It is known that given a Riemannian manifold, then the tangent cone (as a metric space) at any point $p$ is isometric to the tangent space at $p$, with the metric given by the metric tensor.

Is there a converse, or an additional condition to have a converse, in the following form?

Given a metric space $X$ such that the tangent cone at any point is a Euclidean space of dimension $n$, and (possibly additional condition), then $X$ is a Riemannian manifold of dimension $n$.

  • 2
    $\begingroup$ e.g., you may assume that $X$ has bounded curvature (since "at the end" it will). Then $X$ is a manifold by known theorems (see "Alexandrov spaces"). $\endgroup$
    – valeri
    Mar 13 '17 at 12:32
  • 1
    $\begingroup$ @valeri So is there an actual statement "A metric space with curvature bounded below is a Riemannian manifold iff its tangent cone is an Euclidean vector space"? $\endgroup$
    – geodude
    Mar 13 '17 at 12:37
  • 2
    $\begingroup$ note, that if $X$ has bounded (below and above here) curvature and no boundary, then you may have tangent cone=euclidean space for free. Then, if I remember correctly, yes, it was proved that Al.space with bounded curv is ($C^{1,\alpha}$ ?) manifold. Can not help with references though :( So, bounded curvature is a very strong condition, you might like smth else. $\endgroup$
    – valeri
    Mar 13 '17 at 13:10
  • 1
    $\begingroup$ According to Burago-Burago-Ivanov (see Theorem 10.10.13) The result @valeri mentioned is due to Nikolaev (the proof is not given, but a reference is given to V. Berestovskii and I. Nikolaev, Multidimensional generalized Riemannian spaces, in Geometry IV. Non-regular Riemannian geometry. Encyclopaedia of Mathematical Sciences, Springer-Verlag, Berlin, 1993, 165–244.) $\endgroup$ Mar 13 '17 at 14:58
  • 3
    $\begingroup$ Let $X$ be any Riemannian manifold, and $x$ a point. Choose a sequence $(x_n)$ with $d(x_n,x)=2^{-n}$. Let $V_n$ be a subset contained in the ball of radius $2^{-2^{n}}$ around $x_n$. Then it is immediate that the tangent cone of $Y=X\smallsetminus \bigcup V_n$ at $x$ is the same as that of $(X,x)$. But (if every $V_n$ is nonempty), $Y$ is not a topological manifold at $X$. If moreover $V_n$ are chosen closed (e.g., the whole closed ball), $X$ is Riemannian at every other point, so every tangent space is isometric to a Euclidean space ("is a Euclidean space" is an ambiguous formulation) $\endgroup$
    – YCor
    Mar 13 '17 at 16:54

To provide some context the subsets of a Euclidean space that can be approximated by affine planes on every scale are known as Reifenberg-flat sets after E. R. Reifenberg who proved in the 1960s that such sets are bi-Holder to a Euclidean space. There is a substantial literature on the subject (search on "Reifenberg-flat").

Now regarding your specific question: T. Colding and A. Naber construct in Lower Ricci Curvature, Branching, and Bi-Lipschitz Structure of Uniform Reifenberg Spaces a metric space $Y$ such that

  1. The tangent cones of $Y$ are all isometric to $\mathbb R^n$ (which by an earlier work of J. Cheeger and Colding implies that $Y$ is bi-Holder homeomorphic to $\mathbb R^n$).

  2. Every bounded set of $Y$ is bi-Lipschitz embeddable in some Euclidean space.

  3. There is no homeomorphism of $Y$ onto $\mathbb R^n$ such that the pullback geometry is induced by some $C^{0,\beta}$ Riemannian metric where $0<\beta<1$.

To show (3) they prove that geodesic in $Y$ branch in the sense of Theorem 1.3 of the linked paper.

Moreover, $Y$ occurs ``in nature'' as a pointed Gromov-Hausdorff limit of a noncollapsing sequence of Riemannian $n$-manifolds with a common lower bound on Ricci curvature.

I am not sure whether the metric on $Y$ can be induced by a $C^0$ Riemannian metric but in any case branching of geodesics is not the property a decent $C^0$ metric should be proud of.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.