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Let $M$ be a smooth compact manifold, let $p\in M$ and let $g_0$ be a fixed Riemannian metric on $M$.

Does there exists a constant $C>0$ such that for any Riemannian metric $g\ge g_0$, the volume of the geodesic ball with respect to $g$ satisfies $Vol_g(B_g(1,p))\ge C$?

In other words, can we expect that the volume of these balls does not shrink even after making the Riemannian metric really large?

Attempt: All of the results I've researched on this topic involve a fixed Riemannian metric on $M$ and bounding volumes in terms of the injectivity/filling radius. Nothing I've seen so far allows for the Riemannian metric to change as well.

I suppose the result has to be true for $\mathbb{S}^1$ because volume is the same as length, so for any large Riemannian metric, any geodesic ball will have a volume of $1$ as long as the volume of $\mathbb{S}^1$ itself is above $1$.

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cantstop
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1 Answers1

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No - a lower bound $g \ge g_0$ is not sufficient to prevent thin necks. For example, if $M$ is a surface and $g_0$ is flat on some small ball $U$, then we can think of the $g_0$ geometry on $U$ as a flat disc, and replace it in $g$ with a tall peak of width roughly $\epsilon$:peak

Since this is a graph over the flat disc, we have $g \ge g_0$; but the area of $B_1(p)$ will be roughly $2 \pi \epsilon$, which we can make as small as we like by letting $\epsilon \to 0$.

Anthony Carapetis
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