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Let $\{M^n_i\}_{i=1}^\infty$ be a sequence of closed smooth Riemannian $n$-dimensional manifolds with uniformly bounded below Ricci curvature and uniformly bounded above diameter. The Gromov compactness theorem says that there exists a subsequence which converges in the Gromov-Hausdorff sense to a compact metric space $X$.

Questions. (1) Is the Hausdorff dimension of $X$ necessarily integer?

(2) Is it at most $n$?

Remark. If one assumes a stronger condition that the sectional (rather than Ricci) curvature of $M_i$ is uniformly bounded below then the answers to both questions are positive as it is shown in the theory of Alexandrov spaces.

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  • $\begingroup$ The paper "Manifold with ideal boundaries of different dimensions" by N. L. Santos in Osaka J. Math.40 (2003), 455–467; might be related to the questions at hand. $\endgroup$ – valeri Jul 25 '16 at 19:08
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    $\begingroup$ Cheeger and Colding obtained a condition on the sequences of Riemannian manifolds with uniformly lower bounded Ricci curvature (no assumption on the diameters) which guarantees that the Hausdorff dimension of the limit is an integer (see On the structure of spaces with Ricci curvature bounded below. II Theorem 1.38). While this is not exactly what you are asking it might be helpful. $\endgroup$ – Sak Jul 26 '16 at 1:07
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I cannot say anything about question (1), but the answer to question (2) is yes:

By the work of Cheeger-Colding, we can assume that $(M_i)$ endowed with the normalized volume measure converges in the measured Gromov-Hausdorff sense. Thus, the limit is a $\mathrm{CD}(K,n)$-space in the sense of Lott-Sturm-Villani (where $K$ is the uniform lower bound on the Ricci curvatures of $M_i$). These spaces have Hausdorff dimension $\leq n$.

You should find these results (or references to the original sources) in K.-T. Sturm. On the geometry of metric measure spaces I. and II.

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  • $\begingroup$ is this the right link I've added? $\endgroup$ – Amir Sagiv Aug 3 '16 at 13:37
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I think your question (1) is a conjecture by K. Fukaya.

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