Prove that diffeomorphisms are between manifolds with the same dimension

  • Thread starter feynman137
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My definition of diffeomorphism is a one-to-one mapping f:U->V, such that f and f^{-1} are both continuously differentiable. Now, how to prove that if f is a diffeomorphism between euclidean sets U and V, then U and V must be in spaces with equal dimension (using the implicit function theorem)?
 

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  • #2
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Hi feynman137! :smile:

What about following reasoning:
If U and V are diffeomorphic, then the tangents spaces at some points must be isomorphic. But the tangent space at U is [itex]\mathbb{R}^n[/itex] and the tangent space at V is [itex]\mathbb{R}^m[/itex], and these are only isomorphic when n=m.
 
  • #3
quasar987
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By contradiction.
Assume U sits in R^n and V sits in R^m, and assume without loss of generality that n>m.
By Sard's theorem, there is a point p in U such that f satisfies the hypothesis of the the implicit fct theorem (Df(p) is surjective, so modulo a reindexing of the coordinates, the nxn submatrix of the last partial derivatives is nonsingular). Set v:=f(p).
The implicit function theorem says that locally, the level set f^{-1}(v) looks like the graph of some function g: There is an open set R = A x B in R^{n-m} x R^{n} and g:A-->B such that R n f^{-1}(v) = graph(g).

However, if f is a diffeomorphism, f^{-1}(v) = {p}: contradiction.
 
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Looking for a book!

Hi,

I am looking for a book " Transversal Mappings and Flows" written by Ralph Abraham and J. Robbin. I can not find it at any library in my country. Also if you have a Pdf or Djvu format, can you send it to me. It will help me by reading Hutchings lecture notes. This is my e-mail adress: [email protected]
 
  • #5
lavinia
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My definition of diffeomorphism is a one-to-one mapping f:U->V, such that f and f^{-1} are both continuously differentiable. Now, how to prove that if f is a diffeomorphism between euclidean sets U and V, then U and V must be in spaces with equal dimension (using the implicit function theorem)?

try showing that charts on one manifold are carried into charts on the other.
 
  • #6
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Yet another proof:
The Inverse Function Theorem would give you an mxn matrix for the differential of the map between R^n and R^m; in a diffeomorphism, the matrix would be invertible, so that you need m=n.
 

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