Existence of diffeomorphism


by losiu99
Tags: diffeomorphism, pugh, starlike
losiu99
losiu99 is offline
#1
Jun29-10, 11:32 AM
P: 134
While reading C.C.Pugh's "Real Mathematical Analysis" I've encountered a following statement:

"A starlike set [tex]U \subset \mathbb{R}^n[/tex] contains a point [tex]p[/tex] such that the line segment from each [tex]q\in U[/tex] to [tex]p[/tex] lies in [tex]U[/tex]. It is not hard to construct a diffeomorphism from [tex]U[/tex] to [tex]\mathbb{R}^n[/tex]."

It's little embarassing, but despite my best effort, I cannot figure out how to do it.
I appreciate any help.
Thanks in advance!
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Pere Callahan
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#2
Jun29-10, 02:24 PM
P: 588
U can map the intervall [-pi/2,pi/2] diffeomorphically to R via
[tex]
x\mapsto \tan x
[/tex]

You can apply that to every line in your star-shapped set. It's just "blowing up" the set.
losiu99
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#3
Jun30-10, 02:31 AM
P: 134
Thank you for your response.
I'm still not sure about how to make it work. Assuming bounded set, we can find the "ends" of the line and treat it like ends of [-pi/2, pi/2] interval with [tex]p[/tex] as 0, and "blow" the line to infinite length.

I'm more concerned about unbounded sets, where some lines doesn't have "end".

Landau
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#4
Jun30-10, 03:23 PM
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Existence of diffeomorphism


Probably U is assumed to be open? If n=2, you can use that star shaped sets (sorry, I am used to "star shaped" instead of "starlike") are simply-connected and then use the Riemann mapping theorem from complex analysis. For general n, I think it is much harder. E.g. see here at MathOverflow, and here. I am curious to hear if Pere Callahan has an easy proof!
some_dude
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#5
Jun30-10, 03:48 PM
P: 93
I don't get it. If U isn't required to be open, then I don't think it's true. And if it is open, then wouldn't the identity function satisfy this?
Landau
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#6
Jun30-10, 03:53 PM
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Quote Quote by some_dude View Post
I don't get it. If U isn't required to be open, then I don't think it's true.
Yeah, see the first sentence of my post :)
And if it is open, then wouldn't the identity function satisfy this?
Huh? Are you asserting that the identity map is a diffeo? This is obiously not true since it is isn't even bijective (unless of course U happens to be R^n itself) ...
some_dude
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#7
Jun30-10, 03:54 PM
P: 93
Ohhh, stupid me, you want the image of the function to be R^n I bet...

Why map P to the origin, and then for ray originating at p approaching a point on the boundary of U, map that to a corresponding ray originating at 0 (= f(p)) approaching infinity. Apply the MVT to projections of the line segments to show it has maximal rank everywhere. Then, I think, you can use the inverse function theorem to show its a diffeo?
Hurkyl
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#8
Jun30-10, 03:59 PM
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Quote Quote by losiu99 View Post
I'm more concerned about unbounded sets, where some lines doesn't have "end".
Then find a way to reduce the general problem to the bounded problem.

It might be worth making sure your method really doesn't work in this case first....
Pere Callahan
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#9
Jun30-10, 05:49 PM
P: 588
Quote Quote by losiu99 View Post
I'm more concerned about unbounded sets, where some lines doesn't have "end".

Maybe you can first shrink every direction by applying an arctan function to get something bounded. It should even be possible to contract everything to a ball around the center of the star. Then expand it again. This would certainly (at least I think so) be a bijection. I don't know if there are exotic cases where this function would not be a diffeomorphism.
losiu99
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#10
Jul1-10, 01:17 AM
P: 134
Thank you very much for your responses. At the first glance it looks like the links contain everything I need. I shall think about concracting-expanding idea as well, since it looks promising.
Thank you all once again.
Pere Callahan
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#11
Jul1-10, 01:25 AM
P: 588
These links are indeed quite interesting. I'm often surprised about how exotic objects of low dimensional topology can be. Maybe I just know too little about it.


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