Exponential growth and Novikov conjecture Stefan Haller joint with - PDF document

Exponential growth and Novikov conjecture Stefan Haller joint with Dan Burghelea Conference: Topology of closed one-forms ETH Z¨ urich December 15th, 2004

M . . . closed manifold g . . . Riemannian metric ω . . . closed one form X = − grad g ( ω ) = − ♯ g ( ω ) Morse–Smale X ∗ . . . zero set of X , all non-degenerate In a nbh. of x ∈ X k , ind( x ) = k , we assume: x i dx i + x i dx i � � = − ω i ≤ k i>k dx i ⊗ dx i � g = i x i ∂ x i ∂ � � X = ∂x i − ∂x i i ≤ k i>k W − i x : ˆ ˆ x → M . . . completed unstable manifold i x ) ∗ ω = d ˆ h ω h ω (ˆ x and ˆ x ( x ) = 0. h ω W − ˆ x : ˆ x → R is proper 1

� e sh ω � � � x ∈ L 1 ( W − ρ ( X, [ ω ]) := inf s ≥ 0 x ) ∀ x ∈ X � Suppose ρ ( X, [ ω ]) < ∞ If α ∈ Ω ∗ ( M ; R ) we have absolutely converging � e s ˆ h ω i x ) ∗ α, x (ˆ Int s ( α )( x ) := ℜ ( s ) > ρ ( X, [ ω ]) ˆ W − x Int s : Ω ∗ ( M ; R ) → Maps( X ∗ , R ) Integration is onto. W − y T ( x, y ) × W − ∂ 1 ˆ x = � and Stokes theorem y applied to d s α = dα + sω ∧ α imply convergence of I γ ( x, y ) e sω ( γ ) , ˜ � I s ( x, y ) := ℜ ( s ) > ρ ( X, [ ω ]) γ ∈P x,y ∂ s : Maps( X ∗ ; R ) → Maps( X ∗ +1 ; R ) ˜ � ( ∂ s f )( x ) := I ( x, y ) f ( y ) y ∂ 2 Int ◦ d s = ∂ s ◦ Int s = 0 2

Lyapunov classes: � � [ ω ] ∈ H 1 ( M ; R ) � L ( X ) = � ∃ g : X = − grad g ( ω ) � L ( X ) ⊆ H 1 ( M ; R ) is open, convex Lemma. cone. Proof. Indeed, if X = − grad g 0 ( ω ) on nbh. of X and ω ( X ) < 0 on M \ X then ∃ g with g = g 0 near X and X = − grad g ( ω ). � If ξ 0 is Lyapunov for X and Proposition. ρ ( X, ξ 0 ) < ∞ then ρ ( X, ξ ) < ∞ for all ξ which are Lyapunov for X . Proof. ξ 0 ∈ L ( X ) and ρ ( X, ξ 0 ) < ∞ . Every ray in L ( X ) intersects ξ 0 + L ( X ) since L ( X ) is open cone. Hence sξ = ξ 0 + [ ω ] ∈ ξ 0 + L ( X ). But | e h ω x | ≤ 1 if ω ( X ) ≤ 0 hence ρ ( X, sξ ) ≤ ρ ( X, ξ 0 ). � Remark. H¨ older’s inequality implies that ρ ( X, tξ 1 + (1 − t ) ξ 0 ) ≤ max { ρ ( X, ξ 0 ) , ρ ( X, ξ 1 ) } 3

i x : W − x → M , g x := ( i x ) ∗ g r = dist g x ( x, · ) : W − x → [0 , ∞ ) B s ( x ) ⊆ W − x ball of radius s centered at x . X is said to have exponential growth if ∃ C ≥ 0 vol g x ( B s ( x )) ≤ e Cs , ∀ s ≥ 0 ∀ x ∈ X . This property is independent of g . Lemma. The following are equivalent: (i) X has exponential growth with respect to one (and hence every) Riemannian metric. (ii) For one (and hence every) Riemannian met- ric ∃ C ≥ 0 with e − Cr ∈ L 1 ( W − x ). 4

x | ) on W − Lemma. ∃ C 1 ≥ 0 s.t. r ≤ C 1 (1 + | h ω x . Lemma. ∃ C 2 ≥ 0 s.t. | h ω x | ≤ C 2 r on W − x . Proof. γ path in W − x , γ (0) = x . Then: � 1 � � 0 ( dh )( γ ′ ( t )) dt | h ( γ (1)) | = � ≤ || ω || ∞ length( γ ) � � � So | h ( γ (1)) | ≤ C dist( x, γ (1)), C 2 := || ω || ∞ . � ξ Lyapunov for X . Then the Proposition. following are equivalent: (i) X has exponential growth with respect to one (and hence every) Riemannian metric. (ii) ρ ( X, ξ ) < ∞ . Question. Does a Morse–Smale vector field which admits a Lyapunov class have exponen- tial growth? 5

x ∈ X q , ind( x ) = q B ⊆ W − x small ball centered at x Gr( W − x \ B ) ⊆ Gr q ( TM ) For y ∈ X define: y ) ∩ Gr( W − K x ( y ) := Gr q ( T y W − x \ B ) “If K x ( y ) is non-empty then W − badly accu- x mulates at y .” (-) K x ( y ) does not depend on B . (-) We may have K x ( x ) � = ∅ . (-) If q = ind ( x ) > ind ( y ) then K x ( y ) = ∅ . (-) If ind ( x ) = dim( M ) then K x ( y ) = ∅ ∀ y ∈ X . (-) First non-trivial case is dim( M ) = 3 and ind ( x ) = ind ( y ) = 2 . Suppose ξ is Lyapunov for X Proposition. and suppose K x ( y ) = ∅ for all x, y ∈ X . Then ρ ( X, ξ ) < ∞ . 6

S ⊆ W − Proof. small sphere. Consider the x parametrization via the flow of X ϕ : S × [0 , ∞ ) → W − x , ϕ ( x, t ) = ϕ t ( x ) g . . . Riem. metric on M , to be chosen later. g x . . . induced metric on W − x . τ ≥ 0 . . . to be chosen later. [ ω ] = ξ , ω ( X ) < 0 on M \ X . � S × [0 ,t ] e τh ω ψ : [0 , ∞ ) → R , ψ ( t ) := x vol g x Have to show lim t →∞ ψ ( t ) < ∞ . We are free to use any g and τ . One easily shows � ϕ t ( S ) e τh ω ψ ′ ( t ) = X vol g x x i ˜ and � e τh ω ψ ′′ ( t ) = � � X · h ω τ ˜ x + div g x ( ˜ X ) x i ˜ X vol g x ϕ t ( S ) and where ˜ X denotes the restriction of X to W − x . 7

Suppose we can choose g and τ s.t. � � X · h ω τ ˜ x + div g x ( ˜ X ) ≤ − ǫ (1) Then (ln ◦ ψ ′ ) ′ ( t ) ≤ − ǫ hence ψ ′ ( t ) ≤ ψ ′ (0) e − ǫt and integrating again ψ ( t ) ≤ ψ (0) + ψ ′ (0)(1 − e − ǫt ) /ǫ ≤ ψ ′ (0) /ǫ So e τh ω x ∈ L 1 ( W − x ) and thus ρ ( X, ξ ) ≤ τ < ∞ . Using K x ( y ) = ∅ we will construct g s.t. div g x ( ˜ X ) ≤ − ǫ (2) on an open nbh. U of X . Then we easily find τ satisfying (1) since X · h ω = ω ( X ) < 0 on M \ X x div g x ( ˜ tr g x ( ∇ ˜ X ) = X ) ind( x ) ||∇ ˜ ≤ X || g x ≤ ind( x ) ||∇ X || g 8

Let κ > 0 and choose chart s.t. x i ∂ x i ∂ � � X | U y = κ ∂x i − ∂x i i ≤ ind( y ) i> ind( y ) dx i ⊗ dx i � g | U y = i dx i ⊗ ∂ dx i ⊗ ∂ � � ∇ X | U y = ∂x i − κ ∂x i i ≤ ind( y ) i> ind( y ) div g x ( ˜ tr g x ( ∇ ˜ X ) = X ) � p ⊥ � x ◦ ∇ X ◦ i TW − = tr g TW − x ( V, g ) Euclidean vector space, V = Lemma. V + ⊕ V − orthogonal decomposition. For κ ≥ 0 consider A κ := ( κ id) ⊕ − id ∈ End( V ) and the function δ A κ : Gr q ( V ) → R , δ A κ ( W ) := tr g | W ( p ⊥ W ◦ A κ ◦ i W ) Suppose compact K ⊆ Gr q ( V ) with K ∩ Gr q ( V + ) = ∅ . Then there exists κ > 0 and ǫ > 0 s.t. δ A κ ≤ − ǫ on K . 9

Suppose ξ Lyapunov for X and Theorem. assume K x ( y ) = ∅ for all x, y ∈ X with 1 < ind( x ) ≤ ind( y ) < dim( M ). Then X has expo- nential growth, and ρ ( X, ξ ) < ∞ . If moreover X satisfies the Smale condition then I γ ( x, y ) e ξ ( γ ) � γ ∈P x,y converges for all x, y ∈ X . 10