5
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ How about the hyperbolic spaces? $\endgroup$ Jan 3 '20 at 13:51
  • 1
    $\begingroup$ But hyperbolic space has negative sectional curvature. $\endgroup$ Jan 3 '20 at 14:37
3
$\begingroup$

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 $\gamma$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.