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Diameter bounded from above is usually needed in the finiteness theorem or other convergence theorems in Riemannian Geometry. Let $M^n$ be a closed manifold and {$g_i$} be a family of smooth Riemannian metrics on it with $Inj_{g_i}\geq \alpha>0$ and $Vol_{g_i}\leq \beta$. Are those two conditions enough to imply $Diam_{g_i}\leq \gamma$? If so, does the $\gamma$ depend only on $\alpha,\beta$ and $n$?

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    – Will Sawin
    Jan 23, 2020 at 23:16

1 Answer 1

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By Croke, Some isoperimetric inequalities and eigenvalue estimates, Proposition 14, on an $n$-dimensional Riemannian manifold with injectivity radius $\alpha$, a ball of radius $\alpha/2$ has volume at least $C\alpha^n/ (2n)^n$ for a constant $C$. Specifically, by their theorem 11, $C=2^{n-1}V_{n-1}^{n\phantom{1}} / V_{n\phantom{1}}^{n-1}$, where $V_n$ is the volume of the $n$-dimensional sphere.

Let $x_1,\dots,x_m$ be a maximal collection of points such that the balls of radius $\alpha/2$ centered at $x_1,\dots,x_m$ do not overlap. Then clearly $$m\frac{ C\alpha^n }{ (2n)^n} \leq \beta.$$ By maximality, the balls of radius $\alpha$ centered at $x_1,\dots, x_m$ cover $M$. Hence because $M^n$ is connected, the graph with vertices $x_1,\dots,x_m$ and edges connecting $x_i$ to $x_j$ if the distance from $x_i$ to $x_j$ is at most $2\alpha$ is connected and thus has diameter at most $m-1$.

Thus the diameter is at most $$\alpha + (m-1) 2\alpha+ \alpha = 2m\alpha \leq 2 \beta \frac{ (2n)^n}{ C \alpha^n} \alpha = \frac{ \beta}{\alpha^{n-1} } \frac{2^{n+1} n^n} {C} $$ as for the distance between $x$ and $y$, it takes a distance of $\alpha$ to get from $x$ to one of the $x_i$, then $(m-1)$ distances of $2\alpha$ to travel to an appropriate $x_j$, and then $\alpha$ to get from $x_j$ to $y$.

This is sharp up to a constant depending on $n$, as demonstrated by a suitable long thin torus.

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