1

Summary. The ζ-regularized of the Dirac operator and of its square are computed on spherical space forms. On S2 the of Dirac operators twisted by a complex line bundle is also calculated.

Key words: Dirac operator, determinant, η-invariant, ζ-function, spherical space form

2000 Subject Classification: 58J52, 53C27, 53C25

The Dirac Determinant of Spherical Space Forms

Christian Bar¨ and Sven Schopka

Universitat¨ Hamburg, FB Mathematik, Bundesstr. 55, 20146 Hamburg, [email protected], [email protected]

1 Introduction

In classical field theory the physical fields ϕ have to satisfy certain field equations or, equiv- alently, have to be critical points of some action functional [ϕ]. When passing to quantum field theory this requirement is discarted and one instead looksS at partition functions defined as a functional integral Z = Z exp( [ϕ]) ϕ −S D over the space of all fields. It is now a serious problem that in most cases the space of fields is infinite dimensional and the “measure” ϕ does not exist. If the space of fields D 1 is a Hilbert space and the action is of the form [ϕ] = 2 (L ϕ, ϕ) where L is a positive self-adjoint operator, then one can define the functionalS integral by

1/2 Z = (det L)− .

This is motivated by the fact that if the Hilbert space is of finite dimension N, then

1 1 N N/2 1/2 Z exp( (L x, x))dx dx = (2π) det(L)− . RN −2 ···

1 N One can then regard x = dx dx as a renormalized Lebesgue measure. At first it D √2π ··· √2π may seem that one has simply shifted the problem since in the physical case the operator L will typically have unbounded spectrum and the product of its eigenvalues will diverge. So one has to find a reasonable definition for the determinant of L. We will look at two closely related definitions for the determinant, the one most commonly used is coming from the ζ-function of L while the proper time regularized determinant makes essential use of the asymptotic heat kernel expansion. This will be explained in detail in the next section. In order to distinguish these regularized determinants from the usual ones in finite dimensions we will denote them by DET and by DETp.t.. Even though these concepts are standard in quantum field theory not many of these de- terminants have been computed explicitly. Due to the somewhat involved nature of their definition an explicit computation can be achieved only in cases of high symmetry. In this paper we provide such explicit computation for the Dirac operator and its square on spher- ical space forms. The square of the Dirac operator D2 is a non-negative self-adjoint elliptic acting on spinor fields. In Theorem 4.1 we present a formula for the de- terminant of D2 on the n-dimensional sphere, n 2, with its standard Riemannian metric of constant curvature 1. The determinant is given≥ by a of the Riemann ζ-function and its first derivative evaluated at certain non-positive integers. The proper time regularized determinant is given in Corollary 4.3. Since the Dirac operator D itself has a spectrum unbounded from above and from below one has to find a reasonable definition for its determinant. Since regularized determinants are in general not multiplicative,

DET(AB) = DET(A) DET(B), 6 4 Christian Bar¨ and Sven Schopka it is not sufficient to simply set DET(D) := DET(D2). In (3) we define DET(D) to be of p the form exp(iϕ) DET(D2) where the ζ-invariant of D2 and the η-invariant of D enter p into the phase ϕ. This definition is motivated in Section 2. We give a simple expression for the multiplicative anomaly exp(iϕ) in Theorem 4.2. It turns out to be trivial for odd- dimensional spheres. The case of the 1-dimensional sphere needs special treatment since S1 has two different spin structures. The results for S1 are contained in Theorem 4.4 and turn out to depend on the choice of spin structure. The numerical values strongly suggest that the Dirac determinant on the n-dimensional sphere tends to 1 as the dimension n goes to , ∞ n lim DET(D; S ) = 1. n →∞ We have no rigorous proof for this conjecture. We then pass to spherical space forms, i. e. to quotients of the sphere. We first look at the determinant of D2 and compute the covering anomaly

DET(D2; G Sn) \ 1 2 n G DET(D ; S ) | | in Theorem 5.1. Here only odd dimensions n have to be considered. For the example of real-projective space it turns out that the covering anomaly is trivial,

2 n DET(D ; RP ) = DET(D2; Sn). p In Theorem 5.2 we give the formula for the determinant of D itself on spherical space forms. For the computation of the multiplicative anomaly we need to recall a formula for the η-invariant of spherical space forms which we do in Theorem 3.2. In the final section we restrict our attention to S2 but we twist the Dirac operator with a complex line bundle. The Chern number of such a bundle can be interpreted as a topological charge. In Theorem 6.1 we compute the eigenvalues of these twisted Dirac operators and in Theorem 6.2 we compute their determinants. As a special case this includes the determinant of the Laplace-Beltrami operator acting on functions. It is always understood that the spherical space forms are equipped with their standard metrics of constant curvature 1. On S2, S4, and S6 the standard metric is known to have an interesting criticality property for the determinant of D2 [4, 5, 13, 15]. Acknowledgement. The first author has been partially supported by the Research and Training Networks “EDGE” and “Geometric Analysis” funded by the European Commis- sion.

2 Determinant of the Dirac Operator

Throughout this paper let L be a non-negative self-adjoint elliptic differential operator of second order acting on sections of a Riemannian or Hermitian vector bundle over a com- pact n-dimensional Riemannian manifold M. See [10] and [14] for the basics on spin and spectral geometry. The most common scheme for making sense of the deter- minant of L is the definition via its ζ-function which reads

DET(L; M) := exp ( ζ0 (0)) (1) − L with 1 ζ (s) := . L X λs λ Spec(L) 0 ∈ −{ } The Dirac Determinant of Spherical Space Forms 5

This way of defining the determinant is motivated by the observation that for positive num- d s bers λk we have ln λk = λ− s=0 and exp ( ln λk) = λk. The series defining − ds k | Pk Qk ζL(s) converges for (s) > n/2. It can be shown that ζL extends meromorphically to C and that it has no pole< at s = 0. Note that this definition of the determinant excludes all information on the possible eigen- value zero. It has to be taken into account separately.

There is the following expression of ζL(s) in terms of its Mellin transform:

1 ∞ s 1 tL ζL(s) = Z t − tr(e− P0)dt (2) Γ (s) 0 − where P0 denotes the projector onto the kernel of L. So far we have a definition of the determinant only in case the operator L is non-negative. This can be applied to the square of the Dirac operator, L = D2, but not to the Dirac operator itself. We adopt the following convention for the determinant of D [20]:

π ζD0 2 (0) DET(D; M) := exp i (ζD2 (0) ηD(0)) exp   (3)  2 −  · − 2 where the η-function is given by

sgnλ η (s) := D X λ s λ Spec(D) 0 ∈ −{ } | | for (s) >> 0. What enters into the phase of the determinant is the η-invariant η(M) := < ηD(0). It measures the spectral asymmetry and is zero in case of a symmetric spectrum.

Why is the definition in (3) reasonable? Denote the positive eigenvalues of D by λk and the negative ones by νk. We can formally write down a ζ-function for D itself as follows: − s s s ζD(s) = λ− + ( 1)− ν− X k X − k k k s s s s s s s s λk− + νk− λk− νk− s λk− + νk− λk− νk− =  + −  + ( 1)−  −  X 2 2 − X 2 − 2 k k s s ζD2 ( 2 ) + ηD(s) s ζD2 ( 2 ) ηD(s) = + ( 1)− − . 2 − 2

s iπs This is a meromorphic function well-defined up to the sign ambiguity in ( 1)− = e∓ . s iπs − Choosing ( 1)− = e− we get −

π ζD0 2 (0) exp ( ζ0 (0)) = exp i (ζD2 (0) ηD(0)) exp   − D  2 −  · − 2 thus yielding (3). We want to compare the above considerations with another way of defining a regularized determinant. For ε > 0 define

∞ 1 tL ln DETε(L; M) := Z t− tr(e− P0)dt. (4) − ε − Since the integrand decays exponentially fast the integral is finite for every positive ε. One says that DETε(L; M) is obtained from the (infinite) determinant of L by cutoff in proper time [17, p. 170]. To motivate this definition note that (2) gives for (s) >> 0 < 6 Christian Bar¨ and Sven Schopka

Γ 0(s) ∞ s 1 tL ζL0 (s) = 2 Z t − tr(e− P0)dt −Γ (s) 0 −

1 ∞ s 1 tL + Z ln(t)t − tr(e− P0)dt. Γ (s) 0 − After replacing the lower integral boundary 0 by ε > 0 we can take the limit s 0. The Laurent expansion → 1 Γ (s) = + O(1) (5) s

1 Γ 0(s) d 1 gives lims 0 = 0 and lims 0 2 = Γ (s)− s=0 = 1. This way we obtain → Γ (s) → − Γ (s)  ds | the right hand side of (4).

The proper time regularized determinant is now defined as the “finite part” of DETε(L; M) for ε 0. To make this more precise we look at the asymptotic expansion of the heat kernel& of L = D2 for t 0 &

tL ∞ k n tr e− Φk n (L) t − 2 . ∼ X − 2 k=0 In particular, we can plug

[n/2] tL k n tr e− = Φk n (L)t − 2 + R(t) (6) X − 2 k=0 into (4) where the remainder term R(t) is of order O(t) if n is even and of order O(√t) if n is odd. After splitting the integral in (4) into one over [ε, 1] and one over [1, ] we get ∞ 1 [n/2]  k n 1 1 1 ln DETε(L; M) = Z Φk n (L)t − 2 − + R(t)t− dim ker(L)t− dt − X − 2 − ε  k=0 

∞ 1 tL Z t− tr(e− P0)dt − 1 − [(n 1)/2] [(n 1)/2] k n − Φ n (L) − Φ n (L)ε 2 k 2 k 2 − = − + − + (Φ0(L) − X k n X k n k=0 − 2 k=0 − 2 1 1 ∞ 1 tL dim ker(L)) ln ε Z R(t)t− dt Z t− tr(e− P0)dt. − − ε − 1 −

Φk n (L) n − 2 k 2 Here we use the convention Φ0(L) = 0 if n is odd. The terms k n ε − , k [(n − 2 ≤ − 1)/2], and (Φ0(L) dim ker(L)) ln ε explode for ε 0 unless they vanish. We now abandon the terms divergent− in the limit ε 0 and we are& led to &

Definition 1. The proper time regularized determinant DETp.t.(L; M) is defined by

[(n 1)/2] 1 − Φ n (L) k 2 1 ln DETp.t.(L; M) := − Z R(t)t− dt X n k − k=0 2 − 0 ∞ 1 tL Z t− tr(e− P0)dt (7) − 1 − where Φk n (L) and R(t) are as in (6). − 2 The following relation (cf. [17, (28.11), p. 171]) shows that the two regularizations do not differ significantly. The Dirac Determinant of Spherical Space Forms 7

Proposition 2.1 The ζ-regularized determinant DET(L; M) and the proper time regular- ized determinant DETp.t.(L; M) are related by

ln DET(L; M) ln DETp.t.(L; M) = Γ 0(1)(Φ0(L) dim ker L) = Γ 0(1)ζL(0). − −

Proof. In order to compare (7) with the ζ-determinant we insert the heat kernel expansion (6) into (2) and do the corresponding integral splitting into one over [0, 1] and one over [1, ]. ∞ [(n 1)/2] 1 − Φ n (L) 1 k 2 Φ0(L) dim ker L s 1 ζ (s) = − + − + R(t)t − dt L Γ (s) X s k + n s Z k=0 − 2 0

∞ s 1 tL + Z t − tr(e− P0)dt!. (8) 1 −

Using (5) we find ζL(0) = Φ0(L) dim ker L. − Equation (8) gives us for the determinant

[(n 1)/2] 1 − Φ n (L) k 2 1 ln DET(L; M) = ζL0 (0) = − + Γ 0(1)ζL(0) Z R(t)t− dt − X n k − k=0 2 − 0 ∞ 1 tL Z t− tr(e− P0)dt (9) − 1 − Comparing (7) and (9) we get Proposition 2.1.

3 Dirac Spectrum of Spherical Space Forms

The Dirac spectrum of the sphere Sn, n 2, with constant curvature 1 has been computed by different methods in [2, 7, 18, 19]. The≥ eigenvalues are n + k , (10) ±  2 

[n/2]  k + n 1  k N0, with multiplicity 2 − . For n 2 the sphere is simply connected, ∈ ≥  k  hence has only one spin structure. It is given by the standard projection Spin(n + 1) Spin(n + 1)/Spin(n) = Sn. → We now look at spherical space forms M = G Sn where G is a finite fixed point free subgroup of SO(n+1). Spin structures correspond\ to homomorphisms ε : G Spin(n+1) such that the diagram → Spin(n + 1) u: ε uu uu uu uu  G u / SO(n + 1) commutes. The corresponding spin structure is given by ε(G) Spin(n+1) G (Spin(n+ 1)/Spin(n)) = M. Thus spinors on M correspond to ε(G)-invariant\ spinors→ on\Sn. 8 Christian Bar¨ and Sven Schopka

Since any eigenspinor on M can be lifted to Sn all eigenvalues of M are also eigenvalues of Sn, hence of the form (10). To know the spectrum of M one must compute the multiplicities n n µk of 2 + k and µ k of ( 2 + k). It is convenient to encode them into two power series, so-called Poincare´ − series−

∞ F (z) := µ zk, + X k k=0

∞ k F (z) := µ kz . − X − k=0 To formulate the result recall that in even dimension 2m the spinor representation is re- ducible and can be decomposed into two half spinor representations

Spin(2m) Aut(Σ± ), → 2m + Σ2m = Σ Σ− . Denote their characters by χ± : Spin(2m) C. 2m ⊕ 2m → Theorem 3.1 ([2]) Let M = G Sn, n = 2m 1, be a spherical space form with spin structure given by ε : G Spin\ (2m). Then the− eigenvalues of the Dirac operator are ( n + k), k 0, with multiplicities→ determined by ± 2 ≥ + 1 χ−(ε(g)) z χ (ε(g)) F+(z) = − · , G X det(12m z g) g G | | ∈ − · + 1 χ (ε(g)) z χ−(ε(g)) F (z) = − · . − G X det(12m z g) g G | | ∈ − ·

Note that only odd-dimensional spherical space forms are of interest because in even di- mensions real projective space is the only quotient and in this case it is not even orientable. This kind of encoding the spectrum of M is in fact well-suited for computation of the η-invariant.

2m 1 Theorem 3.2 ([3]) Let M = G S − be a spherical space form with spin structure given by ε : G Spin(2m). Then\ the η-invariant of M is given by → + 2m 1 2 (χ− χ )(ε(g)) η(G S − ) = − . \ G X det(12m g) g G 12m | | ∈ −{ } −

For the convenience of the reader and since we will need the arguments again we briefly sketch the proof. We define the θ-functions

n t t θ (t) := e− 2 F (e− ) ± · ± (m+ 1 )t t e− 2 χ∓(ε(g)) e− χ±(ε(g)) = − · t . (11) G X det(12m e g) g G − | | ∈ − · Since G acts freely the non-trivial elements g G do not have 1 as an eigenvalue and thus ∈ det(12m g) = 0. Therefore only the summand for g = 12m contributes to the pole of θ − 6 ± at t = 0. Hence for θ := θ+ θ the poles at t = 0 cancel, θ is holomorphic at t = 0 with − − + 2 (χ− χ )(ε(g)) θ(0) = − . (12) G X det(12m g) g G 12m | | ∈ −{ } − The Dirac Determinant of Spherical Space Forms 9

Next we observe that

∞ θ (t) = µ e (n/2+k)t = e λt, + X k − X − k=0 λ>0 where the last sum is taken over all positive eigenvalues. Similarly for θ , −

∞ (n/2+k)t λt θ (t) = µ ke− = e . − X − X k=0 λ<0 Application of the Mellin transformation yields

1 ∞ s 1 η(s) = Z θ(t)t − dt. Γ (s) 0 Therefore

2m 1 1 ∞ s 1 ∞ s 1 η(G S − ) = lim Z θ(t)t − dt = Ress=0 Z θ(t)t − dt . s 0 Γ (s) \ → 0 0

∞ s 1 Since θ decays exponentially fast for t the function s 1 θ(t)t − dt is holomor- phic at s = 0. Thus → ∞ 7→ R

1 2m 1 s 1 η(G S − ) = Ress=0 Z θ(t)t − dt = θ(0) \ 0 which together with (12) proves Theorem 3.2. Compare this to Goette’s computation of the equivariant η-invariant of spheres in [11, Satz 6.10]. See also [9, 10] where the η-invariant of all twisted signature operators on spher- ical space forms is determined and used to compute their K-theory. In [8] the η-invariant of all twisted Dirac operators on 3-dimensional spherical space forms is computed. Using Theorem 3.2 it is easy to discuss real projective space.

Corollary 3.3 ([3]) For n 2 real projective space RPn is spin if and only if n 3 mod 4, in which case it has exactly≥ two spin structures. The η-invariant for the Dirac≡ operator is given by n m η(RP ) = 2− , n = 2m 1, ± − where the sign depends on the spin structure chosen.

Let us now look at lens spaces. For α R define the rotation ∈ cos(2πα) sin(2πα) R(α) :=  −  .  sin(2πα) cos(2πα) 

For natural numbers q, p1, . . . , pm where pj and q are coprime for all j, let G be the cyclic subgroup of SO(2m) generated by

 R(p1/q) 0  ..  .  .      0 R(pm/q) 

2m 1 We denote the resulting lens space by L(q, p1, . . . , pm) := G S − . Now Theorem 3.1 takes the form (compare [1, Satz 5.3]): \ 10 Christian Bar¨ and Sven Schopka a) If q is odd, then L(q, p1, . . . , pm) has exactly one spin structure and the Poincare´ series are given by

F+(z) =

exp(πik(q + 1) εjpj/q) z exp(πik(q + 1) εjpj/q) q 1 ε1 εm − · ε1 εm P··· Pj P··· m Pj 1 − =( 1)m+1 =( 1) − − , q X m k=0 (exp(2πikpj/q) z)(exp( 2πikpj/q) z) jQ=1 − − − F (z) = − exp(πik(q + 1) εjpj/q) z exp(πik(q + 1) εjpj/q) q 1 ε1 εm − · ε1 εm P··· m Pj P··· Pj 1 − =( 1) =( 1)m+1 − − . q X m k=0 (exp(2πikpj/q) z)(exp( 2πikpj/q) z) jQ=1 − − −

b) If q is even and p1 + ... + pm is odd (i. e. if m is odd), then L(q, p1, . . . , pm) has no spin structures. c) If q is even and p1 + ... + pm is even (i. e. if m is even), then L(q, p1, . . . , pm) has two different spin structures. The Poincare´ series for the first spin structure are given by

exp(πik εjpj/q) z exp(πik εjpj/q) q 1 ε1 εm − · ε1 εm P··· Pj P··· m Pj 1 − =( 1)m+1 =( 1) F (z) = − − , + q X m k=0 (exp(2πikpj/q) z)(exp( 2πikpj/q) z) jQ=1 − − −

exp(πik εjpj/q) z exp(πik εjpj/q) q 1 ε1 εm − · ε1 εm P··· m Pj P··· Pj 1 − =( 1) =( 1)m+1 − − F (z) = m , − q X k=0 (exp(2πikpj/q) z)(exp( 2πikpj/q) z) jQ=1 − − − while for the second spin structure they are given by

exp(πik εjpj/q) z exp(πik εjpj/q) q 1 ε1 εm − · ε1 εm P··· Pj P··· m Pj − =( 1)m+1 =( 1) 1 k − F+(z) = ( 1) − , q X − m k=0 (exp(2πikpj/q) z)(exp( 2πikpj/q) z) jQ=1 − − −

exp(πik εjpj/q) z exp(πik εjpj/q) q 1 ε1 εm − · ε1 εm P··· m Pj P··· Pj 1 − =( 1) =( 1)m+1 k − − F (z) = ( 1) m . − q X − k=0 (exp(2πikpj/q) z)(exp( 2πikpj/q) z) jQ=1 − − −

Here the notation indicates that the sum is taken over all choices of εj = 1 for ε εm 1P··· ± =( 1)m+1 − m+1 m which ε1 εm = ( 1) and similarly for ( 1) . For the η-invariant this yields in case a) ··· − −

η = θ(0)

exp(πik(q + 1) εjpj/q) exp(πik(q + 1) εjpj/q) q 1 ε1 εm − ε1 εm P··· Pj P··· m Pj 2 − =( 1)m+1 =( 1) = − − q X m k=1 (exp(2πikpj/q) 1)(exp( 2πikpj/q) 1) jQ=1 − − − The Dirac Determinant of Spherical Space Forms 11

m q 1 εj exp(πik(q + 1)εjpj/q) 2 − ε ,...,εP jQ=1 = ( 1)m+1 1 m − q X m k=1 (2 2 cos(2πkpj/q)) jQ=1 − m q 1 (exp(πik(q + 1)pj/q) exp( πik(q + 1)pj/q)) − m+1 2 jQ=1 − − = ( 1) m − q X m k=1 2 (1 cos(2πkpj/q)) jQ=1 − q 1 m 2 − sin(πk(q + 1)p /q) = ( 1)m+1 im j . − q X Y 1 cos(2πkpj/q) k=1 j=1 − If m is odd we see that η is imaginary, but on the other hand the η-invariant is always real. Hence it must vanish in this case. In fact, the η-invariant of the Dirac operator always vanishes unless the dimension of the manifold is n 3 mod 4. Case c) is treated similarly. Case b) cannot occur for even m. We summarize: ≡

Corollary 3.4 Let m N be even and let q, p1, . . . , pm N be such that q and pj are coprime for all j. ∈ ∈

Then if q is odd L(q, p1, . . . , pm) has exactly one spin structure and the η-invariant is given by q 1 m − 3m/2+1 2 sin(πk(q + 1)pj/q) η(L(q, p1, . . . , pm)) = ( 1) . − q X Y 1 cos(2πkpj/q) k=1 j=1 −

If q is even, then L(q, p1, . . . , pm) has two different spin structures. For the first spin struc- ture the η-invariant is given by

q 1 m − 3m/2+1 2 sin(πkpj/q) η(L(q, p1, . . . , pm)) = ( 1) − q X Y 1 cos(2πkpj/q) k=1 j=1 − while for the second it is

q 1 m − 3m/2+1 2 k sin(πkpj/q) η(L(q, p1, . . . , pm)) = ( 1) ( 1) . − q X − Y 1 cos(2πkpj/q) k=1 j=1 −

The case q = 2 and p1 = ... = pm = 1 recovers Corollary 3.3. We collect a few examples in Tables 1 and 2. For even q we denote the η-invariants for the two spin structures by η1 and η2.

4 Dirac Determinant of the Sphere

In this section we compute the determinant of the square of the Dirac operator and of the Dirac operator itself over the sphere of constant sectional curvature 1. For this task one could use the general machinery developped in [6] for homogeneous operators on compact locally symmetric spaces. We found a direct approach more convenient. Recall the Riemann ζ-function ζR and the Hurwitz ζ-function ζH defined by meromorphic extension of the series ∞ 1 ∞ 1 ζ (s) = and ζ (s, b) = R X js H X (j + b)s j=1 j=0 12 Christian Bar¨ and Sven Schopka

Table 1.

M dim(M) η

4 L(3, 1, 1) 9 L(3, 1, 2) 4 − 9 4 L(5, 1, 1) 3 5 L(5, 1, 2) 0 L(5, 2, 3) 4 − 5 L(3, 1, 1, 1, 1) 4 − 27 4 L(3, 1, 1, 1, 2) 7 27 L(5, 1, 1, 1, 1) 12 − 25 4 L(3, 1, 1, 1, 1, 1, 1) 81 L(3, 1, 1, 1, 1, 1, 2) 11 4 − 81 8 L(5, 1, 1, 1, 1, 1, 1) 25 L(3, 1, 1, 1, 1, 1, 1, 1, 1) 4 − 243 4 L(3, 1, 1, 1, 1, 1, 1, 1, 2) 15 243 L(5, 1, 1, 1, 1, 1, 1, 1, 1) 28 − 125 4 L(3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 729 L(3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2) 19 4 − 729 4 L(5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 25

Table 2.

M dim(M) η1 η2

L(4, 1, 1) 5 3 8 − 8 L(4, 1, 3) 3 5 8 − 8 L(6, 1, 1) 35 19 36 − 36 L(6, 1, 5) 19 35 36 − 36 L(8, 1, 1) 3 21 11 16 − 16 3 3 L(8, 1, 3) 16 16 L(8, 1, 5) 3 3 − 16 − 16 L(8, 3, 5) 11 21 16 − 16 L(10, 1, 1) 33 17 20 − 20 L(4, 1, 1, 1, 1) 9 7 − 32 32 L(4, 1, 1, 1, 3) 7 9 − 32 32 L(6, 1, 1, 1, 1) 7 329 265 − 432 432 L(8, 1, 1, 1, 1) 105 87 − 64 64 L(10, 1, 1, 1, 1) 1221 1029 − 400 400 L(4, 1, 1, 1, 1, 1, 1) 17 15 128 − 128 L(6, 1, 1, 1, 1, 1, 1) 11 3611 3355 5184 − 5184 L(8, 1, 1, 1, 1, 1, 1) 657 623 256 − 256 The Dirac Determinant of Spherical Space Forms 13

x+n 1 (x+n 1) (x+1) with b > 0. Observe that the binomial coefficient Mn(x) := − = − ··· x  (n 1)! is a polynomial in x of degree n 1. Hence we can define − − 1 dk A(k, n) := Mn(x) x= n (13) k! dxk | − 2 or, equivalently, n 1 − n k M (x) = A(k, n) x + . (14) n X  2  k=0 Moreover, we set

m 1 k  j=1− j ln(j), if n = 2m is even, C(k, n) := P (15)  m 1 1 k 1 j=0− (j + 2 ) ln(j + 2 ), if n = 2m + 1 is odd.  P Now we can formulate the result for D2 which in different notation is also contained in [4, Section 8].

Theorem 4.1 The determinant of the square of the Dirac operator on the n-dimensional sphere Sn, n 2, with constant sectional curvature 1 is given by the following formulas: ≥ If n = 2m is even, then

n 1 2 2m m+2 − ln DET(D ; S ) = 2 A(k, n)(ζ0 ( k) + C(k, n)), (16) − X R − k=0 and if n = 2m + 1 is odd, then

n 1 2 2m+1 m+2 − 1 ln 2 ln DET(D ; S ) = 2 A(k, n) ζ0 ( k)( 1) + ζR( k) + C(k, n) − X R − 2k − 2k − k=0 (17) where ζR is the Riemann ζ-function, A(k, n) is as in (13) and C(k, n) is as in (15).

Proof. From (10) we see that on Sn

j+n 1 − [n/2]+1 ∞ j ζ 2 (s) = 2  . D X (j + n )2s j=0 2

By (14) we obtain

n 1 [n/2]+1 − ∞ n k 2s ζ 2 (s) = 2 A(k, n) (j + ) D X X 2 − k=0 j=0 n 1 [n/2]+1 − n = 2 A(k, n) ζH (2s k, ). (18) X − 2 k=0 In case n = 2m is even we use

m 1 − s ζH (s, m) = ζR(s) j− − X j=1 valid for m N to get ∈ 14 Christian Bar¨ and Sven Schopka

n 1 m 1 m+2 −  − k  ζ0 2 (0) = 2 A(k, n) ζ0 ( k) + j ln(j) . D X R − X k=0  j=1  This proves (16). For n = 2m + 1 odd we obtain

n 1 m+2 − 1 ζ0 2 (0) = 2 A(k, n)ζ0 ( k, m + ). D X H − 2 k=0 From m 1 1 1 − 1 s ζH (s, m + ) = ζH (s, ) (j + )− , 2 2 − X 2 j=0 valid for m N and ∈ 1 s ζH (s, ) = (2 1)ζR(s) 2 − (see e.g. [12, 9.535]) we get (17).

Example 1. From this theorem we get

2 2 DET(D ; S ) = exp( 8 ζ0 ( 1)), − R − 2 3 1 DET(D ; S ) = 2− 2 exp (3 ζ0 ( 2))) , R − 2 4 8 8 DET(D ; S ) = exp  ζ0 ( 1) ζ0 ( 3) , 3 R − − 3 R −

2 5 3 5 5 DET(D ; S ) = 2 16 exp  ζ0 ( 2) + ζ0 ( 4) , −4 R − 8 R −

2 6 16 4 4 DET(D ; S ) = exp  ζ0 ( 1) + ζ0 ( 3) ζ0 ( 5) , −15 R − 3 R − − 15 R −

2 7 5 259 35 7 DET(D ; S ) = 2− 64 exp  ζ0 ( 2) ζ0 ( 4) + ζ0 ( 6) , 480 R − − 96 R − 160 R −

2 8 16 28 8 4 DET(D ; S ) = exp  ζ0 ( 1) ζ0 ( 3) + ζ0 ( 5) ζ0 ( 7) , 35 R − − 45 R − 45 R − − 315 R −

2 9 35 3229 47 21 DET(D ; S ) = 2 1024 exp  ζ0 ( 2) + ζ0 ( 4) ζ0 ( 6) −13440 R − 256 R − − 640 R − 17 + ζ0 ( 8) , 10752 R −

2 10 64 164 13 2 DET(D ; S ) = exp  ζ0 ( 1) + ζ0 ( 3) ζ0 ( 5) + ζ0 ( 7) −315 R − 567 R − − 135 R − 189 R − 1 ζ0 ( 9) . −2835 R −

For n = 2, 4, 6 this reproduces the values computed by Branson in [4, Thm. 8.1]. For the Dirac operator itself we obtain

Theorem 4.2 The determinant of the Dirac operator on the n-dimensional sphere Sn, n 2, with constant sectional curvature 1 is given by the following: ≥ If n = 2m is even, then

2m 1 2m m − Bk+1(m) DET(D; S ) = exp iπ2 A(k, 2m) ! DET(D2; S2m), − X k + 1 p k=0 The Dirac Determinant of Spherical Space Forms 15 and if n is odd, then n DET(D; S ) = DET(D2; Sn) p 2 n where DET(D ; S ) is as in Theorem 4.1, A(k, n) as in (13) and Bk(x) are the Bernoulli polynomials.

Proof. By (3) we have

n π 2 n DET(D; S ) = exp i (ζ 2 (0) ηD(0)) DET(D ; S ).  2 D −  p The η-invariant vanishes because the Dirac spectrum of Sn is symmetric. It remains to compute ζD2 (0). Plugging Bk+1(b) ζH ( k, b) = − − k + 1 (see [12, 9.531 and 9.623(3)]) into (18) we get

n 1 n [n/2]+1 − Bk+1( 2 ) ζ 2 (0) = 2 A(k, n) . (19) D − X k + 1 k=0 2 This proves the even-dimensional case. If n is odd we observe that Φ0(D ) = 0 and ker D2 = 0 and hence { } ζD2 (0) = 0. (20) This concludes the proof.

Using Proposition 2.1 we can combine Theorem 4.1, (19) and (20) to obtain the proper time regularized determinant of D2 on Sn.

Corollary 4.3 The proper time regularized determinant of the square of the Dirac operator on the n-dimensional sphere Sn, n 2, with constant sectional curvature 1 is given by the following formulas: ≥ If n = 2m is even, then

2m 1 2 2m m+1 − Bk+1(m) ln DETp.t.(D ; S ) = 2 A(k, 2m) 2 ζ0 ( k) + 2 C(k, 2m) Γ 0(1)  − X R − − k + 1 k=0 and if n is odd, then 2 n 2 n DETp.t.(D ; S ) = DET(D ; S ) where ζR is the Riemann ζ-function, A(k, n) is as in (13), C(k, n) is as in (15) and Bk(x) are the Bernoulli polynomials.

The 1-dimensional case needs to be treated separately because S1 has two spin structures. Let us call the spin structure which extends to the unique spin structure of the 2-dimensional disk the bounding spin structure, the other one the non-bounding spin structure. For the first spin structure formula (10) still holds while the non-bounding spin structure has Z as its Dirac spectrum. We get

Theorem 4.4 The determinants of the Dirac operator and its square on the 1-dimensional sphere S1 with length 2π take the values 1 2 1 DET(D; S ) = 2 and DET(D ; S ) = 4 for the bounding spin structure and 1 2 1 2 DET(D; S ) = 2πi and DET(D ; S ) = 4π − for the non-bounding spin structure. 16 Christian Bar¨ and Sven Schopka

Recall that by our convention for the determinant the eigenvalue 0 is ignored. Otherwise the determinant on S1 would vanish for the non-bounding spin structure.

Proof. From (10) we get for the bounding spin structure

∞ 1 1 2s ζ 2 (s) = 2 = 2 ζH (2s, ) = 2(2 1)ζR(2s). D X (j + 1 )2s 2 − j=0 2

1 Using ζR(0) = we obtain − 2

ζ0 2 (0) = 4 ln 2 ζR(0) = 2 ln 2 D · − and hence 2 1 DET(D ; S ) = exp ( ζ0 2 (0)) = 4. − D The computation for the non-bounding spin structure is even easier. For ζD2 we get

∞ 1 ζ 2 (s) = 2 = 2 ζ (2s). D X j2s R j=1

1 Now ζ0 (0) = ln(2π) yields R − 2

ζ0 2 (0) = 4ζ0 (0) = 2 ln(2π) D R − and therefore 2 1 2 DET(D ; S ) = 4π .

This proves the formulas for DET(D2; S1). The η-invariant vanishes in both cases due to spectral symmetry. The result for DET(D; S1) follows from (3) and

 0 for the bounding spin structure ζD2 (0) =  1 for the non-bounding spin structure. −  In the following tables we give some results obtained by numerical approximation. The phase ϕ is defined by n n iϕ DET(D; S ) = DET(D; S ) e . | | Here are the values of the determinant on the first few even dimensional spheres. 2 n We do not need the DETp.t.(D ; S )–column in the odd dimensional case because of Corollary 4.3. The phase vanishes according to Theorem 4.2. A view at the numerical values leads us to

Conjecture 1. The determinant of the Dirac operator on the n-dimensional sphere Sn tends to 1 for n , → ∞ n lim DET(D; S ) = 1. n →∞ For example, we have

200 DET(D; S ) 0.9999999999999999999999999999999023251476, | | ≈ and for the phase of DET(D; S200) we have

31 ϕ 0.8894434790878104795255101869645846589176 10− . ≈ · We have no explanation for this phenomenon. The Dirac Determinant of Spherical Space Forms 17

Table 3.

n 2 n n DET(D; S ) ϕ DETp.t.(D ; S ) | | 2 1.938054383626833 -0.523598775598299 3.098641928647827 4 0.796336885184459 0.191986217719376 0.680506885334576 6 1.096240639130369 -0.0793709255073612 1.167199468652179 8 0.961240935820022 0.0345879931923003 0.935802870851431 10 1.017761616865532 -0.0155276797389130 1.029945138033507 12 0.992018956602283 0.00710558492613086 0.986674523357944 14 1.003708484448039 -0.00329528874347978 1.006211552565484 16 0.998272959376817 0.00154335558060539 0.997114236587699 18 1.000814152574965 -0.000728325704396029 1.001360932301507 20 0.999614434504206 0.000345772030034635 0.999355987552311

Table 4.

n DET(D)

3 0.803354268824629 5 1.090359845142337 7 0.963796369884191 9 1.016473922384390 11 0.992614518464762 13 1.003422630166412 15 0.998408322304586 17 1.000749343263366 19 0.999645452552308 21 1.000168795852563

5 Dirac Determinant of Spherical Space Forms

2m 1 Next we determine the Dirac determinant of spherical space forms M = G S − where G SO (2m) is a fixed point free subgroup. The spin structure is given by\ a homomor- phism⊂ ε : G Spin (2m) lifting the inclusion of G into SO (2m) as explained in Sec- → tion 3. In order to distinguish geometric data for different manifolds we write ζD2 (s; M) for the ζ-function of the square of the Dirac operator on M.

2m 1 Theorem 5.1 Let M = G S − be a spherical space form with spin structure given by ε : G Spin (2m). Then the\ square of the Dirac operator on M has the determinant → 1 2 2m 1 2 2m 1 G DET (D ; G S − ) = DET D ; S − | | \  · (m 1 )t t 2χ(ε(g)) e 2 1 e ∞ − − − ! exp Z t − dt · Y − G · 0 det(12m e− g) t g G 12m ∈ −{ } | | −

2 2m 1 where DET (D ; S − ) is given by Theorem 4.1 and χ is the character of the spinor module for Spin (2m) . 18 Christian Bar¨ and Sven Schopka

2m 1 Proof. To compute the ζ-function ζ 2 (s ; G S − ) we recall the θ-functions D \ (m 1 )t t 1 e− − 2 χ∓(ε(g)) e− χ±(ε(g)) (m 2 )t t θ (t) = e− − F (e− ) = − · t , ± · ± G X det(12m e g) g G − | | ∈ − · see (11). Hence by the following Mellin transformation

2m 1 2s ∞ 2s 1 u Γ (2s) ζD2 (s; G S − ) = λ − Z u − e− du \ X | | λ Spec(D) 0 ∈ = ∞ t2s 1 e λ tdt X Z − −| | λ Spec(D) 0 ∈ ∞ ∞ 2s 1 (m 1 +k)t = (µk + µ k) Z t − e− − 2 dt X − k=0 0

∞ 2s 1 = Z t − (θ+ (t) + θ (t)) dt 0 − 2s 1 (m 1 )t t 1 ∞ t − e− − 2 (1 e− ) = χ(ε(g)) Z · · t − dt G X det(12m e g) g G 0 − | | ∈ − · we obtain (m 1 )t t 2m 1 1 ∞ 2s e− − 2 1 e− 2 ζD (s; G S − ) = χ(ε(g)) Z t t − dt. \ G Γ (2s) X det(12m e g) t g G 0 − | | · ∈ −

The special case G = 12m yields { } (m 1 )t t 2m 1 1 ∞ 2s e− − 2 1 e− 2 ζD (s; S − ) = χ(1) Z t t − dt Γ (2s) · · det(12m e 12m) t 0 − − · and therefore 2m 1 2m 1 ζD2 (s; S − ) ζ 2 s; G S − = D \  G | | (m 1 )t t 1 ∞ 2s e− − 2 1 e− + χ(ε(g)) Z t t − dt . G Γ (2s) X 0 det(12m e− g) t g G 12m | | · ∈ −{ } −

Since g G 12m does not have any positive real eigenvalues the integral defines a function in∈ s holomorphic− { } at 0. Thus

2m 1 2m 1 ζD2 (0; S − ) ζ 2 (0; G S − ) = = 0 (21) D \ G | | by (20) and

2m 1 2m 1 ζD0 2 (0; S − ) ζ0 2 (0; G S − ) = D \ G | | (m 1 )t t 2 ∞ e− − 2 1 e− + χ (ε(g)) Z t − dt G X 0 det(12m e− g) t g G 12m | | ∈ −{ } −

2 2m 1 by (5). This implies the formula for DET(D ; G S − ) . \ 2 2m 1 2 2m 1 Equality of DET (D ; G S − ) and DETp.t.(D ; G S − ) follows from Proposi- tion 2.1 and (21). \ \ The Dirac Determinant of Spherical Space Forms 19

2m 1 Example 2. Let us look at real projective space M = RP − . In this case G = 12m, 12m . We assume m to be odd so that M is spin and has two spin structures. { − } They are given by ε( 12m) = ω where ω = e1 e2m is the volume element consid- ered as an element in− Spin(2m) ± Cl(R2m). The volume··· element acts by multiplication by ⊂ + 1 on the two half spinor spaces so that χ (ω) = χ−(ω). Hence for both spin structures ± − we have χ(ε( 12m)) = 0. It follows that − 2 2m 1 2m 1 DET(D ; RP ) = DET(D2; S2m 1) = DET(D; S ). − p − −

Example 3. Let us now look at the lens space M = L(q, p1, . . . , pm). We assume that q is odd so that M has exactly one spin structure. For

 R(k p1/q) 0  .. g =  .       0 R(k pm/q)  we compute

m χ(ε(g)) = exp πik(q + 1) ε p /q X X j j ε1, ,εm j=1 ···   m = exp(πik(q + 1)ε p /q) X Y j j ε1, ,εm j=1 ··· m π k (q + 1) pj = 2m cos   Y q j=1 and m t t det 12m e− g = det 12m e− R(k pj/q) −  Y −  j=1 m t 2t = 1 2 e− cos(2 π k pj/q) + e− . Y −  j=1 Now Theorem 5.1 yields 2 2 2m 1 1/q DET(D ; L(q, p1, . . . , pm)) = DET(D ; S − ) exp(A) · where the covering anomaly exp(A) is given by

m+1 q 1 t (m 1/2)t m 2 − ∞ (1 e− ) e− − dt π k (q + 1) pj A = Z − t 2t cos   . − q X t (1 2 e cos(2 π k pj/q) + e ) Y q k=1 0 Qj − − − j=1

2m 1 Theorem 5.2 Let M = G S − be a spherical space form with spin structure given by ε : G Spin(2m) . Then the\ Dirac operator on M has the determinant → 2m 1 DET (D; G S − ) \ +  iπ (χ χ−)(ε(g)) 2 2m 1 = exp − DET(D ; G S − ) , G X det(12m g) · p \  g G 12m  | | ∈ −{ } −

2 2m 1 where DET (D ; G S − ) is given by Theorem 5.1 and χ± are the characters of the half spinor modules\ of Spin (2m) .

Proof. Follows directly from (3), Theorem 3.2, and (21). 20 Christian Bar¨ and Sven Schopka 6 Spectrum and Determinant of twisted Dirac operators on S2

In this section we compute the spectrum and determinant of certain twisted Dirac operators on the two-sphere S2. Recall that complex line bundles over S2 are in 1-1 correspondence with Z via their Chern numbers. Hence each complex line bundle is of the form k for 2 L some k Z where c1( ) = 1. The spinor bundle ΣS decomposes into a sum of two ∈ L 2 + 2 2 complex line bundles, ΣS = Σ S Σ−S , the bundles of spinors of positive and 2 ⊕ + 2 of negative chirality. Since c1(Σ±S ) = 1 we have = Σ S . The metric and the Levi-Civita connection on Σ+S2 induce Hermitian± metricsL and metric connections on all k the line bundles . Using these connections we can form the twisted Dirac operator Dk acting on sectionsL of the bundle ΣS2 k. ⊗ L

Theorem 6.1 The eigenvalues of the twisted Dirac operator Dk acting on sections of the spinor bundle tensored with the line bundle of Chern number k over S2 are given by j( k + j), j = 0, 1, 2,... ±p | | with multiplicity 2j + k . | |

2 Proof. We start by computing the relevant curvatures. The curvature tensor RΣS of the 2 spinor bundle can be expressed in terms of the curvature tensor RTS of the tangent bundle by 2 ΣS2 1 TS2 R (X,Y ) = R (X,Y )ei, ej eiej 4 X D E · i,j=1 where e1, e2 is an orthonormal tangent frame acting on spinors via Clifford multiplication. Since in our case S2 is of constant sectional curvature 1 we have

2 RTS (X,Y )Z = Y,Z X X,Z Y h i − h i and therefore 2 1 RΣS (X,Y ) = (YX XY ) 4 − · Hence ΣS2 1 R (e1, e2) = e1 e2. −2 · 2 Since the area element e1 e2 acts on Σ±S by multiplication by i we get · ± Σ+S2 i RL(e1, e2) = R (e1, e2) = −2 and thus m im RL (e1, e2) = . − 2 2 1 2 k The decomposition ΣS = − yields the parallel decomposition ΣS = k+1 k 1 L ⊕ L ⊗ L − with respect to which the curvature tensor has the form L ⊕ L

ΣS2 k i  k + 1 0  R ⊗L (e1, e2) = . −2 0 k 1  −  The Dirac operator interchanges the chirality of spinors and hence has the form

+  0 Dk  Dk =  Dk− 0  The Dirac Determinant of Spherical Space Forms 21

+ m where Dk− = (Dk )∗. Denote the connection Laplace operator on by ∗ =: ∆m. 2 L ∇ ∇ The curvature endomorphism in the Weitzenbock¨ formula for Dk is given by

ΣS2 k k = e1 e2 R ⊗L (e1, e2) K · · i i 0 k + 1 0 =     −2 0 i 0 k 1  −   −  1 k + 1 0 =   . 2 0 k + 1  −  2 The Weitzenbock¨ formula D = ∗ + k now says k ∇ ∇ K k + 1 D+D = ∆ + , (22) k k− k+1 2 + k + 1 Dk−Dk = ∆k 1 + − . (23) − 2 Taking the difference of (22) and (23) with k + 2 instead of k we obtain

+ + Dk Dk− = Dk−+2Dk+2 + k + 1. (24)

+ + + + Since Dk Dk− = (Dk−)∗Dk− and Dk−Dk = (Dk )∗Dk are non-negative operators they + do not have negative eigenvalues. Denote the positive eigenvalues by λj(Dk Dk−) and + + + λj(Dk−Dk ), j = 1, 2, 3,..., and their multiplicities by µj(Dk Dk−) and µj(Dk−Dk ) re- + + + + spectively. Write λ0(Dk Dk−) = λ0(Dk−Dk ) = 0 and let µ0(Dk Dk−) and µ0(Dk−Dk ) be + + the multiplicities of the eigenvalue 0. Here µ0(Dk Dk−) = 0 or µ0(Dk−Dk ) = 0 is not excluded. From now on assume k 0. The case of negative k can be treated similarly by interchang- + ≥ ing the roles of Dk and Dk−. For non-negative k equation (22) shows

+ µ0(Dk Dk−) = 0. (25) The Atiyah-Singer index formula yields

+ + + k µ0(D−D ) µ0(D D−) = ind(D ) = c1( ) = k k k − k k k L and thus + µ0(Dk−Dk ) = k. (26) + + + Since Dk intertwines the operators Dk−Dk and Dk Dk− it induces isomorphisms on the eigenspaces for non-zero eigenvalues. Hence for j 1 ≥ + + + + λj(Dk Dk−) = λj(Dk−Dk ) and µj(Dk Dk−) = µj(Dk−Dk ). (27) Combining equations (24) and (27) yields for j 1 ≥ + + + + λj(Dk−Dk ) = λj 1(Dk−+2Dk+2) + k + 1 and µj(Dk−Dk ) = µj 1(Dk−+2Dk+2). (28) − − The parameter shift from j to j 1 is due to the fact that 0 is counted as an eigenvalue (of + − multiplicity 0) for Dk Dk−. Using (28) inductively j times we get

j λ (D D+) = λ (D D+ ) + (k + 2m 1) = j(k + j) j k− k 0 k−+2j k+2j X m=1 − and by (26) 22 Christian Bar¨ and Sven Schopka

+ + µj(Dk−Dk ) = µ0(Dk−+2jDk+2j) = k + 2j.

+ D D− 0 We have computed the spectrum of D2 =  k k , namely 0 is an eigenvalue k +  0 Dk−Dk  of multiplicity k and j(k + j) has multiplicity 2(k + 2j), j = 1, 2, 3,....

It remains to observe that the spectrum of Dk is symmetric about 0 because for each 2 k eigenspinor of Dk of the form ψ = ψ+ + ψ with respect to the splitting ΣS = + 2 k 2 k ˆ − ⊗ L Σ S Σ−S the spinor ψ = ψ+ ψ is an eigenspinor for the opposite eigenvalue.⊗ L Taking⊕ square⊗ L roots proves the theorem.− −

For k = 0 Theorem 6.1 gives the spectrum of the classical Dirac operator which we used in the previous sections. Similarly, for k = 1 we recover the spectrum of the Laplace operator + acting on functions ∆ = D1−D1 . Theorem 6.1 can also be proved by trivialising the twisted spinor bundle over S2 minus a point and then solving the eigenvalue equation explicitly using spin-weighted spherical harmonics. See [16, Sec. 3.1] for this approach.

Knowing the eigenvalues of Dk explicitly we can compute its determinant.

Theorem 6.2 The determinant of the twisted Dirac operator Dk acting on sections of the spinor bundle tensored with the line bundle of Chern number k over S2 is given by

k 2 | | 2 π 1 4 ζ0 ( 1)+k /2 k 2m DET(Dk; S ) = exp i ( k ) e− R − m| |− , 2 3 Y −| | − · m=1 the determinant of its square is

k | | 2 2 8 ζ ( 1)+k2 2 k 4m DET(D ; S ) = e R0 m k − − Y | |− · m=1 and the proper time regularized determinant

k | | 2 2 8 ζ ( 1)+k2+Γ (1)( k + 1 ) 2 k 4m DET (D ; S ) = e R0 0 3 m . p.t. k − − | | Y | |− · m=1

Proof. By Theorem 6.1 we obtain for the ζ-function

∞ 2j + k ζD2 (s) = 2 | | k X js(j + k )s j=1 | |

∞ s 1 s 1 s s = 2 (j− (j + k ) − + j − (j + k )− ). X | | | | j=1

k α For j > k we can expand (1 + | | ) into a binomial series and we get | | j

ζ 2 (s) = Dk k i | | 2j + k ∞ 1 2s ∞ 1 s s k = 2 | | + 2 j −  −  + −  | | X js(j + k )s X X i i j j=1 j= k +1 i=0 | | | | k | | 2j + k ∞ 1 s s i = 2 | | + 2  −  + −  k ζH (2s + i 1, k + 1) X js(j + k )s X i i | | − | | j=1 | | i=0 The Dirac Determinant of Spherical Space Forms 23

Now we partly follow a similar calculation of Weisberger [21, Appendix C]. For the coef- ficients 1 s s di(s) :=  −  + −  i i we have

d0(s) = 2,

d1(s) = 2s + 1, −2 d2(s) = s ,

i i 2 2 di(s) = ( 1) − s + O(s ), i 3. − i(i 1) ≥ − Hence

k | | ζD2 (0) = 2 (2j + k ) + 2 (2ζH ( 1, k + 1) + k ζH (0, k + 1)) k X | | − | | | | | | j=1

= 4ζR( 1) + 2 k ζR(0) − | | 1 = k (29) −3 − | | and since ζH (s, k + 1) has only one pole of first order at s = 1 with residue 1 we have | |

ζ0 2 (0) = Dk k | | = 2 (2j + k )(ln j + ln(j + k )) + 8ζ0 ( 1, k + 1) + 4 k ζ0 (0, k + 1) − X | | | | H − | | | | H | | j=1

2 ∞ i i 2 i 4 k ζH (0, k + 1) + k + 2 ( 1) − k ζH (i 1, k + 1) − | | | | | | X − i(i 1)| | − | | i=3 − k | | = 2 (2j + k )(ln j + ln(j + k )) + 8ζ0 ( 1, k + 1) + 4 k ζ0 (0, k + 1) − X | | | | H − | | | | H | | j=1

2 ∞ i i 2 i +2 k + 5 k + 2 ( 1) − k ζH (i 1, k + 1). (30) | | | | X − i(i 1)| | − | | i=3 − For the evaluation of the last sum we use the Mellin transform of the Hurwitz ζ-function

1 ∞ s 1 at t 1 ζH (s, a) = Z t − e− (1 e− )− dt. (31) Γ (s) 0 − This yields

∞ i i 2 i I := ( 1) − k ζH (i 1, k + 1) X − i(i 1)| | − | | i=3 −

∞ i i 2 i 1 ∞ i 2 ( k +1)t t 1 = ( 1) − k Z t − e− | | (1 e− )− dt X − i(i 1)| | (i 2)! − i=3 − − 0 ∞ 1 (2 k +1)t ( k +1)t t 1 = k Z t− ( e− | | + e− | | (1 k t))(1 e− )− dt | | 0 − − | | −

∞ 2 (2 k +1)t ( k +1)t 1 2 t 1 2 Z t− (e− | | e− | | (1 k t + ( k t) ))(1 e− )− dt. − 0 − − | | 2 | | − To compute these integrals we multiply the integrand by ts, evaluate the integral using (31) and then let s 0. → 24 Christian Bar¨ and Sven Schopka

I = lim k Γ (s)( ζH (s, 2 k + 1) + ζH (s, k + 1)) s 0 → {| | − | | | | 2 k Γ (s + 1)ζH (s + 1, k + 1) 2[Γ (s 1)(ζH (s 1, 2 k + 1) −| | | | − − − | | ζH (s 1, k + 1)) + k Γ (s)ζH (s, k + 1) − − | | | | | | 1 2 k Γ (s + 1)ζH (s + 1, k + 1)] −2| | | | } = lim k Γ (s)( ζH (s, 2 k + 1) ζH (s, k + 1)) + 2Γ (s 1)h(s 1) s 0 → {| | − | | − | | − − } with the definition

k | | s h(s) := (j + k )− = ζH (s, k + 1) ζH (s, 2 k + 1). X | | | | − | | j=1

( 1)l Using 2ζH (0, k + 1) = 1 + 2 k , h(0) = k , and Ress= lΓ (s) = − we conclude − | | | | | | − l! I =

ζH (s, k + 1) ζH (0, k + 1) h(s) h(0) 1 + 3 k = lim ( k sΓ (s)  2 | | − | | + − + | | s 0 s s s → | | −

+2Γ (s 1)h(s 1)) − −

= 2 k ζ0 (0, k + 1) + k h0(0) − | | H | | | | k + lim 2Γ (s 1)[h(s 1) + (s 1)| |(1 + 3 k )] s 0 2 → { − − − | | } = 2 k ζ0 (0, k + 1) + k h0(0) 2h0( 1) k (1 + 3 k ) − | | H | | | | − − − | | | | k k | | | | = 2 k ζ0 (0, k + 1) k ln(j + k ) + 2 ln(j + k )(j + k ) − | | H | | − | | X | | X | | | | j=1 j=1 k (1 + 3 k ) −| | | | k | | = 2 k ζ0 (0, k + 1) + ln(j + k )(2j + k ) k (1 + 3 k ). − | | H | | X | | | | − | | | | j=1 Inserting this into (30) we get

k | | 2 ζD0 2 (0) = 2 (2j + k ) ln j + 8ζH0 ( 1, k + 1) + 2 k + 5 k k − X | | − | | | | | | j=1 2 k (1 + 3 k ) − | | | | k | | 2 = 8ζ0 ( 1) k + 2 (2j k ) ln j, R − − | | X − | | j=1 therefore k | | 2 2 8ζ ( 1)+k2 2 k 4m DET(D ; S ) = e R0 m . k − − Y | |− · m=1 2 The formula for DET(Dk; S ) now follows from (29) and from the vanishing of the η- 2 2 invariant. Proposition 2.1 yields the formula for DETp.t.(Dk; S ).

Remark 1. The case k = 0 yields Theorem 4.1, 4.2 and 4.3 for n = 2. Since the Laplace- + Beltrami operator acting on functions is given by ∆ = D1−D1 it has the same spec- 2 trum as D1 except that all non-zero eigenvalues have only half the multiplicity. Hence DET(∆; S2) = DET(D2; S2) and we get (compare [4, Thm. 8.1]) p 1 The Dirac Determinant of Spherical Space Forms 25

Corollary 6.3 The determinant of the Laplace-Beltrami operator ∆ acting on functions on S2 is given by 1 2 4ζ0 ( 1)+ DET(∆; S ) = e− R − 2 .

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