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I have a question which I am still don't know the correct way to ask. So let me describe it and if someone can comeup with a good way to ask, please let me know.

For any inner metric space or length space $(X, d)$, the natural way to define the topological basis is considering all the open metric balls. So we can consider a $n$-dimensional topological maifold with compatiable metric structure $(X^n, d)$.

My first question is: Is there any way to ask whether there exists a differentiable structure on $X^n$ 'compatible' with the metric? i.e. what exactly could 'compatible' mean? I am thinking that: there exists a differentiable structure on $X^n$ and a Riemannian metric $g$ on $X^n$ such that the metric is induced by the Riemannian metric $g$. (Of course we need some assumption on the differentiability of $g$.) This probably too restrictive. (The example would be the Euclidean cone over round sphere which admits a metric structure, but the metric is not smooth at the apex). So what is the correct way to ask "where there exists an Alexandorv metric with curvature $\ge 1$ on exotic sphere"?

My second question is an example: Suppose $S^{n-1}$ is the $(n-1)$ standard sphere with smooth Riemannian metric such that the sectional curvature is $\ge 1$. Let $f: S^{n-1}\to S^{n-1}$ be an isometry. Is it ture that it always extend to a diffeomorphism of the disk $D^n$?

We know there exist diffeomorphisms $g$ such that it cann't extend to diffeomorphisms of the disk $D^n$. Thus we can glue the two $n$-disk along the boundary using $g$ to get a twist sphere which is exotic sphere.

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I believe you meant to write "compatible" and edited accordingly. – Lee Mosher Jun 13 '12 at 16:09
I think it makes sense to ask whether a topological space can be metrized as a Busemann G-space (in dimensions 2, 3, and 4 it is known that it must then be a topological manifold and in higher dimensions this is still open, I think). – alvarezpaiva Jun 13 '12 at 20:27

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