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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 $\mathbf{p}(u, v): \mathbb{R}^2 \to \mathbb{R}^3$? $$ \forall u, v, \mathbf{p}(u, v)^\intercal \mathbf{Q}\mathbf{p}(u, v) = 0 $$

Servaes
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Henricus V.
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1 Answers1

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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.

Servaes
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  • To clarify, by "omit one point from the parametrization," do you mean the point $p_L$? – Nathan Jul 02 '18 at 18:42
  • also @Servaes that line $L$ may intersect the surface $Q$ at multiple points. Doesn't that make this transformation non-homeomorphic? – Nathan Jul 02 '18 at 22:44
  • @frank No, I mean the point $p$. As for intersecting $Q$ at multiple points; because the surface is quadratic, any line intersects it in at most two points. For every line $L$ through $p$, one of these points is $p$. Hence there is at most one other point $p_L$ in which $L$ intersects $Q$. Whether or not the parametrisation is a homeomorphism depends on whether the matrix is singular or not, amongst other things. – Servaes Jul 03 '18 at 13:49