# Existence of a non-compact Riemannian manifold with infinite injective radius

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

• How about the hyperbolic spaces? Jan 3 '20 at 13:51
• But hyperbolic space has negative sectional curvature. 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.