Scattering Theory with Unitary Twists

We study the spectral properties of the Laplace operator associated to a hyperbolic surface in the presence of a unitary representation of the fundamental group. Following the approach by Guillop\'e and Zworski, we establish a factorization formula for the twisted scattering determinant and describe the behavior of the scattering matrix in a neighborhood of $1/2$.


Introduction
We consider a finitely generated Fuchsian group Γ ⊂ PSL(2, R) and denote the associated hyperbolic surface by X.Thus X = Γ \H, where H denotes the hyperbolic upper half-plane and PSL(2, R) acts via Möbius transformations on H. Throughout this article, we will suppose that X is non-elementary, geometrically finite and of infinite volume.However, we allow that X has orbifold singularities or, equivalently, that Γ has torsion.
We further consider a finite-dimensional unitary representation on a Hermitian vector space V .The representation χ induces a Hermitian vector orbibundle with typical fiber V .It is well-known that the (smooth) sections of E χ are in bijection with the smooth functions f : H → V that obey the twisting equivariance f (g.z) = χ(g)f (z) , z ∈ H, g ∈ Γ .
(1) See, for example, [DFP,Lemma 3.3] for details.On smooth maps f : H → V , the hyperbolic Laplacian is given by where z = x + iy ∈ H. Using the identification of twisted functions (see (1)) and sections of E χ and the fact that χ is unitary, the Laplacian ∆ H gives rise to a non-negative self-adjoint operator ∆ X,χ : L 2 (X, E χ ) → L 2 (X, E χ ) .
For Re s > 1/2 and s ∈ [1/2, 1], the resolvent of ∆ X,χ is defined by As shown in [DFP, Theorem A], the resolvent R X,χ admits a meromorphic continuation to s ∈ C as an operator The poles of R X,χ (s) are the resonances of ∆ X,χ .The multiplicity of the pole s ∈ C is the rank of the residue at s.
In [DFP, Theorem B], we showed that the resonance counting function grows at most quadratically, i.e., s∈R X,χ |s|≤r m X,χ (s) = O(r 2 ) as r → ∞ , where R X,χ denotes the set of resonances and m X,χ (s) the multiplicity of s ∈ R X,χ .Hence, by the Weierstrass factorization theorem, there exists an entire function, P X,χ , such that its zeros coincide with the resonances, and the multiplicity of a zero s of P X,χ is equal to m X,χ (s).We also define the Weierstrass product P X f ,χ (s) associated to the resonances of the disjoint union of funnel ends X f (see Section 5.5 for details).We consider the scattering matrix, which is a certain operator defined on the boundary of a suitable compactification of E χ (see Sections 3 and 5).For each ψ ∈ C ∞ (∂ ∞ X, E χ ) there exists u ∈ C ∞ (X, E χ ) such that (∆ X,χ − s(1 − s))u = 0 and where ρ f and ρ c are the boundary defining functions in the funnel and cusp ends, respectively.Even though the scattering matrix is not trace class, we can define a regularized determinant of S X,χ (s), that we will call the relative scattering determinant, τ X,χ (s).
As the first main result of this article, we prove a factorization of the relative scattering determinant in terms of the Weierstrass product over the resonances.
Theorem A. The scattering determinant admits the factorization τ X,χ (s) = e q(s) P X,χ (1 − s) P X,χ (s) P X f ,χ (s) where q : C → C is a polynomial of degree at most 4.
For dim V = 1 and χ = id, Theorem A reduces to [GZ97,Proposition 3.7].This latter result plays a crucial role in the proof of the factorization of the Selberg zeta function by Borthwick-Judge-Perry [BJP05].
We remark that Theorem A implies that the scattering determinant has no pole or zero at s = 1/2.However, s = 1/2 might be a resonance.The second main result of this article shows that we are able to describe the behavior of the scattering matrix S X,χ (s) in some (small) neighborhood of 1/2.For this, we set Then S X,χ (s) = − id +2P + (2s − 1)T X,χ (s) with T X,χ being an operator family that is holomorphic in a small neighborhood of s = 1/2.Theorem B. The operator P is an orthogonal projection of rank m X,χ (1/2) onto the space of elements in C ∞ (∂ ∞ X, E χ ) that are invariant under the map S X,χ (1/2).
Structure of this article.In Section 3, we discuss the scattering matrices for the model funnel and the parabolic cylinder.In Section 4, we obtain a decomposition of the resolvent, study the structure of the resolvent close to a resonance and obtain that there are no resonances on the line Re(s) = 1/2 except for, maybe, s = 1/2.In Section 5, we introduce the scattering matrix, the relative scattering determinant and prove Theorems A and B.
Acknowledgements.AP's research is funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) -project no.441868048 (Priority Program 2026 "Geometry at Infinity").MD was partially funded by a Universität Bremen ZF 04-A grant.

Preliminaries and Notation
We let X and E χ be as above.We denote by (•, •) Eχ the Hermitian bundle metric on E χ that is induced from the sesquilinear inner product (•, •) V on V .We denote by •, • Eχ the bilinear metric on E χ corresponding to the bundle metric (•, •) Eχ .We abbreviate the norm |v| Eχ = (v, v) Eχ of any v ∈ E χ by |v|.By Selberg's Lemma [Sel60, Lemma 8], there is a finite cover X = Γ \H of X such that the Fuchsian group Γ is a torsion-free subgroup of Γ .We denote the pull-back of E χ under the covering map X → X by E, which becomes a vector bundle over X.We call an operator A acting on the sections of E χ a pseudodifferential operator of order m ∈ R if its pull-back, A, under the map X → X is a pseudodifferential operator of order m, acting on the sections of E.
In the case of the 1-sphere S 1 , pseudodifferential operators have a very simple characterization using Fourier series, which we recall now.To that end let A : C ∞ (S 1 ) → C ∞ (S 1 ) be a continuous linear operator.As proven by McLean [McL91,Theorem 4.4], A is a pseudodifferential operator of order m ∈ R if and only if is a periodic symbol of order m.This means that a ∈ C ∞ (S 1 × Z), and for all b, c ∈ N 0 , we have Here, △ ξ denotes the discrete derivative, i.e., If the constant, C, depends on additional parameters, we indicate the dependence in the subscript.
Let H be a Hilbert space and let B : H → H be a bounded operator.The non-zero eigenvalues of (B * B) 1/2 are called the singular values of the operator B. We denote these singular values by µ k (B), k ∈ N, listed in decreasing order.
For z ∈ C we define the Japanese bracket z := 1 + |z| 2 1/2 .We use the convention to call a function, f , meromorphic on an open set U ⊆ C if there exists a discrete subset, P , of U such that f , considered as a function, is defined on U \ P only, and f is holomorphic on U \ P and has poles (of finite order, which might be zero) at the points in P .

The Scattering Matrix for the Model Cylinders
In this section we present the structure of the twisted scattering matrix for the model ends.We discuss the model funnel in Section 3.1 and the model cusp in Section 3.2.The analysis was originally done in [DFP,Section 4].
Here we restrict to presenting the main results only.
3.1.Model funnel.Let ℓ ∈ (0, ∞) and set ω := 2π/ℓ.We define the hyperbolic cylinder as the quotient C ℓ := h ℓ \H, where h ℓ .z= e ℓ z.We may change coordinates via such a way that the induced metric from the hyperbolic plane becomes g C ℓ (r, φ) := dr 2 + ℓ 2 4π 2 cosh 2 r dφ 2 .We define the model funnel as Taking the boundary defining function as a coordinate function, we may rewrite the funnel metric as The volume form is We also define the metric restricted to the boundary at infinity 4π 2 dφ 2 and denote the corresponding measure by The Laplacian acting on functions F ℓ → C takes the form Let χ : h ℓ → U(V ) be a finite-dimensional unitary representation.As above, we denote by ∆ C ℓ ,χ the Laplacian acting on sections of the vector bundle E χ = h ℓ \(H × V ) over C ℓ .The Laplacian ∆ F ℓ ,χ is the restriction of the Laplacian ∆ C ℓ ,χ to F ℓ with Dirichlet boundary conditions at r = 0. We will also denote by E χ the restriction of E χ to F ℓ .It was shown in [DFP,Proposition 4.12] that the resolvent of the model funnel, (∆ F ℓ ,χ −s(1−s)) −1 , admits a meromorphic continuation to C as an operator where EV(χ(h ℓ )) denotes the multiset of eigenvalues of χ(h ℓ ).See [DFP,Proposition 4.12].
Let ψ ∈ C ∞ (F ℓ × F ℓ ) such that ψ is supported away from the diagonal and s ∈ C is not a pole of the resolvent R F ℓ ,χ .By [DFP,Proposition 4.12], we have where E χ ⊠ E ′ χ is the exterior tensor product of E χ and its dual E ′ χ defined by is well-defined.This allows us to introduce the Poisson operator We now recall the Fourier expansion of the Poisson operator.Let (ψ j ) dim V j=1 be an eigenbasis of χ(h ℓ ) with eigenvalues λ j = e 2πiϑ j , j = 1, . . ., dim V .Let κ ∈ R and s ∈ C \ (−1 − 2N 0 ± iωκ).We define We recall that the regularized hypergeometric function F(a, b; c; z) is defined for a, b, c ∈ C and z ∈ C, |z| < 1, by the power series (see [Olv97,Theorem 9.1]) For arbitrary κ ∈ R, s ∈ C and r ≥ 0, we define It was shown in [DFP,Remark 4.13] that where .
By [DFP,Lemma 6.15], we have for ε ∈ (0, 1/2), ϕ ∈ C ∞ c (F ℓ ), there exist C, c > 0 such that for all s ∈ C with Re s > ε we have the estimate Moreover, the scattering matrix was defined in [DFP,(66)] via the Fourier coefficients of its Schwartz kernel, where From the Fourier expansion, we obtain that 2 ) and therefore by a connection formula (see [Bor16,p. 93]).We note that by [DFP,(67)], where the coefficient functions a m , b m for m ∈ N 0 are meromorphic, with the leading coefficient functions being From this, we obtain for Re s < 1/2 that In what follows we will argue that the scattering matrix is a pseudodifferential operator on ∂ ∞ F ℓ and calculate its principal symbol.From [DLMF, Eq. 5.11.13], we have that for a, b ∈ C and |arg(z)| < π − ε for fixed ε > 0, for some G k (a, b) ∈ C.This immediately implies that for y → ∞, Taking y = ωκ/2, a = 1/2 + s/2 and b = 1 − s/2, we obtain We note that the terms in (14) with odd k vanish, since the left hand side is even as a function of ωκ.The definition in (9) combined with (14) shows that β κ (s) By [DLMF, Eq. 5.11.15], the leading coefficient is given by G 0 (a, b) 2 = 1.Combining this with (11), we obtain that The full asymptotic expansion now implies that S F ℓ ,χ (s) j k satisfies the symbol estimates for global pseudodifferential operators on the torus, as stated in (3).Hence, We define the reduced scattering matrix SF ℓ ,χ (s) as follows: we consider the invertible elliptic pseudodifferential operator and define SF ℓ ,χ (s A straightforward calculation shows that the Fourier coefficients of SF ℓ ,χ (s) are Since the right-hand side of the last equation is defined for all s ∈ R F ℓ ,χ , the scattering matrix is defined as an operator SF ℓ ,χ (s) Proof.By definition of β κ , we have that is holomorphic and non-zero for Re s < 1/2.Therefore, the poles counted with multiplicities of SF ℓ ,χ (s) j k are given by the multiset Hence, the poles of SF ℓ ,χ (s) are given by the multiset (6).
Finally, we recall the singular value estimate for the scattering matrix from [DFP,Lemma 6.14].For this, we will define functions d k : C → C for k ∈ N, which have poles contained in the set of resonances of ∆ F ℓ ,χ .We set where we denote by m ϑ the multiplicity of the eigenvalue λ = e 2πiϑ of χ(h ℓ ).We can assume without loss of generality that ϑ ∈ [0, 1).Denote by d C the Euclidean distance on C. For k ∈ N we define d k,ϑ (s) as follows: for ϑ ∈ (0, 1) \ {1/2}, we set and for ϑ ∈ {0, 1/2}, we set Moreover, we define the function dk,0 by dk,0 (s Finally, we set It is shown in [DFP,Lemma 6.14] that for any ε ∈ (0, 1/2) there exists C > 0 such that for s ∈ C with Re s < 1/2 − ε, 3.2.Parabolic cylinders.We now turn to the parabolic cylinder, where the structure of the resolvent is slightly simpler than for the hyperbolic cylinder.
The parabolic cylinder is given by C ∞ := T \H, where T.z := z + 1.We can choose as fundamental domain the set With the coordinates (ρ, φ) = (y −1 , (2π) −1 x), the induced Riemannian metric reads and ρ c (ρ, φ) = ρ, where ρ c is the canonical boundary defining function.In the (x, y)-coordinates the Laplacian is given by x + ∂ 2 y ) .Let χ : T → U(V ) be a finite-dimensional unitary representation.We denote by E 1 (χ(T )) the eigenspace of χ(T ) for eigenvalue 1, and we set The meromorphically continued resolvent R C∞,χ (s) defines a continuous map provided that s = 1/2, where ψ is any element of C ∞ (C ∞ ) that is supported away from {y = 0}.The only pole of R C∞,χ (s) is at the point s = 1/2 and its multiplicity is equal to n χ c .The integral kernel of the resolvent R C∞,χ (s) admits a Fourier decomposition.For any j ∈ {1, . . ., dim V }, the Fourier decomposition of the non-vanishing matrix coefficients R C∞,χ (s; z, z ′ ) j is given by Here, the maps u κ for κ ∈ R are defined as follows: for κ ∈ R \ {0}, we set where I s−1/2 and K s−1/2 is the modified Bessel function of the first and the second kind, respectively (see [Wat66, § 3.7]).Moreover, for κ = 0 and s = 1/2 we set The Poisson operator E C∞,χ (s) is given by by the Fourier decomposition in (21).In particular,

Analysis of the Resolvent
In this section, we discuss fine-structure properties of the resolvent of ∆ X,χ .We start, in Theorem 4.1, with a decomposition of its resolvent into interior and residual terms, which are then discussed separately in more detail.In Section 4.1, we give a description of the resolvent near a resonance.In Section 4.2, we prove that on the line Re(s) = 1/2 there are no resonances except for potentially s = 1/2.Moreover, in Proposition 4.7, we prove that if the hyperbolic surface X has infinite volume, then ∆ X,χ has no eigenvalues larger than 1/4.
As in [DFP, Section 3.2.3],we take advantage of the decomposition where K is compact and X f and X c are finite collections of funnels and cusps, respectively.For We fix s 0 ∈ C with sufficiently large real part (such that s(1−s) is sufficiently far away from the spectrum of ∆ X,χ ) and denote by n f and n c the number of connected components of X f and X c , respectively.As in [DFP, Section 5], we set s) and, as in [DFP,(85)], we define where and It was proven in [DFP, Section 5] that (id −L(s)) −1 exists as a meromorphic family in s ∈ C.
Theorem 4.1.Let X = Γ \H be geometrically finite and let χ : where • Mi (s) is a compactly supported pseudodifferential operator of order −2, • M f (s) and M c (s) are as in (24) and (25), respectively, and Remark 4.2.The product X × X is not a smooth manifold (even in the absence of orbifold points).The reason is that the geodesic boundary at infinity of a cusp end is a single point.Blowing up each parabolic fixed point to a 1-sphere, we obtain a orbifold with boundary X ′ .We define smooth functions on X × X as the set of function that pullback to smooth functions on X ′ × X ′ .
Note that blowing up a parabolic fixed point amounts to introducing coordinates (ρ, φ) as in Section 3.2, where {ρ = 0} ∼ = S 1 is the blowup of the parabolic fixed point.
Proof of Theorem 4.1.We set For notational simplicity, define η 3 := η f,3 η c,3 .We now split M (s)K(s) as We now have to show that Mi (s) and Q(s) have the claimed properties.
Interior term.The operator M i is a compactly supported pseudodifferential operator by definition, so it suffices to show that η 3 M (s)K(s)η 3 is a pseudodifferential operator of order at most −2.By (26) we have

Now equation (31) directly implies that
for s close to s 0 .By the identity theorem for holomorphic functions, the equality in (32) is valid for all s ∈ C. Formally, we can also obtain this from the geometric series, id +K(s) We have that where Q(s) is compactly supported and smoothing.Thus, L(s)η 3 is a pseudodifferential operator of order −2 and therefore is also a pseudodifferential operator of order −2.By the definition of M (s), the operator η 3 M (s) is a pseudodifferential operator of order −2 and hence η 3 M (s)K(s)η 3 is a pseudodifferential operator of order −4.
Residual term.To study the operator Q(s), we start by considering the operator M (s)K(s)(1 − η 3 ).We use (30) to show that Since L(s) maps to compactly supported smooth sections, we use the explicit calculations for the model resolvents to obtain that, for any ϕ ∈ C ∞ c (X, E χ ), we have Moreover, the property and is compactly supported in the left-most variable.Using that M i is a compactly supported pseudodifferential operator and M f (s) and M c (s) are given by the model resolvents, we conclude that the integral kernel of the operator M (s)K(s)(1 − η 3 ) satisfies For (1 − η 3 )M (s)K(s), we use that K(s)η 3 is compactly supported and that the integral kernel of the operator (1 . This proves the theorem. We will now provide formula for Q(s) restricted to the boundary that will be useful later on.Let ϕ ∈ C ∞ (X, E χ ) such that η 3 ϕ = 0.In this case Q(s)ϕ simplifies to Using that L(s) = η 3 L(s), we obtain Hence, we have 4.1.Resolvent at a Resonance.Let s 0 ∈ C be a resonance of ∆ X,χ .As in [DFP, Section 6], we define the multiplicity of the resonance s 0 as the number encloses exactly one resonance (namely s 0 ).We denote the multiset of resonances by and the multiset of resonances of the model funnel ends by where the multiset R X f,j ,χ j is given as in (6).
In a small neighborhood of the resonance s 0 , the resolvent admits an expansion for some p ∈ N, further referred to as the order of the resonance, where, for j = 1, . . ., p, the coefficient A j (s 0 ) is a finite rank operator, and the map s → H(s, s 0 ) is holomorphic in a small neighborhood of s 0 .Now let s 0 = 1/2 and fix j = 1, . . ., p.We multiply (35) by and integrate both sides along the path γ ε,s 0 .We substitute λ = s(1 − s) and use dλ = (1 − 2s)ds.The path of the integration changes to γε,s For s 0 = 1/2 and ε small enough, γε,s 0 (t) winds around s 0 (1 − s 0 ) once.
Applying the Cauchy integration formula, we get for any j = 1, . . ., p.We note that for j = 1, the equality (36) was obtained in the proof of [GZ97, Lemma 2.4].

Absence of Poles with
Re s = 1/2.In this section, we will show that for s ∈ C with Re s = 1/2 there is at most one resonance at s = 1/2.This will imply that there are no eigenvalues larger than 1/4.
Proposition 4.3.Let F ℓ ⊂ C ℓ = h ℓ \H be a hyperbolic funnel and let χ : h ℓ → U(V ) be a unitary finite-dimensional representation.Denote by and is supported in {r ≥ r 0 }, where r denotes the distance to the geodesic boundary.For r 0 and k sufficiently large there exists C > 0 independent of k such that The Carleman estimate implies the following result on unique continuations (see [Bor16,Proposition 7.4] for the untwisted case).
We adapt the proof of [Bor16, Proposition 7.4] to the twisted case.
Proof of Proposition 4.4.Without loss of generality, we assume that X has only one funnel end, that is n f = 1 and X f = X f,1 .We prove the proposition in two steps.
Step 1: We want to show by induction that The base case is true by (39).Suppose that u|
Therefore we arrive at Letting k → ∞, we obtain that u L 2 (X f ∩{ρ f ≤r 0 },Eχ) = 0 and consequently u = 0 on X f ∩ {ρ f ≤ r 0 }.By standard uniqueness results of elliptic differential operators, we conclude that u = 0 everywhere.
In the case Re s = 1/2 and s = 1/2, we can prove a better result following [Bor16, Lemma 7.7].
Proof.Without loss of generality, we may suppose that n f = 1 and X f = X f,1 .We take local coordinates (r, φ) ∈ R + × R/2πZ ∼ = X f .We have that u(r, φ + 2π) = (χ(h ℓ )u)(r, φ), where h ℓ ∈ Γ is the unique (up to inversion) hyperbolic element associated to the funnel end X f and ℓ ∈ (0, ∞) is the length of the central geodesic of X f (see [DFP, Section 3.2.3]for details).
Let ε > 0 and let ψ ∈ C ∞ (R + ) be real-valued with ψ(t) = 0 for t ≤ 1 and The function u can be written as u = ρ s v, where v ∈ C ∞ (X f , E χ ).By assumption, we have that Re s = 1/2, therefore |u| 2 = ρ|v| 2 .Writing x = ρ/ε, we obtain, using (5), that as ε → 0. It follows from (4), that the measure dµ X restricted to X f is given by Therefore we have, as ε → 0, that We calculate that Together with Proposition 4.4 this implies the claim.Proposition 4.5 implies almost no resonances of the critial line.
Corollary 4.6.For Re s = 1/2 and s = 1/2, the resolvent R X,χ has no pole at s. Proof.By (35), we have that where p ∈ N is the order of the resonance, A j (s 0 ), j = 1, . . ., p are finite rank operators and H(s, s 0 ) is holomorphic in s near s = s 0 .Let ψ ∈ C ∞ c (X, E χ ) and write u = A p (s 0 )ψ.By the definition of the resolvent, we have that For Re s 0 = 1/2 and s = 1/2, Proposition 4.5 implies that u = 0 and consequently A p = 0.This shows that R X,χ (s) is holomorphic near s 0 .
Since X has infinite volume, there is at least one funnel end, which we will denote by X f .We choose coordinates (r, φ) ∈ X f as in Section 3.1.Choose Let ε > 0. Integrating by parts, we have that By the Cauchy-Schwarz inequality, we have that and using (7) and (24), we obtain that Therefore, we can estimate where the first factor in the right-hand side is bounded by a constant and the second factor is O(ε −1/2 ) by a direct calculation.This implies that By the fundamental lemma of calculus of variations, this implies that for z ∈ X f ∩ {r ≥ 2}.By the definition of L f (s), ( 27), we have that and Re s = 1/2, it follows that u 0 (s) ≡ 0 and therefore . Proposition 4.4 now finishes the proof.

Scattering Determinant
In this section, we prove Theorems A and B. We start with introducing the Poisson operator and studying its properties in Section 5.1.In Section 5.2, we define the scattering matrix and show the correspondence of resonances and poles of the scattering matrix for Re s < 1 and s = 1/2.In Section 5.3, we study the behavior of R X,χ (s) near s = 1/2 and prove Theorem B. In Section 5.4, we recall the basics of the Gohberg-Sigal theory and obtain a relation of scattering poles and resonances for Re(s) ≤ 1.In Section 5.5, we introduce the relative scattering matrix and the relative scattering determinant and, finally, prove Theorem A. 5.1.Poisson Operator.Before we define the scattering matrix, we introduce the Poisson operator, which maps sections C ∞ (∂ ∞ X, E χ | ∂∞X ) to solutions of the equation (∆ X,χ − s(1 − s))u = 0 with prescribed asymptotics at the boundary at infinity.The construction is similar to the one in the untwisted case [GZ97, (2.23)-(2.25)],but in our case the Poisson operator acts on sections of vector bundles and we have be more careful due to the compactification in the cusp, which depends on the representation χ.
We recall that the ideal boundary at infinity ∂ ∞ X is a disjoint union of circles (representing funnel ends) and points (representing cusp ends) and that we have the decomposition For j ∈ {1, . . ., n f } and s ∈ R X,χ we define the map The restriction is well-defined by Theorem 4.1 and (7).Similarly, for j ∈ {1, . . ., n c } and s ∈ R X,χ we define The restriction is well-defined by Theorem 4.1 and (20).Further, by (33), (28), and (25), the map E c,j X,χ (s, z, θ ′ ) is independent of θ ′ and defines an operator We denote this two-variable function by E c,j X,χ as well.We obtain the Poisson operator defined by its Schwartz kernel as follows: where By the same arguments as in the proof of [DFP,Lemma 4.14], we can express the difference of resolvents in terms of the Poisson operator for general hyperbolic surfaces.
Proposition 5.1.Let X = Γ \H be a geometrically finite hyperbolic surface and χ : Proof.We follow the proof of [DFP, Lemma 4.14], but we have to take care of the multiple ends.We fix a fundamental domain F ⊂ H of X.Then the bundle E χ ⊠ E ′ χ is trivial and can be identified with End(V ).We fix z, w ∈ F. We define the coefficients of R X,χ (s; z, w) as R jk (s; z, w) := R X,χ (s; z, w)e j , e k V .
We also set R T jk (s; z, w) := R T X,χ (s; z, w)e j , e k V , where R T X,χ (s; z, w) denotes the Schwartz kernel of the operator R X,χ (s) T .
We calculate Here, X ε := {z ∈ X : ρ(z) = ε}, and dσ Xε is the induced measure on X ε .If we pick ε > 0 sufficiently small, then the area of integration splits into a disjoint union of funnel and cusp ends.Without loss of generality, we suppose that X f = X f,j and we set X f,ε := X f ∩ X ε .From (4), we see that and where ℓ ∈ (0, ∞) is the length of the central geodesic associated to X f .Therefore, we obtain that Letting ε → 0 proves the claim for the funnel ends.
For the cusp ends, we also suppose without loss of generality that n c = 1 and X c = X c,1 is a single cusp end.We set X c,ε := X c ∩ X ε .We take coordinates (ρ, φ) ∈ R + × R/2πZ ∼ = X c as in Section 3.2 and calculate g Xc (∂ ρ , ∂ ρ ) = ρ −2 and therefore ∂ ν = −ρ ′ ∂ ρ ′ .By the definition of the Poisson operator, we have for z ∈ X with ρ(z) > ε and z ′ ∈ X c,ε and as The Poisson operator E X,χ (s) provides generalized eigenfunctions in the following sense.
then we have the asymptotics Remark 5.3.In the case of the model funnel, this result follows directly from (13).
Proof of Proposition 5.2.It is straightforward to see that E X,χ (s)ψ solves the equation (41).To obtain (42), we use the result on the structure of the resolvent, Theorem 4.1.We have that and by (24), lim where E X f ,χ (s) is defined by (8).From the asymptotics of Q(s), Theorem 4.1, we obtain For the cusp ends, we have to be more careful, because the compactification at the cusp of the bundle E χ depends on the multiplicity of the eigenvalue 1 of χ(γ j ), where γ j ∈ Γ is a representative of the conjugacy class [γ j ], associated to the cusp X c,j .Similar to the funnel case, we have Using the notation of Section 3.2, we have lim where E 1 (χ(γ j )) is the 1-eigenspace of χ(γ j ).Let .
By (33) we have lim Therefore, we obtain that lim By the definition of the compactification of E χ at the cusp, we have for

Scattering Matrix.
The scattering matrix intertwines the asymptotics of solutions of the equation (∆ X,χ − s(1 − s))u = 0 as described in Proposition 5.2.
Definition 5.4.For s ∈ R X,χ ∪ Z/2, the scattering matrix is given by where φ s is defined by (42).
Proposition 5.5.For any Multiplying this equation from the left with ρ −s f ρ 1−s c and restricting to the boundary yields Using that S X,χ (s) T = S X,χ (s) we obtain the first claim.In order to obtain (44), we calculate By (42), E X,χ (s) is injective.This proves the claim.Proposition 5.5 together with Proposition 5.1 implies that where n χ c = nc j=1 n χ c,j .Using the decomposition into funnel and cusp ends, we can write the scattering matrix as , where For Re s < 1/2, we have that For j = 1, . . ., n f let S X f,j ,χ (s) be the scattering matrix for the funnel end X f,j as described in Section 3.1.The scattering matrix for funnel ends S X f ,χ (s) is diagonal with respect to the decomposition of the boundary ∂ ∞ X and given by As it was already in the case for the resolvent, the scattering matrix S X,χ (s) is closely related to scattering matrix for the funnel ends, S X f ,χ (s).
Proof.By (33) we have that .
From the definition of M (s) and L(s), we see that for instance Using that the integral kernel of E X f ,χ (s) T is given by and the integral kernel of E X f ,χ (s) is given by (8), we obtain that Proposition 5.7.The two scattering matrices, S X,χ (s) and S X f ,χ (s), are related by where 0 : C n χ c → C n χ c is the zero-map and Q # (s) is given by Lemma 5.6.In particular, Proof.For Re s < 1/2, this follows directly from the characterization of the scattering matrix as a limit of the resolvent, (46), Theorem 4.1.For Re s ≥ 1/2 we use meromorphic continuation.Note that Q # (s) ff is smoothing and hence a pseudodifferential operator of order −∞.The second part then follows from S ff X,χ (s) = S X f ,χ (s) + Q # (s) ff and (15).
Remark 5.8.The appearance of the map 0 : 48) is due to the fact that for Re s > 1/2, we have that As in the case of the resolvent, we want to investigate the structure of the scattering matrix near a resonance.For this we consider where φ ℓ is as in (37).Let Lemma 5.9.Let s 0 ∈ C with Re s 0 < 1 and s 0 = 1/2.The scattering matrix has a pole at s 0 if and only if R X,χ (s) has a pole at s 0 .In this case we have that where for some n, k j > 0 with for each j ∈ {1, . . ., n} the matrices P j are rank-1-projections from C m X,χ (s 0 ) to mutually orthogonal subspaces, E(•, s 0 ) and F (•, s 0 ) are holomorphically invertible matrices of dimension m X,χ (s 0 ), and Proof.Using (46) and (35) we have that for some (unique) p ∈ N 0 such that H # (•, s 0 ) is holomorphic.For each k ∈ {1, . . ., p}, the operator A # k (s 0 ) is determined by the integral kernel Recall from (38) that This implies , where d(s 0 ) is nilpotent.Hence, S X,χ (s) can be written as in a sufficiently small neighborhood of s 0 .Denote by N k a Jordan block of dimension k with eigenvalue 0. The Jordan normal form of d(s 0 ) is given by where n j=1 k j = m X,χ (s 0 ) and J is unitary.Using linear algebra, we immediately obtain that for each j ∈ {1, . . ., n}, where E k j and F k j are polynomials in x, and P j , P are diagonal matrices, and each P j has rank one.Putting x = s(1 − s) − s 0 (1 − s 0 ) and applying the argumentation above to every Jordan block, we obtain matrices E(s, s 0 ), F (s, s 0 ) depending polynomially on s and mutually orthogonal projections P j of rank 1 such that where H(•, s 0 ) is holomorphic.This proves the claim.
Lemma 5.10.The resolvent satisfies where H is holomorphic near 1/2, and, for each k ∈ {1, . . ., m X,χ (1/2)}, the function Proof.We note that Im(s Using the self-adjointness of ∆ X,χ , we obtain the estimate Therefore, we have Hence, the order of the resonance at s = 1/2 is at most 2.This implies that where h is holomorphic near 1/2, and A and B are suitable operators, independent of s. Using the resolvent equation, we see that every element u in the range of A and B satisfies (∆ X,χ − 1/4)u = 0. We note that (51) implies that A : L 2 cpt (X, E χ ) → L 2 (X, E χ ).Hence, the range of A consists of eigenfunctions of ∆ X,χ with eigenvalue 1/4.By Proposition 4.7 there are no eigenfunctions if X has infinite volume, hence A = 0.
By the definition of a multiplicity, we have that rank B = m X,χ (1/2).Using the decomposition of the resolvent from (37), we can write R X,χ (s) as (2s − 1) −1 B(s) + H(s), where for some symmetric invertible matrix a 1 (1/2) = (a ℓ,m 1 (s 0 )) , φk ∈ ρ s f ρ s−1 c C ∞ (X, E χ ) and H(s) is holomorphic near s = 1/2.Since the resolvent at 1/2 is self-adjoint and non-negative, a 1 (1/2) is a positive matrix.Therefore we can find a matrix (d k,ℓ ) where Proof of Theorem B. From Proposition 5.2 and Definition 5.4, we obtain that u .At first glance, this does not make any sense for s = 1/2, but we will see that E X,χ (s) has a simple pole at s = 1/2 and hence (2s − 1)E X,χ (s) = 0 for s = 1/2.By Theorem 4.1, we have the decomposition and we recall that Mi (s) and M f (s) are holomorphic near s = 1/2.We write the remainder term Q(s) as where Q hol is holomorphic near s = 1/2.By ( 22) and (25), the term M c (s) is given by where where Q # hol (s) is holomorphic near s = 1/2.This implies that From the Fourier decomposition of S X f ,χ (s), we see that S X f ,χ (1/2) = − id.This implies that is a compact operator.Using (43) and Proposition 5.5, we calculate P 2 = P and P * = P .The residue of the resolvent at s = 1/2 is given by This implies that With φ k given by (49), we set which defines a function φ # k (s) ∈ C ∞ (∂ ∞ X, E χ ) by (50).We note that φ # k (s) is holomorphic in s for s close to 1/2.Further, the functions φ # k (1/2) are linearly independent since, otherwise, a non-trivial linear combination would lead to an L 2 -integrable solution of the eigenvalue equation in contradiction to ∆ X,χ having no eigenvalues at λ = 1/4.From ( 52) and ( 53) we obtain that Thus the restriction of B(1/2) to the boundary at infinity still has rank m X,χ (1/2).This finishes the proof.
Remark 5.11.The proof also shows that 5.4.Scattering Poles.Let s 0 ∈ C be a resonance, let ε > 0 and let γ s 0 ,ε be the path We suppose that ε is small enough such that there is no other resonance inside γ s 0 ,ε rather than s 0 .Recall that the resonance multiplicity of s 0 is given as The analogues of resonances for a scattering matrix are scattering poles.The definition of the multiplicity of a scattering pole is more involved.
We start with briefly recalling some definitions from the Gohberg-Sigal theory [GS71]: let B be a Banach space and let λ 0 ∈ C. We further denote by A the algebra of all linear bounded operators from B to B. We denote by M(λ 0 ) the germ of A-valued functions that are holomorphic in some punctured neighborhood of λ 0 and have either a pole or a removable singularity at λ 0 .In any concrete situation we will pick a suitable neighborhood.
Let B ∈ M(λ 0 ) be holomorphic at least in Ω B \ {λ 0 }, where Ω B is some open neighborhood of λ 0 , and suppose that there exists a function ψ : Ω B → B such that ψ(λ 0 ) = 0, the functions ψ and Bψ are holomorphic at λ 0 ; moreover, we suppose that Bψ(λ 0 ) = 0. We refer to ψ(λ 0 ) as a root vector and to ψ as a root function of B at λ 0 .The rank of a root vector ψ(λ 0 ), further denoted as rank(ψ(λ 0 )), is the maximal order of vanishing of B(λ)φ(λ) at λ = λ 0 among all root functions φ with φ(λ 0 ) = ψ(λ 0 ).If these orders of vanishing are unbounded, we define rank(ψ(λ 0 )) := ∞.The set of all root vectors of B at λ 0 is a vector space.We refer to its closure in B as the kernel of B(λ 0 ) and denote it by ker B(λ 0 ).In what follows, we suppose that m := dim ker B(λ 0 ) < ∞ and rank(v) < ∞ for all v ∈ ker B(λ 0 ).We define a basis, {v (1) , . . ., v (m) }, of ker B(λ 0 ) as follows: the rank of v (1) equals the maximal rank of all root vectors corresponding to λ 0 and the rank of v (j) for j = 2, . . ., m is the maximal rank of root vectors in some direct complement of the span {v (1) , . . ., v (j−1) }.Let r j := rank v (j) .We set We also recall (see, e.g., [Bor16, Definition 6.6]) that a set of bounded operators A(λ) from B to B, parametrized by λ ∈ U ⊂ C, is a finitely meromorphic family if at each point λ ′ ∈ U , we have a Laurent series representation, converging (in the operator topology) in some neighborhood of λ ′ , where for k < 0, the coefficients A k are finite rank operators.
The main result of Gohberg-Sigal [GS71, Theorem 2.1] is the following argument principle: Let B ∈ M(λ 0 ) be such that B is invertible in some neighborhood of λ 0 .Suppose that B and B −1 are finitely meromorphic families of operators in this neighborhood of λ 0 .Suppose that all points where Sff X,χ (s) := The multiplicity of a scattering pole s 0 ∈ C is defined as By (57) it follows that ν X,χ (s) is independent of the specific choice of the operator Λ ∂ f X .
Lemma 5.12.For s 0 ∈ R X,χ with Re s 0 < 1, s 0 = 1/2, we have that Moreover, for a resonance s 0 ∈ R X,χ there exists n # > 0 and k # j ∈ Z, such that we have the decomposition near s where G 1 , G 2 are holomorphically invertible near s 0 ∈ R X,χ and and P 0 is a projection.
Proof.The first part of the statement for s 0 ∈ 1 2 − N follows from (58) and the remark afterwards.Now let us consider s 0 ∈ 1 2 − N for which we follow [Bor16, Lemma 8.12].We set and note that it is well-defined near 1 − s 0 .We can then write The first and last factors on the right hand side of the previous equation are both invertible near 1 − s 0 .Together with [GS71, Section 1] this implies that We note that the first and third factors of the right hand side of the equality above are invertible near s = 1 − s 0 .Hence, 2 is singular as well and thus has no root vectors.Therefore and proves the result.
The second part of the statement follows from the application of the Gohberg-Sigal Logarithmic Residue Theorem, (55), to ΛS X,χ Λ.
We have that Lemma 5.9 implies that Proposition 5.13 (Relation between scattering poles and resonances).For s 0 ∈ C with Re s 0 ≤ 1 we have Proof.First, we note that S X,χ (1/2) = SX,χ (1/2) is unitary, and therefore ν X,χ (1/2) = 0 by (60).Moreover, m X,χ (1/2) − m X,χ (1 − 1/2) = 0, which implies the claimed equality for s 0 = 1/2.Therefore it suffices to consider a resonance s 0 ∈ C with Re s 0 < 1 and s 0 = 1/2.By (55), we have that It remains to show that m X,χ (s 0 ) = N 1−s 0 ( SX,χ ).Note that the inequality m X,χ (s 0 ) ≥ N 1−s 0 ( SX,χ ) follows from Since the operator Φ # in Lemma 5.9 might not have full rank, we cannot directly deduce equality.To prove m X,χ (s 0 ) ≤ N 1−s 0 ( SX,χ ), we have to use (45).Assume that s 0 (1 − s 0 ) does not belong to the discrete spectrum where R hol is holomorphic.To calculate the residue, we note that and hence We define the Laurent expansions The principal parts of these Laurent expansions are given by The residue at s 0 can be calculated as Conjugating by Π yields 5.5.Relative Scattering Matrix.The relative scattering matrix, defined by is a smoothing operator on C ∞ (∂ ∞ X, E χ | ∂∞X ).Therefore it makes sense to define the relative scattering determinant τ X,χ (s) := det S rel X,χ (s) .
The relation (44) implies that By [DFP, Theorem B] and [Boa54, Theorem 2.6.5], the Weierstrass product is well-defined and holomorphic of order 2.
The Weierstrass product P X f ,χ (s) for X f is defined analogously, only exchanging X for X f in (65), i.e., P X f ,χ (s) := s m X f ,χ (0) We recall that R X f ,χ is given by (34) and for one funnel end, the resonances are given by (6).As in the untwisted case (see [GZ97, Proposition 2.14]) we prove the following result.
Proof.We set (68) h(s) := P X,χ (1 − s) P X,χ (s) P X f ,χ (s) P X f ,χ (1 − s) for any s ∈ C, for which the map on the right hand side is defined.Then h is meromorphic on all of C, as is the map τ X,χ .It suffices to show that the zeros and poles of the two maps h and τ X,χ coincide, including their multiplicities.We first consider s ∈ C with Re s = 1/2.If s is a resonance of X (or X f ), and hence contributes to the divisor of some of the Weierstrass products in (68), then also 1 − s is a resonance of X (or X f , respectively) with the same multiplicity as s.Therefore the total contribution of s to the divisor of the quotient of the Weierstrass functions in (68) cancels.Thus, h does not have a zero or pole at s. From (64) it follows that the same is true for τ X,χ .
We consider now s ∈ C with Re s < 1/2 and show that the multiplicities of s as a zero or pole of h and τ X,χ coincide.Since τ X,χ (1 − s) = 1/τ X,χ (s) by (63) as well as h(1 − s) = 1/h(s), this equality of multiplicities then extends immediately to the right half plane {Re s > 1/2}.We now pick ε > 0 such that the ball of radius ε around s contains no zeros of the Weierstrass Proof.By Proposition 5.7, we have the decomposition S X,χ (s) = S X f ,χ (s) ⊕ 0 + (2s − 1)Q # (s) .
From (48) and (47), we have that Hence, the matrix coefficients of S rel X,χ (s) are given by S rel X,χ (s) = S ff rel (s) S f c rel (s) By (12), we have that S X f ,χ (s) −1 E X f ,χ (s) T = −E X f ,χ (1 − s) T and together with (47), we obtain Without loss of generality, we suppose that X f is a single funnel, that is contained in the hyperbolic cylinder C ℓ = h ℓ \H.If Re s > ε > 0, we can directly use (10) to estimate the singular values of AE X f ,χ (s), where A ∈ Diff 1 (X f , E χ | X f ) is compactly supported.For Re s < 1/2 − ε, we use (18) and (10) together with (12), E X f ,χ (s)S X f ,χ (s) −1 = −E X f ,χ (1 − s).Hence, for all s ∈ C, we obtain the estimate where m ϑ denotes the multiplicity of the eigenvalue λ = e 2πiϑ of χ(h ℓ ), the function d k (s) was defined by (17), and A ∈ Diff 1 (X f , E χ ) is compactly supported.We note that for every δ > 0 and s ∈ B(δ), we have that due to the fact that there are only finitely many resonances in a ball of radius 1 around s.
The estimate on the determinant D(s) in [DFP, Section 6] implies-as in the untwisted case (see [GZ97, Lemma 3.6])-that for δ > 0 large enough and any s ∈ for s ∈ B(δ), where we have used that all matrix components involving cusp terms are finite rank operators.So in particular, µ k (S rel X,χ (s) • ) = 0 for k > N for some N ∈ N.
For the funnel term, we estimate µ k (S rel X,χ (s) ff − id) From the remark above, we obtain that for s ∈ B(δ), µ k (S rel X,χ (s) ff − id) ≤ e C s 2+ε µ k ([∆ X,χ , η 0 ]E X f ,χ (s)) .If k ≤ max{m 0 , 2m ϕ }, then can apply the same argument to obtain that µ k (S rel X,χ (s) ff − id) ≤ e C s 2+ε .For k ≥ max{m 0 , 2m ϕ }, we have that With all these results at our disposal, the proof of Theorem A is analogous to the corresponding statement in the untwisted setting.For the convenience of the reader, we provide the details.
Proof of Theorem A. In Proposition 5.14 we established the factorization τ X,χ (s) • P X,χ (s)P X f ,χ (1 − s) P X,χ (1 − s)P X f ,χ (s) = e q(s) (74) with q being an entire function.It remains to show that q is polynomial with degree bounded by 4, for which we will take advantage of the Hadamard factorization theorem [Tit58,8.24].To that end we let ϕ : C → C, ϕ(s) := τ X,χ (s) • P X,χ (s)P X f ,χ (1 − s) P X,χ (1 − s)P X f ,χ (s) , denote the function on the left hand side of the equation in (74) and note that ϕ is entire and has no zeros (as q is entire).Therefore e q(•) is the by the maximum modulus principle (see for instance [Tit58, 5.1]).Thus, ϕ is of order 4 and hence q is a polynomial of degree at most 4.
Further, indicates an upper bound with implied constants.More precisely, for any set Y and any functions a, b : Y → R, we write a b or a(y) b(y) if there exists a constant C > 0 such that for all y ∈ Y we have |a(y)| ≤ C|b(y)| .
χ .Taking advantage of this property, we can characterize the resonances in terms of the scattering matrix.