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As is well-known, in dimension 2, a linear map $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a direct similarity if, once we identify $\mathbb{R}^2$ with $\mathbb{C}$, $f$ is of the form $$\forall z \in \mathbb{C}, \quad f(z) = a z + b$$ with $a \in \mathbb{C}\backslash \{0\}$ and $b \in \mathbb{C}$. This gives an especially appealing way of describing and parameterizing similarities. By writing $a = r\mathrm{e}^{\mathrm{i} \theta}$ with $r >0$ and $\theta \in [0,2\pi)$, we recover that a similarity is the combination of a rotation (of angle $\theta$), a homothety (of parameter $r$), and a translation (of $b$).

I am curious about possible extension of this result in dimension 3. Of course, there is no three-dimensional space such as the complex numbers. However, it is possible to describe 3D direct similarities in terms of a combination of homotheties (around a certain point, possibly non-zero), rotations (idem), and translations. Since we can represent 3D-rotations with unit-quaternions (see this nice Youtube video), I am wondering if there is a nice comparable relation to $f(z) = az + b$ above in the 3D case.

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  • $\begingroup$ One problem is that the rotation action of the quaternions isn't given by $q\mapsto(f_q(z): z\mapsto qz)$ as it is with complex numbers representing 2d rotations, but rather $q\mapsto(f_q(z): z\mapsto q^{-1}zq)$ (possibly with the specific conjugation inverted depending on your conventions). $\endgroup$
    – Steven Stadnicki
    Commented Jun 8, 2020 at 1:46
  • $\begingroup$ This question is, more or less, what led Hamilton to discover the quaternions in the first place. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jun 8, 2020 at 2:22
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    $\begingroup$ What about $v \mapsto qvq^* + u$ where $v$ is a vector quaternion, $u$ is a vector quaternion, and $q$ is a quaternion $\endgroup$
    – wlad
    Commented Jun 8, 2020 at 9:04
  • $\begingroup$ In dimension $4$, any orthogonal transformation is of the form $v\mapsto a\cdot v\cdot b$ with $a,b\in \mathbb S^3\subset H^\times$. This defines the double covering $SU(2)\times SU(2)\to SO(4)$. This generalizes immediately to similarities, the group of which is $\mathbb (H^\times \mathbb H^\times / \big((\lambda,1)\sim (1,\lambda)\big)\ltimes \mathbb H$ acting by $(a,b,x)\circ v = avb + v$. To get to $SO(3)$, restrict to those transformations preserving $\mathbb R^\perp\subset \mathbb H$, which is equivalent to $x\in \mathbb R^\perp,ab\in\mathbb R$, i.e. $b$ is proportional to $\overline a$. $\endgroup$
    – Bertram Arnold
    Commented Jun 8, 2020 at 9:05
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    $\begingroup$ In terms of matrices, the group of similarity transformations is a subgroup of the group of affine transformations, the latter of which has a matrix representation $\endgroup$
    – wlad
    Commented Jun 8, 2020 at 9:09

1 Answer 1

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Call a quaternion whose scalar part is zero a vector quaternion. We shall denote the vector quaternions as $\mathbb R^3$. Given $q = w + xi + yj + zk$, we shall define $q^*$ (called the "conjugate" of $q$) to be $w - xi - yj - zk$.

If $q$ is a unit quaternion, then $v \in \mathbb R^3\mapsto qvq^*$ is a rotation. All rotations about the origin in 3D can be given in this form.

If $q$ is a general (i.e. not necessarily a unit) quaternion, then $v \mapsto qvq^*$ is a rotation followed by a dilation from the origin by $|q|^2$ units.

Finally, add a vector quaternion $x$ to perform a translation. We thus get $v \mapsto qvq^* + x$ as a general form for similarity transformations.

Postscript: This can be generalised to higher dimensions, and some non-Euclidean geometries, using Clifford algebras.

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  • $\begingroup$ That's nice, using the "vector quaternions" for $\mathbb{R}^3$ and then describing the geometric transformations as exposed does the job. Thanks a lot! $\endgroup$
    – Goulifet
    Commented Jun 8, 2020 at 13:33
  • $\begingroup$ Is it obvious that we capture all the direct similarities doing so? $\endgroup$
    – Goulifet
    Commented Jun 8, 2020 at 14:25
  • $\begingroup$ Yes -- every orientation-preserving similarity is expressible uniquely as an element of SO(n) followed by a scaling followed by a translation. $\endgroup$
    – Adam P. Goucher
    Commented Jun 8, 2020 at 14:42

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