Smooth Homotopy, Regular Values (Milnor)

In summary, people neglect regular values whose inverse image is the empty set. This creates a problem because it is not clear what the range of a smooth map is.
  • #1
Sina
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0
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
 
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  • #2
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.
 
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  • #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
 
  • #4
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.
 
  • #5
Ha I actually never thought of empty sets as such thanks for the enlightening comment :) I will think and research more about it
 

1. What is smooth homotopy?

Smooth homotopy is a mathematical concept that refers to the continuous deformation of one smooth function into another. In other words, it is a way of transforming one function into another while maintaining its smoothness. This is often represented by a family of functions that vary continuously between the starting and ending functions.

2. What is the significance of smooth homotopy?

Smooth homotopy is an important tool in differential topology and differential geometry, as it allows for the study of smooth functions and their properties. It also helps in understanding the global behavior of functions by studying their local behavior.

3. What is a regular value in Milnor's theorem?

In Milnor's theorem, a regular value is a point in the domain of a smooth function where the differential of the function is surjective. In other words, the tangent space at that point has the same dimension as the codomain of the function. Regular values are important in the study of smooth manifolds and critical points of functions.

4. How does smooth homotopy relate to regular values in Milnor's theorem?

Smooth homotopy is used in the proof of Milnor's theorem, which states that the set of regular values of a smooth map between two manifolds is dense. This means that for almost all points in the codomain, the function has a surjective differential. Smooth homotopy is used to show that any two regular values can be connected by a path of regular values, which shows that the set of regular values is connected and therefore dense.

5. What are some applications of smooth homotopy and regular values?

Smooth homotopy and regular values have many applications in mathematics, including in differential topology, differential geometry, and algebraic topology. They are also used in physics, particularly in the study of gauge theories and topological defects. They also have applications in computer graphics and animation, where they are used to create smooth and realistic shapes and movements.

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