A general quadric is a surface of the form: $$ Ax^2 + By^2 + Cz^2 + 2Dxy + 2Eyz + 2Fxz + 2Gx + 2Hy + 2Iz + J = 0$$ It can be written as a matrix expression $$ [x, y, z, 1]\begin{bmatrix} A && D && F && G \\ D && B && E && H \\ F && E && C && I \\ G && H && I && J \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} = \mathbf{p}^\intercal \mathbf{Q} \mathbf{p} = 0 $$ Is it possible to represent this quadric as a parametric surface $\mathbf{p}(u, v): \mathbb{R}^2 \to \mathbb{R}^3$? $$ \forall u, v, \mathbf{p}(u, v)^\intercal \mathbf{Q}\mathbf{p}(u, v) = 0 $$
Is there a way to parametrise general quadrics?
423 Views Asked by Henricus V. https://math.techqa.club/user/henricus-v/detail AtThere are 2 best solutions below
Yes. Since $M$ is symmetric, for an appropriate choice $P$ we can factor $M$ as $$M = P^T D P$$ where $D$ is diagonal and has entries in $\{-1, 0, 1\}$, and in fact, we can choose $P$ so that its columns are orthogonal, which is convenient for some purposes. Then, in the new coordinates defined by transforming the original ones via $P$, the quadric has equation $$a x^2 + b y^2 + c z^2 = d,$$ where $a, b, c \in \{-1, 0, 1\}$, and by changing signs if necessary, we may assume that the first nonzero coefficient among $d, a, b, c$ (if any) is $1$. This leaves, up to relabeling of the variable names, just a few possibilities for nonempty quadrics (these thus serve as normal forms for the various types of quadrics, including degenerate cases): \begin{align} x^2 + y^2 + z^2 = 1 & \qquad \textrm{sphere} \\ x^2 + y^2 - z^2 = 1 & \qquad \textrm{hyperboloid of one sheet} \\ x^2 - y^2 - z^2 = 1 & \qquad \textrm{hyperboloid of two sheets} \\ x^2 + y^2 + z^2 = 0 & \qquad \textrm{point} \\ x^2 + y^2 - z^2 = 0 & \qquad \textrm{cone} \\ x^2 + y^2 = 1 & \qquad \textrm{cylinder} \\ x^2 - y^2 = 1 & \qquad \textrm{hyperbolic cylinder} \\ x^2 + y^2 = 0 & \qquad \textrm{line} \\ x^2 - y^2 = 0 & \qquad \textrm{union of two distinct, intersecting planes} \\ x^2 = 1 & \qquad \textrm{disjoint union of two distinct planes} \\ x^2 = 0 & \qquad \textrm{plane} \end{align} All of these can be readily parameterized. Hence, we can produce an explicit parameterization in the original coordinates by changing variables in our parameterization using $P^{-1}$
Given a point $p$ on the quadratic surface $Q$, every line $L$ through $p$ is either tangent to $Q$, or it intersects $Q$ in another point $p_L$. In this way the lines not tangent to $Q$ parametrize $Q-\{p\}$. These lines are in turn parametrized by $\Bbb{R}^2$, so this is possible if you omit one point from the parametrization.