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Integrality of the higher Rademacher symbols

Cormac  O’Sullivan111Date: July 15, 2023.
   2020 Mathematics Subject Classification: 11F20, 11F67, 11B68, 11R11
   Key words and phrases. Rademacher symbols, integrality, Bernoulli numbers, zeta function values.
   Support for this project was provided by a PSC-CUNY Award, jointly funded by The Professional Staff Congress and The City
   University of New York.
Abstract

Rademacher symbols may be defined in terms of Dedekind sums, and give the value at zero of the zeta function associated to a narrow ideal class of a real quadratic field. Duke extended these symbols to give the zeta function values at all negative integers. Here we prove Duke’s conjecture that these higher Rademacher symbols are integer valued, making the above zeta value denominators as simple as the corresponding Riemann zeta value denominators. The proof uses detailed properties of Bernoulli numbers, including a generalization of the Kummer congruences.

1 Introduction

For Γ=SL​(2,ℤ){\Gamma}=\mathrm{SL}(2,{\mathbb{Z}}), the Rademacher symbol Ψ:Γ→ℤ\Psi:{\Gamma}\to{\mathbb{Z}} is not quite a homomorphism, but possesses many important properties that link it to modular forms, continued fractions, knots and much more, as discussed in the introduction to the compelling paper [Duk22]. The value at s=0s=0 of the zeta function associated to a narrow ideal class of a real quadratic field may also be expressed in terms of Ψ\Psi. Duke formulated the higher Rademacher symbols Ψn\Psi_{n} in [Duk22] so that they provide the corresponding values at s=1−ns=1-n for n=2,3,4,…n=2,3,4,\dots. This incorporates the work of Siegel in [Sie68].

After setting up some notation, we give the definition of Ψn\Psi_{n} next in terms of general Dedekind sums. Let T=(1101)T=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}}\right), V=(1011)V=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}1&0\\ 1&1\end{smallmatrix}}\right) and for A=(abcd)A=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}}\right) in Γ{\Gamma}, set

u​(A):=gcd⁡(b,c,d−a)⩾1,v​(A):=sgn⁡(a+d)​cu​(A).{\text{\rm u}}(A):=\gcd(b,c,d-a)\geqslant 1,\qquad{\text{\rm v}}(A):=\operatorname{sgn}(a+d)\frac{c}{{\text{\rm u}}(A)}.

The Bernoulli numbers BnB_{n} play an important role from the start. Define  in{\text{\,\rm i}}_{n},  jn{\text{\,\rm j}}_{n} by means of

 in jn=−B2​n2​n=ζ​(1−2​n)(n∈ℤ⩾1),\frac{{\text{\,\rm i}}_{n}}{{\text{\,\rm j}}_{n}}=-\frac{B_{2n}}{2n}=\zeta(1-2n)\qquad\qquad(n\in{\mathbb{Z}}_{\geqslant 1}), (1.1)

with  in/ jn{\text{\,\rm i}}_{n}/{\text{\,\rm j}}_{n} in lowest terms and  jn>0{\text{\,\rm j}}_{n}>0. For m⩾0m\geqslant 0, the periodized Bernoulli polynomials B¯m​(x)\,\overline{\!{B}}_{m}(x) are Bm​(x−⌊x⌋)B_{m}(x-\lfloor x\rfloor), with the stipulation that B¯1​(x)=0\,\overline{\!{B}}_{1}(x)=0 for x∈ℤx\in{\mathbb{Z}}. The general Dedekind sum, [Car54, LJC17], is

Sr,s(a,c):=∑hmod|c|B¯r(a​hc)B¯s(hc)(r,s∈ℤ⩾0,c∈ℤ≠0).S_{r,s}(a,c):=\sum_{h\bmod{|c|}}\,\overline{\!{B}}_{r}\mathopen{}\mathclose{{}\left(\frac{ah}{c}}\right)\,\overline{\!{B}}_{s}\mathopen{}\mathclose{{}\left(\frac{h}{c}}\right)\qquad\qquad(r,s\in{\mathbb{Z}}_{\geqslant 0},\ c\in{\mathbb{Z}}_{\neq 0}). (1.2)

Finally, for r,s∈ℤ⩾0r,s\in{\mathbb{Z}}_{\geqslant 0}, r+s⩾2r+s\geqslant 2, define the following hypergeometric polynomial of degree at most r+s−1r+s-1,

Fr,s​(z):=(r+s−2)!r!​s!​F12​(1−r,1−s;3−r−s2;z+24).F_{r,s}(z):=\frac{(r+s-2)!}{r!s!}{}_{2}F_{1}\mathopen{}\mathclose{{}\left(1-r,1-s;\frac{3-r-s}{2};\frac{z+2}{4}}\right). (1.3)
Definition 1.1.

[Duk22, Sect. 2] Let nn be a positive integer. For A=(abcd)A=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}}\right) in Γ{\Gamma} with c≠0c\neq 0,

Ψn​(A):=−sgn⁡(c)⋅v​(A)n−1⋅ jn​∑r+s=2​nFr,s​(a+d)⋅Sr,s​(a,c).\Psi_{n}(A):=-\operatorname{sgn}(c)\cdot{\text{\rm v}}(A)^{n-1}\cdot{\text{\,\rm j}}_{n}\sum_{r+s=2n}F_{r,s}(a+d)\cdot S_{r,s}(a,c). (1.4)

If n=1n=1 then 3​sgn⁡(c​(a+d))3\operatorname{sgn}(c(a+d)) should be subtracted from the right of (1.4). When c=0c=0 and hence A=±TkA=\pm T^{k}, set Ψn​(±I):=0\Psi_{n}(\pm I):=0 and Ψn​(±Tk):=Ψn​(±V−k)\Psi_{n}(\pm T^{k}):=\Psi_{n}(\pm V^{-k}).

Then Ψn​(A)\Psi_{n}(A) may be computed for any AA as a finite sum of rational numbers. Here, Ψ1\Psi_{1} equals the original Rademacher symbol Ψ\Psi. Since Ψn​(−A)=Ψn​(A)\Psi_{n}(-A)=\Psi_{n}(A), these symbols are well-defined on the modular group Γ/{±I}{\Gamma}/\{\pm I\}. The fundamental properties of Ψn\Psi_{n} are given after [Duk22, Eq. (2.4)]. These include that Ψn​(A)\Psi_{n}(A) does not change under conjugation, A↦B​A​B−1A\mapsto BAB^{-1}, and for n⩾2n\geqslant 2, Ψn\Psi_{n} sends elliptic elements to 0 and parabolic elements to integer multiples of  in{\text{\,\rm i}}_{n}.

Associated to each narrow ideal class 𝒜\mathcal{A} there is a hyperbolic A∈ΓA\in{\Gamma}. Duke gives an elegant direct proof in [Duk22, Thm. 4] of the connection

ζ​(1−n,𝒜)=Ψn​(A) jn,\zeta(1-n,\mathcal{A})=\frac{\Psi_{n}(A)}{{\text{\,\rm j}}_{n}}, (1.5)

between values of the corresponding zeta function and the higher Rademacher symbols. Equivalent identities were proved by Hecke for n=1n=1 and by Siegel and Shintani in general. Comparing the form of (1.5) with (1.1) and the Riemann zeta function value ζ​(1−2​n)\zeta(1-2n), it is natural to ask if Ψn​(A)\Psi_{n}(A) is an integer. In 1976, Zagier gave a related formula for ζ​(1−n,𝒜)\zeta(1-n,\mathcal{A}), but was unable to find a good estimate for the denominator. Zagier’s formula is rederived in Corollary 6.3.

Conjecture 1.2.

[Duk22, Sect. 2] For every n⩾2n\geqslant 2, the higher Rademacher symbol satisfies Ψn:Γ→ℤ\Psi_{n}:{\Gamma}\to{\mathbb{Z}}.

Our main result is that Duke’s above conjecture is true; this is Corollary 5.2. We prove a slightly stronger reformulation of Conjecture 1.2, given in [Duk22] in terms of a certain polynomial Gn​(x)G_{n}(x) with coefficients involving Bernoulli numbers. This polynomial is described in the next section, where we use reciprocity relations to show that it is a period polynomial. To prove that Gn​(x)G_{n}(x) has integer coefficients requires further Bernoulli number properties, described in Section 4, and including Cohen’s extension of the classical Kummer congruences. As reviewed in Section 3, Gn​(x)G_{n}(x) having integer coefficients implies Conjecture 1.2.

For Rademacher symbols described in the wider context of general Fuchsian groups, see the overview in [Bur22]. The ideas in this paper should be applicable there too.

Acknowledgements. Many thanks to Karen Taylor for helpful conversations about this work and Siegel’s paper [Sie68].

2 Preliminaries

2.1 The spaces 𝒫n\mathcal{P}_{n} and 𝒲n\mathcal{W}_{n}

A positive integer nn is fixed in the background. Write x=(x1,x2)x=(x_{1},x_{2}) and define

Qr​(x):=[zr]​((z−x1)​(z−x2))n−1,Q_{r}(x):=\mathopen{}\mathclose{{}\left[z^{r}}\right]\mathopen{}\mathclose{{}\left((z-x_{1})(z-x_{2})}\right)^{n-1}, (2.1)

where [zr][z^{r}] extracts the coefficient of zrz^{r}. Then Qr​(x)Q_{r}(x) is a symmetric polynomial in x1x_{1} and x2x_{2} of homogeneous degree r∗:=2​n−2−rr^{*}:=2n-2-r for 0⩽r⩽2​n−20\leqslant r\leqslant 2n-2. Explicitly,

Qr​(x)=(−1)r​(x1​x2)n−1−r​∑u+v=r(n−1u)​(n−1v)​x1u​x2v.Q_{r}(x)=(-1)^{r}(x_{1}x_{2})^{n-1-r}\sum_{u+v=r}\binom{n-1}{u}\binom{n-1}{v}x_{1}^{u}x_{2}^{v}. (2.2)

Let 𝒫n\mathcal{P}_{n} be the ℂ{\mathbb{C}} vector space with basis Qr​(x)Q_{r}(x) for 0⩽r⩽2​n−20\leqslant r\leqslant 2n-2.

The group Γ{\Gamma} is generated by T=(1101)T=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}}\right) and S=(0−110)S=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}0&-1\\ 1&0\end{smallmatrix}}\right) with S2=−IS^{2}=-I. Also write

U=T​S=(1−110),V=T​S​T=−S​T−1​S=(1011),U=TS=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}1&-1\\ 1&0\end{smallmatrix}}\right),\qquad V=TST=-ST^{-1}S=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}1&0\\ 1&1\end{smallmatrix}}\right),

with U3=−IU^{3}=-I. For A=(abcd)A=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}}\right), its entries may be specified with a=aAa=a_{A}, b=bAb=b_{A} and so on. As described in [Duk22, Sect. 5], there is a natural action of Γ{\Gamma} on f​(x)∈𝒫nf(x)\in\mathcal{P}_{n} with, for example,

f|T​(x)=f​(x1+1,x2+1),f|S​(x)=(x1​x2)n−1​f​(−1/x1,−1/x2),f|T(x)=f(x_{1}+1,x_{2}+1),\qquad f|S(x)=(x_{1}x_{2})^{n-1}f(-1/x_{1},-1/x_{2}), (2.3)

for the generators. We have

Qr|T​(x)=∑ℓ=02​n−2(−1)ℓ+r​(ℓr)​Qℓ​(x),Qr|S​(x)=(−1)r​Qr∗​(x).Q_{r}|T(x)=\sum_{\ell=0}^{2n-2}(-1)^{\ell+r}\binom{\ell}{r}Q_{\ell}(x),\qquad Q_{r}|S(x)=(-1)^{r}Q_{r^{*}}(x). (2.4)

With U=T​SU=TS, it follows that

Qr|U​(x)=(−1)r​∑ℓ=02​n−2(ℓ∗r)​Qℓ​(x),Qr|U2​(x)=(−1)r​∑ℓ=02​n−2(ℓr∗)​Qℓ​(x).Q_{r}|U(x)=(-1)^{r}\sum_{\ell=0}^{2n-2}\binom{\ell^{*}}{r}Q_{\ell}(x),\qquad Q_{r}|U^{2}(x)=(-1)^{r}\sum_{\ell=0}^{2n-2}\binom{\ell}{r^{*}}Q_{\ell}(x). (2.5)

Any ff in 𝒫n\mathcal{P}_{n} may be written as

f​(x)=∑r=02​n−2cr​Qr​(x).f(x)=\sum_{r=0}^{2n-2}c_{r}Q_{r}(x). (2.6)

The next result follows directly from (2.4) and (2.5). See [KZ84, p. 199] for similar relations.

Proposition 2.1.

We have that ff in (2.6) satisfies f|(1+S)=0f|(1+S)=0 if and only if cr∗+(−1)r​cr=0c_{r^{*}}+(-1)^{r}c_{r}=0 for 0⩽r⩽2​n−20\leqslant r\leqslant 2n-2. We have f|(1+U+U2)=0f|(1+U+U^{2})=0 if and only if

∑r=02​n−2[δr,ℓ+(−1)r​(ℓ∗r)+(−1)r​(ℓr∗)]​cr=0\sum_{r=0}^{2n-2}\mathopen{}\mathclose{{}\left[\delta_{r,\ell}+(-1)^{r}\binom{\ell^{*}}{r}+(-1)^{r}\binom{\ell}{r^{*}}}\right]c_{r}=0

for all ℓ\ell with 0⩽ℓ⩽2​n−20\leqslant\ell\leqslant 2n-2. Here, δr,ℓ\delta_{r,\ell} is the Kronecker delta.

Let 𝒲n\mathcal{W}_{n} be the subspace of 𝒫n\mathcal{P}_{n} consisting of those ffs satisfying f|(1+S)=0f|(1+S)=0 and f|(1+U+U2)=0f|(1+U+U^{2})=0; these are the two-variable period polynomials of [Duk22, Sect. 6].

2.2 The polynomial Gn​(x)G_{n}(x)

For integers m⩾0m\geqslant 0 define

βm:=ζ​(−m)=(−1)m​Bm+1m+1,\beta_{m}:=\zeta(-m)=(-1)^{m}\frac{B_{m+1}}{m+1}, (2.7)

(following the convention in the references that B1=−1/2B_{1}=-1/2). Set ξ0=ξ2​n−2=0\xi_{0}=\xi_{2n-2}=0 and for 1⩽r⩽2​n−31\leqslant r\leqslant 2n-3,

ξr:=β2​n−1​(βr+βr∗+δr,1+δr∗,12​n−2)−βr​βr∗.\xi_{r}:=\beta_{2n-1}\mathopen{}\mathclose{{}\left(\beta_{r}+\beta_{r^{*}}+\frac{\delta_{r,1}+\delta_{r^{*},1}}{2n-2}}\right)-\beta_{r}\beta_{r^{*}}. (2.8)

Note that ξr∗=ξr\xi_{r^{*}}=\xi_{r} and also ξr=0\xi_{r}=0 for all even indices rr. With (1.1) and (2.1), define

Gn​(x):= jn​∑r=02​n−2ξr⋅Qr​(x)(n∈ℤ⩾1),G_{n}(x):={\text{\,\rm j}}_{n}\sum_{r=0}^{2n-2}\xi_{r}\cdot Q_{r}(x)\qquad\qquad(n\in{\mathbb{Z}}_{\geqslant 1}), (2.9)

and this is equivalent to [Duk22, Eq. (2.19)]. Each Gn​(x)G_{n}(x) is in ℚ​[x1,x2]{\mathbb{Q}}[x_{1},x_{2}] with degree at most 2​n−32n-3 and only odd degree terms. These polynomials are symmetric and, by the right identities in (2.3), (2.4), reciprocal:

Gn​(x1,x2)=Gn​(x2,x1),(x1​x2)n−1​Gn​(1/x1,1/x2)=Gn​(x1,x2).G_{n}(x_{1},x_{2})=G_{n}(x_{2},x_{1}),\qquad(x_{1}x_{2})^{n-1}G_{n}(1/x_{1},1/x_{2})=G_{n}(x_{1},x_{2}).

The first nn for which GnG_{n} is not identically 0 is n=6n=6:

G6​(x)=72​(x15​x24+x14​x25+x1+x2)−75​(x15​x22+x12​x25+x13+x23)+6​(x15+x25)−375​(x14​x23+x13​x24+x12​x2+x1​x22)+150​(x14​x2+x1​x24)+600​(x13​x22+x12​x23).G_{6}(x)=72(x_{1}^{5}x_{2}^{4}+x_{1}^{4}x_{2}^{5}+x_{1}+x_{2})-75(x_{1}^{5}x_{2}^{2}+x_{1}^{2}x_{2}^{5}+x_{1}^{3}+x_{2}^{3})+6(x_{1}^{5}+x_{2}^{5})\\ -375(x_{1}^{4}x_{2}^{3}+x_{1}^{3}x_{2}^{4}+x_{1}^{2}x_{2}+x_{1}x_{2}^{2})+150(x_{1}^{4}x_{2}+x_{1}x_{2}^{4})+600(x_{1}^{3}x_{2}^{2}+x_{1}^{2}x_{2}^{3}).
Proposition 2.2.

For all n⩾1n\geqslant 1 we have Gn​(x)G_{n}(x) in 𝒲n\mathcal{W}_{n}.

This important property of Gn​(x)G_{n}(x) is required for the proof of (3.3) below and stated in [Duk22, Lemma 13]. We may give a relatively simple proof based on the reciprocity relations for Bernoulli numbers discussed in [AD08]. These relations are interesting in their own right.

The first is due to Saalschütz in 1892: for all kk, mm in ℤ⩾0{\mathbb{Z}}_{\geqslant 0},

∑j(−1)j​(kj)​βm+j+∑j(−1)j​(mj)​βk+j=−k!​m!(k+m+1)!.\sum_{j}(-1)^{j}\binom{k}{j}\beta_{m+j}+\sum_{j}(-1)^{j}\binom{m}{j}\beta_{k+j}=-\frac{k!m!}{(k+m+1)!}. (2.10)

Note that our βm\beta_{m} equals (−1)m(-1)^{m} times the βm\beta_{m} appearing in [AD08]. Where the range of summation is not specified, as in (2.10), we mean that it is over the finite set of integers where the contained binomial coefficients are defined and non-zero.

The second reciprocity relation follows from a Bernoulli polynomial identity of Nielsen from 1923, as described in [AD08]. For all kk, mm in ℤ⩾0{\mathbb{Z}}_{\geqslant 0},

∑j(−1)j​(kj)​βm+j​βk−j+∑j(−1)j​(mj)​βk+j​βm−j=βk​βm−(k!​m!(k+m+1)!+(−1)kk+1+(−1)mm+1)​βk+m+1.\sum_{j}(-1)^{j}\binom{k}{j}\beta_{m+j}\beta_{k-j}+\sum_{j}(-1)^{j}\binom{m}{j}\beta_{k+j}\beta_{m-j}\\ =\beta_{k}\beta_{m}-\mathopen{}\mathclose{{}\left(\frac{k!m!}{(k+m+1)!}+\frac{(-1)^{k}}{k+1}+\frac{(-1)^{m}}{m+1}}\right)\beta_{k+m+1}. (2.11)

Zagier also proved the underlying Bernoulli polynomial identity in the appendix to [AIK14]. A generalization of (2.11) is provided in [AD08, Thm. 1].

Proof of Proposition 2.2.

Since G1=0G_{1}=0 we may assume that n⩾2n\geqslant 2. Proposition 2.1 implies that Gn|(1+S)=0G_{n}|(1+S)=0 and that proving Gn|(1+U+U2)=0G_{n}|(1+U+U^{2})=0 is equivalent to showing

∑r=02​n−2[(−1)r​(ℓ∗r)+(−1)r​(ℓr∗)]​ξr=−ξℓ,\sum_{r=0}^{2n-2}\mathopen{}\mathclose{{}\left[(-1)^{r}\binom{\ell^{*}}{r}+(-1)^{r}\binom{\ell}{r^{*}}}\right]\xi_{r}=-\xi_{\ell}, (2.12)

for all ℓ\ell with 0⩽ℓ⩽2​n−20\leqslant\ell\leqslant 2n-2. The left side of (2.12) has the form

−β2​n−12​n−2​S1+β2​n−1​S2−S3,-\frac{\beta_{2n-1}}{2n-2}S_{1}+\beta_{2n-1}S_{2}-S_{3},

for

S1\displaystyle S_{1} =(ℓ1)+(ℓ1∗)+(ℓ∗1)+(ℓ∗1∗),\displaystyle=\binom{\ell}{1}+\binom{\ell}{1^{*}}+\binom{\ell^{*}}{1}+\binom{\ell^{*}}{1^{*}},
S2\displaystyle S_{2} =∑r=12​n−3(−1)r​[(ℓr∗)​βr∗+(ℓ∗r)​βr+(ℓr∗)​βr+(ℓ∗r)​βr∗],\displaystyle=\sum_{r=1}^{2n-3}(-1)^{r}\mathopen{}\mathclose{{}\left[\binom{\ell}{r^{*}}\beta_{r^{*}}+\binom{\ell^{*}}{r}\beta_{r}+\binom{\ell}{r^{*}}\beta_{r}+\binom{\ell^{*}}{r}\beta_{r^{*}}}\right],
S3\displaystyle S_{3} =∑r=12​n−3(−1)r​[(ℓr∗)​βr​βr∗+(ℓ∗r)​βr​βr∗].\displaystyle=\sum_{r=1}^{2n-3}(-1)^{r}\mathopen{}\mathclose{{}\left[\binom{\ell}{r^{*}}\beta_{r}\beta_{r^{*}}+\binom{\ell^{*}}{r}\beta_{r}\beta_{r^{*}}}\right].

Starting with S2S_{2}, extending the sum to all rr gives

S2=1+δℓ,0+δℓ∗,02+∑r(−1)r​[(ℓr∗)​βr∗+(ℓ∗r)​βr+(ℓr∗)​βr+(ℓ∗r)​βr∗],S_{2}=1+\frac{\delta_{\ell,0}+\delta_{\ell^{*},0}}{2}+\sum_{r}(-1)^{r}\mathopen{}\mathclose{{}\left[\binom{\ell}{r^{*}}\beta_{r^{*}}+\binom{\ell^{*}}{r}\beta_{r}+\binom{\ell}{r^{*}}\beta_{r}+\binom{\ell^{*}}{r}\beta_{r^{*}}}\right], (2.13)

since β2​n−2=0\beta_{2n-2}=0. Taking m=0m=0 in (2.10) means

∑j(−1)j​(kj)​βj=−βk−1k+1,\sum_{j}(-1)^{j}\binom{k}{j}\beta_{j}=-\beta_{k}-\frac{1}{k+1},

giving the first two sums in (2.13). Next, with k=ℓk=\ell and m=ℓ∗m=\ell^{*} in (2.10) we find, after replacing the first index jj by ℓ−r\ell-r and the second jj by ℓ∗−r\ell^{*}-r,

∑r(−1)r​[(ℓr∗)​βr+(ℓ∗r)​βr∗]=−(−1)ℓ​ℓ!​ℓ∗!(2​n−1)!.\sum_{r}(-1)^{r}\mathopen{}\mathclose{{}\left[\binom{\ell}{r^{*}}\beta_{r}+\binom{\ell^{*}}{r}\beta_{r^{*}}}\right]=-\frac{(-1)^{\ell}\ell!\ell^{*}!}{(2n-1)!}. (2.14)

Therefore

S2=1+δℓ,0+δℓ∗,02−βℓ−1ℓ+1−βℓ∗−1ℓ∗+1−(−1)ℓ​ℓ!​ℓ∗!(2​n−1)!.S_{2}=1+\frac{\delta_{\ell,0}+\delta_{\ell^{*},0}}{2}-\beta_{\ell}-\frac{1}{\ell+1}-\beta_{\ell^{*}}-\frac{1}{\ell^{*}+1}-\frac{(-1)^{\ell}\ell!\ell^{*}!}{(2n-1)!}.

The sum S3S_{3} is obtained from (2.11), similarly to (2.14):

S3=(−1)ℓ​βℓ​βℓ∗−β2​n−1​((−1)ℓ​ℓ!​ℓ∗!(2​n−1)!+1ℓ+1+1ℓ∗+1).S_{3}=(-1)^{\ell}\beta_{\ell}\beta_{\ell^{*}}-\beta_{2n-1}\mathopen{}\mathclose{{}\left(\frac{(-1)^{\ell}\ell!\ell^{*}!}{(2n-1)!}+\frac{1}{\ell+1}+\frac{1}{\ell^{*}+1}}\right).

The identity (2.12) that we are trying to prove becomes

−β2​n−12​n−2​[(ℓ1)+(ℓ1∗)+(ℓ∗1)+(ℓ∗1∗)]+β2​n−1​[1+δℓ,0+δℓ∗,02−βℓ−1ℓ+1−βℓ∗−1ℓ∗+1−(−1)ℓ​ℓ!​ℓ∗!(2​n−1)!]+(−1)ℓ+1​βℓ​βℓ∗+β2​n−1​((−1)ℓ​ℓ!​ℓ∗!(2​n−1)!+1ℓ+1+1ℓ∗+1)=−β2​n−1​(βℓ+βℓ∗+δℓ,1+δℓ∗,12​n−2)+βℓ​βℓ∗,-\frac{\beta_{2n-1}}{2n-2}\mathopen{}\mathclose{{}\left[\binom{\ell}{1}+\binom{\ell}{1^{*}}+\binom{\ell^{*}}{1}+\binom{\ell^{*}}{1^{*}}}\right]\\ +\beta_{2n-1}\mathopen{}\mathclose{{}\left[1+\frac{\delta_{\ell,0}+\delta_{\ell^{*},0}}{2}-\beta_{\ell}-\frac{1}{\ell+1}-\beta_{\ell^{*}}-\frac{1}{\ell^{*}+1}-\frac{(-1)^{\ell}\ell!\ell^{*}!}{(2n-1)!}}\right]\\ +(-1)^{\ell+1}\beta_{\ell}\beta_{\ell^{*}}+\beta_{2n-1}\mathopen{}\mathclose{{}\left(\frac{(-1)^{\ell}\ell!\ell^{*}!}{(2n-1)!}+\frac{1}{\ell+1}+\frac{1}{\ell^{*}+1}}\right)\\ =-\beta_{2n-1}\mathopen{}\mathclose{{}\left(\beta_{\ell}+\beta_{\ell^{*}}+\frac{\delta_{\ell,1}+\delta_{\ell^{*},1}}{2n-2}}\right)+\beta_{\ell}\beta_{\ell^{*}}, (2.15)

where the right side is replaced by 0 if ℓ\ell or ℓ∗\ell^{*} equal 0. After cancelling, the left side of (2.15) reduces to

−β2​n−12​n−2​[(ℓ1)+(ℓ1∗)+(ℓ∗1)+(ℓ∗1∗)]\displaystyle-\frac{\beta_{2n-1}}{2n-2}\mathopen{}\mathclose{{}\left[\binom{\ell}{1}+\binom{\ell}{1^{*}}+\binom{\ell^{*}}{1}+\binom{\ell^{*}}{1^{*}}}\right]
+β2​n−1\displaystyle+\beta_{2n-1} [1+δℓ,0+δℓ∗,02−βℓ−βℓ∗]−(−1)ℓ​βℓ​βℓ∗,\displaystyle\mathopen{}\mathclose{{}\left[1+\frac{\delta_{\ell,0}+\delta_{\ell^{*},0}}{2}-\beta_{\ell}-\beta_{\ell^{*}}}\right]-(-1)^{\ell}\beta_{\ell}\beta_{\ell^{*}},

and it may be verified that this is 0 when ℓ=0\ell=0 or ℓ∗=0\ell^{*}=0, as desired. Cancelling further in (2.15) when 1⩽ℓ⩽1∗1\leqslant\ell\leqslant 1^{*} yields

12​n−2​[(ℓ1)+(ℓ1∗)+(ℓ∗1)+(ℓ∗1∗)]=1+δℓ,1+δℓ∗,12​n−2,\frac{1}{2n-2}\mathopen{}\mathclose{{}\left[\binom{\ell}{1}+\binom{\ell}{1^{*}}+\binom{\ell^{*}}{1}+\binom{\ell^{*}}{1^{*}}}\right]=1+\frac{\delta_{\ell,1}+\delta_{\ell^{*},1}}{2n-2},

which is indeed true. ∎

3 Expressing Ψn\Psi_{n} in terms of GnG_{n}

Any hyperbolic conjugacy class in Γ/{±I}{\Gamma}/\{\pm I\} has a representative

A=Tn1​Vm1​⋯​Tnr​Vmr,A=T^{n_{1}}V^{m_{1}}\cdots T^{n_{r}}V^{m_{r}}, (3.1)

with all powers positive and a total of p=n1+m1+⋯+nr+mrp=n_{1}+m_{1}+\cdots+n_{r}+m_{r} group elements. Cyclic permutations of (3.1) are made by removing elements from the left and placing them on the right. We obtain A~1=A\tilde{A}_{1}=A, A~2=T−1​A​T\tilde{A}_{2}=T^{-1}AT, and so on. Also let c~j\tilde{c}_{j} and b~j\tilde{b}_{j} be the integers cA~j/u​(A~j)c_{\tilde{A}_{j}}/{\text{\rm u}}(\tilde{A}_{j}) and bA~j/u​(A~j)b_{\tilde{A}_{j}}/{\text{\rm u}}(\tilde{A}_{j}), respectively.

Using V=T​S​TV=TST in (3.1) and replacing AA by T−1​A​TT^{-1}AT gives

A=Tk1​S​Tk2​S​⋯​S​Tkq​S,A=T^{k_{1}}ST^{k_{2}}S\cdots ST^{k_{q}}S, (3.2)

with all powers at least 22. Construct cyclic permutations of (3.2) slightly differently by removing blocks Tki​ST^{k_{i}}S from the left and placing them on the right. This gives A1=AA_{1}=A, A2=(Tk1​S)−1​A​Tk1​SA_{2}=(T^{k_{1}}S)^{-1}AT^{k_{1}}S, etc. Let ωj\omega_{j} and ωj′\omega^{\prime}_{j} be the fixed points of AjA_{j}. Also set cjc_{j} to be cAj/u​(Aj)c_{A_{j}}/{\text{\rm u}}(A_{j}). Note that since u is invariant under conjugation, [Duk22, Lemma 4], we have u​(Aj)=u​(A~j)=u​(A){\text{\rm u}}(A_{j})={\text{\rm u}}(\tilde{A}_{j})={\text{\rm u}}(A).

The key formula from Lemma 8 and Theorem 2 of [Duk22] is established there using an interesting two-variable version of Eichler-Shimura cohomology. With the above notation and for n⩾2n\geqslant 2, it says

Ψn​(A)= in​∑j=1p((c~j)n−1−(−b~j)n−1)+∑j=1qcjn−1​Gn​(ωj,ωj′).\Psi_{n}(A)={\text{\,\rm i}}_{n}\sum_{j=1}^{p}\mathopen{}\mathclose{{}\left((\tilde{c}_{j})^{n-1}-(-\tilde{b}_{j})^{n-1}}\right)+\sum_{j=1}^{q}c_{j}^{n-1}G_{n}(\omega_{j},\omega_{j}^{\prime}). (3.3)

Duke mentions in [Duk22, Sect. 2] that for each n⩾2n\geqslant 2, Ψn​(A)\Psi_{n}(A) always being an integer is implied by the polynomial GnG_{n} having integer coefficients. For completeness we prove this next.

Proposition 3.1.

For each n⩾2n\geqslant 2, we have that Ψn\Psi_{n} is integer valued if

Gn​(x)∈ℤ​[x1,x2].G_{n}(x)\in{\mathbb{Z}}[x_{1},x_{2}]. (3.4)
Proof.

Suppose that (3.4) is true. We may assume that AA is hyperbolic and, by (3.3), we need only show that cjn−1​Gn​(ωj,ωj′)c_{j}^{n-1}G_{n}(\omega_{j},\omega_{j}^{\prime}) is always an integer. The symmetric polynomial Gn​(x1,x2)G_{n}(x_{1},x_{2}) may be expressed as an element of ℤ​[e1,e2]{\mathbb{Z}}[e_{1},e_{2}] for the elementary symmetric polynomials e1=x1+x2e_{1}=x_{1}+x_{2} and e2=x1​x2e_{2}=x_{1}x_{2} . To see this explicitly, note that Gn​(x1,x2)G_{n}(x_{1},x_{2}) has terms of the form an integer times (x1​x2)m​(x1ℓ+x2ℓ)(x_{1}x_{2})^{m}(x_{1}^{\ell}+x_{2}^{\ell}) for m+ℓ⩽n−1m+\ell\leqslant n-1 and ℓ⩾1\ell\geqslant 1. This follows from (2.2) since GnG_{n} is a linear combination of the polynomials QrQ_{r} for rr odd. Then

(x1​x2)m​(x1ℓ+x2ℓ)=∑k=0ℓ/2(−1)k​ℓℓ−k​(ℓ−kk)​(x1+x2)ℓ−2​k​(x1​x2)m+k,(x_{1}x_{2})^{m}(x_{1}^{\ell}+x_{2}^{\ell})=\sum_{k=0}^{\ell/2}(-1)^{k}\frac{\ell}{\ell-k}\binom{\ell-k}{k}(x_{1}+x_{2})^{\ell-2k}(x_{1}x_{2})^{m+k}, (3.5)

by a classic result from symmetric functions [Gou99, Eq. (1)]. The coefficients in (3.5) are integers since

ℓℓ−k​(ℓ−kk)=(ℓ−kk)+(ℓ−k−1k−1)(ℓ>k>0).\frac{\ell}{\ell-k}\binom{\ell-k}{k}=\binom{\ell-k}{k}+\binom{\ell-k-1}{k-1}\qquad(\ell>k>0).

Write aja_{j}, bjb_{j} and djd_{j} for the other entries of AjA_{j} divided by u​(Aj){\text{\rm u}}(A_{j}). Along with cjc_{j}, these are all in ℤ{\mathbb{Z}} and we have

ωj+ωj′=aj−djcj,ωj​ωj′=−bjcj.\omega_{j}+\omega_{j}^{\prime}=\frac{a_{j}-d_{j}}{c_{j}},\qquad\omega_{j}\omega_{j}^{\prime}=-\frac{b_{j}}{c_{j}}.

Hence,

(ωj​ωj′)m​(ωjℓ+(ωj′)ℓ)=∑k=0ℓ/2(−1)k​ℓℓ−k​(ℓ−kk)​(aj−djcj)ℓ−2​k​(−bjcj)m+k.(\omega_{j}\omega_{j}^{\prime})^{m}(\omega_{j}^{\ell}+(\omega_{j}^{\prime})^{\ell})=\sum_{k=0}^{\ell/2}(-1)^{k}\frac{\ell}{\ell-k}\binom{\ell-k}{k}\mathopen{}\mathclose{{}\left(\frac{a_{j}-d_{j}}{c_{j}}}\right)^{\ell-2k}\mathopen{}\mathclose{{}\left(-\frac{b_{j}}{c_{j}}}\right)^{m+k}.

The largest possible power of cjc_{j} in the denominator of the right side is m+ℓm+\ell and this is cancelled by the outside factor cjn−1c_{j}^{n-1}. ∎

Conjecture 3.2.

[Duk22, Sect. 2] For every n⩾2n\geqslant 2 we have Gn​(x)∈ℤ​[x1,x2]G_{n}(x)\in{\mathbb{Z}}[x_{1},x_{2}].

So we see that Conjecture 3.2 implies Conjecture 1.2. It seems that Conjecture 3.2 is strictly stronger since a polynomial may send ℤ{\mathbb{Z}} to ℤ{\mathbb{Z}} without having integer coefficients, (for example (x3)\binom{x}{3}). Now, by (2.2), (2.9) we have Gn​(x)∈ℤ​[x1,x2]G_{n}(x)\in{\mathbb{Z}}[x_{1},x_{2}] if and only if

 jn​(n−1u)​(n−1r−u)​ξr∈ℤfor all u, r with 0⩽u⩽r⩽2​n−2.{\text{\,\rm j}}_{n}\binom{n-1}{u}\binom{n-1}{r-u}\xi_{r}\in{\mathbb{Z}}\qquad\text{for all $u$, $r$ with $0\leqslant u\leqslant r\leqslant 2n-2$}. (3.6)

As the expression in (3.6) is unchanged as r→r∗r\to r^{*} we need only consider odd rr in the range 1⩽r⩽n−11\leqslant r\leqslant n-1. Substituting (2.8), (2.7) and simplifying a little finds:

Proposition 3.3.

For all n⩾3n\geqslant 3, Gn​(x)∈ℤ​[x1,x2]G_{n}(x)\in{\mathbb{Z}}[x_{1},x_{2}] is equivalent to

 jn​(n−1u)​(n−1v)​(B2​n2​n​(B2​w2​w+B2​n−2​w2​n−2​w)−B2​w2​w​B2​n−2​w2​n−2​w)+δw,12∈ℤ,{\text{\,\rm j}}_{n}\binom{n-1}{u}\binom{n-1}{v}\mathopen{}\mathclose{{}\left(\frac{B_{2n}}{2n}\mathopen{}\mathclose{{}\left(\frac{B_{2w}}{2w}+\frac{B_{2n-2w}}{2n-2w}}\right)-\frac{B_{2w}}{2w}\frac{B_{2n-2w}}{2n-2w}}\right)+\frac{\delta_{w,1}}{2}\in{\mathbb{Z}}, (3.7)

for all nonnegative integers uu, vv, ww with 1⩽w⩽n/21\leqslant w\leqslant n/2 and u+v=2​w−1u+v=2w-1.

Studying (3.7) numerically reveals that the binomial coefficients are needed to cancel denominators in a small proportion of cases. For r∈ℚ≠0r\in{\mathbb{Q}}_{\neq 0}, let νp​(r)\nu_{p}(r) be the usual pp-adic valuation: νp​(r)=m∈ℤ\nu_{p}(r)=m\in{\mathbb{Z}} when r=pm​α/βr=p^{m}\alpha/\beta for p∤α,βp\nmid\alpha,\beta.

Lemma 3.4.

Let uu, vv, ww be integers with 0⩽u,v⩽n−10\leqslant u,v\leqslant n-1 and u+v=2​w−1u+v=2w-1 for 1⩽w⩽n−11\leqslant w\leqslant n-1. Then

νp​((n−1u)​(n−1v))\displaystyle\nu_{p}\mathopen{}\mathclose{{}\left(\binom{n-1}{u}\binom{n-1}{v}}\right) ⩾{νp​(2​w)+2−(2​n−2​w)p−1,νp​(2​n−2​w)+2−2​wp−1,\displaystyle\geqslant\begin{cases}\nu_{p}(2w)+\frac{2-(2n-2w)}{p-1},\\ \nu_{p}(2n-2w)+\frac{2-2w}{p-1},\end{cases} (3.8)
for pp an odd prime, and
ν2​((n−1u)​(n−1v))\displaystyle\nu_{2}\mathopen{}\mathclose{{}\left(\binom{n-1}{u}\binom{n-1}{v}}\right) ⩾{ν2​(2​w)−(n−w),ν2​(2​n−2​w)−w.\displaystyle\geqslant\begin{cases}\nu_{2}(2w)-(n-w),\\ \nu_{2}(2n-2w)-w.\end{cases} (3.9)
Proof.

It is easy to verify (3.8) and (3.9) when w=1w=1, so we assume w⩾2w\geqslant 2. Without losing generality, suppose u⩾vu\geqslant v. Then u⩾wu\geqslant w and it follows that n−wn-w divides the numerator (n−1)​(n−2)​⋯​(n−u)(n-1)(n-2)\cdots(n-u) of (n−1u)\binom{n-1}{u}, written with denominator u!u!. Hence, the left sides of (3.8) and (3.9) are at least νp​(n−w)−νp​(u!)−νp​(v!)\nu_{p}(n-w)-\nu_{p}(u!)-\nu_{p}(v!). The well-known formula (p−1)​νp​(m!)=m−sp​(m)(p-1)\nu_{p}(m!)=m-s_{p}(m) has sp​(m)s_{p}(m) equalling the sum of the base pp digits of mm. Therefore

νp​(n−w)−νp​(u!)−νp​(v!)\displaystyle\nu_{p}(n-w)-\nu_{p}(u!)-\nu_{p}(v!) =νp​(n−w)−u+v−sp​(u)−sp​(v)p−1\displaystyle=\nu_{p}(n-w)-\frac{u+v-s_{p}(u)-s_{p}(v)}{p-1}
⩾νp​(n−w)−(2​w−1)−1p−1,\displaystyle\geqslant\nu_{p}(n-w)-\frac{(2w-1)-1}{p-1},

proving the bottom inequality in (3.8).

For p=2p=2 we may improve this slightly. Note that s2​(u)+s2​(v)=1s_{2}(u)+s_{2}(v)=1 only for v=0v=0 and uu a power of 22. But if v=0v=0 then u=2​w−1⩾3u=2w-1\geqslant 3 is not a power of 22. Therefore s2​(u)+s2​(v)⩾2s_{2}(u)+s_{2}(v)\geqslant 2. Also, considering the numerators of (n−1u)\binom{n-1}{u} and (n−1v)\binom{n-1}{v}, excluding the n−wn-w factor, they must contain at least ⌊u/2⌋+⌊v/2⌋−1=w−2\lfloor u/2\rfloor+\lfloor v/2\rfloor-1=w-2 twos. Then the left side of (3.9) is at least

ν2​(n−w)+(w−2)−(2​w−1)+2=ν2​(2​(n−w))−w.\nu_{2}(n-w)+(w-2)-(2w-1)+2=\nu_{2}(2(n-w))-w.

Under the maps u↦n−1−uu\mapsto n-1-u and v↦n−1−vv\mapsto n-1-v we have w↦n−ww\mapsto n-w. This symmetry gives the top inequalities in (3.8) and (3.9). ∎

4 Divisibility of Bernoulli numbers

The Bernoulli number results we need in the next section are described here. The following one is due to Clarke in [Cla89, Prop. 8], explicitly determining the fractional part of Bn/nB_{n}/n.

Theorem 4.1.

For nn even and positive,

Bnn=z1​(n)+∑p−1∣nzp​(n)p1+νp​(n),\frac{B_{n}}{n}=z_{1}(n)+\sum_{p-1\mid n}\frac{z_{p}(n)}{p^{1+\nu_{p}(n)}}, (4.1)

where we are summing over primes pp. Here z1​(n)z_{1}(n) and zp​(n)z_{p}(n) are integers with 1⩽zp​(n)<p1+νp​(n)1\leqslant z_{p}(n)<p^{1+\nu_{p}(n)}. Precisely,

zp​(n)≡(2​δp,2​δn,2+(n(p−1)​pνp​(n))−1)modp1+νp​(n).z_{p}(n)\equiv\mathopen{}\mathclose{{}\left(2\delta_{p,2}\delta_{n,2}+\mathopen{}\mathclose{{}\left(\frac{n}{(p-1)p^{\nu_{p}(n)}}}\right)^{-1}}\right)\bmod p^{1+\nu_{p}(n)}. (4.2)

Our formulation in (4.1) and (4.2) slightly simplifies [Cla89, Prop. 8] and includes the case n=2n=2. Since gcd⁡(p,zp​(n))=1\gcd(p,z_{p}(n))=1, Von Staudt’s second theorem is a corollary: for even n⩾2n\geqslant 2, the denominator of Bn/nB_{n}/n is

 jn/2=∏p−1∣np1+νp​(n).{\text{\,\rm j}}_{n/2}=\prod_{p-1\mid n}p^{1+\nu_{p}(n)}. (4.3)

The next result gives detailed information about differences of Bernoulli numbers and is contained in [Coh07, Prop. 11.4.4]. First set

Hp​(n,m):={(1−1p)​(1n−1m)−(n−m)​δp,3if p−1∣n,0if p−1∤n.H_{p}(n,m):=\begin{cases}\Bigl{(}1-\frac{1}{p}\Bigr{)}\Bigl{(}\frac{1}{n}-\frac{1}{m}\Bigr{)}-(n-m)\delta_{p,3}&\text{if \ $p-1\mid n$},\\ 0&\text{if \ $p-1\nmid n$}.\end{cases}
Theorem 4.2.

Let nn and mm be even and positive. If pp is prime and n≡mmod(p−1)​pNn\equiv m\bmod(p-1)p^{N} then

(1−pn−1)​Bnn−(1−pm−1)​Bmm=Hp​(n,m)+αβ​pN+1,\mathopen{}\mathclose{{}\left(1-p^{n-1}}\right)\frac{B_{n}}{n}-\mathopen{}\mathclose{{}\left(1-p^{m-1}}\right)\frac{B_{m}}{m}=H_{p}(n,m)+\frac{\alpha}{\beta}p^{N+1}, (4.4)

for some rational α/β\alpha/\beta with p∤βp\nmid\beta.

The cases of this theorem when p−1∤np-1\nmid n are known as the classical Kummer congruences. They were extended by Cohen to p−1∣np-1\mid n using the theory of the Kubota-Leopoldt pp-adic zeta function in Chapter 11 of [Coh07].

5 Proof of Conjecture 3.2

Theorem 5.1.

Let kk and ℓ\ell be positive and even with k⩽ℓk\leqslant\ell and m=k+ℓ⩾6m=k+\ell\geqslant 6. Then for all nonnegative integers uu, vv with u+v=k−1u+v=k-1,

 jm/2​(m/2−1u)​(m/2−1v)​(Bmm​(Bℓℓ+Bkk)−Bℓℓ​Bkk)+δk,22∈ℤ.{\text{\,\rm j}}_{m/2}\binom{m/2-1}{u}\binom{m/2-1}{v}\mathopen{}\mathclose{{}\left(\frac{B_{m}}{m}\mathopen{}\mathclose{{}\left(\frac{B_{\ell}}{\ell}+\frac{B_{k}}{k}}\right)-\frac{B_{\ell}}{\ell}\frac{B_{k}}{k}}\right)+\frac{\delta_{k,2}}{2}\in{\mathbb{Z}}. (5.1)
Proof.

Let PP be the set of primes pp such that p−1p-1 divides all three of kk, ℓ\ell and mm. Then {2,3}⊆P\{2,3\}\subseteq P. Let KK, LL and MM be the primes with p−1p-1 dividing only kk, ℓ\ell and mm, respectively. (If p−1p-1 divides any two of kk, ℓ\ell and mm, then it must divide the third.) With Theorem 4.1, write

Bmm\displaystyle\frac{B_{m}}{m} =a1+∑p∈Papp1+νp​(m)+∑p∈Mapp1+νp​(m),\displaystyle=a_{1}+\sum_{p\in P}\frac{a_{p}}{p^{1+\nu_{p}(m)}}+\sum_{p\in M}\frac{a_{p}}{p^{1+\nu_{p}(m)}},
Bℓℓ\displaystyle\frac{B_{\ell}}{\ell} =b1+∑p∈Pbpp1+νp​(ℓ)+∑p∈Lbpp1+νp​(ℓ),\displaystyle=b_{1}+\sum_{p\in P}\frac{b_{p}}{p^{1+\nu_{p}(\ell)}}+\sum_{p\in L}\frac{b_{p}}{p^{1+\nu_{p}(\ell)}},
Bkk\displaystyle\frac{B_{k}}{k} =c1+∑p∈Pcpp1+νp​(k)+∑p∈Kcpp1+νp​(k).\displaystyle=c_{1}+\sum_{p\in P}\frac{c_{p}}{p^{1+\nu_{p}(k)}}+\sum_{p\in K}\frac{c_{p}}{p^{1+\nu_{p}(k)}}.

Our goal is to show that the left side of (5.1) is pp-integral, (has pp-adic valuation at least 0), for all pp in PP, MM, LL and KK. Note that, by (4.3),

 jm/2=∏p∈P∪Mp1+νp​(m).{\text{\,\rm j}}_{m/2}=\prod_{p\in P\,\cup\,M}p^{1+\nu_{p}(m)}.

Primes in M\bm{M}. The case p∈Mp\in M is easily dealt with: the p1+νp​(m)p^{1+\nu_{p}(m)} factor from  jm/2{\text{\,\rm j}}_{m/2} cancels the same factor only appearing in the denominator of Bm/mB_{m}/m. So the left side of (5.1) is pp-integral.

Primes in L\bm{L}, K\bm{K}. Next take p∈Lp\in L. Express the Bernoulli number part of (5.1) as

Bℓℓ​(Bmm−Bkk)+Bmm​Bkk.\frac{B_{\ell}}{\ell}\mathopen{}\mathclose{{}\left(\frac{B_{m}}{m}-\frac{B_{k}}{k}}\right)+\frac{B_{m}}{m}\frac{B_{k}}{k}. (5.2)

Then p−1p-1 does not divide mm and kk, while p−1p-1 does divide m−k=ℓm-k=\ell with

m≡kmod(p−1)​pνp​(ℓ).m\equiv k\bmod(p-1)p^{\nu_{p}(\ell)}.

By Theorem 4.2,

(1−pm−1)​Bmm−(1−pk−1)​Bkk=Hp​(m,k)+αβ​p1+νp​(ℓ),\mathopen{}\mathclose{{}\left(1-p^{m-1}}\right)\frac{B_{m}}{m}-\mathopen{}\mathclose{{}\left(1-p^{k-1}}\right)\frac{B_{k}}{k}=H_{p}(m,k)+\frac{\alpha}{\beta}p^{1+\nu_{p}(\ell)}, (5.3)

for rational α/β\alpha/\beta with p∤βp\nmid\beta and Hp​(m,k)=0H_{p}(m,k)=0. Hence

1p1+νp​(ℓ)​(Bmm−Bkk)=αβ+pm−2−νp​(ℓ)​Bmm−pk−2−νp​(ℓ)​Bkk.\frac{1}{p^{1+\nu_{p}(\ell)}}\mathopen{}\mathclose{{}\left(\frac{B_{m}}{m}-\frac{B_{k}}{k}}\right)=\frac{\alpha}{\beta}+p^{m-2-\nu_{p}(\ell)}\frac{B_{m}}{m}-p^{k-2-\nu_{p}(\ell)}\frac{B_{k}}{k}. (5.4)

If k−2−νp​(ℓ)<0k-2-\nu_{p}(\ell)<0 then multiplication by the binomial coefficients ensures that the left side of (5.1) is pp-integral, since

νp​((m/2−1u)​(m/2−1v))⩾νp​(ℓ)+2−kp−1⩾νp​(ℓ)+2−k,\nu_{p}\mathopen{}\mathclose{{}\left(\binom{m/2-1}{u}\binom{m/2-1}{v}}\right)\geqslant\nu_{p}(\ell)+\frac{2-k}{p-1}\geqslant\nu_{p}(\ell)+2-k, (5.5)

by (3.8) in Lemma 3.4. The case p∈Kp\in K is handled in the same way, using the rearrangement

Bkk​(Bmm−Bℓℓ)+Bmm​Bℓℓ.\frac{B_{k}}{k}\mathopen{}\mathclose{{}\left(\frac{B_{m}}{m}-\frac{B_{\ell}}{\ell}}\right)+\frac{B_{m}}{m}\frac{B_{\ell}}{\ell}. (5.6)

Primes in P−{𝟐}\bm{P-\{2\}} with νp​(m)\bm{\nu_{p}(m)} small. That leaves the primes in PP to check. First take pp in the case where N:=νp​(ℓ)N:=\nu_{p}(\ell) is strictly larger than νp​(m)\nu_{p}(m) and νp​(k)\nu_{p}(k). Then necessarily νp​(m)=νp​(k)\nu_{p}(m)=\nu_{p}(k) and let xx be this common valuation. Assume that p≠2p\neq 2. The only terms of (5.2) that can possibly be non pp-integral after multiplication by  jm/2{\text{\,\rm j}}_{m/2} are

bppN+1​(Bmm−Bkk)+appx+1​cppx+1.\frac{b_{p}}{p^{N+1}}\mathopen{}\mathclose{{}\left(\frac{B_{m}}{m}-\frac{B_{k}}{k}}\right)+\frac{a_{p}}{p^{x+1}}\frac{c_{p}}{p^{x+1}}.

Here  jm/2{\text{\,\rm j}}_{m/2} contains a px+1p^{x+1} factor, so with Theorem 4.2 and (5.3) again, we wish to show a nonnegative valuation for

px+1​(bppN+1​Hp​(m,k)+bp​αβ+bp​pm−2−N​Bmm−bp​pk−2−N​Bkk+ap​cpp2​x+2)p^{x+1}\mathopen{}\mathclose{{}\left(\frac{b_{p}}{p^{N+1}}H_{p}(m,k)+b_{p}\frac{\alpha}{\beta}+b_{p}p^{m-2-N}\frac{B_{m}}{m}-b_{p}p^{k-2-N}\frac{B_{k}}{k}+\frac{a_{p}c_{p}}{p^{2x+2}}}\right) (5.7)

multiplied by the binomial coefficients. The terms containing Bm/mB_{m}/m and Bk/kB_{k}/k have total valuation at least k−2−Nk-2-N. If k−2−N<0k-2-N<0 then the binomial coefficients, with valuation at least

νp​(ℓ)+2−kp−1⩾N+1−k/2\nu_{p}(\ell)+\frac{2-k}{p-1}\geqslant N+1-k/2

by (3.8), ensure pp-integrality. Write the integers

m′=m(p−1)​px,k′=k(p−1)​px,ℓ′=ℓ(p−1)​pN,m^{\prime}=\frac{m}{(p-1)p^{x}},\qquad k^{\prime}=\frac{k}{(p-1)p^{x}},\qquad\ell^{\prime}=\frac{\ell}{(p-1)p^{N}},

so that

Hp​(m,k)=−ℓ′m′​k′​pN−1−2​x−ℓ′​(p−1)​pN​δp,3.H_{p}(m,k)=-\frac{\ell^{\prime}}{m^{\prime}k^{\prime}}p^{N-1-2x}-\ell^{\prime}(p-1)p^{N}\delta_{p,3}.

Then (5.7) without the α/β\alpha/\beta, Bm/mB_{m}/m and Bk/kB_{k}/k terms equals

−bp​ℓ′​(p−1)​px​δp,3+ap​cppx+1−bp​ℓ′m′​k′​px+1-b_{p}\ell^{\prime}(p-1)p^{x}\delta_{p,3}+\frac{a_{p}c_{p}}{p^{x+1}}-\frac{b_{p}\ell^{\prime}}{m^{\prime}k^{\prime}p^{x+1}}

and it remains to show νp​(m′​k′​ap​cp−bp​ℓ′)⩾x+1\nu_{p}\mathopen{}\mathclose{{}\left(m^{\prime}k^{\prime}a_{p}c_{p}-b_{p}\ell^{\prime}}\right)\geqslant x+1. But by (4.2) we know m′​ap≡k′​cp≡ℓ′​bp≡1modpx+1m^{\prime}a_{p}\equiv k^{\prime}c_{p}\equiv\ell^{\prime}b_{p}\equiv 1\bmod p^{x+1}, and hence

m′​ap⋅k′​cp−ℓ′​bp≡1⋅1−1≡0modpx+1,m^{\prime}a_{p}\cdot k^{\prime}c_{p}-\ell^{\prime}b_{p}\equiv 1\cdot 1-1\equiv 0\bmod p^{x+1},

as we wanted.

The case 𝟐∈P\bm{2\in P} with ν𝟐​(m)\bm{\nu_{2}(m)} small. We continue the above analysis for p=2p=2, and first suppose that k≠2k\neq 2. The arguments go through as before, with the only difference being that the binomial coefficients now just have valuation at least N−k/2N-k/2 by (3.9). This ensures pp-integrality of the terms in (5.7) containing Bm/mB_{m}/m and Bk/kB_{k}/k for k⩾4k\geqslant 4.

If k=2k=2 then x=1x=1 and we must deal with the Bk/kB_{k}/k term in (5.7) as well as the extra terms δk,2/2\delta_{k,2}/2 from (5.1) and a2/2a_{2}/2 from the 2​δp,2​δk,22\delta_{p,2}\delta_{k,2} part of cpc_{p} in (4.2). Altogether, including the binomial coefficients factor, this is

(m2−1)​(a22−b23⋅2N)+12\displaystyle\mathopen{}\mathclose{{}\left(\frac{m}{2}-1}\right)\mathopen{}\mathclose{{}\left(\frac{a_{2}}{2}-\frac{b_{2}}{3\cdot 2^{N}}}\right)+\frac{1}{2} =ℓ′​2N−1​a22−ℓ′​2N−1​b23⋅2N+12\displaystyle=\ell^{\prime}2^{N-1}\frac{a_{2}}{2}-\ell^{\prime}2^{N-1}\frac{b_{2}}{3\cdot 2^{N}}+\frac{1}{2}
=ℓ′​2N−2​a2+3−ℓ′​b26.\displaystyle=\ell^{\prime}2^{N-2}a_{2}+\frac{3-\ell^{\prime}b_{2}}{6}.

This is 22-integral since N>x=1N>x=1 and ℓ′​b2≡1mod4\ell^{\prime}b_{2}\equiv 1\bmod 4.

It follows that the left side of (5.1) is pp-integral for all p∈Pp\in P where νp​(ℓ)\nu_{p}(\ell) is strictly greater than νp​(m)\nu_{p}(m) and νp​(k)\nu_{p}(k). Clearly the same argument works if νp​(k)\nu_{p}(k) is strictly greater than νp​(m)\nu_{p}(m) and νp​(ℓ)\nu_{p}(\ell).

Primes in P\bm{P} with νp​(m)\bm{\nu_{p}(m)} large. The final primes to check are p∈Pp\in P with νp​(ℓ)=νp​(k)=x\nu_{p}(\ell)=\nu_{p}(k)=x and νp​(m)=N⩾x\nu_{p}(m)=N\geqslant x. Assume that pp and kk are not both 22. Now  jm/2{\text{\,\rm j}}_{m/2} contains the factor pN+1p^{N+1}, and the only terms of (5.1) that can be non pp-integral are

pN+1​(appN+1​bppx+1+appN+1​cppx+1−bppx+1​cppx+1),p^{N+1}\mathopen{}\mathclose{{}\left(\frac{a_{p}}{p^{N+1}}\frac{b_{p}}{p^{x+1}}+\frac{a_{p}}{p^{N+1}}\frac{c_{p}}{p^{x+1}}-\frac{b_{p}}{p^{x+1}}\frac{c_{p}}{p^{x+1}}}\right), (5.8)

omitting the binomial coefficient factors. This time we need the integers

k′=k(p−1)​px,ℓ′=ℓ(p−1)​px,m′=m(p−1)​pN.k^{\prime}=\frac{k}{(p-1)p^{x}},\qquad\ell^{\prime}=\frac{\ell}{(p-1)p^{x}},\qquad m^{\prime}=\frac{m}{(p-1)p^{N}}.

Then m′​pN−x=k′+ℓ′m^{\prime}p^{N-x}=k^{\prime}+\ell^{\prime} and m′​ap≡ℓ′​bp≡k′​cp≡1modpx+1m^{\prime}a_{p}\equiv\ell^{\prime}b_{p}\equiv k^{\prime}c_{p}\equiv 1\bmod p^{x+1}. We claim that px+1p^{x+1} divides the numerator of

ap​bp+ap​cp−bp​cp​pN−xpx+1\frac{a_{p}b_{p}+a_{p}c_{p}-b_{p}c_{p}p^{N-x}}{p^{x+1}} (5.9)

from (5.8). Evidently,

m′​ℓ′​k′​(ap​bp+ap​cp−bp​cp​pN−x)\displaystyle m^{\prime}\ell^{\prime}k^{\prime}\mathopen{}\mathclose{{}\left(a_{p}b_{p}+a_{p}c_{p}-b_{p}c_{p}p^{N-x}}\right) =m′​ap⋅ℓ′​bp⋅k′+m′​ap⋅k′​cp⋅ℓ′−ℓ′​bp⋅k′​cp⋅m′​pN−x\displaystyle=m^{\prime}a_{p}\cdot\ell^{\prime}b_{p}\cdot k^{\prime}+m^{\prime}a_{p}\cdot k^{\prime}c_{p}\cdot\ell^{\prime}-\ell^{\prime}b_{p}\cdot k^{\prime}c_{p}\cdot m^{\prime}p^{N-x}
≡1⋅1⋅k′+1⋅1⋅ℓ′−1⋅1⋅(k′+ℓ′)≡0modpx+1.\displaystyle\equiv 1\cdot 1\cdot k^{\prime}+1\cdot 1\cdot\ell^{\prime}-1\cdot 1\cdot(k^{\prime}+\ell^{\prime})\equiv 0\bmod p^{x+1}.

This implies our claim, since m′​ℓ′​k′m^{\prime}\ell^{\prime}k^{\prime} is relatively prime to pp, and so (5.8) is pp-integral, as desired.

The case 𝟐∈P\bm{2\in P}, k=𝟐\bm{k=2} with ν𝟐​(m)\bm{\nu_{2}(m)} large. The last case of p=2p=2 and k=2k=2 has some further terms to examine. Here x=1x=1 and ℓ>2\ell>2. The extra possibly non 22-integral piece of (5.1), corresponding to the 2​δp,2​δk,22\delta_{p,2}\delta_{k,2} part of cpc_{p} in (4.2), and the δk,2/2\delta_{k,2}/2 term, is

(m2−1)​a2−b2⋅2N−x2+12,\mathopen{}\mathclose{{}\left(\frac{m}{2}-1}\right)\frac{a_{2}-b_{2}\cdot 2^{N-x}}{2}+\frac{1}{2}, (5.10)

for x=1x=1. Since m/2−1=ℓ/2m/2-1=\ell/2 is odd, then (5.10) is an integer if we can show that a2−b2⋅2N−xa_{2}-b_{2}\cdot 2^{N-x} is odd. To verify this, use the congruences before (5.9) to show

m′​ℓ′​(a2−b2⋅2N−x)\displaystyle m^{\prime}\ell^{\prime}(a_{2}-b_{2}\cdot 2^{N-x}) =m′​a2⋅ℓ′−ℓ′​b2⋅m′​2N−x\displaystyle=m^{\prime}a_{2}\cdot\ell^{\prime}-\ell^{\prime}b_{2}\cdot m^{\prime}2^{N-x}
≡1⋅ℓ′−1⋅(k′+ℓ′)mod4\displaystyle\equiv 1\cdot\ell^{\prime}-1\cdot(k^{\prime}+\ell^{\prime})\bmod 4
≡−k′mod4,\displaystyle\equiv-k^{\prime}\bmod 4,

for the odd numbers m′m^{\prime}, ℓ′\ell^{\prime} and k′k^{\prime}. This completes the proof of Theorem 5.1. ∎

Corollary 5.2.

Conjectures 1.2 and 3.2 are true.

Proof.

Theorem 5.1 and Proposition 3.3 imply Conjecture 3.2. This in turn implies Conjecture 1.2 via Proposition 3.1. ∎

6 Further formulas for Ψn\Psi_{n}

For A=(abcd)A=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}}\right) in Γ{\Gamma}, set jA​(z):=c​z+dj_{A}(z):=cz+d and define the quadratic form

qA​(z):=(A​z−z)​jA​(z)=c​z2+(d−a)​z−b.q_{A}(z):=(Az-z)j_{A}(z)=cz^{2}+(d-a)z-b.

Then

qA​(B​z)⋅jB​(z)2=qB−1​A​B​(z)q_{A}(Bz)\cdot j_{B}(z)^{2}=q_{B^{-1}AB}(z)

and hence

qA​(z+k)=qT−k​A​Tk​(z),qA​(−1/z)⋅z2=qS−1​A​S​(z).q_{A}(z+k)=q_{T^{-k}AT^{k}}(z),\qquad q_{A}(-1/z)\cdot z^{2}=q_{S^{-1}AS}(z). (6.1)

Next, for fixed nn, set

fr​(A):=[zr]​qA​(z)n−1,{f}_{r}(A):=[z^{r}]q_{A}(z)^{n-1}, (6.2)

so that, for example,

f0​(A)=(−b)n−1,\displaystyle{f}_{0}(A)=(-b)^{n-1}, f1​(A)=(n−1)​(−b)n−2​(d−a),\displaystyle\qquad{f}_{1}(A)=(n-1)(-b)^{n-2}(d-a), (6.3)
f0∗​(A)=cn−1,\displaystyle{f}_{0^{*}}(A)=c^{n-1}, f1∗​(A)=(n−1)​cn−2​(d−a).\displaystyle\qquad{f}_{1^{*}}(A)=(n-1)c^{n-2}(d-a). (6.4)

Following from (6.1),

fr​(T−k​A​Tk)=∑m=02​n−2(mr)​km−r​fm​(A),fr​(S−1​A​S)=(−1)r​fr∗​(A).{f}_{r}(T^{-k}AT^{k})=\sum_{m=0}^{2n-2}\binom{m}{r}k^{m-r}{f}_{m}(A),\qquad{f}_{r}(S^{-1}AS)=(-1)^{r}{f}_{r^{*}}(A). (6.5)

Set

Ψn(0)​(A):=u​(A)1−n​∑j=1p(f0∗​(A~j)−f0​(A~j))=∑j=1p((c~j)n−1−(−b~j)n−1),\Psi_{n}^{(0)}(A):={\text{\rm u}}(A)^{1-n}\sum_{j=1}^{p}\mathopen{}\mathclose{{}\left({f}_{0^{*}}(\tilde{A}_{j})-{f}_{0}(\tilde{A}_{j})}\right)=\sum_{j=1}^{p}\mathopen{}\mathclose{{}\left((\tilde{c}_{j})^{n-1}-(-\tilde{b}_{j})^{n-1}}\right),

as in the first part of (3.3) and using its notation. The next result is stated in [Duk22, Eq. (2.21)] and we give a direct proof.

Theorem 6.1.

We have

Ψn(0)​(A)=u​(A)1−n​∑j=1q∑r=02​n−2(Br+1r+1+Br∗+1r∗+1−kjr+1r+1+δ​(r))​fr​(Aj),\Psi_{n}^{(0)}(A)={\text{\rm u}}(A)^{1-n}\sum_{j=1}^{q}\sum_{r=0}^{2n-2}\mathopen{}\mathclose{{}\left(\frac{B_{r+1}}{r+1}+\frac{B_{r^{*}+1}}{r^{*}+1}-\frac{k_{j}^{r+1}}{r+1}+\delta(r)}\right){f}_{r}(A_{j}), (6.6)

for δ​(r)=(δr,0+δr∗,0)/2−(δr,1+δr∗,1)/(2​n−2)\delta(r)=(\delta_{r,0}+\delta_{r^{*},0})/2-(\delta_{r,1}+\delta_{r^{*},1})/(2n-2).

Proof.

To translate between (3.1) and (3.2), rewrite (3.1) as

A=Tk1−2​V​Tk2−2​V​⋯​V​Tkq−2​VA=T^{k_{1}-2}VT^{k_{2}-2}V\cdots VT^{k_{q}-2}V

for ki⩾2k_{i}\geqslant 2. Then T​A​T−1TAT^{-1} equals the right side of (3.2) and we see

A~1=T−1​A1​T,A~(k1−1)+1=T−1​A2​T,A~(k1−1)+(k2−1)+1=T−1​A3​T,\tilde{A}_{1}=T^{-1}A_{1}T,\qquad\tilde{A}_{(k_{1}-1)+1}=T^{-1}A_{2}T,\qquad\tilde{A}_{(k_{1}-1)+(k_{2}-1)+1}=T^{-1}A_{3}T,

and so on. Set s0:=0s_{0}:=0 and sj:=(k1−1)+(k2−1)+⋯+(kj−1)s_{j}:=(k_{1}-1)+(k_{2}-1)+\cdots+(k_{j}-1). Then

A~sj−1+u=T−u​Aj​Tufor1⩽j⩽q,1⩽u⩽kj−1.\tilde{A}_{s_{j-1}+u}=T^{-u}A_{j}T^{u}\qquad\text{for}\quad 1\leqslant j\leqslant q,\quad 1\leqslant u\leqslant k_{j}-1.

Hence

∑j=1p(f0∗​(A~j)−f0​(A~j))\displaystyle\sum_{j=1}^{p}\mathopen{}\mathclose{{}\left({f}_{0^{*}}(\tilde{A}_{j})-{f}_{0}(\tilde{A}_{j})}\right) =∑j=1q∑u=1kj−1(f0∗​(T−u​Aj​Tu)−f0​(T−u​Aj​Tu))\displaystyle=\sum_{j=1}^{q}\sum_{u=1}^{k_{j}-1}\mathopen{}\mathclose{{}\left({f}_{0^{*}}(T^{-u}A_{j}T^{u})-{f}_{0}(T^{-u}A_{j}T^{u})}\right)
=∑j=1q∑u=1kj−1(f0∗​(Aj)−∑r=02​n−2ur​fr​(Aj)),\displaystyle=\sum_{j=1}^{q}\sum_{u=1}^{k_{j}-1}\mathopen{}\mathclose{{}\left({f}_{0^{*}}(A_{j})-\sum_{r=0}^{2n-2}u^{r}{f}_{r}(A_{j})}\right),

employing the left side of (6.5). Let

Sr​(k):=1r+2r+⋯+kr=1r+1​∑ℓ=0r(r+1ℓ)​Bℓ⋅(k+1)r+1−ℓ−δr,0,S_{r}(k):=1^{r}+2^{r}+\cdots+k^{r}=\frac{1}{r+1}\sum_{\ell=0}^{r}\binom{r+1}{\ell}B_{\ell}\cdot(k+1)^{r+1-\ell}-\delta_{r,0},

using our last required property of the Bernoulli numbers, as in [AD08, Eq. (2.2)]. Then

∑j=1p(f0∗​(A~j)−f0​(A~j))=∑j=1q((kj−1)​f0∗​(Aj)−∑r=02​n−2Sr​(kj−1)​fr​(Aj)).\sum_{j=1}^{p}\mathopen{}\mathclose{{}\left({f}_{0^{*}}(\tilde{A}_{j})-{f}_{0}(\tilde{A}_{j})}\right)=\sum_{j=1}^{q}\mathopen{}\mathclose{{}\left((k_{j}-1){f}_{0^{*}}(A_{j})-\sum_{r=0}^{2n-2}S_{r}(k_{j}-1){f}_{r}(A_{j})}\right).

Focussing on the sum over rr finds

−∑r=02​n−2Sr​(kj−1)​fr​(Aj)=f0​(Aj)−∑r=02​n−21r+1​∑ℓ=0r(r+1ℓ)​Bℓ​kjr+1−ℓ​fr​(Aj)=f0​(Aj)−∑r=02​n−2kjr+1r+1​fr​(Aj)−∑r=02​n−21r+1​∑ℓ=1r(r+1ℓ)​Bℓ​kjr+1−ℓ​fr​(Aj).-\sum_{r=0}^{2n-2}S_{r}(k_{j}-1){f}_{r}(A_{j})={f}_{0}(A_{j})-\sum_{r=0}^{2n-2}\frac{1}{r+1}\sum_{\ell=0}^{r}\binom{r+1}{\ell}B_{\ell}k_{j}^{r+1-\ell}{f}_{r}(A_{j})\\ ={f}_{0}(A_{j})-\sum_{r=0}^{2n-2}\frac{k_{j}^{r+1}}{r+1}{f}_{r}(A_{j})-\sum_{r=0}^{2n-2}\frac{1}{r+1}\sum_{\ell=1}^{r}\binom{r+1}{\ell}B_{\ell}k_{j}^{r+1-\ell}{f}_{r}(A_{j}).

Write the third term above as

−∑ℓ=02​n−2Bℓ+1ℓ+1​∑r=ℓ+12​n−2(rℓ)​kjr−ℓ​fr​(Aj)=∑ℓ=02​n−2Bℓ+1ℓ+1​fℓ​(Aj)−∑ℓ=02​n−2Bℓ+1ℓ+1​∑r=ℓ2​n−2(rℓ)​kjr−ℓ​fr​(Aj).-\sum_{\ell=0}^{2n-2}\frac{B_{\ell+1}}{\ell+1}\sum_{r=\ell+1}^{2n-2}\binom{r}{\ell}k_{j}^{r-\ell}{f}_{r}(A_{j})=\sum_{\ell=0}^{2n-2}\frac{B_{\ell+1}}{\ell+1}{f}_{\ell}(A_{j})-\sum_{\ell=0}^{2n-2}\frac{B_{\ell+1}}{\ell+1}\sum_{r=\ell}^{2n-2}\binom{r}{\ell}k_{j}^{r-\ell}{f}_{r}(A_{j}).

With (6.5), the last term is recognized as

−∑ℓ=02​n−2Bℓ+1ℓ+1​∑r=ℓ2​n−2(rℓ)​kjr−ℓ​fr​(Aj)\displaystyle-\sum_{\ell=0}^{2n-2}\frac{B_{\ell+1}}{\ell+1}\sum_{r=\ell}^{2n-2}\binom{r}{\ell}k_{j}^{r-\ell}{f}_{r}(A_{j}) =−∑ℓ=02​n−2Bℓ+1ℓ+1​fℓ​(T−kj​Aj​Tkj)\displaystyle=-\sum_{\ell=0}^{2n-2}\frac{B_{\ell+1}}{\ell+1}{f}_{\ell}(T^{-k_{j}}A_{j}T^{k_{j}})
=−∑ℓ=02​n−2Bℓ∗+1ℓ∗+1​fℓ∗​(T−kj​Aj​Tkj)\displaystyle=-\sum_{\ell=0}^{2n-2}\frac{B_{\ell^{*}+1}}{\ell^{*}+1}{f}_{\ell^{*}}(T^{-k_{j}}A_{j}T^{k_{j}})
=−∑ℓ=02​n−2Bℓ∗+1ℓ∗+1​(−1)ℓ​fℓ​(S−1​T−kj​Aj​Tkj​S)\displaystyle=-\sum_{\ell=0}^{2n-2}\frac{B_{\ell^{*}+1}}{\ell^{*}+1}(-1)^{\ell}{f}_{\ell}(S^{-1}T^{-k_{j}}A_{j}T^{k_{j}}S)
=−∑ℓ=02​n−2Bℓ∗+1ℓ∗+1​(−1)ℓ​fℓ​(Aj+1)\displaystyle=-\sum_{\ell=0}^{2n-2}\frac{B_{\ell^{*}+1}}{\ell^{*}+1}(-1)^{\ell}{f}_{\ell}(A_{j+1})
=f0∗​(Aj+1)+∑ℓ=02​n−2Bℓ∗+1ℓ∗+1​fℓ​(Aj+1).\displaystyle={f}_{0^{*}}(A_{j+1})+\sum_{\ell=0}^{2n-2}\frac{B_{\ell^{*}+1}}{\ell^{*}+1}{f}_{\ell}(A_{j+1}).

Altogether we have shown

∑j=1p(f0∗(A~j)−f0(A~j))=∑j=1q((kj−1)f0∗(Aj)+f0(Aj)−∑r=02​n−2kjr+1r+1fr(Aj)+∑ℓ=02​n−2Bℓ+1ℓ+1fℓ(Aj)+f0∗(Aj+1)+∑ℓ=02​n−2Bℓ∗+1ℓ∗+1fℓ(Aj+1)).\sum_{j=1}^{p}\mathopen{}\mathclose{{}\left({f}_{0^{*}}(\tilde{A}_{j})-{f}_{0}(\tilde{A}_{j})}\right)=\sum_{j=1}^{q}\mathopen{}\mathclose{{}\left((k_{j}-1){f}_{0^{*}}(A_{j})+{f}_{0}(A_{j})-\sum_{r=0}^{2n-2}\frac{k_{j}^{r+1}}{r+1}{f}_{r}(A_{j})}\right.\\ \mathopen{}\mathclose{{}\left.+\sum_{\ell=0}^{2n-2}\frac{B_{\ell+1}}{\ell+1}{f}_{\ell}(A_{j})+{f}_{0^{*}}(A_{j+1})+\sum_{\ell=0}^{2n-2}\frac{B_{\ell^{*}+1}}{\ell^{*}+1}{f}_{\ell}(A_{j+1})}\right).

Since Aq+1=A1A_{q+1}=A_{1} for these cyclic permutations, this is

∑j=1q(kj​f0∗​(Aj)+f0​(Aj))+∑j=1q∑r=02​n−2(Br+1r+1+Br∗+1r∗+1−kjr+1r+1)​fr​(Aj).\sum_{j=1}^{q}\mathopen{}\mathclose{{}\left(k_{j}{f}_{0^{*}}(A_{j})+{f}_{0}(A_{j})}\right)+\sum_{j=1}^{q}\sum_{r=0}^{2n-2}\mathopen{}\mathclose{{}\left(\frac{B_{r+1}}{r+1}+\frac{B_{r^{*}+1}}{r^{*}+1}-\frac{k_{j}^{r+1}}{r+1}}\right){f}_{r}(A_{j}).

To get the first sum above into the final form, note first that

fr​(Aj+1)\displaystyle{f}_{r}(A_{j+1}) =fr​(S−1​T−kj​Aj​Tkj​S)\displaystyle={f}_{r}(S^{-1}T^{-k_{j}}A_{j}T^{k_{j}}S)
=(−1)r​fr∗​(T−kj​Aj​Tkj)\displaystyle=(-1)^{r}{f}_{r^{*}}(T^{-k_{j}}A_{j}T^{k_{j}})
=(−1)r​∑m=02​n−2(mr∗)​kjm−r∗​fm​(Aj),\displaystyle=(-1)^{r}\sum_{m=0}^{2n-2}\binom{m}{r^{*}}k_{j}^{m-r^{*}}{f}_{m}(A_{j}),

by (6.5). In the cases r=0r=0, 11 we obtain

f0​(Aj+1)=f0∗​(Aj),(2​n−2)​kj​f0∗​(Aj)=−f1​(Aj+1)−f1∗​(Aj).{f}_{0}(A_{j+1})={f}_{0^{*}}(A_{j}),\qquad(2n-2)k_{j}{f}_{0^{*}}(A_{j})=-{f}_{1}(A_{j+1})-{f}_{1^{*}}(A_{j}).

Hence

∑j=1q(kj​f0∗​(Aj)+f0​(Aj))=∑j=1q(12​(f0​(Aj)+f0∗​(Aj))−12​n−2​(f1​(Aj)+f1∗​(Aj))),\sum_{j=1}^{q}\mathopen{}\mathclose{{}\left(k_{j}{f}_{0^{*}}(A_{j})+{f}_{0}(A_{j})}\right)=\sum_{j=1}^{q}\mathopen{}\mathclose{{}\left(\frac{1}{2}\mathopen{}\mathclose{{}\left({f}_{0}(A_{j})+{f}_{0^{*}}(A_{j})}\right)-\frac{1}{2n-2}\mathopen{}\mathclose{{}\left({f}_{1}(A_{j})+{f}_{1^{*}}(A_{j})}\right)}\right),

and (6.6) follows. ∎

More generally, the same reasoning gives formulas for the integer-valued spanning symbols

Ψn(m)​(A):=u​(A)1−n​∑j=1p(fm∗​(A~j)−(−1)m​fm​(A~j)),\Psi_{n}^{(m)}(A):={\text{\rm u}}(A)^{1-n}\sum_{j=1}^{p}\mathopen{}\mathclose{{}\left({f}_{m^{*}}(\tilde{A}_{j})-(-1)^{m}{f}_{m}(\tilde{A}_{j})}\right),

from [Duk22, Sect. 6]. The higher Rademacher symbol Ψn\Psi_{n} may be expressed as linear combinations of these, though not uniquely.

Theorem 6.2.

Suppose n⩾2n\geqslant 2 and 0⩽m⩽2​n−20\leqslant m\leqslant 2n-2. For hyperbolic AA, in the notation of (2.7), (3.3),

Ψn(m)(A)=u(A)1−n∑j=1q∑r=02​n−2[(rm∗)βr−m∗−(r∗m)βr∗−m−(−1)m(rm)βr−m+(−1)m(r∗m∗)βr∗−m∗−(−1)m(rm)kjr−m+1r−m+1+(rm∗)kjr−m∗+1r−m∗+1]fr(Aj).\Psi_{n}^{(m)}(A)={\text{\rm u}}(A)^{1-n}\sum_{j=1}^{q}\sum_{r=0}^{2n-2}\mathopen{}\mathclose{{}\left[\binom{r}{m^{*}}\beta_{r-m^{*}}-\binom{r^{*}}{m}\beta_{r^{*}-m}-(-1)^{m}\binom{r}{m}\beta_{r-m}}\right.\\ \mathopen{}\mathclose{{}\left.+(-1)^{m}\binom{r^{*}}{m^{*}}\beta_{r^{*}-m^{*}}-(-1)^{m}\binom{r}{m}\frac{k_{j}^{r-m+1}}{r-m+1}+\binom{r}{m^{*}}\frac{k_{j}^{r-m^{*}+1}}{r-m^{*}+1}}\right]{f}_{r}(A_{j}). (6.7)

If AA fixes ω\omega and ω′\omega^{\prime} then, from a short calculation,

cn−1​Qr​(ω,ω′)\displaystyle c^{n-1}Q_{r}(\omega,\omega^{\prime}) =fr​(A),\displaystyle={f}_{r}(A),
cn−1​Gn​(ω,ω′)\displaystyle c^{n-1}G_{n}(\omega,\omega^{\prime}) = jn​∑rξr⋅fr​(A).\displaystyle={\text{\,\rm j}}_{n}\sum_{r}\xi_{r}\cdot{f}_{r}(A).

So formula (3.3) may be rephrased with the fr{f}_{r} notation (6.2) as

Ψn​(A)= in​Ψn(0)​(A)+u​(A)1−n​ jn​∑j=1q∑r=02​n−2ξr⋅fr​(A).\Psi_{n}(A)={\text{\,\rm i}}_{n}\Psi_{n}^{(0)}(A)+{\text{\rm u}}(A)^{1-n}{\text{\,\rm j}}_{n}\sum_{j=1}^{q}\sum_{r=0}^{2n-2}\xi_{r}\cdot{f}_{r}(A). (6.8)

Inserting the expressions from Theorem 6.1 and (2.8) into (6.8) finds a lot of cancellation, giving the following compact result.

Corollary 6.3.

For n⩾2n\geqslant 2 and hyperbolic A∈ΓA\in{\Gamma} we have

Ψn​(A)=u​(A)1−n​ jn​∑j=1q∑r=12​n−1(B2​n2​n​kjrr−Brr​B2​n−r2​n−r)​fr−1​(Aj).\Psi_{n}(A)={\text{\rm u}}(A)^{1-n}{\text{\,\rm j}}_{n}\sum_{j=1}^{q}\sum_{r=1}^{2n-1}\mathopen{}\mathclose{{}\left(\frac{B_{2n}}{2n}\frac{k_{j}^{r}}{r}-\frac{B_{r}}{r}\frac{B_{2n-r}}{2n-r}}\right){f}_{r-1}(A_{j}). (6.9)

This is also discussed in [Duk22, Sect. 3]. The left side of (6.9) may be replaced with  jn​ζ​(1−n,𝒜){\text{\,\rm j}}_{n}\zeta(1-n,\mathcal{A}) by (1.5). In this way we have established another proof of Zagier’s formula in [Zag77, p. 149]. Similar formulas are shown in [VZ13] and as the main theorem of [JL16].

References

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Department of Mathematics, The CUNY Graduate Center, 365 Fifth Avenue, New York, NY 10016-4309, U.S.A.

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