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Why does removing a submanifold of codim 2 preserve connectivity?

  1. Sep 26, 2010 #1

    HMY

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    Let M be a connected manifold. Let E be a submanifold of M of codimension at least 2.
    Show M\E is connected.

    I know examples of this result but how can one generally prove it?
     
  2. jcsd
  3. Sep 27, 2010 #2

    quasar987

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    Locally, it is true: Pick a point p in M. If p is not in E, take a coordinate nbhd U_p of p not intersecting E. Then U_p is path connected since M is. If p is in E, pick a coordinate nbhd U_p of p "adapted" to E, meaning U_p maps to R^n={(x_1,...,x_{n-2},y,z)} and E maps to R^{n-2}={(x_1,...,x_{n-2},0,0)}. So two points P,Q in U_p\E map to points whose last 2 coordinates are not both 0. Clearly we can find a path between P and Q that does not intersect E. For instance, if

    P=(x_1,...,x_{n-2},2,-1), Q = (x'_1,...,x'_{n-2},0,3),

    consider the "rectangular path" that first brings the z coordinate of P from -1 up to 3 while leaving all the other coordinates fixed, then brings the y coordinate of P from 2 down to 0 while leaving all the other coordinates fixed, and then brings the nonprimed coordinates to the primed coordinates in any way. (Note that we cannot do something like this if E has codimension 1, but if E has codim >2, then a similar argument works)

    Ok, so now we're practically done: Pick any 2 points in M\E, and a path (in M) between them. Cover the path by coordinate charts of the form considered above and extract a finite subcover. Then forget the initial path and use the local argument above to construct a path between the two points which never crosses E and hence lies in M\E, thus proving M\E is path connected.
     
  4. Sep 27, 2010 #3
    I think Mayer-Vietoris gives it immediately, but the poster may not know that theorem.
     
  5. Sep 27, 2010 #4

    lavinia

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    In a tubular neighborhood each point of the submanifold looks like a point at the center of a disk of dimension at least 2.
     
  6. Sep 27, 2010 #5

    mathwonk

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    why does removing 0 from R^2 preserve connectivity?
     
  7. Sep 28, 2010 #6

    lavinia

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    I am not sure what the spirit of your question is. Euclidean space minus a point is path connected except in dimension 1. Given 2 points you can explicitly construct the path.
     
  8. Feb 1, 2011 #7

    HMY

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    The spirit of my question was is in the sense that M could be infinite dimensional.

    Could this be proven using Sard's theorem? I also spoke with some other math people
    a while back and that is what they had suggested.

    I did look up that there is an infinite dimensional version of Sards theorem (for infinite
    dimensional Banach manifolds) but I don't see how to use this to prove M\E is connected.
     
  9. Feb 3, 2011 #8
    The idea here would be to use transversality (which is basically the same as Sard's thm). Take two points in the complement of E. A generic path between them intersects E transversally (i.e. not at all). Thus your space is path connected.
     
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