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Smooth Homotopy, Regular Values (Milnor)

  1. May 10, 2012 #1
    Hello, I have a question regular values and smooth homotopies. Usually in giving the definition of regular value, they disregard the regular values whose inverse image is empty set (although they should be called regular values if we want to be able to say that set of regular values is dense for smooth functions :)). Also given a map f: M [itex]\rightarrow [/itex]N a map from some m-dimensional manifold to n-dimensional manifold, and y a regular value [itex]f^{-1}(y)[/itex] is said to be m-n dimensional submanifold and when stating this, they also disregard regular values whose inverse image is the empty set. Now this seems to create a problem, or maybe I am wrong.

    Consider the example in milnor's book where he says a orientation reversing diffeomorphism of a compact boundaryless manifold can be smoothly homotopic to the identity (because homotopic maps have same index).

    As M and N I take the unit disk [itex]D^2[/itex]. Consider the "homotopy" F: [0,1] x [itex]D^2[/itex] [itex]\rightarrow[/itex] [itex]D^2[/itex]

    F(x,y,t) = (x, -1(1-t)y +ty). F(x,y,0) = (x,-y) and F(x,y,1) = (x,y) . First of all why is this not a smooth homotopy. Yes as t changes, range becomes a subset of unit disk and infact at t=1/2 it is the intersection of the unit disk with the y=0 line but nothing is said about the range of maps in smooth homotopy.

    Moreover, DF = [1 0 0 ; 0 2t-1 2y] so that for instance z=(0,-1) is a regular value whose inverse image is the two points (0, 1,0) and (0,-1,0). But since z is a regular value shouldn't this be a 1-dimensional submanifold? I think the problem here arises because intersection of [itex]f^{-1}(z)[/itex] with the sets t x [itex]D^2[/itex] is empty for t not equal to 0 or 1.

    Thanks
     
  2. jcsd
  3. May 10, 2012 #2

    quasar987

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    In your second paragraph, do you mean to say

    "Consider the example in milnor's book where he says a orientation reversing diffeomorphism of a compact boundaryless manifold cannot be smoothly homotopic to the identity (because homotopic maps have same degree)." ?

    If so, then perhaps what fails in your example is not so much the smoothness of your map as it is the fact that D² has a boundary. And if you remove the boundary, you lose compactness.

    In your last paragraph, you seem to be again applying a thm about manifolds without boundary to a manifold with boundary.
     
    Last edited: May 10, 2012
  4. May 10, 2012 #3
    Yes indeed I meant degree (and Brouwer degree). and yes I later realized disk is with boundary. and if you try to take circle or torus it does not work so everything seems okay =) thanks for the reply
     
  5. May 10, 2012 #4

    mathwonk

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    People do not disregard empty fibers. rather note that by definition the empty set is a manifold of every dimension, and also a regular value. i.e. whatever has to be true for every point is certainly true for the set with no counterexamples.
     
  6. May 11, 2012 #5
    Ha I actually never thought of empty sets as such thanks for the enlightening comment :) I will think and research more about it
     
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