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I’m interested in a closed Riemannian manifold $(M^n,g)$ with $sec<0$ and $diam(M)\leq D$. My question is:

If at some point $p\in M^n$, the injectivity radius $injrad(p)\geq1$, then can we get $injrad(M^n)\geq C(n,D)$, for some constant depending on dimension and diameter?

I guess this is true, since if a negatively curved manifold has a very thin part ($injrad$ very small), then the thin part seems to be far away from the thick part ($injrad\geq1$). Does anyone know whether this is true or not?

Any comments are welcomed.

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    $\begingroup$ I don’t think one can say anything like this without a bound on curvature. I think I can construct counterexample with the curvature going to $-\infty$. $\endgroup$
    – Ian Agol
    Commented Apr 25 at 1:38
  • $\begingroup$ So, if the manifold is negatively pinched, such as $-1<\sec<0$, then this is true? $\endgroup$
    – Xin Qian
    Commented Apr 25 at 2:26
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    $\begingroup$ If $-1<K<0$, then this follows from the generalized Margulis lemma. See Theorem 9.5 doi.org/10.1007/978-1-4684-9159-3 and the Cheeger-Gromov compactness theorem. mathoverflow.net/a/258518/1345 From compactness, one gets that the pinched nonpositively curved manifolds with volume bounded below are compact in the Hausdorff topology, and hence one gets a lower bound on injectivity radius everywhere. If the volume approaches zero, then the max injectivity radius approaches $0$. By 9.5, the fundamental group is virtually nilpotent, and hence not negatively curved, a contradiction. $\endgroup$
    – Ian Agol
    Commented Apr 25 at 5:43
  • $\begingroup$ An explicit lower bound injrad(𝑀)≥𝐶(𝑛,𝐷) under the assumption −1<sec<0 is in Buser-Karcher, Gromov's almost flat manifolds, on page 28. $\endgroup$
    – user127309
    Commented Apr 25 at 8:56

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This is false even in two dimensions. Take a right-angled regular hyperbolic $2n$-gon (this may be achieved by taking a triangle with angles $\pi/4,\pi/4,\pi/n$ and reflecting around $2n$ copies at the $\pi/n$ corner). The inscribed circle and circumscribed circle will have radii of bounded difference independent of $n$.

enter image description here

Reflect along the even edges, then the odd edges, to get a surface made of four copies of this polygon. The injectivity radius at the center will be large, but the edge lengths are uniformly bounded indepedent of $n$, and hence the injectivity radius remains bounded above. Scale so that the injectivity radius at the center is $1$, then the injectivity radius will approach $0$ and the diameter will remain bounded. Since the metric is getting rescaled, the curvature approaches $-\infty$.

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