Is injective the differential of $\varphi$?

Question

We consider $U = (- 1 , 1) \times (- 1 , 1)$ and $\varphi : U \to {\mathbb{R}}^3$ given by $\varphi (u , v) = (u^3 , v^3 , u v)$ for all $(u , v) \in U$. Is $d {\varphi}_p : {\mathbb{R}}^2 \to {\mathbb{R}}^3$ injective for all $p \in U$? Note: given $p \in U$, $d {\varphi}_p$ is the differential of $\varphi$ evaluated in $p$; for example, if we denote $x , y , z : U \to \mathbb{R}$ by $x(u , v) = u^3$, $y(u , v) = v^3$ and $z(u , v) = u v$ for all $(u , v) \in U$, then $$ d {\varphi}_p = \left( \begin{matrix} \displaystyle \frac{\partial x}{\partial u}(p) & \displaystyle \frac{\partial x}{\partial v}(p) \\ \displaystyle \frac{\partial y}{\partial u}(p) & \displaystyle \frac{\partial y}{\partial v}(p) \\ \displaystyle \frac{\partial z}{\partial u}(p) & \displaystyle \frac{\partial z}{\partial v}(p) \end{matrix} \right)\mbox{.} $$ I think it is not injective because $d {\varphi}_0 = 0$. Then, calling $q_1 = (1 , 0)$ and $q_2 = (0 , 1)$, we have $q_1 \neq q_2$ and $$ d {\varphi}_0(q_1) = \left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = d {\varphi}_0(q_2)\mbox{.} $$ What do you think? Thank you very much.

Answer 1

As you wrote:

$$ d {\varphi}_p = \left( \begin{matrix} \displaystyle \frac{\partial x}{\partial u}(p) & \displaystyle \frac{\partial x}{\partial v}(p) \\ \displaystyle \frac{\partial y}{\partial u}(p) & \displaystyle \frac{\partial y}{\partial v}(p) \\ \displaystyle \frac{\partial z}{\partial u}(p) & \displaystyle \frac{\partial z}{\partial v}(p) \end{matrix} \right)=\left( \begin{matrix} \displaystyle 3p_x^2 & 0 \\ \displaystyle 0& \displaystyle 3p_y^2 \\ \displaystyle p_y & \displaystyle p_x \end{matrix} \right) $$ Indeed choosing $p=(p_x,p_y)=(0,0)\ $ gives the zero matrix, which is clearly not injective.

But choosing either $p_x$ or $p_y$ non zero makes the matrix injective, as if $p_x\neq 0\neq p_y$ the top $2\times 2$ minor is non-zero, and if one of the coefficients is zero, the you can pick the non-zero rows, which form a non-zero diagonal matrix, thus of rank $2$.