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Diffeomorphism Definition Trouble.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello MHB.

I am not sure what the definition of a diffeomorphism is.

In Spivak's 'Calculus on Manifolds', it is defined as:

A diffeomorphism is a function $h:U\to V$, $U$ and $V$ are open sets in $\mathbb R^n$, such that $h$ is differentiable and has a differentiable inverse.

I have seen the following in Lee's 'Introduction to Smooth Manifolds':

Let $U$ and $V$ be open subsets in $\mathbb R^n$.
A function $F:U\to V$ is smooth if each component of $F$ has continuous partial derivatives of all orders.
A function $h:U\to V$ is a diffeomorphism if it is smooth and has a smooth inverse.

The one in Lee's book is far stronger than the one in Spivak.

I am not sure which one is 'more' standard.
Does this puzzle you too?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello caffeinemachine,

I have moved this topic here as it seems a better fit than the Calculus sub-forum. :D
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello caffeinemachine,

I have moved this topic here as it seems a better fit than the Calculus sub-forum. :D
I thought it was more appropriate for calculus. But I am fine wither way. :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,854
Hello MHB.

I am not sure what the definition of a diffeomorphism is.

In Spivak's 'Calculus on Manifolds', it is defined as:

A diffeomorphism is a function $h:U\to V$, $U$ and $V$ are open sets in $\mathbb R^n$, such that $h$ is differentiable and has a differentiable inverse.

I have seen the following in Lee's 'Introduction to Smooth Manifolds':

Let $U$ and $V$ be open subsets in $\mathbb R^n$.
A function $F:U\to V$ is smooth if each component of $F$ has continuous partial derivatives of all orders.
A function $h:U\to V$ is a diffeomorphism if it is smooth and has a smooth inverse.

The one in Lee's book is far stronger than the one in Spivak.

I am not sure which one is 'more' standard.
Does this puzzle you too?
Lee's book is called 'Introduction to Smooth Manifolds'.
The title suggests that he is limiting himself to smooth manifolds.
If a manifold is smooth to begin with, a diffeomorphism has to be smooth as well.

Rather then specifying everywhere that it is smooth, likely he chose to leave that out.
He should make a note somewhere that he did this though.
Perhaps he did in the introduction?
Or at the start of the chapter where he defines the diffeomorphism?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Lee's book is called 'Introduction to Smooth Manifolds'.
The title suggests that he is limiting himself to smooth manifolds.
If a manifold is smooth to begin with, a diffeomorphism has to be smooth as well.

Rather then specifying everywhere that it is smooth, likely he chose to leave that out.
He should make a note somewhere that he did this though.
Perhaps he did in the introduction?
Or at the start of the chapter where he defines the diffeomorphism?
I checked wiki. It seems to agree with Spivak's definition. I think Lee specifically needs only smooth manifolds so he did that to the definition of diffeomorphism.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
They are the same. Lee just happens to say explicitly all that is involved. When Spivak says that $h$ is differentiable and has a differentiable inverse he means that all component functions are infinitely continuously differentiable, is invertible and the same is true for all component functions of the inverse function. I'm pretty sure Spivak also means that differentiable means having partial derivatives continuous of all orders. :eek: So they are speaking about the same thing but Lee is explicit about what is going on.

Cheers.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
They are the same. Lee just happens to say explicitly all that is involved. When Spivak says that $h$ is differentiable and has a differentiable inverse he means that all component functions are infinitely continuously differentiable, is invertible and the same is true for all component functions of the inverse function. I'm pretty sure Spivak also means that differentiable means having partial derivatives continuous of all orders. :eek: So they are speaking about the same thing but Lee is explicit about what is going on.

Cheers.
Hey Fantini.

I think wiki says different. Diffeomorphism - Wikipedia, the free encyclopedia
There they talk about a $C^r$-diffeomorphism so I think a diffeo doesn't necessarily need to be a$C^\infty$ function with a $C^\infty$ inverse.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
It doesn't have to be $C^{\infty}$, but differentiability usually is not a concern in geometry, so we include that into the definition.