Zeta Functions and Regularized Determinants for Elliptic Operators. Elmar Schrohe Institut für Analysis

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1 Zeta Functions and Regularized Determinants for Elliptic Operators Elmar Schrohe Institut für Analysis

2 PDE: The Sound of Drums How Things Started If you heard, in a dark room, two drums playing, a large one and a small one, you could probably distinguish the two: Both drums produce many different frequencies, but those of the smaller one are higher than those of the larger one. Question: Can you make this precise?

3 PDE: The Sound of Drums Ω R 2 smooth, bounded domain. (Elementary) Physics A membrane, held fixed at the boundary, is set in motion. Its vertical displacement u = u(x, t) (in the direction perpendicular to the membrane) satisfies the wave equation. Mathematical Formulation: A Boundary Value Problem Wave Equation: 2 t u u = 0 in Ω. Clenching: Dirichlet Boundary Condition: u Ω = 0.

4 PDE: The Sound of Drums Ω R 2 smooth, bounded domain. (Elementary) Physics A membrane, held fixed at the boundary, is set in motion. Its vertical displacement u = u(x, t) (in the direction perpendicular to the membrane) satisfies the wave equation. Mathematical Formulation: A Boundary Value Problem Wave Equation: 2 t u u = 0 in Ω. Clenching: Dirichlet Boundary Condition: u Ω = 0.

5 PDE: The Sound of Drums Separation Ansatz Find time harmonic solutions Pure tones Write u(x, t) = v(x)e iωt. Wave Equation: 2 t u u = 0 in Ω. Dirichlet Boundary Condition: ω 2 v(x)e iωt v(x)e iωt = 0. v Ω = 0. Eigenvalue Problem for the Dirichlet-Laplace Operator D Write λ = ω 2. Then we obtain the eigenvalue problem v = λv in Ω and v Ω = 0.

6 PDE: The Sound of Drums Separation Ansatz Find time harmonic solutions Pure tones Write u(x, t) = v(x)e iωt. Wave Equation: 2 t u u = 0 in Ω. Dirichlet Boundary Condition: ω 2 v(x)e iωt v(x)e iωt = 0. v Ω = 0. Eigenvalue Problem for the Dirichlet-Laplace Operator D Write λ = ω 2. Then we obtain the eigenvalue problem v = λv in Ω and v Ω = 0.

7 PDE: The Sound of Drums Separation Ansatz Find time harmonic solutions Pure tones Write u(x, t) = v(x)e iωt. Wave Equation: 2 t u u = 0 in Ω. Dirichlet Boundary Condition: ω 2 v(x)e iωt v(x)e iωt = 0. v Ω = 0. Eigenvalue Problem for the Dirichlet-Laplace Operator D Write λ = ω 2. Then we obtain the eigenvalue problem v = λv in Ω and v Ω = 0.

8 PDE: The Sound of Drums First Observation (Functional Analysis) There is an infinite number of eigenvalues. 0 < λ 1 < λ 2 λ 3... H.A. Lorentz Lorentz Conjecture On the occasion of a series of lectures in Göttingen in 1910, the Dutch physicist H.A. Lorentz stated the conjecture that λ k 2π vol Ω k.

9 PDE: The Sound of Drums The Conjecture λ k 2π vol Ω k. Hilbert s Opinion Supposedly Hilbert considered the problems to be very difficult and did not expect a solution during his lifetime.

10 PDE: The Sound of Drums The Conjecture λ k 2π vol Ω k. Weyl s Solution Only one year later, Hermann Weyl solved the problem. In fact, for a smooth bounded domain in R n λ k (2π)n ω n vol Ω k2/n as k, where ω n is the volume of the unit sphere in R n.

11 Extensions Weyl s Formula in n Dimensions λ k Weyl s formula provides a link between (2π)n ω n vol Ω k2/n (1) Analysis (namely the eigenvalues of the Laplace operator) and Geometry (namely the volume of the domain). It shows that one can hear the volume of a drum. Question: Can one hear more then just the volume? Or: Are there more terms for the asymptotic expansion (1)? In general, this is not possible, but:

12 Extensions Weyl s Formula in n Dimensions λ k Weyl s formula provides a link between (2π)n ω n vol Ω k2/n (1) Analysis (namely the eigenvalues of the Laplace operator) and Geometry (namely the volume of the domain). It shows that one can hear the volume of a drum. Question: Can one hear more then just the volume? Or: Are there more terms for the asymptotic expansion (1)? In general, this is not possible, but:

13 Zeta Functions and Heat Trace Expansions Partition Function ( Heat Trace ) Z(t) = e λ k t. Since λ k c n k 2/n : The sum converges for all t > 0. As the λ k are the eigenvalues of the Dirichlet Problem, we can write k=1 Z(t) = Tr(e t D ). Zeta Function ζ D (s) = Since λ k c n k 2/n : The sum converges for all Re s < n/2 and defines a holomorphic function in this half-plane. k=1 λ s k ζ D (s) = Tr(( D ) s ).

14 Example Laplacian on S 1 (no Boundary) On S 1, has the eigenfunctions e ikt, k Z, eigenvalues 0 with multiplicity 1, and 1, 4, 9..., with multiplicity 2. Thus ζ (s) = 2 k 2s = 2ζ( 2s) k=1 with the Riemann zeta function. Holomorphic for Re s < 1/2. Extends meromorphically to C. Simple pole in s = 1/2 Regular in s = 0.

15 Trace Expansions Asymptotic Expansion for the Partition Function As t 0 +, Tr e t D j=0 a j t j n 2, where the coefficients carry geometric information: a 0 volume of Ω a 1 volume of Ω a j, j 2 curvature terms. (Minakshisundaram&Pleijel, McKean&Singer, Gilkey,... )

16 Trace Expansions Meromorphic Extension of the Zeta Function ζ D extends to a meromorphic function on C: At most simple poles in the points s j = j n 2. No pole in s = 0. Residue in s j coincides with coefficient a j (up to universal factor). Connection L(Tr e t D )(s) = 0 t s 1 Tr(e t D )dt = Γ(s)ζ D ( s) i.e. the zeta function is up to sign and a Γ-factor the Laplace transform of the partition function. Indeed 0 t s 1 e tλ dt = λ s (λt) s 1 e λt d(λt) = λ s Γ(s). 0

17 Complex Powers Dunford Integral Let P be an unbounded operator in a Banach space with a ray of minimal growth, i.e. C does not intersect spec(p) R = {re iφ : r 0}, φ [0, 2π) (P λi ) 1 = O(λ 1 ) as λ. Define P s = i 2π C λ s (P λi ) 1 dλ, Re s < 0. C: From to re iφ, r > 0, along R. Clockwise about origin on S(0, r). Back to on R.

18 Complex Powers Properties of P s s P s is an analytic family of bounded operators P s P t = P s+t, Re s, Re t < 0. P 1 = P 1 (inverse of P). Define P s = { P s, Re s < 0 P k P s k, k integer, 1 Re s k < 0.

19 Pseudodifferential Operators (ψdo s) Fourier Transform on Rapidly Decreasing Functions Let f S(R n ), ξ R n. (Ff )(ξ) = ˆf (ξ) = (2π) n/2 e ixξ f (x) dx, f S(R n ) (F 1 f )(x) = (2π) n/2 e ixξ f (ξ) dξ, Fourier Transform and Differentiation F 1 ξ α Ff (x) = D α f (x) For P = α m a α(x)d α and p(x, ξ) = α m a α(x)ξ α, Pu(x) = (2π) n/2 e ixξ p(x, ξ)fu(ξ) dξ. ψdo s: Replace polynomial p by more general functions.

20 Pseudodifferential Operators (continued) Symbol Classes S µ = {p = p(x, ξ) C : D α ξ Dβ x p(x, ξ) = O((1 + ξ ) µ α ) } Definition: Symbol defines Operator p S µ defines the operator op p : S(R n ) S(R n ) by (op p)u(x) = (2π) n/2 e ixξ p(x, ξ)fu(ξ) dξ.

21 Pseudodifferential Operators (continued) Asymptotic Expansions p S µ has the asymptotic expansion p j=0 p µ j with symbols p µ j S µ j if, for each N, p j<n p µ j S µ N Classical Symbols p is classical, if p p µ j with p µ j S µ j positively homogeneous of degree µ j in ξ for all ξ 1, i.e. p µ j (x, tξ) = t µ j p(x, ξ), x R n, ξ R n, ξ 1, t 1. Theorem Given a sequence (p j ) j=0 with p j S µ j, there exists a symbol p S µ such that p p j.

22 Pseudodifferential Operators (continued) Theorem: Composition Let p and q be symbols of orders µ 1 and µ 2. Then op p op q is a ψdo with a symbol r S µ 1+µ 2. It has the asymptotic expansion r(x, ξ) α 1 α! α ξ p(x, ξ)dα x q(x, ξ) Write: r = p#q. If p and q are classical, then so is p#q.

23 Pseudodifferential Operators (continued) Ellipticity A classical p p µ j S µ is elliptic of order µ, if p µ (x, ξ) is invertible for x R n, ξ = 1. Theorem: Parametrices If p S µ is classical and elliptic, then there exists q S µ with p#q 1 and q#p 1. q is also classical and elliptic of order µ. q is called a parametrix to p.

24 Pseudodifferential Operators (continued) Let X be a closed manifold, dim X = n. Definition: ψdo s on Manifolds An operator P : C (X ) C (X ) is a ψdo of order µ on X, provided that, in all local coordinates, it can be written as a ψdo on R n with a symbol in S µ. Theorem: Mapping Properties If P is a ψdo of order µ, then P extends to a bounded operator for each s R. Properties of Sobolev Spaces P : H s (X ) H s µ (X ) H s (X ) H t (X ) is compact for s > t. H s (X ) H t (X ) is trace class for s t > n.

25 Seeley s Construction of Complex Powers The Situation X closed manifold of dimension n P ψdo of order µ > 0, classical, elliptic Local symbols p p µ j Assume p µ has a ray R of minimal growth: p µ (x, ξ) λ is invertible for all x, ξ, ξ 1, λ R and the inverse is O(λ 1 ). The Plan Define local symbols q = q(x, ξ; λ), depending on λ R, with an asymptotic expansion q(x, ξ; λ) q µ j (x, ξ; λ) such that q(x, ξ; λ)(p(x, ξ) λ) 1.

26 Parametrix Construction The plan requires that q µ (x, ξ, λ) = (p µ (x, ξ) λ) 1 q µ j (x, ξ, λ) = 1 α! α ξ q µ k(x, ξ, λ) D α x p µ l (x, ξ)(p µ (x, ξ) λ) 1, j = 1, 2,..., where the sum extends over all α, k and l such that k + l + α = j and k < j.

27 Parametrix Construction q µ j is a linear combination of terms of the form (p µ λ) 1 α 1 ξ β 1 with Moreover x p µ k1 (p µ λ) 1... αr k k r + α α r = j. ξ βr x p µ kr (p µ λ) 1 q µ j (x, tξ, t µ λ) = t µ j q µ α j (x, ξ, λ).

28 Parametrix Construction Theorem We then find a symbol q = q(x, ξ; λ) with q q µ j. These local symbols can be patched to a parameter-dependent pseudodifferential operator Q(λ) on X such that Q(λ)(P λ) I = R 1 (λ) where R 1 is of arbitrarily negative order and R 1 (λ) = O(λ 1 ) as λ on the ray R, say in L(L 2 (X )). Similarly, one obtains Q(λ) with (P λ) Q(λ) I = R 2 (λ) R 2 (λ) = O(λ 1 ) as λ.

29 Parametrix Construction Corollary (P λ) 1 exists for large λ on R and is O(λ 1 ). (P λ) 1 Q(λ) = O(λ 2 ). (P λ) 1 Q(λ) is of arbitrarily negative order.

30 Complex Powers P s = i 2π C λ s (P λi ) 1 dλ, Re s < 0. (P λi ) 1 Q(λ) is of arbitrarily negative order. Recall Q(λ) has local symbol q(x, ξ; λ) q µ j (x, ξ; λ). Hence (P λ) 1 is a ψdo with same expansion as Q(λ). Theorem P s is a ψdo of order µ Re s. Its symbol p (s) has an expansion p (s) (x, ξ) c j (x, ξ; s), where c j (x, ξ; s) = 1 λ s q µ j (x, ξ; λ) dλ, 2πi C

31 Complex Powers Homogeneity Substituting λ = t µ σ and using the homogeneity c j (x, tξ; s) = i λ s q µ j (x, tξ; λ) dλ 2π C = t µs+µ i σ s q µ j (x, tξ, t µ σ) dσ 2π C = t µs+µ µ j i σ s q µ j (x, ξ, σ) dσ 2π C = t µs j c j (x, ξ; s). Note that the deformation of the contour is irrelevant as λ q(x, ξ; λ) is holomorphic in the area in between.

32 Pseudodifferential Operators and Their Trace Let r S m (R n R n ) for some m < n. Then (op r)u(x) = (2π) n/2 e ixξ r(x, ξ)fu(ξ)dξ = (2π) n e i(x y)ξ r(x, ξ)u(y)dydξ = k(x, y)u(y)dy, where k(x, y) = (2π) n e i(x y)ξ r(x, ξ)dξ. The integral exists, since m < n and defines a continuous function k. If r has compact support in x, then op r is trace class and Tr(op r) = k(x, x)dx.

33 The Trace of P s P s is a ψdo of order µ Re s. Hence it is trace class for Re s < n/µ. Moreover P s is a holomorphic operator family. We therefore easily obtain the first part of Theorem s Tr(P s ) is a holomorphic function on Re s < n/µ. It extends to a meromorphic function on C with at most simple poles in s j = (j n)/µ. s = 0 never is a pole.

34 Idea of the Proof The local symbols p (s) have the asymptotic expansion p (s) (x, ξ) c j (x, ξ; s). The function k j (x, x; s) associated with c j (x, ξ; s) is k j (x, x; s) = (2π) n c j (x, ξ; s)dξ. We split the integral into the part, where ξ 1 and where ξ 1. The integral over ξ 1 gives an entire function of s no contribution to the singularities. We only have to consider the integral over ξ 1.

35 Idea of the Proof For ξ 1, we have c j (x, tξ; s) = t µs j c(x, ξ; s) Using polar coordinates c j (x, ξ; s)dξ = c j (x, tη)t n 1 ds(η)dt ξ 1 = = 1 S c j (x, η)ds(η) t µs j+n 1 dt S 1 1 c j (x, η)t n 1 ds(η) µs j + n So we obtain at most a simple pole in s j = (j n)/µ. It is more work to show that the residue at s = 0 vanishes. S

36 Regularized Determinant Matrices Suppose P is a selfadjoint (n n)-matrix with positive eigenvalues λ 1,..., λ n. Then n det P = λ k, k=1 Since dλ s /ds = λ s ln λ, this yields ln det P = ln λ k = d ds λ s k = d s=0 ds Tr P s. s=0 Ray-Singer Regularized Determinant (1971) P be an unbounded operator on Hilbert space; Tr P s exists for Re s << 0, has a meromorphic extension to C, regular at s = 0. Then define ( ) d det P = exp ds Tr P s. s=0