How to make homotopy smooth

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  • #1
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Main Question or Discussion Point

Let [itex]G\subset\mathbb{R}^n[/itex] be some open set, and [itex]x_1, x_2:[0,1]\to G[/itex] be differentiable paths with the same starting and ending points. Assume that there exists a homotopy [itex]f:[0,1]^2\to G[/itex] between the two paths. That means that the f is continuous, and the following conditions hold.

[tex]
f(t,0)=x_1(t),\quad f(t,1)=x_2(t),\quad f(0,s) = x_1(0) = x_2(0),\quad f(1,s) = x_1(1) = x_2(1)
[/tex]

How do you prove, that there also exists a homotopy [itex]g:[0,1]^2\to G[/itex] between the two paths so, that for all [itex]s\in [0,1][/itex], the mapping
[tex]
t\mapsto g(t,s)
[/tex]
is differentiable?

This seems a clear claim, but doesn't seem to come easily from the definitions.
 

Answers and Replies

  • #2
Hurkyl
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I know in one dimension, there's a technique called smoothing. Choose a [itex]C^{\infty}[/itex] function s, and then you can "smooth" any function f by defining:

[tex]
\hat{f}(x) = \int_{-\infty}^{+\infty} f(t) s(x - t) \, dt
[/tex]

[itex]\hat{f}[/itex], I think, is a [itex]C^{\infty}[/itex] function. Generally, you'd probably choose an s that is only nonzero on a finite interval, like [-1, 1].

Maybe you can use a similar technique here to smooth your homotopy into a differentiable one.
 
  • #3
mathwonk
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there is a vast theoiry of approximating continuous functions by smoothere ones. going back now ov er 40 years mentally, is eem tor ecall the weierstrass theorem as generalized by marshall stone in this context.

the opriginal theorem said any ciontinuosu fucntion on a clsoed bounded interval can be uniformly apporoximated by polynomials, generalizing one says that a subslgebra of continuous functions on a compact hausdorff space is nuiformly dense if it is constant containing, and separates points?

since the coordinate functions separate points, the algebra they generate, i.e. the polynomials, is dense in any square in euclidean space.

so you can approximate your homotopy by one that is smooth, but it is harder for em to see how to get it to equal your original functions on the nose at the extremes.

milnor considers such a problem in his little "topology from the differentiable viewpoint", when he deduces the continuous brouwer fix point theorem from the smooth one.
 
  • #4
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I was aware of something like this, but I thought it would still be problematic because the paths [itex]t\mapsto f(t,s)[/itex] could be very close to the boundary [itex]\partial G[/itex], and a careless smoothing could perhaps push the path out of G.
 
  • #5
mathwonk
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a smothing can be chosen to move the path arbitrarily little, so if the distance from the boundary is finite, e.g. if the path and the boundary are compact and do not meet, then that is not a problem.

the problem you raise is the problem. e.g. in bott-tu, they prove the theorem that any continuous map of manifolds is approximable by a smooth one and simply announce as a corollary that continuous homotopy classes equal smooth homotopy classes. this is not so immediate as your question shows.

i.e. it does show that the map from smooth maps to continuous ones is surjective on homotopy classes, but for injectivity one must show that two smooth maps which are homotopic as continuous maps, are also homotopic as smooth maps, and this does not follow instantly from the approximation result, as one needs to choose the homotopy so as not to move the initial and terminal maps at all.

i guess you have to prove that two smooth maps which are very close to each other are also smoothly homotopic. you probably use some trick like a mapping cylinder, using an auxiliary space which includes the domain space so as to separate points better.......hmmmmm....look at the graphs?

i.e. if the graphs are near each other, then just smoothly project one graph down onto the other?

i.e. given two smooth maps of a rectangle into a rectangle, the graphs live in the product of two rectangles, so we can slide both graphs together along lines in the target rectangle.

then maybe patch these rectangles by a partition of unity?..mumble mumble......
 
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