Let $f: U \to \mathbb{R}^{m}$ differentiable function on open $U \subset \mathbb{R}^{m}$. If $|f(x)|$ is constant when $x$ varies in $U$, then Jacobian determinant of $f$ is zero.
I'm a little lost. I know that $f(x)$ is constant, the partial derivatives are zero. But, how do I deal with $|f|$? To show that the determinant is zero is sufficient to show that the partial derivatives are zero?
On
If $|f(x)|=c$, then $f(x)$ is contained in the sphere of radius $c$. This implies that for every $x\in U$, and $u_1,...,u_n$ a basis, $df_x(u_i)$ is in $T_{S_c}f(x)$, the tangent space at $f(x)$ of the sphere $S_c$ of radius $c$, thus $df_x(u_1),...,df_x(u_n)$ is not independent and the Jacobian determinant of $f$ is zero.
On
If the determinant of the Jacobian is non-zero at some point $p$, then, by the inverse function theorem, $f$ would be a local diffeomorphism ar $p$ and therefore $f(U)$ would contain an open ball centered at $f(p)$. That's impossible, since $|f|$ is constante and therefore $f(D)$ is contained on the surface of some sphere centered at $0$.
On
If $|f|$ is constant so is $g(x) = |f(x)|^2 = \langle f(x), f(x) \rangle$ and thus
$$ 0 = (Dg)_{x}h = 2\langle(Df)_{x}h, f(x)\rangle $$
for all $x \in U$ and $h \in \mathbb{R}^m$. Hence $f(x) \in \left(\operatorname{range} (Df)_{x}\right)^\perp = \operatorname{kernel} ((Df)_{x})^*$ for all $x \in U$. If $f(x) = 0$ for some $x$, then clearly $f = 0$ and the result follows. Otherwise, the previous equality shows that $$ \det (Df)_x = \det ((Df)_x)^* = 0 $$ for all $x \in U$.
If $|f| \equiv 0$ then $f\equiv 0$ hence $Df \equiv 0$.
If $|f| \equiv c \not=0$ then $f(x) \not =0$ for any $x$. Fix $x \in U$ and $h\in \mathbb R^m$, for $t >0$ small we have $x + t h\in U$ and $$\sum_{i=1}^m f_i^2(x + th) = c.$$ Differentiating at $t=0$ we have $$ \sum_{i,j=1}^m \partial_j f_i(x) f_i(x) h_j = 0. $$ In other word $Df(x) f(x) \cdot h = 0$ for any $h\in \mathbb R^m$ hence $Df(x) f(x) =0$. Since $f(x) \not=0$ then $0$ is an eigenvalue of $Df$ which implies $det(Df) =0$.