# Euclidean tangent cone implies Riemannian manifold

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 $n$.

• 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"). Mar 13 '17 at 12:32
• @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"? Mar 13 '17 at 12:37
• 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. Mar 13 '17 at 13:10
• 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.) Mar 13 '17 at 14:58
• 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)
– 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.