On a closed manifold whose curvature is close to "hyperbolic"

Question

$\DeclareMathOperator\Vol{Vol}$In 1.7 on p.224 of the following paper, there is a rigidity result for compact manifolds whose sectional curvature is almost $-1$.

Gromov, M.. Manifolds of negative curvature. J. Differential Geometry 13 (1978), no. 2, 223-230.

Statement: Given $n \geq 4$ and $C>0$, there is an $\epsilon_0=\epsilon_0(n, C)>0$ so that for any $0<\epsilon<\epsilon_0$, if a compact Riemannian manifold $(M^n,g)$ satisfies $-1 \leq K \leq -1+\epsilon$ and the volume $\Vol(M,g) \leq C$, then it must admit a metric of constant negative sectional curvature.

The above result is mentioned as a standard application from the main "non-collapsing" result (in Sec 1.2 on p.223 of the above paper). Roughly speaking, under the above assumption the volume has a lower bound, and the diameter has an upper bound. However I could not figure out the proof of Statement.

Assume by contradiction, we have a sequence of $(M_i, g_i)$ whose curvature is $-1<K<-1+\epsilon_i$ and $\Vol(M_i, g_i) \leq C$, but each of $M_i$ does not admit a hyperbolic metric. then we may apply the $C^{1, \alpha}$ convergence theory of Cheeger-Gromov. Now we get a smooth manifold $N$ and diffeomorphisms $f_i: N \rightarrow M_i$ so that the pull back $f_i^{\ast} g_i$ converges to a $C^{1, \alpha}$ metric $h$ on $N$ under the $C^{1, \alpha}$ norm. Even if $\alpha$ could be $1$, the convergence is still weaker than $C^2$.

Question: Why is $(N, h)$ a hyperbolic metric?

Answer 1

I know two ways to complete the argument.

1. The lower curvature bound is preserved under GH convergence, and so is the upper curvature bound provided there is a lower bound on convexity radius. In the universal cover the convexity radius is infinite. So the universal cover converges to an Alexandrov space of curvature both $\ge -1$ and $\le -1$. Then there is a rigidity result saying that the metric must be Riemannian and hyperbolic. For the latter one usually refers to an old paper of Alexandrov [Uber eine Verallgemeinerung der Riemannschen Geometrie, Schriften Forschungsinst. Math (1957), no. 1, 33–84]. I suspect this is what Gromov had in mind.

2. One can also use smoothing (e.g. by Ricci flow run for short time) to get control of the covariant derivatives of the curvature tensor in the sequence without changing curvature bound very much. This will ensure that the convergence is $C^\infty$, and so the limit is hyperbolic.

EDIT: I take back method 2, but wish to offer a variation on 1. Instead of quoting Alexandrov's paper, (which I must confess I have not read because my German is very poor), one can argue that in the limit there is a weak curvature tensor that satisfies the Einstein equation of the hyperbolic metric, and hence by a standard regularity argument the limiting metric is $C^\infty$, and because it has curvature $\ge -1$ and $\le -1$ in the comparison sense, it is hyperbolic. The weak curvature tensor argument can be extracted from Convergence of Riemannian manifolds; Ricci and $L^{n/2}$-curvature pinching L. Zhiyong Gao J. Differential Geom. 32(2): 349-381 (1990).