If $M$ is a Riemannian manifold the inverse function theorem tells us that for any $p \in M$ the exponential map gives us a nieghborhood $U$ of $p$ and normal coordinates $(x^i)$ in which the components of the metric are $g_{ij}=\delta_{ij}$ and the Christoffel symbols vanish at $p$. Why is this not the same as saying $M$ is locally flat?
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Depends on what you mean by locally flat. I think the idea is that a singleton point is not (in general) open set. – Zhen Lin May 23 '11 at 21:36
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5The metric is only $\delta_{ij}$ at the point $p$. And a point of a manifold is only an open set when the manifold is 0-dimensional. – Eric O. Korman May 23 '11 at 21:50
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@Eric: could you turn your comment into an answer so that the OP can accept it? – Tim van Beek May 24 '11 at 08:44
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The metric is only $\delta_{ij}$ at the point $p$. And a point of a manifold is only an open set when the manifold is 0-dimensional.
Eric O. Korman
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And the physicists always say $M$ is locally flat, which makes me crazy for days... – anonymous67 Jan 23 '15 at 15:58