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Let $(M^n,g)$ be a smooth compact Riemannian manifold. It is well known that any sequence $\{X_i\}$ of compact subsets of $M$ has a subsequence which converges in the Hausdorff metric to a compact subset $X\subset M$.

Assume now that $\{X_i\}$ are, in addition, smooth connected submanifolds with a uniform lower bound on the sectional curvature and a uniform upper bound on the diameter with respect to the induced intrinsic (!) metric.

Question. What is known about a limit space $X$? E.g. should $X$ have an integer Hausdorff dimension?

Remark. Under the above assumptions, the Gromov compactness theorem implies that there is a subsequence converging to a compact Alexandrov space $Y$ in the Gromov-Hausdorff sense (thus $Y$ is not a subset of $M$ a priori). Is there any relation between $X$ and $Y$? (I believe there is a 1-Lipschitz map onto $Y\to X$.)

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You are right about 1-Lipschitz map, but that is about all you can expect. In particular the dimension of $X$ might be not integer.

Assume your $X_i$ all isometric to to a flat torus $\mathbb{T}$ and $M=\mathbb{E}^N$; that is your manifold is the Euclidean space of high enough dimensions $N$.

By Nash and Whitney, any short map $\mathbb{T}\to \mathbb{E}^N$ can be arbitrary well approximated by an isometric embedding.

It reamins to construct a short map $\mathbb{T}\to \mathbb{E}^N$ with non-integer Hausdorff dimension. An image of that type is shown on the picture. It can be arranged that the Cantor set on the top has Hausdorff higher than 1 and lower than 2. FRACTAL-3d-FLOWER,

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  • $\begingroup$ But by Nash and Whitney, the isometric approximation of the short map is only $C^1$-smooth. That probably means that $X_i$ are not infinitely smooth submanifolds. Can one make them to be infinitely smooth? $\endgroup$
    – asv
    Commented Jan 21, 2016 at 13:56
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    $\begingroup$ @sva No, I use the other theorem of Nash --- if $N$ is large, you can do the same for $C^\infty$ embeddings. $\endgroup$
    – Anton Petrunin
    Commented Jan 21, 2016 at 14:42

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