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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.

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I am curious about how you would answer the question for abstract manifolds. It seems to me that in the current answer we are talking about the degree of smoothness of the embedding, but I guess that viewpoint is hard to defend if the abstract manifold doesn't have a differentiable structure in the first place. –  yasmar Apr 7 '11 at 5:33
    
I'm slightly embarassed to ask this question, but can you explain what you exactly mean by your space being a topological submanifold? I know this example as something which is not a manifold, using the argument that any nbrhd of 0 minus 0 has four connected components instead of two. –  Gerben Apr 8 '11 at 7:32
    
@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
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@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
    
thanks for your answer. Let's just say I hadn't had my morning coffee while writing that post! –  Gerben Apr 8 '11 at 20:30

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