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Assume the trace Ricci curvature satisfies $\mathrm{Ric}\ge (n-1)kg$, with $k>0$. If points $p,q$ have Riemannian distance $D=\pi /\sqrt{k}$, the balls $B(p,r)$ and $B(q,D-r)$ are disjoint. Bishop volume comparison with a positive-curvature model bounds their volumes below by $\mathrm{Vol}(M)v_k(r)/v_k(D)$ and $\mathrm{Vol}(M)v_k(D-r)/v_k(D)$. Since $v_k(r)+v_k(D-r)=v_k(D)$, both inequalities must be equalities. Taking $r\downarrow 0$ gives $\mathrm{Vol}M=v_k(D)$, and spherical rigidity of maximal total volume yields the round sphere.

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.

Again assume the intended trace Ricci curvature bound $\mathrm{Ric}\ge (n-1)kg$, and put $D=\pi /\sqrt{k}$. Compactness from the Bonnet-Myers theorem ensures that the diameter is attained: choose $p,q$ with $d(p,q)=D$. For $0<r<D$, the open balls $B(p,r)$ and $B(q,D-r)$ are disjoint, because a point in both would violate the triangle inequality.

The clearly stated Bishop volume comparison with a positive-curvature model from part (b), applied between radius $r$ and radius $D$, givesHere both radius-$D$ ball volumes equal the total volume, as shown in part (b). The symmetry $s_k(D-t)=s_k(t)$ givesThe two lower bounds therefore sum to $\mathrm{Vol}M$, while disjointness bounds that same sum above by $\mathrm{Vol}M$. Both inequalities must be equalities. In particularLetting $r\downarrow 0$, the local Euclidean volume asymptotic gives one on the left. Thus $\mathrm{Vol}M=v_k(D)$, and the detailed equality proof in part (b) applies. We obtainThis proves maximal diameter rigidity from disjoint comparison balls directly, including the required equality argument.

The literal trace interpretation again has a counterexample. On $M=S^2(r)\times S^2(r)$ with $r^2=1/(2k)$, the Ricci curvature equals $2kg\ge kg$. For the product Riemannian metric, distances satisfy $d^2=d_1^2+d_2^2$: every product curve has length at least $\sqrt{d_1^2+d_2^2}$ by the integral triangle inequality, and a pair of constant-speed minimizing factor geodesics attains it. ConsequentlyThe mixed sectional curvature is zero, so this manifold is not $S_k^4$. The stronger bound, or normalized Ricci curvature convention, is essential.

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.