Assume that $f(x) = (p_1(x),p_2(x),p_3(x))$ is a homogeneous polynomial inducing a diffeomorphic mapping of $\mathbb{R}^3\setminus\{0\}$ onto itself. Homegeneous means $f(tx)=t^n f(x)$, for $t>0$ and some $n\in \mathbb{N}$. Why $n$ should be 1?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ $f(x)=(x_1^3,x_2^3,x_3^3)$ is counterexample. $\endgroup$– YCorCommented Feb 23, 2020 at 9:30
-
1$\begingroup$ @YCor. This is not a diffeomorphism of $\mathbb{R}^3\setminus\{0\}$ onto itself. Its Jacobian is $3^3x_1^2 x_2^2 x_3^2=0$ not only on $(0,0,0)$. $\endgroup$– LiraCommented Feb 23, 2020 at 9:41
-
1$\begingroup$ Thanks, you're right... sorry! I'm leaving it for other users... $\endgroup$– YCorCommented Feb 23, 2020 at 9:42
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
6
$f(x)= |x|^2 x$ is a counterexample?
-
-
2$\begingroup$ @ZachTeitler I think it is - $|x|^2 = x_1^2 + x_2^2 + x_3^2$ is a polynomial. $\endgroup$ Commented Feb 23, 2020 at 16:28
-
-
$\begingroup$ @KevinCasto: Isn't the Jacobian $3(x^2+y^2+z^2)^3$? $\endgroup$ Commented Feb 23, 2020 at 18:55
-
$\begingroup$ @MateuszKwaśnicki oops you're right! $\endgroup$ Commented Feb 23, 2020 at 19:26