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For a complete connected $n$-dimensional Riemannian manifold with nonnegative Ricci curvature, the displayed ratio is nonincreasing, tends to one at zero, and is at most one. In particular $\mathrm{Vol}B(p,r)\le {\omega }_n{r}^n$. In geodesic polar coordinates, the radial Jacobi field determinant is at most the Euclidean density $r^{n-1}$ until the cut time. The trace of the radial Riccati equation for distance spheres, with Cauchy-Schwarz inequality, proves this comparison. Equality for every radius forces the Euclidean polar metric and absence of a cut locus.

The Bishop-Gromov inequality forces every volume ratio to equal one if the asymptotic ratio is one. Equality in the radial Jacobian comparison forces the radial operator $A(X)={\nabla }_X{\partial }_r$ of each distance sphere to be $r^{-1}I$ (it is the negative of the outward shape operator in the convention $S=-\nabla \nu$), by the traced radial Riccati equation for distance spheres and equality in Cauchy-Schwarz inequality. Any finite cut time would remove a positive cone of radial volume at a later radius, contradicting equality. Thus the Riemannian exponential map is global and one-to-one, and the polar metric is $d{r}^2+r^2{g}_{S^{n-1}}$, the Euclidean metric.

Use the intended trace bound $\mathrm{Ric}\ge (n-1)kg$ and write $D=\pi /\sqrt{k}$. We state precisely the allowed Bishop volume comparison with a positive-curvature model. Setwhere ${\omega }_{n-1}$ is the area of the unit $(n-1)$-sphere. For $V_p(r)=\mathrm{Vol}B(p,r)$, the ratio $V_p(r)/v_k(r)$ is nonincreasing on $0<r<D$, tends to one at zero, and is at most one. Before the cut time, the radial volume density $J(r,\theta )$ in geodesic polar coordinates is at most $s_k(r{)}^{n-1}$.

By the Bonnet-Myers theorem, $\mathrm{diam}M\le D$. Therefore $V_p(D)=\mathrm{Vol}M$. If distance-$D$ endpoints exist, they do not affect this volume equality: the distance sphere is contained in the smooth image ${\exp }_p(D{S}_p^{n-1})$, which has zero $n$-dimensional Riemannian volume. Taking limits in the comparison ratio at $D$ is legitimate. The assumed equality of total volume gives $V_p(D)/v_k(D)=1$, so monotonicity forcesfor every $p$.

We now extract the equality geometry, rather than merely naming a rigidity theorem. On a small normal neighborhood of $p$, the continuous nonnegative difference $s_k(r{)}^{n-1}-J(r,\theta )$ has integral zero. It is therefore identically zero. Let $A(X)={\nabla }_X{\partial }_r$ on a distance sphere and $h=\mathrm{tr}A={\partial }_r\log J$. Equality of the density gives $h=(n-1)s_k^{\prime }/s_k$. The traced radial Riccati equation for distance spheres readsSince $s_k^{\prime \prime }=-k{s}_k$, substituting $h$ makes the sum of the following two nonnegative quantities zero:Equality in the Cauchy-Schwarz inequality for the eigenvalues of the self-adjoint operator $A$ forces $A=(s_k^{\prime }/s_k)I$. The angular metric $g_r$ in geodesic polar coordinates consequently satisfies ${\partial }_r{g}_r=2(s_k^{\prime }/s_k)g_r$. Smoothness at the center fixes its integration constant, since $r^{-2}{g}_r\to g_{S^{n-1}}$. Henceon the normal ball: the Riemannian metric is locally the round curvature-$k$ metric. As $p$ was arbitrary, $M$ has constant sectional curvature $k$.

The classification of complete positive constant-curvature manifolds now gives $M=S_k^n/\Gamma$ with $\Gamma$ a finite freely acting group of isometries. Covering degree gives $\mathrm{Vol}M=\mathrm{Vol}{S}_k^n/|\Gamma |$. Volume equality forces $|\Gamma |=1$, provingThis establishes spherical rigidity of maximal total volume under the correct normalization.

Under the literal trace bound, the assertion is false in dimension four. Take the Riemannian product $M=S^2(r)\times S^2(r)$ with $r^2=1/(\sqrt{6}k)$. Its Ricci curvature is $r^{-2}g=\sqrt{6}kg\ge kg$, andBut a mixed tangent two-plane has sectional curvature zero, whereas a factor plane has curvature $\sqrt{6}k$. Thus this manifold is not the round sphere despite having the literal hypotheses.

Put $\theta (r)=\mathrm{Vol}B(p,r)/({\omega }_n{r}^n)$. The Bishop-Gromov inequality says that $\theta$ is nonincreasing and at most one, while smoothness gives $\theta (r)\to 1$ as $r\downarrow 0$. The prescribed limit at infinity is also one. Hence $\theta (r)=1$ for every $r>0$.

Here is how equality gives the global Euclidean rigidity of maximal asymptotic volume ratio. Assume first $n\ge 2$. In geodesic polar coordinates centered at $p$, let $c(\xi )$ be the cut time along unit direction $\xi$, and $J(r,\xi )$ the positive radial volume density before that time. The comparison giving the Bishop-Gromov inequality says $J(r,\xi )\le r^{n-1}$, andNo cut time can be finite. If $c({\xi }_0)<t$, then the radial segment at time $t$ is not minimizing, so $d(p,{\exp }_p(t{\xi }_0))<t$. This strict inequality persists for nearby directions by continuity, and their cut times are also at most $t$. At any radius $R>t$, that open set of directions omits all radial volume from $t$ to $R$. Even the maximal density $r^{n-1}$ cannot then give Euclidean volume, contradicting $\theta (R)=1$. Thus $c(\xi )=\infty$ for all directions. Equality of all ball volumes and continuity now force $J(r,\xi )=r^{n-1}$ everywhere.

Let $A(r)X={\nabla }_X{\partial }_r$ on the distance sphere. This is the negative of the outward shape operator in the convention $S=-\nabla \nu$. Put $h=\mathrm{tr}A={\partial }_r\log J$. The traced radial Riccati equation for distance spheres isSince $h=(n-1)/r$, it givesBut $\mathrm{tr}(A^2)\ge h^2/(n-1)=(n-1)/r^2$ by the Cauchy-Schwarz inequality, and the Ricci term is nonnegative. Equality therefore holds in both, so $A=r^{-1}I$. Writing the polar metric as $d{r}^2+g_r$, its evolution is ${\partial }_r{g}_r=2g_r/r$. Its small-radius Euclidean limit then givesThe Riemannian exponential map is onto by the Hopf-Rinow theorem, has no conjugate points, and is injective because there is no finite cut time. Explicitly, two distinct minimizing radial geodesics reaching the same point would, after extending one past that point, create a length-minimizing broken curve with a genuine corner; smoothing the corner shortens it. That contradicts global minimization of every radial segment. Thus ${\exp }_p$ is a global diffeomorphism, and the displayed metric provesIn dimension one, a connected complete manifold without boundary is a line or a circle. The circle has limiting volume ratio zero, so the hypothesis leaves only the Euclidean line.

Under the hypotheses of Bishop volume comparison with a positive-curvature model, the Bonnet-Myers theorem gives diameter at most $D$. If total Riemannian volume equals $v_k(D)$, every comparison ball ratio is one. In a small normal ball the pointwise radial density bound therefore becomes equality. Writing $h={\partial }_r\log J$, the traced radial Riccati equation for distance spheres and Cauchy-Schwarz inequality force the radial operator to equal $(s_k^{\prime }/s_k)I$. The angular metric is consequently $s_k(r{)}^2{g}_{S^{n-1}}$, so the metric is locally round. The classification of complete positive constant-curvature manifolds gives a quotient $S_k^n/\Gamma$, whose volume is $v_k(D)/|\Gamma |$. Equality forces $\Gamma$ trivial.