Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 16 / 2

Consider the subsetIt is nonempty by the given initial data. A local isometry preserves the Levi-Civita connection and therefore affinely parametrized geodesics. On a sufficiently small normal neighborhood of $x\in A$, this givesThe right sides agree, so $f_1=f_2$ on that neighborhood and their differentials agree there as well. Consequently $A$ is open.

It is also closed. If $x_j\in A$ converges to $x$, continuity first gives $f_1(x)=f_2(x)$. Choose coordinate charts around $x$ and this common image. The coordinate matrices of the two differentials depend continuously on the base point; their equalities at $x_j$ therefore pass to $x$. Thus $x\in A$. Since $M$ is connected, the nonempty subset $A$ that is both open and closed equals $M$. Hence $f_1=f_2$ everywhere. This proves local isometries are determined by first-order data without requiring completeness.

Let $\pi :\tilde{M}\to M$ be the universal cover and equip it with the pullback Riemannian metric $\tilde{g}={\pi }^{\ast }g$. The covering map is a local isometry, so $\tilde{M}$ has constant sectional curvature one.

We must check completeness of the lifted Riemannian metric. A geodesic on $\tilde{M}$ projects to a geodesic on $M$. By geodesic completeness of $M$, the projected geodesic extends for all real time. Its unique lift with a specified initial point extends the original lifted geodesic, because $\pi$ is a local isometry. Hence $\tilde{M}$ is geodesically complete, and the Hopf-Rinow theorem also gives metric completeness.

The allowed constant-curvature form of Cartan-Ambrose-Hicks theorem says that a linear tangent-space isometry between complete simply connected Riemannian manifolds of the same constant sectional curvature extends to a global isometry. The unit round sphere $S^n$ is complete, simply connected for $n\ge 2$, and has sectional curvature one. Comparing it with $\tilde{M}$ therefore yieldsThis is the covering step in the classification of complete positive constant-curvature manifolds. As usual for this assertion, $n\ge 2$ is understood: in dimension one there are no sectional two-planes, and the universal cover of the circle is the line, not the circle.

By part (b), $M=S^n/\Gamma$, where the group of deck transformations acts freely by round isometries. Every round isometry is the restriction of an orthogonal transformation of ${\mathbb{R}}^{n+1}$: it preserves the inner product $⟨x,y⟩=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.

Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal eigenvalues come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm 1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the deck transformation action consequently implies that the determinant map $\Gamma \to \{\pm 1\}$ has trivial kernel. Thus $|\Gamma |\le 2$.

If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm 1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal involution. We obtainThis proves the even-dimensional part of the classification of complete positive constant-curvature manifolds.

The conclusion fails in odd dimensions. For $n=2m-1\ge 3$, identify $S^{2m-1}$ with the unit sphere in ${\mathbb{C}}^m$. Scalar multiplication by $\zeta =e^{2\pi i/r}$, with $r\ge 3$, generates a freely acting cyclic group of isometries: ${\zeta }^{j}z=z$ for nonzero $z$ forces ${\zeta }^j=1$. The quotient is complete with constant sectional curvature one and has fundamental group $C_r$, so it is neither the simply connected sphere nor Real projective space with fundamental group $C_2$. In particular $S^3/C_3$ is a lens space counterexample.