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 $n$.

 A: 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 


*

*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 $\mathbb R^n$).

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

*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.  
