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 = −gradg(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: ω = −

• i≤k

xidxi +

• i>k

xidxi g =

• i

dxi ⊗ dxi X =

• i≤k

xi ∂ ∂xi −

• i>k

xi ∂ ∂xi ˆ ix : ˆ W −

x → M . . . completed unstable manifold

(ˆ ix)∗ω = dˆ hω

x and ˆ

hω

x(x) = 0.

ˆ hω

x : ˆ

W −

x → R is proper

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ρ(X, [ω]) := inf

• s ≥ 0
• eshω

x ∈ L1(W −

x ) ∀x ∈ X

• Suppose ρ(X, [ω]) < ∞

If α ∈ Ω∗(M; R) we have absolutely converging Ints(α)(x) :=

• ˆ

W −

x

esˆ

hω

x(ˆ

ix)∗α, ℜ(s) > ρ(X, [ω]) Ints : Ω∗(M; R) → Maps(X ∗, R) Integration is onto. ∂1 ˆ W −

x = y T (x, y) × W − y

and Stokes theorem applied to dsα = dα+sω ∧α imply convergence

• f

˜ Is(x, y) :=

• γ∈Px,y

Iγ(x, y)esω(γ), ℜ(s) > ρ(X, [ω]) ∂s : Maps(X ∗; R) → Maps(X ∗+1; R) (∂sf)(x) :=

• y

˜ I(x, y)f(y) Int ◦ ds = ∂s ◦ Int ∂2

s = 0

2

Lyapunov classes: L(X) =

• [ω] ∈ H1(M; R)
• ∃g : X = −gradg(ω)
• Lemma.

L(X) ⊆ H1(M; R) is open, convex cone.

• Proof. Indeed, if X = −gradg0(ω) on nbh. of

X and ω(X) < 0 on M \ X then ∃g with g = g0 near X and X = −gradg(ω). Proposition. If ξ0 is Lyapunov for X and ρ(X, ξ0) < ∞ then ρ(X, ξ) < ∞ for all ξ which are Lyapunov for X.

• Proof. ξ0 ∈ L(X) and ρ(X, ξ0) < ∞. Every ray

in L(X) intersects ξ0+L(X) since L(X) is open

• cone. Hence sξ = ξ0 + [ω] ∈ ξ0 + L(X). But

|ehω

x| ≤ 1 if ω(X) ≤ 0 hence ρ(X, sξ) ≤ ρ(X, ξ0).

• Remark. H¨
• lder’s inequality implies that

ρ(X, tξ1 + (1 − t)ξ0) ≤ max{ρ(X, ξ0), ρ(X, ξ1)}

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ix : W −

x → M, gx := (ix)∗g

r = distgx(x, ·) : W −

x → [0, ∞)

Bs(x) ⊆ W −

x ball of radius s centered at x.

X is said to have exponential growth if ∃C ≥ 0 volgx(Bs(x)) ≤ eCs, ∀s ≥ 0 ∀x ∈ X. This property is independent of g.

• Lemma. The following are equivalent:

(i) X has exponential growth with respect to

• ne (and hence every) Riemannian metric.

(ii) For one (and hence every) Riemannian met- ric ∃C ≥ 0 with e−Cr ∈ L1(W −

x ).

4

• Lemma. ∃C1 ≥ 0 s.t. r ≤ C1(1 + |hω

x|) on W − x .

• Lemma. ∃C2 ≥ 0 s.t. |hω

x| ≤ C2r on W − x .

• Proof. γ path in W −

x , γ(0) = x. Then:

|h(γ(1))| =

• 1

0 (dh)(γ′(t))dt

• ≤ ||ω||∞length(γ)

So |h(γ(1))| ≤ Cdist(x, γ(1)), C2 := ||ω||∞. Proposition. ξ Lyapunov for X. Then the following are equivalent: (i) X has exponential growth with respect to

• ne (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?

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x ∈ X q, ind(x) = q B ⊆ W −

x small ball centered at x

Gr(W −

x \ B) ⊆ Grq(TM)

For y ∈ X define: Kx(y) := Grq(TyW −

y ) ∩ Gr(W − x \ B)

“If Kx(y) is non-empty then W −

x

badly accu- mulates at y.” (-) Kx(y) does not depend on B. (-) We may have Kx(x) = ∅. (-) If q = ind(x) > ind(y) then Kx(y) = ∅. (-) If ind(x) = dim(M) then Kx(y) = ∅ ∀y ∈ X. (-) First non-trivial case is dim(M) = 3 and ind(x) = ind(y) = 2. Proposition. Suppose ξ is Lyapunov for X and suppose Kx(y) = ∅ for all x, y ∈ X. Then ρ(X, ξ) < ∞.

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Proof. S ⊆ W −

x

small sphere. Consider the parametrization via the flow of X ϕ : S × [0, ∞) → W −

x ,

ϕ(x, t) = ϕt(x) g . . . Riem. metric on M, to be chosen later. gx . . . induced metric on W −

x .

τ ≥ 0 . . . to be chosen later. [ω] = ξ, ω(X) < 0 on M \ X. ψ : [0, ∞) → R, ψ(t) :=

• S×[0,t] eτhω

xvolgx

Have to show limt→∞ ψ(t) < ∞. We are free to use any g and τ. One easily shows ψ′(t) =

• ϕt(S) eτhω

xi ˜

Xvolgx

and ψ′′(t) =

• ϕt(S)
• τ ˜

X · hω

x + divgx( ˜

X)

• eτhω

xi ˜

Xvolgx

and where ˜ X denotes the restriction of X to W −

x .

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Suppose we can choose g and τ s.t.

• τ ˜

X · hω

x + divgx( ˜

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 eτhω

x ∈ L1(W −

x ) and thus ρ(X, ξ) ≤ τ < ∞.

Using Kx(y) = ∅ we will construct g s.t. divgx( ˜ X) ≤ −ǫ (2)

• n an open nbh. U of X. Then we easily find

τ satisfying (1) since X · hω

x

= ω(X) < 0

• n M \ X

divgx( ˜ X) = trgx(∇ ˜ X) ≤ ind(x)||∇ ˜ X||gx ≤ ind(x)||∇X||g

8

Let κ > 0 and choose chart s.t. X|Uy = κ

• i≤ind(y)

xi ∂ ∂xi −

• i>ind(y)

xi ∂ ∂xi g|Uy =

• i

dxi ⊗ dxi ∇X|Uy = κ

• i≤ind(y)

dxi ⊗ ∂ ∂xi −

• i>ind(y)

dxi ⊗ ∂ ∂xi divgx( ˜ X) = trgx(∇ ˜ X) = trg

• p⊥

TW −

x ◦ ∇X ◦ iTW − x

• Lemma.

(V, g) Euclidean vector space, V = V + ⊕ V − orthogonal decomposition. For κ ≥ 0 consider Aκ := (κid) ⊕ −id ∈ End(V ) and the function δAκ : Grq(V ) → R, δAκ(W) := trg|W (p⊥

W ◦ Aκ ◦ iW)

Suppose compact K ⊆ Grq(V ) with K∩Grq(V +) = ∅. Then there exists κ > 0 and ǫ > 0 s.t. δAκ ≤ −ǫ on K.

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Theorem. Suppose ξ Lyapunov for X and assume Kx(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

• γ∈Px,y

Iγ(x, y)eξ(γ) converges for all x, y ∈ X.

10