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Does there exist a Riemannian manifold $(M,g)$ such that for every $S>0$ there is a geodesic unit ball $B\subset M$ with $\text{Vol}(B)>S$?

If I understand the Bishop-Gromov-inequality correctly, then this cannot happen for complete $(M,g)$. However I am mostly interested in the non-complete case.

My attempt to find statements about $\sup_B \text{Vol}(B)$ were not very succesful, because most people seem to be interested in lower bounds on manifolds with finite volume.

Jan Bohr
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    Of course this can happen even for complete manifolds. As a warm up, take the disjoint union of infinitely many hyperbolic spaces $X_k$ of constant curvature $-k$, $k=1, w, 3....$ Then modify this example to get a connected manifold if you wish. – Moishe Kohan Feb 09 '18 at 21:17
  • Oh, you are absolutely right, I completely ignored the curvature bound that you need in order to apply Bishops result. Thank you, if you write an answer I can accept it. – Jan Bohr Feb 11 '18 at 15:00

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Let $(M,g)$ be the disjoint union of complete simply-connected $n$-dimensional Riemannian manifolds $X_i$ of curvature $-i^2$, where $i\in {\mathbb N}$ (each component of $M$ is, therefore, a rescaled copy of the hyperbolic $n$-space). Then if $B_i\subset X_i\subset M$ is the sequence of unit balls, we obtain $$ \lim_{i\to\infty} Vol(B_i)=\infty. $$ One can easily modify this example to make it connected (and, with more work, even contractible).

Moishe Kohan
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