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A critical cluster point for a compact gradient trajectory

Statement

Assume countable choice. Let $(M, g)$ be a nonempty connected compact manifold with analytic potential $U$ and smooth Riemannian metric as in Transfer of the analytic gradient inequality to a smooth metric. Every downward gradient trajectory $q$ , satisfying $q^{'} = - {grad}_{g} U$ , exists for all forward times. Set $E (t) = U (q (t))$ . There are $L \in R$ , $p \in M$ and times $t_{j} \to \infty$ such that $E (t) ↓ L$ , $q (t_{j}) \to p$ , $U (p) = L$ and ${grad}_{g} U (p) = 0$ .

Facts & Assumptions

[F1]

Smooth vector fields on compact manifolds are complete under countable choice. Riemannian distance is a finite metric on a connected manifold (Riemannian distance is a metric), induces its topology and is at most the length of any joining curve. Compact metric spaces are sequentially compact, and continuous real functions on them are bounded. Gradient dissipation gives $E^{'} = - ∥ q^{'} ∥_{g}^{2}$ and, for energy bounded below, a finite limit $L$ and $\int_{0}^{\infty} ∥ q^{'} ∥_{g}^{2} d t = E (0) - L$ .

[F2]

A continuous real function on a closed nondegenerate real interval is Riemann integrable; its integral is additive over adjacent intervals and is nonnegative when the integrand is nonnegative (Continuous functions have an oriented Riemann integral).

[F3]

If $f$ is continuous on $[a, b]$ with $a < b$ , the first mean-value theorem for integrals with the nonnegative integrable weight identically one gives $s \in [a, b]$ such that $\int_{a}^{b} f (t) d t = f (s) (b - a)$ (If f is continuous on [a,b] and g is integrable with g ≥ 0, there is ξ ∈ [a,b] with ∫ₐᵇ fg = f(ξ)∫ₐᵇ g).

[F9]

Exact suppliers: Transfer of the analytic gradient inequality to a smooth metric, Energy dissipation along a downward gradient curve, Every smooth vector field on a compact manifold is complete, The Axiom of Countable Choice (AC_ω), A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle, Riemannian distance on a connected manifold, Riemannian distance is a metric, The riemannian distance topology is the manifold topology, Riemannian speed and length.

Proof

Given: The data and hypotheses of the Statement.

Goal: Establish its conclusion.

1.1F1F9given

The smooth metric and analytic potential give a smooth vector field $- {grad}_{g} U$ . Compact-vector-field completeness gives its global trajectories. The metric theorem makes $d_{g}$ a finite metric, and the distance-topology theorem makes $(M, d_{g})$ a compact metric space. Boundedness below of $U$ and dissipation therefore supply $L$ and the finite squared-speed integral.

2.1F2step 1.1algebrachoose

Put $f (t) = ∥ q^{'} (t) ∥_{g}^{2}$ , a continuous nonnegative function, and $I (T) = \int_{0}^{T} f (t) d t$ . By step 1.1, $I (T) \to I : = E (0) - L$ as $T \to \infty$ . For each integer $n \geq 0$ , set $a_{n} = \int_{n}^{n + 1} f (t) d t$ . Additivity and nonnegativity give $a_{n} = I (n + 1) - I (n) \geq 0$ . Given $ε > 0$ , choose $N$ so $∣ I (T) - I ∣ < ε / 2$ for $T \geq N$ . For integers $n \geq N$ , $0 \leq a_{n} \leq ∣ I (n + 1) - I ∣ + ∣ I (n) - I ∣ < ε$ , so $a_{n} \to 0$ .

3.1F3F9step 2.1choose

On each interval $[n, n + 1]$ , $f$ is continuous and the constant weight one is integrable and nonnegative. Applying [F3] with $a = n$ , $b = n + 1$ gives a nonempty set of $s_{n} \in [n, n + 1]$ for which $f (s_{n}) = a_{n}$ . Countable choice permits these selections simultaneously for all integers $n \geq 0$ . Thus $∥ q^{'} (s_{n}) ∥_{g}^{2} \to 0$ .

4.1F1F9step 1.1step 3.1choose∎

Sequential compactness supplies a subsequence $q (s_{n_{j}}) \to p$ . The subsequence indices are strictly increasing, hence $n_{j} \to \infty$ , and $t_{j} : = s_{n_{j}} \geq n_{j}$ gives $t_{j} \to \infty$ . Continuity of $U$ and $E (t) \to L$ give $U (p) = L$ . Continuity of the scalar function $x \mapsto ∥ {grad}_{g} U (x) ∥_{g}^{2}$ , the gradient equation, and step 3.1 give $∥ {grad}_{g} U (p) ∥_{g}^{2} = 0$ . Positive definiteness therefore gives ${grad}_{g} U (p) = 0$ , as required.

Depends on

Used by · 2

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