Spherical rigidity of maximal total volume ($\mathrm{Vol}M=\mathrm{Vol}{S}_k^n⟹M\cong S_k^n$)

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.

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.