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Bishop gromov volume ratio is nondecreasing under a ricci lower bound

Statement

Assume the inherited Axiom of Countable Choice ${A C}_{ω}$ . False claim: if $(M, g)$ is a complete, connected, boundaryless Riemannian manifold of dimension $n \geq 2$ with $Ric \geq (n - 1) k g$ for a real number $k$ , $p \in M$ and $V_{k}^{⋆}$ is the saturated model ball volume, then the Bishop–Gromov ratio $R_{p} (r) = \frac{{vol}_{g} (B (p, r))}{V_{k}^{⋆} (r)}, r > 0,$ is nondecreasing on $(0, \infty)$ . The claim is refuted below by an explicit manifold satisfying the Ricci lower bound on which the ratio strictly decreases; the correct statement is the opposite monotonicity, as asserted by Bishop gromov volume comparison.

Facts & Assumptions

Given: The inherited ${A C}_{ω}$ of [A1]; the unit round sphere $S_{1}^{n} \subseteq R^{n + 1}$ of dimension $n \geq 2$ with the induced Riemannian metric $g$ ; a point $p \in S_{1}^{n}$ ; the constant $k = 0$ and the saturated model volume $V_{0}^{⋆}$ ; and the Bishop–Gromov ratio $R_{p} (r) = {vol}_{g} (B (p, r)) / V_{0}^{⋆} (r)$ of the false claim.

[A1]

The countable-choice premise is the inherited ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried by the sphere, cut-locus and Bishop–Gromov suppliers below.

[F1]

Round sphere (The round sphere has positive constant sectional curvature, Round sphere model geometry, Curvature tensor of constant sectional curvature, Ricci curvature): for $k = 1$ and radius $R = 1$ the sphere $S_{1}^{n}$ has constant sectional curvature $1$ , so the curvature four-tensor is $Rm (X, Y, Z, W) = g (Y, Z) g (X, W) - g (X, Z) g (Y, W)$ ; tracing $Ric (X, Y) = \sum_{i} Rm (e_{i}, X, Y, e_{i})$ in an orthonormal basis $(e_{1}, \dots, e_{n})$ gives $Ric (X, Y) = \sum_{i} (g (X, Y) g (e_{i}, e_{i}) - g (e_{i}, Y) g (X, e_{i})) = (n - 1) g (X, Y)$ , that is $Ric = (n - 1) g$ . By the round-sphere model proposition the sphere $S_{1}^{n}$ is complete, connected and boundaryless of dimension $n \geq 2$ , with diameter $π$ . In particular $Ric = (n - 1) g \geq 0 = (n - 1) \cdot 0 \cdot g$ , so the Ricci lower bound of the false claim holds with $k = 0$ , and the centre $p$ is unrestricted.

[F2]

Diameter and large balls (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): every $q \in S_{1}^{n}$ satisfies $d_{g} (p, q) \leq diam (S_{1}^{n}, g) = π$ , and therefore $B (p, r) = S_{1}^{n}$ for every $r > π$ .

[F3]

Finite volume (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, Riemannian volume is the radon measure of the riemannian density, Riemannian volume density): $S_{1}^{n}$ is a closed and bounded subset of $R^{n + 1}$ , hence compact; the Riemannian volume ${vol}_{g}$ is compact-finite, so ${vol}_{g} (S_{1}^{n}) < + \infty$ . Consequently ${vol}_{g} (B (p, r)) \leq {vol}_{g} (S_{1}^{n}) < + \infty$ for every $r > 0$ .

[F4]

Bishop–Gromov with $k = 0$ (Bishop gromov volume comparison): on $S_{1}^{n}$ , which satisfies $Ric \geq 0$ , the ratio $R_{p}$ is well defined on $(0, \infty)$ , is nonincreasing there, and $\lim_{r ↓ 0} R_{p} (r) = 1$ .

[F5]

Model volume for $k = 0$ (Model space radial area and ball volume, Comparison sine, cosine and cotangent functions): $V_{0}^{⋆} (r) = V_{0} (r) = ω_{n - 1} \int_{0}^{r} {sn}_{0} (t)^{n - 1} d t = ω_{n - 1} \int_{0}^{r} t^{n - 1} d t$ with $ω_{n - 1} > 0$ the total surface measure of the unit sphere; for $r \geq 1$ the integrand satisfies $t^{n - 1} \geq 1$ , so $V_{0}^{⋆} (r) \geq ω_{n - 1} (r - 1)$ and $V_{0}^{⋆} (r) \to + \infty$ as $r \to + \infty$ .

Refutation

Proof technique: direct: take the unit round sphere with $k = 0$ , where the Ricci lower bound holds; the ratio tends to one at the origin but to zero at infinity because the ball volume saturates at the finite total volume while the Euclidean model volume grows without bound; two such radii violate monotonicity.

1.1F1F3F5

The witness satisfies the hypothesis. [F1] The unit sphere $S_{1}^{n}$ is complete, connected, boundaryless and of dimension $n \geq 2$ , and its Ricci curvature is $Ric = (n - 1) g \geq 0$ , which is the lower bound $Ric \geq (n - 1) k g$ at $k = 0$ [F1]. The ratio $R_{p} (r) = {vol}_{g} (B (p, r)) / V_{0}^{⋆} (r)$ is defined for every $r > 0$ because $V_{0}^{⋆} (r) > 0$ [F5] and the ball volume is finite [F3].

1.2F2F3F4F5

The ratio tends to $1$ at the origin and to $0$ at infinity. [F2, F3, F4, F5] Near the origin, $\lim_{r ↓ 0} R_{p} (r) = 1$ by [F4]. At infinity, [F2] gives $B (p, r) = S_{1}^{n}$ for every $r > π$ , so $R_{p} (r) = \frac{{vol}_{g} (S_{1}^{n})}{V_{0}^{⋆} (r)} (r > π);$ here the numerator is finite by [F3] and the denominator satisfies $V_{0}^{⋆} (r) \geq ω_{n - 1} (r - 1) \to + \infty$ by [F5], whence $\lim_{r \to + \infty} R_{p} (r) = 0$ .

2.1F4step 1.1step 1.2∎

The ratio is not nondecreasing. [step 1.2] Choose $r_{1} > 0$ with $R_{p} (r_{1}) > \frac{1}{2}$ , possible because $\lim_{r ↓ 0} R_{p} (r) = 1$ ; then choose $r_{2} > \max {r_{1}, π}$ with $R_{p} (r_{2}) < \frac{1}{2}$ , possible because $\lim_{r \to + \infty} R_{p} (r) = 0$ . Then $0 < r_{1} < r_{2}$ while $R_{p} (r_{1}) > R_{p} (r_{2})$ , so the ratio is not nondecreasing on $(0, \infty)$ . The manifold satisfies every hypothesis of the false claim (step 1.1), so the claim is false; the correct monotonicity is the nonincreasing one asserted by [F4], and the failure here is precisely the saturation of the ball volume at the finite total volume of the compact sphere against the unbounded Euclidean model volume. No direction or radius family beyond the two explicit radii $r_{1}, r_{2}$ is selected, so the inherited ${A C}_{ω}$ of [A1] is not drawn on beyond its declaration.

Source locator

Datar §§27.2 and 28.1, pp.200–209, and Eschenburg §§4–5, pp.15–20, prove that the Bishop–Gromov ratio is nonincreasing, with value one at the origin and saturation once the ball exhausts the manifold. The witness is the unit round sphere at $k = 0$ , read off from the published round-sphere curvature and the round-sphere model proposition recording completeness and diameter $π$ ; the Ricci trace is computed above from the constant-curvature tensor identity.

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Sources

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