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Complete noncompact manifolds with nonnegative ricci curvature have at most euclidean volume growth

Statement

Assume the inherited Axiom of Countable Choice ${A C}_{ω}$ . Let $(M, g)$ be a complete, connected, noncompact, boundaryless Riemannian manifold of dimension $n \geq 2$ whose Ricci curvature satisfies $Ric \geq 0$ , let $p \in M$ , and let $B (p, r) = {q \in M : d_{g} (p, q) < r}$ be the open metric ball. Then ${vol}_{g} (B (p, r)) \leq \frac{ω_{n - 1} r^{n}}{n} for every r > 0,$ where $ω_{n - 1}$ is the total surface measure of the unit sphere $S^{n - 1} \subseteq R^{n}$ of Model space radial area and ball volume. In other words, a complete noncompact manifold of nonnegative Ricci curvature has at most Euclidean volume growth. Neither compactness nor noncompactness enters the inequality itself: the noncompactness in the hypothesis records the situation to which the growth statement is applied, and the bound holds for every complete, connected, boundaryless manifold of dimension $n \geq 2$ with $Ric \geq 0$ . No choice beyond the inherited ${A C}_{ω}$ is used.

Facts & Assumptions

Given: The inherited ${A C}_{ω}$ of [A1]; a complete, connected, noncompact Riemannian manifold $(M, g)$ of dimension $n \geq 2$ with $Ric \geq 0$ ; a point $p \in M$ ; the ball volume ${vol}_{g} (B (p, r))$ ; and the model functions ${sn}_{0}$ , $A_{0}$ , $V_{0}$ and $V_{0}^{⋆}$ of the flat model.

[A1]

The countable-choice premise is the inherited ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried by the Bishop–Gromov comparison and the model-volume interface cited below; no further selection is made.

[F1]

Bishop volume upper bound (Bishop volume upper bound): if $(M, g)$ is complete, connected and boundaryless of dimension $n \geq 2$ with $Ric \geq (n - 1) k g$ for a real number $k$ , then ${vol}_{g} (B (p, r)) \leq V_{k}^{⋆} (r)$ for every $r > 0$ . No compactness of $M$ is assumed.

[F2]

Flat model volumes (Model space radial area and ball volume, Comparison sine, cosine and cotangent functions): the comparison sine is ${sn}_{0} (t) = t$ , the model radial area is $A_{0} (t) = ω_{n - 1} {sn}_{0} (t)^{n - 1} = ω_{n - 1} t^{n - 1}$ , the model ball volume is $V_{0} (r) = \int_{0}^{r} A_{0} (t) d t$ , and for $k = 0$ the saturated model volume of [F1] is $V_{0}^{⋆} (r) = V_{0} (r)$ for every $r \geq 0$ .

[F3]

Power integral (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion): for every integer $m \geq 0$ and every $0 \leq s \leq r$ , $\int_{s}^{r} t^{m} d t = \frac{r^{m + 1} - s^{m + 1}}{m + 1},$ because $t \mapsto t^{m + 1} / (m + 1)$ has derivative $t^{m}$ and $t^{m}$ is continuous, hence integrable, on $[s, r]$ .

Proof

technique · direct: specialise the Bishop upper bound to $k=0$, where $\operatorname{Ric}\ge0=(n-1)\cdot0\cdot g$, and evaluate the integral defining the flat model volume

1.1F1F2given

The Bishop bound with $k = 0$ . [F1, F2, given] The Ricci hypothesis $Ric \geq 0$ is exactly $Ric \geq (n - 1) k g$ for $k = 0$ , and $(M, g)$ is complete, connected and boundaryless of dimension $n \geq 2$ . Hence [F1] with $k = 0$ gives ${vol}_{g} (B (p, r)) \leq V_{0}^{⋆} (r) (r > 0),$ and by [F2] the flat saturated model volume is the integral $V_{0}^{⋆} (r) = V_{0} (r) = \int_{0}^{r} A_{0} (t) d t = ω_{n - 1} \int_{0}^{r} t^{n - 1} d t .$

2.1F2F3step 1.1∎

Evaluation and conclusion. [F2, F3, step 1.1] The power integral [F3] with $m = n - 1 \geq 1$ and $s = 0$ gives $\int_{0}^{r} t^{n - 1} d t = r^{n} / n$ . Substituting into step 1.1 yields ${vol}_{g} (B (p, r)) \leq V_{0}^{⋆} (r) = \frac{ω_{n - 1} r^{n}}{n}, r > 0,$ which is the asserted Euclidean-growth bound. Two boundary conventions should be noted: the value $r = 0$ is excluded, where $B (p, 0) = \emptyset$ and $V_{0} (0) = 0$ ; and the dimension hypothesis $n \geq 2$ makes the exponent $n - 1 \geq 1$ , so the integrand is the continuous monomial $t^{n - 1}$ of [F3]. The noncompactness hypothesis is not used in the argument, so the bound also holds in the compact case. The comparison parameter $k = 0$ is supplied directly by $Ric \geq 0$ ; no positive uniform Ricci lower bound is assumed. No direction, ball, chart or family is selected anywhere: the argument is the single specialisation $k = 0$ of [F1] together with an explicit integral, and the only choice principle used is the inherited [A1].

Source locator

Datar §§27.2 and 28.1, pp.200–209, and Eschenburg §§4–5, pp.15–20, give the Bishop–Gromov upper bound by the model ball volume; the flat model volume is the integral of $ω_{n - 1} t^{n - 1}$ , namely $ω_{n - 1} r^{n} / n$ . The content of this corollary is that specialisation, together with the observation that the inequality itself does not use noncompactness.

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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)