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Suppose I have a smooth, noncompact manifold $M$ with metrics $g_i$ for $i = 1,2$. Suppose there exists a $C \geq 1$ such that $$ C^{-1} d_1(x,y) \leq d_2(x,y) \leq C d_1(x,y)$$ where $d_i$ are the induced distance metric of $g_i$. Then, if the injectivity radius with respect to $g_1$ satisfies $\mathrm{inj}(M, g_1) \geq \kappa > 0$, does it follow that $\mathrm{inj}(M, g_2) \geq \kappa' > 0$ for some $\kappa'$?

I think I've confused and misled some people by introducing the completely irrelevant $\kappa$ and $\kappa'$. Sorry about that. What I mean is that if $\mathrm{inj}(M,g_1) > 0$ then does it follow that $\mathrm{inj}(M,g_2) > 0$?

Note also that $\mathrm{inj}(M,g)$ is the infimum of the injectivity radius with respect to $g$ at each point.

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  • $\begingroup$ do you mean injectivity radius around every point is positive (and finite) in metric 1? $\endgroup$
    – Will Jagy
    Commented Dec 1, 2013 at 3:03
  • $\begingroup$ Do you mean that $M$ is not (necessarily) compact and you care about the existence of SOME $\kappa^\prime,$ or do you want $\kappa^\prime$ to depend only on $\kappa$ and $C?$ $\endgroup$
    – Igor Rivin
    Commented Dec 1, 2013 at 3:41
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    $\begingroup$ No, the injectivity radius can be made arbitrary small by arbitrary small bi-Lipschitz perturbation. You need to produce a short geodesic diangle; it is easy. $\endgroup$
    – Anton Petrunin
    Commented Dec 1, 2013 at 3:45
  • $\begingroup$ I mean a noncompact manifold. I don't care whether $\kappa'$ depends on $\kappa$. The $\mathrm{inj}(M,g_i)$ is the infimum of the $\mathrm{inj}(M,g_i,x)$ over each $x$. $\endgroup$
    – Lashi
    Commented Dec 1, 2013 at 16:14
  • $\begingroup$ I presume what you mean by saying that the injectivity radius can be made arbitrarily small is that for every $\epsilon > 0$ there is a bi-Lipschitz transform such that the injectivity radius of the metric induced by the bi-Lipschitz transform is now less than $\epsilon$? If that's the case, it doesn't exclude it from being $0$. $\endgroup$
    – Lashi
    Commented Dec 2, 2013 at 0:45

2 Answers 2

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This is an expansion of Anton Petrunin's comment.

Let me describe how to perturb the standard metric of the plane so that the resulting metric is bi-Lipschitz to the original with Lipschitz constant arbitrarily close to 1 and has arbitrarily small injectivity radius.

Locate the plane in $\mathbb R^3$ as the $xy$-plane. Remove the unit disc centered at 0. Pick a point $(0,0,\delta)\in\mathbb R^3$ slightly above the origin and connect it by segments with the points of the boundary circle of the removed disc. This gives you a singular surface consisting of a planar part and a conical part. On the cone, there are arbitrarily short geodesic digons near the apex. Fix such a digon of small length $\rho>0$ and smoothen the surface in a so small neighborhood of the apex that it does not reach the digon. Also smoothen the surface near the circle. Now we have a smooth surface with injectivity radius $\le\rho$ because the geodesic digon is still there.

The projection of this surface to the plane is a bi-Lipschitz diffeomorphism with bi-Lipschitz constant $C=\sqrt{1+\delta^2}$. This map sends the surface's metric to a new Riemannian metric on the plane, and it is $C$-bi-Lipschitz to the original one.

To construct a metric with zero injectivity radius, just repeat this construction in a countable collection of disjoint discs $B_i\subset\mathbb R^2$, $i=1,2,\dots$, using the parameter $\rho=\rho_i\to 0$.

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In a picture, glue in a very small sphere via a much smaller small tube attached at one end to the small sphere and at the other end to your manifold. The injectivity radius is now very small, as most geodesics on the sphere are still periodic, unaffected, because the tube is so small by comparison. The distances are almost unchanged. So for any Riemannian manifold, there is another Riemannian manifold with arbitrarily close metric which has injectivity radius arbitrarily close to zero. It will take some work to make a proof, because it isn't easy to write out an explicit diffeomorphism between, for example, Euclidean space and Euclidean sphere with a very small sphere glued to it.

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  • $\begingroup$ Thanks for the answer, but I'm having trouble understanding it. The manifold we start with is the Euclidean sphere? We now use a small tube and attach a smaller sphere to it? But now you're suggesting to pull back the Euclidean metric through a diffeomorphism and compare the two metrics? Are you suggesting this is a counterexample which shows the question has a negative answer? $\endgroup$
    – Lashi
    Commented Dec 2, 2013 at 0:48
  • $\begingroup$ Just do everything as surfaces in 3-space and used the metric induced by the 3-dimensional Euclidean metric. $\endgroup$
    – Deane Yang
    Commented Dec 2, 2013 at 2:18

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