# Thread: show the Jacobian is zero

1. ## show the Jacobian is zero

$\displaystyle F: R^{2}\rightarrow R^{2} , \ \ \ J(F)$ is Jacobian.

Suppose F maps into a curve in the plane, in the sense tht F is a composition of two smooth mappings, $\displaystyle R^{2}\rightarrow R\rightarrow R^{2}$

show that the Jacobian of F is identically zero.

2. Originally Posted by silversand
$\displaystyle F: R^{2}\rightarrow R^{2} , \ \ \ J(F)$ is Jacobian.

Suppose F maps into a curve in the plane, in the sense tht F is a composition of two smooth mappings, $\displaystyle R^{2}\rightarrow R\rightarrow R^{2}$

show that the Jacobian of F is identically zero.
so $\displaystyle F(x,y)=fg(x,y),$ where $\displaystyle g: \mathbb{R}^2 \longrightarrow \mathbb{R}$ and $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}^2.$ let $\displaystyle g(x,y)=z.$ then $\displaystyle F(x,y)=f(z)=(u(z),v(z)).$ thus: $\displaystyle J(F)=\begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix}= \begin{vmatrix} u_z z_x & u_z z_y \\ v_z z_x & v_z z_y \end{vmatrix}= u_zz_xv_zz_y - u_zz_yv_zz_x = 0.$