−1
2(s−s0)(s+s0−1)
[g(z,h,s)]=0. (4.4) Hence, by a standard regularity argument,gextends to a smooth function on all of XK. By the hypothesis g belongs to L2(XK). Since is a negative operator on L2(XK)and(s) >s0 = n4 + 12, the eigenvalue equation (4.4) implies thatg=0.
Remark 4.6 Similar characterizations of m,μ(z,h,s) can also be obtained forn ≤2 and(s) >s0 by imposing additional conditions on the growth at the boundary of XK, see e.g. Sect.6.1and [49, Sect. 3.4].
4.2 Positive integral values of the spectral parameter
Recall thatk =1−n/2 ands0 = n4 +12. In the present paper we are mainly interested in the Green functionm,μ(z,h,s)when the spectral parameters is specialized tos0 + j for j ∈ Z≥0. These special values can be described using lifts ofharmonicMaass forms.
Proposition 4.7 Letμ∈ L/L and m∈Z+Q(μ)with m>0. For j ∈Z>0
let Fm,μ(τ,k−2j) ∈ Hk−2j,ρ¯L be the unique harmonic Maass form whose principal part is given by
Fm,μ(τ,k−2j)=e(−mτ)φμ+e(−mτ)φ−μ+O(1) asv→ ∞. Then
m,μ(z,h,s0+ j)= 1
(4πm)jj!
z,h,Rkj−2jFm,μ(τ,k−2j) .
Proof The assertion is an immediate consequence of (4.1) and Corollary3.5.
5 CM values of higher Green functions
Recall thatk =1−n/2. Proposition4.7suggests to study for any harmonic Maass form f ∈ Hk−2j,ρ¯L its ‘higher’ regularized theta lift
j(z,h, f):= 1
(4π)j(z,h,Rkj−2j f). (5.1) Denote the Fourier coefficients of f by c±(m, μ) for μ ∈ L/L andm ∈ Z−Q(μ). Thenj(z,h, f)is a higher Green function for the divisor
Zj(f)=
m>0 μ∈L/L
c+(−m, μ)mjZ(m, μ) (5.2)
on XK. Here we compute the values of such regularized theta lifts at small CM cycles.
LetU ⊂ V be a subspace of signature(0,2)which is defined overQand consider the corresponding CM cycle defined in (2.2). For(z,h)∈ Z(U)we want to compute the CM valuej(z,h, f). Moreover, we are interested in the average over the cycle Z(U),
j(Z(U), f):= 2 wK,T
(z,h)∈supp(Z(U))
j(z,h, f). (5.3)
According to [38, Lemma 2.13], we have j(Z(U), f):= deg(Z(U))
2
h∈SO(U)(Q)\SO(U)(Af)j(zU+,h, f)dh (5.4)
and deg(Z(U))= vol(4KT).
The spaceU determines two definite lattices N =L∩U, P =L∩U⊥.
The direct sum N ⊕ P ⊂ L is a sublattice of finite index. For z = z±U and h ∈T(Af), the Siegel theta functionθL(τ,z,h)splits as a product
θL(τ,zU±,h)=θP(τ)⊗θN(τ,z±U,h).
In this case Lemma 3.1 of [12] implies for theC-bilinear pairings on SL and SN⊕P, that
f, θL = fP⊕N, θP⊗θN,
where fP⊕N is defined by Lemma3.7. Hence we may assume in the following calculation that L = P ⊕N if we replace f by fP⊕N. Forh ∈ T(Af)the CM value we are interested in is given by the regularized integral
j(zU±,h, f)= 1 (4π)j
reg
F Rkj−2j f, θP(τ)⊗θN(τ,z±U,h)dμ(τ).
To compute it, we first replace the regularized integral by a limit of truncated integrals. If S(q) =
n∈Zanqn is a Laurent series inq (or a holomorphic Fourier series inτ), we write
CT(S)=a0 (5.5)
for the constant term in theq-expansion.
Lemma 5.1 If we define A0=(−1)jCT
(f+)(j)(τ), θP(τ)⊗φ0+N ,
where f+is the holomorphic part of f and g(j):= (2π1i)j ∂j
∂τjg, we have j(z±U,h, f)
= lim
T→∞
1 (4π)j
FT
Rkj−2j f(τ), θP(τ)⊗θN(τ,z±U,h)dμ(τ)− A0log(T)
.
Proof This result is proved in the same way as [12, Lemma 4.5]. In addition
By means of (5.4) and the Siegel–Weil formula in Proposition3.8, we obtain the following corollary.
Here the Petersson scalar product is antilinear in the second argument. The meromorphic continuation of the Eisenstein seriesEN(τ,s;1)can be used to obtain a meromorphic continuation ofL(g,U,s)to the whole complex plane.
Ats=0, the center of symmetry,L(g,U,s)vanishes because the Eisenstein series EN(τ,s;1)is incoherent.
Lemma 5.3 Let g ∈ S1+n/2+2j,ρL and denote by g =
m,μb(m, μ)qmφμ
the Fourier expansion. WriteθP =
m,μr(m, μ)qmφμ. Then L(g,U,s)can Morerover, it has the Dirichlet series representation
L(g,U,s)= 1
Proof We rewrite the Petersson scalar product (5.6) using Lemmas3.3and 3.2. We obtain
L(g,U,s)= 1 (4π)j
2j−k j
θP ⊗(R1jEN(τ,s;1)), g
Pet. According to (3.10) we have
RjEN(τ,s;1)= (s2 +1+ j)
(2s +1) EN(τ,s;1+2j), and therefore
L(g,U,s)= 1 (4π)j
2j−k j
(s2+1+ j) (s2 +1)
θP⊗EN(τ,s;1+2j),g
Pet. (5.7) The latter scalar product can be computed by means of the usual unfolding argument. We obtain
θP⊗EN(τ,s;1+2j), g
Pet
=
∞\HθP⊗vs2−jφ0+N, gv1+n2+2j dμ(τ)
=
μ∈P/P m>0
r(m, μ)b(m, μ) ∞
0
e−4πmvv2s+n2+j−1dv
=
μ∈P/P m>0
r(m, μ)b(m, μ)(4πm)−s2−n2−js 2 +n
2 + j .
Inserting this into (5.7), we obtain the claimed Dirichlet series representation.
Theorem 5.4 Let f ∈ Hk−2j,ρ¯L. The value of the higher Green function j(z,h, f)at the CM cycle Z(U)is given by
1
deg(Z(U))j(Z(U), f)=CT
f+,[θP,EN+]j
−L
ξk−2j(f),U,0 .
Here EN+ denotes the holomorphic part of the harmonic Maass form EN(τ,0;1), see (3.14). The Rankin–Cohen bracket is taken for the weights (1,1).
Proof According to Corollary5.2we have
j(Z(U), f)= degZ(U) 2
× lim
T→∞
1 (4π)j
FT
Rkj−2j f(τ), θP(τ)⊗EN(τ,0; −1)dμ(τ)
−2A0log(T) .
We use the ‘self-adjointness’ of the raising operator (which is a consequence of (3.5) and Stokes’ theorem) to rewrite this as
j(Z(U), f)= degZ(U)
2 (5.8)
× lim
T→∞
1 (−4π)j
FT
f(τ),R−jk(θP(τ)⊗EN(τ,0; −1))dμ(τ)
−2A0log(T) .
Here the vanishing of the boundary terms in the limit T → ∞ fol-lows from Lemma 4.2 of [8] by inserting the Fourier expansions. Since R−1EN(τ,0; −1)=0 and because of (3.13), we have
R−jk(θP(τ)⊗EN(τ,0; −1))=
R1j−kθP(τ)
⊗EN(τ,0; −1)
=2
R1j−kθP(τ)
⊗
L1EN(τ,0;1) .
By Proposition3.6, we find
R−jk(θP(τ)⊗EN(τ,0; −1))=2(−4π)jL[θP(τ),EN(τ,0;1)]j. (5.9) Hence, we obtain for the integral
IT(f):= 1 (−4π)j
FT
f(τ), R−jk(θP(τ)⊗EN(τ,0; −1))dμ(τ) (5.10)
that
The second summand on the right hand side can be interpreted as the Petersson scalar product of [θP,EN(·,0;1)]j and the cusp form ξk−2j(f). The first summand can be computed by means of Stokes’ theorem. We get
As in the proof of [12, Theorem 4.7] we find that the first term on the right hand side is equal to
2 CT
f+, [θP,EN+]j . Putting this into (5.8) and inserting (5.6), we obtain
j(Z(U), f)=deg(Z(U)) CT
f+, [θP,E+N]j
−L(ξk−2j(f),U,0) . This concludes the proof of the theorem.
In the same way one proves the following result for the values at individual CM points.
Theorem 5.5 Let f ∈ Mk!−2j,ρ¯
L and h∈T(Af). Then we have j(z±U,h, f)=CT
f, [θP,G+N(τ,h)]j .
Here, G+N(τ,h) denotes holomorphic part of any harmonic Maass form GN(τ,h)∈ H1!,ρ
N satisfying L1GN(τ,h)=θN(τ,h).
By taking the particulary nice preimagesGN(τ,h)from Theorem3.9, we obtain the following algebraicity statement. We use the same notation as in Theorem 3.9, in particular, D < 0 denotes the discriminant of N, kD = Q(√
D), and we write HD for the ring class field of the orderOD ⊂ kD of discriminant D.
Corollary 5.6 Assume that f ∈ Mk!−2j,ρ¯
L has integral Fourier coefficients and that Zj(f) is disjoint from Z(U). For every h ∈ Cl(OD) there is an αU,f(h)∈ HD×such that:
(1) We have
Aj ·j(z±U,h, f)= −1
r ·log|αU,f(h)|,
where A∈Z>0is the least common multiple of the levels of the lattices P and N , and r ∈ Z>0 is a constant that only depends on L and D (but not on f , h or j ).
(2) The algebraic numbersαU,f(h)satisfy the Shimura reciprocity law αU,f(h)=αU,f(1)[h,kD].
Proof Recall that we assume thatL= P⊕Nholds. According to [34, Lemma 2.6], the group GSpin(Lˆ)is the maximal subgroup of H(Af)that preservesL and acts trivially onL/L. Hence f is GSpin(Lˆ)-invariant and we can assume thatK =GSpin(Lˆ). We now show thatKT ∼= ˆO×D. Consider the embedding ι:T → H, whereT acts trivially onU⊥. Since GSpin(Nˆ)= ˆO×D, we have ι(Oˆ×D)⊂ KT. However, by the maximality of GSpin(Nˆ), the other inclusion follows as well. Now the assertion follows from Theorem3.9.