[proofplan] We first exhibit the three standard simply connected complete models: Euclidean space, the round sphere of radius \begin{align*} \frac{1}{\sqrt{k}}, \end{align*} and hyperbolic space scaled to curvature $k$. For uniqueness, we compare any two complete simply connected manifolds with the same constant sectional curvature by choosing a linear isometry between tangent spaces and checking that it preserves the curvature tensor under parallel translation. The Cartan--Ambrose--Hicks theorem then extends this infinitesimal curvature-preserving datum to a global Riemannian isometry. [/proofplan]

[step:Construct the standard complete simply connected models] For $k = 0$, let $M_0$ be $\mathbb{R}^n$ with its Euclidean metric $g_0$. This manifold is complete, simply connected, and has sectional curvature $0$. For $k > 0$, define the radius $r \in (0,\infty)$ by \begin{align*} r := \frac{1}{\sqrt{k}}, \end{align*} and let $M_k$ be the round sphere \begin{align*} S^n(r) := \{x \in \mathbb{R}^{n+1} : |x| = r\} \end{align*} with the metric $g$ induced by the Euclidean metric on $\mathbb{R}^{n+1}$. By the Gauss equation for the round hypersphere in Euclidean space, if $X,Y \in T_xS^n(r)$ are orthonormal with respect to $g_x$, then the second fundamental form is $\operatorname{II}(X,Y) = r^{-1}g_x(X,Y)\nu$, where $\nu \in T_x\mathbb{R}^{n+1}$ is the outward unit normal. Hence \begin{align*} K(X,Y) = |\operatorname{II}(X,X)|\,|\operatorname{II}(Y,Y)| - |\operatorname{II}(X,Y)|^2 = \frac{1}{r^2} = k. \end{align*} Thus $S^n(r)$ has constant sectional curvature $k$. Since $n \geq 2$, the sphere $S^n(r)$ is simply connected, and compactness implies geodesic completeness. For $k < 0$, let $a := \sqrt{|k|}$ and define $M_k$ as the upper sheet of the hyperboloid in Minkowski space $\mathbb{R}^{n,1}$: \begin{align*} M_k := \left\{x \in \mathbb{R}^{n+1} : -x_0^2 + x_1^2 + \cdots + x_n^2 = -\frac{1}{a^2},\ x_0 > 0\right\}. \end{align*} Equip $M_k$ with the Riemannian metric induced by the Lorentzian form \begin{align*} \langle x,y\rangle_{n,1} := -x_0y_0 + \sum_{i=1}^n x_i y_i. \end{align*} The hyperboloid is diffeomorphic to $\mathbb{R}^n$ by radial projection onto the spatial coordinates, so it is simply connected. To verify completeness, let \begin{align*} \gamma: I &\to M_k \end{align*} be a unit-speed geodesic on its maximal interval $I \subset \mathbb{R}$. Viewing $\gamma$ as a map into $\mathbb{R}^{n,1}$, the constraint $\langle \gamma,\gamma\rangle_{n,1} = -a^{-2}$ and the geodesic equation for the induced metric give \begin{align*} \gamma''(t) = a^2\gamma(t) \end{align*} for all $t \in I$, where prime denotes ordinary differentiation in the ambient [vector space](/page/Vector%20Space). Hence \begin{align*} \gamma(t) = \cosh(at)\gamma(0) + \frac{\sinh(at)}{a}\gamma'(0), \end{align*} a formula defined for all $t \in \mathbb{R}$. Thus every unit-speed geodesic extends for all real time, and the Hopf--Rinow theorem implies that the induced Riemannian metric is complete. The Gauss equation for this hyperquadric gives, for every tangent two-plane $\sigma \subset T_xM_k$, \begin{align*} K(\sigma) = -a^2 = k. \end{align*} Thus $M_k$ is a complete simply connected $n$-dimensional model of constant sectional curvature $k$. [/step]

[step:Reduce uniqueness to a curvature-preserving tangent-space isometry]Let $(M,g)$ and $(N,h)$ be complete simply connected $n$-dimensional Riemannian manifolds with constant sectional curvature $k$. Choose points $p \in M$ and $q \in N$. Since $T_pM$ and $T_qN$ are $n$-dimensional real [inner product](/page/Inner%20Product) spaces, choose a linear isometry \begin{align*} A: T_pM &\to T_qN. \end{align*} We will verify the curvature-compatibility hypothesis needed for the Cartan--Ambrose--Hicks theorem. For a Riemannian manifold of constant sectional curvature $k$, the curvature tensor is determined by the metric through \begin{align*} R(X,Y)Z = k\big(g(Y,Z)X - g(X,Z)Y\big) \end{align*} for all tangent vectors $X,Y,Z$ at the same point. Therefore, because $A$ preserves the inner products $g_p$ and $h_q$, we have \begin{align*} A\big(R^M_p(X,Y)Z\big) &= k\big(g_p(Y,Z)A X - g_p(X,Z)A Y\big) \\ &= k\big(h_q(A Y,A Z)A X - h_q(A X,A Z)A Y\big) \\ &= R^N_q(A X,A Y)A Z \end{align*} for all $X,Y,Z \in T_pM$.[/step]

[guided]The uniqueness theorem we need is global, but its input is infinitesimal: it starts with an isometry between one tangent space of $M$ and one tangent space of $N$. We choose points $p \in M$ and $q \in N$, and because both tangent spaces are $n$-dimensional real inner product spaces, there exists a linear isometry \begin{align*} A: T_pM &\to T_qN. \end{align*} The Cartan--Ambrose--Hicks theorem requires this map to preserve curvature in the appropriate parallel-transport sense. In the constant-curvature case, the verification begins already at the base point. The curvature tensor of a Riemannian manifold with constant sectional curvature $k$ is the algebraic tensor determined by the metric: \begin{align*} R(X,Y)Z = k\big(g(Y,Z)X - g(X,Z)Y\big). \end{align*} Thus, for $X,Y,Z \in T_pM$, the linear isometry property of $A$ gives \begin{align*} h_q(A Y,A Z) &= g_p(Y,Z), \\ h_q(A X,A Z) &= g_p(X,Z). \end{align*} Substituting these identities into the constant-curvature formula gives \begin{align*} A\big(R^M_p(X,Y)Z\big) &= A\big(k(g_p(Y,Z)X - g_p(X,Z)Y)\big) \\ &= k\big(g_p(Y,Z)A X - g_p(X,Z)A Y\big) \\ &= k\big(h_q(A Y,A Z)A X - h_q(A X,A Z)A Y\big) \\ &= R^N_q(A X,A Y)A Z. \end{align*} This computation is the point where the common value of the sectional curvature is used: the same scalar $k$ appears on both sides, so the curvature tensors match after applying $A$. The classification theorem will then follow once this local curvature-matching computation is transported along geodesics and fed into the global Cartan--Ambrose--Hicks theorem.[/guided]

[step:Verify the curvature tensors match along geodesics] Let $\gamma: [0,a] \to M$ be a geodesic segment with $\gamma(0)=p$. Let $P^M_\gamma(t): T_pM \to T_{\gamma(t)}M$ denote parallel transport along $\gamma$ from $0$ to $t$. Let $\eta: [0,a] \to N$ be the geodesic with $\eta(0)=q$ and initial velocity $\eta'(0)=A\gamma'(0)$, and let $P^N_\eta(t): T_qN \to T_{\eta(t)}N$ denote parallel transport along $\eta$ from $0$ to $t$. Define, for each $t \in [0,a]$, the linear isometry \begin{align*} A_t: T_{\gamma(t)}M &\to T_{\eta(t)}N, \\ V &\mapsto P^N_\eta(t)\, A\, \big(P^M_\gamma(t)\big)^{-1}V. \end{align*} Parallel transport preserves the Riemannian metric, so $A_t$ is a linear isometry. Since both manifolds have constant sectional curvature $k$, the same calculation as at the base point gives \begin{align*} A_t\big(R^M_{\gamma(t)}(X,Y)Z\big) = R^N_{\eta(t)}(A_tX,A_tY)A_tZ \end{align*} for all $X,Y,Z \in T_{\gamma(t)}M$ and all $t \in [0,a]$. [/step]

[step:Apply Cartan--Ambrose--Hicks to obtain a global isometry] We use the following form of the Cartan--Ambrose--Hicks theorem: if $M$ and $N$ are complete simply connected Riemannian manifolds, $A:T_pM \to T_qN$ is a linear isometry, and for every geodesic $\gamma$ starting at $p$ the parallel-transported map $A_t$ identifies the curvature tensor of $M$ along $\gamma$ with the curvature tensor of $N$ along the corresponding geodesic starting at $q$, then $A$ is realised as the differential at $p$ of a global Riemannian isometry. These hypotheses are satisfied: $M$ and $N$ are complete and simply connected by assumption, $A:T_pM \to T_qN$ is a linear isometry, and the previous step verifies the required curvature preservation along each such geodesic. Hence the theorem produces a Riemannian isometry \begin{align*} F: M &\to N \end{align*} with $F(p)=q$ and $dF_p=A$. Therefore any two complete simply connected $n$-dimensional Riemannian manifolds with constant sectional curvature $k$ are isometric. Applying this uniqueness to the model $M_k$ constructed above gives the asserted classification: $\mathbb{R}^n$ for $k=0$, the round sphere of radius \begin{align*} \frac{1}{\sqrt{k}} \end{align*} for $k>0$, and hyperbolic space of curvature $k$ for $k<0$. [/step]