Is this parameterization a regular surface?

Question

Let us consider the parameterization $\phi: \mathbb{R}^2 \to \mathbb{R}^3$ given by:

$$ \phi(u,v)=(u-\frac{u^3}{3}+uv^2,v-\frac{v^3}{3}+vu^2,u^2-v^2) $$

Is this a global parameterization for some regular surface?

Answer 1

Global issues such as self-intersection and self-crossing aside, the surface described by the parametrization

$\phi(u, v) = \left ( u - \dfrac{u^3}{3} + uv^2, v - \dfrac{v^3}{3} + u^2 v, u^2 - v^2 \right ) \tag 1$

is regular precisely at those points where the Jacobean matrix

$D\phi(u, v) = \begin{bmatrix} \phi_u(u, v) & \phi_v(u, v) \end{bmatrix}, \tag 2$

where

$\phi_u(u, v) = \begin{pmatrix} 1 - u^2 + v^2 \\ 2uv \\ 2u \end{pmatrix} \tag 3$

and

$\phi_v(u, v) = \begin{pmatrix} 2uv \\ 1 - v^2 + u^2 \\ -2v \end{pmatrix}, \tag 4$

is of maximal rank, i.e. $2$; and this will be precisely at those parameter values $(u, v) \in \Bbb R^2$ where $\phi_u(u, v)$ and $\phi_v(u, v)$ are linearly independent, i.e. where

$\phi_u(u, v) \times \phi_v(u, v) \ne 0; \tag 5$

we thus compute

$\phi_u(u, v) \times \phi_v(u, v)$ $= \begin{pmatrix} -4uv^2 - 2u(1 - v^2 + u^2) \\ 4u^2v - (-2v)(1 - u^2 + v^2) \\ (1 - u^2 + v^2)(1 - v^2 + u^2) - 4u^2v^2 \end{pmatrix}$ $= \begin{pmatrix} -4uv^2 - 2u + 2uv^2 - 2u^3 \\ 4u^2v + 2v - 2u^2v + 2v^3 \\ (1 - (u^2 - v^2))(1 + (u^2 - v^2)) - 4u^2v^2 \end{pmatrix} = \begin{pmatrix} -2uv^2 - 2u - 2u^3 \\ 2u^2v + 2v + 2v^3 \\ 1 - (u^2 - v^2)^2 - 4u^2v^2 \end{pmatrix}$ $= \begin{pmatrix} -2uv^2 - 2u - 2u^3 \\ 2u^2v + 2v + 2v^3 \\ 1 - u^4 +2u^2 v^2 - v^4- 4u^2v^2 \end{pmatrix} = \begin{pmatrix} -2uv^2 - 2u - 2u^3 \\ 2u^2v + 2v + 2v^3 \\ 1 - (u^2 + v^2)^2 \end{pmatrix}; \tag 6$

if

$\phi_u(u, v) \times \phi_v(u, v) = 0, \tag 7$

then

$ -2uv^2 - 2u - 2u^3 = 0, \tag 8$

$2u^2v + 2v + 2v^3 = 0, \tag 9$

$1 - (u^2 + v^2)^2 = 0; \tag{10}$

from (8) and (9):

$-2u^2v^2 - 2u^2 - 2u^4 = 0, \tag{11}$

$2u^2v^2 + 2v^2 + 2v^4 = 0; \tag{12}$

adding,

$2(v^2 - u^2) + 2(v^4 - u^4) = 0, \tag{13}$

which may in fact be re-written as

$2(u^2 - v^2) + 2(u^2 - v^2)(u^2 + v^2) = 0; \tag{14}$

now suppose that

$u^2 = v^2; \tag{15}$

then (12) becomes

$2v^2( 1 + 2v^2) = 0, \tag{16}$

which, since $2v^2 + 1 \ne 0$ for all real $v$, forces

$u = v = 0, \tag{17}$

which is incompatible with (10); therefore we rule out the possibility that $\phi_u \times \phi_v = 0$ where $u^2 = v^2$; if $u^2 \ne v^2$, then (14) yields

$1 + u^2 + v^2 = 0, \tag{18}$

clearly impossible for real $u$, $v$. We thus see that there is no

$(u, v) \in \Bbb R^2 \tag{19}$

for which (7) binds; it follows that the vectors $\phi_u(u, v)$ and $\phi_v(u, v)$ are linearly independent for all pairs as in (19); thus the surface $\phi(u, v)$ is everywhere regular, locally, but differing values of $(u, v) \in \Bbb R^2$ may map to the same point; but the tangent maps will be regular in a neighborhood of each such $(u, v)$.