# degree of differentiability of a manifold at a point

I am neither aware fully nor have studied differential geometry, but i'd like to learn it if i get to know the answer for this question. I am asking this question based on the very superficial knowledge of differential geometry i got know reading wikipedia.

Does a manifold exist whose degree of differentiability is different at different points ?

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Yes, certainly. For example the solutions $C$ to the equation $y=|x|$ is a topological submanifold of $\mathbb R^2$. Away from the origin it's a $C^\infty$-manifold, meaning that for sufficiently small neighbourhoods $U$ of points, $U \cap C$ is $C^\infty$. But if you intersect $C$ with any neighbourhood of the origin, it's never even a $C^1$-manifold. So you can make sense of "degree of differentiability near a point". For abstract manifolds a sheafy language would be the most natural way to phrase your question.
@Gerben: the graph of any continuous function $f: \mathbb R \to \mathbb R$ is a topological submanifold of $\mathbb R^2$. In this case the function is $f(x)=|x|$. From the flavour of your argument it sounds like you're talking about the set $\{(x,y)\in \mathbb R^2 : |x|=|y| \}$ ?? This is a different kind of object. – Ryan Budney Apr 8 '11 at 17:52
@yasmar: Most definitions of differentiable structures make sense for topological manifolds -- topological manifolds are smooth $C^0$-manifolds. That means there is an atlas and the transition maps are simply homeomorphisms. But given a neighbourhood $U$ of a point $p$ in the manifold you can intersect all your charts with $U$ and then look at all the transition maps for those charts, and ask what their order of differentiability is. If it's larger than $0$, then you could say your topological manifold is $C^k$ for some $k>0$ near $p$. This is equivalent to the above submanifold def. – Ryan Budney Apr 8 '11 at 18:04