0
$\begingroup$

I have the following function: $$f: \mathbb{R^2}/{(0,0)} \rightarrow \mathbb{R^2}$$ where $$f(x,y) = (\frac{x²-y²}{2}, xy).$$

I've shown that $f$ is a local $C^1$-diffeomorphism for all $(a,b) \in \mathbb{R^2}/{(0,0)}$.

Is f a class $C^1$-homeomorphism?

How can I prove that f is invertible in a neighborhood of $(0,0)$?

I tried using the differentiability of the inverse homeomorphism and the inverse function theorem, but the hypothesis fails because the jacobian matrix on $(0,0)$ is non-invertible.

$\endgroup$
  • $\begingroup$ The point $(0,0)$ does not belong to the domain of definition of your function. $\endgroup$ – Tomas Jun 6 '16 at 11:29
  • $\begingroup$ Does that imply I can't take a neighborhood of it where f is invertible? Isn't there a way around it by using an accumulation point? $\endgroup$ – DrHAL Jun 6 '16 at 11:56
  • $\begingroup$ determinant of Jacobi-Matrix $\ne 0$ => invertible $\endgroup$ – user90369 Jun 6 '16 at 11:57
1
$\begingroup$

$f$ is not invertible in any neighborhood of $(0,0)$. I'm actually going to prove this for the function $g(x,y) = 2 \cdot f(x,y) = (x^2-y^2,2xy),$ but we'll get the result for $f$.

Think of $g$ as a function $\mathbb{C} \to \mathbb{C}$. Then it's clear that $g(z) = z^2$, since $(a+ib)^2 = (a^2-b^2)+i(2ab).$ Hence $g$ is a two-sheeted cover $\mathbb{C} \to \mathbb{C}$ branched at $z = 0$, and cannot be a homeomorphism in a neighborhood of $0$.

Essentially, any neighborhood of $0$ must contain both $z'$ and $-z'$ for some small $z'$, and since $g(z') = g(-z')$ it cannot be a homeomorphism in such a neighborhood. For any nonzero point $z_0 \in \mathbb{C}$ we can find a small neighborhood (take radius less than $|z_0|$) that does not contain any antipodal pairs, and $g$ will be a smooth (in fact holomorphic) homeomorphism on such a neighborhood. But it cannot even be a homeomorphism in any neighborhood of 0.


Note that this is not implied by the vanishing of the Jacobian. The function $h: \mathbb R^2 \to \mathbb R^2$, $h(x,y) = (x^3,y)$ has vanishing Jacobian at $(0,0)$ but is a smooth homeomorphism there. But the Jacobian vanishing generally indicates that things are going wrong.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.