REGULARIZED DETERMINANTS OF LAPLACE TYPE OPERATORS, ANALYTIC SURGERY AND RELATIVE DETERMINANTS

JÖRN MÜLLER AND WERNER MÜLLER

1. Introduction

In this paper we study the behaviour of regularized determinants of Laplace type operators with respect to certain singular deformations which are related to analytic surgery. Analytic surgery is a method developed by Mazzeo and Melrose [MM] to study the behaviour of global spectral invariants of Dirac- and Laplace operators with respect to decompositions of the underlying Riemannian manifolds.

The singular deformations that we consider in this paper are defined in the following way. Let M be a compact Riemannian manifold and let Y be an embedded hypersurface in M such that M −Y consists of two components M_{1} and M_{2}. Assume that the metric in a collar neighborhood N of Y is a product. Then by ”analytic surgery” we mean the stretching of the collar neighborhood N to a cylinder of infinite length. In this way we get a family of Riemannian manifolds (M_{r}, g_{r}), r ≥ 1. The singular limit of this family is the disjoint union of two manifolds with cylindrical ends M_{1,∞} and M_{2,∞}. Let Δ: C^{∞}(M, E) → C^{∞}(M, E) be a Laplace type operator on M which is adapted to the product structure on N. Then we can define an associated family of Laplace type operators Δ_{r} on M_{r} and the main purpose of this paper is to study the behaviour of det(Δ_{r}) as r → ∞. We will show that log det Δ_{r} has an asymptotic expansion and the main ingredient of the constant term of this expansion are the relative determinants det(Δ_{i,∞}, Δ_{0}), i = 1, 2, associated to the Laplacian on the manifolds with cylindrical ends M_{1,∞} and M_{2,∞}, respectively. Here the relative determinants are defined as in [Mu1]. For surfaces our results are related to the work of Bismut and Bost [BB] who studied the Quillen metric on the determinant line bundle associated to a family of complex curves with singular fibers.

We also consider the analogous problem for a compact manifold with boundary where we stretch a collar neighborhood of the boundary to an infinite half-cylinder. The singular limit of the associated family of Riemannian manifolds with boundary is a manifold with a cylindrical end. The relative determinant of the Laplacian on the manifold with cylindrical end arises in the same manner as above in the asymptotic expansion of the determinant of the Dirichlet Laplacians. This gives a new interpretation of relative determinants.

If we compare the asymptotic expansions of the determinants of the Laplacians on the manifolds M_{r}, M_{1,r} and M_{2,r}, respectively, obtained by stretching the corresponding collar neighborhoods of Y, we recover the adiabatic decomposition formulas of Park and Wojciechowski [PW1], [PW3]. We also establish a gluing formula for relative determinants of Laplace type operators on manifolds with cylindrical ends.

Date: November 23, 2004.

In the present paper we consider Laplace operators of two types. First we assume that the induced Laplace operator on Y is invertible and that the Laplacians Δ_{i,∞}, i = 1, 2, on M_{i,∞} have no nonzero L^{2}-solutions. This simplifies the constructions. The second case that we consider are Bochner-Laplace operators. In a followup paper we will study the case of Dirac-Laplace operators Δ = D^{2}.

Now we describe the content of the paper in more detail. Let (X, g) be a Riemannian manifold and let E → X be a Hermitian vector bundle. First recall that a Laplace type operator

Δ: C^{∞}(X, E) → C^{∞}(X, E)

is a second order elliptic differential operator which is symmetric, nonnegative and whose principal symbol is given by

σ_{Δ}(x, ξ) =∥ ξ ∥^{2} Id_{E_{x}} .

Suppose that X is a compact manifold with boundary ∂X, which may be empty. We impose Dirichlet boundary conditions at ∂X and denote the corresponding selfadjoint extension by Δ_{D}. This is a selfadjoint nonnegative operator in L^{2}(X, E). The regularized determinant det Δ_{D} of Δ_{D} is defined in the usual way by

det Δ_{D} = exp (−\frac{d}{ds}ζ_{Δ_{D}}(s)|_{s=0}) ,

where ζ_{Δ_{D}}(s) is the zeta function of Δ_{D}.

Our first result is a gluing formula for relative determinants of Laplace type operators on a manifold X with a cylindrical end. By definition, X has a decomposition

X = M ∪_{Y} Z, Z = ℝ^{+} × Y,

where M is a compact manifold with boundary Y and the metric g^{X} of X is a product on ℝ^{+} × Y. Let E → X be a hermitian vector bundle. We assume that there exist a hermitian vector bundle E_{0} → Y such that E|Z ≅ pr_{Y}^{∗} E_{0} and that the fiber metric h^{E} of E is a product on ℝ^{+} × Y. Let Δ: C^{∞}(X, E) → C^{∞}(X, E) be a Laplace type operator on X. We assume that the restriction of Δ to Z satisfies

(1.1)  Δ|_{Z} = −\frac{∂^{2}}{∂u^{2}} + Δ_{Y} ,

where Δ_{Y} is a Laplace type operator on Y. This implies that Δ_{X} is essentially selfadjoint in L^{2}. We will denote the unique selfadjoint extension of Δ_{X} by the same letter. Consider the operator

−\frac{∂^{2}}{∂u^{2}} + Δ_{Y} : C_{c}^{∞}(ℝ^{+} × Y, E) → L^{2}(ℝ^{+} × Y, E).

and impose Dirichlet boundary conditions at {0} × Y. Let Δ_{0} be the corresponding selfadjoint extension. Then the relative regularized determinant det(Δ, Δ_{0}) is defined as in [Mu1].

Let Δ_{M} denote the restriction of Δ to M and let Δ_{M,D} be the selfadjoint extension obtained by imposing Dirichlet boundary conditions at ∂M. We assume that Δ_{M,D} is invertible. This assumption is satisfied in many cases. Suppose, for example, that D: C^{∞}(X, E) → C^{∞}(X, E) is a Dirac operator and Δ = D^{2}. Then it follows from [Ba] that Δ_{D} is invertible. In particular, if Δ_{p}: Λ^{p}(X) → Λ^{p}(X) is the Laplacian on p-forms on a compact manifold with boundary, then Δ_{p,D} is invertible. Other examples are Bochner-Laplace operators.

If Δ_{M,D} is invertible, then the Dirichlet-to-Neumann operator R with respect to the hypersurface Y ≅ {0} × Y ⊂ X can be defined in the usual way. This is a pseudo-differential operator of order 1 on Y which is selfadjoint and nonnegative. So R has a well-defined determinant det R.

The last ingredient of the gluing formula is defined in terms of the space H of extended L^{2}-solutions of Δ. Recall that a section φ ∈ C^{∞}(X, E) is called an extended L^{2}-solution of Δ, if φ is a bounded solution of Δφ = 0 and its restriction to ℝ^{+} × Y has the form

φ(u, y) = ϕ(y) + ψ(u, y),

where ψ is in L^{2} and ϕ ∈ ker Δ_{Y}. In this case, ϕ is called the limiting value of φ. Let V^{+} ⊂ ker Δ_{Y} be the space of all limiting values of extended L^{2}-solutions of Δ. Given ϕ ∈ V^{+}, let E(ϕ, λ) be the associated generalized eigensection of Δ (cf. [Mu4]). Then E(ϕ, λ) is holomorphic at λ = 0 and E(ϕ, 0) is an extended L^{2} solution of Δ with limiting value 2ϕ. Let ρ_{Y} : C^{∞}(X, E) → C^{∞}(Y, E|Y) denote the restriction map and set H_{Y} := ρ_{Y} (H). We show that ρ_{Y} : H → H_{Y} is an isomorphism. Let φ_{1}, ..., φ_{k} be an orthonormal basis of ker Δ and let ϕ_{1}, ..., ϕ_{l} be an orthonormal basis of V^{+}. Put ψ_{i} = ρ_{Y} (φ_{i}), if 1 ≤ i ≤ k, and ψ_{k+j} = \frac{1}{2}ρ_{Y} (E(ϕ_{j}, 0)), if 1 ≤ j ≤ l. Put a_{ij} = ⟨ψ_{i}, ψ_{j}⟩_{Y}, 1 ≤ i, j ≤ k + l and let A be the (k +l)×(k +l)-matrix with entries a_{ij}. Then our first main result is the following theorem.

Theorem 1.1. Assume that Δ_{M,D} is invertible. Let h_{Y} = dim ker Δ_{Y} and denote by ζ_{Y} (s) is the zeta function of Δ_{Y}. Then

\frac{det(Δ, Δ_{0})}{det(Δ_{M,D})} = 2^{−ζ_{Y} (0)−h_{Y}} \frac{det R}{det A}.

The same result has been proved independently by Loya and Park [LP].

Now assume that (M, g) is an oriented closed connected n-dimensional Riemannian manifold and let Y be a hypersurface of M such that M − Y consists of two components. We denote the closure of the components of M − Y by M_{1} and M_{2}. Thus M_{1} and M_{2} are compact manifolds with common boundary Y such that

(1.2)  M = M_{1} ∪_{Y} M_{2}, ∂M_{1} = ∂M_{2} = Y.

Let E → M be a Hermitian vector bundle and let

Δ_{M}: C^{∞}(M, E) → C^{∞}(M, E)

be a Laplace type operator. We assume that there exists a tubular neighborhood N of Y which is diffeomorphic to [−1, 1] × Y such that all geometric structures are products over N, i.e., g|_{N} = du^{2} + g^{Y}, there exists a Hermitian vector bundle E_{0} → Y such that E|N = pr_{Y}^{∗} (E_{0}) and

(1.3)  Δ_{M} |_{N} = −\frac{∂^{2}}{∂u^{2}} + Δ_{Y} ,

where Δ_{Y} : C^{∞}(Y, E_{0}) → C^{∞}(Y, E_{0}) is a Laplace type operator on Y. Let Δ_{M_{i}} be the restriction of Δ_{M} to M_{i}, i = 1, 2. We assume that Δ_{M_{1},D} and Δ_{M_{2},D} are invertible (see the above remark).

We define a family of Riemannian manifolds (M_{r}, g_{r}), r > 0, as follows. Given r > 0, let N_{r} = [−r, r] × Y and set

(1.4)  M_{r} = M_{1} ∪_{Y} N_{r} ∪_{Y} M_{2},

where ∂M_{1} is identified with {−r} × Y and ∂M_{2} with {r} × Y. Since g is a product in a neighborhood of Y, it has a canonical extension to a metric g_{r} on M_{r} such that g_{r}|_{N_{r}} = du^{2} + g^{Y}. Similarly, E → M and Δ_{M} have natural extensions E_{r} → M_{r} and Δ_{M_{r}} to M_{r}. Our main purpose is to study the asymptotic behavior of det(Δ_{M_{r}}) as r → ∞. To describe the result we need some more notation. Set

M_{i,∞} = M_{i} ∪_{Y} (ℝ^{+} × Y), i = 1, 2.

This is a manifold with a cylindrical end Z = ℝ^{+} × Y. The disjoint union of M_{1,∞} and M_{2,∞} may be regarded as the singular limit of M_{r} as r → ∞. Let Δ_{i,∞} be the canonical extension of Δ_{M} |_{M_{i}} to M_{i,∞} which is defined by

Δ_{i,∞}|_{M_{i}} = Δ_{M} |_{M_{i}}, Δ_{i,∞}|_{ℝ^{+}×Y} = −\frac{∂^{2}}{∂u^{2}} + Δ_{Y} .

Then Δ_{i,∞} is essentially selfadjoint in L^{2}. We denote the unique selfadjoint extension of Δ_{i,∞} by the same letter. Let Δ_{0} be as in Theorem 1.1 and let det(Δ_{i,∞}, Δ_{0}) be the relative determinant [Mu1].

Let

(1.5)  ξ_{Y} (s) := \frac{Γ(s − 1/2)}{\sqrt{π}Γ(s)}ζ_{Y} (s − 1/2),

where ζ_{Y} (s) is the zeta function of Δ_{Y}. Our first result concerning the asymptotic behaviour of the determinant of a Laplace type operator is obtained under the assumption that all involved operators are invertible.

Theorem 1.2. Suppose that ker Δ_{Y} = {0} and ker Δ_{i,∞} = {0}, i = 1, 2. Then Δ_{M_{r}} is invertible for r ≥ r_{0} and

(1.6)  lim_{r→∞} e^{rξ′_{Y} (0)} det Δ_{M_{r}} = (det Δ_{Y})^{−1/2} ∏_{i=1}^{2} det(Δ_{i,∞}, Δ_{0}).

In particular, the assumption of Theorem 1.2 are satisfied for the operator Δ_{M} + λ, where λ > 0. Let ξ_{Y} (s, λ) be defined as in (1.5) with ζ_{Y} (s) replaced by the zeta function ζ_{Y} (s, λ) of Δ_{Y} + λ. Then we get

Corollary 1.3. Let λ > 0. Then

lim_{r→∞} e^{rξ′_{Y} (0,λ)} det(Δ_{M_{r}} + λ) = det(Δ_{Y} + λ)^{−1/2} ∏_{i=1}^{2} det(Δ_{i,∞} + λ, Δ_{0} + λ).

We note that (1.6) also holds if M has a nonempty boundary ∂M. In this case we impose Dirichlet boundary conditions at ∂M.

In particular, we may consider a separating hypersurface which is parallel to the boundary. This is a special case which we consider separately. Let X_{0} be a compact manifold with boundary Y and assume that all geometric structures are products in a collar neighborhood of Y. Let X_{r} = X_{0} ∪_{Y} ([0, r] × Y) and let Δ_{X_{r},D} be the selfadjoint extension of the corresponding Laplace operator with respect to Dirichlet boundary conditions. Then the analogous statement to Theorem 1.2 is

Proposition 1.4. Assume that Δ_{Y} and Δ_{∞} are invertible. Then Δ_{X_{r},D} is invertible for r ≥ r_{0} and

(1.7)  lim_{r→∞} e^{rξ′_{Y} (0)/2} det Δ_{X_{r},D} = (det Δ_{Y})^{−1/2} det(Δ_{∞}, Δ_{0}), r → ∞.

Especially consider the manifolds with boundary M_{1} and M_{2} of the decomposition (1.2) of M. Let M_{i,r} = M_{i} ∪_{Y} ([0, r] × Y), i = 1, 2. If we apply (1.7) to Δ_{M_{i,r},D} and compare it to (1.6) , we obtain

Corollary 1.5. Let M be closed. Assume that Δ_{Y} and Δ_{i,∞}, i = 1, 2, are invertible. Then Δ_{M_{r}} and Δ_{M_{i,r},D}, i = 1, 2, are invertible for r ≥ r_{0} and

(1.8)  lim_{r→∞} \frac{det Δ_{M_{r}}}{det Δ_{M_{1,r},D} det Δ_{M_{2,r},D}} = (det Δ_{Y})^{1/2}.

This is the ”adiabatic decomposition formula” established by Park and Wojciechowski in [PW1].

Next we study the case of a Bochner-Laplace operator. Let ∇ is a metric connection on E which is a product on N. Let Δ_{M} = ∇^{∗}∇ be the associated Bochner-Laplace operator. Then ∇ has canonical extensions to a connection ∇^{r} on E_{r} → M_{r} and ∇^{i,∞} on E_{i,∞} → M_{i,∞}, respectively, and Δ_{M_{r}} and Δ_{i,∞} are the corresponding Bochner-Laplace operators. We need to introduce some further notation. Let

S_{i}(0): ker Δ_{Y} → ker Δ_{Y} , i = 1, 2,

denote the on-shell scattering operator at energy zero associated to (Δ_{i,∞}, Δ_{0}) (see e.g. [Mu4]). This operator satisfies S_{i}(0)^{2} = Id. Let

ker Δ_{Y} = V_{i}^{+} ⊕ V_{i}^{−}, i = 1, 2,

be the decomposition of ker Δ_{Y} into the ±1-eigenspaces of S_{i}(0). Let C_{12} denote the restriction of S_{1}(0)S_{2}(0) to the orthogonal complement of (V_{1}^{+} ∩ V_{2}^{+}) ⊕ (V_{1}^{−} ∩ V_{2}^{−}) in ker Δ_{Y}. Then our next result is

Theorem 1.6. Let Δ_{M} = ∇^{∗}∇ be a Bochner-Laplace operator. Let h_{Y} = dim ker Δ_{Y} and h = dim V_{1}^{+} + dim V_{2}^{+} − 2 dim V_{1}^{+} ∩ V_{2}^{+}. Then

lim_{r→∞} r^{h−h_{Y}} e^{rξ′_{Y} (0)} det Δ_{M_{r}} =2^{2h_{Y} −h}(det Δ_{Y})^{−1/2}

· det ((Id −C_{12})/2) ∏_{i=1}^{2} det(Δ_{i,∞}, Δ_{0}).

(1.9)

If we specialize Theorem 1.6 to the case of the Laplacian Δ = d^{∗}d on functions on a closed surface M, we obtain

det Δ_{r} ∼ 2 det(Δ_{1,∞}, Δ_{0}) det(Δ_{2,∞}, Δ_{0})re^{−πr/3}

as r → ∞. This is Theorem 13.7 of [BB] with an explicit constant expressed in terms of relative determinants.

As in Proposition 1.4, we may also consider the case of a compact Riemannian manifold X_{0} with boundary Y. For a Bochner-Laplace operator on X_{0} it follows from [Ba] that Δ_{X_{r},D} is invertible. Let V^{+} ⊂ ker Δ_{Y} be the +1-eigenspace of the scattering operator S(0) and let h^{+} = dim V^{+}. The analogous result to (1.7) is

(1.10)  lim_{r→∞} r^{h^{+}−h_{Y}} e^{rξ′_{Y} (0)/2} det Δ_{X_{r},D} = 2^{h_{Y}} (det Δ_{Y})^{−1/2} det(Δ_{∞}, Δ_{0}).

Now we apply this again to the manifolds with boundary M_{1} and M_{2} of the decomposition (1.2) of M and compare it to (1.9). In this way we get

Theorem 1.7. Let the notation be as in Theorem 1.6 and let h_{12} = dim V_{1}^{+} ∩ V_{2}^{+}.Then

lim_{r→∞} \frac{r^{h_{Y} −2h_{12}} det Δ_{M_{r}}}{det Δ_{M_{1,r},D} det Δ_{M_{2,r},D}} = 2^{−h} (det Δ_{Y})^{1/2} det ((Id −C_{12})/2) .

Remark. This result was first proved by by Park and Wojciechowski [PW3] under an additional assumption, called Condition A [PW3, p.4], which rules out the existence of exponentially decreasing eigenvalues of Δ_{M_{r}}. As pointed out by Park and Wojciechowski, their assumption implies that 1 is not an eigenvalue of S_{1}(0)S_{2}(0). This has the consequence that V_{1}^{±} ∩ V_{2}^{±} = {0}, which in turn implies that h = h_{Y} and h_{12} = 0 and Theorem 1.7 specializes to Theorem 0.1 of [PW3].

Next we explain some of the main ideas of the proofs. The strategy to prove Theorem 1.1 is analogous to the proof of the surgery formula in [HZ]. Let z ∈ ℂ−ℝ_{−}. Then the relative determinant det(Δ + z, Δ_{0} + z) and the determinant det(Δ_{M,D} + z) are defined. Moreover the Dirichlet-to-Neumann operator R(z) with respect to Δ+z and the hypersurface Y ⊂ X exists and the determinant det R(z) can be defined. Then by Theorem 4.2 of [Ca] there is a polynomial P (z) with real coefficients of degree ≤ (n − 1)/2 such that

(1.11)  \frac{det(Δ + z, Δ_{0} + z)}{det(Δ_{M,D} + z)} = e^{P (z)} det(R(z)).

Both sides of this equality have an expansion in z as z → 0. We determine these expansions and compare the constant terms. This proves Theorem 1.1.

To prove Theorems 1.2 and 1.6, we apply the Mayer-Vietoris formula of [BFK] to det(Δ_{M_{r}}+ λ), λ > 0, with respect to the decomposition (1.4) and take the limit λ → 0. To this end we assume that Δ_{M_{1},D} and Δ_{M_{2},D} are invertible. Under this assumption the Dirichlet-to-Neumann operator R_{r} with respect to the hypersurface Σ_{r} := ({−r}×Y)⊔({r}×Y) exists and we get a splitting formula for det Δ_{M_{r}}. We compare this splitting formula with the splitting formulas for det(Δ_{i,∞}, Δ_{0}) given by Theorem 1.1. Finally we study the limit of det R_{r} as r → ∞ and compare it to det R_{1,∞} det R_{2,∞}. Let Δ_{N_{r},D} be the Laplace operator on N_{r} with Dirichlet boundary conditions. Under the assumptions of Theorem 1.2 or 1.6 the limit as r → ∞ of r^{h} det Δ_{M_{r}}(det Δ_{N_{r},D})^{−1} exists and

lim_{r→∞} \frac{r^{h} det Δ_{M_{r}}}{det Δ_{N_{r},D}} = 2^{−h} det ((Id −C_{12})/2) ∏_{i=1}^{2} det(Δ_{i,∞}, Δ_{0}).

Finally we determine the asymptotic behaviour of det Δ_{N_{r},D} as r → ∞. This completes the proof of Theorem 1.2 and Theorem 1.6.

2. Expansion of relative determinants

Let X be a manifold with a cylindrical end and let Δ be a Laplace type operator on X as above. In this section we consider the asymptotic expansion of log det(Δ + z, Δ_{0} + z) as z → 0. We use the framework introduced in [Mu1]. Let H, H_{0} be two self-adjoint nonnegative linear operators in a separable Hilbert space H such that e^{−tH} − e^{−tH_{0}} is a trace class operator for all t > 0. Suppose that the following two conditions are satisfied:

1) As t → 0+, there exists an asymptotic expansion of the form

Tr(e^{−tH} − e^{−tH_{0}}) ∼ ∑_{j=0}^{∞} a_{j}t^{α_{j}},

where −∞ < α_{0} < α_{1} < ⋯ and α_{j} → ∞.

2) There exist b_{0} ∈ ℂ, ρ > 0 such that

Tr(e^{−tH} − e^{−tH_{0}}) ∼ b_{0} + O(t^{−ρ})

as t → ∞.

Set

ζ_{1}(s, H, H_{0}) = \frac{1}{Γ(s)} ∫_{0}^{1} t^{s−1} Tr(e^{−tH} − e^{−tH_{0}}) dt, Re(s) > −α_{0};
ζ_{2}(s, H, H_{0}) = \frac{1}{Γ(s)} ∫_{1}^{∞} t^{s−1} Tr(e^{−tH} − e^{−tH_{0}}) dt, Re(s) < 0.

Then ζ_{1}(s, H, H_{0}) admits a meromorphic extension to ℂ which is holomorphic at s = 0. Similarly ζ_{2}(s, H, H_{0}) has a meromorphic extension to the half-plane Re(s) < ρ which is also holomorphic at s = 0. It is given by

(2.1)  ζ_{2}(s, H, H_{0}) = −\frac{b_{0}}{Γ(s + 1)} + \frac{1}{Γ(s)} ∫_{1}^{∞} t^{s−1} (Tr(e^{−tH} − e^{−tH_{0}}) − b_{0}) dt.

The relative zeta function ζ(s, H, H_{0}) is then defined by

ζ(s, H, H_{0}) = ζ_{1}(s, H, H_{0}) + ζ_{2}(s, H, H_{0}),

and the relative determinant by

det(H, H_{0}) = exp (−\frac{d}{ds}ζ(s, H, H_{0})|_{s=0}) .

Let λ > 0 and define det(H + λ, H_{0} + λ) similarly.

Proposition 2.1. As λ → 0+, we have

log det(H + λ, H_{0} + λ) = b_{0} log λ + log det(H, H_{0}) + o(1).

Proof. From the construction of the analytic continuation of ζ_{1}(s, H + λ, H_{0} + λ) and ζ_{1}(s, H, H_{0}), respectively, it follows immediately that

lim_{λ→0} ζ′_{1}(0, H + λ, H_{0} + λ) = ζ′_{1}(0, H, H_{0}).

Let Re(s) < ρ. Using (2.1) we get

ζ_{2}(s, H + λ, H_{0} + λ) = \frac{1}{Γ(s)} ∫_{1}^{∞} t^{s−1}e^{−tλ} Tr(e^{−tH} − e^{−tH_{0}}) dt

= \frac{1}{Γ(s)} ∫_{1}^{∞} t^{s−1}e^{−tλ} (Tr(e^{−tH} − e^{−tH_{0}}) − b_{0}) dt

+ \frac{b_{0}}{Γ(s)} ∫_{1}^{∞} t^{s−1}e^{−tλ} dt

= ζ_{2}(s, H, H_{0}) + \frac{b_{0}}{Γ(s + 1)} + b_{0}λ^{−s} − \frac{b_{0}}{Γ(s)} ∫_{0}^{1} t^{s−1}e^{−tλ} dt + o(1)

as λ → 0+. This implies that

ζ′_{2}(0, H + λ, H_{0} + λ) = ζ′_{2}(0, H, H_{0}) − b_{0} log λ + o(1)

as λ → 0+. □

In order to apply this result to our case, we need to compute b_{0}. Let ξ(λ) be the spectral shift function of (Δ, Δ_{0}) [Mu1, pp. 315]. By (2.16) of [Mu1], we have

(2.2)  b_{0} = −ξ(0+).

So we are reduced to the study of the spectral shift function near zero. Recall that the spectral shift function is a real valued function in L2_{loc}(ℝ) which is uniquely determined by the following two properties

(1) ξ(λ) = 0 for all λ < 0.

(2) For every f ∈ C_{c}^{∞}(ℝ), f(Δ) − f(Δ_{0}) is a trace class operator and

Tr(f(Δ) − f(Δ_{0})) = ∫_{R} f′(λ)ξ(λ) dλ.

Let Δ_{d} and Δ_{ac} denote the restriction of Δ to the subspace of L^{2}(X, E) corresponding to the point spectrum and the absolutely continuous spectrum of Δ, respectively. By [Do], the eigenvalues of Δ have no finite point of accumulation. Hence f(Δ_{d}) is a trace class operator for every f ∈ C_{c}^{∞}(ℝ). This implies that f(Δ_{ac}) − f(Δ_{0}) is also a trace class operator for every f ∈ C_{c}^{∞}(ℝ). Let ξ_{c}(λ) be the spectral shift function of (Δ_{ac}, Δ_{0}) and let N(λ) denote the counting function of the eigenvalues of Δ. Then it follows from (1) and (2) that

(2.3)  ξ(λ) = −N(λ) + ξ_{c}(λ).

The spectral shift function ξ_{c}(λ) can be determined in the same way as in Chapter IX of [Mu3]. The manifolds considered in [Mu3] are manifolds with fibered cusps which are different from the manifolds in the present paper. However, the structure of the continuous spectrum is similar and everything said about the continuous spectrum in [Mu3] applies with minor modifications in our case as well. Let μ_{1} > 0 be the smallest positive eigenvalue of Δ_{Y}. Let

S(s): ker Δ_{Y} → ker Δ_{Y} , |s| < \sqrt{μ_{1}},

be the scattering matrix [Mu4]. It is an analytic function. Then it follows as in the proof of Theorem 9.25 of [Mu3] that

ξ_{c}(λ) = −\frac{1}{4}(Tr(S(0)) + dim ker Δ_{Y}) + \frac{i}{2π} ∫_{0}^{λ} Tr(S′(s)S(−s)) dλ

for 0 ≤ λ < \sqrt{μ_{1}}. Hence we get

ξ_{c}(0+) = −\frac{1}{4}(Tr(S(0)) + dim ker Δ_{Y}).

Together with (2.3) we obtain

ξ(0+) = − dim ker Δ − \frac{1}{4}(Tr(S(0)) + dim ker Δ_{Y})

and by (2.2) it follows that

b_{0} = dim ker Δ + \frac{1}{4}(Tr(S(0)) + dim ker Δ_{Y}).

Now observe that S(0) satisfies S(0)^{2} = Id. Hence

Tr(S(0)) + dim ker Δ_{Y} = 2 dim ker(S(0) − Id).

Combined with Proposition 2.1 we obtain the following corollary.

Corollary 2.2. Let k = dim ker Δ and l = dim ker(S(0) − Id). Then

log det(Δ + λ, Δ_{0} + λ) = (k + l/2) log λ + log det(Δ, Δ_{0}) + o(1)

as λ → 0+.

3. Expansion of the Dirichlet-to-Neumann operator

Let X = M ∪_{Y} Z be a manifold with a cylindrical end Z = ℝ^{+} ×Y and let Δ: C^{∞}(Z, E) → C^{∞}(Z, E) be a Laplace type operator on X with properties as above. For z ∈ ℂ − ℝ_{−} let R(z) be the Dirichlet-to-Neumann operator with respect to Δ + z and the hypersurface Y = {0} × Y ⊂ X. In this section we study the expansion of det(R(z)) as z → 0. To begin with we recall the definition of the Dirichlet-to-Neumann operator. Let z ∈ ℂ − R_{−} and φ ∈ C^{∞}(Y, E|Y). There exists a unique section ψ ∈ C^{∞}(X − Y, E) ∩ L^{2}(X, E) such that

\begin{aligned}(Δ + z)ψ & = 0 on X − Y; \\ ψ & = φ on Y. \end{aligned}

The solution ψ is obtained as follows. Let φ̃ ∈ C_{c}^{∞}(X, E) be any extension of φ. Let Δ_{D} be the operator Δ with Dirichlet boundary conditions along Y. Then

(3.1)  ψ = φ̃ − (Δ_{D} + z)^{−1}((Δ + z)(φ̃)).

Furthermore, ψ is continuous on X and smooth on \overline{M} and \overline{Z}. Its normal derivative has a jump along Y. Then R(z)φ is defined by

(3.2)  R(z)φ = \frac{∂}{∂u}(ψ|_{M})|_{∂M} − \frac{∂}{∂u}(ψ|_{Z})|_{∂Z}.

By Theorem 2.1 of [Ca], R(z) is an invertible pseudo-differential operator of order 1. Its principal symbol is given by

σ(R(z))(x, ξ) = 2\sqrt{g_{x}(ξ, ξ)} Id_{E_{x}}, ξ ∈ T_{x}^{∗}Y.

Furthermore, z ∈ ℂ − R_{−} ↦ R(z) is a holomorphic function with values in the space of pseudo-differential operators. Let G(x, y, z) denote the kernel of (Δ+z)^{−1}. Then G(x, y, z) is smooth in the complement of the diagonal and for x ≠ y, G(x, y, z) ∈ Hom(E_{y}, E_{x}). As shown in the proof of Theorem 2.1 of [Ca], we have

(3.3)  R(z)^{−1}φ(x) = ∫_{Y} G(x, y, z)φ(y) dy, x ∈ Y, φ ∈ C^{∞}(Y, E|Y).

In other words

R(z)^{−1} = ρ_{Y} ∘ (Δ + z)^{−1}(· ⊗ δ_{Y}),

where ρ_{Y} is the restriction map to Y and δ_{Y} is the Dirac δ-function along Y. Especially, if λ > 0 then R(λ) is an elliptic pseudodifferential operator of order 1 which is selfadjoint and positive definite. Hence its regularized determinant det(R(λ)) is defined.

Under the assumption that Δ_{M,D} is invertible, we can also define the Dirichlet-to-Neumann operator with respect to Δ and Y. For this purpose we need the following lemma.

Lemma 3.1. For every φ ∈ C^{∞}(Y, E|Y) there exists a unique ψ ∈ C^{∞}(X − Y, E) ∩ C^{0}(X, E), which is bounded and satisfies

(3.4)  \begin{aligned}Δψ & = 0 on X − Y; \\ ψ|_{Y} & = φ. \end{aligned}

Proof. Since Δ_{M,D} is invertible, the Dirichlet problem on M has a unique solution, i.e., for every φ ∈ C^{∞}(Y, E|Y) there exists a unique ψ_{1} ∈ C^{∞}(M, E) ∩ C^{0}(\overline{M}, E) such that

\begin{aligned}Δ_{M}ψ_{1} & = 0 in M; \\ ψ_{1}|_{Y} & = φ. \end{aligned}

Next we show that the Dirichlet problem on Z has also a unique solution. Let {ϕ_{i}}_{i∈ℕ} be an orthonormal basis of L^{2}(Y, E|Y) consisting of eigenfunctions of Δ_{Y} with eigenvalues 0 ≤ λ_{0} ≤ λ_{1} ≤ · · ·. Let φ ∈ C^{∞}(Y, E|Y). Then φ has an expansion of the form

φ = ∑_{i=1}^{∞} a_{j}ϕ_{j}.

Set

ψ_{2}(u, y) = ∑_{j=1}^{∞} a_{j}e^{−u\sqrt{λ_{j}}}ϕ_{j}(y).

Then ψ_{2} ∈ C^{∞}(Z, E) is bounded and satisfies

(3.5)  Δψ_{2} = 0 and ψ_{2}(0, y) = φ(y), y ∈ Y.

This proves existence. Now suppose that ψ̃_{2} is a second bounded solution of (3.4). Set g = ψ_{2} − ψ̃_{2}. Then g ∈ C^{∞}(Z, E) is bounded and satisfies

\begin{aligned}(−\frac{∂^{2}}{∂u^{2}} + Δ_{Y} ) g & = 0; \\ g(u, y) & = 0, y ∈ Y. \end{aligned}

If we expand g in the orthonormal basis {ϕ_{j}}_{j∈ℕ} it follows that

g(u, y) = ∑_{j=1}^{m}(b_{j}u + a_{j})ϕ_{j}(y) + ∑_{j=m+1}^{∞}(b_{j}e^{\sqrt{λ_{j}}u} + a_{j}e^{−\sqrt{λ_{j}}u})ϕ_{j}(y),

where m = dim ker Δ_{Y}. Since g is bounded, it follows that b_{j} = 0 for all j ∈ ℕ. Using that g(0, y) = 0, we obtain a_{j} = 0 for all j ∈ ℕ. This proves uniqueness. □

Now we can proceed as above. Given φ ∈ C^{∞}(Y, E|Y), let ψ ∈ C^{∞}(X − Y, E) ∩ C^{0}(X, E) be the unique solution of (3.4). Then the Dirichlet-to-Neumann operator is defined by

(3.6)  Rφ = \frac{∂}{∂u}(ψ|_{M})|_{∂M} − \frac{∂}{∂u}(ψ|_{Z})|_{∂Z}.

Next we establish some properties of R.

Lemma 3.2. There exist a smoothing operator K such that

R = 2\sqrt{Δ_{Y}} + K.

Proof. Since X − Y = M ⊔ Z, R can be written as

R = R_{int} + R_{ext},

where R_{int} is the Neumann jump operator on M. It is defined as follows. Given φ ∈ C^{∞}(Y, E|Y), let ψ_{1} ∈ C^{∞}(M, E) ∩ C^{0}(\overline{M}, E) be the unique solution of

Δψ_{1} = 0 on M, ψ_{1}|_{Y} = φ.

Then R_{int} is defined as

R_{int}φ := \frac{∂ψ_{1}}{∂u} |_{Y} .

Similarly let ψ_{2} ∈ C^{∞}(Z, E) ∩ C^{0}(\overline{Z}, E) be the unique bounded solution of

Δψ_{2} = 0 on Z, ψ_{2}|_{Y} = φ.

Set

R_{ext}(φ) := −\frac{∂ψ_{2}}{∂u} |_{Y} .

As explained above, ψ_{2} is given by

ψ_{2}(u, y) = ∑_{j=1}^{∞}⟨φ, ϕ_{j}⟩e^{−\sqrt{λ_{j}}u}ϕ_{j}(y).

Hence we get

−\frac{∂ψ_{2}}{∂u} (0, y) = ∑_{j=1}^{∞} \sqrt{λ_{j}}⟨φ, ϕ_{j}⟩ϕ_{j}(y) = (\sqrt{Δ_{Y}} φ)(y).

Thus R_{ext} = \sqrt{Δ_{Y}}. By Theorem 2.1 of [Le3] it follows that R_{int} = \sqrt{Δ_{Y}} + K, where K is a smoothing operator. This proves the lemma. □

In particular, it follows that R is an elliptic pseudodifferential operator of order 1.

Lemma 3.3. For every ϕ ∈ C^{∞}(Y, E|Y), R(λ)ϕ is a continuous function of λ ∈ [0, ∞) and

lim_{λ→0+} R(λ)ϕ = Rϕ.

Proof. Let λ ≥ 0. As above, R(λ) can be written as

R(λ) = R_{int}(λ) + R_{ext}(λ).

Given ϕ ∈ C^{∞}(Y, E|Y), let ψ_{1}(λ) ∈ C^{∞}(M −Y, E)∩C^{0}(M, E) be the unique section which satisfies (Δ + λ)ψ_{1}(λ) = 0 and ψ_{1}(λ)|_{Y} = ϕ. Let ϕ̃ ∈ C^{∞}(M, E) be any extension of ϕ which is smooth up to the boundary. Then

ψ_{1}(λ) = ϕ̃ − (Δ_{M,D} + λ)^{−1}((Δ_{M} + λ)(ϕ̃)).

Since Δ_{M,D} is invertible, this formula also holds for λ = 0. From this representation of ψ_{1}(λ) it follows immediately that R_{int}(λ)ϕ converges to R_{int}ϕ as λ → 0+. Next observe that the unique bounded solution ψ_{2}(λ) ∈ C^{∞}(Z, E) ∩ C^{0}(Z, E) of

(Δ + λ)ψ_{2}(λ) = 0 on Z, ψ_{2}(λ)|_{Y} = ϕ

is given by

ψ_{2}(λ, u, y) = ∑_{j=1}^{∞}⟨φ, ϕ_{j}⟩e^{−(λ_{j}+λ)^{1/2}u}ϕ_{j}(y).

Then R_{ext}(λ)ϕ := ∂ψ_{2}(λ, u.y)/∂u|_{u=0} and it follows that R_{ext}(λ)ϕ is continuous in λ ∈ [0, ∞) and R_{ext}(λ)ϕ converges to R_{ext}ϕ as λ → 0+. □

Corollary 3.4. The operator R is formally selfadjoint and nonnegative.

Proof. As explained above, for every λ > 0, the operator R(λ) is formally selfadjoint and positive, and therefore the claim follows immediately from Lemma 3.3. □

Together we have proved that R is a first order elliptic pseudo-differential operator which is formally selfadjoint and nonnegative. Hence the regularized determinant det R is welldefined.

Our next purpose is to study the bahaviour of the bounded operator R(λ)^{−1} as λ → 0. First we recall some facts about the spectral resolution of Δ. For more details we refer to [Mu4]. We have

L^{2}(X, E) = L_{d}^{2}(X, E) ⊕ L_{c}^{2}(X, E),

where

L_{d}^{2}(X, E) = ⨁_{j} E(λ_{j})

is the discrete sum of the eigenspaces of Δ with eigenvalues 0 ≤ λ_{1} < λ_{1} < · · · . Each eigenspace is finite dimensional. The orthogonal complement L_{c}^{2}(X, E) of L_{d}^{2}(X, E) is the absolutely continuous subspace for Δ. It can be described in terms of generalized eigensections E(ϕ_{j}, λ) attached to the eigensections ϕ_{j} of Δ_{Y} . Each E(ϕ_{j}, λ) is a smooth section of E and satisfies

ΔE(ϕ_{j}, λ) = λE(ϕ_{j}, λ).

Of particular importance for our purpose are the generalized eigensections E(ϕ, λ) attached to ϕ ∈ ker Δ_{Y} . Let μ_{1} > 0 be the smallest positive eigenvalue of Δ_{Y} . If we put λ = s^{2} and regard E(ϕ, λ) as a function of s, then E(ϕ, s) has an analytic continuation to the disc |s| < μ_{1}. Let

S(s) : ker Δ_{Y} → ker Δ_{Y} , |s| < μ_{1},

be the corresponding scattering matrix. It is also holomorphic for |s| < μ_{1} and on ℝ^{+} × Y we have

(3.7)  E(ϕ, s, (u, y)) = e^{ius}ϕ(y) + e^{−isu}(S(s)ϕ)(y) + ψ(s),

where ψ(s) is in L^{2}. Let 0 < μ < μ_{1} and let P_{μ} be the spectral projection of Δ onto [0, μ]. By (3.3) we have

(3.8)  R(λ)^{−1} = ρ_{Y} ∘ P_{μ}(Δ + λ)^{−1}(· ⊗ δ_{Y}) + ρ_{Y} ∘ (Id −P_{μ})(Δ + λ)^{−1}(· ⊗ δ_{Y}).

First we study the second operator on the right. Let

i_{Y} : L^{2}(Y, E|Y) → H^{−1}(X, E)

be the map which is defined by i_{Y} (φ) = φδ_{Y} . Then i_{Y} is continuous. Furthermore the restriction map ρ_{Y} defines a continuous map

ρ_{Y} : H^{1}(X, E) → L^{2}(Y, E|Y).

Since (Δ + λ)^{−1} : H^{−1}(X, E) → H^{1}(X, E) is continuous, we get a continuous map

ρ_{Y} ∘ (Id −P_{μ})(Δ + λ)^{−1} ∘ i_{Y} : L^{2}(Y, E|Y) → L^{2}(Y, E|Y).

Lemma 3.5. There exists C > 0 such that

∥ ρ_{Y} ∘ (Id −P_{μ})(Δ + λ)^{−1} ∘ i_{Y} ∥_{L^{2}}≤ C

for all λ ≥ 0.

Proof. Let φ ∈ H^{−1}(X, E). Then ∥ φ ∥_{H^{−1}}=∥ (Δ + Id)^{−1/2}φ ∥_{L^{2}}. Hence we get

∥ (Id −P_{μ})(Δ + λ)^{−1}φ ∥_{H^{1}} =∥ (Δ + Id)(Id −P_{μ})(Δ + λ)^{−1}(Δ + Id)^{−1/2}φ ∥_{L^{2}}
≤∥ (Δ + Id)(Id −P_{μ})(Δ + λ)^{−1} ∥_{L^{2}} · ∥ φ ∥_{H^{−1}} .

Using the spectral theorem we get

∥ (Δ + Id)(Id −P_{μ})(Δ + λ)^{−1} ∥_{L^{2}}≤ 1 + 1/μ

for λ ≥ 0. This implies

∥ (Id −P_{μ})(Δ + λ)^{−1} ∥_{L(H^{−1},H^{1})}≤ 1 + 1/μ

for λ ≥ 0. Since i_{Y} and ρ_{Y} are continuous, the lemma follows. □

It remains to consider the first operator on the right hand side of (3.8). This is a smoothing operator whose kernel R(y_{1}, y_{2}, λ) can be described as follows. Let {φ_{j}} be an orthonormal basis of eigensections of Δ with eigenvalues 0 ≤ λ_{1} ≤ λ_{2} ≤ ⋯ and let ϕ_{1}, …, ϕ_{m} be an orthonormal basis of ker Δ_{Y} . Then it follows from the explicite description of the spectral resolution of Δ (see [Gu], [Mu4]) that

\begin{aligned}R(y_{1}, y_{2}, λ) = & ∑_{λ_{j}≤μ}(λ_{j} + λ)^{−1}φ_{j}(y_{1}) ⊗ φ_{j}(y_{2}) \\ & + \frac{1}{2π} ∑_{j=1}^{m} ∫_{0}^{μ}(s^{2} + λ)^{−1}E(ϕ_{j}, s, y_{1}) ⊗ E(ϕ_{j}, −s, y_{2})ds. \end{aligned} ; (3.9)

We shall now determine the behaviour of this kernel as λ → 0. The behaviour of the first sum is obvious and we only need to investigate the second sum.

Lemma 3.6. Let ϕ_{1}, ..., ϕ_{m} be an orthonormal basis of ker Δ_{Y}. Then

∑_{j=1}^{m} E(ϕ_{j}, s, y_{1}) ⊗ E(ϕ_{j}, −s, y_{2})

is an even function of s, |s| < μ_{1}.

Proof. We recall that the generalized eigensections and the scattering matrix satisfy the following functional equations. Let ϕ ∈ ker Δ_{Y}. Then

(3.10)  \begin{aligned}E(ϕ, −s) & = E(S(−s)ϕ, s), \\ S(s)S(−s) & = Id, S(s)^{t} = S(s), |s| < μ_{1}. \end{aligned}

Let ϕ_{1}, ..., ϕ_{m} be an orthonormal basis of ker Δ_{Y}. Then there exist analytic functions a_{ij}(s), i, j = 1, ..., m, defined in |s| < μ_{1}, such that

(3.11)  S(s)ϕ_{i} = ∑_{j=1}^{m} a_{ij}(s)ϕ_{j}, i = 1, ..., m.

Using (3.10) and (3.11) we get

\begin{aligned}∑_{j=1}^{m} E(ϕ_{j}, −s, y_{1}) ⊗ E(ϕ_{j} & , s, y_{2}) = ∑_{j=1}^{m} E(S(−s)ϕ_{j}, s, y_{1}) ⊗ E(S(s)ϕ_{j}, −s, y_{2}) \\ & = ∑_{j=1}^{m} ∑_{k,l=1}^{m} a_{jk}(−s)a_{jl}(s)E(ϕ_{k}, s, y_{1}) ⊗ E(ϕ_{l}, −s, y_{2}). \end{aligned}

By (3.10) the matrix A(s) = (a_{ij}(s))_{i,j} is symmetric and satisfies A(−s)A(s) = Id. This implies

∑_{j=1}^{m} E(ϕ_{j}, −s, y_{1}) ⊗ E(ϕ_{j}, s, y_{2}) = ∑_{j=1}^{m} E(ϕ_{j}, s, y_{1}) ⊗ E(ϕ_{j}, −s, y_{2})

as claimed. □

By Lemma 3.6 there exists a smooth section Ẽ(s) of E⊠E over X×X which is holomorphic for |s| < μ such that

\begin{aligned}∑_{j=1}^{m} E(ϕ_{j}, s, y_{1}) ⊗ E(ϕ_{j}, −s, y_{2}) & = ∑_{j=1}^{m} E(ϕ_{j}, 0, y_{1}) ⊗ E(ϕ_{j}, 0, y_{2}) \\ & + s^{2}Ẽ(s, (y_{1}, y_{2})), |s| < μ. \end{aligned} ; (3.12)

Note that

∫_{0}^{μ} \frac{ds}{s^{2} + λ} = \frac{π}{2\sqrt{λ}} − \frac{1}{μ} + O(λ)

as λ → 0. Together with (3.12) we get

\begin{aligned}\frac{1}{2π} ∑_{j=1}^{m} ∫_{0}^{μ}(s^{2} + λ)^{−1}E(ϕ_{j}, s, y_{1}) & ⊗ E(ϕ_{j}, −s, y_{2})ds \\ & = \frac{1}{4\sqrt{λ}} ∑_{j=1}^{m} E(ϕ, 0, y_{1}) ⊗ E(ϕ_{j}, 0, y_{2}) + O(1) \end{aligned}

as λ → 0. To continue we consider the the scattering matrix S(0) at zero energy. It satisfies

S(0)^{2} = Id .

Let ϕ ∈ ker Δ_{Y}. If S(0)ϕ = ϕ then it follows from (3.7) that on ℝ^{+} × Y we have

E(ϕ, 0) = 2ϕ + ψ,

where ψ ∈ L^{2}(ℝ^{+} × Y, E). If S(0)ϕ = −ϕ, then E(ϕ, 0) = 0 [Mu2, p. 209]. Let

ker Δ_{Y} = V^{+} ⊕ V^{−}

be the decomposition of ker Δ_{Y} in the ±1- eigenspaces of S(0). Then V^{+} equals the space of limiting values of extended solutions of Δ [Mu2]. Let ϕ_{1}, …, ϕ_{l} be an orthonormal basis of V^{+} and let φ_{1}, ..., φ_{m} be an orthonormal basis of ker Δ. Define the kernel R_{1} by

(3.13)  R_{1}(y_{1}, y_{2}, λ) = \frac{1}{λ} ∑_{j=1}^{m} φ_{j}(y_{1}) ⊗ φ_{j}(y_{2}) + \frac{1}{4\sqrt{λ}} ∑_{j=1}^{l} E(ϕ_{j}, 0, y_{1}) ⊗ E(ϕ_{j}, 0, y_{2}).

Let R_{1}(λ): L^{2}(Y, E|Y) → L^{2}(Y, E|Y) be the operator defined by this kernel. Together with Lemma 3.4 we obtain

Proposition 3.7. There exists a bounded operator R_{2}(λ) : L^{2}(Y, E|Y) → L^{2}(Y, E|Y) such that

R(λ)^{−1} = R_{1}(λ) + R_{2}(λ), λ > 0,

and ∥ R_{2}(λ) ∥ is uniformly bounded as λ → 0.

Let H ⊂ C^{∞}(X, E) be the subspace spanned by ker Δ and E(ϕ_{1}, 0), …, E(ϕ_{l}, 0). Then H is the subspace of all bounded sections ϕ ∈ C^{∞}(X, E) such that Δϕ = 0. Set

H_{Y} = ρ_{Y} (H).

Lemma 3.8. The restriction map ρ_{Y} : H → H_{Y} is an isomorphism.

Proof. Let ϕ ∈ H. Then Δϕ = 0 and ϕ is bounded. Suppose that ρ_{Y} (ϕ) = 0. This means that ϕ|_{Y} = 0. By the uniqueness of the Dirichlet problem, it follows that ϕ = 0. Thus ρ_{Y} is injective and hence an isomorphism. □

Lemma 3.9. ker R = H_{Y} .

Proof. Let φ ∈ H_{Y} . Then there exists ψ ∈ H with ψ|_{Y} = φ. Moreover ψ is bounded and Δψ = 0. Thus ψ is a solution of the Dirichlet problem (3.4). Since ψ is smooth on X, it follows that Rφ = 0. Now suppose that φ ∈ ker R. Then there exists a bounded solution ψ of (3.4) such that

\frac{∂}{∂u}(ψ|_{M})|_{∂M} = \frac{∂}{∂u}(ψ|_{Z})|_{∂Z}.

This implies that Δψ = 0 in the sense of distributions. By elliptic regularity it follows that ψ ∈ C^{∞}(X, E) and Δψ = 0. If we expand ψ|_{Z} in the orthonormal basis {ϕ_{j}}_{j∈ℕ} we get

ψ(u, y) = ∑_{j=1}^{m} a_{j}ϕ_{j}(y) + ∑_{j=m+1}^{∞} a_{j}e^{−u\sqrt{λ_{j}}}ϕ_{j}(y),

where m = dim ker Δ_{Y}. Let ϕ = ∑_{j=1}^{m} a_{j}ϕ_{j}. Then we get

ψ|_{Z} = ϕ + ψ_{1},

where ψ_{1} ∈ L^{2}. Put ψ̃ = ψ − \frac{1}{2}E(ϕ, 0). Then it follows that ψ̃ ∈ ker Δ. This implies that ψ ∈ H. □

Let ⟨·, ·⟩_{Y} be the inner product in H_{Y} induced by the inner product in L^{2}(Y, E|Y). Let φ_{1}, …, φ_{k} be an orthonormal basis of ker Δ. Set ψ_{i} = ρ_{Y} (φ_{i}), if 1 ≤ i ≤ k, and ψ_{k+j} = \frac{1}{2}ρ_{Y} (E(ϕ_{j}), 0)), if 1 ≤ j ≤ l. Set a_{ij} = ⟨ψ_{i}, ψ_{j}⟩_{Y}, 1 ≤ i, j ≤ k + l and let A be the (k + l) × (k + l)-matrix with entries a_{ij}, i, j = 1, ..., k + l. Then the main result of this section is the following theorem.

Theorem 3.10. Let k = dim ker Δ and l = dim V^{+}. Then

log det R(λ) = (k + l/2) log λ − log det A + log det R + O(λ)

as λ → 0+.

Proof. The proof is analogous to the proof of Theorem B of [Le1]. Let

0 ≤ μ_{1}(λ) ≤ ⋯ ≤ μ_{k+l}(λ) < μ_{k+l+1}(λ) ≤ ⋯

be the eigenvalues of R(λ). By Lemma 3.9 it follows that

lim_{λ→0} μ_{j}(λ) = 0 for 1 ≤ j ≤ k + l,

and μ_{j}(λ) ≥ c > 0 for j > k + l. Then

(3.14)  log det R(λ) = log(μ_{1}(λ) ⋯ μ_{k+l}(λ)) + log det R + O(λ)

as λ → 0. So it remains to determine the behaviour of log(μ_{1}(λ) ⋯ μ_{k+l}(λ)) as λ → 0. Let η_{1}(λ), ..., η_{k+l}(λ) be an orthonormal set of eigensections of R(λ) corresponding to the eigenvalues μ_{1}(λ), ..., μ_{k+l}(λ). Let 1 ≤ j ≤ k + l. By Proposition 3.7 we get

μ_{i}(λ)^{−1}δ_{ij} = ⟨R(λ)^{−1}η_{i}(λ), η_{j}(λ)⟩ = ⟨R_{1}(λ)η_{i}(λ), η_{j}(λ)⟩ + ⟨R_{2}(λ)η_{i}(λ), η_{j}(λ)⟩,

and the second term on the right remains bounded as λ → 0+. By (3.13) the first term equals

(3.15)  \begin{aligned}⟨R_{1}(λ)η_{i}(λ), η_{j}(λ)⟩ & = \frac{1}{λ} ∑_{p=1}^{k}⟨φ_{p}, η_{i}(λ)⟩_{Y} ⟨φ_{p}, η_{j}(λ)⟩_{Y} \\ & + \frac{1}{4\sqrt{λ}} ∑_{q=1}^{l}⟨E(ϕ_{q}, 0), η_{i}(λ)⟩_{Y} ⟨E(ϕ_{q}, 0), η_{j}(λ)⟩_{Y} . \end{aligned}

Set

ψ̃_{i}(λ) = \begin{cases}ρ_{Y} (φ_{i}), & if 1 ≤ i ≤ k, \\ \frac{λ^{1/4}}{2} ρ_{Y} (E(ϕ_{i−k}, 0)), & if k + 1 ≤ i ≤ k + l. \end{cases}

Let ã_{ij}(λ) = ⟨ψ̃_{i}(λ), η_{j}(λ)⟩ and let Ã(λ) be the matrix with entries ã_{ij}(λ), 1 ≤ i, j ≤ k + l. Then (3.15) can be written as

⟨R_{1}(λ)η_{i}(λ), η_{j}(λ)⟩ = \frac{1}{λ}(Ã(λ)^{t}Ã(λ))_{ij}

and we get

μ_{i}(λ)^{−1}δ_{ij} = \frac{1}{λ}(Ã(λ)^{t}Ã(λ))_{ij} + O(1)

as λ → 0+. Note that (Ã(λ)^{t}Ã(λ))_{ij} is bounded as λ → 0+. Hence for i ≠ j we get (Ã(λ)^{t}Ã(λ))_{ij} = O(λ) as λ → 0+. This implies

(3.16)  (μ_{1}(λ) ⋯ μ_{k+l}(λ))^{−1} = λ^{−(k+l)} det(Ã(λ)^{t}Ã(λ))(1 + O(λ))

as λ → 0+. Now observe that Ã(λ)Ã(λ)^{t} is equal to the matrix with entries ⟨ψ̃_{i}(λ), ψ̃_{j}(λ)⟩, 1 ≤ i, j ≤ k + l. Let

C(λ) = (\begin{matrix}Id_{k} & 0 \\ 0 & λ^{1/4} Id_{l} \end{matrix}) .

Then it follows from the definition of A that

Ã(λ)Ã(λ)^{t} = C(λ) · A · C(λ).

Together with (3.16) we obtain

(μ_{1}(λ) ⋯ μ_{k+l}(λ))^{−1} = λ^{−(k+l/2)} det(A)(1 + O(λ)).

Taking the logarithm and inserting the result in (3.14), the theorem follows.

□

4. Proof of Theorem 1.1

Let λ > 0. By Theorem 4.2 of [Ca] there is a polynomial P (λ) with real coefficients of degree ≤ (n − 1)/2 such that

(4.1)  \frac{det(Δ + λ, Δ_{0} + λ)}{det(Δ_{M,D} + λ)} = e^{P (λ)} det(R(λ)).

All terms have asymptotic expansions as λ → 0. Since Δ_{M,D} is invertible, det(Δ_{M,D} +λ) is continuous at λ = 0 and lim_{λ→0} det(Δ_{M,D} +λ) = det(Δ_{M,D}). Next consider the polynomial P (λ). In the proof of Proposition 4.7 of [Ca], Carron has shown that the polynomial P (λ) can be computed in terms of the coefficients of the asymptotic expansion of Tr(e^{−tΔ_{Y}}) as t → 0. Let

Tr(e^{−tΔ_{Y}}) ∼ ∑_{j=0}^{∞} a_{j}t^{−(n−1)/2+j}, t → 0+,

be the heat expansion. If n is even, we have P = 0, and if n = 2p + 1 then

P (λ) = − log(2) ∑_{j=0}^{p} \frac{(−1)^{p−j}}{(p − j)!} a_{j}λ^{p−j}.

In particular, it follows that

P (0) = − log(2)(h_{Y} + ζ_{Y} (0)),

where h_{Y} = dim ker Δ_{Y} and ζ_{Y} (s) is the zeta function of Δ_{Y}. Together with Corollary 2.2 and Theorem 3.10, Theorem 1.1 follows.

5. Regularized determinants on a finite cylinder

In this section we study the regularized determinant of a Laplace type operator on a finite cylinder over a closed Riemannian manifold Y. Let Δ_{Y} : C^{∞}(Y, E_{0}) → C^{∞}(Y, E_{0}) be a Laplace type operator on Y. For r > 0 set

Z_{r} = [0, r] × Y.

Let E → Z_{r} be the pull back bundle of E_{0}, i.e., E = [0, r] × E_{0}. Let

Δ = Δ_{Z_{r}} := −\frac{∂^{2}}{∂u^{2}} + Δ_{Y} : C^{∞}(Z_{r}, E) → C^{∞}(Z_{r}, E).

Then Δ is a Laplace type operator on Z_{r}. Impose Dirichlet boundary conditions at ∂Z_{r} and let Δ_{D} be the corresponding self-adjoint extension. Then Δ_{D} is positive definite. Let

0 ≤ μ_{1} ≤ μ_{2} ≤ ⋯ → +∞

be the eigenvalues of Δ_{Y} , counted with multiplicity. Let ζ_{Y} (s) be the zeta function of Δ_{Y} and set

(5.1)  ξ_{Y} (s) = \frac{Γ(s − 1/2)}{\sqrt{π}Γ(s)}ζ_{Y} (s − 1/2).

20 JÖRN MÜLLER AND WERNER MÜLLER

Sine ζ_{Y} (s) has at most a simple pole at s = −1/2, ξ_{Y} (s) is holomorphic at s = 0. The main result of this section is the following proposition. Proposition 5.1. Let h_{Y} = dim ker Δ_{Y} . Then

(5.2)  det(Δ_{D}) = (2r)^{h_{Y}} e^{−rξ′_{Y} (0)/2}(det Δ_{Y})^{−1/2} · ∏_{μ_{j}>0}(1 − e^{−2r\sqrt{μ_{j}}}).

Proof. The eigenvalues of Δ_{D} are given by

λ_{k,l} = μ_{l} + (\frac{π}{r} )^{2} k^{2}, k, l ∈ ℕ.

Hence the zeta function of Δ_{D} equals

ζ_{Δ_{D}}(s) = ∑_{k,l∈ℕ}(μ_{l} + (\frac{π}{r} )^{2} k^{2})^{−s}, Re(s) > \frac{d + 1}{2},

where d = dim Y. Let ζ(s) denote the Riemann zeta function. Then

(5.3)  ζ_{Δ_{D}}(s) = h_{Y} (\frac{π}{r} )^{−2s} ζ(2s) + ∑_{k∈ℕ} ∑_{μ_{l}>0}(μ_{l} + (\frac{π}{r} )^{2} k^{2})^{−s}, Re(s) > \frac{d + 1}{2}.

Recall that ζ(0) = −1/2 and ζ′(0) = −1/2 log(2π). Hence we get

(5.4)  \frac{d}{ds} \{(\frac{π}{r} )^{−2s} ζ(2s)\} |_{s=0} = − log 2 − log r.

Set

ζ_{1}(s) := ∑_{k∈ℕ} ∑_{μ_{l}>0}(μ_{l} + (\frac{π}{r} )^{2} k^{2})^{−s}, Re(s) > \frac{d + 1}{2}.

By the Poisson summation formula we get

\begin{aligned}Γ(s)ζ_{1}(s) & = ∑_{μ_{l}>0} ∫_{0}^{∞} e^{−μ_{l}t} ∑_{k∈ℕ}e^{−(π/r)^{2}k^{2}t}t^{s−1}dt \\ & = ∑_{μ_{l}>0} ∫_{0}^{∞} e^{−μ_{l}t} (\frac{r}{\sqrt{πt}} ∑_{k∈ℕ}e^{−r^{2}k^{2}/t} + \frac{1}{2} (\frac{r}{\sqrt{πt}} − 1)) t^{s−1}dt \\ & = \frac{r}{\sqrt{π}} ∑_{k∈ℕ} ∑_{μ_{l}>0} ∫_{0}^{∞} e^{−(μ_{l}t+r^{2}k^{2}/t)}t^{s−3/2}dt \\ & + \frac{1}{2} \frac{r}{\sqrt{π}} Γ(s − 1/2)ζ_{Y} (s − 1/2) − \frac{1}{2}Γ(s)ζ_{Y} (s). \end{aligned} ; (5.5)

Denote by T (s) the integral-series on the right hand side. For a, b, c ≠ 0 and s ∈ ℂ set K_{s}(a, b) = ∫_{0}^{∞} e^{−(a^{2}t+b^{2}/t)}t^{s−1}dt and K_{s}(c) = ∫_{0}^{∞} e^{−c(t+1/t)}t^{s−1}dt.

It is proved in [La, p.270f] that the following relations hold

(5.6)  K_{s}(c) = K_{−s}(c), K_{s}(a, b) = (\frac{b}{a})^{s} K_{s}(ab), K_{1/2}(c) = \sqrt{\frac{π}{c}} e^{−2c}.

Furthermore, for every x_{0} > 0 and σ_{0} < σ_{1} there exists C = C(x_{0}, σ_{0}, σ_{1}) such that

(5.7)  |K_{s}(x)| ≤ Ce^{−2x}

for all x ≥ x_{0} and Re(s) ∈ [σ_{0}, σ_{1}] [La]. With this notation we have

T (s) = \frac{r}{\sqrt{π}} ∑_{k∈ℕ} ∑_{μ_{l}>0} K_{s−1/2}(\sqrt{μ_{l}}, rk).

Using (5.6) and (5.7) it follows that T (s) is an entire function of s.. Especially it is holomorphic at s = 0. Since by (5.6) we have K_{−1/2}(a, b) = \frac{\sqrt{π}}{b}e^{−2ab}, we get

(5.8)  T (0) = \frac{r}{\sqrt{π}} ∑_{k∈ℕ} ∑_{μ_{l}>0} \frac{\sqrt{π}}{rk} e^{−2r\sqrt{μ_{l}}k} = − ∑_{μ_{l}>0} log(1 − e^{−2r\sqrt{μ_{l}}}).

Thus by (5.5) we have

ζ_{1}(s) = \frac{1}{Γ(s)}T (s) + \frac{1}{2} \frac{r}{\sqrt{π}} \frac{Γ(s − 1/2)}{Γ(s)}ζ_{Y} (s − 1/2) − \frac{1}{2}ζ_{Y} (s).

Using that ξ_{Y} (s) is holomorphic at s = 0, we obtain

ζ′_{1}(0) = T (0) + rξ′_{Y} (0) − \frac{1}{2}ζ′_{Y} (0).

Together with (5.3), (5.4) and (5.8) we get

ζ′_{Δ_{D}}(0) = − ∑_{μ_{k}>0} log (1 − e^{−2r\sqrt{μ_{k}}}) + rξ′_{Y} (0) − \frac{1}{2}ζ′_{Y} (0) − h_{Y} (log 2 + log r).

This implies the claimed equality. □

6. The decomposition formula

Let (M, g) be a closed connected n-dimensional Riemannian manifold and let Y ⊂ M be a separating hypersurface as in the introduction such that

M = M_{1} ∪_{Y} M_{2}, M_{1} ∩ M_{2} = Y.

We assume that the metric g is a product on a tubular neighborhood N of Y. For r ≥ 0 let

M_{1,r} = M_{1} ∪ ([−r, 0] × Y), M_{2,r} = M_{2} ∪ ([0, r] × Y),

where we identify Y with {−r} × Y in the first case and with {r} × Y in the second case. Set

M_{r} = M_{1,r} ∪_{{0}×Y} M_{2,r}, N_{r} = [−r, r] × Y.

Then

(6.1)  M_{r} = M_{1} ∪_{Y} N_{r} ∪_{Y} M_{2},

where ∂M_{1} is identified with {−r} × Y and ∂M_{2} with {r} × Y. The metric g on M has an obvious extension to a metric on M_{r}. Furthermore, let

M_{i,∞} = M_{i} ∪_{Y} (ℝ^{+} × Y), i = 1, 2.

Let Δ_{M} : C^{∞}(M, E) → C^{∞}(M, E) be a Laplace type operator as in the introduction. and let Δ_{M_{r}} be its canonical extension to a Laplace type operator on M_{r}, i.e. Δ_{M_{r}} is uniquely defined by

Δ_{M_{r}} |_{M_{i}} = Δ|_{M_{i}}, Δ_{M_{r}} |_{N_{r}} = −\frac{∂^{2}}{∂u^{2}} + Δ_{Y} .

Let Δ_{M_{i}} = Δ|_{M_{i}} and let Δ_{M_{i},D} be the selfadjoint extension of Δ_{M_{i}} : C_{c}^{∞}(M_{i}, E) → L^{2}(M_{i}, E) with respect to Dirichlet boundary conditions. We assume that Δ_{M_{1},D} and Δ_{M_{2},D} are invertible. Let Δ_{N_{r},D} denote the selfadjoint extension of

−\frac{∂^{2}}{∂u^{2}} + Δ_{Y} : C_{c}^{∞}(N_{r}, E) → L^{2}(N_{r}, E)

with respect to Dirichlet boundary conditions. Let Y_{±r} := {±r}×Y and denote by Σ_{r} ⊂ M_{r} the hypersurface

Σ_{r} := Y_{−r} ⊔ Y_{r}.

Given z ∈ ℂ−ℝ^{−}, let R_{r}(z) be the Dirichlet-to-Neumann operator associated to (Δ_{M_{r}} +z) and the hypersurface Σ_{r}. We recall the definition of R_{r}(z). Let ϕ ∈ C^{∞}(Σ_{r}, E_{r}|Σ_{r}). There exists a unique section φ ∈ C^{∞}(M_{r} − Σ_{r}, E_{r}) ∩ C^{0}(M_{r}, E_{r}) such that

(6.2)  \begin{aligned}(Δ_{M_{r}} + z)φ & = 0 on M_{r} − Σ_{r}; \\ φ & = ϕ on Σ_{r}. \end{aligned}

Then R_{r}(z)(ϕ) is given by

(6.3)  \begin{aligned}R_{r}(z)(ϕ)|_{Y_{−r}} & = \frac{∂}{∂u}(φ|_{M_{1}})|_{∂M_{1}} − \frac{∂}{∂u}(φ|_{N_{r}})|_{Y_{−r}}, \\ R_{r}(z)(ϕ)|_{Y_{r}} & = \frac{∂}{∂u}(φ|_{N_{r}})|_{Y_{r}} − \frac{∂}{∂u}(φ|_{M_{2}})|_{∂M_{2}}. \end{aligned}

Now we apply the Mayer-Vietoris formula of [BFK], specialized to our case. We note that Theorem 1.4 of [Ca] also holds in our case, where M_{r} − Σ_{r} consists of three components. Thus there exists a polynomial P (z) with real coefficients of degree < (n − 1)/2 such that for every z ∈ ℂ − ℝ^{−} :

\frac{det(Δ_{M_{r}} + z)}{det(Δ_{N_{r},D} + z) det(Δ_{M_{1},D} + z) det(Δ_{M_{2},D} + z)} = e^{P (z)} det R_{r}(z).

Since we assume that the metric of M_{r} is a product on a tubular neighborhood of Σ_{r}, the polynomial depends only on Y and can be computed as follows. Let ζ_{Y} (s, z) be the zeta function of Δ_{Y} + z. Then it follows from [PW1, Theorem 6.3] and also from the proof of Proposition 4.7 of [Ca] that

P (z) = −2ζ_{Y} (0, z).

Thus

(6.4)  \frac{det(Δ_{M_{r}} + z)}{det(Δ_{N_{r},D} + z) det(Δ_{M_{1},D} + z) det(Δ_{M_{2},D} + z)} = 2^{−2ζ_{Y} (0,−z)} det R_{r}(z).

Now take z = λ > 0 and consider the limit as λ → 0 of the left and right hand side of (6.4). Since Δ_{M_{i},D}, i = 1, 2, and Δ_{N_{r},D} are invertible, it follows that

(6.5)  lim_{z→0} det(Δ_{M_{i},D} + λ) = det Δ_{M_{i},D}, lim_{z→0} det(Δ_{N_{r},D} + λ) = det Δ_{N_{r},D}.

Let h_{r} = dim ker Δ_{M_{r}}. Then

det(Δ_{M_{r}} + λ) = λ^{h_{r}} det(Δ_{M_{r}}|_{(ker Δ_{Mr})^{⊥}} + λ)

and therefore we get

(6.6)  lim_{λ→0} det(Δ_{M_{r}} + λ)λ^{−h_{r}} = det Δ_{M_{r}}.

Also note that

(6.7)  lim_{λ→0} ζ_{Y} (0, λ) = ζ_{Y} (0) + h_{Y} ,

where h_{Y} = dim ker Δ_{Y} . It remains to consider the limit of det R_{r}(λ) as λ → 0. Let

ρ_{r}: C^{∞}(M_{r}, E_{r}) → C^{∞}(Σ_{r}, E|Σ_{r})

denote the restriction operator. Let H_{r} := ρ_{r}(ker Δ_{M_{r}}).

Lemma 6.1.

ρ_{r} : ker Δ_{M_{r}} → H_{r}

is an isomorphism.

Proof. Let ϕ ∈ ker Δ_{M_{r}} and suppose that ϕ|_{Σ_{r}} = 0. Let ψ = ϕ|_{N_{r}}. Then Δ_{N_{r}}ψ = 0 and ψ|_{∂N_{r}} = 0. Since Δ_{N_{r},D} is invertible, it follows that ψ = 0. In the same way we get ϕ|_{M_{i}} = 0, i = 1, 2, and hence ϕ = 0. Thus ρ_{r} is injective and therefore an isomorphism. □

Let Δ_{M_{r},D} be the selfadjoint extension of

Δ_{M_{r}} : C_{c}^{∞}(M_{r} − Σ_{r}, E_{r}) → L^{2}(M_{r}, E_{r})

with respect to Dirichlet boundary conditions. By our assumption, Δ_{M_{r},D} is invertible and hence, the Dirichlet-to-Neumann operator R_{r} associated to Δ_{M_{r}} with respect to Σ_{r} ⊂ M_{r} can be defined in the same way as R_{r}(z).

Lemma 6.2. We have

ker R_{r} = ρ_{r}(ker Δ_{M_{r}}).

Proof. Let φ ∈ ker Δ_{M_{r}} and let ϕ = ρ_{r}(φ). Then φ is a solution of the Dirichlet problem with boundary value ϕ. Since φ is smooth on M_{r}, it follows that R_{r}(ϕ) = 0. Now suppose that ϕ ∈ ker R_{r}. Then there exists φ ∈ C^{∞}(M_{r} − Σ_{r}) ∩ C^{0}(M_{r}) such that Δ_{M_{r}}φ = 0 on M_{r} − Σ_{r}, φ|_{Σ_{r}} = ϕ and

\frac{∂}{∂u}(φ|_{M_{1}})|_{∂M_{1}} = \frac{∂}{∂u}(φ|_{N_{r}})|_{Y_{−r}}, \frac{∂}{∂u}(φ|_{N_{r}})|_{Y_{r}} = \frac{∂}{∂u}(φ|_{M_{2}})|_{∂M_{2}}.

This implies that Δ_{M_{r}}φ = 0 in the distributional sense. By elliptic regularity we conclude that φ ∈ ker Δ_{M_{r}} and ρ_{r}(φ) = ϕ. □

Let φ_{1}, ..., φ_{p} be an orthonormal basis of ker Δ_{M_{r}}. Set

b_{ij} = ⟨ρ_{r}(φ_{i}), ρ_{r}(φ_{j})⟩_{Σ_{r}}, i, j = 1, ..., p

and let

B_{r} = (b_{ij})_{i,j=1}^{p}.

Then B_{r} is a symmetric invertible matrix.

Proposition 6.3. Let h_{r} = dim ker Δ_{M_{r}}. Then

log det R_{r}(λ) = h_{r} log λ − log det B_{r} + log det R_{r} + O(λ)

as λ → ∞.

Proof. We use Lemma 6.2 and proceed in the same way as in the proof of Theorem (3.10).

□

Combining (6.4)-(6.7) and Proposition (6.3) we obtain

log det(Δ_{M_{r}}) = log det Δ_{N_{r},D} + log det Δ_{M_{1},D} + log det Δ_{M_{2},D}
− log det B_{r} + log det R_{r} − 2(ζ_{Y} (0) + h_{Y}) log 2.

(6.8)

Let Z = ℝ^{+} × Y and let Δ_{0} be the selfadjoint extension of the symmetric operator

−\frac{∂^{2}}{∂u^{2}} + Δ_{Y} : C_{c}^{∞}(Z, E) → L^{2}(Z, E)

with respect to Dirichlet boundary conditions at ∂Z = {0} × Y. Let R_{i,∞} be the Dirichletto-Neumann operator for Δ_{i,∞} with respect to the hypersurface Y = {0} × Y ⊂ M_{i,∞}. Let A_{i} be the Grahm matrix defined by the restrictions of the extended L^{2}-solutions of Δ_{i,∞} to Y as in Theorem 3.10. By Theorem (1.1) we have

log det(Δ_{i,∞}, Δ_{0}) = log det R_{i,∞} + log det Δ_{M_{i},D} − log det A_{i}
− (ζ_{Y} (0) + h_{Y}) log 2.

Together with (6.8) we get Proposition 6.4. Let the notation be as above. Then

log det Δ_{M_{r}} = log det Δ_{N_{r},D} − log det B_{r} + log det R_{r}
+ ∑_{i=1}^{2}(log det(Δ_{i,∞}, Δ_{0}) − log det R_{i,∞} + log det A_{i}).

(6.9)

Our next purpose is to study the behaviour of the various terms in this equality as r → ∞. This, of course, will require additional assumptions. We begin with the consideration of det R_{r}.

To this end we need to describe the operator R_{r} more explicitely. Let Q_{i} denote the Neumann jump operator on M_{i}. In the proof of Lemma (3.2) we established the following equality

(6.10)  R_{i,∞} = Q_{i} + \sqrt{Δ_{Y}} , i = 1, 2.

Let P_{0} : L^{2}(Y, E|Y) denote the orthogonal projection onto ker Δ_{Y} . Let

h_{r}(x) = \frac{\sqrt{x}}{sinh(2\sqrt{x}r)}.

Define

K_{r} : L^{2}(Y, E|Y) ⊕ L^{2}(Y, E|Y) → L^{2}(Y, E|Y) ⊕ L^{2}(Y, E|Y)

by

(6.11)  K_{r} := (\frac{1}{2r}P_{0} + h_{r}(Δ_{Y})P_{0}^{⊥}) (\begin{matrix}e^{−2\sqrt{Δ_{Y}} r} & −Id \\ −Id & e^{−2\sqrt{Δ_{Y}} r} \end{matrix}) .

Set

(6.12)  R_{∞} = (\begin{matrix}R_{1,∞} & 0 \\ 0 & R_{2,∞} \end{matrix}) .

Recall that Σ_{r} ≅ Y ⊔ Y. Using (5.8) and the formula at the bottom of p. 4104 of [Le3], it follows that

R_{r} : C^{∞}(Y, E|Y) ⊕ C^{∞}(Y, E|Y) → C^{∞}(Y, E|Y) ⊕ C^{∞}(Y, E|Y)

is given by

(6.13)  R_{r} = R_{∞} + K_{r}.

Next observe that K_{r} is a trace class operator and its trace norm ∥ K_{r} ∥_{1} satisfies

(6.14)  ∥ K_{r} ∥_{1}⟶_{r→∞} 0.

By Corollary 3.4, R_{i,∞}, i = 1, 2, are selfadjoint nonnegative operators in L^{2}(Y, E|Y).

Lemma 6.5. Suppose that R_{i,∞} > 0, i = 1, 2. Then

lim_{r→∞} det R_{r} = det R_{1,∞} · det R_{2,∞}.

Proof. This is proved in [Le3, Lemma 4.1 ]. For the convenience of the reader we recall the proof. It follows from (6.12) and the assumptions that R_{∞} > 0. By (6.14) it follows that there exists r_{0} > 0 such that the operator R_{∞} + tK_{r} is invertible for 0 ≤ t ≤ 1 and r ≥ r_{0}. Thus

log det(R_{∞} + K_{r}) − log det R_{∞} = ∫_{0}^{1} \frac{d}{dt} log det(R_{∞} + tK_{r})dt

= ∫_{0}^{1} Tr((R_{∞} + tK_{r})^{−1}K_{r})dt ≤ \frac{1}{2λ_{0}} ∥ K_{r} ∥_{1},

where λ_{0} > 0 is the smallest eigenvalue of R_{∞}. The lemma follows from (6.14). □

Let H_{i}, i = 1, 2, be the space of extended L^{2}-solutions of Δ_{i,∞}. By Lemma (3.8) and Lemma (3.9) it follows that R_{i,∞} is invertible if and only if H_{i} = {0}, and the latter condition is a consequence of ker Δ_{Y} = {0} and ker Δ_{i,∞} = {0}. Furthermore, if R_{∞} is invertible, it follows from (6.13) and (6.14) that R_{r} is invertible for r ≥ r_{0}. By Lemma 6.1 and Lemma 6.2, R_{r} is invertible if and only if ker Δ_{M_{r}} = {0}.

Using these observation together with Proposition (6.4) and Lemma (6.5), we obtain

Corollary 6.6. Suppose that ker Δ_{Y} = {0} and ker Δ_{i,∞} = 0, i = 1, 2. Then Δ_{M_{r}} is invertible for r ≥ r_{0} and

lim_{r→∞} \frac{det Δ_{M_{r}}}{det Δ_{N_{r},D}} = det(Δ_{1,∞}, Δ_{0}) · det(Δ_{2,∞}, Δ_{0}).

The asymptotic behaviour of det Δ_{N_{r},D} as r → ∞ is described by Proposition 5.1. Using this result, Theorem 1.2 follows.

Next we consider a compact Riemannian manifold (X_{0}, g) with a nonempty boundary Y. We assume that the metric is a product on a collar neighborhood N = (−ϵ, 0] × Y of Y in X_{0}. Let

Δ_{X_{0}} : C^{∞}(X_{0}, E) → C^{∞}(X_{0}, E)

be Laplace type operator as above such that on N it equals −∂^{2}/∂u^{2} + Δ_{Y} . For r > 0 set

Z_{r} = [0, r] × Y, and X_{r} = X_{0} ∪_{Y} Z_{r},

where {0} × Y ⊂ Z_{r} is identified with ∂X_{0} = Y. Let X_{∞} = X_{0} ∪_{Y} (ℝ^{+} × Y) be the corresponding manifold with a cylindrical end. We extend Δ_{X_{0}} in the obvious to Laplace type operators Δ_{X_{r}} and Δ_{∞} on X_{r} and X_{∞}, respectively. Let Δ_{X_{r},D} and Δ_{Z_{r},D} denote the Dirichlet Laplacians associated to Δ_{X_{r}} and Δ_{Z_{r}}, respectively. Furthermore, let R_{r} be the Dirichlet-to-Neumann operator associated to the decomposition X_{r} = X_{0} ∪ Z_{r}. Let λ > 0. Then by Theorem 4.2 of [Ca] we have

(6.15)  \frac{det(Δ_{X_{r},D} + λ)}{det(Δ_{X_{0},D} + λ) det(Δ_{Z_{r},D} + λ)} = 2^{−ζ_{Y} (0,λ)} det R_{r}(λ).

As above, we let λ → 0. Note that Δ_{Z_{r},D} is invertible. Assume that Δ_{X_{r},D} is invertible. Then as λ → 0, the determinants converge to the determinants of Δ_{X_{r}} and Δ_{Z_{r}}, respectively. Furthermore as in Lemma 6.2 it follows that

(6.16)  ker R_{r} = ρ_{Y} (ker Δ_{M_{r},D}).

Hence R_{r} is also invertible and

log det R_{r}(λ) = log det R_{r} + O(λ)

as λ → 0. Thus taking the limit λ → 0 of both sides of (6.15), we get

(6.17)  \frac{det Δ_{X_{r},D}}{det Δ_{X_{0},D} det Δ_{Z_{r},D}} = 2^{−ζ_{Y} (0)} det R_{r}.

Now recall that by Theorem 1.1 we have

log det(Δ_{∞}, Δ_{0}) = log det R_{∞} + log det Δ_{X_{r},D} − log det A − log(2) (ζ_{Y} (0) + h_{Y}).

Combining this equality with (6.17) we obtain

log det Δ_{X_{r},D} = log det Δ_{N_{r},D} + log det R_{r} − log det R_{∞}
+ det(Δ_{∞}, Δ_{0}) + log det A.

(6.18)

To study the behaviour of det R_{r} as r → ∞, we proceed as above. Let

f_{r}(x) = \frac{\sqrt{x}}{sinh(r\sqrt{x})}, x ∈ ℝ.

Define the operator

L_{r}: L^{2}(Y, E|Y) → L^{2}(Y, E|Y)

by

(6.19)  L_{r} := (\frac{1}{r}P_{0} + f_{r}(Δ_{Y})P_{0}^{⊥}) e^{−r\sqrt{Δ_{Y}}} ,

where P_{0} denotes the orthogonal projection onto ker Δ_{Y}. Then

R_{r} = R_{∞} + L_{r}.

Suppose that ker Δ_{Y} = {0} and ker Δ_{∞} = {0}. Then it follows from Lemma 3.9 that ker R_{∞} = {0} and by Lemma 4.1 of [Le3] we get

lim_{r→∞} det R_{r} = det R_{∞}.

Since ∥ L_{r} ∥→ 0 as r → ∞, it follows that R_{r} is invertible for r ≥ r_{0}. By (6.16) this implies that Δ_{M_{r},D} is invertible for r ≥ r_{0}. Under the same assumptions we have det A = 1. Together with (6.18) we get

Proposition 6.7. Suppose that ker Δ_{Y} = {0} and ker Δ_{∞} = {0}. Then

lim_{r→∞} \frac{det Δ_{X_{r},D}}{det Δ_{Z_{r},D}} = det(Δ_{∞}, Δ_{0}).

Using Proposition 5.1, it follows that as r → ∞,

(6.20)  det Δ_{X_{r},D} ∼ e^{−rξ′_{Y} (0)/2} (det Δ_{Y})^{−1/2} det(Δ_{∞}, Δ_{0}).

We apply (6.20) to det Δ_{M_{i,r},D}, i = 1, 2, and compare the asymptotic behaviour with (1.11). In this way we get

lim_{r→∞} \frac{det Δ_{M_{r}}}{det Δ_{M_{1,r},D} det Δ_{M_{2,r},D}} = (det Δ_{Y})^{1/2}

which is the statement of Corollary 1.5.

7. Bochner-Laplace operators

In this section we study the case where Δ is a connection Laplacian. To begin with we consider a manifold with a cylindrical end X = M ∪_{Y} Z, Z = ℝ^{+} × Y. Let F → X be a Hermitian vector bundle over X such that F |_{ℝ^{+}×Y} = pr_{Y}^{∗} (F_{0}) for some Hermitian vector bundle F_{0} over Y. Let ∇ be a metric connection in F such that on ℝ^{+} × Y it has the form

(7.1)  ∇ = d_{u} ⊗ Id +d ⊗ ∇^{Y} ,

where ∇^{Y} is a metric connection on F_{0}. Let

Δ = ∇^{∗}∇, Δ_{Y} = (∇^{Y})^{∗}∇^{Y}

be the associated Bochner-Laplace operators. Then

(7.2)  Δ = −\frac{∂^{2}}{∂u^{2}} + Δ_{Y} on ℝ^{+} × Y.

Let \overline{Δ} be the unique selfadjoint extension of Δ_{X} in L^{2}.

Let {ϕ_{i}}_{i∈ℕ} be an orthonormal basis of L^{2}(Y, F |Y) consisting of eigensections of Δ_{Y} with eigenvalues

0 ≤ μ_{1} ≤ μ_{2} ≤ ⋯ → +∞.

Lemma 7.1. We have

ker \overline{Δ} = \{0\}.

Proof. Let φ ∈ C^{∞}(X, F) be a square integrable solution of Δφ = 0. Then

0 = ⟨∇^{∗}∇φ, φ⟩ =∥ ∇φ ∥^{2} .

Thus ∇φ = 0. Since φ is square integrable, it has the following expansion on ℝ^{+} × Y in terms of the orthonormal basis {φ_{i}}_{i∈ℕ} :

(7.3)  φ(u, y) = ∑_{μ_{i}>0} c_{i}e^{−\sqrt{μ_{i}}u}ϕ_{i}(y), u ∈ ℝ^{+}, y ∈ Y.

Furthermore,

\frac{∂φ}{∂u}(u, y) = ∇_{\frac{∂}{∂u}} φ = 0.

Using (7.3), it follows that the restriction of φ to ℝ^{+} × Y vanishes. Since ∇φ = 0 and ∇ is a metric connection, it follows that d ∥ φ ∥^{2}= 0 and hence φ = 0. □

Let

ker Δ_{Y} = V^{+} ⊕ V^{−}

be the decomposition into the ±1-eigenspaces of the scattering matrix S(0) (cf. §2) and let

R : C^{∞}(Y, F |Y) → C^{∞}(Y, F |Y)

be the Dirichlet-to-Neumann operator with respect to the hypersurface

Y = {0} × Y ⊂ X.

Lemma 7.2. We have

ker R = V^{+}.

Proof. Let H ⊂ C^{∞}(X, E) be the space of bounded solutions of Δφ = 0. By Lemma (3.9) we have ker R = ρ_{Y} (H). So it suffices to prove that

ρ_{Y} (H) = V^{+}.

Let φ ∈ H. For r > 0 let X_{r} = M ∪_{Y} ([0, r] × Y). Using integration by parts, we get

\begin{aligned}0 & = ∫_{X_{r}}⟨∇^{∗}∇φ(x), φ(x)⟩dx \\ & = ∫_{X_{r}} |∇φ(x)|^{2}dx − ∫_{Y} ⟨\frac{∂}{∂u}φ(u, y), φ(u, y)⟩|_{u=r}dy. \end{aligned} ; (7.4)

Since φ is bounded and satisfies Δφ = 0, it has the following expansion on ℝ^{+} × Y:

(7.5)  φ(u, y) = ∑_{i=1}^{h_{Y}} a_{i}ϕ_{i}(y) + ∑_{i=h_{Y} +1}^{∞} b_{i}e^{−\sqrt{μ_{i}}u}ϕ_{i}(y),

where h_{Y} = dim ker Δ_{Y}. This implies that the second integral on the right of (7.4) is exponentially decreasing as r → ∞. Hence ∇φ = 0. In particular, it follows that

\frac{∂}{∂u}φ(u, y) = 0, u ∈ ℝ^{+}, y ∈ Y.

Together with (7.5) we get

(7.6)  φ|_{Z} = ϕ ∈ ker Δ_{Y} .

Thus ρ_{Y} (H) ⊂ ker Δ_{Y} . Now recall that φ ∈ H if and only if there exist ϕ ∈ V^{+} and ψ ∈ L^{2}(Z, F) such that φ|_{Z} = ϕ + ψ. By (7.6) it follows that ψ = 0. This proves that ρ_{Y} (H) = V^{+}. □

Let A be the matrix that occurs in Theorem 1.1.

Corollary 7.3. We have

det A = 1.

Proof. Recall the definition of A. Given ϕ ∈ V^{+}, let \frac{1}{2}E(ϕ, 0) be the extended solution of Δ_{X} with limiting value ϕ. Let ϕ_{1}, …, ϕ_{p} be an orthonormal basis of V^{+}. Let ψ_{j} = \frac{1}{2}ρ_{Y} (E(ϕ, 0)). Since by Lemma (7.1) ker \overline{Δ} = \{0\}, it follows that the entries of A are a_{ij} = ⟨ψ_{i}, ψ_{j}⟩_{Y} . By Lemma (7.2), we have \frac{1}{2}ρ_{Y} (E(ϕ, 0)) = ϕ for ϕ ∈ V^{+}. Hence a_{ij} = δ_{ij}. □

Now consider a compact Riemannian manifold (M, g) and a Hermitian vector bundle E → M as in the previous section. Let ∇ be a metric connection on E such that on the tubular neighborhood N = [−1, 1] × Y of Y in M

(7.7)  ∇ = d_{u} ⊗ Id + Id ⊗∇^{Y} ,

where ∇^{Y} is a metric connection on E_{0} = E|Y. Let

Δ_{M} = ∇^{∗}∇.

Let E_{r} → M_{r} and E_{i,∞} → M_{i,∞} be the canonical extensions of the vector bundle E → M to vector bundles over M_{r} and M_{i,∞}, respectively. By (7.7), ∇ has a canonical extension to a connection ∇^{r} on E_{r} and ∇^{i,∞} on E_{i,∞,} i = 1, 2, respectively. Then

Δ_{M_{r}} = (∇^{r})^{∗}∇^{r}, Δ_{i,∞} = (∇^{i,∞})^{∗}∇^{i,∞}, i = 1, 2.

Recall that the Dirichlet-to-Neumann operator R_{r} is a selfadjoint operator in C^{∞}(Y, E|Y)⊕ C^{∞}(Y, E|Y).

Next we determine ker R_{r}. Let V_{i}^{+} ⊂ ker Δ_{Y} be the subspace of limiting values of extended L^{2}-sections of Δ_{i,∞}, i = 1, 2.

Lemma 7.4. We have

ker R_{r} = {(ϕ, ϕ) | ϕ ∈ V_{1}^{+} ∩ V_{2}^{+}}.

Proof. By Lemma 6.2 we have

ker R_{r} = ρ_{r}(ker Δ_{M_{r}}).

Let φ ∈ ker Δ_{M_{r}}. Then

0 = ⟨∇^{∗}∇φ, φ⟩ =∥ ∇φ ∥^{2} .

Thus

(7.8)  ∇φ = 0 for all φ ∈ ker Δ_{M_{r}}.

Next observe that the restriction of φ to N_{r} satisfies

(−\frac{∂^{2}}{∂u^{2}} + Δ_{Y} ) φ(u, y) = 0, u ∈ [−r, r], y ∈ Y,

and hence the expansion of φ|_{N_{r}} in the orthonormal basis {ϕ_{i}}_{i∈ℕ} is of the form

φ(u, y) = ∑_{i=1}^{h_{Y}}(a_{i} + b_{i}u)ϕ_{i}(y) + ∑_{i=h_{Y} +1}^{∞}(c_{i}e^{−\sqrt{μ_{i}}u} + d_{i}e^{\sqrt{μ_{i}}u})ϕ_{i}(y).

By (7.8) it follows that \frac{∂}{∂u}φ(u, y) = 0, u ∈ [−r, r]. Hence we get

(7.9)  φ(u, y) = ∑_{i=1}^{h_{Y}} a_{i}ϕ_{i}(y), (u, y) ∈ N_{r}.

Actually, by our assumptions this holds on a slightly larger collar neighborhood of Y. Denote the right hand side by ϕ. Then ϕ ∈ ker Δ_{Y} and it follows that

ρ_{r}(φ) = (φ(−r, ·), φ(r, ·)) = (ϕ, ϕ).

Furthermore, let φ_{i} = φ|M_{i}, i = 1, 2. Then φ_{i} satisfies

Δ_{M_{i}}φ_{i} = 0 and φ_{i}|_{(−ε,0]×Y} = ϕ.

Thus φ_{i}, i = 1, 2, has a unique extension φ̂_{i} to an extended L^{2}-solution of Δ_{i,∞} with limiting value ϕ. This implies ϕ ∈ V_{1}^{+} ∩ V_{2}^{+}. On the other hand, suppose that ϕ ∈ V_{1}^{+} ∩ V_{2}^{+}. Let φ̂_{i} ∈ C^{∞}(M_{i,∞}, E_{i,∞}), i = 1, 2, be an extended L^{2}-solution with limiting value ϕ. By (7.6) we have

φ̂_{i}|_{(−ε,0]×Y} = ϕ.

Thus we can patch together φ̂_{1}|M_{1} and φ̂_{2}|M_{2} to a section φ ∈ ker Δ_{M_{r}} with ρ_{r}(φ) = (ϕ, ϕ). □

Lemma 7.5. For all r > 0 there exists an isomorphism

j_{r} : ker Δ_{M_{r}} → ker Δ_{M}.

Proof. Let φ ∈ ker Δ_{M_{r}}. By (7.9), there exists ϕ ∈ ker Δ_{Y} such that

(7.10)  φ(u, y) = ϕ(y), (u, y) ∈ N_{r}.

Note that by our assumption, ∇ is the product connection on a slightly larger tubular neighborhood N_{r+ϵ} of Y and (7.10) continues to hold on N_{r+ϵ}. Set

ψ_{i} = φ|_{M_{i}}, i = 1, 2.

By (7.10), it follows that

(7.11)  ψ_{1}|_{∂M_{1}} = ψ_{2}|_{∂M_{2}}.

Define ψ ∈ C^{∞}(M − Y, E) ∩ C^{0}(M, E) by

ψ(x) = \begin{cases}ψ_{1}(x), & x ∈ M_{1}, \\ ψ_{2}(x), & x ∈ M_{2}. \end{cases}

By the above observation, there exists a tubular neighborhood N_{ϵ} = (−ϵ, ϵ) × Y of Y such that ψ|_{N_{ϵ}} = ϕ. Hence ψ ∈ C^{∞}(M, E) and Δ_{M}ψ = 0. By construction, the map

j_{r} : φ ∈ ker Δ_{M_{r}} ⟼ ψ ∈ ker Δ_{M}

is injective and the inverse map can be defined in the same way. This proves that j_{r} is surjective. □

Corollary 7.6. The dimension of ker_{M_{r}} is independent of r.

Put

q := dim ker Δ_{M_{r}}.

Let the matrix B_{r} be defined as in the previous section. Our next purpose is to study the behaviour of det B_{r} as r → ∞. To this end we need some auxiliary result. Let

P_{i} : C^{∞}(Y, E|Y) → C^{∞}(M_{i}, E), i = 1, 2

be the Poisson operator. Recall that for ϕ ∈ C^{∞}(Y, E|Y), P_{i}(ϕ) is the unique solution of the Dirichlet problem

Δ_{M_{i}}ψ = 0, ψ|_{∂M_{i}} = ϕ.

Lemma 7.7. There exists C > 0 such that

∥ P_{i}(ϕ) ∥≤ C ∥ ϕ ∥, ϕ ∈ ker Δ_{Y} , i = 1, 2.

Proof. There exists a collar neighborhood (−ϵ, 0] × Y of Y in M_{i} such that

(7.12)  Δ_{M_{i}} = −\frac{∂^{2}}{∂u^{2}} + Δ_{Y} on (−ϵ, 0] × Y.

Let f ∈ C^{∞}(ℝ) be such that f(u) = 1 for u ≥ −ϵ/4 and f(u) = 0 for u ≤ −ϵ/2. Given ϕ ∈ C^{∞}(Y, E|Y), set

ϕ̃(u, y) = f(u)ϕ(y), u ∈ (−ϵ, 0], y ∈ Y,

and extend ϕ̃ by zero to a smooth section of E → M_{i}. Then

(7.13)  P_{i}(ϕ) = ϕ̃ − (Δ_{M_{i},D})^{−1}(Δ_{M_{i}} ϕ̃).

Let ϕ ∈ ker Δ_{Y} . By (7.12) we get

Δ_{M_{i}} ϕ̃ = −g′′ϕ.

Let λ_{1} > 0 be the smallest eigenvalue of Δ_{M_{i},D}. Then by (7.13) we get

∥ P_{i}(ϕ) ∥≤ C_{1} ∥ ϕ ∥ +\frac{1}{λ_{1}} C_{2} ∥ ϕ ∥≤ C ∥ ϕ ∥,

where C > 0 is independent of ϕ ∈ ker Δ_{Y} . □

Lemma 7.8. Let q = dim ker R_{r}. Then

r^{q} det B_{r} = 1 + O(r^{−1})

as r → ∞.

Proof. Let ψ_{r,1}, …, ψ_{r,q} ∈ ker Δ_{M_{r}} be an orthonormal basis of ker Δ_{M_{r}}. Then B_{r} is defined as

B_{r} = (⟨ρ_{r}(ψ_{r,i}), ρ_{r}(ψ_{r,k})⟩)_{i,k=1}^{q} .

By (7.9), for each r > 0 and k, k = 1, …, q, there exists ϕ_{r,k} ∈ ker Δ_{Y} such that

(7.14)  ψ_{r,k}(u, y) = ϕ_{r,k}(y), u ∈ [−r, r], y ∈ Y.

Let M_{0} = M_{1} ⊔ M_{2}. Then

(7.15)  δ_{ik} = ⟨ψ_{r,i}, ψ_{r,k}⟩_{M_{r}} = ⟨ψ_{r,i}|_{M_{0}}, ψ_{r,k}|_{M_{0}}⟩_{M_{0}} + 2r⟨ϕ_{r,i}, ϕ_{r,k}⟩.

By (7.14) we have

ρ_{r}(ψ_{r,k}) = (ϕ_{k,r}, ϕ_{k,r}).

Hence by (7.15) we get

(7.16)  ⟨ρ_{r}(ψ_{r,i}), ρ_{r}(ψ_{r,k})⟩ = 2⟨ϕ_{r,i}, ϕ_{r,k}⟩ = \frac{1}{r} (1 − ⟨ψ_{r,i}|_{M_{0}}, ψ_{r,k}|_{M_{0}}⟩) .

Furthermore, by (7.15)

(7.17)  ∥ ϕ_{r,k} ∥^{2}≤ \frac{1}{2r}.

Now observe that by (7.14) we have

ψ_{k,r}|_{∂M_{i}} = ϕ_{k,r}, k = 1, …, q.

Moreover Δ_{M_{i}}ψ_{k,r} = 0. Thus

ψ_{r,k}|_{M_{0}} = ψ_{r,k}|_{M_{1}} + ψ_{r,k}|_{M_{2}} = P_{1}(ϕ_{k,r}) + P_{2}(ϕ_{k,r}).

Together with Lemma (7.7) and (7.17) it follows that there exists C > 0 such that

∥ ψ_{k,r}|_{M_{0}} ∥≤ C ∥ ϕ_{k,r} ∥≤ \frac{c}{2\sqrt{r}}

for all r > 0 and k = 1, …, q. Hence by (7.15) we get

⟨ρ_{r}(ψ_{r,i}), ρ_{r}(ψ_{r,k})⟩ = \frac{1}{r} (δ_{ik} + O (\frac{1}{r}))

as r → ∞. This implies r^{q} det B_{r} = 1 + O(r^{−1}). □

Our next purpose is to study the behaviour of det R_{r} as r → ∞. Recall that by Lemma 7.2

(7.18)  ker R_{∞} = V_{1}^{+} ⊕ V_{2}^{+}.

Furthermore, by Lemma 7.4 we have

(7.19)  ker R_{r} = {(ϕ, ϕ) | ϕ ∈ V_{1}^{+} ∩ V_{2}^{+}}.

To study R_{r} on the orthogonal complement of ker R_{r} we need to introduce some auxiliary subspaces of L^{2}(Y, E|Y) ⊕ L^{2}(Y, E|Y). First put

(7.20)  L = (V_{1}^{+} ∩ V_{2}^{+}) ⊕ (V_{1}^{+} ∩ V_{2}^{+}).

By (7.18) we have L ⊂ ker R_{∞}. Furthermore, it follows from (6.11) that on ker Δ_{Y} ⊕ker Δ_{Y} the operator K_{r} is given by

(7.21)  K_{r} = \frac{1}{2r} (\begin{matrix}Id & − Id \\ − Id & Id \end{matrix}) .

This implies that L is invariant under K_{r}. Therefore, L is an invariant subspace for R_{r} = R_{∞} + K_{r}. Let

W = {(ϕ, −ϕ) | ϕ ∈ V_{1}^{+} ∩ V_{2}^{+}}.

Then by (7.19) we get an orthogonal decomposition

L = ker R_{r} ⊕ W

and it follows from (7.20) that W is an invariant subspace of K_{r} and hence of R_{r}. Moreover

R_{r}|_{W} = \frac{1}{r} Id .

Set

h_{12} := dim(V_{1}^{+} ∩ V_{2}^{+}).

Note that h_{12} = q = dim ker R_{r}. Let L^{⊥} be the orthogonal complement of L in L^{2}(Y, E|Y)⊕ L^{2}(Y, E|Y). Then it follows that

(7.22)  det R_{r} = r^{−h_{12}} det (R_{r}|L^{⊥}) .

So we can continue with the investigation of R_{r}|L^{⊥}. Let L_{1} ⊂ V_{1}^{+} ⊕ V_{2}^{+} be the orthogonal complement of L in V_{1}^{+} ⊕ V_{2}^{+} and (ker R_{∞})^{⊥} the orthogonal complement of ker R_{∞} in L^{2}(Y, E|Y) ⊕ L^{2}(Y, E|Y). Then

(7.23)  L^{⊥} = L_{1} ⊕ (ker R_{∞})^{⊥}

with L_{1} ⊂ ker R_{∞}. This decomposition is invariant under R_{∞}, however, it is not invariant under K_{r} and hence, it is not invariant under R_{r}. In fact, with respect to (7.23) we may write

R_{r}|_{L^{⊥}} = (\begin{matrix}A(r) & B(r) \\ C(r) & D(r) \end{matrix}) ,

where the operators A(r), ..., D(r) are defined as follows. Let Π_{1} denote the orthogonal projection of L^{⊥} onto L_{1}. Then

A(r) = Π_{1}K_{r}Π_{1}, B(r) = Π_{1}K_{r}(Id −Π_{1}), C(r) = (Id −Π_{1})K_{r}Π_{1},

and

(7.24)  D(r) = R_{∞}|_{(ker R_{∞})^{⊥}} + (Id −Π_{1})K_{r}(Id −Π_{1}).

Recall that K_{r} is a trace class operator whose trace norm ∥ K_{r} ∥_{1} satisfies

∥ K_{r} ∥_{1}= O(r^{−1})

as r → ∞. Thus

(7.25)  K_{r,1} := (Id −Π_{1})K_{r}(Id −Π_{1})

is also a trace class operator with trace norm satisfying

(7.26)  ∥ K_{r,1} ∥_{1}= O(r^{−1}), r → ∞.

Furthermore, B(r) and C(r) are finite rank operators with

(7.27)  ∥ B(r) ∥_{1}, ∥ C(r) ∥_{1}= O(r^{−1}).

Finally, A(r) is a linear operator in the finite dimensional vector space L_{1} whose norm is also O(r^{−1}). This operator can be described more explicitely as follows. First note that L_{1} ⊂ ker Δ_{Y} ⊕ ker Δ_{Y} and hence we can replace Π_{1} by the orthogonal projection Π_{2} of ker Δ_{Y} ⊕ ker Δ_{Y} onto L_{1}. Let (V_{1}^{+} ∩ V_{2}^{+})_{i}^{⊥} ⊂ V_{i}^{+} denote the orthogonal complement of V_{1}^{+} ∩ V_{2}^{+} in V_{i}^{+}, i = 1, 2, and let

P_{i} : ker Δ_{Y} → (V_{1}^{+} ∩ V_{2}^{+})_{i}^{⊥}

be the orthogonal projection of ker Δ_{Y} onto (V_{1}^{+} ∩V_{2}^{+})_{i}^{⊥}. Then Π_{2} = (P_{1}, P_{2}) and by (7.21) it follows that

A(r) = \frac{1}{2r} (\begin{matrix}P_{1} & 0 \\ 0 & P_{2} \end{matrix}) ∘ (\begin{matrix}Id & − Id \\ − Id & Id \end{matrix}) ∘ (\begin{matrix}P_{1} & 0 \\ 0 & P_{2} \end{matrix}) = \frac{1}{2r} (\begin{matrix}P_{1} & −P_{1}P_{2} \\ −P_{2}P_{1} & P_{2} \end{matrix}) .

Regarded as operator in (V_{1}^{+} ∩ V_{2}^{+})_{1}^{⊥} ⊕ (V_{1}^{+} ∩ V_{2}^{+})_{2}^{⊥}, we get

(7.28)  A(r) = \frac{1}{2r} (\begin{matrix}Id & −P_{1} \\ −P_{2} & Id \end{matrix}) .

Suppose that (ϕ, ψ) ∈ L_{1} is in the kernel of A(r). Then it follows that

ϕ = P_{1}ψ, ψ = P_{2}ϕ.

Since ϕ ∈ V_{1}^{+} and ψ ∈ V_{2}^{+}, it follows that ϕ, ψ ∈ V_{1}^{+} ∩ V_{2}^{+} and therefore ϕ = ψ = 0. Thus A(r) is invertible and its norm satisfies

(7.29)  ∥ A(r) ∥= cr^{−1}, r > 0.

for some constant c > 0. Let

S: (V_{1}^{+} ∩ V_{2}^{+})_{1}^{⊥} ⊕ (V_{1}^{+} ∩ V_{2}^{+})_{2}^{⊥} → (V_{1}^{+} ∩ V_{2}^{+})_{1}^{⊥} ⊕ (V_{1}^{+} ∩ V_{2}^{+})_{2}^{⊥}

denote the restriction of the operator

(\begin{matrix}Id & −P_{1} \\ −P_{2} & Id \end{matrix}) : ker Δ_{Y} ⊕ ker Δ_{Y} → ker Δ_{Y} ⊕ ker Δ_{Y}

to the subspace (V_{1}^{+} ∩ V_{2}^{+})_{1}^{⊥} ⊕ (V_{1}^{+} ∩ V_{2}^{+})_{2}^{⊥}. Set

(7.30)  h := dim V_{1}^{+} + dim V_{2}^{+} − 2 dim V_{1}^{+} ∩ V_{2}^{+} and h_{12} := dim V_{1}^{+} ∩ V_{2}^{+}.

Lemma 7.9. We have

lim_{r→∞} r^{h+h_{12}} det R_{r} = 2^{−h} det(S) det R_{1,∞} det R_{2,∞}.

Proof. Let

T_{0}(r) = (\begin{matrix}A(r) & 0 \\ 0 & D(r) \end{matrix}) .

Since A(r) is an invertible operator in a finite-dimensional vector space and D(r) is invertible for r ≥ r_{0}, it follows that T_{0}(r) is invertible for r ≥ r_{0} and

det T_{0}(r) = det A(r) det D(r).

36 JÖRN MÜLLER AND WERNER MÜLLER

Let

T_{1}(r) = (\begin{matrix}0 & B(r) \\ C(r) & 0 \end{matrix}) .

Then T_{1}(r) is a trace class operator with ∥ T_{1}(r) ∥_{1}= O(r^{−1}) as r → ∞, and

R_{r} = T_{0}(r) + T_{1}(r)

Moreover T_{0}(r) + tT_{1}(r) is invertible for 0 ≤ t ≤ 1 and r ≥ r_{0}. Put

T_{2}(r) = T_{1}(r)T_{0}(r)^{−1}.

Then for r ≥ r_{0} we get

log det R_{r} − log det T_{0}(r) = ∫_{0}^{1} \frac{d}{dt} log det(T_{0}(r) + tT_{1}(r)) dt
= ∫_{0}^{1} Tr (T_{1}(r)(T_{0}(r) + tT_{1}(r))^{−1}) dt
= ∫_{0}^{1} Tr (T_{2}(r)(Id +tT_{2}(r))^{−1} dt.

(7.31)

Set

B̃(r) = B(r)D(r)^{−1}, C̃(r) = C(r)A(r)^{−1}.

Using the definition of T_{2}(r), we get

T_{2}(r) = (\begin{matrix}0 & B̃(r) \\ C̃(r) & 0 \end{matrix})

By (7.27) and (7.29) it follows that

(7.32)  ∥ B̃(r) ∥= O(r^{−1}), ∥ C̃(r) ∥= O(1)

as r → ∞. Thus

T_{2}(r)^{2} = (\begin{matrix}B̃(r)C̃(r) & 0 \\ 0 & C̃(r)B̃(r) \end{matrix})

and by (7.32) we have

∥ T_{2}(r)^{2} ∥= O(r^{−1}), r → ∞.

Let r_{1} > 0 be such that

∥ T_{2}(r)^{2} ∥≤ \frac{1}{2}

for r ≥ r_{1}. Then

∑_{k=0}^{∞} T_{2}(r)^{k} = (Id +T_{2}(r)) ∑_{k=0}^{∞} T_{2}(r)^{2k}

is absolutely convergent and hence, Id +tT_{2}(r) is invertible for 0 ≤ t ≤ 1 and r ≥ r_{1} with

(Id +tT_{2}(r))^{−1} = ∑_{k=0}^{∞} t^{k}T_{2}(r)^{k}.

Moreover it follows that

(Id +tT_{2}(r))^{−1} = Id +tT_{2}(r) + O(r^{−1}), r → ∞.

Thus

T_{2}(r)(Id +tT_{2}(r))^{−1} = T_{2}(r) + tT_{2}(r)^{2} + O(r^{−1}) = T_{2}(r) + O(r^{−1}).

Since Tr(T_{2}(r)) = 0, we get

Tr(T_{2}(r)(Id +tT_{2}(r))^{−1}) = O(r^{−1}).

Together with (7.31) this implies

| log det R_{r} − log det T_{0}(r)| ≤ Cr^{−1}.

Hence we get

\frac{det(R_{r}|L^{⊥})}{det T_{0}(r)} = 1 + O(r^{−1}), r → ∞.

As observed above, det T_{0}(r) = det A(r) det D(r). Using the definition of D(r) by (7.24) and that R_{∞}|(ker R_{∞})^{⊥} is invertible, it follows as in Lemma 6.5 that

lim_{r→∞} det D(r) = det R_{1,∞} det R_{2,∞}.

Let h and h_{12} be defined by (7.30). Note that

h = dim(V_{1}^{+} ∩ V_{2}^{+})_{1}^{⊥} + dim(V_{1}^{+} ∩ V_{2}^{+})_{2}^{⊥}.

Then by definition of A(r)

det A(r) = (2r)^{−h} det S.

So combined with (7.22) we get

lim_{r→∞} r^{h+h_{12}} det R_{r} = 2^{−h} det(S) det R_{1,∞} det R_{2,∞}.

□

Next we express det(S) in terms of the scattering matrices S_{1}(0) and S_{2}(0). Let V = ker Δ_{Y} and set

V_{2} = V ⊖ ((V_{1}^{+} ∩ V_{2}^{+}) ⊕ (V_{1}^{−} ∩ V_{2}^{−})).

Lemma 7.10. Let C_{12} = S_{1}(0)S_{2}(0)|V_{2}. We have

det(S) = det ((Id −C_{12})/2) .

Proof. First we consider the following special case: Assume that

(1) V_{1}^{+} ∩ V_{2}^{+} = {0}, V_{1}^{−} ∩ V_{2}^{−} = {0},

(2) dim V = 2p and dim V_{i}^{+} = dim V_{i}^{−} = p, i = 1, 2,

(3) P_{1}^{+} : V_{2}^{+} → V_{1}^{+} is an isomorphism.

Let e_{1}, ..., e_{2p} be an orthonormal basis of ker Δ_{Y} such that e_{1}, ..., e_{p} is an orthonormal basis of V_{1}^{+} and e_{p+1}, ..., e_{2p} is an orthonormal basis of V_{1}^{−}. Let f_{1}, ..., f_{p} ∈ V_{2}^{+} be such that

P_{1}^{+}(f_{i}) = e_{i}, i = 1, ..., p.

Then there exists a symmetric matrix A = (a_{ij}) ∈ GL(p, ℝ) such that

f_{i} = e_{i} + ∑_{j=1}^{p} a_{ij}e_{p+j}, i = 1, ..., p.

Let A^{−1} = (b_{kl}) and put

f_{p+k} = e_{k} + ∑_{l=1}^{p} b_{kl}e_{p+k}, k = 1, ..., p.

Then ⟨f_{i}, f_{p+j}⟩ = 0, i, j = 1, ..., p. Thus f_{p+j} ∈ V_{2}^{−}, j = 1, ..., p. Furthermore P_{1}^{+}(f_{p+j}) = e_{j}. Thus f_{p+1}, ..., f_{2p} is a basis of V_{2}^{−}. By definition the matrix T which transforms the basis (e_{1}, ..., e_{2p}) into (f_{1}, ..., f_{2p}) is given by

T = (\begin{matrix}Id & A \\ Id & −A^{−1} \end{matrix}) .

Since A is symmetric, it follows that (A^{2} + Id) is invertible and one immediately verifies that the inverse of T is given by

T^{−1} = (\begin{matrix}(A^{2} + Id)^{−1} & A^{2}(A^{2} + Id)^{−1} \\ A(A^{2} + Id)^{−1} & −A(A^{2} + Id)^{−1} \end{matrix}) .

Now note that the matrix of S_{1}(0)S_{2}(0) with respect to the basis (e_{1}, ..., e_{2p}) is given by

(7.33)  (\begin{matrix}Id & 0 \\ 0 & − Id \end{matrix}) ∘ T^{−1} ∘ (\begin{matrix}Id & 0 \\ 0 & − Id \end{matrix}) ∘ T

Hence the matrix of Id −S_{1}(0)S_{2}(0) in the basis (e_{1}, ..., e_{2p}) is equal to

(\begin{matrix}2A^{2}(A^{2} + Id)^{−1} & −2A(A^{2} + Id)^{−1} \\ 2A(A^{2} + Id)^{−1} & 2A^{2}(A^{2} + Id)^{−1} \end{matrix}) .

This implies

det(Id −S_{1}(0)S_{2}(0)) = 2^{h_{Y}} det(A^{2}) det(A^{2} + Id)^{−1}.

On the other hand P_{2}^{+} = 1/2(Id +S_{2}(0)). So it follows from (7.33) that in the basis (e_{1}, ..., e_{2p}), P_{2}^{+} is represented by the matrix

(\begin{matrix}(A^{2} + Id)^{−1} & A(A^{2} + Id)^{−1} \\ A(A^{2} + Id)^{−1} & A^{2}(A^{2} + Id)^{−1} \end{matrix}) .

Thus, with respect to the bases (e_{1}, ..., e_{p}) and (f_{1}, ..., f_{p}), the operator P_{2}^{+} : V_{1}^{+} → V_{2}^{+} is represented by the matrix (A^{2} + Id)^{−1}. Hence the matrix of S with respect to the basis (e_{1}, ..., e_{p}, f_{1}, ..., f_{p}) is given by

(\begin{matrix}Id & − Id \\ −(A^{2} + Id)^{−1} & Id \end{matrix}) .

Thus

det(S) = det (\begin{matrix}Id −(A^{2} + Id)^{−1} & 0 \\ −(A^{2} + Id)^{−1} & Id \end{matrix}) = det(A^{2}) det(A^{2} + Id)^{−1}.

Next we reduce the general case to this special one. If we restrict S_{1}(0) and S_{2}(0) to V_{2}, it follows immediately that we can assume condition 1). Now suppose that

dim V_{2}^{+} ≤ dim V_{1}^{+} and P_{1}^{+} : V_{2}^{+} → V_{1}^{+}

is injective. Let W_{1} := P_{1}^{+}(V_{1}^{+}) and let W_{2} ⊂ V_{1}^{+} denote the orthogonal complement of W_{1} in V_{1}^{+}. We claim that W_{2} ⊂ V_{2}^{−}. To prove this claim let w ∈ W_{2} and v ∈ V_{2}^{+} be given. Write v = v_{1} + v_{2}, v_{1} ∈ V_{1}^{+}, v_{2} ∈ V_{1}^{−}. By definition we have ⟨w, v_{1}⟩ = 0. Since w ∈ W_{2} ⊂ V_{1}^{+}, we have ⟨w, v_{2}⟩ = 0. Thus ⟨w, v⟩ = 0, which shows that W_{2} is orthogonal to V_{2}^{+}, and hence W_{2} ⊂ V_{2}^{−}. Now

(7.34)  S_{1}(0)|W_{2} = Id, S_{2}(0)|W_{2} = − Id .

Thus S_{1}(0)S_{2}(0)|W_{2} = − Id. Let

Ṽ = V ⊖ W_{2}.

Then by (7.34), Ṽ is an invariant subspace for S_{1}(0) and S_{2}(0). Let S̃_{i} = S_{i}(0)|Ṽ, i = 1, 2. Then

Id −S_{1}(0)S_{2}(0) = (\begin{matrix}2 Id & 0 \\ 0 & Id −S̃_{1}S̃_{2} \end{matrix}) .

Hence we get

(7.35)  det(Id −S_{1}(0)S_{2}(0)) = 2^{dim W_{2}} det(Id −S̃_{1}S̃_{2}).

Let Ṽ_{i}^{±} ⊂ Ṽ be the ±1-eigenspaces of S̃_{i}, i = 1, 2. Then it follows that

Ṽ_{1}^{+} = P_{1}^{+}(V_{2}^{+}) = W_{1}, Ṽ_{2}^{+} = V_{2}^{+}.

In particular, P̃_{1}^{+} : Ṽ_{2}^{+} → Ṽ_{1}^{+} is an isomorphism. Thus dim Ṽ_{1}^{±} = Ṽ_{2}^{±}. Since Ṽ_{1}^{±} ∩ Ṽ_{2}^{±} = {0}, it follows that dim Ṽ_{i}^{+} = 1/2 dim Ṽ. Thus conditions 2) and 3) are also satisfied and hence, by the first part of the proof we get

(7.36)  det(Id −S̃_{1}S̃_{2}) = 2^{dim Ṽ} det (\begin{matrix}Id & −P̃_{1}^{+} \\ −P̃_{2}^{+} & Id \end{matrix}) .

Finally note that with respect to the decomposition V_{1}^{+} = W_{1} ⊕ W_{2},

P_{1}^{+} : V_{2}^{+} → V_{1}^{+} and P_{2}^{+} : V_{1}^{+} → V_{2}^{+}

are of the form

P_{1}^{+} = (P̃_{1}^{+}, 0), P_{2}^{+} = P̃_{2}^{+} ⊕ 0.

40 JÖRN MÜLLER AND WERNER MÜLLER

Hence

(\begin{matrix}Id & −P_{1}^{+} \\ −P_{2}^{+} & Id \end{matrix}) = (\begin{matrix}Id & 0 & −P̃_{1}^{+} \\ 0 & Id & 0 \\ −P̃_{2}^{+} & 0 & Id \end{matrix}) ,

which shows that

det (\begin{matrix}Id_{V_{1}^{+}} & −P_{1}^{+} \\ −P_{2}^{+} & Id_{V_{2}^{+}} \end{matrix}) = det (\begin{matrix}IdṼ_{1}^{+} & −P̃_{1}^{+} \\ −P̃_{2}^{+} & IdṼ_{2}^{+} \end{matrix}) .

Together with (7.35) and (7.36), the lemma follows. □ Combining Proposition 6.4 with Corollary 7.3 and Lemmas 7.8, 7.9 and 7.10, we obtain

lim_{r→∞} r^{h} \frac{det Δ_{M_{r}}}{det Δ_{N_{r},D}} = ∏_{i=1}^{2} \frac{det(Δ_{i,∞}, Δ_{0})}{det R_{i,∞}} lim_{r→∞} \frac{r^{h+h_{12}} det R_{r}}{r^{h_{12}} det B_{r}}
= 2^{−h} det ((Id −C_{12})/2) ∏_{i=1}^{2} det(Δ_{i,∞}, Δ_{0}).

(7.37)

Using Proposition (5.1), we get

det Δ_{M_{r}} ∼ r^{h_{Y} −h}e^{−rξ′_{Y} (0)}2^{2h_{Y} −h}(det Δ_{Y})^{−1/2}

· det ((Id −C_{12})/2) ∏_{i=1}^{2} det(Δ_{i,∞}, Δ_{0}).

(7.38)

As an example, we consider the case of a closed surface M. Let

M_{L} = M_{1} ∪_{Y} ([0, L] × Y) ∪_{Y} M_{2}, Y = ℝ/ℤ, L > 0.

Then

ζ_{Y} (s) = (2π)^{−2s}2ζ(2s),

where ζ(s) denotes the Riemann zeta function. Recall that

ζ(−1) = −\frac{1}{12}, ζ(0) = −\frac{1}{2}, ζ′(0) = −\frac{1}{2} log 2π.

Since ^{·}Γ(s − 1/2) and ζ(2s) are analytic at s = 0, we get

ξ′_{Y} (0) = \frac{1}{\sqrt{π}} \frac{d}{ds} (\frac{Γ(s − 1/2)}{Γ(s)}ζ_{Y} (s − 1/2)) |_{s=0} = −\frac{\sqrt{π}}{3} \frac{d}{ds} (\frac{Γ(s − 1/2)}{Γ(s)}) |_{s=0} = \frac{2}{3}π.

Similarly

ζ′_{Y} (0) = −4 log(2π)ζ(0) + 4ζ′(0) = 0.

Thus

det Δ_{Y} = e^{−ζ′_{Y} (0)} = 1.

Furthermore note that h_{Y} = h_{12} = 1, h = 0, and det ((Id −C_{12})/2) = 1/2. Inserting this into (7.38), we get

(7.39)  det Δ_{M_{L}} ∼ 2Le^{−πL/3} det(Δ_{1,∞}, Δ_{0}) · det(Δ_{2,∞}, Δ_{0})

as L → ∞. Bismut and Bost proved in [BB] that det Δ_{M_{L}} ∼ cLe^{−πL/3}, L → ∞, with some constant c. Our result expresses the constant c explicitely as c = 2 det(Δ_{1,∞}, Δ_{0}) · det(Δ_{2,∞}, Δ_{0}).

Next consider a compact Riemannian manifold (X_{0}, g) with boundary Y as at the end of the previous section. We assume that the connection ∇^{E} is a product on the collar neighborhood N = (−ϵ, 0] × Y of Y in X_{0}. By (6.18) and Corollary 7.3 we have

(7.40)  log det Δ_{X_{r},D} = log det Δ_{N_{r},D} + log det R_{r} − log det R_{∞} + det(Δ_{∞}, Δ_{0}).

Furthermore by Lemma 7.2 we have ker R_{∞} = V^{+}. By (6.19) it follows that ker R_{∞} is invariant under L_{r} and hence under R_{r}, and

R_{r}|_{ker R_{∞}} = \frac{1}{r} Id .

Let h^{+} = dim V^{+}. Then

det R_{r} = r^{−h^{+}} det (R_{r}|(ker R_{∞})^{⊥})

and by Lemma 4.1 of [Le3] it follows that

lim_{r→∞} r^{h^{+}} det R_{r} = det R_{∞}.

Using (7.40) we obtain

lim_{r→∞} \frac{r^{h^{+}} det Δ_{X_{r},D}}{det Δ_{N_{r},D}} = det(Δ_{∞}, Δ_{0}).

Together with Proposition 5.1 we get

det Δ_{X_{r},D} ∼ r^{−h^{+}+h_{Y}} e^{−rξ′_{Y} (0)/2}2^{h_{Y}} (det Δ_{Y})^{−1/2} det(Δ_{∞}, Δ_{0})

as r → ∞. If we apply this to det Δ_{M_{i,r},D} and use (1.9), we get

lim_{r→∞} \frac{r^{h_{Y} −2h_{12}} det Δ_{M_{r}}}{det Δ_{M_{1,r},D} det Δ_{M_{2,r},D}} = 2^{−h} (det Δ_{Y})^{1/2} det ((Id −C_{12})/2) ,

which proves Theorem 1.7.

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Universität Bonn, Mathematisches Institut, Beringstrasse 1, D – 53115 Bonn, Germany

Universität Bonn, Mathematisches Institut, Beringstrasse 1, D – 53115 Bonn, Germany

E-mail address: joern@uni-bonn.de
E-mail address: mueller@math.uni-bonn.de