Theorem: A Riemannian manifold with transitive isometry group has constant curvature.

Proof Sketch:

Step 1: Transitivity means for any points $p, q$, there exists an isometry $\phi$ with $\phi(p) = q$. This makes the geometry homogeneous.

Step 2: If the isometry group also acts transitively on unit tangent vectors at each point (isotropy), the curvature must be constant—no direction is distinguished.

Step 3: For constant curvature $K$, the isometry group is maximal: dimension $\frac{n(n+1)}{2}$ (rigid motions plus rotations). These groups are $SO(n+1)$ ($K > 0$), $E(n)$ ($K = 0$), $SO^+(n,1)$ ($K < 0$).

Conclusion: Homogeneity and isotropy force constant curvature, and different curvature signs give the three classical geometries. ∎