1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The S^infinity sphere

  1. Feb 9, 2007 #1
    Hi,

    I was able to enumerate all the subcomplexes of S^{infinity}, where S^{infinity} has two 0-cells, two 1-cells, two 2-cells, etc. But how do I show that S^{infinity} is contractible?

    Can anyone point me in the right direction? X is contractible if and only if the identity map of X is homotopic to a constant map.

    I guess I don't see what kind of homotopy (a shrinking map) I need to set up between X and a point.

    Thank you!
     
  2. jcsd
  3. Feb 9, 2007 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    My best guess would be to construct your homotopy piece wise.

    Let P be a point.

    For each of your 0-cells, can you find a homotopy that maps your 0-cell to P?

    For each of your 1-cells, can you find a homotopy that maps your 1-cell to P, that restricts to the aforementioned homotopies on the two endpoints?

    For each of your 2-cells...
     
  4. Feb 11, 2007 #3
    I worked on it yesterday and I think the following might work: choose an arbitrary point p =(x_1, x_2, x_3,...) in S^infinity. Then show that at time t=1, this point is located at q =(y_1,0,0,0,0,....). This method should work.

    The above advice should work as well (I thought about it before), but remember that we have two 0-cells, two 1-cells, two 2-cells, etc. We know that the unit circle S^1 is not contractible, and neither does the unit sphere. So say we assume that S^infinity = S^2. Then if we look at one cell at a time, sure, we can find a homotopy between a cell and a point, but if we look at S^2 as a whole, it certainly is not contractible.

    So I'm not sure if looking at each of the cells and then find a contracting map works, otherwise, this will imply that S^1, S^2, S^3, etc. are contractible. But this certainly is not true.
     
  5. Feb 11, 2007 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Your reasoning about 'cell at a time' is flawed - there is nothing that says that the map must stay within the cell at each point ("in time"). In fact the homotopy *must* map each cell to the point.
     
  6. Feb 11, 2007 #5

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    what is clear is that each copy of S^n shrinks to a point in S^(n+1).

    what you must do is see whetehr this allows you to write down a global homotopy inductively.

    by the way what is your definition of S(infinity)?
     
  7. Feb 11, 2007 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    If it's mine then that is trivially the homotopy you need (since it is surely a limit in some appropriate sense).
     
  8. Feb 12, 2007 #7
    "what is clear is that each copy of S^n shrinks to a point in S^(n+1)."

    That makes sense. However, writing down the global homotopy inductively might be a little difficult, I think.

    The definition of S^{infinity} is the union of S^n, where n=0,1,2,3,..., where each of the subspheres S^k is a subcomplex. It is obtained inductively from the equatorial S^(k-1) by attaching two k-cells, which are the components of S^k - S^(k-1).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: The S^infinity sphere
  1. Infinity in equations (Replies: 1)

  2. Limit on infinity (Replies: 7)

  3. Integrating Infinities (Replies: 5)

Loading...