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Open Access Published by De Gruyter March 26, 2022

# Differential operators, exact pullback formulas of Eisenstein series, and Laplace transforms

Tomoyoshi Ibukiyama
From the journal Forum Mathematicum

# Abstract

For a variable Z=(zij) of the Siegel upper half space Hn of degree n, put Z=(1+δij2zij)1i,jn. For a polynomial P(T) in components of n×n symmetric matrix T, we have P(Z)det(Z)s=det(Z)sQ(Z-1) for some polynomial Q(T). We show that the correspondence of P and Q are bijective for most s, and give a formula of P for any Q. In particular, when Q is a monomial, we show that such P corresponds exactly to the descending basis developed in a joint work with D, Zagier, for which an explicit generating series is known. By using the above results and the generating series, we give an exact formula for differential operators 𝔻 such that for any Siegel modular forms F of weight k, the restriction ResHn1×Hn2(𝔻F) to the diagonal blocks Hn1×Hn2Hn1+n2=Hn is a vector-valued Siegel modular forms of weight detkρ, where ρ is a fixed representation of GLn1()×GLn2(). These results are applied to give an exact Garrett–Böcherer-type pullback formula for any ρ that describes the restriction of 𝔻Ekn to Hn1×Hn2 for holomorphic Siegel Eisenstein series Ekn of weight k of degree n.

## 1 Introduction

There are two themes in this paper. One is a theory of spherical polynomials on symmetric matrices and the other is a theory of explicit vector-valued differential operators on Siegel modular forms that preserve automorphy under the restriction to the diagonal blocks, and of exact pullback formulas of Siegel Eisenstein series. A theory on spherical polynomials has been developed in [23, 24], and several explicit formulas for the polynomials are known in [15, 22, 23, 18, 19]. The relation of the spherical polynomials to differential operators of the above type is initiated in [15] and such differential operators have been used in many papers that treat the Garrett–Böcherer-type pullback formulas of the Siegel Eisenstein series (e.g. [3, 2, 4, 5, 6, 34, 35, 28, 29]). In this paper, we will show that these two theories are merged into one good theory in a very concrete level.

Let T be an n×n symmetric matrices. We denote by [T] the ring of polynomials P(T) in the components of T. We denote by Hn the Siegel upper half space of degree n defined as usual as

Hn={Z=X+iY=ZtMn():X=Xt,Y=YtMn(),Y>0},

where Y>0 means that Y is positive definite. Let Sp(n,) be the symplectic group of rank n defined as

Sp(n,)={gM2n():gJgt=J},J=(0n-1n1n0n).

For Z=(zij)Hn, we denote by Z the n×n symmetric matrix of operators defined by

Z=(1+δij2zij)1i,jn(where δij is the Kronecker delta).

We fix a complex number k. The first main concern of this paper is to describe P(Z)(det(CZ+D)-k) for g=(ABCD)Sp(n,) for any P as neatly as possible. The behavior does not depend on g when g is generic enough, and it is essentially the same to ask this only for C=1n and D=0. Shimura in his paper [33] asked the following question (said to be a naive question in his writing).

## Question.

Can one compute h(Z)det(Z)s in a reasonable way for arbitrary homogeneous polynomial function h on T?

His answer is that if we take h in an irreducible subspace (σ,V) of the representation π of [T] defined by (π(g)P)(T)=P(gtTg) for gGLn(), then we have

h(Z)det(Z)s=βn(s)det(Z)sh(Z-1),

where βn(s) is a sort of Gamma factor depending σ and s. But we often need P which does not belong to any irreducible subspace in the above sense. Indeed, in order to apply this problem to the pullback formula for example, it is indispensable to answer the question for much more general P. (Of course, any P is a linear combination of those belonging to irreducible subspaces in the sense of Shimura, but if you ask for example what is P for a given monomial Q, you cannot answer right away by his results.) We can easily show that for any homogeneous polynomial P[T], there exists a polynomial Q(T)[T] such that P(Z)det(Z)-k=det(Z)-kQ(Z-1). We first give a general formula for Q for any given homogeneous P[T] in Theorem 1. Next we show that elements in a certain canonical basis of [T] defined essentially in [23] are characterized by the fact that the corresponding Q are monomials times explicit constants (Theorem 2). The matrix T has 12n(n+1) independent variables, so monomials in components of T are labelled by multi-indices consisting of 12n(n+1) non-negative integers, and so are the polynomials P in the canonical basis. (A precise definition will be given later.) This canonical basis is a natural extension of the descending basis of higher spherical polynomials in [T] defined in [23]. If we restrict the previous action π(g) of GLn() on [T] to the diagonal subgroup (×)n of GLn(), then isotypic parts of (×)n are labelled by a certain set of multi-degrees 𝐚=(a1,,an) (ai0), and given by polynomials P(T) such that P(gtTg)=c1a1cnanP(T) for diagonal matrices g=diag(c1,,cn). The space of such P of multi-degree 𝐚 is in general not one-dimensional, and has some special basis labelled by several multi-indices. Now such P does not belong to any single irreducible subrepresentation of the representation π of GLn() in general. This is an essential difference from the case of Shimura.

To characterize P for any monomial Q, we employ here the theory of weighted Laplace transforms (first for enough big k for convergence) developed in [9] (see also [10]). By this tool we will show that such P is characterized as an element consisting of some canonical basis of [T] obtained by extending the descending basis of higher spherical polynomials defined in [23]. The higher spherical polynomials are a sort of generalization of Gegenbauer polynomials to many variables and span a subspace of elements of [T] satisfying certain harmonicity conditions. From the theory in [23], it is obvious that we can construct a natural extended basis of [T] containing the descending basis. The above theory of correspondence between canonical basis and monomial images can be generalized without change to any symmetric tube domains associated with formally real Jordan algebras. For simplicity, we will state the theory only for the case of Siegel upper half space here. Then for the symplectic case, by virtue of the existence theorem of descending basis in [23] (see also [19]), we can generalize this characterization to any general complex number k except for the case when 2k is an integer with 2k<n. By using the universal generating series of the descending basis in [23] (see also [19]), we can give a formula for P for any given polynomial Q (not necessarily a monomial) in Theorem 3. Universal generating series in case of general symmetric tube domains have not been known yet and this is a part of the reasons why we did not write a theory for general domains.

Next, we apply the above results to the Garrett–Böcherer-type pullback formula of Siegel Eisenstein series. This is also called the doubling method. For the pullback formula, we need holomorphic differential operators 𝔻 with some special property. We put n=n1+n2 for integers n1, n21. We call a sequence λ=(λ1,,λi,) of integers a dominant integral weight if λiλi+10 and λi0 only for a finitely many i. We denote by depth(λ) the biggest integer i such that λi0, and call it the depth of λ. We fix λ and assume that depth(λ)=mmin(n1,n2). We denote by (ρn1,λ,Vn1,λ) and (ρn2,λ,Vn2,λ) the irreducible representation of GLn1() and GLn2() corresponding to the same λ, respectively. Roughly speaking, we consider a linear holomorphic Vn1,λVn2.λ-valued differential operator 𝔻 with constant coefficients on holomorphic Siegel modular forms F of weight k on Hn such that the restriction ResHn1×Hn2𝔻F of 𝔻F to the diagonal blocks (Z100Z2) (Z1Hn1, Z2Hn2) is a Siegel modular form of weight detkρn1,λ for Z1 and detkρn2,λ for Z2. Now we denote by Ekn(Z) the holomorphic Siegel Eisenstein series of even weight k of degree n of level 1, assuming k>n+1. The pullback formula is a formula to describe ResHn1×Hn2𝔻Ekn by a linear combination of tensors of vector-valued Siegel modular forms of degree n1 and n2. Such formula is initiated by Garrett in [12] when 𝔻 is trivial (see also [30]) and by Böcherer [3, 2] for non-trivial 𝔻 when ρni,λ are scalar valued. One of the interesting features here is that the critical values of the standard L functions of Siegel cusp forms appear as coefficients. So explicit pullback formulas can be used to calculate the critical values explicitly or to give algebraicity results on critical values (see for example [21, 20, 27]). Here the effect of 𝔻 is that we can change the appearing critical points by that. Several explicit pullback formulas are known also for those λ with dimVni,λ>1, notably in [4, 34, 35]. For general λ, Kozima gave a pullback formula in [28] but there are two problems in his results. First of all, his formula is not quite explicit, since the formula depends on 𝔻(det(CZ+D)-k) that is not calculated in his paper. Secondly, he assumed that the differential operators have some special shape realized in the tensor of natural representation. The existence of 𝔻 preserving automorphy under the restriction of domains is known in [15] for any fixed k, n=n1+n2 and ρ. But it is not trivial at all that our differential operators can be realized in the way as Kozima described. He did not say either anything how to realize such differential operators concretely.

There are a lot of formulas or theories for such differential operators 𝔻 (e.g. in [2, 15, 22, 18, 19, 26], and by some authors the operator is called a symmetric breaking operator), but we still need a new formula for 𝔻 for our purpose in order to fix an idea what operators we are treating. Such 𝔻 is unique for each set {k,n1,n2,ρ} with 2kn=n1+n2 up to a constant as shown in [15], but we must fix the constant too to give an exact pullback formula. In this paper, we give a new simple formula for 𝔻 for any ρ in Theorem 4. Since we have 𝔻=P(Z) for some Vn1,λVn2,λ-valued polynomial P, it is enough to describe the polynomial P. We know by [15] that the components of such P satisfy some harmonicity conditions and belong to a representation subspace of the diagonal subgroup GLn1()×GLn2()GLn() acting on [T] by π(g)P(T)=P(gtTg). By using bideterminants realization of the representation space and the universal generating series of the descending basis we mentioned, we can give a simple formula to write down such vectors of differential operators 𝔻 that we need. This is a much simpler formula than those given in [18] and [19] before. (Several far simpler examples for the case when depth(λ)=m2 will be given in the last section.) An explicit formula for 𝔻(det(CZ+D)-k) can be also obtained by virtue of the results in the first half of this paper. So under the assumption that k>n1+n2+1, we give an exact pullback formula for any general λ (or any representation) for each concretely fixed differential operator (see Theorem 6). An explicit pullback formula for real analytic Siegel Eisenstein series including parameter s, or in the case of kn1+n2+1 through meromorphic continuation, is remained as a work in future.

The paper is organized as follows. In the next section, after preparing notation and giving a general preliminary for the paper, we give a formula of P(Z)det(Z)-k for general homogeneous polynomial P in Theorem 1. In Section 3, we first review the descending basis in [23], then for each monomial Q, we show in Theorem 2 that the polynomial P such that Q(Z-1)=det(Z)kP(Z)det(Z)-k is given by the descending basis. Then after giving a concrete example, we will give formula for P for any given Q in Theorem 3 by using the universal generating series in [23]. This is a kind of the converse of Theorem 1. For readers’ convenience, here we review the formula for the universal generating series. In Section 4, after reviewing the general setting of our differential operators, we give a formula for 𝔻 in Theorem 4. In Section 5, we give an exact pullback formula for this 𝔻 in Theorem 6. In Section 6, in case of scalar-valued 𝔻, we give Theorem 7 and 8 that are similar to Theorem 3 and 6 but in slightly different formulation by a much simpler alternative proof. In particular, for representations corresponding to λ=(,0,,0) and (,,0,,0), we explain completely explicit differential operators and their associated pullback formulas which are used to show Harder’s conjecture in [17] and [1]. The former case is an alternative formulation of [4].

## 2 First formulas on derivatives

In this section, by a direct calculation, we will write down a formula for

P(Z)(det(CZ+D)-k),g=(ABCD)Sp(n,),

for any homogeneous polynomial P(T)[T] and gSp(n,). More theoretical formulas will be given in the later sections.

The above expression apparently depends on a choice of g, but actually we have a unified formula independent of g. We will explain the reason. For later use, we give more general formulation. Let 𝕌 be an n×n matrix of variables and T an n×n symmetric matrix of variables. For another n×n symmetric matrix S of variables, let P(S) be a polynomial in [S]. For ZHn, we define Z as before. We fix g=(ABCD)Sp(n,). We write

(2.1)δ=det(CZ+D),Δ𝕌=𝕌(CZ+D)-1C𝕌t.

It is well known and easy to see that (CZ+D)-1C is a symmetric matrix (and hence Δ𝕌 also). We write components of Δ𝕌 as Δ𝕌=(Δij𝕌)1i,jn. When n=n and 𝕌=1n (the unit matrix), we write Δ𝕌=Δ and Δij𝕌=Δij. We define an n×n matrix Z𝕌 of operators and its components Z,ijU by

Z𝕌=𝕌Z𝕌t=(Z,ijU)1i,jn.

If n=n and 𝕌=1n (the unit matrix), then Z𝕌=Z.

Now we have the following formulas that can be easily proved. (The statement is in [28] and a precise proof is in [16] for example.) For a fixed gSp(n,), we define δ and Δ𝕌 by (2.1). Then we have

(D1)Z,ij𝕌(fh)=(Z,ij𝕌f)h+f(Z,ij𝕌h)for any functions fh on Hn,
(D2)Z,ij𝕌(δ-k)=-kδ-kΔij𝕌,
(D3)Z.ij𝕌(Δpq𝕌)=-12(Δip𝕌Δjq𝕌+Δiq𝕌Δjp𝕌).

Here for the sake of completeness, we give a proof of equality (D2) for arbitrary complex number k, assuming that we know (D2) already for integers k. It is enough to show it for n=n and 𝕌=1n. Since we are taking a complex power of complex numbers, we must explain how to take a branch. We follow the way that Freitag explained in [11]. We fix g as above. For a real number t with 0t1, we put

α(t)=det(C((1-t)i1n+tZ)+D).

Since (1-t)i1n+tZHn for 0t1, it is well known that α(t)0 for any ZHn. So α(t)α(t) is continuous for 0t1, where α(t) is a derivative of α(t) with respect to t. For 0a1 and ZHn, we define H(a,Z) by

H(a,Z)=0aα(t)α(t)𝑑t.

We put F(a)=eH(a,Z) and G(a)=F(a)α(a). Then we have

G(a)a=eH(a,Z)α(a)α(a)2-F(a)α(a)α(a)2=0.

So G(a) is a constant, and we have G(1)=G(0)=α(0)-1=det(Ci+D)-1. So by F(1)=α(1)G(1), we have

det(CZ+D)=α(1)=det(Ci+D)eH(1,Z).

We choose c0 such that det(Ci+D)=ec0 and define the -k-th power of det(CZ+D) by

det(CZ+D)-k=e-c0k-H(1,Z)k.

Once we fix c0, this is a well-defined holomorphic function on ZHn. In this sense, we have

Zdet(CZ+D)-k=e-c0k-H(1,Z)k(-k)ZH(1,Z).

If we put Δ(t)=(C((1-t)i1n+tZ)+D)-1C, then using (D2) for k=-1, we have Zα(t)=tα(t)Δ(t). By the smoothness of the functions, we have Z(α(t))=(Zα(t)), where means to take the derivative with respect to t. Here we have

(Zα(t))=(α(t)tΔ(t))=α(t)(tΔ(t))+α(t)(tΔ(t)).

So we have

ZH(1,Z)=01(Z(α(t))α(t)-α(t)α(t)tΔ(t)α2(t))
=01(tΔ(t))𝑑t=Δ(1)=Δ.

So we have (D2), assuming we take the same branch in both sides.

For a polynomial P(T)[T] and non-negative integer ν, we say that P(T) is homogeneous of degree ν if P(cT)=cνP(T) for any c. By the above formulas (D1), (D2), (D3), it is obvious that the following lemma holds.

## Lemma 1.

Notation being as above, for any polynomial P(S)C[S] and any complex number k, there exists a polynomial Q(S)C[S] such that

(2.2)P(Z𝕌)(δ-k)=δ-kQ(Δ𝕌).

If components ΔijU of ΔU are algebraically independent for 1ijn, then the polynomial Q depends only on k and P. If P is homogeneous of degree ν, then so is Q.

## Proof.

This is almost obvious by (D1), (D2), (D3), but we remark that we can take Q independent of gSp(n,). If we regard k and components Δij𝕌=Δji𝕌 of Δ𝕌 for 1ijn are algebraically independent variable, then by a formal calculation, using (D1), (D2), (D3), we obtain a polynomial Q^(k,Δ𝕌) in k and Δij𝕌 satisfying P(Z𝕌)δ-k=δ-kQ^(k,Δ𝕌). So (2.2) holds for Q=Q^(k,Δ𝕌) by substituting variables in Q by Δ𝕌 corresponding to any g and the value of k. ∎

Note that a polynomial Q[S] satisfying (2.2) for some special g is not unique at all. For example, if we take g=12n, then we have δ=1 and Δ=0, so any pair of polynomials P and Q with P(0)=0 and Q(0)=0 satisfies relation (2.2), On the other hand, if we define Q=Q^(k,Δ𝕌) as in the above proof by δ and Δ𝕌 for some gSp(n,), then this is uniquely determined for any P, k and g. We mostly fix k in later discussions and regard this Q^ as a polynomial Q(S)[S] by Q(S)=Q^(k,S). Defining Q in this sense for P, we denote by ϕk the linear homomorphism from [S]P to ϕk(P)=Q[S]. The first half of this paper is devoted to studying precise properties of ϕk. We will see that except for a minor exception of k, ϕk is an isomorphism, and also commute with the representation P(S)P(AtSA) for AGLm(). Later we need finer irreducible decompositions of representations of a certain subgroup G of GL(m,) on [S], which is important for the pullback formula.

Now we come back to the problem how to obtain Q for a given P. For example, if we put n=n, U=1n, g=(0n-1n1n0n), then ΔU=Z-1 and this is generic in the above sense. So to obtain Q(S), it is enough to calculate P(Z)(det(Z)-k). So we will give the formula for this. To calculate this, we use an integral expression of det(Z)-k. It is well known that for any positive integer d>n-1, we have

det(Zi)-d2=(2π)-dn2Mn,d()ei2Tr(XtZX)𝑑X
=cn(d)T>0ei2Tr(TZ)det(T)d-n-12dT,

where dX=1in, 1jddxij, dT=1ijndtij and

(2.3)cn(d)=2-nd2π-n(n-1)4i=0n-1Γ(d-i2)-1.

The first equality is known by Siegel and the second is just a coordinate change for XXt=T (e.g. see [10, 23].)

Now assume that P(T)[T] is homogeneous of degree ν and put d=2k. Then differentiating under integral, we have

(2.4)P(Z)(det(Zi)-k)=cn(2k)(i2)νT>0ei2Tr(TZ)P(T)det(T)k-n+12dT.

Since the final formula should be holomorphic on ZHn, it is enough to calculate the right-hand side for Z=iY with 0<Y=YtMn(). We write

(2.5)(k*P)(Y)=T>0e-Tr(TY2)P(T)det(T)k-n+12dT.

This is essentially the same as the weighted Laplace transform k(P) of P(T) given in [10, p. 323]. (A subtle difference will be explained later.) By using this notation, the right-hand side of (2.4) for Z=iY is given by

(i2)νcn(2k)(k*P)(Y).

By Lemma 1, it is obvious that we have

(2.6)(k*P)(Y)=det(Y)-kR*(Y-1)

for some polynomial R*[Y]. We define a representation π of GLn() on [T] by

(π(A)P)(T)=P(AtTA).

We choose 0<A=AtGLn() such that Y=A2 and put T1=ATA. Then by dT1=det(A)n+1dT, we have

(2.7)(k*P)(Y)=det(Y)-kk*(π(A-1)P)(1n).

For an n×n symmetric matrix W=(wij) with variable components, we denote by W an n×n matrix of differential operators defined by

W=(1+δij2wij)1i,jn.

## Theorem 1.

Notation being as above, for any homogeneous polynomial P(T)C[T] of degree ν, we have

(2.8)P(Z)(det(CZ+D)-k)=det(CZ+D)-kϕk(P),

where

(2.9)ϕk(P)=(-1)ν(P(W)(det(1n-W(CZ+D)-1C)-k))|W=0

for any g=(**CD)Sp(n,R).

## Proof.

We put d=2k and first we assume that d>n-1. As in [23], we define Ed[P] by

(2.10)E2k[P]=cn(d)(k*P)(1n).

The values of Ed[P] for monomials P has been calculated in [23] and the generating function of such values is written down there. To explain it, we introduce several notations which will be used later again. We define a set 𝒩n of n×n integral symmetric matrices as

𝒩n={𝝂=𝝂t=(νij)Mn():νij0 for any ij, and νii20}.

We call an element of 𝒩n a multi-index. We define

𝝂!=1inνii!!1i<jnνij!,

where νii!! is defined as 2νii2(νii2)!. For any 𝝂=(νij)𝒩n and any symmetric matrix T=(tij), we write

T𝝂=1i,jntijνij2=i=1ntiiνii2×1i<jntijνij.

We denote by deg(𝝂) the total degree of T𝝂, given by deg(𝝂)=2-11i,jnνij.

## Lemma 2 ([23, Theorem 3]).

Let W be an n×n symmetric matrix of variables. Then we have

𝝂𝒩nEd[T𝝂]W𝝂𝝂!=det(1n-W)-d2.

Now we come back to the proof of Theorem 1. For any symmetric matrix AGLn() and 𝝂, 𝝁𝒩n, we define r𝝂,𝝁(A) by

(A-1TA-1)𝝂=𝝁𝒩nr𝝂,𝝁(A)T𝝁𝝁!.

If we write

P(T)=𝝂𝒩nc𝝂T𝝂,

then we have

(2.11)E2k[P(A-1TA-1)]=𝝂𝒩nc𝝂E2k[(A-1TA-1)𝝂]=𝝂,𝝁𝒩nc𝝂r𝝂,𝝁(A)Ed[T𝝁]𝝁!.

If we put Y=A2, then by (2.7), we also have

(2.12)E2k[P(A-1TA-1)]=E2k[π(A-1)P]=det(Y)kk*(P).

On the other hand, for any 𝝂, 𝝁𝒩n, we have

2ν(W)𝝂(W𝝁)|W=0=δ𝝂,𝝁𝝂!.

So we have

2νP(W)(𝝁𝒩nE2k[T𝝁](A-1WA-1)𝝁𝝁!)|W=0=𝝁,𝝂𝒩nc𝝂r𝝁,𝝂(A)Ed[T𝝁]𝝁!.

Here if r𝝂,𝝁(A)=r𝝁,𝝂(A), then by (2.12) and (2.11), this is equal to det(Y)kk*(P). Here, by Lemma 2, we have

𝝂𝒩nE2k[T𝝁](A-1WA-1)𝝁𝝁!=det(1n-A-1WA-1)-k=det(1n-WY-1)-k.

So now we will show r𝝂,𝝁(A)=r𝝁,𝝂(A). For any P(T) and Q(T)[T], we define the (hermitian) inner product by

(P(T),Q(T))0=(P(T)Q¯)(0),

where the bar means to take the complex conjugation of the coefficients of Q. We have

(P(T),Q(T))=(Q(T),P(T))¯.

For any AGLn(), we have

(P(AtTA),Q(T))=(P(T),Q(ATAt)).

So P(T)P(A-1TA-1) is the self-adjoint operator with respect to the inner metric. Since {T𝝂𝝂!}𝝂𝒩n is an orthonormal basis of [T], and since

(A-1TA-1)𝝂𝝂!=𝝁r𝝂,𝝁(A)1𝝂!𝝁!×T𝝁𝝁!,

we have

r𝝂,𝝁(A)𝝂!𝝁!=r𝝁,𝝂(A)𝝂!𝝁!,

so r𝝂,𝝁(A)=r𝝁,𝝂(A). So we proved that

(2.13)(k*P)(Y)=2νcn(2k)-1det(Y)-k(P(W)det(1n-WY-1)-k)|S=0.

Finally, note that if we replace Y in (2.13) by Zi, then det(1n-WY-1)=det(1n-iWZ-1). Since we assumed P is homogeneous of degree ν, we have

P(W)det(1n-iWZ-1)-k|W=0=iνP(W)det(1n-WZ-1)-k|W=0.

So multiplying (i2)ν to 2νiν, we have (-1)ν in (2.9). Since n(n+1)2 components of upper half of Z-1 are algebraically independent, we can replace Z-1 in (2.9) by (CZ+D)-1C. This proves (2.8) for d>n-1. Now for any monomial in [T], it is obvious by (D1), (D2), (D3) that ϕk(P) is a polynomial in k. Since (2.8) holds for any k>n-12, the relation holds for any k for any monomial. Since P is a linear combination of monomials and ϕk is linear, relation (2.8) holds for any P and k. (It does not matter if P has coefficients depending on k or not.) ∎

## 3 Spherical polynomials and the descending basis

Although the formula in Theorem 1 gives a direct way to calculate P(Z)(δ-k), it is not very structural. On the other hand, several nice bases of [T] are known. For example, in [33] or in [10, Chapter XI], some representation theoretical decomposition of [T] is given. But these bases do not fit our motivation for differential operators on Siegel modular forms. In fact, we have two other canonical bases in [23] of spherical polynomials associated with mixed Laplacians. Here we use a basis obtained as a natural extension of one of these bases (the descending basis in [23]), which is very different from the decomposition in [33] or [10]. For any polynomial P(T) that is a member of the descending basis, we can show that an exact formula of P(Z)δ-k is very simple and explicit. To give these formulas, we first explain the definition of the descending basis of [T]. For any positive integers d and n, we consider the space of n×d matrices X=(xiν) and put T=XXt. For each i, j with 1i,jn, we define a mixed Laplacian by

Δij(X)=ν=1d2xiνxjν.

We have Δij(X)=Δji(X) by definition. For P(T)[T], we put P~(X)=P(XXt). If we rewrite operators Δij(X) by the coordinate of the symmetric matrix T=(tij)=XXt, then we have Δij(X)P~(X)=(DijP)(XXt) for the operator Dij defined as

(3.1)Dij=dij+k,l=1ntklikjl,

where we put ij=(1+δij)tij (so (ij)1i,jn=2T). Here we write tij=tji, and we have Dij=Dji. Now once we define Dij as (3.1), we may forget that d is a positive integer and we may regard it as a general complex number, or even a variable. We denote by eij the n×n matrix whose (i,j) component is 1 and the other components are 0. We put

𝐞ij=eij+eji.

Since integers d with d<n are a bit exceptional, we put

^n={d:d<n}.

## Proposition 1 ([23, Section 8.2 and Theorem 11]).

Assume that dC^n. Then we have a basis of C[T] consisting of the set of polynomials P𝛎D(T) for 𝛎Nn of degree deg(𝛎) such that

P𝟎D(T)=1,
DijP𝝂D(T)=P𝝂-𝐞ijD(T)for any (i,j) with 1i,jn,

where for 𝛎=(νij), we understand P𝛎(T)=0 if any νij<0.

In the paper [23], apparently we defined P𝝂D(T) only for 𝝂𝒩n such that νii=0 for i=1,,n. We called a polynomial P(T) such that DiiP=0 for all i with 1in a higher spherical polynomial, and called the basis of higher spherical polynomials consisting of P𝝂D(T) for νii=0 with 1in the descending basis. But we can prove the above theorem completely in the same way as in [23] without the restriction to νii=0. We we omit the proof here since the argument is the same. So we will also call the set {P𝝂D(T):𝝂𝒩n} in Proposition 1 the descending basis of [T]. The coefficients of P𝝂D(T) can be regarded as rational functions of d that can be singular only for integers d<n.

The proof of this proposition in [23], in particular, of the claim of the non-singularity for any d^n, is not easy at all. We treat P𝝂D(T) in a slightly different formulation in the present paper, but that does not mean that we have an alternative proof of this claim on the range of non-singularity. By the way, we note that the non-singularity for d>n-1 can be proved easily in various ways (both in [23] and in the present paper).

Now one of our main results here is the following theorem. We fix g=(**CD)Sp(n,) and defined δ and Δ=(Δij) for g and ZHn as (2.1). For 𝝂𝒩n, we define Δ𝝂 as before by

Δ𝝂=1i,jnΔiiνii2=i=1nΔiiνii2×1i<jnΔijνij.

## Theorem 2.

The following statements hold:

1. For any 𝝂𝒩n and d=2k^n, we have

(3.2)P𝝂D(Z)(δ-k)=δ-k×(-1)deg(𝝂)2deg(𝝂)𝝂!Δ𝝂(deg(𝝂)=121i,jnνij).
2. For 2k^n, the linear map ϕk:[T][T] defined by P(Z)δ-k=δ-kϕk(P) is an linear isomorphism over and commutes with the representation π of GLn() given by (π(A)P)(T)=P(AtTA), that is, we have

ϕk(π(A)P)=π(A)(ϕk(P)),P[T].

We give an obvious corollary below. We denote by S an n×n symmetric matrix and by 𝕌 an n×n matrix of variables.

## Corollary 1.

For ZHn, g=(**CD)Sp(n,R) and for the descending basis {P𝛎D(S):𝛎Nn} of C[S] for d=2kC^n, we have

P𝝂D(𝕌Z𝕌t)det(CZ+D)-k=det(CZ+D)-k(-1)deg(𝝂)2deg(𝝂)𝝂!(𝕌(CZ+D)-1C𝕌t)𝝂

for any 𝛎Nn.

## Proof of Theorem 2.

For any ZHn, the inverse Z-1 is also a symmetric matrix and the (i,j) components of Z-1 for ij are obviously algebraically independent, so it is sufficient to prove (3.2) for δ=det(Z).

We assume first that k>n-12 (so automatically 2k is not an integer with 2k<n). Then k*P converges for any P[T]. We prove (3.2) by induction on 𝝂. To compare DijP𝝂D and P𝝂D, we quote a key lemma in [10, Proposition XV.2.4] for a formally real Jordan algebra. Since the coordinate system in [10] is slightly different from the one that we use here, we review the notation in [10]. We regard the space Symn() of real n×n symmetric matrices as a vector space over . This is one of the five types of formally real Jordan algebras. We define an inner product (x|y) for x, ySymn() by (x|y)=Tr(xy). Then an orthonormal basis with respect to this metric can be taken as

fii=eii,(1in),fij=eij+eji2(1i<jn).

If we write an element x of Symn() as

x=i=1nxiifii+1i<jnxijfij

and x=TSymn() as T=(tij)=1ijntijeij, then we have xii=tii and xij=2tij for ij. For any parameter k, we define a Symn()-valued operator k on scalar-valued functions F(x) of Symn() as in [10] by

kF(x)=α,β2Fxαxβ×12(fαxfβ+fβxfα)+kαFxαfα,

where α, β runs over the set {(i,j):1ijn}. Writing the coordinate change carefully, we can show that

k=141i,jneijDij=14(i=1neiiDii+1i<jn𝐞ijDij).

For k>n-12 and a polynomial P(T)[T], we put

kP(Y)=2n(n-1)4T>0e-Tr(TY)P(Y)det(T)k-n+12dT

This is exactly the weighted Laplace transform defined in [10, Proposition XV.2.4 (ii)]. (There they used dx instead of dT and this causes the apparent difference of the power of 2 from the one in the book.) It is trivial from definition that (k*P)(2Y)=2-n(n-1)4(kP)(Y). Now assume that P is homogeneous of degree ν. Then in the definition of (k*P)(2Y) in (2.5), changing T=2T1, we see P(T)=P(T12)=2-νP(T1), dT1=2n(n+1)2dT and det(T1)k-n+12=2nk-n(n+1)2det(T)k-n+12, so we see that

(k*P)(2Y)=2-nk-ν(k*P)(Y)

and we have

(3.3)(kP)(Y)=2-n(n-1)4-ν-nk(k*P)(Y).

For YSymn(), we write as before

Y=(1+δij2yij).

## Lemma 3 ([10, Proposition XV.2.4 (ii)]).

One has

(3.4)14((k(DijF)(Y))1i,jn=-(Y(kF)YY+kYkF(Y)).

By (2.6), writing B=Y-1, we may assume that (k*P)(Y)=det(Y)-kR*(B) and (kP)(Y)=det(Y)-kR(B) for some homogeneous polynomials R* and R of degree ν. By (3.3), we have

R*(B)=2nk+ν+n(n-1)4R(B).

Now we consider the right-hand side of (3.4) for F=P. We have

(det(Y)-kR(B))Y=-kdet(Y)-kY-1R(B)+det(Y)-kR(B)Y,

so we have

-(Ydet(Y)-kR(B)YY+kYdet(Y)-kR(B))=-det(Y)-kYR(B)YY.

We write Y-1=B=(bij). By (D3), we see easily that Y(bkl)=(-2-1(bikbjl+bilbjk))1i,jn. So if we put

cij,kl=-12(bikbjl+bilbjk)

for 1i,jn, 1k,ln, then for 𝝂=(νij)𝒩n. we have

B𝝂Y=(dij),

where

dij=(1k,lnνklcij,klB𝝂bkl)1i,jn.

So for a fixed k, l, the (p,q) component of Y(cij,kl)Y is given by

i,j=1nypicijyjq=-12(i=1nypibikj=1nbjlyjq+i=1nypibilj=1nyqjbjk)
=-12(δpkδlq+δplδqk).

If p=q, this is non-zero (and equal to -1) only for k=l=p=q. If pq, then this is non-zero (and equal to -12) only for (k,l)=(p,q) or (q,p). So we have

-YB𝝂YY=B𝝂B.

By this and Lemma 3, if k(F)(Y)=det(Y)-kR(B) and k(DijF)(Y)=det(Y)-1R1(B), then we have

(3.5)14R1(B)=R(B)B.

Now for an element P𝝂D in the descending basis, we define polynomials R𝝂D(B) by

k(P𝝂D)=det(Y)-kR𝝂D(B).

For any 𝝂𝒩n, by (3.5) we have relations

(3.6)1+δij2R𝝂D(B)bij=14R𝝂-𝐞ijD(B)

for any (i,j) with 1i,jn. Now, we fix a positive number ν, and we assume that for any 𝝁𝒩n with det(𝝁)<ν, we have

(3.7)R𝝁D(B)=2-n(n-1)4-nk-deg(𝝁)cn(2k)-1B𝝁𝝁!.

First we show that this is true for 𝝂=𝟎. By definition, we have P𝟎D(T)=1. We have E2k[1]=1, so by (2.10), we have k*P𝟎D(1n)=cn(2k)-1, and by (2.7), we have

k*P𝟎D(Y)=cn(2k)-1det(Y)-k

and by (3.3), we have

k(P𝟎D)(Y)=2-n(n-1)4-nkcn(2k)-1det(Y)-k.

So (3.7) is true for 𝝂=0. Next we show (3.7) by induction. We can apply the inductive assumption for the right-hand side of (3.6). For the left-hand side, by direct calculation, we have

12deg(𝝂)𝝂!B𝝂B=14(B𝝂-𝐞ij2det(𝝂-𝐞ij)(𝝂-𝐞ij)!)1i,jn.

and so we see that the simultaneous equations (3.6) has the unique solution

R𝝂D(B)=2-n(n-1)4-nk-deg(𝝂)cn(2k)-1B𝝂𝝂!

for any 𝝂𝒩n. So (3.7) is proved. By (3.3), we have

cn(2k)k*(P𝝂D)(Y)=det(Y)-kB𝝂𝝂!.

Now by (2.4), changing Y by Zi, we have

P𝝂D(Z)(det(Zi)-k)=(i2)deg(𝝂)det(Zi)-k(iZ-1)𝝂𝝂!,

so by i2det(𝝂)=(-1)det(𝝂), we have (3.2) for k>n-12. The existence of the descending basis {P𝝂D(T)} for any k with 2k^n has been proved in [23], and the coefficient is a rational function of k non-singular on k with 2k^n. So if we define a polynomial Q𝝂D=ϕk(P𝝂D)[T] by P𝝂D(Z)(δ-k)=δ-kQ𝝂D(Z-1), then by (D1), (D2), (D3), the polynomial Q𝝂D has also coefficients that are rational functions of k non-singular on k with 2k^n. So the validity for any k>n-12 means that the same relation holds for any k with 2k^n. So we prove (i) of Theorem 2. We prove (ii). Since {P𝝂D:ν𝒩n} is a basis of [T] and the monomials consisting of bij of B=(bij) is also a basis of [B], it is clear that ϕk is an isomorphism. Next, if we write (k*P)(Y)=det(Y)-kR(Y-1), then R(T) is proportional to ϕk(P). If we put P1=π(A)P, then changing T to T1=AtTA, we have dT1=det(A)n+1dT and

(k*P1)(Y)=T>0e-12Tr(TY)P1(T)det(T)k-n+12dT
=det(A)-2kT1>0e-12Tr(T1(A-1YA-1t))P(T1)det(T1)kn+12dT1
=det(A)-2k(k*P)(A-1YA-1t)=det(Y)-kR(AtY-1A).

So writing (k*P1)(Y)=det(Y)-kR1(Y-1), we have R1(Y)=R(AtYA)=π(A)R. ∎

## Remark 1.

Assume that n2. Then the linear map ϕk is not injective for any k with 2k^n. Indeed, if k is a non-positive integer, then det(Z)-k is a polynomial, so obviously P(Z)det(Z)-k=0 if the degree of P is big enough. Moreover, because of the Cayley-type formula (see [33, 8]), we have

det(Z)det(Z)-k=(-k)(-k+12)(-k+n-12)det(Z)-k-1

for any k. So if 2k with 02kn-1, then this vanishes. Besides, if k is a negative half integer, then making det(Z) operate on det(Z)-k several times, we obtain det(Z)-12 or 1. so again by the above formula, this vanishes for n2 under det(Z). So for any k with 2k^n, we have ϕk(det(T)l)=0 for some positive number l, and ϕk is not injective. Since ϕk is a linear map on homogeneous polynomials of each fixed degree and the space of such polynomials is finite dimensional, non-injective means non-surjective. When n=1, the map ϕk is an isomorphism if and only if k0.

We have already given a formula for Q=ϕk(P) for any P[T] in Theorem 1. Using Theorem 2, for any given Q[T], we can give a formula for P such that ϕk(P)=Q. This will be written in Theorem 3 below and later we need this for the pullback formula.

But before that, we will explain by an example that the polynomial P for a simple ϕk(P) is not so simple. Assume n=4. We write a block matrices of Z-1H4 as

Z-1=(B1B12tB12B2)(B1,B2H2,B12M2()).

If we write B=(bij), then we have

det(B12)2=b132b142-2b13b14b23b24+b142b232.

So if we want a polynomial P such that

P(Z)(det(CZ+D)-k)=det(CZ+D)-kdet(B12)2,

then by virtue of Theorem 2, we can directly see that the answer is

P(T)=26(P2𝐞13+2𝐞24D(T)-12P𝐞13+𝐞14+𝐞23+𝐞24D(T)+P2𝐞14+2𝐞23D(T)).

So the problem is to write down the descending basis. In this case, as given in the Appendix of [23], if we put d=2k and

f(d)=4(d-3)(d-2)(d-1)d(d+1)(d+2)(d+4)(d+6),

then we have

f(d)P2𝐞13+2𝐞24D(T)=2(d2+12)t122t342+(d4+5d3-10d2-36d+24)t132t242+2(d2+12)t142t232
-4(d-2)d(d+4)t13t14t23t24+4(d-10)dt12t14t23t24
-4(d-2)d(d+4)t12t13t24t34,
f(d)P2𝐞14+2𝐞23D(T)=2(d2+12)t122t342+2(d2+12)t132t242+(d4+5d3-10d2-36d+24)t142t232
-4(d-2)d(d+4)t13t14t23t24-4(d-2)d(d+4)t12t14t23t34
+4(d-10)dt12t13t24t34,
f(d)P𝐞13+𝐞14+𝐞23+𝐞24D(T)=4(d-10)dt122t342-4(d-2)d(d+4)t132t242-4(d-2)d(d+4)t142t232
+4(d-2)2(d+3)(d+4)t13t14t23t24-4(d3-7d2-10d+24)t12t14t23t34
-4(d3-7d2-10d+24)t12t13t24t34.

For several cases when Q and n are explicitly given, we know exact formulas of such P (for example [15, p. 114]), but there is no exact formula for P𝝂D(T) in general. Instead, we have a universal generating series for some constant multiples of P𝝂D(T) in [23] (see also [19]). To explain the generating series, we prepare an n×n dummy symmetric matrix X=(xij) of variables, and writing ν=deg(𝝂), we put

(3.8)P𝝂(T)=2ν(k)ν(2k-2)νP𝝂D(T),

where we write (x)ν=x(x+1)(x+ν-1) (the ascending Pochhammer symbol). We note that the difference of constant only depends on the total degree deg(P)=det(𝝂) of P𝝂D and k.

We put

G(n)(T,X)=𝝂𝒩nP𝝂(T)X𝝂(X𝝂=1i,jnxijνij2=i=1nxiiνii21i<jnxijνij).

We reproduce the explicit formula for the series G(n)(T,X) later for the readers’ convenience. Here first we give a formula for P for any given Q[T] such that ϕk(P)=Q.

## Theorem 3.

Assume that 2kC^n. For a given homogeneous polynomial Q(T)C[T] of degree ν, we put

P(T)=2ν(-1)ν(k)ν(2k-2)νQ(X)G(n)(X,T)|X=0.X=(1+δij2xij).

Then ϕk(P)=Q and

P(Z)det(CZ+D)-k=det(CZ+D)-kQ(Δ),Δ=(CZ+D)-1C.

## Proof.

Using the relations

((X)𝝁(X𝝂))|X=0=δ𝝂,𝝁𝝂!2ν,

the theorem is obvious by Theorem 2 and the definition (3.8) of P𝝂. ∎

The explicit formula for G(n)(T,X) in [23] is given as follows. For each i=0,,n, we define a polynomial σi=σi(T,X) in components of T and X by

det(λ1n-TX)=i=0n(-1)iσiλn-i.

We regard σi as variables for a while and define a differential operator i by

i=0<a,b<iia+bσa+b-iab,

where we write i=σi and we regard σi=0 unless 0in. We note that iσi=σii. For any complex number ν<0, we define a formal power series 𝕁ν(x) by

𝕁ν(x)=r=0xrr!(ν+1)r=1+xν+1+x22(ν+1)(ν+2)+.

Then by [23, Theorem 12], we have the following formula:

G(n)(T,X)=𝕁k-n+12(σnn)𝕁k-n2(σn-1n-1)𝕁k-32(σ22)((1-σ12)2-2k).

For some more algebraic closed formulas for G(n)(T,X) for n=2, 3, 4 has been given in [23]. In particular, when n=2, this is nothing but the generating function of (homogeneous version of) Gegenbauer polynomials and it is given by

G(2)(X,T)=((1-σ12)2-σ2)-k+1.

When n=3, we have

G(3)(X,T)=1Δ02-8σ3(Δ0+Δ02-8σ32)-k+2,Δ0=(1-σ12)2-σ2.

When n4, there are no algebraic expressions of G(n)(X,T) in general, but for concrete algebraic expressions for n=4 and integer k2, see [23].

## 4 Differential operators on Siegel modular forms

We fix integers n, t2, and a partition 𝐧 of n with n=n1++nt (ni>0). For ZHn, we consider the block decomposition Z=(Zij)1i,jt where Zij is an ni×nj matrix. We consider the restriction Res of Hn to Hn1××Hnr by setting Zij=0 for all ij. We fix polynomial representations ρi of GL(ni,). Roughly speaking, we consider a vector-valued differential operator 𝔻 such that for any holomorphic Siegel modular forms F on Hn of weight k, Res(𝔻F(Z)) is again a Siegel modular form of weight detkρi for each Zii. Actually, this description is not quite correct since our formulation has nothing to do with discrete subgroups of Sp(n,), and 𝔻 is characterized as an intertwining operator between holomorphic discrete series. We write

Sp(𝐧,)=Sp(n1,)××Sp(nt,).

This group naturally acts on H𝐧:=Hn1××Hnt. If we consider a diagonal embedding of H𝐧 to Hn, then the action of Sp(𝐧,) on H𝐧 and Hn is equivariant by the following embedding of Sp(𝐧,)(g1,,gt) to ι(g1,,gt)Sp(n,), where

ι(g1,,gt)=(A100B100000000At00BtC100D100000000Ct00Dt)Sp(n,)for gi=(AiBiCiDi)Sp(ni,).

For any polynomial representation (ρi,Vi) of GL(ni,) for i=1,,t, any V1Vt-valued function f(Z11,Z22,,Ztt) on H𝐧, and any element gi=(AiBiCiDi)Sp(ni,), we write

f|ρ1,,ρt[(g1,,gt)]=ρ1(C1Z11+D1)-1ρt(CtZtt+Dt)-1f(g1Z11,,gtZtt).

On the other hand, for any function F on Hn, any element g=(ABCD)Sp(n,), and any representation ρ of GL(n,), we define

F|ρ[g]=ρ(CZ+D)-1F(gZ).

When ρ=detk, we simply write F|ρ[g]=F|k[g]. Notation being as above, for any fixed integer k, we consider a V1Vt-valued polynomial P such that

(4.1)ResHn1××HntP(Z)(F|k[ι(g1,,gn)])=(ResHn1××HntP(Z)F)|detkρ1,,detkρt[(g1,,gt)]

for any holomorphic function F(Z) on Hn and any giSp(ni,). Differential operators 𝔻=P(Z) of this type are used to describe the pullback formula. For example, when t=n and ni=1 for all i, such differential operators are realized by the elements in the descending basis and we have an application to critical values of the triple L functions when n=t=3 besides (see [5, 21]).

A general theory on the characterization of such polynomials P satisfying (4.1) has been given in [15]. We shortly review it for readers’ convenience. We embed A=(A1,,At)GLn1()××GLnt() into GLn() by

ι(A)=(A100000At)GLn().

## Proposition 2 ([15, Theorem 1]).

When kn2, for a V1Vt-valued polynomial P in components of n×n symmetric matrix T, the differential operator P(Z) satisfies Condition (4.1) if and only if the following two conditions are satisfied:

1. DijP=0 for any (i,j) with 1+l=1s-1nli,jl=1snl for some s with 1st.

2. P(ATAt)=ρ1(A1)ρt(At)P(T).

In general, the operator P(Z) satisfying (4.1) is not unique even up to constant. The dimension of such P can be calculated by the Littlewood–Richardson rule as pointed out in [19, Theorem 3.4]. In the present paper, we consider only the case when t=2, so we will explain this case more. It is well known that equivalence classes of irreducible representations ρ of GL(n,) correspond bijectively to the set of dominant integral weights λ with depth(λ)n (see [36]). Since the same λ corresponds to representations of GL(n,) of various n, in order to distinguish the size of GL, we denote by (ρn,λ,Vn,λ) the irreducible representation of GL(n,) corresponding to λ.

When 2k is an integer with 2kn=n1+n2, a non-zero polynomial P which satisfies (4.1) exists if and only if ρ1=ρn1,λ and ρ2=ρn2,λ for the same λ. For each k and each λ with depth(λ)min(n1,n2), the polynomial P is unique up to constant (see [15]). We have already given a one-line formula for such P in [18] and some other formula in [19]. But those formulas are not easy to apply for the pullback formula, so here we consider another way using the descending basis. Before going further, we explain a realization of irreducible representations of GL(n,) to describe our differential operators neatly. In Kozima [28, 29], he uses a differential operator satisfying (4.1) for t=2, but he used the traditional realization of the representations in the tensor of natural representations. This makes that paper considerably complicated and difficult to read, and besides, he assumed that P has some special shape there. It is not trivial at all that P we need can be realized in his way and also he did not say anything on construction of such operators. So here instead, we use bideterminants realization of the representation which is much simpler and suitable for our purpose, and also give an actual formula for P.

First we review the bideterminants realization. For any set I, we denote by |I| the cardinality of I. For any m×n matrix M, a positive integer q, and subsets I{1,,m} and J{1,,n} such that |I|=|J|=q, we denote by MIJ the (I,J)-minor of M, that is, the determinant of the q×q submatrix of M consisting of i-th rows with iI and j-th columns with jJ. We write [q]={1,,q} for simplicity. We fix positive integers n1 and n2. Let λ=(λ1,λ2,) be a dominant integral weight with depth(λ)=mmin(n1,n2). Let U and V be an m×n1 and an m×n2 matrix of variables, respectively. Then the space Vn1,λVn2,λ of the irreducible representation ρn1,λρn2,λ is realized as the linear space Vn1,n2,λ over spanned by polynomials

p(U,V)=q=1m=1λq-λq+1U[q],I(q)V[q],J(q),

where I(q) and J(q) run over subsets of {1,2,,n1} and {1,2,,n2} with |I(q)|=|J(q)|=q, respectively. Here the action of (A1,A2)GLn1()×GLn2() on Vn1,n2,λVn1,λVn2,λ is defined by

p(U,V)p(UA1,VA2).

We denote by W an m×m matrix. We identify GLm()×GLm() with a subgroup of GL2m() by

A=(A100A2),AiGLm().

We denote by W an m×m matrix of variables. If we consider a representation πm(A1,A2) of GLm()×GLm() on [W] by (πm(A)Q)(W)=Q(A1tWA2) for Q[W], then we have the following irreducible decomposition (see [13]):

[W]=depth(λ)mρm,λρm,λ.

We denote by Wi the i×i principal minor of W (i.e. the determinant of the first i rows and columns) and put

(4.2)Wλ=W1λ1-λ2W2λ2-λ3Wmλm.

Then the set {πm(A)Wλ:AGLm()×GLm()} spans the irreducible representation space ρm,λρm,λ. For an 2m×2m symmetric matrix S of variables, we write the m×m matrix blocks Sij of S by

S=(S11S12tS12S22).

We define a polynomial Q0[S] by Q0(S)=(S12)λ. Then by Theorem 2, for a fixed kn2, we have

ϕk(P0)=Q0

for some unique polynomial P0(S)[S]. The formula for P0 is given in Theorem 3. Now taking an m×n1 matrix U and an m×n2 matrix V as before, we define 2m×n matrix 𝕌 as

(4.3)𝕌=(U00V).

## Theorem 4.

Notation being as above, for an n×n symmetric matrix T of variables, we put

P(T,U,V)=P0(𝕌T𝕌t).

Then for ZHn, P(Z,U,V) is a Vn1,λVn2,λ-valued differential operator that satisfies condition (4.1) for t=2, the above fixed k, and ρ1=ρn1,λ, ρ2=λn2,λ.

## Proof.

By the assumption ϕk(P0)=Q0, we have

P0(𝕌Z𝕌t)δ-k=δ-kQ0(𝕌Δ𝕌t)

since, as explained before, this is a direct formal consequence of (D1), (D2), (D3). Here if we write the block decomposition of Δ by

Δ=(Δ(11)Δ(12)Δ(12)tΔ(22)),

where Δ(11), Δ(12), Δ(22) are an n1×n1, an n1×n2 and an n2×n2 matrix, respectively, then by definition we have

Q0(𝕌Δ𝕌t)=(UΔ(12)Vt)λ.

The i-th principal minor (UΔ(12)Vt)i is by the Laplace expansion given by

I[n1],J[n2]|I|=|J|=iU[i],I(Δ(12))IJV[i],J,

where [i]={1,,i}. So the polynomial (UΔ(12)Vt)λ in components of U and V defined by (4.2) is obviously in the irreducible representation space Vn1,λVn2,λ of bideterminants defined before. So if we take a basis {ej(U,V):j=1,,d0} of Vn1,λVn2,λ, then we have

(4.4)Q0(𝕌Δ𝕌t)=Q0(UΔ(12)Vt)=j=1d0Qj(Δ)ej(U,V)

for some polynomials Ql in components of n1×n2 matrix Δ(12). Now for simplicity we write

P(T,U,V)=P0(𝕌T𝕌t)andQ(T,U,V)=Q0(𝕌T𝕌t)

for an n×n symmetric matrix T. Then since ej(U,V) are constants as functions of T, we have obviously

P(T,U,V)=l=1d0Pj(T)ej(U,V)

for some polynomials Pj(T) and we have

Pj(Z)δ-k=δ-kQj(Δ)

for Qj in (4.4), i.e. ϕk(Pj)=Qj. For A=(A100A2)GLn(), where AiGL(ni,) for i=1, 2, we have

Q(ATtA,U,V)=Q(T,UA1,VA2)

as can be seen by definition. Define the representation matrix ρ(A) of ρ=ρn1,λρn2,λ by

(el(UA1,VA2))l=1,,d0=(ej(U,V))ρ(A).

Then we have

(Q1(ATAt)Qd0(ATAt))=ρ(A)(Q1(T)Qd0(T)).

Since we have ϕk(Pj(ATAt))=Qj(ATAt) by Theorem 2 (ii), we have

(4.5)(P1(ATAt)Pd0(ATAt))=ρ(A)(P1(T)Pd0(T))

by the injectivity of ϕk. Besides, since the image of Pl(T) by ϕk belongs to a polynomial generated by components of Δ(12), this means that Pl(T) is in the linear space spanned by P𝝂D(T) with νij=0 for all (i,j) with 1i,jn1 or n1+1i,jn1+n2. This means that

(4.6)DijPl=0for all (i,j) with 1i,jn1, or n1+1jn.

By Proposition 2, these conditions (4.5) and (4.6) are the sufficient (and necessary) condition so that the column vector (Pl(Z)), or equivalently P(Z,U,V), satisfies condition (4.1). ∎

## 5 Exact pullback formula

We write Γn=Sp(n,)M2n()Sp(n,). Let Ekn be the Siegel Eisenstein series of even weight k of degree n belonging to Γn defined by

Ekn(Z)=(C,D)det(CZ+D)-k,

where (C,D) runs over symmetric coprime pairs. We assume that k>n+1 to assure the convergence. Let n=n1+n2 be a partition of n with 1n1n2. For a dominant integral weight λ, let 𝔻=P(Z) be a Vn1,λVn2,λ-valued holomorphic differential operator with constant coefficients which satisfies condition (4.1) for k, the partition n=n1+n2 and ρi=ρni,λ (i=1. 2). The Garrett–Böcherer-type pullback formula means the formula to write down

(P(Z)Ekn)(Z100Z2),

where Z1Hn1 and Z2Hn2. Exact pullback formula for scalar-valued P has been given in [12, 3, 2] and several vector-valued cases by [4, 34, 35]. Most general but not exact formula is in Kozima [28] and [29]. A main unsolved problem in his paper is the exact behavior of P(Z)δ-k. Since we know this now, we can give an exact pullback formula for a vector-valued polynomial P given by an explicit algorithm in the previous sections.

We assume that k is a fixed integer such that k>n1+n2+1. We fix a dominant integral weight λ with m=depth(λ)min(n1,n2). Let r be an integer such that mrmin(n1,n2). We have a linear subspace Vr of Vn1,λ such that VrVr,λ and a representation σ of Gr() isomorphic to ρr,λ such that for vVr and for g=(g1g20g4)GLn1() with g1GLr(), we have

(5.1)ρ(g1g20g4)v=σ(g1)v.

Indeed, if we use the bideterminant realization, the space Vr can be taken as the space spanned by bideteminants of m×r submatrix U(r) of U consisting of the first r columns of U. In the same way, there exists a subspace Vr*Vn2,λ and which satisfies the same property as (5.1). Here Vr* can be taken as an isomorphic image of Vr by mapping the matrix U(r) to V(r), where V(r) is a submatrix of V defined in the same way as U(r). For vVr, we sometimes write the isomorphic image of v in Vr* by v*. For simplicity, we write ρr,λ=ρr from now on.

We denote by Sdetkρr(Γr) and Adetkρr(Γr) the space of Siegel cusp forms and Siegel modular forms of degree r of weight detkρr belonging to Γr, respectively. For fSdetkρr(Γr), we denote by [f]rni the Klingen lift of f to Adetkρni(Γni). As usual, this is defined as follows. We denote by Γn,r the subgroup of Γn consisting of elements whose left lower (n-r,n+r) block is 0. For Zi=(z1z12tz12z2)Hni (z1Hr, z2Hni-r), we write prrni(Zi)=z1. Then for i=1, 2, we put

[f]rni(Zi)=γΓni,r\Γnif(prrni(Zi))|detkρni,λ[γ].

Observe that this converges for k>ni+r+1 (a fortiori for k>n1+n2+1). For r×r diagonal matrices D=diag(d1,,dr) with 0<d1|d2||dr, we denote by T(D) the Hecke operator on Sdetkρr(Γr) defined by Γr(D00D-1)Γr, and assume that fSdetkρr(Γr) is the common Hecke eigen form of all T(D). Denote by λ(D,f) the eigenvalue of T(D) on f. Put

D(f)=Dλ(D,f)det(D)-k.

Then by [3], it is well known that

D(f)=ζ(k)-1i=1rζ(2k-2i)-1L(k-r,f,St),

where L(s,f,St) is the standard L function of f.

We fix an inner product *,* of Vr,ρr and define an inner product of Sdetkρr(Γr) by

(5.2)(f,g)=Γr\Hrρr(Im(z1)12)g(z1),ρr(Im(z1)12)g(z1)det(Im(z1))-r-1dz1

where z1=(xij+-1yij)Hr and dz1=1ijrdxijdyij. Also fix an orthonormal basis {fr,j:j=1,,er} of Sdetkρr(Γr) with respect to the fixed inner product (5.2) consisting of common Hecke eigenforms.

The following theorem is known.

## Theorem 5 ([28, 2, 4, 12, 34, 35]).

We have

(P(Z)Ekn)(Z100Z2)=r=mmin(n1,n2)crj=1erD(fr,j)[fr,j]rn1(Z1)[fr,j]rn2(Z2)

for some constant cr.

Although [fr,j]rni are vector valued, if we realize the representation spaces by bideterminants with respect to U and V, respectively, the tensor product [fr,j]rn1[fr,j]rn2 can be simply written as a product [fr,j]rn1[fr,j]rn2 as a polynomial in components of U and V. Here the constant cr does not depend on a choice of fr,j but depends on P (though P is unique up to constant) and also on a choice of the inner product of Vr used to define (5.2) since we are taking an orthonormal basis.

Now our problem here is to write down cr as explicitly as possible. We will explain this below. Put Q=ϕk(P), where we prolong ϕk linearly to the vectors. Let Dr be the bounded symmetric domain equivalent to Hr defined by

Dr={Z0Mr():Z0=Z0t,Z0Z0¯<1r}.

By definition, the polynomial vector Q(T)=Q0(𝕌Δt𝕌) is Vn1,λVn2,λ-valued. We know already that Q(T) depends only on the right-upper n1×n2 matrix block of T. (This is shown also in [28, Proposition 4.1] for 𝔻=P(Z) satisfying (4.1) for t=2, by using automorphic properties of the operator, but our approach is more direct as proved in Theorem 4. Kozima assumed that the differential operator is realized in some special shape, and we can also show that his realization is possible through polarizations of part of variables of our polynomial. We omit the details here. The converse process does not seem possible.) So if we write

I~r=(1r0r,n2-r0n1-r,r0n1-r,n2-r)

for simplicity, then the polynomial

Q(*I~r**)=Q0(*UI~rVt**)

does not depend on *. We put |λ|=λ1++λr. By Kozima [28, p. 247], identifying v with v*, the endomorphism

Drρr,λ(1r-Z0¯Z0)v,Q0(*UI~rVt**)det(1r-Z0¯Z0)k-r-1dZ0

of Vr is φidVr for some constant φ, and we have

cr=2r(r+1)-(rk+|λ|)+1ikl+|λ|φ.

Now we put

Ir(k,ρr,λ)=Drρr,λ(1r-Z0¯Z0)det(1l-Z0¯Z0)k-r-1dZ0.

By [27, Lemma 2], the integral is a scalar diagonal and Ir(k,ρr,λ)=c(k,ρr,λ)id, where

c(k,ρr,λ)=2rπr(r+1)21μνr(2k+λμ+λν-μ-ν).

For a vector v0Vr,λ, we have

Drdet(1r-Z0Z0¯)k-r-1ρm,λ(1r-Z0Z0¯)v0,v0v0*dZ0=c(k,ρr,λ)v0,v0v0*.

In order to avoid an ambiguity depending on an inner product of Vr, we define it here explicitly. Let λ=(λ1,,λm,0,,) be a dominant integral weight with mr as before. We take the bideterminant realization Vr=Vr,λ of the representation λr,λ of GLr(), We denote by U an m×r matrix of variables and write

U=(uij)1im,1jr.

For polynomials P, Q[U], we define the inner product of Vr by

(5.3)P(U),Q(U)=P(U)Q¯(U)|U=0,P,QVr,

where Q¯ is the polynomial obtained by replacing the coefficients by the complex conjugation. Then for gGLr(), we can easily show that

(5.4)P(Ug),Q(U)=P(U),Q(Utg¯).

We use this inner product to define (5.2). For any positive integer r, we write [r]={1,,r}. For any m×r matrix A and an integer i with 1im, we denote by Ai the i×i principal minor of A as before, and by AIJ the (I,J)-minor of A for I[m] and J[r] with |I|=|J|. We take the highest weight vector vλVr=Vr,λ defined by

vλ=U1λ1-λ2U2λ2-λ3Umλm.

(Note that since we assumed depth(λ)=m, we have automatically λm+1=0.) For =U, we write

λ=1λ1-λ22λ2-λ3mλm.

By definition (5.4), for any uVr, we have

vλ,u=λ(u)|U=0.

Define (UI~rVt)λ as in (4.2). Then by the Laplace expansion, this is a linear combination of the terms

R(U,V,{Ki,j})=i=1mj=1λi-λi+1U[i],Ki,jV[i].Ki,j

where {Ki,j} is a family of subsets of [r] such that |Ki,j|=i for i=1,,m. When we define vλ* to be the isomorphic image of vλ in Vr* obtained by replacing U by V, then in case when Ki,j=[i] for all i, j, we have R(U,V,{Ki,j})=vλvλ*.

## Lemma 4.

Notation being as above, we have

vλ,R(U,V,{Ki,j})0

if and only if Ki,j=[i] for all i, j.

## Proof.

We fix a family {Ki,j} such that Ki,j[r], |Ki,j|=i, where i runs over 1 to r and for each i, the number j runs over 1 to λi-λi+1. We assume that λ(R(U,V,{Ki,j})0 for this fixed family. We fix a positive integer i0 with 1i0m and consider the claim that i0Ki,j for any i0i and any j. We prove by induction on i0 that the above claim holds for any i0. We first prove it for i0=1. By expansion of the determinants, we see that λ is a linear combination of the differential operators such as

(5.5)0ν=1λ1iν,1,

where ab is the (a,b) component of =U for each (a,b) and 0 are products of *,i with 2i, On the other hand, by definition, the polynomial R(U,V,{Ki,j}) is homogeneous as a polynomial in u11, u21,,um1 and its total degree with respect to these variables is equal to

N1=|{(i,j):1Ki,j}|.

The total number of (i,j) is λ1, so we have N1λ1. So if N1<λ1, then the polynomial R(U,V,{Ki,j}) vanishes under any differential operator of the shape (5.5). This means that λR(U,V,{Ki,j})0 only if 1Ki,j for all (i,j). In particular, we have K1,j={1} for any j. Next, fix i0 and assume that Ki,j=[i] for any ii0 and [i0]Ki,j for any ii0. We put

Ni0+1=|{(i,j):i0+1Ki,j}|.

By the inductive assumption, we have i0+1Ki,j if i<i0+1. So we have Ni0+1λi0+1. On the other hand, each monomial in λ is of degree λi0+1 with respect to 1,i0+1,,m,i0+1. So if λ(R(U,V,{Ki,j}))0, then we have Ni0+1=λi0+1. This means that we have i0+1Ki,j for any i0+1i. Continuing the induction, we can conclude that Ki,j=[i] for any (i,j). Since λ(vλ)=vλ,vλ0, this is also a sufficient condition for non-vanishing. ∎

By Lemma 4, if we write

Q0(*UI~rVt**)=(UI~rVt)λ=vλvλ*+R(U,V)

for some polynomial R(U,V), then R(U,V) is orthogonal to vλ as an element in Vr. In other words, we have

vλ,Q0(*UI~rVt**)=vλ,vλvλ*.

## Theorem 6.

We define P as in the previous section by setting Q0=Qλ, ϕk(P0)=Q0 and P(Z)=P0(UZUt) for U=(U00V). Then for the above inner product, the constants cr in Theorem 5 are given by

cr=vλ,vλ×(-1)rk+|λ|22(r+1)2πr(r+1)21μνr(2k+λμ+λν-μ-ν).

By the way, we assumed k is even to define the Siegel Eisenstein series, and if |λ| is odd, then we have ρr,λ(-1r)=-1, so Sdetkρr,λ(Γr)=0. This means that we may assume that |λ| is even and (-1)rk+|λ|2=±1.

For completeness, we give the value of the inner product vλ,vλ for the highest weight vector vλ for the above inner product. In the following argument, we can assume r=m without loss of generality.

## Proposition 3.

We have

vλ,vλ=λ(vλ)=i=1m(λi+m-i)!1i<jm(λi-λj+j-i).

This is an easy corollary of Proposition 4 below. For I={i1,,ip} and J={j1,,jp}{1,,m}, we define the signature of (I,J) as usual by

ϵ(I,J)=(-1)i1++ip+j1++jp.

For I{1,,m}, we denote by Ic the complement of I in {1,,m}, We fix any non-negative integers s1,,sm, and for any integer with 1m, we write

Fm()=UsU+1s+1Umsm.

In particular, we write Fm=Fm(1). For any and p with 1,pm and m-p, we put

c(p,)=(sm+sm-1++s+m-)p-(m-)sm(sm+sm-1+1)(sm++s+1+m--1).

## Proposition 4.

We fix an integer m. Let be an integer with 1m. For subsets I, J{1,,m} with |I|=|J|=p such that

{+1,,m}I,J,

we have

I,J(Fm())=c(p,)ϵ(I,J)UIcJcFm()Um.

For a fixed m, Proposition 4 can be proved by a purely algebraic induction from =m to =1 by using [7, 8, 14], but we omit the proof here since Fumihiro Sato kindly informed me that these propositions are well-known results on the b-functions of the prehomogeneous vector space Mm() with the lower triangular and upper triangular group acting from left and right, respectively (see [32, Example 8] or [31]).

## 6 Some simpler special cases

In some special cases, there is a more direct way to calculate 𝔻(det(Z)-k). Also sometimes we have more explicit formulas for the differential operators and pullback formulas. In this section we give remarks on these. When n=2m, n1=n2=m and ρm,λ=det in Condition (4.1), then an explicit pullback formula is known in Böcherer [2]. There he constructed an explicit differential operator 𝒟~k for =1 which commutes with the action of Sp(m,)×Sp(m,) without restriction of the domain and then iterated it times. The results of this iteration would be very complicated in general and there is no exact formula. Here in the present paper, in the same scalar-valued case, we also have a simpler approach different from the one in the previous sections to the action on det(Z)-k and the pullback formula, still for differential operators that preserve automorphy under the restriction of the domain. Our ingredient of the proof here is the Fourier transform of pluri-harmonic polynomials.

First we assume that k is an integer with km. Let S be a 2m×2m symmetric matrix of variables and ZHn=H2m. Let P0(S) be a non-zero polynomial such that the differential operator 𝔻=P0(Z) satisfies condition (4.1) for k and ρ1=ρ2=det. Such P0 exists and is unique up to constant for km (see [15]). If we put B=Z-1 and denote by B(12) the right upper m×m block of B, then by Theorem 2 and [15], we see that ϕk(P0) is equal to det(B(12)) up to non-zero constant. Multiplying a rational function of k if necessary, we can take P0 so that the coefficients are rational functions of k that is non-singular for k with 2k^2m. We fix such a P0 here. We put

Km=(1m1m1m1m).

## Theorem 7.

The notation and assumption being as above, we can show P0(Km)0. For ZHn and k with 2kC^2m, we have

(6.1)P0(Z)(det(Z)-k)=P0(Km)det(Z)-k(-1)mldet(B(12))j=1m(k-j-12),

where (*) is the ascending Pochhammer symbol. In (6.1), we may also change Z and B by CZ+D and Δ=(CZ+D)-1C, respectively.

## Proof.

The proof is similar to that of Theorem 2, but we use here the pluri-harmonicity of P0. We first assume that k is an integer with km. By the assumption that P0 satisfies condition (4.1) for k and detk+ldetk+l, this means that deg(P0)=m and DijP0=0 for 1i,jm and m+1i,j2m=n by [15]. So if we put

P~(X1,X2)=P0(X1X1tX1X2tX2X1tX2X2t)

for m×2k matrices X1 and X2, then P~ is pluri-harmonic for each X1 or X2. We write

Z=(Z1Z12tZ12Z2)(Z1Z2HmZ12Mm())

and put X=(X1X2). We have

det(Zi)-k=(2π)-knMn,2k()ei2Tr(XtZX)𝑑X,

so we have

(6.2)𝔻(det(Zi)-k)=(2π)-2km(i2)lmM2m,2k()P~(X1,X2)ei2(Tr(X1tZ1X1)+2Tr(X1tZ12X2)+Tr(X2tZ2X2))𝑑X1𝑑X2.

For any pluri-harmonic polynomial P1(X1) of X1, any m×2k matrix X0, and any Z1Hm, we have

(6.3)Mm,2k()eiTr(X1tX0)ei2Tr(X1tZ1X1)P1(X1)𝑑X1=(2π)mkdet(Z1i)-ke12Tr(X0t(-Z1-1)X0)P1(-Z1-1X0)

(see [25, Lemma 4.5]). We apply (6.3) for (6.2) by setting X0=Z12X2 and P1(X1)=P~(X1,X2). Since

P~(BX1,CX2)=det(BC)lP~(X1,X2)

for any B, CGLm() by definition, we have

P~(-Z1-1X0,X2)=det(-Z1)-ldet(Z12)lP~(X2,X2).

Here, if we write T2=X2X2t, we have

P~(X2,X2)=P0(X2X2tX2X2tX2X2tX2X2t)=det(T2)lP0(Km),

by noting that for α=αt with α2=T2, we have

P~(X2,X2)=P0((α00α)Km(α00α)).

So we have

(6.4)𝔻(det(Zi)-k)=(2π)-km(i2)mldet(Z1i)-kdet(-Z1)-det(Z12)P0(Km)I(T2,Z),

where we write

I(T2,Z)=Mm,2k()det(X2X2t)ei2Tr(X2t(Z2-tZ12Z1-1Z12)X2)dX2.

Taking the imaginary part

Y=(Y1Y12tY12Y2)

of Z=(Z1Z12tZ12Z2), and by changing the coordinate from X2 to (Y2-Y12tY1-1Y12)12X2 and then to T2=X2X2t, we have

I(T2,iY)=(2π)mkcm(2k)det(Y2-Y12tY1-1Y12)-k-lT2>0e-12Tr(T2)det(T2)l+k-m+12dT2,

where cm(2k) is given by (2.3). It is well known that

T2>0e-12Tr(T2)det(T2)l+k-m+12dT2=2m(k+l)πm(m-1)4j=1mΓ(k+l-j-12).

By noting det(Y1)det(Y2-Y12tY1-1Y12)=det(Y) and rewriting cm(2k) by the relation

Γ(k+l-j-12)Γ(k-j-12)-1=(k-j-12)l,

we see that the right-hand side of (6.4) for Z=iY is given by

imldet(Y)-kdet(-Y1-1Y12(Y2-Y12tY1-1Y12)-1)j=1m(k-j-12)l.

Since

Z-1=(Z1-1+Z1-1Z12(Z2-Z12tZ1-1Z12)Z12tZ1-1-Z1-1Z12(Z2-Z12tZ1-1Z12)-1-(Z2-Z12tZ1-1Z12)-1Z12tZ1-1(Z2-Z12tZ1-1Z12)-1),

we see that for Z=iY, we have

det(-Y1-1Y12(Y2-Y12tY1-1Y12)-1)=imdet(B(12)).

We note that i part of det(Zi)-k cancels in both sides. Continuing analytically from iY to any ZHn, we have (6.1). The fact that P0(Km)0 is obvious since ϕk is bijective for the present k by Theorem 2. The proof for k with 2k^2m is similar as before. ∎

## Remark 2.

In [2, Lemma 10], S. Böcherer gave a formula that 𝒟~kdet(CZ+D)-k=constdet(CZ+D)-k-1 for some special (C,D), where const is a constant depending on C, D. This claim apparently looks close to the above Theorem but actually very different. The operator 𝒟~k is equal to our operator for l=1 if we restrict it to the diagonal blocks. But his operator have polynomial coefficients in zij and operation on det(CZ+D)-k itself is quite different before the restriction. The advantage of his operator is that it can be iterated. Our operator would be regarded as a main term of his operator. A comparison of two operators is interesting but we do not discuss it here.

The exact pullback formula for the case λ=(,,,0,,0) with depth(λ)=m is easy to write down. We define 𝕌 as in (4.3) and for ZHn, put P(T)=P0(𝕌T𝕌t). To make the formula simpler, we normalize an inner product of Vr,λ by defining

v,w0=v,w1i<jm(j-i)i=1m(l+m-i)!,

where <v,w> is define by (5.3) and use this to define the Petersson inner metric in (5.2). Then for n=n1+n2 with mmin(n1,n2) for this metric, we have:

## Theorem 8.

We have

P(Z)Ekn(Z100Z2)=P0(Km)(-1)mlj=1m(k-j-12)lr=mmin(n1,n2)crj=1erD(fr,j)[fr,j]rn1(Z1)[fr,j]rn2(Z2),

where fr,j runs over orthonormal basis of Sdetkρr,λ(Γr) with respect to the inner product *,*0, and

cr=(-1)k(r+m)22(r+1)2πr(r+1)21μνr(2k+λμ-λν-μ-ν).

Here λκ= if κm and =0 if m<κ.

We can make P0(Km) in Theorem 7 more explicit for the following two kinds of differential operators. The first one is for λ(1)=(,0,,0) and the second one is for λ(2)=(,,0,,0). (The partition n=n1+n2 is general, but we must assume 2n1, n2 in the second case.) First we consider the case λ(1) for the partition n=2=1+1. Let S=(s11s12s12s22) be a 2×2 symmetric matrix. Then the differential operator 𝔻=P(Z) satisfying condition (4.1) is given by P0(S)=P(S) up to constant, where P are defined by

1(1-2s12u+s11s22u2)k-1==0P(S)u

(see [15, p. 114] or [4]). If S=K1, then this is equal to

1(1-2u+u2)k-1==0(2k-2)l!ul.

So for P, we have P(K1)=(2k-2)ll!. Let T be an n×n symmetric matrix and write matrix blocks as

T=(T11T12tT12T22),

where Tii is an ni×ni symmetric matrices and T12 is an n1×n2 matrix.

We write u=(u1,,un1), v=(v1,,vn2). We put

𝕌=(u00v)and𝔻U=P(uT11utuT12vtvT12tutvT22vt).

Then for ZHn, we have

𝔻𝕌(det(Z)-k)=det(Z)-k((Δ𝕌)(12))(-1)(k)(2k-2)!.

This is essentially the same as in [4].

Next we consider a 4×4 symmetric matrix S=(S11S12tS12S22) where Sij are 2×2 blocks. We put

F1(S)=det(S12),
F2(S)=det(S11)det(S22),
F3(S)=det(S).

We have F1(K2)=1, F2(K2)=1, F3(K2)=0. We define G(S) by

G(S)=a+2b+2c=2a(-1)ba!b!c!(k+c-32)a+b+cF1(S)aF2(S)bF3(S)c.

Then using the generating function of G(S) in [15, p. 114], we have

=0G(K2)u==0(2k-3)!u.

Let U, V be a 2×n1 matrix and a 2×n2 matrix, respectively, and put

𝕌=(U00V).

Put n=n1+n2. For ZHn, define Z as before, and put 𝔻(2)=G(𝕌Z𝕌t). Then by [15], this is a differential operator for ρni,λ(2) for t=2 in (4.1), and we have

𝔻(2)(det(Z)-k)=det(Z)-kdet((Δ𝕌)(12))(2k-3)(2k-1)222!.

This is exactly the same as the one given in [1]. The corresponding pullback formulas for the above two cases are obvious by explicit values of Pl(K1) and G(K2) and by Theorem 8. We do not repeat them here (see [4, 17, 1]).

Communicated by Jan Bruinier

Funding source: Japan Society for the Promotion of Science

Award Identifier / Grant number: JP19K03424

Award Identifier / Grant number: JP20H00115

Funding statement: This work was supported by JSPS Kakennhi Grant Numbers JP19K03424 and JP20H00115.

# Acknowledgements

The author thanks H. Katsurada for asking him if we can give a pullback formula of Eisenstein series that is explicit enough to apply for obtaining congruences between modular forms. A part of this paper is an answer to his question.

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