hyperbolic rotation

Let $𝔼$ be the Euclidean plane equipped with the Cartesian coordinate system. Recall that given a circle $C$ centered at the origin $O$, one can define an “ordinary” rotation $R$ to be a linear transformation that takes any point on $C$ to another point on $C$. In other words, $R(C)\subseteq C$.

Similarly, given a rectangular hyperbola (the counterpart of a circle) $H$ centered at the origin, we define a hyperbolic rotation (with respect to $H$) as a linear transformation $T$ (on $𝔼$) such that $T(H)\subseteq H$.

Since a hyperbolic rotation is defined as a linear transformation, let us see what it looks like in matrix form. We start with the simple case when a rectangular hyperbola $H$ has the form $xy=r$, where $r$ is a non-negative real number.

Suppose $T$ denotes a hyperbolic rotation such that $T(H)\subseteq H$. Set

where is the matrix representation of $T$, and $xy=x^{\prime }{y}^{\prime }=r$. Solving for $a,b,c,d$ and we get $ad=1$ and $b=c=0$. In other words, with respect to rectangular hyperbolas of the form $xy=r$, the matrix representation of a hyperbolic rotation looks like

Since the matrix is non-singular, we see that in fact $T(H)=H$.

Now that we know the matrix form of a hyperbolic rotation when the rectangular hyperbolas have the form $xy=r$, it is not hard to solve the general case. Since the two asymptotes of any rectangular hyperbola $H$ are perpendicular, by an appropriate change of bases (ordinary rotation), $H$ can be transformed into a rectangular hyperbola $H^{\prime }$ whose asymptotes are the $x$ and $y$ axes, so that $H^{\prime }$ has the algebraic form $xy=r$. As a result, the matrix representation of a hyperbolic rotation $T$ with respect to $H$ has the form

for some $0\ne a\in ℝ$ and some orthogonal matrix $P$. In other words, $T$ is diagonalizable with $a$ and $a^{-1}$ as eigenvalues ($T$ is non-singular as a result).

Below are some simple properties:

• Unlike an ordinary rotation $R$, where $R$ fixes any circle centered at $O$, a hyperbolic rotation $T$ fixing one rectangular hyperbola centered at $O$ may not fix another hyperbola of the same kind (as implied by the discussion above).

• Let $P$ be the pencil of all rectangular hyperbolas centered at $O$. For each $H\in P$, let $[H]$ be the subset of $P$ containing all hyperbolas whose asymptotes are same as the asymptotes for $H$. If a hyperbolic rotation $T$ fixing $H$, then $T(H^{\prime })=H^{\prime }$ for any $H^{\prime }\in [H]$.

• $[\cdot ]$ defined above partitions $P$ into disjoint subsets. Call each of these subset a sub-pencil. Let $A$ be a sub-pencil of $P$. Call $T$ fixes $A$ if $T$ fixes any element of $A$. Let $A\ne B$ be sub-pencils of $P$. Then $T$ fixes $A$ iff $T$ does not fix $B$.

• Let $A,B$ be sub-pencils of $P$. Let $T,S$ be hyperbolic rotations such that $T$ fixes $A$ and $S$ fixes $B$. Then $T\circ S$ is a hyperbolic rotation iff $A=B$.

• In other words, the set of all hyperbolic rotations fixing a sub-pencil is closed under composition. In fact, it is a group.

• Let $T$ be a hyperbolic rotation fixing the hyperbola $xy=r$. Then $T$ fixes its branches (connected components) iff $T$ has positive eigenvalues.

• $T$ preserves area.

• Suppose $T$ fixes the unit hyperbola $H$. Let $P,Q\in H$. Then $T$ fixes the (measure of) hyperbolic angle between $P$ and $Q$. In other words, if $\alpha$ is the measure of the hyperbolic angle between $P$ and $Q$ and, by abuse of notation, let $T(\alpha )$ be the measure of the hyperbolic angle between $T(P)$ and $T(Q)$. Then $\alpha =T(\alpha )$.

The definition of a hyperbolic rotation can be generalized into an arbitrary two-dimensional vector space: it is any diagonalizable linear transformation with a pair of eigenvalues $a,b$ such that $ab=1$.