CMS winter meeting 2008, Ottawa. The heat kernel on connected sums

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1 CMS winter meeting 2008, Ottawa The heat kernel on connected sums Laurent Saloff-Coste (Cornell University) December

2 p(t, x, y) = The heat kernel ) 1 x y 2 exp ( (4πt) n/2 4t The heat kernel is: The fundamental solution of the most basic parabolic PDE ( t )u = 0 where = ( ) n 2. 1 x i The kernel of the heat semigroup e t. The density at time t of the stochastic process (X t ) driven by (Brownian Motion). u(t, x) = e t u 0 (x) = p(t, x, y)u 0 (y)dy = E x (u 0 (X t )). ( t )u = 0, u(0, x) = u 0 (x).

3 Many variants, even in R n Uniformly elliptic operators: Af = i,j Lower order terms, e.g., V ( x i a i,j (t, x) x j f Domains in R n with the Neumann boundary condition: ), An unbounded 3-d domain with 3 different ends Riemannian manifolds, e.g., on hyperbolic 3-space, p(t, x, y) = ( 1 4πt )3/2 ρ sinh ρ e t e ρ2 /4t, ρ = ρ(x, y). Weighted manifolds: = σ 1 div(σ grad), dµ = σdv.

4 What are some of the fundamental questions? Asymptotic expansion as t 0: p(t, x, y) 1 (4πt) n/2 e d(x,y)2 /4t ( ) A k (x, y)t k. Gaussian behavior: lim t 0 4t log p(t, x, y) = d(x, y) 2. Behavior as t : p(t, x, y)? ϕ(x, t) = sup y {p(t, x, y)}? ϕ(t) = sup x,y {p(t, x, y)}? On which manifolds is the heat kernel similar to that of R n? If two manifolds are similar, are their heat kernels similar? 0

5 What are some of the fundamental questions? Find explicit descriptions (depending on M) of the type p(t, x, y) F (t, x, y)exp ( c ) d(x, y)2. t The function F = F M captures the on-diagonal and mid-range behavior in the time-space region {(t, x, y) : d(x, y) 2 Ct}. For any x, t (and for some ɛ (0, 1)), give a rough description of the hot-spot region A ɛ (x, t) = {y : p(t, x, y) ɛϕ(x, t)} where ϕ(x, t) = sup y {p(t, x, y)}. This question relates to F.

6 On-diagonal behavior: the classical case Let (M, g) be a complete (non-compact) Riemannian manifold. The bound sup x {p(t, x, x)} Ct n/2, t > 0, is equivalent to any one of the following inequalities: Sobolev (n > 2): f C c (M), f 2n/(n 2) C S f 2. Nash: f C c (M), f 2(1+2/n) 2 CN 2 f 2 2 f 4/n 1. Faber-Krahn: Ω M, λ D (Ω) c FK Ω 2/n. Rozenblum-Cwikel-Lieb (n > 2): V L 1 loc(m), N ( + V ) C R V n/2 dλ where N (A) = # of negative eigenvalues of A in L 2.

7 Parabolic Harnack inequality (PHI) We say that (PHI) holds if there is a constant H such that for any positive solution u of the heat equation ( t )u = 0 in Q = (τ, τ + 4T ) B(x, R), T = R 2, we have R/2 T Q+ 4T T Q sup{u} H 0 inf{u}. Q Q + Q R The better known elliptic version reads: any harmonic function u 0 in B = B(x, R) satisfies sup (1/2)B {u} H 1 inf {u}. (1/2)B

8 Example: the Li-Yau inequality Peter Li and Shing Tung Yau (1986) proved that, on any complete n-dimensional Riemannian manifold with non-negative Ricci curvature, if u is a non-negative solution of the heat equation then log u 2 t log u n 2t. This implies the two-sided heat kernel bound c ɛ V (x, t) e d2 /(4(1 ɛ)t) p(t, x, y) where d = d(x, y) and V (x, r) = v(b(x, r)). C ɛ V (x, t) e d2 /(4(1+ɛ)t), The localized version of the top inequality implies the parabolic Harnack inequality (PHI) of the previous slide.

9 Doubling and Poincaré Theorem (Grigor yan, 91; LSC, 92) For fixed R (0, ], the following properties are equivalent: (a) The conjunction of The doubling property: x, r (0, R), V (x, 2r) DV (x, r). The Poincaré inequality: r (0, R), B = B(x, r), f Lip(B), f f B 2 dµ Pr 2 f 2 dµ. B B (b) The two-sided Gaussian bound: for all x, y, t (0, R), p(t, x, y) C ( c V (x, t) exp ) d(x, y)2. t (c) The parabolic Harnack inequality (PHI) up to scale R.

10 Examples of (PHI) with R = : Harnack-type manifolds Complete Riemannian manifolds with Ric 0. Bishop-Gromov and P. Buser, Li-Yau, Lie groups with polynomial volume growth. Gromov 1981, Varopoulos Includes all nilpotent Lie groups. Quotients of any such space by an isometric group action, e.g., homogeneous spaces of nilpotent Lie groups. Spaces that are (measure) quasi-isometric to any of the previous examples, e.g., coverings of compact manifolds which have polynomial volume growth (Kanai, Coulhon, LSC)

11 Examples with R = : Harnack-type manifolds Convex domains in Euclidean space (well-known). and their complements (geodesic distance) (P. Gyrya, LSC). The surface of revolution S α satisfies (PHI) iff α [1, 1). The two-sided surface of revolution S(α, β) satisfies (PHI) iff α = β (1, 1). (Grigor yan/lsc) S S(, ) f(r) r r r r f(r)=r, r>1.

12 Manifolds with ends We will consider manifolds with ends: M = M 1 #M 2 #... #M k where each M i, 1 i k, is of Harnack type. More precisely, M is the disjoint union M = K E 1 E k with K compact with smooth boundary and E i isometric to an open set in M i (we can allow E i = M i ). Curvature conditions that yield such manifolds: (c1) Asymptotically non-negative sectional curvature, (c2) Non-negative Ricci curvature outside a compact set with ends satisfying (RCA)

13 Euclidean domains V 3 (r) r 3 V 2 (r) r 2 V 1 (r) r

14 The heat kernel on manifolds with transient ends Consider M = M 1 # #M k and assume that each M k is of Harnack type, transient. Then the heat kernel is bounded above and below by expressions of the type 1 V ix (x, t)v iy (y, exp ( d (x, y) 2 ) + t) t ( ) H(x, t)h(y, t) V 0 ( H(x, t) + t) V iy ( H(y, t) + t) V ix ( exp ( d +(x, y) 2 ) t) t V 0 (r) = min i {V i (r)} { 1, H(x, t) = min x 2 V ix ( x ) + ( t x 2 ) } ds V ix (, x = 1 + d(k, x). s) +

15 Ingredients of Proof: Hitting probability: ψ K (t, x) := P x ( s t : X s K), ψ K (t, x); Dirichlet heat kernel in each end: p Ei (t, x, y); Central estimate p(t, k, k ) 1/V 0 ( t), k, k K; Gluing technique. R n Ni R x K y R Nj

16 R N 1 # #RN p, 2 < n = min{n i} For fixed x, y and t : p(t, x, y) sup u,v p(t, u, v) t n/2 For x R N i, y R N j, x, y t: p(t, x, y) 1 t n/2 x N i 2 y N j t N i /2 y N j t N j /2 y. N i 2 R n R N i = R N /Z N N i Ni R x K y R Nj

17 Transient manifolds with recurrent ends V 3 (r) r 3 V 2 (r) r 2 V 1 (r) r

18 The harmonic function h (RCA) with respect to o: There exists A > 1 such that any two points x 1, x 2 with d(o, x i ) = r, R > a, are connected in B(o, Ar) \ B(o, A 1 R). Let M = M 1 #... #M k. Assume that M is transient and that, for each i = 1,..., k, M i is Harnack type and satisfies (RCA). Then there exists a positive harmonic function h on M such that, for all x M, ( ) x 2 ds h(x) V ix (. s) + (Sung,Tam and Wang Grigor yan and LSC)

19 What about transient manifolds M with recurrent ends? V 3 (r) r 3 h(x) 1 V 2 (r) r 2 h(x) 1 + log x V 1 (r) r h(x) 1 + x

20 The general transient case Consider M = M 1 # #M k, transient. Assume that each M k is of Harnack type, and satisfies (RCA). Then the heat kernel is bounded above and below by expressions of the type ( h(x)h(y) Ṽix (x, t)ṽiy (y, exp d ) 2 (x, y) t t) ( H(x, t) H(y, t) + h(x)h(y) Ṽ 0 ( + H(x, t) t) Ṽ iy ( t) + H(y, ) ( t) Ṽ ix ( exp d 2 ) +(x, y) t) t The tilde means relative to (M, h 2 dµ)!

21 The general transient case Corollary Assume that M = M 1 # #M k is transient with each M k of Harnack type and satisfying (RCA). Then sup x,y {p(t, x, y)} max i {V i ( t) 1 } sup y {p(t, x, y)} max i {[η i ( t)v i ( t)] 1 } p(t, x, y) max i {[η i ( t) 2 V i ( t)] 1 } η i (r) := 1 + ( r 2 1 ) ds V i (. s) +

22 p(t, x, y) 1/(t log 2 t) R 1 #R 2 #R 3 sup x,y p(t, x, y) 1/ t sup y p(t, x, y) 1/t

23 R 1 #R 2 #R 3 p(t, x, y) 1/(t log 2 t) H(y) = p(t,x,y) sup z p(t,x,z). sup y p(t, x, y) 1/t 3 R y 1 x H(y) 1/ log 2 t _ y t / log t H(y) 1/ log t R 1 _ y t, H(y) 1 log y log t R 2

24 References Heat kernels and analysis on manifolds, graphs, and metric spaces. Contemporary Mathematics, 338. AMS Grigor yan, Alexander: Analytic and geometric background of recurrence and non-explosion of the Brownian motion on Riemannian manifolds. Bull. AMS 36 (1999) Saloff-Coste, Laurent: Aspects of Sobolev-type inequalities. LMS Lect. Note Ser CUP, Grigor yan, A. and Saloff-Coste, L.: Stability results for Harnack inequalities Ann. Inst. Fourier, heat kernel on manifolds with ends. To appear in Ann. Inst. Fourier.

25 Thank you for your attention!