Let $M$ be a complete non-compact Riemannian manifold with non-negative sectional curvature. Please tell is it possible that $M$ has infinite injective radius expect Euclidean space?
Thank you

  • $\begingroup$ How about the hyperbolic spaces? $\endgroup$ Jan 3 '20 at 13:51
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    $\begingroup$ But hyperbolic space has negative sectional curvature. $\endgroup$ Jan 3 '20 at 14:37

EDIT: I'm not sure this answer is correct - I said $|J|^2|J'|^2 \le g(J',J)^2$ when of course the inequality goes the other way. Don't have time to think about it again right now, but if someone else could check if there's a fix or if this is just wrong, that'd be nice.

No, this is not possible - if the curvature is everywhere non-negative, then any small pocket of positive curvature will lense some geodesics together. We can prove this using Jacobi fields:

If $M$ is non-negatively curved and not Euclidean, then there must be some $u,v \in T_pM$ with $R(u,v,u,v) > 0.$ By completeness there is a geodesic $\gamma : \mathbb R \to M$ with $\gamma(0) = p$ and $\gamma'(0) = u.$ Let $J$ be the Jacobi field along $\gamma$ with $J(0) = v$ and $J'(0) = 0,$ so that $J$ generates a family of geodesics that are roughly parallel near $p.$

Let $f = |J|$. Differentiating, we find $f' = \frac1fg(J',J)$ and thus $$f'' = -\frac1{f^3}g(J',J)^2 + \frac1f(g(J'',J)+|J'|^2), \tag1$$ and the initial conditions for $J$ tell us that $f'(0)=0.$ Substituting the Jacobi equation $J'' = R(\gamma', J)\gamma'$ into $(1)$ we obtain

$$f'' = -\frac1{f^3}g(J',J)^2 + \frac1f(-R(\gamma', J, \gamma', J)+|J'|^2).$$

Our curvature assumption then tells us that $$f'' \le \frac1{f^3}\left(|J|^2|J'|^2-g(J',J)^2\right) \le 0,$$ with this inequality being strict at $t=0$ (where the given curvature is exactly the one we assumed was positive).

Thus we have established that $f$ is strictly concave-down at $t=0$ and weakly concave-down everywhere; so $f$ must have zeroes on both sides of the origin, which correspond to conjugate points on $\gamma$.


Take the paraboloid of revolution in $R^3$ with the induced Riemannian metric. Since the paraboloid is strictly convex, the metric is positively curved. The exponential map from the tip of the paraboloid is a diffeomorphism.

Edit. It is possible that I misunderstood the question and you were asking about a manifold where at every point the injectivity radius is infinite. In this case, the answer is that the manifold has to be flat, see Jason's answer here.


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