Quaternions

Quaternions, $\mathbb{H}$ are a number system like real or complex numbers but with 4 dimensions. In particular, $\mathbb{H}$ is nothing but ${\mathbb{R}}^4$ together with a multiplication law. The identification of $\mathbb{H}$ and ${\mathbb{R}}^4$ is given by:
$$\mathbb{H}=\{x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}|x_0,x_1,x_2,x_3\in \mathbb{R}\}.$$
With the usual addition and scalar multiplication inherited from ${\mathbb{R}}^4$, $\mathbb{H}$ becomes a four dimensional  $\mathbb{R}$ vector space with canonical basis $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}.$ The quaternion multiplication is defined by:
$$\begin{array}{rl}& (x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k})\cdot (y_0+y_1\mathbf{i}+y_2\mathbf{j}+y_3\mathbf{k})\\ :=& \left(x_0{y}_0–x_1{y}_1–x_2{y}_2–x_3{y}_3\right)\\ +& (x_0{y}_1+x_1{y}_0+x_2{y}_3–x_3{y}_2)\mathbf{i}\\ +& (x_0{y}_2–x_1{y}_3+x_2{y}_0+x_3{y}_1)\mathbf{j}\\ +& (x_0{y}_3+x_1{y}_2–x_2{y}_1+x_3{y}_0)\mathbf{k}.\end{array}$$

A more simple way to define the quaternion multiplication is to say that the multiplication with $\alpha \in \mathbb{R}$ works as usually and that $\mathbf{i},\mathbf{j},\mathbf{k}$ satisfy the following  multiplication rules:

$${\mathbf{i}}^2={\mathbf{j}}^2={\mathbf{k}}^2=-1,\mathbf{i}\mathbf{j}=-\mathbf{j}\mathbf{i}=\mathbf{k},\mathbf{j}\mathbf{k}=-\mathbf{k}\mathbf{j}=\mathbf{i},\mathbf{k}\mathbf{i}=-\mathbf{i}\mathbf{k}=\mathbf{j}.$$

$\mathrm{R}\mathrm{e}(\mathbb{H}):=span\{1\}$ is called the real part of  $\mathbb{H}$ and $\mathrm{I}\mathrm{m}(\mathbb{H}):=span\{\mathbf{i},\mathbf{j},\mathbf{k}\}$  the imaginary part and there holds:  $\mathbb{H}=\mathrm{R}\mathrm{e}(\mathbb{H})\oplus \mathrm{I}\mathrm{m}(\mathbb{H})$. If we identify $\mathrm{R}\mathrm{e}(\mathbb{H})$ with $\mathbb{R}$ and $\mathrm{I}\mathrm{m}(\mathbb{H})$ with ${\mathbb{R}}^3$, we can write any $\mathbf{x}\in \mathbb{H}$ as: $\mathbf{x}=\alpha +\mathbf{v},$ for some  $\alpha \in \mathbb{R}$ and $\mathbf{v}\in {\mathbb{R}}^3.$ This notation gives us a new expression of the quaternion multiplication using the scalar and cross product of ${\mathbb{R}}^3$:
$$\mathbf{p}\mathbf{q}=(\alpha +\mathbf{v})(\beta +\mathbf{w})=\alpha \beta -⟨\mathbf{v},\mathbf{w}⟩+\alpha \mathbf{w}+\beta \mathbf{v}+\mathbf{v}\times \mathbf{w}.$$

Due to the fact that the cross product is skew symmetric, we immediately see that the quaternion multiplication is not commutative, i.e. in general there holds $\mathbf{p}\mathbf{q}\ne \mathbf{q}\mathbf{p}$.

Proposition: The quaternion multiplication is associative i.e. for $\mathbf{p},\mathbf{q},\mathbf{r}\in \mathbb{H}$ there holds:
$$(\mathbf{p}\mathbf{q})\mathbf{r}=\mathbf{p}(\mathbf{q}\mathbf{r})$$

Proof: Let $$I:=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),X:=\left(\begin{array}{cc}\mathbf{i}& 0\\ 0& -\mathbf{i}\end{array}\right),Y:=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),Z:=\left(\begin{array}{cc}0& -\mathbf{i}\\ -\mathbf{i}& 0\end{array}\right).$$

Now we can define a linear map $$\begin{array}{rl}F:\mathbb{H}& \to {\mathbb{C}}^{2\times 2}\\ x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}& \mapsto x_{0}I+x_{1}X+x_{2}Y+x_{3}Z\end{array}$$
A straight forward calculation shows that $I,X,Y,Z$ satisfy the following product rules with respect to the matrix multiplication:
$$\begin{array}{rl}{I}^2=I,& IX=XI=I,XY=-YX=Z,\\ & IY=YI=Y,YZ=-ZY=X,\\ & IZ=ZI=Z,ZX=-XZ=Y.\end{array}$$
Therfore, $F$ is an algebra isomorphism onto its image. Due to the fact that the matrix multiplication is associative we obtain that the quaternion multiplication is associative too.

The conjugate of an quaternion number $\mathbf{x}=\alpha +\mathbf{v}$ is given by $\overline{\mathbf{x}}=\alpha –\mathbf{v}$.

Proposition: For $\mathbf{p},\mathbf{q}\in \mathbb{H}$ there holds : $\overline{\mathbf{p}\mathbf{q}}=\overline{\mathbf{q}}\overline{\mathbf{p}}.$

Proof: $$\begin{array}{rl}\mathbf{p}\mathbf{q}& =\alpha \beta -⟨\mathbf{v},\mathbf{w}⟩+\alpha \mathbf{w}+\beta \mathbf{v}+\mathbf{v}\times \mathbf{w}\\ \overline{\mathbf{p}\mathbf{q}}& =\alpha \beta –⟨\mathbf{v},\mathbf{w}⟩–\alpha \mathbf{w}–\beta \mathbf{v}+\mathbf{w}\times \mathbf{v}\\ & =\overline{\mathbf{q}}\overline{\mathbf{p}}\end{array}$$

$\mathbb{H}$ also inherits the euclidean norm from ${\mathbb{R}}^4$: $|\mathbf{p}|=(\sum _{i=0}^3{p}_i^2{)}^{\frac{1}{2}}$. Similar to the complex numbers we have $|\mathbf{p}{|}^2=\mathbf{p}\overline{\mathbf{p}}=\overline{\mathbf{p}}\mathbf{p}$. Additionally, there holds the following useful formula:

Proposition: For $\mathbf{p},\mathbf{q}\in \mathbb{H}$ there holds: $|\mathbf{p}\mathbf{q}|=|\mathbf{p}||\mathbf{q}|.$

Proof:$$\begin{array}{r}|\mathbf{p}\mathbf{q}|=((\mathbf{p}\mathbf{q})(\overline{\mathbf{p}\mathbf{q}}){)}^{\frac{1}{2}}=(\mathbf{p}\mathbf{q}\overline{\mathbf{q}}\overline{\mathbf{p}}{)}^{\frac{1}{2}}=(\mathbf{p}|\mathbf{q}{|}^2\overline{\mathbf{p}}{)}^{\frac{1}{2}}=(|\mathbf{p}{|}^2|\mathbf{q}{|}^2{)}^{\frac{1}{2}}=|\mathbf{p}||\mathbf{q}|.\end{array}$$

Note that for $\mathbf{q}\in \mathbb{H}$ the inverse element with respect to the quaternion multiplication is given by
$${\mathbf{q}}^{-1}=\frac{\overline{\mathbf{q}}}{|\mathbf{q}{|}^2}.$$
For quaternions with unit length this gives us immediately ${\mathbf{q}}^{-1}=\overline{\mathbf{q}}.$
What makes quaternions so useful is the fact that one can describe rotation in ${\mathbb{R}}^3$ with them in a very elegant way. Therefore, we have to consider ${\mathbb{R}}^3$ as $\mathrm{I}\mathrm{m}(\mathbb{H})$.

Theorem: Let $\mathbf{a}\in {\mathbb{R}}^3$ with $|\mathbf{a}|=1$ , $\alpha \in \mathbb{R}$ and $\mathbf{q}=\cos \left(\frac{\alpha }{2}\right)+\sin \left(\frac{\alpha }{2}\right)\mathbf{a}$. Then for all $\mathbf{y}\in {\mathbb{R}}^3$ we have:

(i)  $\mathbf{q}\mathbf{y}\overline{\mathbf{q}}\in \mathrm{I}\mathrm{m}(\mathbb{H})={\mathbb{R}}^3$

(ii) The map $F:{\mathbb{R}}^3\to {\mathbb{R}}^3$, $\mathbf{y}\mapsto \mathbf{q}\mathbf{y}\overline{\mathbf{q}}$ is a rotation around $\mathbf{a}$ by the angle $\alpha$.

Proof: (i) First note that a quaternion $\mathbf{y}$ is purely imaginary if and only if $\overline{\mathbf{y}}=-\mathbf{y}$. Thus we have to show $\overline{\mathbf{q}\mathbf{y}\overline{\mathbf{q}}}=-\mathbf{q}\mathbf{y}\overline{\mathbf{q}}$.
$$\overline{\mathbf{q}\mathbf{y}\overline{\mathbf{q}}}=\mathbf{q}\overline{\mathbf{y}}\overline{\mathbf{q}}=-\mathbf{q}\mathbf{y}\overline{\mathbf{q}}.$$

(ii) $F$ is a linear map and therefore completely determined by its action on a basis. We extend $\mathbf{a}$ to an positive oriented  orthonormal basis $\{\mathbf{a},\mathbf{b},\mathbf{c}\}$ of ${\mathbb{R}}^3$.
$$\begin{array}{rl}F(\mathbf{a})=\mathbf{q}\mathbf{a}\overline{\mathbf{q}}& =(\cos \left(\frac{\alpha }{2}\right)+\sin \left(\frac{\alpha }{2}\right)\mathbf{a})\mathbf{a}(\cos \left(\frac{\alpha }{2}\right)–\sin \left(\frac{\alpha }{2}\right)\mathbf{a})\\ & =(\cos \left(\frac{\alpha }{2}\right)+\sin \left(\frac{\alpha }{2}\right)\mathbf{a})(\cos \left(\frac{\alpha }{2}\right)\mathbf{a}+\sin \left(\frac{\alpha }{2}\right))\\ & =({\cos }^2\left(\frac{\alpha }{2}\right)+{\sin }^2\left(\frac{\alpha }{2}\right))\mathbf{a}+\cos \left(\frac{\alpha }{2}\right)\sin \left(\frac{\alpha }{2}\right)(1+{\mathbf{a}}^2)\\ & =\mathbf{a}.\end{array}$$

In the last step we used that ${\mathbf{a}}^2=-1$. This gives us that  the $\mathbf{a}$-axis is invariant under $F$. Now we consider the action of $F$ on the plane orthogonal to $\mathbf{a}$, i.e. the plane spanned by $\mathbf{b}$ and $\mathbf{c}$. Since $⟨\mathbf{a},\mathbf{b}⟩=0$ we have:
$$\mathbf{a}\mathbf{b}=\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}=-\mathbf{b}\mathbf{a}.$$

Using this and the addition formulas for cosine and sine we get:
$$\begin{array}{rl}F(\mathbf{b})& =(\cos \left(\frac{\alpha }{2}\right)+\sin \left(\frac{\alpha }{2}\right)\mathbf{a})\mathbf{b}(\cos \left(\frac{\alpha }{2}\right)–\sin \left(\frac{\alpha }{2}\right)\mathbf{a})\\ & =(\cos \left(\frac{\alpha }{2}\right)+\sin \left(\frac{\alpha }{2}\right)\mathbf{a})(\cos \left(\frac{\alpha }{2}\right)+\sin \left(\frac{\alpha }{2}\right)\mathbf{a})\mathbf{b}\\ & =(\cos (\alpha )+\sin (\alpha )\mathbf{a})\mathbf{b}=\cos (\alpha )\mathbf{b}+\sin (\alpha )(\mathbf{a}\times \mathbf{b})\\ & =\cos (\alpha )\mathbf{b}+\sin (\alpha )\mathbf{c}\end{array}$$

Analog we obtain $F(\mathbf{c})=\cos (\alpha )\mathbf{b}–\sin (\alpha )\mathbf{b}$ and the matrix representation of $F$ with respect to the basis  $\{\mathbf{a},\mathbf{b},\mathbf{c}\}$ is given by:
$$F=\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos (\alpha )& -\sin (\alpha )\\ 0& \sin (\alpha )& \cos (\alpha )\end{array}\right)$$.

Now it is easy to see that $F$ describes a rotation around $\mathbf{a}$ by the angle $\alpha$.

Corollary: For every $\mathbf{q}\in \mathbb{H}\mathrm{\setminus }\{0\}$ the map $F:{\mathbb{R}}^3\to {\mathbb{R}}^3$, $\mathbf{y}\mapsto \mathbf{q}\mathbf{y}{\mathbf{q}}^{-1}$ is a rotation.

Proof: With ${\mathbf{q}}^{-1}=\frac{\overline{\mathbf{q}}}{|\mathbf{q}{|}^2}$ we get:
$$\mathbf{q}\mathbf{y}{\mathbf{q}}^{-1}=\frac{\mathbf{q}}{|\mathbf{q}|}\mathbf{y}\overline{\left(\frac{\mathbf{q}}{|\mathbf{q}|}\right)}$$
Since $\frac{\mathbf{q}}{|\mathbf{q}|}$ has unit length there exists $\mathbf{a}\in {\mathbb{R}}^3$ with $|\mathbf{a}|=1$ and  $\alpha \in \mathbb{R}$  such that: $\frac{\mathbf{q}}{|\mathbf{q}|}=\cos \left(\frac{\alpha }{2}\right)+\sin \left(\frac{\alpha }{2}\right)\mathbf{a}$ and we can apply the theorem.

Note that the quaternion multiplication corresponds to the concatenation of rotations. Let $\mathbf{p},\mathbf{q}\in \mathbb{H}$ with $|\mathbf{p}|=|\mathbf{q}|=1$ and $F,G:{\mathbb{R}}^3\to {\mathbb{R}}^3$ the corresponding rotations i.e. $F(\mathbf{y})=\mathbf{q}\mathbf{y}\overline{\mathbf{q}}$ and $G(\mathbf{y})=\mathbf{p}\mathbf{y}\overline{\mathbf{p}}$, then we get for the concatenation:

$$(F\circ G)(y)=\mathbf{q}\left(\mathbf{p}\mathbf{y}\overline{\mathbf{p}}\right)\overline{\mathbf{q}}=\left(\mathbf{q}\mathbf{p}\right)\mathbf{y}\overline{\left(\mathbf{q}\mathbf{p}\right)}.$$