[proofplan] The proof compares the trace of the radial Riccati equation with the scalar Riccati equation satisfied by the model mean curvature $(n-1)\operatorname{ct}_k(r)$. Equality in Laplacian comparison forces the scalar trace to be exactly the model trace, so substituting this equality into the Riccati equation expresses a sum of two non-negative deficits as zero. The first deficit is the Cauchy-Schwarz deficit for the eigenvalues of the shape operator, and the second is the radial Ricci deficit from the lower curvature bound. Since both vanish, the shape operator has one eigenvalue and the radial Ricci curvature is exactly the model value. [/proofplan] [step:Write the radial Riccati equation on the regular set] Since $U \subset \mathcal{R}_p$, the function $r$ is smooth on $U$ and \begin{align*} N: U &\to TM \\ x &\mapsto \nabla r(x) \end{align*} is a smooth unit vector field. For each $x \in U$, the operator \begin{align*} S_x: E_x &\to E_x \\ v &\mapsto \nabla_v N \end{align*} is self-adjoint with respect to $g_x|_{E_x}$, and its trace is the Laplacian of $r$: \begin{align*} \operatorname{tr} S_x = \Delta r(x). \end{align*} Let \begin{align*} h: U &\to \mathbb{R} \\ x &\mapsto \operatorname{tr}S_x \end{align*} denote the radial mean curvature. The trace of the radial Riccati equation gives, at every $x \in U$, \begin{align*} N(h)(x)+\operatorname{tr}(S_x^2)+\operatorname{Ric}_g(N_x,N_x)=0. \end{align*} [/step] [step:Substitute the equality hypothesis into the traced Riccati equation] Define \begin{align*} c: (0,R_k) &\to \mathbb{R} \\ t &\mapsto \operatorname{ct}_k(t). \end{align*} The equality hypothesis says \begin{align*} h(x)=(n-1)c(r(x)) \end{align*} for every $x \in U$. Since $N(r)=|\nabla r|^2=1$ on $U$, the chain rule gives \begin{align*} N(h)(x)=(n-1)c'(r(x)). \end{align*} The model function $c=\operatorname{ct}_k$ satisfies the scalar Riccati identity \begin{align*} c'(t)+c(t)^2+k=0 \end{align*} for every $t \in (0,R_k)$. Therefore \begin{align*} N(h)(x)=-(n-1)c(r(x))^2-(n-1)k. \end{align*} Substituting this into the traced radial Riccati equation yields \begin{align*} \operatorname{tr}(S_x^2)-(n-1)c(r(x))^2 +\operatorname{Ric}_g(N_x,N_x)-(n-1)k=0. \end{align*} [guided] We now use the equality assumption in the only place where it is needed. The function \begin{align*} h: U &\to \mathbb{R} \\ x &\mapsto \operatorname{tr}S_x \end{align*} is the mean curvature of the distance sphere through $x$, and on the regular set it equals $\Delta r$. The hypothesis says that this mean curvature is exactly the model value: \begin{align*} h(x)=(n-1)\operatorname{ct}_k(r(x)). \end{align*} To differentiate this identity in the radial direction, define \begin{align*} c: (0,R_k) &\to \mathbb{R} \\ t &\mapsto \operatorname{ct}_k(t). \end{align*} Because $N=\nabla r$ and $|\nabla r|=1$ on $\mathcal{R}_p$, we have \begin{align*} N(r)=g(\nabla r,\nabla r)=1. \end{align*} The chain rule therefore gives \begin{align*} N(h)(x)=(n-1)c'(r(x))\,N(r)(x)=(n-1)c'(r(x)). \end{align*} The model quantity $c=\operatorname{ct}_k$ satisfies \begin{align*} c'(t)+c(t)^2+k=0. \end{align*} This follows directly from $\operatorname{ct}_k=\operatorname{sn}_k'/\operatorname{sn}_k$ and the identity $\operatorname{sn}_k''+k\operatorname{sn}_k=0$: \begin{align*} c'(t) &=\frac{\operatorname{sn}_k''(t)\operatorname{sn}_k(t)-(\operatorname{sn}_k'(t))^2}{(\operatorname{sn}_k(t))^2} \\ &=\frac{-k(\operatorname{sn}_k(t))^2-(\operatorname{sn}_k'(t))^2}{(\operatorname{sn}_k(t))^2} \\ &=-k-c(t)^2. \end{align*} Thus \begin{align*} N(h)(x)=-(n-1)c(r(x))^2-(n-1)k. \end{align*} Putting this expression into \begin{align*} N(h)(x)+\operatorname{tr}(S_x^2)+\operatorname{Ric}_g(N_x,N_x)=0 \end{align*} gives \begin{align*} \operatorname{tr}(S_x^2)-(n-1)c(r(x))^2 +\operatorname{Ric}_g(N_x,N_x)-(n-1)k=0. \end{align*} This is the key algebraic identity: it writes equality in the scalar comparison as equality in the sum of the two geometric inequalities used to prove comparison. [/guided] [/step] [step:Split the identity into non-negative Cauchy-Schwarz and Ricci deficits] Fix $x \in U$. Since $S_x$ is self-adjoint on the $(n-1)$-dimensional [inner product space](/page/Inner%20Product%20Space) $(E_x,g_x|_{E_x})$, choose an [orthonormal basis](/page/Orthonormal%20Basis) of $E_x$ consisting of eigenvectors of $S_x$, and let \begin{align*} \lambda_1,\dots,\lambda_{n-1} \in \mathbb{R} \end{align*} be the corresponding eigenvalues. Then \begin{align*} \operatorname{tr}S_x=\sum_{i=1}^{n-1}\lambda_i, \qquad \operatorname{tr}(S_x^2)=\sum_{i=1}^{n-1}\lambda_i^2. \end{align*} The finite-dimensional [Cauchy-Schwarz inequality](/theorems/432) applied to $(\lambda_1,\dots,\lambda_{n-1})$ and $(1,\dots,1)$ gives \begin{align*} \operatorname{tr}(S_x^2) =\sum_{i=1}^{n-1}\lambda_i^2 \geq \frac{1}{n-1}\left(\sum_{i=1}^{n-1}\lambda_i\right)^2 =\frac{h(x)^2}{n-1}. \end{align*} Using $h(x)=(n-1)c(r(x))$, this becomes \begin{align*} \operatorname{tr}(S_x^2)-(n-1)c(r(x))^2 \geq 0. \end{align*} The curvature hypothesis gives \begin{align*} \operatorname{Ric}_g(N_x,N_x)\geq (n-1)k\,g_x(N_x,N_x)=(n-1)k, \end{align*} because $|N_x|=1$. Hence \begin{align*} \operatorname{Ric}_g(N_x,N_x)-(n-1)k\geq 0. \end{align*} The two non-negative quantities have sum zero, so both are zero: \begin{align*} \operatorname{tr}(S_x^2)-(n-1)c(r(x))^2=0, \qquad \operatorname{Ric}_g(N_x,N_x)-(n-1)k=0. \end{align*} [/step] [step:Use equality in Cauchy-Schwarz to identify the shape operator] The equality condition in the Cauchy-Schwarz inequality for the vectors $(\lambda_1,\dots,\lambda_{n-1})$ and $(1,\dots,1)$ implies that there exists $a \in \mathbb{R}$ such that \begin{align*} \lambda_i=a \end{align*} for every $i \in \{1,\dots,n-1\}$. Since \begin{align*} \sum_{i=1}^{n-1}\lambda_i=h(x)=(n-1)c(r(x)), \end{align*} we get $a=c(r(x))$. Therefore every eigenvalue of $S_x$ equals $c(r(x))$, and hence \begin{align*} S_x=c(r(x))\,\operatorname{id}_{E_x} =\operatorname{ct}_k(r(x))\,\operatorname{id}_{E_x}. \end{align*} The second vanished deficit already gives \begin{align*} \operatorname{Ric}_g(N_x,N_x)=(n-1)k. \end{align*} Since $x \in U$ was arbitrary, both conclusions hold throughout $U$. This proves the equality rigidity statement. [/step]