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Homeomorphism and diffeomorphism

  • Thread starter rayman123
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  • #1
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Homework Statement



I have some problems with understanding these two things.

Homeomoprhism is a function f [tex] f: M\rightarrow N[/tex] is a homeomorphism if if is bijective and invertible and if both [tex] f, f^{-1}[/tex] are continuous.
Here comes an example, lets take function
[tex] f(x) = x^{3}[/tex] it is clear that f(x) is continuous and bijective and at the same [tex] f^{-1}(x)\rightarrow x^{\frac{1}{3}}[/tex] is also continuous which indicates that f is a homeomorphism. No strange things here

Diffeomorphism- given two manifolds M, N and a bijective map f is called a diffeomorphism if both
[tex] f: M\Rightarrow N[/tex]
[tex] f^{-1}: N\Rightarrow M[/tex] are of class [tex] C^{\infty}[/tex] (if these functions have derivatives of all orders) f is called diffeomorphism


Going back to my example.

f is clearly a homeomorphism but :

why is [tex] f(x)=x^{3}[/tex] of a class [tex] C^{\infty}[/tex]?? what do they mean by 'all orders'
[tex] \frac{df}{dx}=3x^{2}[/tex]
[tex] \frac{d^2f}{dx^2}=6x [/tex]
[tex]\frac{d^3f}{dx^3}=6[/tex]
[tex]\frac{d^4f}{dx^4}=0...[/tex]


and its inversion
[tex] f^{-1}=x^{\frac{1}{3}}[/tex] is not of a [tex] C^{\infty}[/tex]....because
[tex] \frac{df^{-1}}{dx}=\frac{1}{3}x^{\frac{-2}{3}}[/tex]
it is not definied at x=0
 

Answers and Replies

  • #2
Dick
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So f is a homeomorphism but not a diffeomorphism. What's wrong with that?
 
  • #3
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nothing is wrong with that, you did not read my message carefully and overlooked what I was asking for. The problem was that I did not exactly understand the concept of a function which has derivatives of all orders, what do they mean by that?
Apparently f has 4 derivatives until we get 0 but why its inversion is not a smooth function (because it is not definied at x=0?)

We could differentiate the inversion function as many times as we wanted to....
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
nothing is wrong with that, you did not read my message carefully. The problem was that I did not exactly understand the concept of a function which has derivatives of all orders, what do they mean by that?
Apparently f has 4 derivatives until we get 0 but why its inversion is not a smooth function (because it is not definied at x=0?)

We could differentiate the inversion function as many times as we wanted to....
f is C^infinity. It has derivatives of all orders. f^(-1) isn't. It isn't even differentiable at 0. On the other hand if you define them on a domain that doesn't include 0, like [1,infinity) then they are both C^infinity and it's also a diffeomorphism.
 

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