# Is hyperbolic rotation really a rotation?

We define a $2\times 2$ Givens rotation matrix as:

$${\bf G}(\theta) = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) &\cos(\theta) \end{bmatrix}.$$

On the other hand, we define a $2\times 2$ hyperbolic rotation matrix as:

$${\bf H}(y)=\begin{bmatrix} \cosh( y) & \sinh( y) \\ \sinh( y) &\cosh( y) \end{bmatrix}.$$

I don't see why do we qualify matrix ${\bf H}$ as a rotation!

Suppose we take a 2-D vector $x=[-3, 1]^T$ and we transform it using ${\bf G}(\theta), \theta = 0,\dots, \pi/2$, and ${\bf H}, y = -2,\dots, 2.5$. See below for the result.

For me Givens rotation does clearly rotate the initial point around the point $[0,0]^T$ but for the hyperbolic rotation, we see a bending but not a rotation, at least not around a fixed point (I checked for other points and its the same behavior with different bending angles). am I missing something?

• An hyperbolic rotation should correspond to sliding points along an hyperbola instead of a circle, so this makes sense, no? Feb 26, 2018 at 23:58
• @AlexProvost Guess so. But what does not make sense is the term "rotation" I think! Feb 27, 2018 at 0:00
• A rotation is something which preserves a distance to a given center. And the clue lies in the term distance. You are pretty free in defining what distance means. For the Euclidea distance you get the usual rotation. For other distance measures (as e.g. in special relativity) you get hyperbolic rotation. It preseves some distance to some center, just not the obvious one. Feb 27, 2018 at 14:53
• One way to view the hyperbolic transformation is as a rotation through an imaginary angle.
– amd
Feb 27, 2018 at 21:27
• Yes, that's why scientists call call space "imaginary time." Multiply both sides by i to see that, equivalently, (elapsed) time is negative real distance. You can see this in the interval metric, where the (square of) the elapsed time is SUBTRACTED from the sum of the space distances (squared). [Using the Einstein metric signature notation]. Oct 21, 2022 at 10:39

You can say that ${\bf H}(y)$ is a rotation, not for the usual inner product $$\langle {\bf x}, {\bf y}\rangle_E = x_1y_1 + x_2y_2,$$but for the Lorentz-Minkowski product $$\langle {\bf x}, {\bf y}\rangle_L = x_1y_1 - x_2y_2.$$ In the same way that orthogonal transformations are linear maps preserving $\langle\cdot,\cdot\rangle_E$, we call the linear maps preserving $\langle \cdot,\cdot\rangle_L$ Lorentz transformations.

The point is that ${\bf H}(y)$ is not an orthogonal map, but a Lorentz transformation. You can see it as a "rotation" moving points along hyperbolas $xy = {\rm constant}.$

In $\Bbb R^n$, consider $$\langle {\bf x},{\bf y}\rangle_L = x_1y_1+\cdots+x_{n-1}y_{n-1} - x_ny_n.$$Call ${\bf x} \neq {\bf 0}$ spacelike, timelike or lightlike if $\langle {\bf x}, {\bf x}\rangle_L$ is positive, negative, or zero.

In the Euclidean case, we call the elements of ${\rm SO}(n,\Bbb R)$ (orthogonal maps with unit determinant) rotations (with respect to the Euclidean inner product). You can mimic that and say that Lorentz transformations with unit determinant are rotations with respect to the Lorentz-Minkowski product. You must be careful in odd dimensions, though. For example, in $n=3$ Lorentz transformations with unit determinant always have an eigenvector. The rotation will be called hyperbolic (resp. elliptic, parabolic) if said eigenvector is spacelike (resp. timelike, lightlike).

In your case we consider ${\bf H}(y)$ a hyperbolic rotation seeing the plane $\Bbb R^2$ as the $xz$ (or $yz$) plane in $\Bbb R^3$, so that the eigenvector $(1,0,0)$ of $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cosh y & \sinh y \\ 0 & \sinh y & \cosh y\end{pmatrix}$$is spacelike.

• What's the characteristic that defines a "rotation", i.e., the thing that allows us to say that something can be considered a rotation and something else cannot? Feb 27, 2018 at 0:03
• I added a bit more of an explanation in the answer, since it would get a bit long for a comment. Feb 27, 2018 at 0:11
• My point is: there is no reason whatsoever for ${\bf H}(y)$ to preserve anything Euclidean, since ${\bf H}(y)$ is essentially a Lorentzian object. Feb 27, 2018 at 0:24
• You're right. I guess I have to switch to a non-Euclidean mindset :-) Feb 27, 2018 at 0:27
• @MedNait: That's the key to the whole question imo. Hyperbolic rotations indeed aren't really rotations in Euclidian space. But they're used in Minkowski space, and there they make sense as rotations that involve the time axis. Feb 27, 2018 at 9:45

A hyperbolic rotation is a rotation because of its effect on hyperbolic angles!

Like the fact circular angles relate to the area of a (circular) wedge, hyperbolic angle is related to the area of a hyperbolic wedge:

(source: WolframAlpha)

Just like how the point on the unit circle making a circular angle $\theta$ with the positive $x$-axis is given by $\mathbf{c}(\theta) = (\cos(\theta), \sin(\theta))$, the point on the unit hyperbola making hyperbolic angle $u$ with the positive $x$-axis is given by $\mathbf{h}(u) = (\cosh(u), \sinh(u))$.

And you can check:

$$\mathbf{h}(u+v) = \mathbf{H}(u) \mathbf{h}(v)$$

so applying $\mathbf{H}(u)$ does have the effect of adding a constant hyperbolic angle to the points it acts on.

The plot you want to make is the one where radial lines are labelled by hyperbolic angle (note the asymptotes have angles $\pm \infty$!), and the lines of constant magnitude should be the hyperbolic arcs $x^2 - y^2 = \mathrm{const}$.

(similar statements can be made with the hyperbolas opening upwards, leftwards, and downwards)

Hyperbolic rotations are useful when working with hyperbolic geometry, in the Minkowski hyperboloid model.

Euclidean plane is represented as the set of points $[x,y,z]^T$ such that $z=1$. Isometries which do not change orientation are generated by rotations $\left(\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right)$ and translations $\left(\begin{array}{ccc}1&0&x\\0&1&0\\0&0&1\end{array}\right)$.

Sphere is represented as the set of points $[x,y,z]^T$ such that $x^2+y^2+z^2=1$. Rotations do not change, while translations get replaced by rotations which affect $(x,z)$.

Hyperbolic plane is represented as the set of points $[x,y,z]^T$ such that $x^2+y^2-z^2=-1$. Note that the Minkowski hyperboloid model lives in the Minkowski space, so the isometries preserve the Minkowski metric rather than the usual Euclidean distance. Rotations do not change, translations get replaced by hyperbolic rotations in $(x,z)$. HyperRogue uses hyperbolic rotations to transform the world coordinates when the player moves, and it feels very natural indeed to call them rotations.

Notice that

$${\bf det\ H}(y)= {\bf det}\begin{bmatrix} \cosh( y) & \sinh( y) \\ \sinh( y) &\cosh( y) \end{bmatrix} = 1$$

This immediately tells us hyperbolic rotations preserve areas. This is a hint to consider rotating areas rather than single vectors.

Also notice that $${\bf H}$$ preserves the quadratic form $$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$ and that the level sets of this form are hyberbolas: $$\begin{bmatrix}t&x\end{bmatrix}^T\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}t\\x\end{bmatrix} = t^2 - x^2 = c$$ So rotations by $${\bf H}$$ also keep vectors on the same hyperbola.

Given any vector $${\hat v}$$, construct $${\hat v_2}$$ by reflecting $${\hat v}$$ over the $$t = x$$ line. This construction ensures $$v -v_2 \perp v_2 - v$$, and so the quadrilateral defined by the four corners $${\hat v}, {\hat v_2}, -{\hat v}, -{\hat v_2}$$ is a rectangle whose corners lie on the same level set $$\pm c$$. Call this rectangle $$R_{v}$$ It can be shown by symmetry or direct computation that $${\bf area}(R_v) = {\bf area}(R_{\bf H}(y)v)$$ In this way we may say $${\bf H}(y)$$ rotates rectangles along hyperbolic level sets. The rectangle's length/width change during the rotation, but the area is preserved.

Note that this mirrors trigonometry, where the analogous construction rotates squares inside of a circle.