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Trace riccati inequality

Statement

Assume the inherited Axiom of Countable Choice ${A C}_{ω}$ . Let $(M, g)$ be a Riemannian manifold of dimension $n \geq 2$ , let $γ : I \to M$ be a unit-speed geodesic with $0$ in the interior of $I$ , let $A$ be the radial Jacobi tensor of Radial Jacobi tensor, let $S = D_{t} A \circ A^{- 1}$ be the radial Riccati operator of Radial riccati equation on the interval $0 < t < τ$ , where $τ$ is the first conjugate instant of $γ (0)$ along $γ$ , and let $h (t) : = tr S (t)$ be its trace on the normal space $N_{t}$ . Then $h$ is differentiable and $h^{'} (t) + \frac{h (t)^{2}}{n - 1} + {Ric}_{γ (t)} (\dot{γ} (t), \dot{γ} (t)) \leq 0 (0 < t < τ) .$

In dimension $n = 2$ the normal space is one-dimensional and the inequality is an identity: $h^{'} = S^{'}$ and $h^{2} / (n - 1) = S^{2}$ . The endpoint $t = 0$ is excluded because $S$ is undefined there; here $S$ is defined only on $(0, τ)$ , regardless of whether the radial tensor becomes invertible again later. No choice beyond the inherited ${A C}_{ω}$ is used.

Facts & Assumptions

Given: The inherited ${A C}_{ω}$ of [A1], a Riemannian manifold $(M, g)$ of dimension $n \geq 2$ , a unit-speed geodesic $γ$ , the radial Jacobi tensor $A$ , the Riccati operator $S = D_{t} A \circ A^{- 1}$ defined on $0 < t < τ$ , and its trace $h = tr S$ .

[A1]

The countable-choice premise is the inherited ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried by the curvature-symmetry and Wronskian inputs of Radial riccati equation and by the Ricci-curvature interface [F2]; no further selection is made below.

[F1]

Riccati equation: on $0 < t < τ$ the operator $S$ is self-adjoint on the normal space and satisfies $S^{'} + S^{2} + R_{γ} = 0$ , where $R_{γ} (X) = R (X, \dot{γ}) \dot{γ}$ (Radial riccati equation). The normal space $N_{t}$ is the orthogonal complement of $\dot{γ} (t)$ in $T_{γ (t)} M$ , of dimension $n - 1$ (Radial Jacobi tensor).

[F2]

Ricci curvature: ${Ric}_{p} (X, Y) = tr (Z \mapsto R_{p} (Z, X) Y)$ is the basis-independent trace of that endomorphism (Ricci curvature, Ricci curvature is symmetric and basis independent). In particular $tr (Z \mapsto R (Z, T) T) = Ric (T, T)$ .

[F3]

Trace: for a finite-dimensional vector space the trace is the sum of the diagonal entries in any ordered basis, is basis-independent and is additive (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[F4]

Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal basis of eigenvectors, with real eigenvalues (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis).

[F5]

Cauchy–Schwarz: for real numbers $s_{1}, \dots, s_{m}$ , $(\sum_{i = 1}^{m} s_{i})^{2} \leq m \sum_{i = 1}^{m} s_{i}^{2}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Proof

technique · direct: trace the Riccati equation, identify the curvature trace with the Ricci curvature, and apply Cauchy–Schwarz to the real eigenvalues of the self-adjoint operator $S$

1.1F1F2F3given

Tracing the Riccati equation. [F1, F2, F3, given] Differentiating the trace and using the Riccati equation of [F1], $h^{'} = (tr S)^{'} = tr (S^{'}) = tr (- S^{2} - R_{γ}) = - tr (S^{2}) - tr (R_{γ}),$ the second equality because the trace is linear and basis-independent and the derivative acts entrywise in any fixed basis of the normal space [F3]. The operator $R_{γ}$ is $Z \mapsto R (Z, \dot{γ}) \dot{γ}$ , so by [F2] $tr (R_{γ}) = tr (Z \mapsto R (Z, \dot{γ}) \dot{γ}) = Ric (\dot{γ}, \dot{γ}) .$ Hence $h^{'} + tr (S^{2}) + Ric (\dot{γ}, \dot{γ}) = 0.$

1.2F4F5F1given

The Cauchy–Schwarz bound on the quadratic trace. [F4, F5, F1, given] By [F1] the operator $S (t)$ is self-adjoint on the $(n - 1)$ -dimensional inner product space $N_{t}$ , so by [F4] there is an orthonormal basis $e_{1}, \dots, e_{n - 1}$ of $N_{t}$ with $S e_{i} = s_{i} e_{i}$ and $s_{i} \in R$ . In this basis, which is orthonormal and hence admissible for the trace of [F3], $h = tr S = \sum_{i = 1}^{n - 1} s_{i}, tr (S^{2}) = \sum_{i = 1}^{n - 1} ⟨ S^{2} e_{i}, e_{i} ⟩ = \sum_{i = 1}^{n - 1} s_{i}^{2} .$ Applying [F5] with $m = n - 1$ to the real numbers $s_{1}, \dots, s_{n - 1}$ gives $h^{2} = (\sum_{i = 1}^{n - 1} s_{i})^{2} \leq (n - 1) \sum_{i = 1}^{n - 1} s_{i}^{2} = (n - 1) tr (S^{2}),$ that is $tr (S^{2}) \geq h^{2} / (n - 1)$ , since $n - 1 \geq 1 > 0$ .

2.1step 1.1step 1.2given∎

Conclusion. [step 1.1, step 1.2, given] Substituting the bound of step 1.2 into the identity of step 1.1, $h^{'} = - tr (S^{2}) - Ric (\dot{γ}, \dot{γ}) \leq - \frac{h^{2}}{n - 1} - Ric (\dot{γ}, \dot{γ}),$ which is the asserted inequality $h^{'} + h^{2} / (n - 1) + Ric (\dot{γ}, \dot{γ}) \leq 0$ on $0 < t < τ$ . In dimension $n = 2$ the sum in [F5] has the single term $s_{1}$ , so the Cauchy–Schwarz step is an equality: $h^{'} = - h^{2} - Ric (\dot{γ}, \dot{γ})$ . At $t = 0$ the operator $S$ is not defined, and its domain here is $(0, τ)$ ; the radial tensor may become invertible again later. Neither completeness of $M$ nor any choice beyond [A1] enters.

Source locator

Eschenburg §4 (printed p.15) traces the Riccati equation $A^{'} + A^{2} + R_{V} = 0$ to $trace (A)^{'} + trace (A^{2}) + Ric (V) = 0$ , the starting point of the average comparison theorems; the step from this identity to the scalar inequality $h^{'} + h^{2} / (n - 1) + Ric \leq 0$ is the $a = trace (A) / (n - 1)$ substitution of the same section, whose Cauchy–Schwarz input is the one proved above. Datar §§26.1–26.2 and §28.1, pp.191–197 and 205–209, contains the same Riccati calculus. The proof above is carried out from the in-run Riccati equation and the published Ricci-curvature suppliers.

Depends on

Used by

Laplacian comparison for distance under a ricci lower bound Theorem
Relative volume density comparison Theorem

Dependency tree · two levels

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Sources

J.-H. Eschenburg, Comparison Theorems in Riemannian Geometry (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry (2025) (standard reference, not scraped)