3.1 or from Equation (5-8).

The master polynomial for $J^{(D)}(x)$ is given by the formulas

$$ \begin{align*} \Psi_s(t,x) &= t^{(p^s-2)/3}(t-1)^{(p^s-2)/3}(t-x)^{(2p^s-1)/3}, \quad\, \operatorname{odd}\,s,\\ \Psi_s(t,x) &= t^{2(p^s-1)/3}(t-1)^{2(p^s-1)/3}(t-x)^{(p^s-1)/3}, \quad \!\operatorname{even}\,s. \end{align*} $$

The $p^s$ -approximation polynomial $R_s(x)$ is defined as the coefficient of $t^{p^s-1}$ in $\Phi _s(t,x)$ ,

$$ \begin{align*} R_s(x) &= (-1)^{(p^s-2)/3}\sum_{k} \binom{(2p^s-1)/3}{k} \binom{(p^s-2)/3}{k}\, x^k, \quad \operatorname{odd}\,s, \\ R_s(x) &= (-1)^{2(p^s-1)/3}\sum_{k} \binom{2(p^s-1)/3}{k} \binom{(p^s-1)/3}{k}\, x^k, \quad\!\! \operatorname{even}\,s. \notag \end{align*} $$

Define $R_0(x)=1$ . We have

(5-9) $$ \begin{align} \begin{aligned} &\kern2pt R_s(x) = -\, {}_2F_1\bigg(\frac{2-p^s}3, \frac{1-2p^s}3; 1; x\bigg), \quad \operatorname{odd}\,s,\\ &R_s(x) = {}_2F_1\bigg(\frac{2-2p^s}3, \frac{1-p^s}3; 1; x\bigg),\quad \ \quad \!\!\operatorname{even}\,s. \end{aligned}\qquad \end{align} $$

Here we use the fact that for $p=3\ell +2>2$ , we have $(-1)^{(p^s-2)/3}=-1$ for odd s and $(-1)^{(2p^s-1)/3}=1 $ for even s.

The polynomial $R_s(x)$ is a solution of the hypergeometric equation in Equation (5-1) modulo $p^s$ . This follows from Theorem 3.1 or from Equation (5-9).

Lemma 5.2. The master polynomials satisfy the baby congruences,

(5-10) $$ \begin{align} &\Phi_{s+1}(t,x) \Psi_{s-1}(t^p,x^p) \equiv \Phi_{s}(t,x) \Psi_{s}(t^p,x^p) \,(\mathrm{mod}\,{p^s}), \end{align} $$

(5-11) $$ \begin{align} &\Psi_{s+1}(t,x) \Phi_{s-1}(t^p,x^p) \equiv \Psi_{s}(t,x) \Phi_{s}(t^p,x^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

Proof. We prove Equation (5-10) for an odd s. The case of an even s and the congruence in Equation (5-11) are proved similarly. The left-hand side of Equation (5-10) for an odd $s=2k+1$ equals

$$ \begin{align*} t^{(p^{s+1}-1)/3}(t-1)^{(p^{s+1}-1)/3}(t-x)^{2(p^{s+1}-1)/3} t^{2p(p^{s-1}-1)/3}(t^p-1)^{2(p^{s-1}-1)/3}(t^p-x^p)^{(p^{s-1}-1)/3}, \end{align*} $$

while the right-hand side equals

$$ \begin{align*} t^{(2p^s-1)/3}(t-1)^{(2p^s-1)/3}(t-x)^{(p^s-2)/3} t^{p(p^s-2)/3}(t^p-1)^{(p^s-2)/3}(t^p-x^p)^{(2p^s-1)/3}. \end{align*} $$

Now the congruence in Equation (5-10) for an odd s follows from Equation (4-5).

Theorem 5.3. For $p=3\ell +2>2$ , the approximation polynomials $R_s(x)$ and $Q_s(x)$ satisfy the congruences

(5-12) $$ \begin{align} & Q_{s+1}(x) R_{s-1}(x^p) \equiv Q_{s}(x) R_{s}(x^p) \,(\mathrm{mod}\,{p^s}), \end{align} $$

(5-13) $$ \begin{align} & R_{s+1}(x) Q_{s-1}(x^p) \equiv R_{s}(x) Q_{s}(x^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

Proof. We prove Equation (5-12) for an odd s. The case of an even s and the congruence in Equation (5-13) are proved similarly. Denote

$$ \begin{align*} f(t,x) &= (t-1)^{(2p-1)/3}(1-x/t)^{(p-2)/3}=(t-1)^{2\ell+1}(1-x/t)^{\ell}, \\ g(t,x) &= (t-1)^{(p-2)/3}(1-x/t)^{(2p-1)/3}= (t-1)^{\ell}(1-x/t)^{2\ell+1}. \end{align*} $$

It is easy to see that for an odd s,

$$ \begin{align*} Q_{s+1}(x) &=\operatorname{CT}_t[(t-1)^{(p^{s+1}-1)/3}(1-x/t)^{2(p^{s+1}-1)/3}] \\ &=\operatorname{CT}_t[f(t,x) g(t,x)^{p} \cdots f(t,x)^{p^{s-1}} g(t,x)^{p^s} ], \\ R_{s-1}(x) &=\operatorname{CT}_t[(t-1)^{2(p^{s-1}-1)/3}(1-x/t)^{(p^{s-1}-1)/3}]\\ &=\operatorname{CT}_t[g(t,x) f(t,x)^{p} \cdots g(t,x)^{p^{s-3}}f(t,x)^{p^{s-2}} ], \end{align*} $$

$$ \begin{align*} Q_{s}(x) &= \operatorname{CT}_t[(t-1)^{(2p^{s}-1)/3}(1-x/t)^{(p^{s}-2)/3}]\\ &=\operatorname{CT}_t[f(t,x) g(t,x)^{p} \cdots g(t,x)^{p^{s-2}} f(t,x)^{p^{s-1}} ],\\ R_{s}(x) &=\operatorname{CT}_t[(t-1)^{(p^{s}-2)/3}(1-x/t)^{(2p^{s}-1)/3}]\\ &=\operatorname{CT}_t[g(t,x) f(t,x)^{p} \cdots f(t,x)^{p^{s-2}}g(t,x)^{p^{s-1}} ]. \end{align*} $$

Observe that the $(s+1)$ -tuple of Laurent polynomials $(f(t,x),g(t,x),\ldots , f(t,x), g(t,x))$ is admissible in the sense of Definition 2.6. Now the application of Theorem 2.9 gives the congruence in Equation (5-12) for an odd s.

Using Equations (5-8) and (5-9), we may reformulate the congruences in Equations (5-12) and (5-13) as

(5-14) $$ \begin{align} &{}_2F_1\bigg(\dfrac{2-p^{s+1}}3, \dfrac{1-2p^{s+1}}3; 1; x\bigg) {}_2F_1\bigg(\dfrac{2-p^{s-1}}3, \dfrac{1-2p^{s-1}}3; 1; x^p\bigg) \notag \\ &\quad \equiv {}_2F_1\bigg(\dfrac{2-2p^s}3, \dfrac{1-p^s}3; 1; x\bigg) {}_2F_1\bigg(\dfrac{2-2p^s}3, \dfrac{1-p^s}3; 1; x^p\bigg) \,(\mathrm{ mod}\,{p^s}), \ \ \operatorname{odd} \,s, \end{align} $$

(5-15) $$ \begin{align} &{}_2F_1\bigg(\dfrac{2-2p^{s+1}}3, \dfrac{1-p^{s+1}}3; 1; x\bigg) {}_2F_1\bigg(\dfrac{2-2p^{s-1}}3, \dfrac{1-p^{s-1}}3; 1; x^p\bigg) \notag\\ &\quad \equiv {}_2F_1\bigg(\dfrac{2-p^s}3, \dfrac{1-2p^s}3; 1; x\bigg) {}_2F_1\bigg(\dfrac{2-p^s}3, \dfrac{1-2p^s}3; 1; x^p\bigg) \,(\mathrm{ mod}\,{p^s}), \ \ \operatorname{even} \,s. \end{align} $$

Recall that in these congruences, we have $p=3\ell +2$ .

Consider the p-adic presentations

(5-16) $$ \begin{align} -1/3 &= 2\ell +1 + \ell p+(2\ell+1)p^2+\ell p^3+\cdots , \end{align} $$

(5-17) $$ \begin{align} -2/3 &= \ell + (2\ell+1)p+\ell p^2+ (2\ell+1)p^3+\cdots. \end{align} $$

Recall that $[-1/3]_s$ (respectively, $[-2/3]_s$ ) is the sum of the first s summands in Equation (5-16) (respectively, (5-17)). Then the congruences in Equations (5-14) and (5-15) imply that for $p=3\ell +2$ , $s\geqslant 1$ ,

(5-18) $$ \begin{align} &{}_2F_1\big([-\tfrac23]_{s+1}, [-\tfrac13]_{s+1};1;x\big)\, {}_2F_1\big([-\tfrac23]_{s-1}, [-\tfrac13]_{s-1};1;x^p\big) \notag\\ &\quad \equiv {}_2F_1\big([-\tfrac 23]_{s}, [-\tfrac13]_{s};1;x\big)\, {}_2F_1\big([-\tfrac23]_{s}, [-\tfrac13]_{s};1;x^p\big) \,(\mathrm{mod}\,{p^s}). \end{align} $$

5.4 Limits of $\overline {Q}_s(x)$

Define

$$ \begin{align*} \bar Q_s(x) = {}_2F_1\big(\big[\!-\tfrac 23\big]_{s}, \big[\!-\tfrac13\big]_{s};1;x\big). \end{align*} $$

Then for any prime $p>3$ ,

$$ \begin{align*} \bar Q_{s+1}(x)\bar Q_{s-1}(x^p) \equiv \bar Q_{s}(x)\bar Q_{s}(x^p) \,(\mathrm{mod}\,{p^s}) \end{align*} $$

by Equations (5-7) and (5-18).

Theorem 5.5. For any prime $p>3$ and integer $s\geqslant 1$ , the rational function ${\bar Q_{s+1}(x)}/{\bar Q_s(x^p)}$ is regular on the domain

$$ \begin{align*} \mathfrak D = \{ x\in{\mathbb Z}_p\ |\ |\bar Q_1(x)|_p=1\}. \end{align*} $$

The sequence $({\bar Q_{s+1}(x)}/{\bar Q_s(x^p)})_{s\geqslant 1}$ uniformly converges on $\mathfrak D$ . The limiting analytic function $f(x)$ equals the ratio ${F(x)}/{F(x^p)}$ on the disc $D_{0,1}$ , where $F(x):={}_2F_1(2/3,1/3;1;x)$ is defined by the corresponding convergent power series.

5.5 Remark

Although the congruences in Equations (5-7) and (5-18) look the same for $p=3\ell +1$ and $p=3\ell +2$ , the proofs of them are different as already presented. The proof of Equation (5-7) for $p=3\ell +1$ uses just any one of the two master polynomials: $\Phi _s(t,x)$ or $\Psi _s(t,x)$ , while the proof of Equation (5-18) for $p=3\ell +2$ uses the interaction of the two master polynomials $\Phi _s(t,x)$ and $\Psi _s(t,x)$ . See the baby congruences in Equations (5-4), (5-10), (5-11).

5.6 Congruences related to $-\tfrac 15$ , $-\tfrac 25$ , $-\tfrac 35$ , $-\tfrac 45$

In this section, we formulate the congruences related to the above rational numbers. The proof of these congruences is similar to the corresponding proofs in Sections 4 and 5.

For $p=5\ell \pm 2$ and any $s\geqslant 1$ ,

$$ \begin{align*} &{}_2F_1\big([-\tfrac45]_{s+1}, [-\tfrac15]_{s+1};1;x\big)\, {}_2F_1\big([-\tfrac35]_{s-1}, [-\tfrac25]_{s-1};1;x^p\big) \notag \\ &\quad \equiv {}_2F_1\big([-\tfrac 45]_{s}, [-\tfrac15]_{s};1;x\big)\, {}_2F_1\big([-\tfrac35]_{s}, [-\tfrac25]_{s};1;x^p\big) \,(\mathrm{mod}\,{p^s}), \end{align*} $$

$$ \begin{align*} &{}_2F_1\big([-\tfrac35]_{s+1}, [-\tfrac25]_{s+1};1;x\big)\, {}_2F_1\big([-\tfrac45]_{s-1}, [-\tfrac15]_{s-1};1;x^p\big) \notag\\ &\quad \equiv {}_2F_1\big([-\tfrac 35]_{s}, [-\tfrac25]_{s};1;x\big)\, {}_2F_1\big([-\tfrac45]_{s}, [-\tfrac15]_{s};1;x^p\big) \,(\mathrm{mod}\,{p^s}).\end{align*} $$

For $p=5\ell \pm 1$ and any $s\geqslant 1$ ,

$$ \begin{align*} &{}_2F_1\big([-\tfrac45]_{s+1}, [-\tfrac15]_{s+1};1;x\big)\, {}_2F_1\big([-\tfrac45]_{s-1}, [-\tfrac15]_{s-1};1;x^p\big) \notag\\ &\quad \equiv {}_2F_1\big([-\tfrac 45]_{s}, [-\tfrac15]_{s};1;x\big)\, {}_2F_1\big([-\tfrac45]_{s}, [-\tfrac15]_{s};1;x^p\big) \,(\mathrm{mod}\,{p^s}), \end{align*} $$

$$ \begin{align*} & {}_2F_1\big([-\tfrac35]_{s+1}, [-\tfrac25]_{s+1};1;x\big)\, {}_2F_1\big([-\tfrac35]_{s-1}, [-\tfrac25]_{s-1};1;x^p\big) \notag \\ &\quad \equiv {}_2F_1\big([-\tfrac 35]_{s}, [-\tfrac25]_{s};1;x\big)\, {}_2F_1\big([-\tfrac35]_{s}, [-\tfrac25]_{s};1;x^p\big) \,(\mathrm{mod}\,{p^s}). \end{align*} $$

Similar congruences hold for rational numbers of the form $a/b$ , where b is a prime and $1-b\leqslant a \leqslant -1$ . These congruences are described somewhere else.

6.1 KZ equations

Let ${\mathfrak g}$ be a simple Lie algebra with an invariant scalar product. The Casimir element is

$$ \begin{align*} \Omega = \sum_i \,h_i\otimes h_i \ \ \in \ {\mathfrak g} \otimes {\mathfrak g}, \end{align*} $$

where $(h_i)\subset {\mathfrak g}$ is an orthonormal basis. Let $V=\otimes _{i=1}^n V_i$ be a tensor product of ${\mathfrak g}$ -modules, $\kappa \in {\mathbb C}^\times $ a nonzero number. The KZ equations are the system of differential equations on a V-valued function $I(z_1,\ldots ,z_n)$ ,

$$ \begin{align*} \frac{\partial I}{\partial z_i}\ =\ \frac 1\kappa\,\sum_{j\ne i}\, \frac{\Omega_{i,j}}{z_i-z_j} I, \quad i=1,\ldots,n, \end{align*} $$

where $\Omega _{i,j}:V\to V$ is the Casimir operator acting in the i th and j th tensor factors; see [Reference Knizhnik and ZamolodchikovKZ, Reference Etingof, Frenkel and KirillovEFK].

This system is a system of Fuchsian first-order linear differential equations. The equations are defined on the complement in ${\mathbb C}^n$ to the union of all diagonal hyperplanes.

The object of our discussion is the following particular case. We consider the following system of differential and algebraic equations for a column $3$ -vector $I=(I_1,I_2,I_3)$ depending on variables $z=(z_1,z_2,z_3)$ :

(6-1) $$ \begin{align} \begin{aligned} \frac{\partial I}{\partial z_1} &= {\frac 12} \bigg( \frac{\Omega_{12}}{z_1 - z_2} +\frac{\Omega_{13}}{z_1 - z_3} \bigg) I , \quad \frac{\partial I}{\partial z_2} = {\frac 12} \bigg( \frac{\Omega_{21}}{z_2 - z_1}+\frac{\Omega_{23}}{z_2 - z_3} \bigg) I ,\quad\ \ \\ \frac{\partial I}{\partial z_3} &= {\frac 12} \bigg( \frac{\Omega_{31}}{z_3 - z_1}+\frac{\Omega_{32}}{z_3 - z_2}\bigg) I, \quad 0= I_1+I_2+I_3, \end{aligned} \end{align} $$

where $\Omega _{ij}=\Omega _{ji}$ and

$$ \begin{align*} \Omega_{12} = \begin{pmatrix} -1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} , \quad \Omega_{13} = \begin{pmatrix} -1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \end{pmatrix} , \quad \Omega_{23} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{pmatrix}. \end{align*} $$

Denote

$$ \begin{align*} &H_1(z) = {\dfrac 12} \bigg( \dfrac{\Omega_{12}}{z_1 - z_2}+\dfrac{\Omega_{13}}{z_1 - z_3} \bigg), \quad H_2(z)={\dfrac 12}\bigg( \dfrac{\Omega_{21}}{z_2 - z_1} +\dfrac{\Omega_{23}}{z_2 - z_3} \bigg), \\ & H_3(z)= {\dfrac 12}\bigg( \dfrac{\Omega_{31}}{z_3 - z_1} +\dfrac{\Omega_{32}}{z_3 - z_2}\bigg), \quad \nabla_i^{\operatorname{KZ}} = \dfrac{\partial}{\partial z_i} - H_i(z), \quad i=1,2,3. \end{align*} $$

Then the KZ equations can be written as the system of equations,

$$ \begin{align*} \nabla_i^{\operatorname{KZ}}I=0, \quad i=1,2,3,\quad I_1+I_2+I_3 =0. \end{align*} $$

System in Equation (6-1) is the system of KZ equations with parameter $\kappa =2$ associated with the Lie algebra ${\mathfrak {sl}}_2$ and the subspace of singular vectors of weight $1$ of the tensor power $({\mathbb C}^2)^{\otimes 3}$ of two-dimensional irreducible ${\mathfrak {sl}}_2$ -modules, up to a gauge transformation; see this example in [Reference VarchenkoV2, Section 1.1].

6.2 Solutions over ${\mathbb C}$

Define the master function

$$ \begin{align*} \Phi(t,z) = (t-z_1)^{-1/2}(t-z_2)^{-1/2}(t-z_3)^{-1/2} \end{align*} $$

and the column 3-vector

(6-2) $$ \begin{align} I^{(C)}(z) = (I_1(z),I_2(z),I_3(z)):= \int_{C} \bigg(\frac {\Phi(t,z)}{t-z_1}, \frac {\Phi(t,z)}{t-z_2}, \frac {\Phi(t,z)}{t-z_3}\bigg)\,dt , \end{align} $$

where $C\subset {\mathbb C}-\{z_1,z_2,z_3\}$ is a contour on which the integrand takes its initial value when t encircles C.

This theorem is a very particular case of the results in [Reference Schechtman and VarchenkoSV1].

Proof. The theorem follows from Stokes’ theorem and the two identities:

(6-3) $$ \begin{align} -\frac 12\, \bigg(\frac {\Phi(t,z)}{t-z_1} + \frac{\Phi(t,z)}{t-z_2} + \frac {\Phi(t,z)}{t-z_3}\bigg)\, =\, \frac{\partial\Phi}{\partial t}(t,z), \end{align} $$

(6-4) $$ \begin{align} \bigg(\frac{\partial }{\partial z_i}-\frac12 \sum_{j\ne i} \frac {\Omega_{i,j}}{z_i-z_j} \bigg) \bigg(\frac {\Phi(t,z)}{t-z_1}, \frac{\Phi(t,z)}{t-z_2},\frac {\Phi(t,z)}{t-z_3}\bigg)\, = \frac{\partial \Psi^i}{\partial t} (t,z), \end{align} $$

where $\Psi ^i(t,z)$ is the column $3$ -vector $(0,\ldots ,0,-{\Phi (t,z)}/{t-z_i},0,\ldots ,0)$ with the nonzero element at the i th place.

6.3 Solutions as vectors of first derivatives

Consider the elliptic integral

$$ \begin{align*} T(z) = T^{(C)}(z) = \int_C \Phi(t,z)\,dt. \end{align*} $$

Then,

$$ \begin{align*} I^{(C)}(z) = \, 2\, \bigg(\frac {\partial T^{(C)}}{\partial z_1}, \frac {\partial T^{(C)}}{\partial z_2}, \frac {\partial T^{(C)}}{\partial z_3}\bigg). \end{align*} $$

Denote $\nabla T = ( {\partial T}/{\partial z_1}, {\partial T}/{\partial z_2}, {\partial T}/{\partial z_3})$ . Then the column gradient vector of the function $T(z)$ satisfies the following system of (KZ) equations:

$$ \begin{align*} \nabla_i^{\operatorname{KZ}} \nabla T =0, \quad i=1,2,3,\quad \frac {\partial T}{\partial z_1} + \frac {\partial T}{\partial z_2}+ \frac {\partial T}{\partial z_3}=0. \end{align*} $$

This is a system of second-order linear differential equations on the function $T(z)$ .

6.4 Solutions modulo $p^s$

For an integer $s\geqslant 1$ , define the master polynomial

$$ \begin{align*} \Phi_s(t,z) = ((t-z_1)(t-z_2)(t-z_3))^{(p^s-1)/2}. \end{align*} $$

Define the column 3-vector

$$ \begin{align*} I_s(z)=(I_{s,1}(z), I_{s,2}(z), I_{s,3}(z)) \end{align*} $$

as the coefficient of $t^{p^s-1}$ in the polynomial

$$ \begin{align*} \bigg(\frac {\Phi_s(t,z)}{t-z_1}, \frac {\Phi_s(t,z)}{t-z_2}, \frac {\Phi_s(t,z)}{t-z_3}\bigg). \end{align*} $$

Proof. We have the following modifications of the identities in Equations (6-3), (6-4):

(6-5) $$ \begin{align}\begin{aligned} &\qquad\quad\frac {p^s-1}2\, \bigg(\frac {\Phi_s(t,z)}{t-z_1} + \frac{\Phi_s(t,z)}{t-z_2} + \frac {\Phi_s(t,z)}{t-z_3}\bigg)\, =\, \frac{\partial\Phi_s}{\partial t}(t,z), \\&\ \bigg(\frac{\partial }{\partial z_i} + \frac {p^s-1}2 \sum_{j\ne i} \frac {\Omega_{i,j}}{z_i-z_j} \bigg) \bigg(\frac {\Phi_s(t,z)}{t-z_1}, \frac{\Phi_s(t,z)}{t-z_2},\frac {\Phi_s(t,z)}{t-z_3}\bigg)\, = \frac{\partial \Psi_s^i}{\partial t} (t,z),\end{aligned}\end{align} $$

where $\Psi _s^i(t,z)$ is the column $3$ -vector $(0,\ldots ,0,-{\Phi _s(t,z)}/({t-z_i}),0,\ldots ,0)$ with the nonzero element at the i th place. Theorem 6.3 follows from these identities.

6.5 $p^s$ -Approximation polynomials of $T(z)$

Define the $p^s$ -approximation polynomial $T_s(z)$ of the elliptic integral $T(z)$ as the coefficient of $t^{p^s-1}$ in the master polynomial $\Phi _s(t,z)$ ,

(6-6) $$ \begin{align} T_s(z) = (-1)^{(p^s-1)/2}\sum_{k_1+k_2+k_3 = (p^s-1)/2} \binom{(p^s-1)/2}{k_1}\binom{(p^s-1)/2}{k_2}\binom{(p^s-1)/2}{k_3} z_1^{k_1}z_2^{k_2}z_3^{k_3}. \end{align} $$

We put $T_0(x)=1$ .

The polynomial $T_s(z_1,z_2,z_3)$ is symmetric with respect to permutations of $z_1,z_2,z_3$ and

$$ \begin{align*} T_s(1,z_2,0) = P_s(z_2), \end{align*} $$

where $P_s(x)$ is defined in Equation (4-3). The gradient vector

$$ \begin{align*}\nabla T_s :=({\partial T_s}/{\partial z_1}, {\partial T_s}/{\partial z_2}, {\partial T_s}/{\partial z_3}) \end{align*} $$

of the $p^s$ -approximation polynomial $T_s(z)$ is a solution modulo $p^s$ of the system in Equation (6-1) since

$$ \begin{align*} \nabla T_s = \frac{1-p^s}2\,(I_{s,1}(z), I_{s,2}(z), I_{s,3}(z)). \end{align*} $$

Lemma 6.4. For $s\geqslant 1$ , the master polynomials satisfy the baby congruences,

$$ \begin{align*} \Phi_{s+1}(t,z) \Phi_{s-1}(t^p,z_1^p,z_2^p,z_3^p) \equiv \Phi_{s}(t,z) \Phi_{s}(t^p,z_1^p,z_2^p,z_3^p) \,(\mathrm{mod}\,{p^s}). \end{align*} $$

Theorem 6.5. For $s\geqslant 1$ ,

(6-7) $$ \begin{align} T_{s+1}(z_1,z_2,z_3) \, T_{s-1}(z_1^p,z_2^p,z_3^p) \,\equiv\, T_{s}(z_1,z_2,z_3)\, T_{s}(z_1^p,z_2^p,z_3^p) \,(\mathrm{mod}\,{p^s}). \end{align} $$

Proof. Let $h(t,z) = t^{1-p} ((t-z_1)(t-z_2)(t-z_3))^{(p-1)/2}$ . Then,

$$ \begin{align*} T_s(z) = \operatorname{CT}_t[h(t,z) h(t,z)^p\cdots h(t,z)^{p^{s-1}}]. \end{align*} $$

The tuple of Laurent polynomials $(h(t,z),h(t,z), \ldots )$ is admissible in the sense of Definition 2.6. Now the application of Theorem 2.9 gives the congruence in Equation (6-7).

6.6 Limits of $T_s(z)$

Denote $\bar T_s(z) := (-1)^{(p^s-1)/2} T_s(z)$ ; see Equation (6-6),

(6-8) $$ \begin{align} \mathfrak D = \{(z_1,z_2,z_3)\in {\mathbb Z}_p^3 \,\|\bar T_1(z_1,z_2,z_3)|_p=1\}. \end{align} $$

Notice that ${\mathbb Z}_p^3 = \bigcup _{\alpha ,\beta ,\gamma \in {\mathbb F}_p} D_{\alpha ,1}\times D_{\beta ,1}\times D_{\gamma ,1}$ . Since $\bar T_1(z)$ has coefficients in ${\mathbb Z}_p$ ,

$$ \begin{align*} \mathfrak D ={\bigcup}^o D_{\alpha,1}\times D_{\beta,1}\times D_{\gamma,1}, \end{align*} $$

where the summation ${\bigcup }^o$ is over all $\alpha ,\beta ,\gamma \in {\mathbb F}_p$ such that $|T(\omega (\alpha ),\omega (\beta ),\omega (\gamma ))|_p=1$ . For the same reason,

$$ \begin{align*} \mathfrak D = \{(z_1,z_2,z_3)\in {\mathbb Z}_p^3 \,\mid |\bar T_1(z_1^p,z_2^p,z_3^p)|_p=1\}. \end{align*} $$

Denote

$$ \begin{align*} \mathfrak E =\{ (1,z_2,0)\in {\mathbb Z}_p^3\ |\ |z_2|_p<1 \}. \end{align*} $$

Proof. Similarly to the proof of Theorem 4.6, we have $\bar T_s(z) \equiv \bar T_1(z) \bar T_1(z^p)\cdots \bar T_1(x^{p^{s-1}}) (\mathrm { mod}\,{p})$ . Hence, $|\bar T_s(z)|_p = |\bar T_s(z^p)|_p=1$ for $s\geqslant 1$ , $z\in \mathfrak D$ . Hence, the rational functions ${\bar T_{s+1}(z)}/{\bar T_s(z^p)}$ are regular on $\mathfrak D$ .

The congruence in Equation (6-7) implies that

$$ \begin{align*} \bigg|\frac{\bar T_{s+1}(z)}{\bar T_{s}(z^p)} - \frac{\bar T_{s}(z)}{\bar T_{s-1}(z^p)}\bigg|_p\leqslant p^{-s} \quad\operatorname{for}\ z\in\mathfrak D. \end{align*} $$

This shows the uniform convergence of $({\bar T_{s+1}(z)}/{\bar T_s(z^p)})_{s\geqslant 1}$ on $\mathfrak D$ . For the limiting function $f(z)$ , we have $|f(z)|_p=1$ for $z\in \mathfrak D$ .

We have $\bar T_s(1,z_2,0) = \bar P_s(z_2) = \sum _{k} \binom {(p^s-1)/2}{k}^2z_2^k$ . Clearly, for any fixed index k, the coefficient $\binom {(p^s-1)/2}{k}^2$ of $z_2^k$ in $\bar T_s(1,z_2,0)$ converges p-adically to the coefficient $\binom {-1/2}{k}^2$ of $z_2^k$ in $F(z_2)$ . Hence, the sequence $(\bar T_s(1,z_2,0))_{s\geqslant 1}$ converges to $F(z_2)$ on $\mathfrak E$ , so that $f(1,z_2,0) ={F(z_2)}/{F(z^p)}$ on $\mathfrak E$ . The theorem is proved.

For $i,j\in \{1,2,3\}$ and $s\geqslant 1$ , denote

$$ \begin{align*} f_s(z)=T_{s}(z)/T_{s-1}(z^p), \quad \eta^{(i)}_s(z) = \frac{\partial T_s}{\partial z_i}(z)/T_s(z), \quad \eta^{(ij)}_s(z) = \frac{\partial^2 T_s}{\partial z_i\partial z_j}(z)/T_s(z). \end{align*} $$

Theorem 6.9. For $s\geqslant 1$ , the rational functions $\eta ^{(i)}_s(z)$ and $\eta ^{(ij)}_s(z)$ are regular on $\mathfrak D$ . The sequences of rational functions $(\eta ^{(i)}_s(z))_{s\geqslant 1}$ and $(\eta ^{(ij)}_s(z))_{\geqslant 1}$ converge uniformly on $\mathfrak D$ to analytic functions. If $\eta ^{(j)}$ and $\eta ^{(ij)}$ denote the corresponding limits, then

(6-9) $$ \begin{align} &\eta^{(1)} + \eta^{(2)} + \eta^{(3)}=0, \qquad\qquad\quad\ \ \,\end{align} $$

(6-10) $$ \begin{align} &\kern1pt\eta^{(j1)} + \eta^{(j2)} + \eta^{(j3)}=0, \quad j=1,2,3, \end{align} $$

(6-11) $$ \begin{align} &\frac{\partial}{\partial z_j}\eta^{(i)}=\eta^{(ji)} - \eta^{(i)}\eta^{(j)}.\ \,\ \ \,\qquad\qquad\end{align} $$

Proof. Denote $\delta _i=z_i({\partial }/{\partial z_i})$ . By Theorem 6.7, the sequence $(f_s)$ uniformly converges to the analytic function f on $\mathfrak D$ . Therefore, the sequence of the derivatives $(({\partial }/{\partial z_i})f_s)$ uniformly converges on $\mathfrak D$ to $({\partial }/{\partial z_i})f$ . Hence, the sequence $((\delta _if_s)/f_s)$ uniformly converges on $\mathfrak D$ to the function $(\delta _if)/f$ . At the same time,

$$ \begin{align*} \frac{\delta_if_s}{f_s}(z) =\frac{\delta_i T_s}{T_s}(z)- p\, \frac{\delta_i T_{s-1}}{T_{s-1}}(z^p) \end{align*} $$

and, more generally,

$$ \begin{align*} \frac{\delta_if_{s-k}}{f_{s-k}}(z^{p^k}) =\frac{\delta_iT_{s-k}}{T_{s-k}}(z^{p^k}) -p\,\frac{\delta_iT_{s-k-1}}{T_{s-k-1}}(z^{p^{k+1}}) \quad\text{for } k=0,1,\ldots,s. \end{align*} $$

Summing the relations up with suitable weights to get telescoping, we obtain, for any $r\leqslant s$ ,

$$ \begin{align*} \sum_{k=0}^{r-1}\,p^k\, \frac{\delta_if_{s-k}}{f_{s-k}}(z^{p^k}) \,=\, \frac{\delta_iT_s}{T_s}(z)\,-\,p^r\, \frac{\delta_i T_{s-r}}{T_{s-r}}(z^{p^r}). \end{align*} $$

Choosing $r=[s/2]$ and taking the limit as $s\to \infty $ on both sides,

$$ \begin{align*} \sum_{k=0}^\infty \,p^k\,\frac{\delta_i f}{f}(z^{p^k}) =\lim_{s\to\infty} \frac{\delta_iT_s}{T_s}(z). \end{align*} $$

The series on the left uniformly converges on $\mathfrak D$ . Hence, there exists the limit on the right-hand side. This means that

$$ \begin{align*} \eta^{(i)}(x) =\lim_{s\to\infty} \frac{\frac{\partial}{\partial z_i}T_s}{T_s}(z) =\frac1{z_i}\sum_{k=0}^\infty \,p^k\,\frac{\delta_i f}{f}(z^{p^k}). \end{align*} $$

One can further differentiate the resulting equality with respect to any of the variables $z_1,z_2,z_3$ to get, by induction, formulas for $\eta ^{(ij)}$ and more generally for $\eta ^{(ijk\ldots )}$ . Note that Equation (6-11) comes out from differentiating logarithmic derivatives.

Equations (6-9) and (6-10) follow from Equation (6-5). The theorem is proved.

Theorem 6.10. We have the following system of equations on $\mathfrak D$ :

(6-12) $$ \begin{align} \begin{aligned} &\hspace{-12pt} \begin{pmatrix} \eta^{(11)} \\ \eta^{(12)} \\ \eta^{(13)} \end{pmatrix} = {\dfrac 12} \bigg( \dfrac{\Omega_{12}}{z_1 - z_2}+\dfrac{\Omega_{13}}{z_1 - z_3} \bigg) \begin{pmatrix} \eta^{(1)} \\ \eta^{(2)} \\ \eta^{(3)} \end{pmatrix}, \quad \begin{pmatrix} \eta^{(21)} \\ \eta^{(22)} \\ \eta^{(23)} \end{pmatrix} = {\dfrac 12} \bigg( \dfrac{\Omega_{21}}{z_2 - z_1}+\dfrac{\Omega_{23}}{z_2 - z_3} \bigg) \begin{pmatrix} \eta^{(1)} \\ \eta^{(2)} \\ \eta^{(3)} \end{pmatrix},\\ &\begin{pmatrix} \eta^{(31)} \\ \eta^{(32)} \\ \eta^{(33)} \end{pmatrix}= {\dfrac 12} \bigg( \dfrac{\Omega_{31}}{z_3 - z_1}+\dfrac{\Omega_{33}}{z_3 - z_2} \bigg) \begin{pmatrix} \eta^{(1)} \\ \eta^{(2)} \\ \eta^{(3)} \end{pmatrix}, \quad \eta^{(1)}+\eta^{(2)}+ \eta^{(3)} =0. \end{aligned} \end{align} $$

Theorem 6.11. The column vector

$$ \begin{align*} \vec \eta(z):=(\eta^{(1)}(z), \eta^{(2)}(z), \eta^{(3)}(z)) \end{align*} $$

is nonzero at every point $z\in \mathfrak D$ .

Proof. On the one hand, if $\vec \eta (a) =0$ for some $a\in \mathfrak D$ , then all derivatives of $\vec \eta (z)$ at a are equal to zero. This follows from the first three equations in Equation (6-12) written as

(6-13) $$ \begin{align} \frac\partial{\partial z_i} \vec \eta = (H_i -\eta^{(i)})\,\vec\eta,\quad i=1,2,3. \end{align} $$

Hence, $\vec \eta (z)$ equals zero identically on $\mathfrak D$ . On the other hand, $\eta ^{(2)}(1,0,0) = F'(0)/F(0) = 1/4$ by Theorem 6.7. This contradiction implies the theorem.

6.7 Subbundle $\mathcal L \,\to \, \mathfrak D$

Denote $W=\{(I_1,I_2,I_3)\in {\mathbb Q}_p^3\ |\ I_1+I_2+I_3=0\}$ . The differential operators $\nabla ^{\operatorname {KZ}}_i$ , $i=1,2,3$ , define a connection on the trivial bundle $W\times \mathfrak D \to \mathfrak D$ , called the KZ connection. The KZ connection is flat,

$$ \begin{align*} [\nabla^{\operatorname{KZ}}_i, \nabla^{\operatorname{KZ}}_j]=0 \quad \text{for all }\,i,j. \end{align*} $$

The flat sections of the KZ connection are solutions of the system in Equation (6-1) of KZ equations.

For any $a\in \mathfrak D$ , let $\mathcal L_a \subset W$ be the one-dimensional vector subspace generated by $\vec \eta (a)$ . Then,

$$ \begin{align*} \mathcal L := \bigcup_{a\in \mathfrak D}\,\mathcal L_a \,\to\, \mathfrak D \end{align*} $$

is an analytic line subbundle of the trivial bundle $W\times \mathfrak D \to \mathfrak D$ .

6.8 Other definitions of subbundle $\mathcal L \to \mathfrak D$

6.8.1 Line subbundle $\mathcal M \to \mathfrak D$

Define a polynomial $U_s(z)$ as the coefficient of $t^{p^s-1}$ in the master polynomial $\Phi _s(t+z_3,z) = ((t-(z_1-z_3))(t-(z_2-z_3))t)^{(p^s-1)/2}$ . It is easy to see that $U_1(z)=T_1(z)\ \,(\mathrm {mod}\,{p})$ . Similarly to Theorem 6.5, we conclude that

$$ \begin{align*} \phantom{aaa} U_{s+1}(z_1,z_2,z_3)\, U_{s-1}(z_1^p,z_2^p,z_3^p) \,\equiv\, U_{s}(z_1,z_2,z_3)\, U_{s}(z_1^p,z_2^p,z_3^p) \,(\mathrm{mod}\,{p^s}). \end{align*} $$

Hence, the sequence $({U_{s+1}(z)}/{U_s(z^p)})_{s\geqslant 1}$ uniformly converges to an analytic function on the domain $\mathfrak D$ defined in Equation (6-8). The vector-valued polynomial $\nabla U_s(z) =({\partial U_s}/{\partial z_1}, {\partial U_s}/{\partial z_2}, {\partial U_s}/{\partial z_3} )$ is a solution modulo $p^s$ of the KZ equations in Equation (6-1); see [Reference VarchenkoV4, Theorem 9.1], and see the proof of Theorem 6.3. Consider the function

$$ \begin{align*} \vec{\mu}=(\mu^{(1)}, \mu^{(2)},\mu^{(3)}):=\lim_{s\to\infty} \frac{\nabla U_s}{U_s} \end{align*} $$

defined on the same domain $\mathfrak D$ . Similarly to the proofs of Theorems 6.9–6.12, we conclude that the function $\vec {\mu }(z)$ is nonzero on $\mathfrak D$ and its values span an analytic line subbundle

$$ \begin{align*} \mathcal M := \bigcup_{a\in \mathfrak D}\,\mathcal M_a \,\to\, \mathfrak D \end{align*} $$

of the trivial bundle $W\times \mathfrak D\to \mathfrak D$ ; here, $\mathcal M_a\subset W$ is the one-dimensional subspace generated by $\vec \mu (a)$ . The line subbundle $\mathcal M \,\to \, \mathfrak D$ is invariant with respect to the KZ connection.

Theorem 6.17. The line bundles $\mathcal M \,\to \, \mathfrak D$ and $\mathcal L \,\to \, \mathfrak D$ coincide.

Proof. The proof rests on the following two lemmas.

Lemma 6.18. The line bundles $\mathcal M \,\to \, \mathfrak D$ and $\mathcal L \,\to \, \mathfrak D$ coincide if there is $a\in \mathfrak D$ such that $\mathcal M_a=\mathcal L_a$ .

Proof. Let $\mathcal M_a=\mathcal L_a$ for some $a\in \mathfrak D$ . Then, $\mathcal M_z = \mathcal L_s$ in some neighborhood of a, since locally the subbundles are generated by the values of the solutions with the same initial condition at $z=a$ . Hence, $\mathcal M_z = \mathcal L_z$ on $\mathfrak D$ .

Lemma 6.19. For $i=1,2,3$ , the functions ${\partial T_s}/{\partial z_i}(z)/ T_s(z)$ and ${\partial U_s}/{\partial z_i}(z)/ U_s(z)$ are equal on the line $z_1=1$ , $z_3=0$ .

Hence, $\mathcal M=\mathcal L$ over the points of that line and, therefore, $\mathcal M_z = \mathcal L_z$ for $z\in \mathfrak D$ .

6.8.2 Line subbundle $\mathcal N\to \hat {\mathfrak D}$

Let $\omega (x) =F'(x)/F(x)$ , where $F(x)\ \mathrm{is\ defined\ as} {}_2F_1(1/2, 1/2; 1;x)$ . We have $\omega (x) = \lim _{s\to \infty }({P_s'(x)}/{P_s(x)})$ on $D_{0,1}$ . Introduce new variables

$$ \begin{align*} u_1=z_1-z_3, \quad u_2=\frac{z_2-z_3}{z_1-z_3}, \quad u_3=z_1+z_2+z_3, \end{align*} $$

and a vector-valued function

$$ \begin{align*} \vec\omega(u) = \frac1{u_1} (\!- 1/2 - \omega(u_2) u_2,\, \omega(u_2),\, 1/2 + \omega(u_2)(u_2-1)). \end{align*} $$

Define

$$ \begin{align*} \hat{\mathfrak D}_0 = \{(z_1,z_2,z_3)\in {\mathbb Q}_p^3\ |\ z_i\ne z_j\quad \text{for all } i\ne j\}. \end{align*} $$

For any $\sigma =(i,j,k)\in S_3$ , define

$$ \begin{align*} &\phantom{aaaaa} \hat {\mathfrak D}_1^\sigma =\bigg\{(z_1,z_2,z_3)\in \hat {\mathfrak D}_0\ \bigg|\ \ \frac{z_j-z_k}{z_i-z_k}\in {\mathbb Z}_p,\ \bigg| g\bigg(\frac{z_j-z_k}{z_i-z_k}\bigg)\bigg|_p = 1\bigg\}, \\ &\hat{\mathfrak D}_2^\sigma = \bigg\{(z_1,z_2,z_3)\in \hat {\mathfrak D}_0\ \bigg|\ \frac{z_i-z_k}{z_j-z_k} \in \hat {\mathfrak D}_1^\sigma\bigg\}, \quad \hat{\mathfrak D}^\sigma= \hat{\mathfrak D}_1^\sigma\cup\hat {\mathfrak D}_2^\sigma, \quad \hat{\mathfrak D} = \sum_{\sigma\in S_3}\hat{\mathfrak D}^\sigma, \end{align*} $$

where $g(\lambda )$ is the Igusa polynomial in Equation (4-13).

Using Dwork’s results in [Reference DworkDw], it is shown in [Reference VarchenkoV4, Appendix] that the values of the analytic continuation of the function $\vec \omega (u)$ generate a line bundle $\mathcal N\to \hat {\mathfrak D}$ invariant with respect to the KZ connection.

Theorem 6.20. The line bundles $\mathcal M\to \mathfrak D$ and $\mathcal N \to \hat {\mathfrak D}$ coincide on $\mathfrak D\cap \hat {\mathfrak D}$ .

Thus, we identify the line bundles $\mathcal L\to \mathfrak D$ , $\mathcal M\to \mathfrak D$ , and $\mathcal N \to \hat {\mathfrak D}$ over $\mathfrak D\cap \hat {\mathfrak D}$ .

Proof. We have

$$ \begin{align*} U_s(z) &= (z_1-z_3)^{(p^s-1)/2} P_s\bigg(\dfrac{z_2-z_3}{z_1-z_3}\bigg), \quad \dfrac{\partial U_s}{\partial z_2} = (z_1-z_3)^{(p^s-1)/2-1} P_s'\bigg(\dfrac{z_2-z_3}{z_1-z_3}\bigg) \\ \dfrac{\partial U_s}{\partial z_1} &= \dfrac{p^s-1}2\,\operatorname{const}\,(z_1-z_3)^{(p^s-1)/2-1} P_s\bigg(\dfrac{z_2-z_3}{z_1-z_3}\bigg)\\ &\quad- \operatorname{const}\,(z_1-z_3)^{(p^s-1)/2-1} \dfrac{z_2-z_3}{z_1-z_3} P_s'\bigg(\dfrac{z_2-z_3}{z_1-z_3}\bigg). \end{align*} $$

Hence,

$$ \begin{align*} \dfrac 1{U_s(z)}\bigg(\dfrac{\partial U_s}{\partial z_1}, \dfrac{\partial U_s}{\partial z_2}, \dfrac{\partial U_s}{\partial z_3}\bigg) = \dfrac 1{u_1}\bigg(\dfrac{p^s-1}2 - u_2 \dfrac{P_s'(u_2)}{P_s(u_2)},\,\dfrac{P_s'(u_2)}{P_s(u_2)}, \,-\dfrac{p^s-1}2 + (u_2-1) \dfrac{P_s'(u_2)}{P_s(u_2)}\bigg). \end{align*} $$

Clearly, the limit of this vector equals $\vec \omega (u)$ as $s\to \infty $ . The theorem is proved.

7.1 Conjectural stronger congruences for $\overline {P}_s(x)$

By Theorem 4.2, we have for polynomials $\bar P_s(x) := (-1)^{(p^s-1)/2} P_s(x)$ ,

$$ \begin{align*} \bar P_{4}(x) \bar P_{2}(x^p) - \bar P_{3}(x) \bar P_{3}(x^p) \equiv 0\,(\mathrm{mod}\,{p^{3}}). \end{align*} $$

In particular, for the coefficient of $x^{N_0+N_1p+N_2p^2 +N_3p^3}$ in $\bar P_4(x) \bar P_2(x^p) -\bar P_3(x)\bar P_3(x^p)$ ,

$$ \begin{align*} &\sum_{\genfrac{}{}{0pt}1{\scriptstyle k_1+l_1=N_1}{\scriptstyle k_2+l_2=N_2}} \Bigg(\binom{(p^4-1)/2}{N_0+k_1p+k_2p^2+N_3p^3}^2 \binom{(p^2-1)/2}{l_1+l_2p}^2 \notag \\ &\qquad - \binom{(p^3-1)/2}{N_0+k_1p+k_2p^2}^2 \binom{(p^3-1)/2}{l_1+l_2p+N_3p^2}^2\,\Bigg) \equiv 0 \,(\mathrm{mod}\,{p^{3}}). \end{align*} $$

Computer experiments show that this sum can be split into subsums with at most four terms so that each subsum is divisible by $p^3$ . More precisely, let $0\leqslant a, b, \,c,c',d,d'\leqslant p-1$ be integers. Define

$$ \begin{align*} A(a,b;c,c';d,d') = \binom{(p^4-1)/2}{a+cp+dp^2+bp^3}^2 \binom{(p^2-1)/2}{c'+d'p}^2 -\binom{(p^3-1)/2}{a+cp+dp^2}^2 \binom{(p^3-1)/2}{c'+d'p+bp^2}^2 \end{align*} $$

and

(7-1) $$ \begin{align} &B(a,b;c,c';d,d') = \operatorname{Sym} A(a,b;c,c';d,d')\notag\\ &\quad :=A(a,b;c,c';d,d') + A(a,b;c',c;d,d')+A(a,b;c,c';d',d)+A(a,b;c',c;d',d). \end{align} $$

We expect that the integer $B(a,b;c,c';d,d')$ is divisible by $p^3$ .

More generally, define

$$ \begin{align*} k=(k^{(1)}, \ldots,k^{(s)}),\quad k^{(i)}=(k^{(i)}_1, k^{(i)}_2), \end{align*} $$

$$ \begin{align*} A(a,b;k) = \binom{(p^{s+2}-1)/2}{a+\sum_{i=1}^sk^{(i)}_1p^i+bp^{s+1}}^2 \binom{(p^{s}-1)/2}{\sum_{i=1}^sk^{(i)}_2p^{i-1}}^2 -\binom{(p^{s+1}-1)/2}{a+\sum_{i=1}^sk^{(i)}_1p^i}^2 \binom{(p^{s+1}-1)/2}{\sum_{i=1}^sk^{(i)}_2p^{i-1}+bp^{s}}^2. \end{align*} $$

Set

$$ \begin{align*} B(a,b;k) = \operatorname{Sym} A(a,b;k), \end{align*} $$

where $\operatorname {Sym}$ denotes the symmetrization with respect to the index j in $k^{(i)}_j$ in each group $k^{(i)}=(k^{(i)}_1,k^{(i)}_2)$ . Thus, the symmetrization has $2^s$ summands; the case $s=2$ of this symmetrization is displayed in Equation (7-1).

This conjecture is supported by computer experiments and is checked for $s=1$ using [Reference GranvilleGr].

7.2 Papers [Reference Beukers and VlasenkoBV, Reference VlasenkoVl]

After reading this paper, Masha Vlasenko was able to invent a new proof of our congruences in Theorems 4.2 and 5.1 (private communication). Her proof was based on the results of [Reference Beukers and VlasenkoBV, Reference VlasenkoVl].