Diffeomorphism of elementary symmetric polynomials

Question

Let $\sigma_1(x,y,z) = x + y + z$, $\sigma_2(x,y,z) = xy + xz + yz$ and $\sigma_3(x,y,z) = xyz$.

When is the map $\Phi: \mathbb{R}^3 \to \mathbb{R}^3, \Phi\,(x,y,z) = \begin{pmatrix} \sigma_1(x,y,z) \\ \sigma_2(x,y,z) \\ \sigma_3(x,y,z)\end{pmatrix}$ a local $C^{\infty}$-diffeomorphism?

Answer 1

We have that $\Phi \in C^{\infty}$. Now:

$$\det [D \Phi(x,y,z)] = x^2(y - z) + y^2(z - x) + z^2(x - y) = (x - z)(y - z)(x - y)$$

So let $\Omega$ be the union of the three planes: $(x = y)$, $(x = z)$, $(y = z)$. We have that $\Phi$ is a local $C^{\infty}$-diffeomorphism near each $(x,y,z) \in \Bbb R^3 \setminus \Omega$

Answer 2

It cannot be, because of the symmetry property. More specifically, if $\epsilon > 0$, the six triples $$(1,2,3) \cdot \epsilon, (1,3,2) \cdot \epsilon, (2,1,3) \cdot \epsilon, (2,3,1) \cdot \epsilon, (3,1,2) \cdot \epsilon \text{ and } (3,2,1) \cdot \epsilon$$ always have the same image through $\Phi$.

Now, if $\Phi$ was a local diffeorphism around $0$, it would be locally injective: sufficiently small triples would not have the same image. But our six elements can be made as small as wished, so this prove that $\Phi$ cannot be a local diffeomorphism.

Note that it's not hard to modify this argument to prove that $\Phi$ isn't locally injective near any triple with repeated coordinates, which may shed some light on Ahmed Hussein's computation.