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Cartan hadamard says exp p is injective without simple connectedness

Statement

Assume the inherited Axiom of Countable Choice ${A C}_{ω}$ . False claim: if $(M, g)$ is a complete, connected, boundaryless Riemannian manifold with $K \leq 0$ , then for every $p \in M$ the exponential map $\exp_{p} : T_{p} M \to M$ is globally injective.

The claim omits exactly one hypothesis of the Cartan–Hadamard theorem, namely simple connectedness. It fails as soon as $M$ carries a closed geodesic without being simply connected; the flat two-torus below is the standard witness.

Facts & Assumptions

Given: The inherited ${A C}_{ω}$ of [A1], the flat two-torus $T^{2} = (R / Z)^{2}$ with its descended Euclidean metric, and the cartesian exponential maps of its tangent spaces.

[A1]

The countable-choice premise is the inherited ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried by the Hopf–Rinow and covering-map suppliers used below; the quotient constructions and the chosen lattice vectors are explicit.

[F1]

The flat torus metric: with $q : R^{n} \to R^{n} / Z^{n}$ the quotient map, the torus carries a Riemannian metric $g$ , unique with $q^{*} g = \sum_{i} d x^{i} \otimes d x^{i}$ , whose charts are boxes with integer-translation transitions; $(T^{n}, g)$ is a connected boundaryless smooth manifold on which $q$ is a surjective local isometry, it is geodesically and metrically complete, and for $n \geq 2$ it has vanishing sectional curvature (Flat torus model geometry, The two-dimensional torus $T^{2} = (R / Z)^{2}$ ).

[F2]

For the lattice quotient $R^{n} / Z^{n}$ with the descended flat metric, identifying $T_{[x]} T^{n}$ with $R^{n}$ by a quotient chart, every fibrewise exponential map has domain all of $T_{[x]} T^{n}$ and satisfies $\exp_{[x]} (v) = [x + v]$ ; it is not injective, since $0 \neq e_{1} \in Z^{n}$ gives two distinct vectors $0$ and $e_{1}$ with the same image (Flat torus model geometry).

[F3]

Hopf–Rinow: for a nonempty connected boundaryless Riemannian manifold, metric completeness, geodesic completeness and the global definition of $\exp_{p}$ on all of $T_{p} M$ for one point are equivalent (Hopf–Rinow theorem).

[F4]

Curvature of a locally Euclidean metric: in a chart whose metric matrix has constant entries the Christoffel symbols vanish (Christoffel formula for the levi civita connection), so the coordinate formula for the curvature tensor gives $R = 0$ (Coordinate formula for the curvature tensor); since $Rm (X, Y, Z, W) = g (R (X, Y) Z, W)$ and $K (σ)$ is the quotient by the positive Gram determinant, every sectional curvature vanishes (Riemann curvature four-tensor, Sectional curvature).

[F5]

Cartan–Hadamard (Cartan hadamard): for a complete connected boundaryless manifold with $K \leq 0$ , the metric $\exp_{p}^{*} g$ on $T_{p} M$ is complete and $\exp_{p}$ is a smooth universal covering map. If $M$ is also simply connected, $\exp_{p}$ is a diffeomorphism. The source completeness is proved there before the covering conclusion; it is not inferred from a lemma which already assumes a covering.

[F6]

The torus is not simply connected: $π_{1} (T^{2}, ([0], [0])) ≅ (Z, +) \times (Z, +)$ ( $π_{1} (T^{2}) ≅ Z \times Z$ ).

Refutation

1.1F1F2F3F4

The flat two-torus is complete with $K \leq 0$ . [F1, F2] By [F1] with $n = 2$ the torus $(T^{2}, g_{f l a t})$ is a connected boundaryless Riemannian surface with $q^{*} g = d x^{2} + d y^{2}$ , and it is geodesically and metrically complete with vanishing sectional curvature: $K (σ) = 0$ for every tangent two-plane, in particular $K \leq 0$ . By [F2] with $n = 2$ the exponential of $T^{2}$ at every point has domain all of the tangent space. The flatness also follows directly from the constant-coefficient charts by [F4], and metric completeness from geodesic completeness by [F3]. Thus $(T^{2}, g_{f l a t})$ is a complete, connected, boundaryless Riemannian surface with $K \leq 0$ .

1.2F1F2

Its exponential is not injective. [F1, F2] Identify $T_{([0], [0])} T^{2}$ with $R^{2}$ . By [F2] with $Λ = Z^{2}$ , $\exp_{([0], [0])} (v) = [v] for every v \in R^{2} .$ The vectors $v = 0$ and $w = (1, 0)$ are distinct, but $[w] = [(1, 0)] = [(0, 0)]$ because $(1, 0) \in Z^{2}$ acts trivially on the quotient, so $\exp_{([0], [0])} (v) = \exp_{([0], [0])} (w)$ . Hence the exponential is not injective.

2.1F5F6step 1.1step 1.2

The false claim fails, and simple connectedness is the missing hypothesis. [F5, F6, step 1.1, step 1.2] Steps 1.1 and 1.2 exhibit a complete, connected, boundaryless Riemannian manifold with $K \leq 0$ whose exponential is not injective, so the claim of the Statement is false. The general conclusion of [F5] is that $\exp_{p} : (T_{p} M, \exp_{p}^{*} g) \to (M, g)$ is a universal covering map with complete source. The diffeomorphism conclusion additionally assumes simple connectedness of $M$ . Consistently, $T^{2}$ is not simply connected by [F6], and $\exp_{([0], [0])}$ is the nontrivial covering $R^{2} \to T^{2}$ rather than a diffeomorphism. The flat torus therefore refutes the claim while remaining fully consistent with Cartan–Hadamard, which correctly asserts a diffeomorphism only in the simply connected case.

3.1step 2.1∎

Dimension and degeneracy conventions. [step 2.1] The witness is two-dimensional; the same computation applies to $R^{n} / Z^{n}$ for every $n \geq 2$ , where the noninjectivity is produced by any nonzero lattice vector. In dimension one the circle $R / Z$ is likewise complete and flat with noninjective exponential, but it is not needed here. The zero-dimensional case has $T_{p} M = {0}$ and plays no role. The claim was asserted for all complete $K \leq 0$ manifolds, so one counterexample suffices, and no choice beyond the inherited [A1] enters the explicit construction.

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Sources

Ved Datar, Lectures on Riemannian Geometry (2025) (standard reference, not scraped)
J.-H. Eschenburg, Comparison Theorems in Riemannian Geometry (standard reference, not scraped)