[proofplan] The proof is the rigidity case of the Bonnet--Myers diameter comparison. We choose points $p,q \in M$ at maximal distance and study the two distance functions from $p$ and $q$. Laplacian comparison and the Calabi barrier maximum principle force the sum of these distance functions to be identically equal to the model value \begin{align*} \frac{\pi}{\sqrt{k}}. \end{align*} Equality in Laplacian comparison then forces equality in the [Riccati equation for geodesic spheres](/theorems/5357), so the metric is the spherical warped product with radial expression \begin{align*} dt^2+k^{-1}\sin^2(\sqrt{k}t)\,g_{S^{n-1}}, \end{align*} which is the round metric of sectional curvature $k$. [/proofplan] [step:Choose antipodal points realizing the Bonnet--Myers diameter] Let \begin{align*} d_g:M\times M&\to [0,\infty) \end{align*} denote the Riemannian distance function induced by $g$. Set \begin{align*} D := \frac{\pi}{\sqrt{k}}. \end{align*} By the [Bonnet--Myers theorem](/page/Bonnet--Myers%20Theorem), the Ricci lower bound $\operatorname{Ric}_g \geq (n-1)k\,g$ and the completeness of $(M,g)$ imply that $M$ is compact and \begin{align*} \operatorname{diam}(M,g) \leq D. \end{align*} The hypothesis gives equality. Since $M$ is compact, there exist points $p,q \in M$ such that \begin{align*} d_g(p,q)=D. \end{align*} Define the distance functions \begin{align*} r_p:M &\to [0,D], & x &\mapsto d_g(p,x),\\ r_q:M &\to [0,D], & x &\mapsto d_g(q,x). \end{align*} The triangle inequality gives, for every $x \in M$, \begin{align*} D=d_g(p,q)\leq d_g(p,x)+d_g(x,q)=r_p(x)+r_q(x). \end{align*} Thus the function \begin{align*} u:M &\to [0,\infty), & x &\mapsto r_p(x)+r_q(x)-D \end{align*} is non-negative. [/step] [step:Apply Laplacian comparison to make the distance sum superharmonic] Let \begin{align*} \Omega := M \setminus \{p,q\}. \end{align*} On the [open set](/page/Open%20Set) where $r_p$ and $r_q$ are smooth, define the auxiliary angle functions \begin{align*} a:M\setminus\{p\} &\to (0,\pi], & x &\mapsto \sqrt{k}\,r_p(x),\\ b:M\setminus\{q\} &\to (0,\pi], & x &\mapsto \sqrt{k}\,r_q(x). \end{align*} Since $0<r_p(x)<D$ and $0<r_q(x)<D$ on $\Omega$, we have \begin{align*} 0<a(x)<\pi, \qquad 0<b(x)<\pi. \end{align*} The inequality $r_p(x)+r_q(x)\geq D$ gives \begin{align*} a(x)+b(x)\geq \pi. \end{align*} Since also $a(x)+b(x)\leq 2\pi$, the identity \begin{align*} \cot a+\cot b=\frac{\sin(a+b)}{\sin a\,\sin b} \end{align*} implies \begin{align*} \cot a(x)+\cot b(x)\leq 0. \end{align*} The [Laplacian comparison theorem for distance functions](/page/Laplacian%20Comparison%20Theorem) under $\operatorname{Ric}_g\geq (n-1)k\,g$ gives, in the barrier sense across the cut loci and classically where the distance functions are smooth, \begin{align*} \Delta r_p(x) &\leq (n-1)\sqrt{k}\,\cot(\sqrt{k}\,r_p(x)),\\ \Delta r_q(x) &\leq (n-1)\sqrt{k}\,\cot(\sqrt{k}\,r_q(x)). \end{align*} Adding the two inequalities gives, in the barrier sense on $\Omega$, \begin{align*} \Delta u(x) &=\Delta r_p(x)+\Delta r_q(x)\\ &\leq (n-1)\sqrt{k}\left(\cot a(x)+\cot b(x)\right)\\ &\leq 0. \end{align*} Thus $u$ is a non-negative superharmonic function on $\Omega$ in the barrier sense. [guided] The purpose of this step is to turn the geometric diameter condition into a maximum-principle statement. The function \begin{align*} u:M &\to [0,\infty), & x &\mapsto r_p(x)+r_q(x)-D \end{align*} measures the excess in the triangle inequality for the triple $(p,x,q)$. Because $d_g(p,q)=D$, the triangle inequality gives \begin{align*} u(x)=r_p(x)+r_q(x)-D\geq 0. \end{align*} To prove rigidity, we want to show that this excess is always zero. We now compute the sign of $\Delta u$. On the set where both distance functions are smooth, define the maps \begin{align*} a:M\setminus\{p\} &\to (0,\pi], & x &\mapsto \sqrt{k}\,r_p(x),\\ b:M\setminus\{q\} &\to (0,\pi], & x &\mapsto \sqrt{k}\,r_q(x). \end{align*} The diameter bound gives $r_p(x)\leq D$ and $r_q(x)\leq D$, while $x \neq p,q$ gives $r_p(x)>0$ and $r_q(x)>0$. Hence \begin{align*} 0<a(x)<\pi, \qquad 0<b(x)<\pi. \end{align*} The triangle inequality gives \begin{align*} a(x)+b(x)=\sqrt{k}\,(r_p(x)+r_q(x))\geq \sqrt{k}\,D=\pi. \end{align*} Since each of $a(x)$ and $b(x)$ is at most $\pi$, we also have $a(x)+b(x)\leq 2\pi$. Therefore \begin{align*} \sin(a(x)+b(x))\leq 0. \end{align*} Because $\sin a(x)>0$ and $\sin b(x)>0$, the trigonometric identity \begin{align*} \cot a+\cot b=\frac{\sin(a+b)}{\sin a\,\sin b} \end{align*} gives \begin{align*} \cot a(x)+\cot b(x)\leq 0. \end{align*} Now apply the [Laplacian comparison theorem for distance functions](/page/Laplacian%20Comparison%20Theorem) under the Ricci lower bound $\operatorname{Ric}_g\geq (n-1)k\,g$. The theorem applies to $r_p$ and $r_q$ because $(M,g)$ is complete and has the stated Ricci curvature lower bound. It gives, in the barrier sense and classically away from the cut loci, \begin{align*} \Delta r_p(x) &\leq (n-1)\sqrt{k}\,\cot(\sqrt{k}\,r_p(x)),\\ \Delta r_q(x) &\leq (n-1)\sqrt{k}\,\cot(\sqrt{k}\,r_q(x)). \end{align*} Adding these inequalities yields \begin{align*} \Delta u(x) &=\Delta r_p(x)+\Delta r_q(x)\\ &\leq (n-1)\sqrt{k}\left(\cot a(x)+\cot b(x)\right)\\ &\leq 0. \end{align*} Thus the triangle-excess function $u$ is superharmonic in the barrier sense on $\Omega=M\setminus\{p,q\}$. This is the key analytic input: a non-negative superharmonic function that attains an interior minimum must be constant. [/guided] [/step] [step:Use the Calabi maximum principle to force equality in the triangle inequality] Let \begin{align*} \gamma:[0,D] &\to M \end{align*} be a unit-speed minimizing geodesic from $p$ to $q$, so $\gamma(0)=p$ and $\gamma(D)=q$. For every $t \in (0,D)$, \begin{align*} r_p(\gamma(t))=t,\qquad r_q(\gamma(t))=D-t, \end{align*} and hence \begin{align*} u(\gamma(t))=0. \end{align*} Since $n\geq 2$, the punctured manifold $\Omega=M\setminus\{p,q\}$ is connected. The [Calabi barrier maximum principle for superharmonic functions](/page/Calabi%20Maximum%20Principle) applies because $u$ is continuous on $\Omega$, non-negative on $\Omega$, superharmonic in the barrier sense on $\Omega$, and attains its minimum value $0$ at the interior point $\gamma(t)$ for every $t\in(0,D)$. The principle gives \begin{align*} u \equiv 0 \quad \text{on } \Omega. \end{align*} By continuity of $r_p$ and $r_q$, the same identity holds on all of $M$: \begin{align*} r_p(x)+r_q(x)=D \end{align*} for every $x \in M$. [/step] [step:Extract equality in the Laplacian comparison inequalities] On the regular set where both $r_p$ and $r_q$ are smooth, the identity $r_p+r_q=D$ gives \begin{align*} \Delta r_p+\Delta r_q=0. \end{align*} The comparison inequalities and the trigonometric identity give \begin{align*} 0 &=\Delta r_p+\Delta r_q\\ &\leq (n-1)\sqrt{k}\left(\cot(\sqrt{k}\,r_p)+\cot(\sqrt{k}\,r_q)\right)\\ &=(n-1)\sqrt{k}\, \frac{\sin(\sqrt{k}(r_p+r_q))} {\sin(\sqrt{k}\,r_p)\sin(\sqrt{k}\,r_q)}\\ &=0. \end{align*} Therefore equality holds throughout this chain. In particular, \begin{align*} \Delta r_p=(n-1)\sqrt{k}\,\cot(\sqrt{k}\,r_p) \end{align*} where $r_p$ is smooth. [claim:Every point between $p$ and $q$ is regular for the distance from $p$] Every point $x\in M\setminus\{p,q\}$ is not in the cut locus of $p$. Consequently $r_p$ is smooth on $M\setminus\{p,q\}$, and the displayed equality for $\Delta r_p$ holds on all of $M\setminus\{p,q\}$. [/claim] [proof] Fix $x\in M\setminus\{p,q\}$. Let \begin{align*} \alpha:[0,r_p(x)]&\to M, & s&\mapsto \alpha(s) \end{align*} be a unit-speed minimizing geodesic from $p$ to $x$, and let \begin{align*} \beta:[0,r_q(x)]&\to M, & s&\mapsto \beta(s) \end{align*} be a unit-speed minimizing geodesic from $x$ to $q$. Since $r_p(x)+r_q(x)=D=d_g(p,q)$, the concatenation of $\alpha$ and $\beta$ has length $D$ and is therefore a minimizing curve from $p$ to $q$. A length-minimizing curve is a geodesic on each interior subinterval and has no corner at the joining point, so the terminal velocity of $\alpha$ equals the initial velocity of $\beta$ at $x$. This proves uniqueness of the minimizing geodesic from $p$ to $x$: if $\alpha_1$ and $\alpha_2$ are two such geodesics, then concatenating each with the same $\beta$ gives two minimizing geodesics from $p$ to $q$ that agree on the non-empty interval traced by $\beta$ after $x$. By uniqueness for the geodesic initial-value problem, the two geodesics agree after reversing time from $x$ to $p$, hence $\alpha_1=\alpha_2$. It remains to exclude conjugacy. If $x$ were conjugate to $p$ along the unique minimizing geodesic from $p$ to $x$, then $x$ would be an interior conjugate point along the minimizing geodesic from $p$ to $q$ obtained by continuing through $x$ to $q$. By the [second variation theorem for geodesics](/page/Second%20Variation%20Formula), an interior conjugate point produces a non-zero variation field along the geodesic vanishing at the endpoints and with non-positive index form. Perturbing this field by a compactly supported field near the conjugate point gives a proper variation through curves from $p$ to $q$ with strictly negative second variation of length; hence the original geodesic cannot be length-minimizing between $p$ and $q$. This contradicts $d_g(p,q)=D$. Therefore $x$ is neither a branching cut point nor a conjugate cut point of $p$, so $x$ is not in the cut locus of $p$. [/proof] [/step] [step:Convert equality in Laplacian comparison into the model shape operator] Let \begin{align*} S_pM:=\{v\in T_pM:g_p(v,v)=1\} \end{align*} be the unit tangent sphere at $p$, where $g_p:T_pM\times T_pM\to \mathbb{R}$ is the [inner product](/page/Inner%20Product) induced by $g$ on $T_pM$. Let \begin{align*} \exp_p:\mathcal{D}_p\subset T_pM&\to M \end{align*} denote the Riemannian exponential map at $p$, defined on its maximal star-shaped domain $\mathcal{D}_p$. For each $v\in S_pM$, define the radial geodesic \begin{align*} \gamma_v:(0,D)&\to M, & t&\mapsto \exp_p(tv). \end{align*} For $t\in(0,D)$, let $A_t:T_{\gamma_v(t)}\{r_p=t\}\to T_{\gamma_v(t)}\{r_p=t\}$ denote the shape operator of the geodesic sphere $\{r_p=t\}$ with respect to the outward unit normal $\nabla r_p$. The [Riccati equation for the shape operator of geodesic spheres](/page/Riccati%20Equation%20For%20Geodesic%20Spheres) gives \begin{align*} \frac{d}{dt}\operatorname{tr}A_t+\operatorname{tr}(A_t^2)+\operatorname{Ric}_g(\dot{\gamma}_v(t),\dot{\gamma}_v(t))=0. \end{align*} Since \begin{align*} \operatorname{tr}A_t=\Delta r_p(\gamma_v(t))=(n-1)\sqrt{k}\,\cot(\sqrt{k}t), \end{align*} we compute \begin{align*} \frac{d}{dt}\operatorname{tr}A_t =-(n-1)k\,\csc^2(\sqrt{k}t). \end{align*} The Cauchy--Schwarz inequality for the eigenvalues of the symmetric endomorphism $A_t$ gives \begin{align*} \operatorname{tr}(A_t^2)\geq \frac{(\operatorname{tr}A_t)^2}{n-1} =(n-1)k\,\cot^2(\sqrt{k}t). \end{align*} The Ricci lower bound gives \begin{align*} \operatorname{Ric}_g(\dot{\gamma}_v(t),\dot{\gamma}_v(t))\geq (n-1)k. \end{align*} Substituting these estimates into the Riccati equation yields equality in both estimates, because \begin{align*} -(n-1)k\,\csc^2(\sqrt{k}t) +(n-1)k\,\cot^2(\sqrt{k}t) +(n-1)k=0. \end{align*} Equality in the Cauchy--Schwarz inequality for the eigenvalues of $A_t$ means all principal curvatures are equal. Therefore \begin{align*} A_t=\sqrt{k}\,\cot(\sqrt{k}t)\,\operatorname{id}_{T_{\gamma_v(t)}\{r_p=t\}} \end{align*} for every $t\in(0,D)$ and every $v\in S_pM$. [/step] [step:Integrate the shape operator equation to obtain the spherical warped product metric] Define the polar-coordinate map \begin{align*} F:(0,D)\times S_pM &\to M\setminus\{p,q\}, & (t,v)&\mapsto \exp_p(tv). \end{align*} The preceding claim implies that $\exp_p$ has no cut point at distance $t\in(0,D)$. Hence $F$ is locally a diffeomorphism everywhere. It is surjective because every $x\in M\setminus\{p,q\}$ lies on the unique minimizing geodesic from $p$ to $x$, and it is injective because that minimizing geodesic is unique. Therefore $F$ is a diffeomorphism. Write \begin{align*} F^*g=dt^2+g_t, \end{align*} where $g_t$ is the Riemannian metric induced on the slice $\{t\}\times S_pM$. The relation between the slice metric and the shape operator is \begin{align*} \frac{1}{2}\partial_t g_t(X,Y)=g_t(A_tX,Y) \end{align*} for smooth vector fields $X,Y$ tangent to $S_pM$ and independent of $t$. Using \begin{align*} A_t=\sqrt{k}\,\cot(\sqrt{k}t)\operatorname{id}, \end{align*} we get \begin{align*} \partial_t g_t(X,Y)=2\sqrt{k}\,\cot(\sqrt{k}t)\,g_t(X,Y). \end{align*} Solving this scalar differential equation in $t$ gives \begin{align*} g_t=\frac{\sin^2(\sqrt{k}t)}{\sin^2(\sqrt{k}t_0)}\,g_{t_0} \end{align*} for every fixed $t_0\in(0,D)$. In geodesic [normal coordinates](/theorems/2713) at $p$, the slice metrics satisfy \begin{align*} \lim_{t\downarrow 0}\frac{1}{t^2}g_t=g_{S_pM}, \end{align*} where $g_{S_pM}$ is the standard round metric on the unit sphere in the Euclidean [inner product space](/page/Inner%20Product%20Space) $(T_pM,g_p)$. Since \begin{align*} \lim_{t\downarrow 0}\frac{\sin^2(\sqrt{k}t)}{t^2}=k, \end{align*} the preceding identity gives \begin{align*} g_t=\frac{1}{k}\sin^2(\sqrt{k}t)\,g_{S_pM}. \end{align*} Consequently \begin{align*} F^*g=dt^2+\frac{1}{k}\sin^2(\sqrt{k}t)\,g_{S_pM} \end{align*} on $(0,D)\times S_pM$. [/step] [step:Identify the warped product with the round sphere of curvature $k$] Let $g_{\mathrm{round}}$ denote the unit round metric on $S^n$. The round metric of constant sectional curvature $k$ is $k^{-1}g_{\mathrm{round}}$. In geodesic polar coordinates around a pole, this metric has the form \begin{align*} dt^2+\frac{1}{k}\sin^2(\sqrt{k}t)\,g_{S^{n-1}}, \qquad 0<t<\frac{\pi}{\sqrt{k}}. \end{align*} Since $(S_pM,g_{S_pM})$ is isometric to the unit round sphere $S^{n-1}$, the formula obtained above identifies $(M\setminus\{p,q\},g)$ with the complement of the two poles in the round sphere of sectional curvature $k$. The metric expression extends smoothly across $t=0$ because \begin{align*} \frac{1}{k}\sin^2(\sqrt{k}t)=t^2+O(t^4) \end{align*} as $t\downarrow 0$, which is exactly the first-order radial behaviour required in geodesic normal coordinates at $p$. At the other endpoint, put $s:=D-t$. Since $\sqrt{k}D=\pi$, we have \begin{align*} \frac{1}{k}\sin^2(\sqrt{k}t)=\frac{1}{k}\sin^2(\sqrt{k}s)=s^2+O(s^4) \end{align*} as $s\downarrow 0$. Together with the identity $r_p+r_q=D$, this is the corresponding geodesic normal-coordinate behaviour at $q$. Therefore the isometry on the punctured manifolds extends continuously and smoothly across the two endpoints. Hence $(M,g)$ is isometric to the round sphere $S^n$ with constant sectional curvature $k$. [/step]