Results 1 to 7 of 7

Math Help - Submersion between manifolds

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    131

    Submersion between manifolds

    Hello,

    I have a question about submersions. A submersion f:M->N between manifolds defines a foliation on M of codimension dimN.
    I agree with this. But in my proof i showed that the origin atlas on M is already a foliation atlas (by using that we have a submersion of course!).

    Now i'm not sure about the implications of this proposition. I think i can define a lot of submersions f:M->N' with dimN' \leq dimM, right? Therefore the given atlas on M can be considered as foliation atlas of codimension \leq dimM (for different submersions with different dimension of the range we get different codimension)

    I.e. i can consider one and the same atlas as a foliation atlas of various codimensions, if i find the corresponding submersion.

    This sounds a little bit odd to me.

    Therefore my question is, whether my thoughts are right? Can you interpret this in a reasonable way?

    Regards
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    Since manifolds are locally diffeomorphic to R^n there can always locally foliated. But having a submersion imposes global topologic restriction. How can you define a submersion from S^2 to R^1?
    Last edited by xxp9; February 9th 2011 at 04:37 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2010
    Posts
    131
    Hello,

    thank you for your answer. I don't know how to define a submersion from S^2 to \mathbb{R}.

    When i was thinking about submersions and foliations i had the simple example of a open cube in mind.

    For instance if we have the manifold (0,1)^n \subset \mathbb{R}^n

    Then we can define various submersions like:

    1) 0,1)^n->(0,1)^n, f=id" alt="f0,1)^n->(0,1)^n, f=id" />
    2) 0,1)^n->\mathbb{R},\; f(x_1,....,x_n):=x_1\; or\;
    3)f0,1)^n->\mathbb{R}^2\; f(x_1,...,x_n):=(x_1,x_2)" alt="f0,1)^n->\mathbb{R},\; f(x_1,....,x_n):=x_1\; or\;
    3)f0,1)^n->\mathbb{R}^2\; f(x_1,...,x_n):=(x_1,x_2)" /> and so on.

    This maps are submersions between manifolds , whereas the image of f has different dimension.

    Ok, this examples are really simple. I don't know whether one can define "various" submersions, if we have complicated manifolds. (even if they are so descriptive like your example S^2
    I have not so much experience with such situations.

    If i have understand your comment correct, it is in general difficult to define submersions, right?

    Regards
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    Since (0,1)^n is diffeomorphic to R^n, so your examples are still the simple (or even trivial) cases.
    As I said, having submersion introduces some topological restriction. That is, not all of the manifolds have non-trivial submersions.
    Take M=S^2, N=R^1 as a simple example. Since M is compact any real function f of M must have maxmimal and minimal values. Thus f must have critical points. This makes f fail to be a submersion.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2010
    Posts
    131
    Hello,

    thanks again for your explanations.
    To consider again our example M=S^2. The image of M under any real function f has to be compact again, i.e.  f(S^2)=[a,b]
    If our maximal (minimal) value is a real maximum (minimum) then we have critical values, as you said.
    But what happens if our maximal and minimal values are a and b?
    Then I think the derivative doesn't vanish anymore!?. Why is it impossoble to have such a case?

    Regards
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    Try to prove that "the derivative doesn't vanish anymore".
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2010
    Posts
    131
    Thank you very much! It must vanish of course. I had the wrong picture in mind.

    Regards
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ok, I know manifolds. What's next?
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 15th 2011, 01:50 AM
  2. Embedding of manifolds
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 22nd 2010, 11:57 PM
  3. Mobius n-manifolds within Complex n manifolds
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: January 20th 2010, 08:15 AM
  4. Manifolds
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 16th 2009, 01:44 PM
  5. [SOLVED] 1-manifolds
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 19th 2008, 03:56 PM

Search Tags


/mathhelpforum @mathhelpforum