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Show there is exist no diffeomorphism from $\mathbb{R}^2 \to \mathbb{R}^3$

PS: Don't say $\mathbb{R}^2,\mathbb{R}^3$ aren't homeomorphic, I need explanation without using topology

dragoboy
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  • I've seen a solution but I couldn't understand it. It says suppose $f$ is some diffeomorphism then $D(f)D(f^{-1}):\mathbb{R}^3 \to \mathbb{R}^3$ (ok) and $rank(D(f))\leq 2$ (ok) so a contradiction (?) – dragoboy Aug 27 '14 at 15:05
  • Maybe explain what is unclear in that proof. A diffeomorphism has everywhere a differential of full rank, and a linear map from $\mathbb R^2$ to $\mathbb R^3$ cannot have rank 3. – frog Aug 27 '14 at 15:07
  • Okay got it now, I forgot that $rank(AB)\leq max{rank(A),rank(B)}$ – dragoboy Aug 27 '14 at 15:14

2 Answers2

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If $$g:\Bbb R^3\longrightarrow \Bbb R^2$$ is a diffeomorphism, $$g\circ g^{-1}=Id:\Bbb R^3\longrightarrow\Bbb R^3.$$ Taking differential and using the chain rule $$Dg(g^{-1}(0))Dg^{-1}(0)=D(g\circ g^{-1})(0)=D Id(0)=Id:R^3\longrightarrow\Bbb R^3.$$ But the dimensions of $Dg(g^{-1}(0))$ and $Dg^{-1}(0)$...

Martín-Blas Pérez Pinilla
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Suppose $f$ were such a diffeomorphism, and $g$ its inverse. Let $P$ be a point of $R^2$ and $Q = f(P)$.

Then $$ f(q(x))= x $$ for every point $x$ in $R^3$.

By the chain rule, $$ df(g(x)) \cdot dg(x) = dI(x) $$ where $I$ is the identity map on $R^3$; its derivative is the 3 by 3 identity matrix. Applying this to the point $x = Q$, we get $$ f'(P) \cdot g'(Q) = I_3 $$ where this is an equality of matrices. The rank of $AB$ is generally no greater than $max(rank(A), rank(B))$. Since $f'(P)$ is a $3 \times 2$ matrix, its rank is at most 2. So the rank of the product on the left is at most 2; the rank of $I_3$ is 3. That's a contradiction.

John Hughes
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