Find all linear transformations of $\mathbb{R}^3$ that preserve the hyperboloid $x^2+y^2− z^2 = 1$

Question

I need to find all linear transformations of $\mathbb{R}^3$ that preserve the hyperboloid $x^2+y^2− z^2 = 1$ . I could not come up with any good idea how to do this. \ The only thing I can do is to substitute in this equation $(ax+by+cz)$ instead of $x$, $(dx+ey+gz)$ instead of $y$, and $(hx+iy+jz)$ instead of $z$, so that these $(x,y,z)$ satisfy $x^2+y^2−z^2=1$, and it does not seem to give any nice result or I am too lazy to compute it properly. I did analogous computations with $x^2-y^2=1$ and that worked, but in this case I am stuck. Could anybody help?

Answer 1

HINT:

Start with parametrization

$$(x,y,z)= { \sec u \cos v, \, \sec u \sin v, \, \tan u\,}$$

where $u$ parametrization caters for the skew straight line generation for given $v$ .. which in turn serves to generate the hyperboloid swept surface of revolution.

Answer 2

Define the matrix $D = \text{diag}(1,1,-1)$ and let $\mathbf{x} = \begin{bmatrix} x\\ y\\ z \\ \end{bmatrix}$. Then the hyperboloid is $H = \{\mathbf{x} ~|~ \mathbf{x}^T D \mathbf{x} = x^2 + y^2 - z^2 = 1 \}$. Then a linear map $T$ preserves $H$ is equivalent to $$\mathbf{x}^T D \mathbf{x} = 1 \iff (T \mathbf{x})^T D (T\mathbf{x}) = 1$$.

A sufficient condition is $T^T D T = D$.

Answer 3

Problem:

The condition for the surface $H$ and the invariance $A(H) = H$ gives \begin{align} 1 &= x^2 + y^2 - z^2 \\ &= u^T (e_1, e_2, -e_3) u \\ &= u^T M_3 u \\ &= (Au)^T M_3 Au \\ &= u^T A^T M_3 Au \end{align} so we need $A$ with $$ M_3 = A^T M_3 A \quad (*) $$ where $M_3=(e_1,e_2,-e_3)$ is the reflection along the $z$-axis. One property is $M_3^2=I$.

Orthogonal Solutions:

Squaring both sides of $(*)$ gives: $$ I = A^T M_3 A A^T M_3 A $$ The middle part looks familiar. So certain orthogonal transformations ($A^TA=AA^T=I$) might be among the solutions of $(*)$, they would fulfill $$ A M_3 = A A^T M_3 A = M_3 A \\ M_3 A^T = A^T M_3 A A^T = A^T M_3 $$ thus commute with $M_3$.

Property of the Determinants: For the determinants of solutions of $(*)$ we have $$ -1 = \det(A) (-1) \det(A) \iff \det(A)^2 = 1 \iff \det(A) = \pm 1 $$ So the solution set of $(*)$ might include certain rotations $R$ (orthogonal, $\det R = 1$) and certain reflections $M$ (orthogonal, $\det(M) = -1$).

Some Solutions:

Candidates are $R_3$ (rotations around the $z$-axis) and $M_3$. $$ M_3^T M_3 M_3 = M_3^T = M_3 $$ So $M_3$ solves $(*)$. Now the rotation: $$ R_3 = \begin{pmatrix} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ so \begin{align} R_3^T M_3 R_3 &= \begin{pmatrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ \\ &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \\ &= M_3 \end{align}

Other Solutions: From \begin{align} A^T M_3 A &= \begin{pmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \\ &= \begin{pmatrix} a_{11} & a_{21} & -a_{31} \\ a_{12} & a_{22} & -a_{32} \\ a_{13} & a_{23} & -a_{33} \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \\ &= \begin{pmatrix} a_{11}^2 + a_{21}^2 - a_{31}^2 & a_{11} a_{12} + a_{21} a_{22} - a_{31} a_{32} & a_{11} a_{13} + a_{21} a_{23} - a_{31} a_{33} \\ a_{12} a_{11} + a_{22} a_{21} - a_{32} a_{31} & a_{12}^2 + a_{22}^2 - a_{32}^2 & a_{12} a_{13} + a_{22} a_{23} - a_{32} a_{33} \\ a_{13} a_{11} + a_{23} a_{21} - a_{33} a_{31} & a_{13} a_{12} + a_{23} a_{22} - a_{33} a_{32} & a_{13}^2 + a_{23}^2 - a_{33}^2 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \end{align} we get the six equations $$ a_{11}^2 + a_{21}^2 - a_{31}^2 = 1 \\ a_{12}^2 + a_{22}^2 - a_{32}^2 = 1 \\ a_{13}^2 + a_{23}^2 - a_{33}^2 = -1 \\ a_{11} a_{12} + a_{21} a_{22} - a_{31} a_{32} = 0 \\ a_{11} a_{13} + a_{21} a_{23} - a_{31} a_{33} = 0 \\ a_{12} a_{13} + a_{22} a_{23} - a_{32} a_{33} = 0 $$

I have tried out a few samples, choosing different free variables, but did not manage to produce a non-orthogonal solution yet.