LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-24 (gpt-6-sol)

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Fredholm index and range of an asymptotically hyperbolic first-order operator

Statement

Let $A : R \to M_{d} (R)$ be continuous with limits $A_{\pm}$ as $t \to \pm \infty$ . Suppose each limit is self-adjoint for some positive-definite inner product and has no zero eigenvalue. Put $D_{A} : C_{0}^{1} (R, R^{d}) \to C_{0}^{0} (R, R^{d}), D_{A} u = u^{'} - A (t) u .$ Let $E_{-}^{u}$ be the initial values at $0$ of homogeneous solutions decaying as $t \to - \infty$ , and let $E_{+}^{s}$ be the corresponding values for solutions decaying as $t \to + \infty$ . Then $D_{A}$ is bounded Fredholm, $\ker D_{A} ≅ E_{-}^{u} \cap E_{+}^{s}, coker D_{A} ≅ R^{d} / (E_{-}^{u} + E_{+}^{s}),$ and $ind D_{A} = \dim E^{u} (A_{-}) - \dim E^{u} (A_{+}),$ where $E^{u} (A_{\pm})$ denotes the positive eigenspace of the limiting matrix. In particular $D_{A}$ is onto exactly when $E_{-}^{u} + E_{+}^{s} = R^{d}$ . The isomorphism for the cokernel is induced by the half-line right inverses constructed in the proof; no canonical identification with a tangent quotient is asserted.

Facts & Assumptions

Given: The matrix path, limits, and function spaces in the statement.

[F1]

On each half-line the restricted operator has a bounded right inverse, the decaying homogeneous initial-value space has the indicated spectral dimension, and right-inverse values of forcings vanishing at $0$ span modulo that space (A half-line first-order operator with a hyperbolic limit has a right inverse).

[F2]

Homogeneous linear matrix equations have unique solutions on finite intervals (Linear matrix ODEs have unique global solutions on a fixed interval).

Proof

technique · direct

1.1

The convergence of $A (t)$ at both ends makes it bounded, so $D_{A}$ is a bounded map from the indicated supremum $C^{1}$ norm to the supremum $C^{0}$ norm. Apply [F1] to $A ∣_{[0, \infty)}$ and, after time reversal, to $A ∣_{(- \infty, 0]}$ . Denote the bounded right inverses by $R_{+}, R_{-}$ . The two homogeneous initial-value spaces are $E_{+}^{s}, E_{-}^{u}$ , with $\dim E_{+}^{s} = d - \dim E^{u} (A_{+})$ and $\dim E_{-}^{u} = \dim E^{u} (A_{-})$ .

F1given

2.1

A homogeneous whole-line solution is determined by its value at zero by [F2] and belongs to $C_{0}^{1}$ exactly when that value lies in both $E_{-}^{u}$ and $E_{+}^{s}$ . These spaces are finite-dimensional, hence closed, and the evaluation map gives $\ker D_{A} ≅ E_{-}^{u} \cap E_{+}^{s}$ .

F2step 1.1algebra

2.2

For $v \in C_{0}^{0} (R, R^{d})$ , write $v_{\pm}$ for its half-line restrictions and define the bounded linear map $T (v) = [R_{+} v_{+} (0) - R_{-} v_{-} (0)] \in R^{d} / (E_{-}^{u} + E_{+}^{s}) .$ Every decaying solution on the positive half-line has the form $R_{+} v_{+} + u_{+}$ , with $u_{+} (0) \in E_{+}^{s}$ ; the analogous form on the negative half-line has $u_{-} (0) \in E_{-}^{u}$ . The two half-line solutions can be matched at zero exactly when $T (v) = 0$ . When matched, their first derivatives also agree there because each satisfies $u^{'} = A u + v$ and $v$ is continuous. Thus $ran D_{A} = \ker T$ , which is closed.

F1F2step 1.1algebra

3.1

The map $T$ is onto. By [F1], $E_{+}^{s}$ together with ${R_{+} w (0) : w (0) = 0}$ spans $R^{d}$ . Extend any such $w$ by zero to the negative half-line; the extension is continuous at zero, belongs to $C_{0}^{0} (R)$ , and has $R_{-} v_{-} (0) = 0$ . Its $T$ -images therefore span the quotient by $E_{-}^{u} + E_{+}^{s}$ . Consequently $C_{0}^{0} / ran D_{A} ≅ R^{d} / (E_{-}^{u} + E_{+}^{s})$ , a finite-dimensional space.

F1step 2.2algebra

4.1

Set $a = \dim E_{-}^{u}$ , $b = \dim E_{+}^{s}$ , and $c = \dim (E_{-}^{u} \cap E_{+}^{s})$ . Steps 2.1 and 3.1 give $\dim \ker D_{A} = c$ and $codim ran D_{A} = d - (a + b - c)$ . Hence $D_{A}$ is Fredholm and $ind D_{A} = c - [d - a - b + c] = a + b - d = \dim E^{u} (A_{-}) - \dim E^{u} (A_{+})$ by step 1.1. The quotient in step 3.1 vanishes exactly when the two initial-value spaces span $R^{d}$ .

step 1.1step 2.1step 3.1algebra∎

Depends on

Used by

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

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alberto Abbondandolo and Pietro Majer, Lectures on the Morse Complex, Proposition 1.8 and proof, pp. 43-44 (standard reference, not scraped)