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Let $(M,g)$ be a closed Riemannian manifold and let $N$ be a closed embedded submanifold. A tube $T(N,r)$ of radius $r$ of $N$ is defined as the set of points of $M$ which can be reached by a geodesic path of length $< r$ starting from $N$ orthogonally. We further require that $T(N,r)$ is open in $M$ and that the map $(p,v)\mapsto \exp_{p}(v)$ defined on $\{(p,v)\in TN^{\perp}\,\,:\,\, g_p(v,v)<r^2\}$ is a diffeomorphism onto $T(N,r)$. Then the normal injectivity radius of $N$ (which we denote by ${\rm nir}(N)$) is defined as the maximal radius of a tube around $N$.

Assume that the sectional curvatures of $M$ and the second fundamental form of $N$ are uniformly bounded by a constant $K$. Is it possible to show that ${\rm nir}(N)\geq c(K)$, for some positive constant $c(K)$ depending only on $K$? Can this bound be obtained explicitly?

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This is false as can be seen already on curves in the plane. Consider a "Dumbbell" shaped curve in the plane.

Assume that the two "weights" are roughly circles of radius 1. If these two circles are far enough, they can be joined smoothly by a neck as thin as you like in the middle while the curvature of the curve stays less than 1.

The normal injectivity radius of these examples are as small as one wants.

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  • $\begingroup$ Thank you for your answer. Maybe the lower bound can be obtained by introducing some other dependency, such as the area of $M$? $\endgroup$ – pedro Oct 4 '16 at 13:46
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    $\begingroup$ I'll have to think about it but I think you can build a counterexample in a fixed compact manifold... $\endgroup$ – Thomas Richard Oct 4 '16 at 14:09
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This is false also if you add some conditions on $M$. It seems that you must add more dependency on $N$ in order to get a lower bound.

Fix a natural $n$ and consider the subtorus $N_n$ in $M=\mathbb{R}^2/\mathbb{Z}^2$ given as the image of $[0,1]$ under $x\mapsto (x,nx)$. Here $M$ is a fixed flat Riemannian manifold of volume 1 and $N_n$ is a sequence of flat submanifolds of arbitrarily small nir. You can also have such examples in higher dimensions, of course (and also in $\dim(M)=1$; I hesitated to give a trivial counterexample...).

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  • $\begingroup$ Thank you very much. Very nice example! Does anyone know if the lower bound on nir can be obtained by assuming that the area of $N$ is bounded by $K$? $\endgroup$ – pedro Oct 5 '16 at 14:00
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    $\begingroup$ @pedro: you can take $M$ a 3-dimensional torus with one very short systole (e.g. $\mathbb{R}/\mathbb{Z}\times \mathbb{R}/A\mathbb{Z}\times \mathbb{R}/\varepsilon\mathbb{Z}$) and $N$ a loop that follows twice (or more) the systole with small lateral displacements. Then it is easy to keep all curvatures bounded while making $nir$ arbitrarily small, and $N$ can have small volume (and $M$ can have any volume). $\endgroup$ – Benoît Kloeckner Oct 5 '16 at 14:41

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