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Morse--Smale transversality and surjectivity of the linearized flow operator

Statement

Let f be Morse and let γ connect critical points p,q for X=gradgf. Let Eγ=C01(R,γTM) be the Banach space of C1 sections for which both ξ and tξ tend to zero at ±, with the supremum C1 norm, and let Fγ=C00(R,γTM) have the supremum norm. For the tangent Levi--Civita connection metric-dual to the cotangent connection, put

Dγξ=tξ+ξgradgf.

Then Dγ:EγFγ is bounded Fredholm of index λ(p)λ(q), and it is surjective if and only if Wu(p) and Ws(q) are transverse at γ(0).

Facts & Assumptions

Given: A Morse function f, a connecting orbit γ between its critical points p,q, and the displayed C01 and C00 Banach spaces.

[F1]

Covariant Hessians define the linearization of the gradient equation (Riemannian metrics, symmetric cotangent-bundle connections, and covariant Hessians).

[F2]

The point-marked trajectory space is the stable--unstable intersection (Parametrized Morse trajectory space).

Proof

technique · direct
1.1

Differentiating γ˙+gradgf(γ)=0 in a decaying variation gives the displayed operator, by [F1]. Its kernel is the tangent space of the point-marked solution set.

F1given
2.1

The hyperbolic Hessians at p and q give exponential dichotomies at the two ends. The standard first-order Fredholm theorem therefore gives index λ(p)λ(q). Its adjoint solvability condition identifies the dual cokernel with the annihilator of Tγ(0)Wu(p)+Tγ(0)Ws(q), rather than canonically identifying the cokernel itself with a tangent-space quotient.

F2step 1.1
3.1

That annihilator vanishes exactly when the two tangent spaces span Tγ(0)M, which is transversality. Since a Fredholm operator is onto exactly when its cokernel vanishes, this proves the biconditional.

step 2.1algebra

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