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MATHEMATICS have length larger than 1,000 except for the three coordinate plane sections. The coordinate sections x = 0 and y = 0 contain the largest axis and fail to minimize between the antipodal points on their second axis. The remaining coordinate section z = 0 is the only half-geodesic. Example 2.3. For each integer n ≥ 2, there exist Riemannian metrics on the 2-sphere with exactly n half-geodesics (6); a sequence of such metrics is constructed converging to a doubled regular 2n-gon in the Gromov–Hausdorff metric. Example 2.4. There exist nonround Riemannian 2-spheres arbitrarily close to a round sphere swept out by a continuum of half- geodesics: Let 0 < c < 1 and consider the oblate ellipsoid

x 2 + y 2 + (z/c)2 = 1.

This ellipsoid is rotationally symmetric about the z-axis. The meridians are half-geodesics of length twice the diameter. The equator is a closed geodesic with length greater than twice the diameter and therefore is not a half-geodesic. 3. Proofs Proof of Theorem 1.1: Geodesics in a Blaschke manifold M are (closed) half-geodesics by (ref. 3, Proposition 5.39). Conversely, assume all geodesics in M are (closed) half-geodesics. Let SM denote the unit-sphere bundle of M and

C : SM → R the cut time function. Then C (v) equals half the prime length of the closed geodesic with initial velocity v ∈ SM . The values of C are discrete by ref. 10. As C is continuous and SM is connected, C is constant. Therefore, all prime closed geodesics have equal length 2C and minimize up to but not beyond length C . Conclude

inj ≤ diam ≤ C ≤ inj .

 A proof of the next proposition is presented at the end of this section. Proposition 3.1. Let M a Besse n-sphere with diameter one in which all prime geodesics have length two. If p ∈ M is a diameter realizing point, then the restriction of expp to the unit sphere in Tp M is a point map. Proof of Theorem 1.2 assuming Proposition 3.1: Rescale the metric, if necessary, so prime geodesics have length two. Then inj ≤ diam ≤ 1. It suffices to prove 1 ≤ inj , or equivalently, each geodesic in M of length one is minimizing. Fix p ∈ M and a unit-length vector v ∈ Tp M . Let γv : [0, 1] → M

denote the length one geodesic segment with γ˙ v (0) = v. By assumption, the point p lies in a half-geodesic of length two. Let q denote the point in this half-geodesic with d(p, q) = 1. By Proposition 3.1, γv (1) = q. As the endpoints of γv are distance one apart and γv has length one, γv is minimizing.  Proof of Theorem 1.4 assuming Theorem 1.2 and Propostion 3.1: Rescale the metric, if necessary, so prime geodesics have length two. Fix p ∈ N and a unit vector v ∈ Tp M that is not tangent to N at p. Let

γv : [0, 1] → M

denote the length one geodesic determined by γ˙ v (0) = v. By ref. 11, it suffices to prove

t ∈ (0, 1) =⇒ γv (t) ∈/ N and γv (1) ∈ N .

A half-geodesic in M tangent to N is also a half-geodesic in N . By Theorem 1.2, N is a round sphere of diameter one. Let q be the point antipodal to p in N . By Proposition 3.1, γv (1) = q ∈ N .

Moreover, as in the previous proof, γv : [0, 1] → M is a minimizing geodesic. As N is round, each geodesic leaving p starting tangent to N is minimizing up to length one. As two minimizing geodesic segments with a common initial point can only intersect again in their endpoints, the requisite condition t ∈ (0, 1) =⇒ γv (t) ∈/ N is satisfied.  Some well-known preliminary material is needed in the proof of Proposition 3.1: Given (x, y) ∈ M × M , let Λ(x, y) denote the pathspace consisting of piecewise smooth maps c : [0, 1] → M with c(0) = x and c(1) = y. Define the length and energy functions

L : Λ(x, y) → [0, ∞) and E : Λ(x, y) → [0, ∞)

by 1 1 Z Z L(c) = kc˙(t)kdt and E(c) = kc˙(t)k2dt.

0 0 The Cauchy–Schwartz inequality implies L2(c) ≤ E(c)

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MATHEMATICS n−1 Let f¯ : S → Λ(p, z) be a continuous map representing a nonzero class in πn−1(Λ(p, z)). By Lemma 3.2, critical points of length greater than two have index at least n. It follows (cf. ref. 13, Theorem 2.5.16) f¯ is homotopic to a map

f : S n−1 → Λ4.1(p, z).

Let τ : [0, ] → M be a minimizing geodesic with τ(0) = z and τ()= q and define

g : S n−1 → Λ(p, q)

by g = τf . For each θ ∈ S n−1, E(f (θ)) 1 E(g(θ)) = +  ≤ 2E(f (θ)) + ≤ 8.7 1 −  2 concluding the proof. 

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