Diffeomorphism of a disk and a square?

  • Thread starter phyalan
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  • #1
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I have trouble in showing a disk and a square is not diffeomorphic.
Intuitively, I know there is smoothness problem occurs at the corner of the square if I suppose there is a diffeomorphism between the two, but how can I explicitly write down the proof? I hope someone can provide me with some hints. Thx!
 

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  • #2
lavinia
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I have trouble in showing a disk and a square is not diffeomorphic.
Intuitively, I know there is smoothness problem occurs at the corner of the square if I suppose there is a diffeomorphism between the two, but how can I explicitly write down the proof? I hope someone can provide me with some hints. Thx!

I think just follow the tangent to the boundary circle of the disk. It will not have a limit at the corners.
 
  • #3
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No, it can have a limit, but that limit will be zero :) Therefore, the inverse isn't differentiable.
 
  • #4
Erland
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Are you here talking about a _closed_ disk and a _closed_ square, i.e. manifolds with boundaries?

Because if we talk about an _open_ disk and an _open_ square, I believed they are diffeomorphic, and that a diffeomorphism could be chosen in some clever way, using text functions. Am I mistaken here?
 
  • #5
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Phyalan:

I don't know what tools you're allowed to use, but (formalizing what I think Lavinia

was trying to say) an isomorphism between

spaces gives rise to (or, the dreaded "induces") an isomorphism between the

respective tangent spaces. The tangent space of the square (which in this case

is just the derivative, being 1-dimensional). Notice that the tangent space of the

square is not defined at the corners (using the coordinates (0,0), (0,1), (1,0), (1,1))

while the tangent space of the disk is defined at each point. I guess by the square

you are referring to the square together with its interior; otherwise, the disk is

contractible, but the square is not, so that would do it.
 
  • #6
lavinia
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No, it can have a limit, but that limit will be zero :) Therefore, the inverse isn't differentiable.

it can not have a limit of zero and be a diffeo - right? So i meant that the vectors approaching the corner along the boundary can not shrink to zero - so no limit
 
  • #7
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To be more specific, using the coordinates with vertices {(0,0),(0,1),(1,0), (1,1)}, find

tangent vectors along (0,y), and (x,1) , as you approach the corner, as x->0 and as

y->1 , so that the tangent vector is not defined, and then this corner point cannot be the

image of the circle under a diffeomorphism, since the tangent space is not defined therein.
 

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