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I'd like to know whether a manifold of constant curvature, which has large injectivity radius at many points, can have points of arbitrary small injectivity radius.

More precisely, for a point $x$ in a Riemann manifold $M$ let $\rho(x)$ denote the injectivity radius at $x$ and let $\rho(M)$ denote the infimum of all $\rho(x)$, $x\in M$.

My question is this: Let $R>0$ be given. Do there exist $c,\varepsilon >0$ such that for every hyperbolic surface $S$, such that $$ \frac{\mathrm{vol}\big(\{x\in S: \rho(x)\le R\}\big)}{\mathrm{vol}(S)}<\varepsilon $$ one has that $\rho(S)\ge c$?

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The answer to the question is "no". To see this, fix $R>0,\ \varepsilon>0$. For any $n\ge2$, let $S_n$ be a compact Riemann surface of genus $n$ with Fenchel-Nielsen coordinates $$L=\left(\frac{1}{n},\underbrace{n,\dots,n}_{3n-2}\right),$$ $$A=(\underbrace{0,\dots,0}_{3n-3}).$$

The sequence $\left\{S_n\right\}_{n\ge2}$ converges to the hyperbolic plane in the sense of Benjamini-Schramm, i.e. $$\frac{vol(\{x\in\ S_n\ :\ \rho(x)\le R\})}{vol(S_n)}\xrightarrow[]{n\to\infty}0.$$

In particular, for $n\in\mathbb{N}$ big enough, $$\frac{vol(\{x\in\ S_n\ :\ \rho(x)\le R\})}{vol(S_n)}<\varepsilon.$$ On the other hand, due to the choice of the Fenchel-Nielsen parameters, the length of the short geodesic of $S_n$ is $\frac{1}{n}$. Therefore, for any $c>0$, for $n\in\mathbb{N}$ big enough, $$\rho(S_n)<c.$$

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  • $\begingroup$ Presumably for these metrics to converge to the hyperbolic plane, you have to choose a sequence of based metrics, i.e., choose a specific point on each of the compact surfaces. Could you comment more fully on how you choose the basepoints? $\endgroup$
    – Mikhail Katz
    Commented Feb 26 at 10:07
  • $\begingroup$ I don't think this answer is complete. Fenchel-Nielsen coordinates themselves are insufficient to control the global geometry of the surface. What could in principle happen is that shorter loops could arise (shorter than the basis elements). $\endgroup$
    – Mikhail Katz
    Commented Feb 27 at 9:17
  • $\begingroup$ I am sorry, but I think I do not understand how there could be loops shorter than any basis element. If I understood Fenchel-Nielsen coordinates properly, then in the above example the length of the shortest simple closed geodesic (not loop in general) is 1/n. Then, by the collar Theorem the injectivity radius of a point is an (increasing) function of its distance from the closest simple short geodesic, so that any geodesic loop is longer than 1/n. $\endgroup$
    – Lille Nordmann
    Commented Feb 27 at 9:45
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    $\begingroup$ That's true, but one needs a better lower bound for the other loops than just $\frac1n$ here :-) $\endgroup$
    – Mikhail Katz
    Commented Feb 27 at 9:57
  • $\begingroup$ I see the problem now, I have to be indeed be more precise on why such a sequence should be Benjamini-Schramm convergent. I'll try and work out the details and come back to that. Coming back to the original question, I think the answer is still "no". asking a uniform lower bound on the injectivity radius is equivalent to asking that the sequence of lattices which generate the surfaces is uniformly discrete. An example of a sequence of lattices which is Benjamini-Schramm convergent but not uniformly discrete can be found in "Benjamini-Schramm and spectral convergence", Example 2.10. $\endgroup$
    – Lille Nordmann
    Commented Feb 27 at 10:48

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