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On the Newton polygons of twisted LL-functions of binomials

Shenxing Zhang School of Mathematics, Hefei University of Technology, Hefei, Anhui 230009, China [email protected]
Abstract.

Let χ\chi be an order cc multiplicative character of a finite field and f(x)=xd+λxef(x)=x^{d}+\lambda x^{e} a binomial with (d,e)=1(d,e)=1. We study the twisted classical and TT-adic Newton polygons of ff. When p>(de)(2d1)p>(d-e)(2d-1), we give a lower bound of Newton polygons and show that they coincide if pp does not divide a certain integral constant depending on pmodcdp\bmod{cd}.

We conjecture that this condition holds if pp is large enough with respect to c,dc,d by combining all known results and the conjecture given by Zhang-Niu. As an example, we show that it holds for e=d1e=d-1.

Key words and phrases:
Newton polygons; exponential sums; L-function
2020 Mathematics Subject Classification:
11L07

1. Introduction

1.1. Background

Fix a rational prime pp. For q=paq=p^{a} a power of pp, denote by 𝔽q{\mathbb{F}}_{q} the finite field with qq elements, q{\mathbb{Q}}_{q} the unramified extension of p{\mathbb{Q}}_{p} of degree aa and q{\mathbb{Z}}_{q} its ring of integers. Let f(x)𝔽q[x]f(x)\in{\mathbb{F}}_{q}[x] be a polynomial of degree dd with Teichmüller lifting f^(x)q[x]\hat{f}(x)\in{\mathbb{Z}}_{q}[x]. Let χ:𝔽q×p×\chi:{\mathbb{F}}_{q}^{\times}\to{\mathbb{C}}_{p}^{\times} be a multiplicative character and ω:𝔽q×q×\omega:{\mathbb{F}}_{q}^{\times}\to{\mathbb{Z}}_{q}^{\times} the Teichmüller lifting. Then we can write χ=ωu\chi=\omega^{-u} for some 0uq20\leq u\leq q-2.

For a non-trivial additive character ψm:pp×\psi_{m}:{\mathbb{Z}}_{p}\to{\mathbb{C}}_{p}^{\times} of order pmp^{m}, define the twisted LL-function

Lu(s,f,ψm)=exp(k=1Sk,u(f,ψm)smm),L_{u}(s,f,\psi_{m})=\exp\left(\sum_{k=1}^{\infty}S_{k,u}(f,\psi_{m})\frac{s^{m}}{m}\right),

where Sk,u(f,ψm)S_{k,u}(f,\psi_{m}) is the twisted exponential sum

Sk,u(f,ψm)=x𝔽qk×ψm(Trqk/p(f^(x^)))ωu(Nm𝔽qk/𝔽q(x)).S_{k,u}(f,\psi_{m})=\sum_{x\in{\mathbb{F}}_{q^{k}}^{\times}}\psi_{m}\left({\mathrm{Tr}}_{{\mathbb{Q}}_{q^{k}}/{\mathbb{Q}}_{p}}\bigl{(}\hat{f}(\hat{x})\bigr{)}\right)\omega^{-u}\left({\mathrm{Nm}}_{{\mathbb{F}}_{q^{k}}/{\mathbb{F}}_{q}}(x)\right).

If pdp\nmid d, then Lu(s,f,ψm)L_{u}(s,f,\psi_{m}) is a polynomial of degree pm1dp^{m-1}d by Adolphson-Sperber [AS87, AS91, AS93], Li [Li99], Liu-Wei [LW07] and Liu [Liu07].

We will use the twisted TT-adic exponential sums developed by Liu-Wan [LW09] and Liu [Liu02, Liu09]. Define the twisted TT-adic LL-function

Lu(s,f,T)=exp(k=1Sk,u(f,T)skk)1+sqTs,L_{u}(s,f,T)=\exp\left(\sum_{k=1}^{\infty}S_{k,u}(f,T)\frac{s^{k}}{k}\right)\in 1+s{\mathbb{Z}}_{q}\llbracket T\rrbracket\llbracket s\rrbracket,

where Sk,u(f,T)S_{k,u}(f,T) is the twisted TT-adic exponential sum

Sk,u(f,T)=x𝔽qk×(1+T)Trqk/p(f^(x^))ωu(Nm𝔽qk/𝔽q(x)).S_{k,u}(f,T)=\sum_{x\in{\mathbb{F}}_{q^{k}}^{\times}}(1+T)^{{\mathrm{Tr}}_{{\mathbb{Q}}_{q^{k}}/{\mathbb{Q}}_{p}}(\hat{f}(\hat{x}))}\omega^{-u}\left({\mathrm{Nm}}_{{\mathbb{F}}_{q^{k}}/{\mathbb{F}}_{q}}(x)\right).

Then Lu(s,f,ψm)=Lu(s,f,πm)L_{u}(s,f,\psi_{m})=L_{u}(s,f,\pi_{m}) where πm=ψm(1)1\pi_{m}=\psi_{m}(1)-1.

Denote by

Cu(s,f,T)=j=0Lu(qjs,f,T)1+sqTsC_{u}(s,f,T)=\prod_{j=0}^{\infty}L_{u}(q^{j}s,f,T)\in 1+s{\mathbb{Z}}_{q}\llbracket T\rrbracket\llbracket s\rrbracket

the characteristic function, which is TT-adic entire in ss. Then

Lu(s,f,T)=Cu(s,f,T)Cu(qs,f,T)1.L_{u}(s,f,T)=C_{u}(s,f,T)C_{u}(qs,f,T)^{-1}.

Since the πma(p1)\pi_{m}^{a(p-1)}-adic Newton polygon of Cu(s,f,πm)C_{u}(s,f,\pi_{m}) does not depend on the choice of ψm\psi_{m}, we denote it by NPu,m(f){\mathrm{NP}}_{u,m}(f). Denote by NPu,T(f){\mathrm{NP}}_{u,T}(f) the Ta(p1)T^{a(p-1)}-adic Newton polygon of Cu(s,f,T)C_{u}(s,f,T). As shown in [LW09] and [Liu07], NPu,m(f){\mathrm{NP}}_{u,m}(f) lies over the infinity uu-twisted Hodge polygon H[0,d],uH_{[0,d],u}^{\infty}, which has slopes

nd+1bd(p1)k=1buk,n.\frac{n}{d}+\frac{1}{bd(p-1)}\sum_{k=1}^{b}u_{k},\ n\in{\mathbb{N}}. (1.1)

If we write 0s0spm1d110\leq s_{0}\leq\cdots\leq s_{p^{m-1}d-1}\leq 1 the qq-adic slopes of Lu(s,f,πm)L_{u}(s,f,\pi_{m}), then the qq-adic slopes of Cu(s,f,πm)C_{u}(s,f,\pi_{m}) are

j+si,0ipm1d1,j.j+s_{i},\quad 0\leq i\leq p^{m-1}d-1,j\in{\mathbb{N}}.

That’s to say, the πma(p1)\pi_{m}^{a(p-1)}-adic Newton polygon of Lu(s,f,πm)L_{u}(s,f,\pi_{m}) is the restriction of NPu,m(f){\mathrm{NP}}_{u,m}(f) on [0,pm1d][0,p^{m-1}d], and it determines NPu,m(f){\mathrm{NP}}_{u,m}(f).

The prime pp is required large enough in the following results. When χ=ωu\chi=\omega^{-u} is trivial, in [Zhu14] and [LLN09], they gave a lower bound of the Newton polygons. They defined a polynomial on the coefficients of ff, called Hasse polynomial. If the Hasse polynomial is nonzero, then the Newton polygons coincide this lower bound.

Assume that f(x)=xd+λxef(x)=x^{d}+\lambda x^{e} is a binomial. Since the exponential sums can be transformed to the twisted case when dd and ee are not coprime, we assume (d,e)=1(d,e)=1 in this paper. When u=0u=0, we list the known cases here.

  • p1moddp\equiv 1\bmod d, it’s well-known that the Newton polygons coincides the Hodge polygon.

  • e=1e=1, see [Yan03, §1, Theorem], [Zhu14, Theorem 1.1] and [OY16, Theorem 1.1].

  • e=d1,p1modde=d-1,p\equiv-1\bmod d, see [OZ16].

  • e=2,p2modde=2,p\equiv 2\bmod d, see [ZN21].

For arbitrary uu, Liu-Niu [LN11] obtained the Newton polygons when e=1e=1. Zhang-Niu [ZN21] also give a conjectural description of the Newton polygons when pemoddp\equiv e\bmod d.

1.2. Notations

We list the notations we will use.

  • i,j,v,w,k,,ni,j,v,w,k,\ell,n indices.

  • f(x)=xd+λxe𝔽q[x]f(x)=x^{d}+\lambda x^{e}\in{\mathbb{F}}_{q}[x] a binomial with d>e1,(d,e)=1,λ0d>e\geq 1,(d,e)=1,\lambda\neq 0.

  • ωu:𝔽q×p×\omega^{-u}:{\mathbb{F}}_{q}^{\times}\to{\mathbb{C}}_{p}^{\times}, where ω\omega is the Teichml̈ler lifting and 0uq20\leq u\leq q-2.

  • H[0,d],uH_{[0,d],u}^{\infty}, the infinity uu-twisted Hodge polygon with slopes in (1.1).

  • c=q1(q1,u)c=\frac{q-1}{(q-1,u)} the order of ωu\omega^{-u}, then u=(q1)μcu=\frac{(q-1)\mu}{c} for some (μ,c)=1(\mu,c)=1.

  • Pu,e,dP_{u,e,d} a polygon with slopes w(i)w(i), defined in (1.2).

  • bb the least positive integer such that pbuumod(q1)p^{b}u\equiv u\bmod{(q-1)} (equivalently, pb1modcp^{b}\equiv 1\bmod c).

  • 0uip10\leq u_{i}\leq p-1 such that u=u0+u1p++ua1pa1u=u_{0}+u_{1}p+\cdots+u_{a-1}p^{a-1}, ui=ub+iu_{i}=u_{b+i}.

  • x¯\overline{x} the minimal non-negative residue of xx modulo dd.

  • δP\delta_{P} takes value 11 if PP happens; 0 if PP does not happen.

  • In={1,,n},In={0,1,,n}I_{n}=\left\{1,\dots,n\right\},I_{n}^{*}=\left\{0,1,\dots,n\right\}.

  • SnS_{n} (resp. SnS_{n}^{*}) the set of permutations of InI_{n} (resp. InI_{n}^{*}).

  • Ct,nC_{t,n} the minimum of i=0ne1(piτ(i)+t)¯\sum_{i=0}^{n}\overline{e^{-1}(pi-\tau(i)+t)} for τSn\tau\in S_{n}^{*} and St,nS_{t,n}^{\circ} the set of τSn\tau\in S_{n}^{*} such that the summation reaches minimal. Set Ct,1=0C_{t,-1}=0 for convention.

  • Ri,α=e1(pi+α)¯,ri,α=e1(tαi)¯R_{i,\alpha}=\overline{e^{-1}(pi+\alpha)},\ r_{i,\alpha}=\overline{e^{-1}(t-\alpha-i)}, see Proposition 2.1. We will the subscript α\alpha if there is no confusion.

  • 𝐂t,n,α{\mathbf{C}}_{t,n,\alpha} the maximal size of {iInRi,α+rτ(i),αd}\left\{i\in I_{n}^{*}\mid R_{i,\alpha}+r_{\tau(i),\alpha}\geq d\right\} for τSn\tau\in S_{n}^{*}. We will the subscript α\alpha if there is no confusion.

  • yt,iτ=e1(piτ(i)+t)¯,xt,iτ=d1(piτ(i)+teyt,iτ)y_{t,i}^{\tau}=\overline{e^{-1}(pi-\tau(i)+t)},\ x_{t,i}^{\tau}=d^{-1}(pi-\tau(i)+t-ey_{t,i}^{\tau}) the unique solution of dx+ey=piτ(i)+tdx+ey=pi-\tau(i)+t with 0yd10\leq y\leq d-1.

  • hn,k,hu,e,dh_{n,k},h_{u,e,d} the Hasse numbers defined in (1.3).

  • 𝐩{\mathbf{p}} the minimal non-negative residue of pp modulo cdcd.

  • Hμ,c,𝐩,e,dH_{\mu,c,{\mathbf{p}},e,d}\in{\mathbb{Z}} a constant defined in (3.1).

  • E(X)E(X) the pp-adic Artin-Hasse series, see (2.1).

  • π\pi a TT-adic uniformizer of pT{\mathbb{Q}}_{p}\llbracket T\rrbracket given by E(π)=1+TE(\pi)=1+T, with a fixed d(q1)d(q-1)-th root π1d(q1)\pi^{\frac{1}{d(q-1)}}.

  • Ef(X)E_{f}(X), see (2.2).

  • Mu=uq1+M_{u}=\frac{u}{q-1}+{\mathbb{N}}.

  • u{\mathcal{L}}_{u} a Banach space, see (2.3).

  • u{\mathcal{B}}_{u} a subspace of u{\mathcal{L}}_{u}, see (2.4).

  • =upupb1u{\mathcal{B}}={\mathcal{B}}_{u}\oplus{\mathcal{B}}_{pu}\oplus\cdots\oplus{\mathcal{B}}_{p^{b-1}u}.

  • ψ:up1u\psi:{\mathcal{L}}_{u}\rightarrow{\mathcal{L}}_{p^{-1}u} defined as ψ(vMubvXv)=vMp1ubpvXv\psi\left(\sum_{v\in M_{u}}b_{v}X^{v}\right)=\sum_{v\in M_{p^{-1}u}}b_{pv}X^{v}.

  • σGal(q/p)\sigma\in{\mathrm{Gal}}({\mathbb{Q}}_{q}/{\mathbb{Q}}_{p}) the Frobenius, which acts on u{\mathcal{L}}_{u} via the coefficients.

  • Ψ=σ1ψEf:up1u\Psi=\sigma^{-1}\circ\psi\circ E_{f}:{\mathcal{B}}_{u}\to{\mathcal{B}}_{p^{-1}u} the Dwork’s TT-adic semi-linear operator.

  • cnc_{n} the coefficients of det(1Ψs)\det(1-\Psi s\mid{\mathcal{B}}), see (2.6).

  • skpkumodq1s_{k}\equiv p^{k}u\bmod{q-1} with 0skq20\leq s_{k}\leq q-2.

  • Γ=(γ(v,skq1+i),(w,sq1+j))\Gamma=\left(\gamma_{(v,\frac{s_{k}}{q-1}+i),(w,\frac{s_{\ell}}{q-1}+j)}\right) the matrix coefficient of Ψ\Psi on {\mathcal{B}}, see (2.7).

  • Γ(k)\Gamma^{(k)} the sub-matrix of Γ\Gamma defined in (2.7).

  • A(k)=AΓ(k)A^{(k)}=A\cap\Gamma^{(k)} the sub-matrix of a principal minor AA of Γ\Gamma.

  • 𝒜n{\mathcal{A}}_{n} the set of all principal minor AA of order bnbn, such that every A(k)A^{(k)} has order nn.

  • ϕ(n){+}\phi(n)\in{\mathbb{N}}\cup\left\{+\infty\right\} the minimal x+yx+y where dx+ey=n,x,ydx+ey=n,x,y\in{\mathbb{N}}.

  • γ(skq1+i,sq1+j)\gamma_{(\frac{s_{k}}{q-1}+i,\frac{s_{\ell}}{q-1}+j)}, see (2.9).

  • (x)[n]:=x(x1)(xn+1),(x)[0]:=1\left(x\right)_{\left[n\right]}:=x(x-1)\cdots(x-n+1),\left(x\right)_{\left[0\right]}:=1 the falling factorial.

1.3. Main results

In this paper, we give an explicit lower bound of Newton polygons of twisted LL-functions of binomial f(x)=xd+λxef(x)=x^{d}+\lambda x^{e}. We reduce the Hasse polynomial to a certain integer (3.1). Then p>(de)(2d1)p>(d-e)(2d-1) does not divide this constant, if and only if this lower bound coincides the Newton polygons. Finally, we show that this condition holds for e=d1e=d-1.

Denote by Pu,e,dP_{u,e,d} the polygon such that

Pu,e,d(n)=n(n1)2d+1bd(p1)k=1b(nuk+(de)Cuk,n1),n.P_{u,e,d}(n)=\frac{n(n-1)}{2d}+\frac{1}{bd(p-1)}\sum_{k=1}^{b}\bigl{(}nu_{k}+(d-e)C_{u_{k},n-1}\bigr{)},\ n\in{\mathbb{N}}. (1.2)

Denote by w(n)=Pu,e,d(n+1)Pu,e,d(n)w(n)=P_{u,e,d}(n+1)-P_{u,e,d}(n). Then

w(n)=nd+1bd(p1)k=1b(uk+(de)(Cuk,nCuk,n1)).w(n)=\frac{n}{d}+\frac{1}{bd(p-1)}\sum_{k=1}^{b}\bigl{(}u_{k}+(d-e)(C_{u_{k},n}-C_{u_{k},n-1})\bigr{)}.

This polygon lies above the Hodge polygon H[0,d],uH_{[0,d],u}^{\infty} with same points at dd{\mathbb{Z}}, and w(n+d)=1+w(n)w(n+d)=1+w(n). Moreover, we have w(n)w(n+1)w(n)\leq w(n+1) if p>(de)(2d1)p>(d-e)(2d-1). See Proposition 2.1.

Theorem 1.1.

Assume that p>(de)(2d1)p>(d-e)(2d-1). Then NPu,T(f){\mathrm{NP}}_{u,T}(f) lies above Pu,e,dP_{u,e,d}. As a corollary, NPu,m(f){\mathrm{NP}}_{u,m}(f) lies above Pu,e,dP_{u,e,d}.

Define

hn,k:=τSuk,nsgn(τ)i=0n1xuk,iτ!yuk,iτ!,hu,e,d:=n=0d2k=1bhn,k.h_{n,k}:=\sum_{\tau\in S_{u_{k},n}^{\circ}}{\mathrm{sgn}}(\tau)\prod_{i=0}^{n}\frac{1}{x_{u_{k},i}^{\tau}!y_{u_{k},i}^{\tau}!},\quad h_{u,e,d}:=\prod_{n=0}^{d-2}\prod_{k=1}^{b}h_{n,k}. (1.3)
Theorem 1.2.

Assume that p>(de)(2d1)p>(d-e)(2d-1). Then

NPu,m(f)=NPu,T(f)=Pu,e,d{\mathrm{NP}}_{u,m}(f)={\mathrm{NP}}_{u,T}(f)=P_{u,e,d} (1.4)

holds if and only if hu,e,dp×h_{u,e,d}\in{\mathbb{Z}}_{p}^{\times}, if and only if pHμ,c,𝐩,e,dp\nmid H_{\mu,c,{\mathbf{p}},e,d}.

Here Hμ,c,𝐩,e,dH_{\mu,c,{\mathbf{p}},e,d}\in{\mathbb{Z}} is a constant defined in (3.1) and 𝐩{\mathbf{p}} is the minimal positive residue of pp modulo cdcd. Thus we have the following corollary.

Corollary 1.3.

Assume that (1.4) holds for

a,m,p,f(x)=xd+λxe𝔽pa[x],u=(pa1)μc,a,m,p,f(x)=x^{d}+\lambda x^{e}\in{\mathbb{F}}_{p^{a}}[x],u=\frac{(p^{a}-1)\mu}{c},

where ba,λ0b\mid a,\lambda\neq 0 and p>(de)(2d1)p>(d-e)(2d-1). Then

  1. (1)

    Hμ,c,𝐩,e,d0H_{\mu,c,{\mathbf{p}},e,d}\neq 0.

  2. (2)

    For any

    a,m,p,f(x)=xd+λxe𝔽pa[x],u=(pa1)μc,a^{\prime},m^{\prime},p^{\prime},f^{\prime}(x)=x^{d}+\lambda^{\prime}x^{e}\in{\mathbb{F}}_{p^{\prime a^{\prime}}}[x],u^{\prime}=\frac{({p^{\prime}}^{a^{\prime}}-1)\mu}{c},

    where ba,λ0b\mid a,\lambda\neq 0 and p>(de)(2d1)p^{\prime}>(d-e)(2d-1), we have (1.4) if ppmodcdp^{\prime}\equiv p\bmod{cd} and p>Hμ,c,𝐩,e,dp^{\prime}>H_{\mu,c,{\mathbf{p}},e,d}.

  3. (3)

    As ppmodcdp^{\prime}\equiv p\bmod cd tends to infinity, the polygons NPu,m(f){\mathrm{NP}}_{u,m}(f) and NPu,T(f){\mathrm{NP}}_{u,T}(f) tend to H[0,d],uH_{[0,d],u}^{\infty}, which only depends on μ,c,𝐩,d\mu,c,{\mathbf{p}},d.

The following result extends [OZ16], as they considered the untwisted case with an additional condition p1moddp\equiv-1\bmod d.

Theorem 1.4.

Assume that e=d1e=d-1. We have NPu,m(f)=NPu,T(f)=Pu,e,d{\mathrm{NP}}_{u,m}(f)={\mathrm{NP}}_{u,T}(f)=P_{u,e,d} if p>c(d2d+1)p>c(d^{2}-d+1).

We give the following conjecture, which generalizes the conjecture in [ZN21]. Note that hu,e,dh_{u,e,d} may be zero since Suk,nS_{u_{k},n}^{\circ} may be empty, so we require that pp is large with respect to cc, as in Corollary 1.3 and Theorem 1.4.

Conjecture 1.5.

If pp is large enough with respect to c,dc,d, then NPu,m(f)=NPu,T(f)=Pu,e,d{\mathrm{NP}}_{u,m}(f)={\mathrm{NP}}_{u,T}(f)=P_{u,e,d}.

2. The lower bound

2.1. The property of the lower bound polygon

For any integer tt, we denote

Ct,n=minτSni=0ne1(piτ(i)+t)¯.C_{t,n}=\min_{\tau\in S_{n}^{*}}\sum_{i=0}^{n}\overline{e^{-1}(pi-\tau(i)+t)}.

We set Ct,1=0C_{t,-1}=0 for convention. For any integer α\alpha, we denote

Ri,α=e1(pi+α)¯,ri,α=e1(tαi)¯R_{i,\alpha}=\overline{e^{-1}(pi+\alpha)},\ r_{i,\alpha}=\overline{e^{-1}(t-\alpha-i)}

and

𝐂t,n,α=max#{iInRi,α+rτ(i),αd}.{\mathbf{C}}_{t,n,\alpha}=\max\#\left\{i\in I_{n}^{*}\mid R_{i,\alpha}+r_{\tau(i),\alpha}\geq d\right\}.
Proposition 2.1.

(1) For any α\alpha, we have

Ct,n=i=0n(Ri,α+ri,α)d𝐂t,n,α.C_{t,n}=\sum_{i=0}^{n}(R_{i,\alpha}+r_{i,\alpha})-d{\mathbf{C}}_{t,n,\alpha}.

(2) For any α\alpha, we have

𝐂t,n+d,α=d1+𝐂t,n,α,Ct,n+d=Ct,n.{\mathbf{C}}_{t,n+d,\alpha}=d-1+{\mathbf{C}}_{t,n,\alpha},\quad C_{t,n+d}=C_{t,n}.

Thus w(n+d)=1+w(n)w(n+d)=1+w(n) and Pu,e,d(dn)=H[0,d],u(dn)P_{u,e,d}(dn)=H_{[0,d],u}^{\infty}(dn).

(3) If p>(de)(2d1)p>(d-e)(2d-1), we have w(n)w(n+1)w(n)\leq w(n+1).

Proof.

We omit the subscript α\alpha in this proof for convention.

(1) It follows from

e1(piτ(i)+t)¯=Ri+rτ(i)dδRi+rτ(i)d.\overline{e^{-1}(pi-\tau(i)+t)}=R_{i}+r_{\tau(i)}-d\delta_{R_{i}+r_{\tau(i)}\geq d}.

(2) We have

𝐂t,n=maxτSn#{iInRidrτ(i)}.{\mathbf{C}}_{t,n}=\max_{\tau\in S_{n}^{*}}\#\left\{i\in I_{n}^{*}\mid R_{i}\geq d-r_{\tau(i)}\right\}.

Note that

{RiiIn+d}={RiiIn}{0,1,,d1},\left\{R_{i}\mid i\in I_{n+d}^{*}\right\}=\left\{R_{i}\mid i\in I_{n}^{*}\right\}\cup\left\{0,1,\dots,d-1\right\},
{driiIn+d}={driiIn}{d,1,,d1}.\left\{d-r_{i}\mid i\in I_{n+d}^{*}\right\}=\left\{d-r_{i}\mid i\in I_{n}^{*}\right\}\cup\left\{d,1,\dots,d-1\right\}.

We may drop the 0 and dd since they do not affect the size. Apple Lemma 2.2 (d1)(d-1) times and we get 𝐂t,n+d=d1+𝐂t,n{\mathbf{C}}_{t,n+d}=d-1+{\mathbf{C}}_{t,n}.

Since

i=n+1n+d(Ri+ri)=2j=0d1j=d(d1),\sum_{i=n+1}^{n+d}(R_{i}+r_{i})=2\sum_{j=0}^{d-1}j=d(d-1),

we have Ct,n+d=Ct,nC_{t,n+d}=C_{t,n}. Thus w(n+d)=1+w(n)w(n+d)=1+w(n).

Note that Ct,n+d=Ct,nC_{t,n+d}=C_{t,n} also holds for n=1n=-1. Hence Ct,dn1=0C_{t,dn-1}=0 and Pu,e,d(dn)=H[0,d],u(dn)P_{u,e,d}(dn)=H_{[0,d],u}^{\infty}(dn).

(3) Denote by δ=δRn+rnd\delta=\delta_{R_{n}+r_{n}\geq d}. For any τSn\tau\in S_{n}^{*}, write i=τ(n)i=\tau(n), j=τ1(n)j=\tau^{-1}(n) and τ1=(ni)τ\tau_{1}=(ni)\tau. Then τ1(n)=n\tau_{1}(n)=n, τ1(j)=i\tau_{1}(j)=i and

δ+#{iIn1Ri+rτ1(i)d}#{iInRi+rτ(i)d}=δ+δRj+ridδRj+rndδRn+rid.\begin{split}&\delta+\#\left\{i\in I_{n-1}^{*}\mid R_{i}+r_{\tau_{1}(i)}\geq d\right\}-\#\left\{i\in I_{n}^{*}\mid R_{i}+r_{\tau(i)}\geq d\right\}\\ =&\delta+\delta_{R_{j}+r_{i}\geq d}-\delta_{R_{j}+r_{n}\geq d}-\delta_{R_{n}+r_{i}\geq d}.\end{split}

If this is 2-2, then 2d>Rn+rn+Rj+ri2d2d>R_{n}+r_{n}+R_{j}+r_{i}\geq 2d, that’s impossible. Thus δ+𝐂t,n1𝐂t,n1\delta+{\mathbf{C}}_{t,n-1}-{\mathbf{C}}_{t,n}\geq-1.

Any σSn1\sigma\in S_{n-1}^{*} can be viewed as an element σ1Sn\sigma_{1}\in S_{n}^{*} fixing nn. Thus

δ+#{iIn1Ri+rσ(i)d}=#{iInRi+rσ1(i)d}.\delta+\#\left\{i\in I_{n-1}^{*}\mid R_{i}+r_{\sigma(i)}\geq d\right\}=\#\left\{i\in I_{n}^{*}\mid R_{i}+r_{\sigma_{1}(i)}\geq d\right\}.

and then δ+𝐂t,n1𝐂t,n\delta+{\mathbf{C}}_{t,n-1}\leq{\mathbf{C}}_{t,n}.

Now

Ct,nCt,n1=Rn+rnd(𝐂t,n𝐂t,n1)=e1(pnn+t)¯+d(δ+𝐂t,n1𝐂t,n)\begin{split}&C_{t,n}-C_{t,n-1}\\ =&R_{n}+r_{n}-d({\mathbf{C}}_{t,n}-{\mathbf{C}}_{t,n-1})\\ =&\overline{e^{-1}(pn-n+t)}+d(\delta+{\mathbf{C}}_{t,n-1}-{\mathbf{C}}_{t,n})\end{split}

lies in [d,d1][-d,d-1]. Therefore,

w(n)w(n1)=1d+debd(p1)k=1b(Cuk,n2Cuk,n1+Cuk,n2)1d+(de)(12d)d(p1)0\begin{split}&w(n)-w(n-1)\\ =&\frac{1}{d}+\frac{d-e}{bd(p-1)}\sum_{k=1}^{b}(C_{u_{k},n}-2C_{u_{k},n-1}+C_{u_{k},n-2})\\ \geq&\frac{1}{d}+\frac{(d-e)(1-2d)}{d(p-1)}\geq 0\end{split}

since p>(de)(2d1)p>(d-e)(2d-1). ∎

Lemma 2.2.

Let A={a0,,am}A=\left\{a_{0},\dots,a_{m}\right\} and B={b0,,bm}B=\left\{b_{0},\dots,b_{m}\right\} be two multi-sets of integers. Assume that a0b0a_{0}\geq b_{0} and for any i>0i>0, bi>a0b_{i}>a_{0} or bib0b_{i}\leq b_{0}. Then

maxτSm#{iImaibτ(i)}=1+maxσSm#{iImaibτ(i)}.\max_{\tau\in S_{m}^{*}}\#\left\{i\in I_{m}^{*}\mid a_{i}\geq b_{\tau(i)}\right\}=1+\max_{\sigma\in S_{m}}\#\left\{i\in I_{m}\mid a_{i}\geq b_{\tau(i)}\right\}.
Proof.

Every permutation in SnS_{n} can be viewed as an permutation in SnS_{n}^{*} fixing 0, thus “\geq” holds trivially. Write i=τ(0)i=\tau(0), j=τ1(0)j=\tau^{-1}(0) and τ1=(0i)τ\tau_{1}=(0i)\tau. Then τ1(0)=0\tau_{1}(0)=0 and τ1(j)=i\tau_{1}(j)=i. Thus

#{iImaibτ1(i)}#{iImaibτ(i)}=1+δajbiδajb0δa0bi.\begin{split}&\#\left\{i\in I_{m}^{*}\mid a_{i}\geq b_{\tau_{1}(i)}\right\}-\#\left\{i\in I_{m}^{*}\mid a_{i}\geq b_{\tau(i)}\right\}\\ =&1+\delta_{a_{j}\geq b_{i}}-\delta_{a_{j}\geq b_{0}}-\delta_{a_{0}\geq b_{i}}.\end{split}

If this is negative, then a0bi>ajb0a_{0}\geq b_{i}>a_{j}\geq b_{0}, which is impossible. Thus “\leq” holds. ∎

2.2. The twisted TT-adic Dwork’s trace formula

This part is almost the same with [LN11, SS2,3]. Denote by

E(X)=exp(i=0piXpi)=n=0λnXnpXE(X)=\exp\left(\sum_{i=0}^{\infty}p^{-i}X^{p^{i}}\right)=\sum_{n=0}^{\infty}\lambda_{n}X^{n}\in{\mathbb{Z}}_{p}\llbracket X\rrbracket (2.1)

the pp-adic Artin-Hasse series. Then λn=1/n!\lambda_{n}=1/n! if n<pn<p. Denote by

Ef(X)=E(πXd)E(πλ^Xe)=n=0γnXn.E_{f}(X)=E(\pi X^{d})E(\pi\hat{\lambda}X^{e})=\sum_{n=0}^{\infty}\gamma_{n}X^{n}. (2.2)

Then

γk=πx+yλxλyλ^y,\gamma_{k}=\sum\pi^{x+y}\lambda_{x}\lambda_{y}\hat{\lambda}^{y},

where (x,y)(x,y) runs through non-negative solutions of dx+ey=kdx+ey=k.

Denote by Mu=uq1+M_{u}=\frac{u}{q-1}+{\mathbb{N}}. Define

u={vMubvπvdXv|bvqπ1d(q1)}{\mathcal{L}}_{u}=\left\{\left.\sum_{v\in M_{u}}b_{v}\pi^{\frac{v}{d}}X^{v}\;\right|\;b_{v}\in{\mathbb{Z}}_{q}\llbracket\pi^{\frac{1}{d(q-1)}}\rrbracket\right\} (2.3)

and

u={vMubvπvdXvu|ordπbv+ as v+}.{\mathcal{B}}_{u}=\left\{\left.\sum_{v\in M_{u}}b_{v}\pi^{\frac{v}{d}}X^{v}\in{\mathcal{L}}_{u}\;\right|\;{\mathrm{ord}}_{\pi}b_{v}\to+\infty\text{ as }v\to+\infty\right\}. (2.4)

Define a map

ψ:up1uvMubvXvvMp1ubpvXv.\begin{split}\psi:{\mathcal{L}}_{u}&\longrightarrow{\mathcal{L}}_{p^{-1}u}\\ \sum_{v\in M_{u}}b_{v}X^{v}&\longmapsto\sum_{v\in M_{p^{-1}u}}b_{pv}X^{v}.\end{split} (2.5)

The power series EfE_{f} defines a map on u{\mathcal{B}}_{u} via multiplication. Let σGal(q/p)\sigma\in{\mathrm{Gal}}({\mathbb{Q}}_{q}/{\mathbb{Q}}_{p}) be the Frobenius, which acts on u{\mathcal{L}}_{u} via the coefficients. Then the Dwork’s TT-adic semi-linear operator Ψ=σ1ψEf\Psi=\sigma^{-1}\circ\psi\circ E_{f} sends u{\mathcal{B}}_{u} to p1u{\mathcal{B}}_{p^{-1}u}. Hence Ψ\Psi acts on

:=i=0b1piu.{\mathcal{B}}:=\bigoplus_{i=0}^{b-1}{\mathcal{B}}_{p^{i}u}.

We have a linear map

Ψa=ψai=0a1Efσi(Xpi)\Psi^{a}=\psi^{a}\circ\prod_{i=0}^{a-1}E_{f}^{\sigma^{i}}(X^{p^{i}})

on {\mathcal{B}} over qπ1d(q1){\mathbb{Z}}_{q}\llbracket\pi^{\frac{1}{d(q-1)}}\rrbracket. Since Ψ\Psi is completely continuous in the sense of [Ser62], the following determinants are well-defined.

Theorem 2.3.

We have

Cu(s,f,T)=det(1Ψas|u/qπ1d(q1)).C_{u}(s,f,T)=\det\left(1-\Psi^{a}s\;\left|\;{\mathcal{B}}_{u}/{\mathbb{Z}}_{q}\llbracket\pi^{\frac{1}{d(q-1)}}\rrbracket\right.\right).

Thus the TT-adic Newton polygon of Cu(s,f,T)C_{u}(s,f,T) is the lower convex closure of

(n,1bordT(cabn)),n,\left(n,\frac{1}{b}{\mathrm{ord}}_{T}(c_{abn})\right),\ n\in{\mathbb{N}},

where

det(1Ψs|/pπ1d(q1))=i=0(1)ncnsn.\det\left(1-\Psi s\;\left|\;{\mathcal{B}}/{\mathbb{Z}}_{p}\llbracket\pi^{\frac{1}{d(q-1)}}\rrbracket\right.\right)=\sum_{i=0}^{\infty}(-1)^{n}c_{n}s^{n}. (2.6)
Proof.

See [LW09, Theorem 4.8], [Liu07], [LLN09, Theorems 2.1, 2.2] and [LN11, Theorems 2.1, 5.3]. ∎

Write skpkumodq1s_{k}\equiv p^{k}u\bmod{q-1} with 0skq20\leq s_{k}\leq q-2. Then sbk=sk=uk+uk+1p++uk+a1pa1s_{b-k}=s_{-k}=u_{k}+u_{k+1}p+\cdots+u_{k+a-1}p^{a-1}. Let ξ1,,ξa\xi_{1},\dots,\xi_{a} be a normal basis of q{\mathbb{Q}}_{q} over p{\mathbb{Q}}_{p}. The space {\mathcal{B}} has a basis

{ξv(π1dX)skq1+i}(i,v,k)×Ia×Ib\left\{\xi_{v}(\pi^{\frac{1}{d}}X)^{\frac{s_{k}}{q-1}+i}\right\}_{(i,v,k)\in{\mathbb{N}}\times I_{a}\times I_{b}}

over pπ1d(q1){\mathbb{Z}}_{p}\llbracket\pi^{\frac{1}{d(q-1)}}\rrbracket. Let Γ=(γ(v,skq1+i),(w,sq1+j))×Ia×Ib\Gamma=\left(\gamma_{(v,\frac{s_{k}}{q-1}+i),(w,\frac{s_{\ell}}{q-1}+j)}\right)_{{\mathbb{N}}\times I_{a}\times I_{b}} be the matrix of Ψ\Psi on {\mathcal{B}} with respect to this basis. Then

Γ=(0Γ(1)0000Γ(2)0000Γ(b1)Γ(b)000),\Gamma=\begin{pmatrix}0&\Gamma^{(1)}&0&\cdots&0\\ 0&0&\Gamma^{(2)}&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&\Gamma^{(b-1)}\\ \Gamma^{(b)}&0&0&\cdots&0\end{pmatrix}, (2.7)

where

Γ(k)=(γ(v,sk1q1+i),(w,skq1+j))×Ia.\Gamma^{(k)}=\left(\gamma_{(v,\frac{s_{k-1}}{q-1}+i),(w,\frac{s_{k}}{q-1}+j)}\right)_{{\mathbb{N}}\times I_{a}}.

Hence we have

det(1Ψs|/pπ1d(q1))=det(1Γs)=n=0(1)bncbnsbn\det\left(1-\Psi s\;\left|\;{\mathcal{B}}/{\mathbb{Z}}_{p}\llbracket\pi^{\frac{1}{d(q-1)}}\rrbracket\right.\right)=\det(1-\Gamma s)=\sum_{n=0}^{\infty}(-1)^{bn}c_{bn}s^{bn}

with cn=det(A)c_{n}=\sum\det(A), where AA runs through all principal minors of order nn, see [LZ05]. Denote by A(k)=AΓ(k)A^{(k)}=A\cap\Gamma^{(k)} as a minor of Γ(k)\Gamma^{(k)}. If AA has order bnbn, but the order of some A(k)A^{(k)} is not nn, then det(A)=0\det(A)=0. Denote by 𝒜n{\mathcal{A}}_{n} the set of all principal minors of order bnbn, such that every A(k)A^{(k)} has order nn. Then

cbn=A𝒜ndet(A)=(1)n(b1)A𝒜nk=1bdet(A(k)).c_{bn}=\sum_{A\in{\mathcal{A}}_{n}}\det(A)=(-1)^{n(b-1)}\sum_{A\in{\mathcal{A}}_{n}}\prod_{k=1}^{b}\det(A^{(k)}). (2.8)
Theorem 2.4.

If p>(de)(2d1)p>(d-e)(2d-1), then

ordπ(det(A))ab(p1)Pu,e,d(n+1){\mathrm{ord}}_{\pi}(\det(A))\geq ab(p-1)P_{u,e,d}(n+1)

for any A𝒜a(n+1)A\in{\mathcal{A}}_{a(n+1)}.

Proof of Theorem 1.1.

By Theorem 2.4 and (2.8), we have

ordπ(cabn)ab(p1)Pu,e,d(n).{\mathrm{ord}}_{\pi}(c_{abn})\geq ab(p-1)P_{u,e,d}(n).

Thus NPu,T(f){\mathrm{NP}}_{u,T}(f) lies above Pu,e,dP_{u,e,d} by Theorem 2.3. Note that NPu,m(f)NPu,T(f){\mathrm{NP}}_{u,m}(f)\geq{\mathrm{NP}}_{u,T}(f) by definition. Therefore, NPu,m(f){\mathrm{NP}}_{u,m}(f) also lies above Pu,e,dP_{u,e,d}. ∎

2.3. Estimation on cnc_{n}

Denote by

ϕ(n)=min{x+ydx+ey=n,x,y}{+}.\phi(n)=\min\left\{x+y\mid dx+ey=n,x,y\in{\mathbb{N}}\right\}\in{\mathbb{N}}\cup\left\{+\infty\right\}.

Here the minimal element in \emptyset is regarded as ++\infty. For i,j,kIbi,j\in{\mathbb{N}},k\in I_{b}, define

γ(sk1q1+i,skq1+j)=πsksk1d(q1)+jidγpij+uk.\gamma_{(\frac{s_{k-1}}{q-1}+i,\frac{s_{k}}{q-1}+j)}=\pi^{\frac{s_{k}-s_{k-1}}{d(q-1)}+\frac{j-i}{d}}\gamma_{pi-j+u_{-k}}. (2.9)

Then

ξwσ1γ(sk1q1+i,skq1+j)σ1=uIaγ(v,sk1q1+i),(w,skq1+j)ξv\xi_{w}^{\sigma^{-1}}\gamma_{(\frac{s_{k-1}}{q-1}+i,\frac{s_{k}}{q-1}+j)}^{\sigma^{-1}}=\sum_{u\in I_{a}}\gamma_{(v,\frac{s_{k-1}}{q-1}+i),(w,\frac{s_{k}}{q-1}+j)}\xi_{v}

and

ordπ(γ(v,sk1q1+i),(w,skq1+j))ordπ(γ(sk1q1+i,skq1+j))=sksk1d(q1)+jid+ϕ(pij+uk).\begin{split}&{\mathrm{ord}}_{\pi}\left(\gamma_{(v,\frac{s_{k-1}}{q-1}+i),(w,\frac{s_{k}}{q-1}+j)}\right)\geq{\mathrm{ord}}_{\pi}\left(\gamma_{(\frac{s_{k-1}}{q-1}+i,\frac{s_{k}}{q-1}+j)}\right)\\ =&\frac{s_{k}-s_{k-1}}{d(q-1)}+\frac{j-i}{d}+\phi(pi-j+u_{-k}).\end{split} (2.10)
Lemma 2.5.

For any τSn\tau\in S_{n}^{*} and integer tt,

i=0nϕ(piτ(i)+t)d1((p1)n(n+1)2+(n+1)t+(de)Ct,n).\sum_{i=0}^{n}\phi(pi-\tau(i)+t)\geq d^{-1}\left(\frac{(p-1)n(n+1)}{2}+(n+1)t+(d-e)C_{t,n}\right).
Proof.

We may assume that piτ(i)+td+epi-\tau(i)+t\in d{\mathbb{N}}+e{\mathbb{N}} for each ii. One can easily show that

ϕ(k)=d1(k+(de)e1k¯)\phi(k)=d^{-1}\left(k+(d-e)\overline{e^{-1}k}\right)

and the minimum arrives at

(x,y)=(d1(kee1k¯),e1k¯).(x,y)=\left(d^{-1}(k-e\overline{e^{-1}k}),\overline{e^{-1}k}\right).

Thus

ϕ(pij+t)=d1(pij+t+(de)e1(pij+t)¯).\phi(pi-j+t)=d^{-1}\left(pi-j+t+(d-e)\overline{e^{-1}(pi-j+t)}\right). (2.11)

The result then follows easily. ∎

Lemma 2.6.

Assume ai=ai+ma_{i}=a_{i+m} and bi=bi+mb_{i}=b_{i+m} for any ii. Then

maxτSmd#{iImdaibτ(i)}=dmaxσSm#{iImaibσ(i)}.\max_{\tau\in S_{md}}\#\left\{i\in I_{md}\mid a_{i}\geq b_{\tau(i)}\right\}=d\max_{\sigma\in S_{m}}\#\left\{i\in I_{m}\mid a_{i}\geq b_{\sigma(i)}\right\}.
Proof.

We may assume that akbka_{k}\geq b_{k} and for any iki\neq k, bi>akb_{i}>a_{k} or bibkb_{i}\leq b_{k}. Otherwise both sides should be zero. We may assume that k=mk=m for simplicity. Apply Lemma 2.2 to (ami,bmi)(a_{mi},b_{mi}), we get

maxτSmd#{iImdaibτ(i)}=d+maxσ#{iImdmaibτ(i)},\max_{\tau\in S_{md}}\#\left\{i\in I_{md}\mid a_{i}\geq b_{\tau(i)}\right\}=d+\max_{\sigma}\#\left\{i\in I_{md}-m{\mathbb{Z}}\mid a_{i}\geq b_{\tau(i)}\right\},

where σ\sigma runs through permutations on ImdmI_{md}-m{\mathbb{Z}}. Since

maxτSm#{iImaibτ(i)}=1+maxσ#{iIm{m}aibτ(i)},\max_{\tau\in S_{m}}\#\left\{i\in I_{m}\mid a_{i}\geq b_{\tau(i)}\right\}=1+\max_{\sigma}\#\left\{i\in I_{m}-\left\{m\right\}\mid a_{i}\geq b_{\tau(i)}\right\},

where σ\sigma runs through permutations on Im{m}I_{m}-\left\{m\right\}, the result then follows by induction on mm. ∎

Lemma 2.7.

For any i×Iai\in{\mathbb{N}}\times I_{a}, we write i=(i,i′′)i=(i^{\prime},i^{\prime\prime}). Then for any permutation τ\tau on In×IaI_{n}^{*}\times I_{a},

iIn×Iaϕ(piτ(i)+t)ad((p1)n(n+1)2+(n+1)t+(de)Ct,n).\sum_{i\in I_{n}^{*}\times I_{a}}\phi(pi^{\prime}-\tau(i)^{\prime}+t)\geq\frac{a}{d}\left(\frac{(p-1)n(n+1)}{2}+(n+1)t+(d-e)C_{t,n}\right).
Proof.

By Eq. (2.11), we only need to show that

minτiIn×Iae1(piτ(i)+t)¯=aCt,n.\min_{\tau}\sum_{i\in I_{n}^{*}\times I_{a}}\overline{e^{-1}(pi-\tau(i)+t)}=aC_{t,n}.

By Proposition 2.1, it can be reduced to

maxτ#{iIn×IaRi,α+rτ(i),αd}=a𝐂t,n,α.\max_{\tau}\#\left\{i\in I_{n}^{*}\times I_{a}\mid R_{i^{\prime},\alpha}+r_{\tau(i)^{\prime},\alpha}\geq d\right\}=a{\mathbf{C}}_{t,n,\alpha}.

This follows from Lemma 2.6. ∎

Proof of Theorem 2.4.

This proof is similar to [ZN21, Theorem 3.2]. Denote by {\mathcal{R}} the set of indices of AA and

(k)×{k}=(×Ia×{k}),(0)=(b).{\mathcal{R}}^{(k)}\times\left\{k\right\}={\mathcal{R}}\cap({\mathbb{N}}\times I_{a}\times\left\{k\right\}),\quad{\mathcal{R}}^{(0)}={\mathcal{R}}^{(b)}.

Then #(k)=a(n+1)\#{\mathcal{R}}^{(k)}=a(n+1),

A(k)=(γ(v,sk1q1+i),(w,skq1+j))(i,v)(k1),(j,w)(k)A^{(k)}=\left(\gamma_{(v,\frac{s_{k-1}}{q-1}+i),(w,\frac{s_{k}}{q-1}+j)}\right)_{(i,v)\in{\mathcal{R}}^{(k-1)},(j,w)\in{\mathcal{R}}^{(k)}}

and

det(A)=k=1bdet(A(k))=τsgn(τ)iγi,τ(i),\det(A)=\prod_{k=1}^{b}\det(A^{(k)})=\sum_{\tau}{\mathrm{sgn}}(\tau)\prod_{i\in{\mathcal{R}}}\gamma_{i,\tau(i)},

where τ\tau runs through permutations of {\mathcal{R}} such that τ((k1))=(k)\tau({\mathcal{R}}^{(k-1)})={\mathcal{R}}^{(k)}. Here,

ordπ(iγi,τ(i))Sτ{\mathrm{ord}}_{\pi}\left(\prod_{i\in{\mathcal{R}}}\gamma_{i,\tau(i)}\right)\geq S_{\mathcal{R}}^{\tau}

by (2.10), where

Sτ=k=1bi(k1)(τ(i)id+ϕ(piτ(i)+uk))d1k=1bi(k1)((p1)i+(de)e1(piτ(i)+uk)¯)\begin{split}S_{\mathcal{R}}^{\tau}&=\sum_{k=1}^{b}\sum_{i\in{\mathcal{R}}^{(k-1)}}\left(\frac{\tau(i)^{\prime}-i^{\prime}}{d}+\phi\bigl{(}pi^{\prime}-\tau(i)^{\prime}+u_{-k}\bigr{)}\right)\\ &\geq d^{-1}\sum_{k=1}^{b}\sum_{i\in{\mathcal{R}}^{(k-1)}}\left((p-1)i^{\prime}+(d-e)\overline{e^{-1}(pi^{\prime}-\tau(i)^{\prime}+u_{-k})}\right)\end{split}

by Eq. (2.11). By Lemma 2.7,

S𝒩σab(p1)Pu,e,d(n+1),S_{\mathcal{N}}^{\sigma}\geq ab(p-1)P_{u,e,d}(n+1),

where 𝒩=In×Ia×Ib{\mathcal{N}}=I_{n}^{*}\times I_{a}\times I_{b}. By (2.8), we only need to show that for any permutation τ\tau of 𝒩{\mathcal{R}}\neq{\mathcal{N}} such that τ((k1))=(k)\tau({\mathcal{R}}^{(k-1)})={\mathcal{R}}^{(k)}, there is a permutation σ\sigma of 𝒩{\mathcal{N}} such that σ(𝒩(k1))=𝒩(k)\sigma({\mathcal{N}}^{(k-1)})={\mathcal{N}}^{(k)} and SτS𝒩σS_{\mathcal{R}}^{\tau}\geq S_{\mathcal{N}}^{\sigma}.

Assume #(\𝒩)=m\#({\mathcal{R}}\backslash{\mathcal{N}})=m. Write T=(𝒩\)τ1(\𝒩)T=({\mathcal{N}}\backslash{\mathcal{R}})\cup\tau^{-1}({\mathcal{R}}\backslash{\mathcal{N}}), then #T=2m\#T=2m and 𝒩\T=𝒩τ1(𝒩){\mathcal{N}}\backslash T={\mathcal{N}}\cap\tau^{-1}({\mathcal{N}}\cap{\mathcal{R}}). Thus τ(𝒩\T)𝒩\tau({\mathcal{N}}\backslash T)\subset{\mathcal{N}}. Note that for i\𝒩,j𝒩\i\in{\mathcal{R}}\backslash{\mathcal{N}},j\in{\mathcal{N}}\backslash{\mathcal{R}}, in+1j+1i^{\prime}\geq n+1\geq j^{\prime}+1. We can choose a permutation σ\sigma of 𝒩{\mathcal{N}} such that σ(𝒩(k1))=𝒩(k)\sigma({\mathcal{N}}^{(k-1)})={\mathcal{N}}^{(k)} and σ=τ\sigma=\tau on 𝒩\T{\mathcal{N}}\backslash T. Then

d(SτS𝒩σ)(i\𝒩i𝒩\)(p1)ik=1biT𝒩(k)(de)e1(piτ(i)+uk)¯m(p1)2m(de)(d1)>0.\begin{split}&d(S_{\mathcal{R}}^{\tau}-S_{\mathcal{N}}^{\sigma})\\ \geq&\left(\sum_{i\in{\mathcal{R}}\backslash{\mathcal{N}}}-\sum_{i\in{\mathcal{N}}\backslash{\mathcal{R}}}\right)(p-1)i^{\prime}-\sum_{k=1}^{b}\sum_{i\in T\cap{\mathcal{N}}^{(k)}}(d-e)\overline{e^{-1}(pi^{\prime}-\tau(i)^{\prime}+u_{-k})}\\ \geq&m(p-1)-2m(d-e)(d-1)>0.\end{split}

The result then follows. ∎

3. The Newton polygons

Lemma 3.1.

The Newton polygon NPm(f){\mathrm{NP}}_{m}(f) lies over NPT(f){\mathrm{NP}}_{T}(f). Moreover, if the equality holds for one mm, then it holds for all mm.

Proof.

See [LW09, Theorem 2.3] and [LN11, Theorem 5.5]. ∎

Proof of Theorem 1.2.

(1) Since w(d+i)=1+w(i)w(d+i)=1+w(i), both of NPu,m(f){\mathrm{NP}}_{u,m}(f) and Pu,e,dP_{u,e,d} across points (di,H[0,d],u(di))\bigl{(}di,H_{[0,d],u}^{\infty}(di)\bigr{)}, we only need to show that NPu,m(f)=Pu,e,d{\mathrm{NP}}_{u,m}(f)=P_{u,e,d} on [1,d1][1,d-1]. By Lemma 3.1, we may assume that m=1m=1.

Assume 0nd20\leq n\leq d-2. Recall that St,nS_{t,n}^{\circ} is the set of τSn\tau\in S_{n}^{*} such that

#{iInRi,α+rτ(i),αd}=𝐂t,n,α\#\left\{i\in I_{n}^{*}\mid R_{i,\alpha}+r_{\tau(i),\alpha}\geq d\right\}={\mathbf{C}}_{t,n,\alpha}

and every piτ(i)+td+epi-\tau(i)+t\in d{\mathbb{N}}+e{\mathbb{N}}. It’s equivalently to say, the equality in Lemma 2.5 holds. Recall that

yt,iτ=e1(piτ(i)+t)¯,xt,iτ=ϕ(piτ(i)+t)yt,iτ.y_{t,i}^{\tau}=\overline{e^{-1}(pi-\tau(i)+t)},\quad x_{t,i}^{\tau}=\phi(pi-\tau(i)+t)-y_{t,i}^{\tau}.

Denote by mm the right hand side in Lemma 2.5. Then we have

det(γpij+t)i,jInπmτSt,nsgn(τ)i=0nλxt,iτλyt,iτλ^yt,iτπmλ^vt,nτSt,nsgn(τ)i=0n1xt,iτ!yt,iτ!modπm+1,\begin{split}&\det(\gamma_{pi-j+t})_{i,j\in I_{n}^{*}}\equiv\pi^{m}\sum_{\tau\in S_{t,n}^{\circ}}{\mathrm{sgn}}(\tau)\prod_{i=0}^{n}\lambda_{x_{t,i}^{\tau}}\lambda_{y_{t,i}^{\tau}}\hat{\lambda}^{y_{t,i}^{\tau}}\\ \equiv&\pi^{m}\hat{\lambda}^{v_{t,n}}\sum_{\tau\in S_{t,n}^{\circ}}{\mathrm{sgn}}(\tau)\prod_{i=0}^{n}\frac{1}{x_{t,i}^{\tau}!y_{t,i}^{\tau}!}\mod\pi^{m+1},\end{split}

where

vt,n:=i=0nyt,iτ=i=1n(Ri,α+ri,α)d𝐂t,n,αv_{t,n}:=\sum_{i=0}^{n}y_{t,i}^{\tau}=\sum_{i=1}^{n}(R_{i,\alpha}+r_{i,\alpha})-d{\mathbf{C}}_{t,n,\alpha}

is independent on τSn\tau\in S_{n}^{\circ}.

Recall that Sτ>S𝒩σS_{\mathcal{R}}^{\tau}>S_{\mathcal{N}}^{\sigma} in the proof of Theorem 2.4. Then modulo πab(p1)Pu,e,d(n+1)+1\pi^{ab(p-1)P_{u,e,d}(n+1)+1}, we have

cab(n+1)=A𝒜a(n+1)det(A)det((γi,j)i,j𝒩)=±Nm(k=1bdet(γ(sk1q1+i,skq1+j))i,jIn)=±Nm(k=1bdet(γpij+uk)i,jIn)±πab(p1)Pu,e,d(n+1)Nm(k=1bλ^vuk,nhn,k)\begin{split}c_{ab(n+1)}&=\sum_{A\in{\mathcal{A}}_{a(n+1)}}\det(A)\equiv\det\bigl{(}(\gamma_{i,j})_{i,j\in{\mathcal{N}}}\bigr{)}\\ &=\pm{\mathrm{Nm}}\left(\prod_{k=1}^{b}\det\left(\gamma_{(\frac{s_{k-1}}{q-1}+i,\frac{s_{k}}{q-1}+j)}\right)_{i,j\in I_{n}^{*}}\right)\\ &=\pm{\mathrm{Nm}}\left(\prod_{k=1}^{b}\det(\gamma_{pi-j+u_{k}})_{i,j\in I_{n}^{*}}\right)\\ &\equiv\pm\pi^{ab(p-1)P_{u,e,d}(n+1)}{\mathrm{Nm}}\left(\prod_{k=1}^{b}\hat{\lambda}^{v_{u_{k},n}}h_{n,k}\right)\end{split}

by (2.8), (2.9), [LLN09, Lemma 4.4] and [LN11, Lemma 3.5]. Hence we get the first assertion by replacing π\pi by π1\pi_{1}.

(2) Denote by tkt_{k} the minimal non-negative residue of pkμp^{-k}\mu modulo cc. Then uk=tk+1ptkcu_{k}=\frac{t_{k+1}p-t_{k}}{c}. Write 𝐩{\mathbf{p}} the minimal positive residue of pp modulo cdcd and p=cd+𝐩p=cd\ell+{\mathbf{p}}. Denote by

𝐮k=tk+1𝐩tkc,𝐲𝐮k,iτ=e1(𝐩iτ(i)+𝐮k)¯,𝐱𝐮k,iτ=𝐩iτ(i)+𝐮ke𝐲𝐮k,iτd.{\mathbf{u}}_{k}=\frac{t_{k+1}{\mathbf{p}}-t_{k}}{c},\ {\mathbf{y}}_{{\mathbf{u}}_{k},i}^{\tau}=\overline{-e^{-1}({\mathbf{p}}i-\tau(i)+{\mathbf{u}}_{k})},\ {\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}=\frac{{\mathbf{p}}i-\tau(i)+{\mathbf{u}}_{k}-e{\mathbf{y}}_{{\mathbf{u}}_{k},i}^{\tau}}{d}.

Then

uk=tk+1d+𝐮k,yuk,iτ=𝐲𝐮k,iτ,xuk,iτ=(ci+tk+1)+𝐱𝐮k,iτ.u_{k}=t_{k+1}d\ell+{\mathbf{u}}_{k},\ y_{u_{k},i}^{\tau}={\mathbf{y}}_{{\mathbf{u}}_{k},i}^{\tau},\ x_{u_{k},i}^{\tau}=(ci+t_{k+1})\ell+{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}.

It’s easy to see that 𝐱𝐮k,iτ<𝐩{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}<{\mathbf{p}} and xuk,iτ<px_{u_{k},i}^{\tau}<p. Since

𝐱𝐮k,iτne(d1)d>e1,{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}\geq\frac{-n-e(d-1)}{d}>-e-1,

we have 𝐱𝐮k,iτe{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}\geq-e. Note that yt,iτy_{t,i}^{\tau} does not depend on \ell. Denote by

Hμ,c,𝐩,e,d=k=1bn=0d2τSnsgn(τ)i=1n(d1)[d1𝐲𝐮k,iτ]×(cd)𝐩1𝐱𝐮k,iτ×(𝐩(ci+tk+1)cd+𝐩1)[𝐩1𝐱𝐮k,iτ].\begin{split}H_{\mu,c,{\mathbf{p}},e,d}=&\prod_{k=1}^{b}\prod_{n=0}^{d-2}\sum_{\tau\in S_{n}^{\circ}}{\mathrm{sgn}}(\tau)\prod_{i=1}^{n}\left(d-1\right)_{\left[d-1-{\mathbf{y}}_{{\mathbf{u}}_{k},i}^{\tau}\right]}\times(cd)^{{\mathbf{p}}-1-{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}}\\ &\times\left(-\dfrac{{\mathbf{p}}(ci+t_{k+1})}{cd}+{\mathbf{p}}-1\right)_{\left[{\mathbf{p}}-1-{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}\right]}\in{\mathbb{Z}}.\end{split} (3.1)

Then

Hμ,c,𝐩,e,dk=1bn=0d2τSnsgn(τ)i=1n(d1)[d1𝐲𝐮k,iτ]×(cd)𝐩1𝐱𝐮k,iτ×((ci+tk+1)+𝐩1)[𝐩1𝐱𝐮k,iτ]=hu,e,dk=1bn=0d2i=1n(d1)!(cd)𝐩1𝐱𝐮k,iτ((ci+tk+1)+𝐩1)!modp\begin{split}&H_{\mu,c,{\mathbf{p}},e,d}\\ \equiv&\prod_{k=1}^{b}\prod_{n=0}^{d-2}\sum_{\tau\in S_{n}^{\circ}}{\mathrm{sgn}}(\tau)\prod_{i=1}^{n}\left(d-1\right)_{\left[d-1-{\mathbf{y}}_{{\mathbf{u}}_{k},i}^{\tau}\right]}\times(cd)^{{\mathbf{p}}-1-{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}}\\ &\times\left((ci+t_{k+1})\ell+{\mathbf{p}}-1\right)_{\left[{\mathbf{p}}-1-{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}\right]}\\ =&h_{u,e,d}\prod_{k=1}^{b}\prod_{n=0}^{d-2}\prod_{i=1}^{n}(d-1)!(cd)^{{\mathbf{p}}-1-{\mathbf{x}}_{{\mathbf{u}}_{k},i}^{\tau}}\bigl{(}(ci+t_{k+1})\ell+{\mathbf{p}}-1\bigr{)}!\mod p\end{split}

Note that d1,(ci+tk+1)+𝐩1<pd-1,(ci+t_{k+1})\ell+{\mathbf{p}}-1<p. Thus

NPu,m(f)=NPu,T(f)=Pu,e,dpHμ,c,𝐩,e,d{\mathrm{NP}}_{u,m}(f)={\mathrm{NP}}_{u,T}(f)=P_{u,e,d}\iff p\nmid H_{\mu,c,{\mathbf{p}},e,d}

for p>(de)(2d1)p>(d-e)(2d-1). ∎

Proof of Corollary 1.3.

Since pHμ,c,𝐩,e,dp\nmid H_{\mu,c,{\mathbf{p}},e,d}, we have Hμ,c,𝐩,e,d0H_{\mu,c,{\mathbf{p}},e,d}\neq 0. Hence pHμ,c,𝐩,e,dp^{\prime}\nmid H_{\mu,c,{\mathbf{p}},e,d} for any p>Hμ,c,𝐩,e,dp^{\prime}>H_{\mu,c,{\mathbf{p}},e,d}. Note that

k=1buk=p1ck=1btk,\sum_{k=1}^{b}u_{k}=\frac{p-1}{c}\sum_{k=1}^{b}t_{k},

thus H[0,d],uH_{[0,d],u}^{\infty} only depends on μ,c,𝐩,d\mu,c,{\mathbf{p}},d. Since

Pu,e,d(n)H[0,d],u(n)=debd(p1)k=1bCuk,n1(de)n¯(d1)d(p1)P_{u,e,d}(n)-H_{[0,d],u}^{\infty}(n)=\frac{d-e}{bd(p-1)}\sum_{k=1}^{b}C_{u_{k},n-1}\leq\frac{(d-e)\overline{n}(d-1)}{d(p-1)}

tends to zero as pp tends to infinity, the result then follows. ∎

Example 3.2.

Assume that p1moddp\equiv 1\bmod d and dukd\mid u_{k} for all kk. Write p=dk+1p=dk+1 and t=ukt=u_{k}. Then

Ri:=Ri,0=e1i¯,Ri:=ri,0=e1i¯,𝐂t,n=n,Sn={1}R_{i}:=R_{i,0}=\overline{e^{-1}i},\quad R_{i}:=r_{i,0}=\overline{-e^{-1}i},\quad{\mathbf{C}}_{t,n}=n,\quad S_{n}^{\circ}=\left\{1\right\}

and xt,i1=(p1)i+td,yt,i1=0x_{t,i}^{1}=\frac{(p-1)i+t}{d},y_{t,i}^{1}=0. Since

hn,k=(i=0n((p1)i+ukd)!)1p×,h_{n,k}=\left(\prod_{i=0}^{n}\left(\frac{(p-1)i+u_{k}}{d}\right)!\right)^{-1}\in{\mathbb{Z}}_{p}^{\times},

we obtain that the Newton polygons coincide H[0,d],uH_{[0,d],u}^{\infty}.

4. The case e=d1e=d-1

If piτ(i)+td+epi-\tau(i)+t\notin d{\mathbb{N}}+e{\mathbb{N}} for some ii, then xt,iτ<0x_{t,i}^{\tau}<0. Set 1/k!=01/k!=0 for negative integer kk. Then

hn,k=τSuk,nsgn(τ)i=1n1xuk,iτ!yuk,iτ!,h_{n,k}=\sum_{\tau\in S_{u_{k},n}^{\bullet}}{\mathrm{sgn}}(\tau)\prod_{i=1}^{n}\frac{1}{x_{u_{k},i}^{\tau}!y_{u_{k},i}^{\tau}!},

where St,nS_{t,n}^{\bullet} the set of τSn\tau\in S_{n}^{*} such that the size of {iInRi,α+rτ(i),αd}\left\{i\in I_{n}^{*}\mid R_{i,\alpha}+r_{\tau(i),\alpha}\geq d\right\} is Ct,n,αC_{t,n,\alpha}.

Lemma 4.1.

Denote by c(j)=(αj+β)[j]c(j)=\left(-\alpha j+\beta\right)_{\left[j\right]}.

(1) If ui=αvi+βu_{i}=\alpha v_{i}+\beta for any ii, then the matrix

((ui)[j](vi+n)[nj])0jn(c(j)vinj)0jn\bigl{(}\left(u_{i}\right)_{\left[j\right]}\cdot\left(v_{i}+n\right)_{\left[n-j\right]}\bigr{)}_{0\leq j\leq n}\implies\left(c(j)v_{i}^{n-j}\right)_{0\leq j\leq n} (4.1)

by third elementary column transformations.

(2) If uiαvi+βmodpu_{i}\equiv\alpha v_{i}+\beta\bmod p for any ii, then (4.1) holds by third elementary column transformations, modulo pp.

Proof.

(1) Write

(αx+β)[j]=t=0jct(j)(x+j)[t],\left(\alpha x+\beta\right)_{\left[j\right]}=\sum_{t=0}^{j}c_{t}(j)\cdot\left(x+j\right)_{\left[t\right]},

then c0(j)=c(j)c_{0}(j)=c(j) and

(ui)[j](vi+n)[nj]=t=0jct(j)(vi+j)[t](vi+n)[nj]=t=0jct(j)(vi+n)[nj+t].\begin{split}&\left(u_{i}\right)_{\left[j\right]}\cdot\left(v_{i}+n\right)_{\left[n-j\right]}\\ =&\sum_{t=0}^{j}c_{t}(j)\cdot\left(v_{i}+j\right)_{\left[t\right]}\cdot\left(v_{i}+n\right)_{\left[n-j\right]}\\ =&\sum_{t=0}^{j}c_{t}(j)\cdot\left(v_{i}+n\right)_{\left[n-j+t\right]}.\end{split} (4.2)

Hence by third elementary column transformations,

((ui)[j](vi+n)[nj])(c(j)(vi+n)[nj])(c(j)vinj).\bigl{(}\left(u_{i}\right)_{\left[j\right]}\cdot\left(v_{i}+n\right)_{\left[n-j\right]}\bigr{)}\implies\bigl{(}c(j)\cdot\left(v_{i}+n\right)_{\left[n-j\right]}\bigr{)}\implies\left(c(j)v_{i}^{n-j}\right).

(2) In this case, (4.2) holds modulo pp. The result then follows easily. ∎

Proof of Theorem 1.4.

Since p>c(d2d+1)p>c(d^{2}-d+1), we have p>(de)(2d1)p>(d-e)(2d-1). Denote by t=ukt=u_{k} and tkt_{k} the minimal non-negative residue of pkμp^{-k}\mu modulo cc. Then t=tk+1ptkct=\frac{t_{k+1}p-t_{k}}{c}. If c>1c>1, then tp(c1)cd(d1)t\geq\frac{p-(c-1)}{c}\geq d(d-1) and t<(c1)pcpd(d1)t<\frac{(c-1)p}{c}\leq p-d(d-1). If c=1c=1, then t=0t=0.

Assume that 0nd20\leq n\leq d-2. Denote by

Ri=Ri,t=e1(pi+t)¯=pit¯=pit+idR_{i}=R_{i,t}=\overline{e^{-1}(pi+t)}=\overline{-pi-t}=-pi-t+\ell_{i}d

and

ri=ri,t=e1i¯=i¯.r_{i}=r_{i,t}=\overline{-e^{-1}i}=\overline{i}.

Then

{driiIn}={d,d1,,dn}.\left\{d-r_{i}\mid i\in I_{n}^{*}\right\}=\left\{d,d-1,\dots,d-n\right\}.

We have

𝐂t,n=#{iInRidn}{\mathbf{C}}_{t,n}=\#\left\{i\in I_{n}^{*}\mid R_{i}\geq d-n\right\}

and

Sn={τSnRi+τ(i)dforRidn}.S_{n}^{\bullet}=\left\{\tau\in S_{n}^{*}\mid R_{i}+\tau(i)\geq d\ \text{for}\ R_{i}\geq d-n\right\}.

For Ri<dnR_{i}<d-n, we have Ri+τ(i)<dR_{i}+\tau(i)<d and

xt,iτ=pi+tieτ(i),yt,iτ=pit+id+τ(i);x_{t,i}^{\tau}=pi+t-\ell_{i}e-\tau(i),\quad y_{t,i}^{\tau}=-pi-t+\ell_{i}d+\tau(i);

for RidnR_{i}\geq d-n, we have Ri+τ(i)dR_{i}+\tau(i)\geq d and

xt,iτ=pi+tie+eτ(i),yt,iτ=pit+idd+τ(i).x_{t,i}^{\tau}=pi+t-\ell_{i}e+e-\tau(i),\quad y_{t,i}^{\tau}=-pi-t+\ell_{i}d-d+\tau(i).

If τSn\tau\notin S_{n}^{\bullet}, there is ii such that yt,iτ<0y_{t,i}^{\tau}<0 or xt,iτ<0x_{t,i}^{\tau}<0. Denote by

(ui,vi)={(pi+tie,pit+id), if Ri<dn;(pi+tie+e,pit+idd), if Ridn.(u_{i},v_{i})=\begin{cases}(pi+t-\ell_{i}e,-pi-t+\ell_{i}d),&\text{ if }R_{i}<d-n;\\ (pi+t-\ell_{i}e+e,-pi-t+\ell_{i}d-d),&\text{ if }R_{i}\geq d-n.\end{cases}

Then

hn,k=det(1(uij)!(vi+j)!).h_{n,k}=\det\left(\frac{1}{(u_{i}-j)!(v_{i}+j)!}\right).

Apply Lemma 4.1(2) with α=d1e,β=t(1d1e)\alpha=-d^{-1}e,\beta=t(1-d^{-1}e), we obtain that

hn,ki=0nui!(vi+n)!j=0n(d1e(jt)+t)[j]det(vinj)j=0n(d1e(jt)+t)[j]0i<jn(vivj)modp.\begin{split}&h_{n,k}\cdot\prod_{i=0}^{n}u_{i}!\cdot(v_{i}+n)!\\ \equiv&\prod_{j=0}^{n}\left(d^{-1}e(j-t)+t\right)_{\left[j\right]}\cdot\det\left(v_{i}^{n-j}\right)\\ \equiv&\prod_{j=0}^{n}\left(d^{-1}e(j-t)+t\right)_{\left[j\right]}\cdot\prod_{0\leq i<j\leq n}(v_{i}-v_{j})\mod p.\end{split}

If Ri<dnR_{i}<d-n, then vi=Ri0v_{i}=R_{i}\geq 0; if RidnR_{i}\geq d-n, then vi+n=Rid+n0v_{i}+n=R_{i}-d+n\geq 0. Hence 0vi+nd10\leq v_{i}+n\leq d-1 are different and (vi+n)!,(vivj)p×(v_{i}+n)!,(v_{i}-v_{j})\in{\mathbb{Z}}_{p}^{\times} if iji\neq j. Note that ui=iRiu_{i}=\ell_{i}-R_{i} or iRi+e\ell_{i}-R_{i}+e. When c=1c=1, we have t=R0=0t=R_{0}=\ell_{0}, u0=0u_{0}=0 or ee, and for i1i\geq 1,

uiiRipi+tdd+1pdd+10.u_{i}\geq\ell_{i}-R_{i}\geq\frac{pi+t}{d}-d+1\geq\frac{p}{d}-d+1\geq 0.

When c>1c>1, we have

uiiRipi+tdd+1tdd+10.u_{i}\geq\ell_{i}-R_{i}\geq\frac{pi+t}{d}-d+1\geq\frac{t}{d}-d+1\geq 0.

Meanwhile,

uiiRi+e=pi+t(d1)Ri+dedp(d2)+t+ded<p,u_{i}\leq\ell_{i}-R_{i}+e=\frac{pi+t-(d-1)R_{i}+de}{d}\leq\frac{p(d-2)+t+de}{d}<p,

hence ui!p×u_{i}!\in{\mathbb{Z}}_{p}^{\times}.

For any 0kj10\leq k\leq j-1, we have

0<e(jt)+d(tk)=d(jk)+tj(d1)j+pd(d1)<p,0<e(j-t)+d(t-k)=d(j-k)+t-j\leq(d-1)j+p-d(d-1)<p,

which means that p(d1e(jt)+t)[j]p\mid\left(d^{-1}e(j-t)+t\right)_{\left[j\right]}. Hence hn,kp×h_{n,k}\in{\mathbb{Z}}_{p}^{\times}. ∎

Acknowledgments. The author would like to thank Chuanze Niu and Daqing Wan for helpful discussions. The author is partially supported by NSFC (Grant No. 12001510), Anhui Initiative in Quantum Information Technologies (Grant No. AHY150200) and the Fundamental Research Funds for the Central Universities (Grant No. WK0010000061).

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