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Example of a topological manifold without smooth transition functions.

  1. Jan 17, 2014 #1
    In the definition of smooth manifolds we require that the transition functions between different charts be infinitely differentiable (a circle is an example of such a manifold). Topological manifolds, however, does not require transitions functions to be smooth (or rather no transition functions at all), but just out of curiosity are there examples of simple topological manifolds (like the square) which have transition functions between different charts, but where these are not smooth?

    I tried to find some for the square, but I can not seem to find charts that overlap (the edges are in the way). Is this a general phenomenon?
     
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  3. Jan 17, 2014 #2

    jgens

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    I am fairly certain the following holds: Given a smooth manifold M we can find a new topological atlas {(U,φ)} whose transition functions are not smooth. So the interesting question here is whether there are topological manifolds which are not smoothable. It turns out the answer to this question is yes, but unfortunately there are no simple examples of this phenomenon, the first known example (to my knowledge) being a manifold of dimension 4.
     
  4. Jan 17, 2014 #3
    I don't consider myself an expert, but let's see if I can think of something.

    Think about a finite set of points in ##\mathbb{R}^n## with the discrete topology. Clearly, it is a topological manifold (each point is homeomorphic to ##\mathbb{R}^0##). Is it smooth?
     
  5. Jan 17, 2014 #4

    jgens

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    This is not the best example. Although the transition functions are not smooth in the usual sense, unless you plan on making myriad addendums in your theorems about smooth manifolds to include the 0-dimensional case, it is best to consider them smooth.
     
  6. Jan 17, 2014 #5
    Okay. Now I'm confused. Can you explain?
     
  7. Jan 17, 2014 #6

    jgens

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    Usually smoothness requires the domain of your transition functions to have non-zero dimension, so the transition functions on finite manifolds fail to satisfy this condition, and one could consider finite manifolds as non-smooth. However most theorems about smooth manifolds apply equally well to finite manifolds, so unless for each theorem about smooth manifolds you want to add caveats about how it also holds for finite manifolds, then it is best to call them smooth as well.
     
  8. Jan 17, 2014 #7

    Office_Shredder

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    A very simple example of a manifold with a non-smooth set of charts (although the manifold is smooth itself) is the interval (-1,1) with the transition maps
    [tex] \phi_1: (-1, 1/2) \to \mathbb{R},\ x\mapsto x [/tex]
    and
    [tex] \phi_2:(-1/2,1)\to \mathbb{R},\ x\mapsto x \text{ if } x > 0,\ 2x \text{ if } x < 0 [/tex]

    On the interval (-1/2,1/2) the composition [itex] \phi_2 \circ \phi_1^{-1} [/itex] is easily calculated (as [itex] \phi_1^{-1} = \phi_1 = x [/itex]) to be [itex] \phi_2[/itex] restricted to the interval (-1/2,1/2). And of course it is not differentiable on this interval.
     
  9. Jan 18, 2014 #8
    So it is actually not possible to find overlapping charts for the square with non-differentiable transition functions?
     
  10. Jan 18, 2014 #9

    Office_Shredder

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    Topologically there is no difference between a square and a circle; the fact that you say you're having trouble with the corners is a conceptual block on your part, not a meaningful statement about the corners of a square. Try adapting my example above to work with a circle.
     
  11. Jan 18, 2014 #10
    I agree. What I was trying to do was rather to find different parametrization of the unit square in R^2 which i could use for charts.

    I guess I could adapt your example to the circle by by defining the same charts on the upper half circle ##U_1=\left\{(x,y) \in \mathbb{R}^2: x^2 + y^2 = 1 \ , x \in (-1, 1/2) \ \text{and} \ y > 0\right\}##
    ##U_2=\left\{(x,y) \in \mathbb{R}^2: x^2 + y^2 = 1 \ , x \in (-1/2, 1) \ \text{and} \ y > 0\right\}##

    $$\phi_1: U_1 \to \mathbb{R},\ (x,y) \mapsto x$$
    $$\phi_2:U_2 \to \mathbb{R},\ (x,y)\mapsto x \text{ if } x > 0,\ 2x \text{ if } x < 0$$

    + some additional charts so that they cover the unit circle. Would one now say that one has defined a non-smooth structure on the unit circle?
     
  12. Jan 18, 2014 #11

    Office_Shredder

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    Yes! You should be able to write down two charts on the square which overlap and have a non-smooth transition as well with a little bit of thinking about how the circle and square are really the same topologically.
     
  13. Jan 18, 2014 #12

    jgens

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    Jesus! Read the first response in this thread! You can find such charts on any manifold!
     
  14. Jan 22, 2014 #13

    WWGD

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    Just to make the point that it is true that the two are the same topologically , e.g., find a circle circumscribing the square ; circle centered at (0,0) . Then draw any half-line L from (0,0) ; the half-line L will hit the circle and the square once each, at, say, p, p' respectfully (there will be four points where there is just one intersection; with this layout, this will be at (1,0),(0,1), (-1,0), (0,-1)). Define a function from the square to the circle that takes p to p'. This is a continuous bijection between compact and Hausdorff, so it is a homeomorphism.

    But the two are not diffeomorphic ( using the subspace topology). If they were, then a diffeomorphism f would give rise to a vector space isomorphism between the respective tangent spaces. But the square does not have a well-defined tangent space at the corners, while the circle has no such problem.
     
    Last edited: Jan 22, 2014
  15. Jan 22, 2014 #14

    jgens

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    There are some nuanced points here that need elaboration. The argument alluded to above is sufficient to establish the non-smoothness of the standard (topological) embedding for the square (i.e. the inclusion). A homeomorphism between the square and circle, however, can be turned into a diffeomorphism rather easily. Simply use this map to pull the charts on S1 back to charts on the square! So the issue is not whether the square or this homeomorphism is smoothable, but rather that the standard embedding is not smooth.
     
  16. Jan 22, 2014 #15

    WWGD

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    Yes, of course, I agree here; the general result is that if X,Y are homeomorphic topological spaces , by, say, h, and Y admits smooth manifold charts, then one can use h to pullback the smooth charts from Y into X, i.e., if X,Y are homeomorphic and Y is a smooth manifold, then X can be made into a smooth manifold using h.
     
  17. Jan 27, 2014 #16

    WWGD

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    Maybe for a quick summary, the circle _ as a subspace_ is not a submanifold (remember that there may be non-compatible charts for a given manifold--viewed as a topological space), and it is not, under this collection of charts, diffeomorphic to the (standard embedding of the)circle, but it can be given smooth charts using the pullback by the/a diffeomorphism with the circle.
     
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