You can represent a vector by a quaternion with no scalar part, and you can also represent the rotation itself as a quaternion. Then the rotation is applied to the vector by conjugation. The situation is similar in 2D, where you can represent both the vector and the rotation by complex numbers -- but in this case, the vector includes the "scalar" (real) part of the complex number. Furthermore, there is no need for conjugation in 2D, and in fact if you tried to conjugate you'd just get the identity rotation due to the commutativity I think.

Now, I know that these cases are both really the even components of their respective Clifford algebras, in which vectors have their own proper representation, and rotations are always applied by conjugation. But it's interesting that representations of vectors can be finagled from within the even subalgebras. For the quaternions, I think it might just be because the vectors and bivectors in 3D can effectively be identified with each other (which is why we can call the cross product a vector). But I'm not sure exactly how to explain why it works, and it's even less clear why it should work for the complex numbers. And I'm also not sure whether you could pull off a vector representation in higher dimensions; I'm guessing not.

Can anyone shed some light on this topic?


2 Answers 2


$ \newcommand\Cl{\mathrm{Cl}} \newcommand\R{\mathbb R} \newcommand\form[1]{\langle#1\rangle} \newcommand\grd\hat \newcommand\rev\widetilde \newcommand\cnj\widebar \newcommand\egrd\check \newcommand\egrdsup\vee \newcommand\erev[1]{#1^\dagger} \newcommand\ecnj[1]{#1^\ddagger} \newcommand\pseud\mathbf \newcommand\ei{\pseud i} $

I will start by describing the specific cases of complex numbers and quaternions, and then continue on to talk about paravectors in general. I will then describe how the paravector idea applies to each of the complex numbers and quaternions. We can find that there is a unifying perspective between complex numbers and quaternions as pseudoparavectors, and their transformation properties follow from this, in particular the fact that imaginary quaternions transform the same as vectors.

The Special Case of Quaternions

As you say, the quaternion case is a bit more straightforward. We consider the real Euclidean Clifford algebra $\Cl_3$, let $e_1,e_2,e_3$ be an orthonormal basis of vectors, and let $I = e_1e_2e_3$ be the right-handed unit pseudoscalar. Pseudoscalar multiplication (closely related to the Hodge star) gives the translation between vectors and bivectors. To match up with quaternions as they are usually defined, however, we need to use the left-handed pseudoscalar $-I$. We define $$ i = -e_1I = e_3e_2,\quad j = -e_2I = e_1e_3,\quad k = -e_3I = e_2e_1. $$ It doesn't matter where we put $I$ since pseudoscalars in odd dimensions commute with the whole algebra. These are three linearly-independent bivectors since $e_1, e_2, e_3$ are independent, and they satisfy the quaternion relations. Since any vector $v$ is a linear combination of $e_1,e_2,e_3$, the corresponding quaternion under this identification is $-vI$. But if $R$ is a rotor (or any even versor, i.e. any quaternion) $$ R^{-1}(-vI)R = -(R^{-1}vR)I $$ since $I$ commutes with $R$.

Complex Numbers as Paravectors, Part 1

You say

there is no explicit need for conjugation in 2D, although you can if you want (doesn't matter because it's commutative anyway).

but as best I can interpret this is incorrect; at least, some more nuance is required. Observe $$ e^{-\theta i/2}ze^{\theta i/2} = ze^{-\theta i/2}e^{\theta i/2} = z, $$ and we haven't actually rotated anything, while $$ e^{\theta i/2}ze^{\theta i/2} = ze^{\theta i}. $$ So let's add some more nuance.

We consider the real Euclidean Clifford algebra $\Cl_2$ and choose an orthonormal basis $e_1, e_2$ with $e_1$ as our real axis. The pseudoscalar $I = e_1e_2$ will be our complex $i$. Then given any vector $v$, there is an associated complex number $z = e_1v$; notice that the $e_1$ component $v\cdot e_1$ of $v$ is exactly the real part of $z$, and that the $e_2$ component is exactly the imaginary part. Since the only bivectors are multiples of $I$, every rotor is of the form $e^{\theta I/2}$. We achieve rotation via $$ v \mapsto e^{-\theta I/2}ve^{\theta I/2} $$ so our complex numbers get mapped as $$ z = e_1v \mapsto e_1e^{-\theta I/2}ve^{\theta I/2} = e^{\theta I/2}e_1ve^{\theta I/2} = e^{\theta I/2}ze^{\theta I/2} = ze^{\theta I}, $$ or in short $z \mapsto ze^{\theta I}$. It is the fact the $I$ does not commute with the whole algebra that allows this to happen. The form $ze^{\theta I}$ is a direct consequence of the fact that 2D space consists of only one plane.


Let $\Cl_{p,q}$ be the real Clifford algebra such that there is an orthonormal basis $\{e_i\}_{i=1}^n$, $n = p+q$, where $$ e_1^2 = e_2^2 = \cdots = e_p^2 = 1,\quad e_{p+1}^2 = e_{p+2}^2 = \cdots = e_{p+q}^2 = -1. $$ The even subalgebra $\Cl^+_{p,q}$ is isomorphic to $\Cl_{p,q-1}$ when $q\not= 0$ and to $\Cl_{q,p-1}$ when $p\not= 0$. But how do we realize such an isomorphism?

The construction above with complex numbers is a special case, and we proceed similarly. Choose any vector $r$ (with $r^2\not=0$), which perhaps you would call a real axis if $r^2 > 0$ or an imaginary axis if $r^2 < 0$. We will assume $r$ is a unit vector. For brevity we will write $r' := r^{-1} = \iota r$ where $\iota = 1/r^2 = \pm1$. Given any other vector $v$, we may form a bivector via $r'\wedge v$. The set $Z_1 := r'\wedge\R^n$ of all such bivectors may be taken as the set of "vectors" for $\Cl^+_{p,q} \cong \Cl_{p,q-1} \cong \Cl_{q,p-1}$, and they generate $\Cl^+_{p,q}$ in the same way $\R^n$ generates $\Cl_{p,q}$. Whether we get $(p,q-1)$ or $(q,p-1)$ as the signature depends on the sign of $r^2$. We will write $\Cl(Z_1) := \Cl^+_{p,q}$ to emphasize $\Cl^+_{p,q}$ as a Clifford algebra over $Z_1$.

We get an embedding of the vectors in $\Cl_{p,q}$ into $\R\oplus Z_1 \subseteq \Cl(Z_1)$ via the Clifford product $r'v$; such elements of $Z := r'\R^n$ are called paravectors. Note that if $z \in Z$ then the corresponding vector is $rz$. The paravectors $Z_1 \subseteq Z$ correspond to the set of all vectors orthogonal to $r$; we will call them (scalar-)free paravectors.

Grade-Involution and Reversal

The map $$ X \mapsto \egrd X = (X)^\egrdsup := r'Xr = rXr' $$ is a grade-preserving involution on $\Cl_{p,q}$ and hence $\Cl(Z_1)$. Choosing an orthornormal basis $\{e_i\}_{i=1}^n$ so that $e_1 = r$, the bivectors $\{r'e_i\}_{i=2}^n$ form a basis for $Z_1$. Conjugation by $r$ yields $$ (r'e_i)^\egrdsup = r'r'e_ir = -r'e_ir'r = -r'e_i $$ so in fact conjugation by $r$ is $\Cl(Z_1)$-grade involution. It is worth noting that $\egrd z = \rev z$ (the reversal) for paravectors $z$.

The "vectors" of $\Cl(Z_1)$ are the free paravectors, i.e. the bivectors $Z_1$; from this, we can surmise that $\Cl(Z_1)$-reversal can be accomplished by $\Cl_{p,q}$-reversal (to reverse everything) followed by $\Cl(Z_1)$-grade involution (to reverse each $Z_1$ "vector"). To illustrate, $$\begin{aligned} (r'\wedge v_1)(r'\wedge v_2)(r'\wedge v_3) &\mathrel{\tilde\longmapsto} (v_3\wedge r')(v_1\wedge r')(v_2\wedge r') \\ &\mathrel{\egrd\longmapsto} (r'\wedge v_3)(r'\wedge v_1)(r'\wedge v_2). \end{aligned}$$ For $X \in \Cl(Z_1)$, putting this together gives $$ \erev X := \egrd{\rev X} = r'\rev Xr, $$ and it follows that $\Cl(Z_1)$-Clifford conjugation is just $\rev X$.

The Paravector Metric

The paravectors inherit a metric via $$ \form{z, z'}_Z := \form{rzrz'}_0 = \iota\form{\egrd zz'}_0. $$ Note that $\Cl_{p,q}$ and $\Cl(Z_1)$ have the same scalar-part operator. When $z = z'$, we clearly have $$ \form{z, z}_Z = \iota\egrd zz. $$

Lipschitz Groups and Orthogonal Transformation of Paravectors

Recall that the Lipshitz group is $$ \Gamma = \{\gamma \in \Cl_{p,q}^\times \;:\; \hat\gamma^{-1}\R^n\gamma \subseteq \R^n\} $$ where $\hat\gamma$ is the $\Cl_{p,q}$-grade involution of $\gamma$ and $\Cl_{p,q}^\times$ is the set of all invertible multivectors; we also define $$ \Gamma^+ := \Gamma\cap\Cl^+_{p,q},\quad \Gamma^{(+)} := \{\gamma \in \Gamma^+ \;:\; \rev\gamma\gamma = 1\}. $$ The elements of $\Gamma$ are called versors; $\Gamma^+$ are the even versors, and $\Gamma^{(+)}$ are even versors with unit norm, i.e. rotors. It turns out that a multivector is a versor iff it is a product of vectors, and it follows that all invertible paravectors are versors: $Z^\times \subseteq \Gamma^+$.

Applying the above definition of the Lipschitz group to $\Cl(Z_1)$ gives $$ \Gamma_{Z_1} := \{\gamma \in \Cl(Z_1) \;:\; \egrd\gamma^{-1}Z_1\gamma \subseteq Z_1\}. $$ But this is equivalent to looking for all $\gamma$ such that $$ r'\gamma^{-1}rr'v\gamma = r'w \iff \gamma^{-1}v\gamma = w $$ where $v, w \in \R^n$ are orthogonal to $r$. Such a $\gamma$ must be in $\Gamma^+$. If $I \in \Cl_{p,q}$ is any pseudoscalar, then this is equivalent to $$ \gamma^{-1}rI\gamma = \pm rI \iff \gamma^{-1}r\gamma = \pm r, $$ the last equality following since $\gamma$ is even and commutes with $I$. Hence $\Gamma_{Z_1}$ is exactly the set of all even versors which fix $r$ or negate $r$ (or equivalently commute or anticommute with it). From the perspective of $\Cl(Z_1)$, however, we note that $$ \gamma^{-1}r\gamma = r \iff \egrd\gamma = \pm\gamma. $$ Hence $$ \Gamma_{Z_1} = \{\gamma \in \Gamma^+\;:\; \egrd\gamma = \pm\gamma\}. $$ The condition $\egrd\gamma = \pm\gamma$ is to be expected; this is just saying that $\gamma$ is $\Cl(Z_1)$-even or -odd.

Paravectors $z \in Z$ inherit additional transformations. For any $\gamma \in \Gamma^+$, we may map $r'v \mapsto r'\gamma^{-1}v\gamma$, or $$ z \mapsto r'\gamma^{-1}rz\gamma = \egrd \gamma^{-1}z\gamma. $$ Such a $\gamma$ is, of course, an element of $\Cl(Z_1)$. But in general $\Gamma^+$ is a proper superset of $\Gamma_{Z_1}$. We note that $$ \gamma^{-1}\R^n\gamma \subseteq \R^n \iff \egrd \gamma^{-1}Z\gamma \subseteq Z $$ so we in fact have $$ \Gamma^+ = \{\gamma \in \Cl(Z_1) \;:\; \egrd\gamma^{-1}Z\gamma \subseteq Z\}. $$

We can say even more: every element of $\Gamma^+$ is a product of invertible paravectors $Z^\times$ and $\Cl(Z_1)$-versors $\Gamma_{Z_1}$. It's worth noting that if $z, w \in Z$ and $w$ is a unit (i.e. $\form{w,w}_Z = \pm1$) then $w^{-1} = \rev w = \egrd w$ so the action of $w$ on $z$ is $$ z \mapsto wzw. $$ If such a $w$ is e.g. complex (discussed below), then $w = e^{\theta w_1/2}$ for some unit $w_1 \in Z_1$ and $\theta \in \R$, whence we get the form $$ z \mapsto e^{\theta w_1/2}ze^{\theta w_1/2}. $$ In terms of vectors, this is a rotation in the plane spanned by $r, rw_1$, going from $r$ to $rw_1$ by angle $\theta$.


Let $I = e_1e_2\cdots e_n$ be the pseudoscalar associated with the orthonormal basis $\{e_i\}_{i=1}^n$. We can compute that the corresponding pseudoscalar $\ei$ of $\Cl(Z_1)$ is $$ \ei = (r'e_2)(r'e_3)\cdots(r'e_n) = \begin{cases} \iota^{\lfloor n/2\rfloor} I &\text{if $n$ is even}, \\ \iota^{\lfloor n/2\rfloor}r'I &\text{if $n$ is odd}. \end{cases} $$ From here on let $I$ be any unit $\Cl_{p,q}$-pseudoscalar and $\ei$ the associated $\Cl(Z_1)$-pseudoscalar. This gives us the pseudoparavectors $\ei Z$. Define $$ \ecnj X := \ei^{-1}X\ei = \ei X\ei^{-1} = \begin{cases} X &\text{if $n$ is even}, \\ \egrd X &\text{if $n$ is odd}, \end{cases} $$ so that $\ei X = \ecnj X\ei$. This means that the action of $\gamma \in \Gamma^+$ on a pseudoparavector is $$ \ei z \mapsto \ei\egrd\gamma^{-1}z\gamma = (\ecnj{\egrd\gamma})^{-1}(\ei z)\gamma. $$ It follows immediately that $\Gamma^+$ acts on $\ei Z$ the same as $\R^n$ iff $\ecnj\gamma = \egrd\gamma$, which happens exactly when $n$ is odd (or $n-1$ is even). We can, of course, extend the paravector metric to pseudoparavectors $\pseud z, \pseud w$ by $$ \form{\pseud z, \pseud w}_{\ei Z} := \form{\ei^{-1}\pseud z, \ei^{-1}\pseud w}_Z = \iota\form{\egrd\ei^{-1}\egrd{\pseud z}\ei^{-1}\pseud w}_0 = \iota\ei^2\form{\ecnj{\egrd{\pseud z}}\pseud w}_0 $$

The Polar Decomposition

Paravectors generalize the polar decomposition of complex numbers. Let $z = z_0 + z_1 \in Z$ be any paravector with $z_0$ a scalar and $z_1$ free. By definition $z_1$ is a bivector blade, so $z_1^2$ is a scalar. When $z_1^2 \not= 0$ let $z_1' = z_1/|z_1^2|$. We may classify paravectors based on the sign of $z_1^2$:

  • When $z_1^2 < 0$ call $z$ complex. In this case there are $\rho, \theta \in \R$ such that $$ z = \rho e^{\theta z_1'},\quad z_0 = \rho\cos\theta,\quad z_1 = (\rho\sin\theta)z_1'. $$
  • When $z_1^2 = 0$ call $z$ nil. In this case, if $z_0 \not= 0$ $$ z = z_0e^{z_1/z_0}, $$ and if $z_0 = 0$ $$ z = z_1e^{z_1} = e^{z_1} - 1. $$
  • When $z_1^2 > 0$ call $z$ hyperbolic. If $z_0^2 < z_1^2$ then there are $\rho,\zeta \in \R$ such that $$ z = \rho z_1'e^{\zeta z_1'},\quad z_0 = \rho\sinh\zeta,\quad z_1 = (\rho\cosh\zeta)z_1'. $$ If $z_0^2 = z_1^2$, we don't get an exponential decomposition; instead $$ z = z_0(1 \pm z_1'). $$ If $z_0^2 > z_1^2$ $$ z = \rho e^{\zeta z_1'},\quad z_0 = \rho\cosh\zeta,\quad z_1 = (\rho\sinh\zeta)z_1'. $$

With appropriate restrictions on the parameters $\rho,\theta,\zeta, z_1'$, these decompositions are unique.

Since we can recover the vector represented by a paravector $z$ by $rz$, these polar decompositions apply to regular vectors as well; for example, for a vector $v$ such that $r'v$ is complex there is unit $v_1 \in \R^2$ orthogonal to $r$ such that $$ v = \rho re^{\theta r'v_1}. $$ In a similar fashion, we can apply the polar decomposition to pseudoparavectors $\pseud z$ $$ \pseud z = \rho\ei e^{\theta\ei^{-1}\pseud z_1} $$ where $\pseud z_1 \in \ei Z_1$ is unital.

Iterated Paravectors

We can take this "polar" decomposition further. Since $\Cl(Z_1)$ is itself a Clifford algebra, we can iterate the paravector construction. Let $\{e_i\}_{i=1}^n$ be a basis for $\R^n$ and define $$ r_1 = e_1,\quad r_2 = r_1'e_2,\quad r_3 = r_2'r_1'e_3,\quad r_4 = r_3'r_2'r_1'e_4,\quad\cdots $$ where as before we are using the shorthand $r_i' = r_i^{-1}$. Expanding these definitions gives $$ r_1 = e_1,\quad r_2 = e_1^{-1}e_2,\quad r_3 = e_2^{-1}e_3,\quad r_4 = e_3^{-1}e_4,\quad\cdots. $$ We define the $i^{\text{th}}$-paravector spaces $Z^i$ and the free $i^{\text{th}}$-paravector spaces $Z^i_1$ by $$ Z^1 = \left\{r_1'v \;:\; v \in \R^n\right\} = \R\oplus Z^1_1,\quad Z^{i+1} = \left\{r_{i+1}'z \;:\; z \in Z^i_1\right\} = \R\oplus Z^{i+1}_1. $$ Defining $R_i = r_1r_2\cdots r_i$, $R_i' = R_i^{-1}$, and $[v]^\perp = \{x \in \R^n \;:\; v\cdot x = 0\}$ we may write $$ Z^1 = \left\{R_1'v \;:\; v \in \R^n\right\},\quad Z^{i+1} = \left\{R_{i+1}'v \;:\; v \in [r_1]^\perp\cap\cdots\cap[r_i]^\perp\right\}, $$$$ Z^{i+1}_1 = \left\{R_{i+1}'v \;:\; v \in [r_1]^\perp\cap\cdots\cap[r_{i+1}]^\perp\right\}. $$ In $\Cl_{n,0}$, every $i^{\text{th}}$-paravector is complex. Furthermore, from the $i^{\text{th}}$-paravector $e^{\theta z}$ ($z \in Z^i_1$ and $|z^2| = 1$) we get the corresponding free $(i-1)^{\text{th}}$-paravector $r_ie^{\theta z}$. Writing $\exp$ for the exponentional function, iterating this construction naturally leads to $$ \rho r_1\exp(\theta_1r_2\exp(\theta_2r_3\exp(\cdots))) $$ for $0 < \rho \in \R$ and angles $\theta_1,\theta_2,\dotsc$. When the basis $\{e_i\}_{i=1}^n$ is orthonormal, this process can be thought of as follows:

  • Start with $e_{n-1}$ and rotate it towards $e_n$ by angle $\theta_{n-1}$, resulting in a vector $f_1$ represented by the $(n-1)^{\text{th}}$-paravector $z_1 = r_{n-1}e^{\theta_{n-1}r_n}$;
  • then rotate $e_{n-2}$ towards $f_1$ by angle $\theta_{n-2}$, resulting in a vector $f_2$ represented by $z_2 = r_{n-2}e^{\theta_{n-2}z_1}$;
  • then rotate $e_{n-3}$ towards $f_2$ by angle $\theta_{n-3}$, resulting in a vector $f_3$ represented by $z_3 = r_{n-3}e^{\theta_{n-3}z_2}$;
  • ...
  • Continue until we have $f_{n-1} = r_1e^{\theta_1z_{n-2}}$ and then scale $f_{n-1}$ by $\rho$ to get $\rho f_{n-1}$.

This is a generalization of spherical coordinates. Using $\Cl_{3,0}$ as an example, every vector may be written $$ \rho e_1\exp\bigl(\theta_1e_1e_2\exp(\theta_2 e_2e_3)\bigr) $$ with $\rho$ the radius, $\theta_1$ the polar angle, and $\theta_2$ the azimuthal angle (and with $e_1$ the pole and the azimuthal angle measured from $e_2$).

Complex Numbers as Paravectors, Part 2

Let's circle back to $\Cl_2 = \Cl_{2,0}$ and apply the paravector framework.

Recall that we choose $r = e_1$ and set $I = e_1e_2$. The associated $\Cl(Z_1)$-pseudoscalar is just $\ei = I$, so we will not notate them differently here. Since there is a unique 1D space orthogonal to $e_1$, namely the span of $e_2$, we have $$ Z_1 = \{aI \;:\; a \in \R\} $$ which is just the set of pseudoscalars. The paravectors are then just $$ Z = \{a + bI \;:\; a, b \in \R\} $$ and it so happens that in this case $Z = IZ = \Cl(Z_1)$ and all three of these are the complex numbers (in particular, they form a commutative algebra). We see that $$ (a + bI)^\egrdsup = a - bI $$ is exactly the complex conjugate; the $\Cl(Z_1)$-Clifford conjugate is again the complex conjugate since $\egrd z = \rev z$ on paravectors, and the $\Cl(Z_1)$-reversal $\erev z$ is then the identity. Since $2-1 = 1$ is odd, the "pseudoconjugate" $\ecnj z = z$, and elements of $\Gamma^+$ act the same on $IZ$ as they do $Z$; as such, we will not mention pseudoparavectors again in this section. The paravector metric is the usual inner product on complex numbers: $$ \form{z, z'}_{Z} = \form{\egrd zz'}_0 = \mathrm{Re}[\egrd zz']. $$

Since every element of $Z = \Cl(Z_1)$ is a versor, $\Gamma^+ = \Cl(Z_1)$. The rotors are $$ \Gamma^{(+)} = \{e^{\theta I/2} \;:\; \theta \in \R\} $$ Since every $\gamma = e^{\theta I/2} \in \Gamma^{(+)}$ is a unit paravector, they act on other paravectors $z$ by $$ z \mapsto \gamma z\gamma = z\gamma^2 = ze^{\theta I}. $$

The group $\Gamma_{Z_1}$ is the multiplicative group of real and imaginary numbers: $$ \Gamma_{Z_1} = \R^\times \cup \R^\times I = \R^\times\cup(Z_1)^\times, $$ which acts on paravectors by either the identity or reflection across the $e_2$-axis (i.e. real-part negation). This is as it should be, since $\Gamma_{Z_1}$ is the Lipschitz group of $\Cl(Z_1) \cong \Cl_{0,1}$ and acts by isometries of $\R$.

Complex numbers are, of course, complex paravectors; since every free paravector is of the form $\theta I$, the polar decomposition is $$ z = \rho e^{\theta I} $$ exactly as you would expect.

Quaternion Paravectors

We now turn our attention to $\Cl_{3,0}$. To be concrete, we again choose $r = e_1$ and let $I = e_1e_2e_3$. The associated $\Cl(Z_1)$-pseudoscalar of the left-handed $-I$ is then $\ei = -r'I = e_3e_2$. As before, we define $$ i = e_3e_2,\quad j = e_1e_3,\quad k = e_2e_1, $$ and we see that $\ei = i$, so we will use the former notation. Paravectors then take the form $$ Z = \{a - bk + cj \;:\; a,b,c \in \R\},\quad Z_1 = \{-bk + cj \;:\; b,c \in \R\}. $$ Note how in this case $Z$ is a proper subset of $\Cl(Z_1)$; not all quaternions are paravectors. The even subalgebra looks like $$ \Cl(Z_1) = \{a - bk + cj + d\ei \;:\; a,b,c,d \in \R\}, $$ noting that $-kj = \ei$. Of particular importance are the pseudoparavectors, for $$ \ei(-k) = j,\quad \ei j = k $$ and so $$ \ei Z = \{a\ei + bj + ck \;:\; a,b,c \in \R\} $$ revealing that the imaginary quaternions are exactly the pseudoparavectors of $\Cl_3$ with the direct correspondence $$ v = ae_1 + be_2 + ce_3 \quad\longleftrightarrow\quad \ei r'v = a\ei + bj + ck. $$ For this reason, we will focus on the pseudoparavectors. It can be easily confirmed that choosing $r = e_2$ or $r = e_3$ results in the same pseudoparavectors. The element $\ei r'$ is $$ \ei r' = e_3e_2e_1 = -I $$ as it should be.

The various (anti)-involutions on $\gamma = d + a\ei + bj + ck$ are $$ \egrd\gamma = d + a\ei - bj - ck,\quad \erev\gamma = d - a\ei + bj + ck,\quad \rev\gamma = d - a\ei - bj - ck,\quad \ecnj\gamma = \egrd\gamma, $$ showing that $\rev\gamma$ is the usual quaternion conjugation. Since $\check z = \rev z$ when $z$ is a paravector, the paravector metric is the usual quaternion inner product: $$ \form{z, w}_Z = \form{\check zw}_0 = \mathrm{Re}[\rev zw]. $$ The pseudoparavector metric is also the quaternion inner product $$ \form{\pseud z, \pseud w}_{\ei Z} = -\form{\pseud z\pseud w}_0 = \mathrm{Re}[\rev{\pseud z}\pseud w] $$ since $\rev{\pseud z} = -\pseud z$. (Of course, this is to be expected by the construction of these metrics as being identical to the vector metric.)

Since $n = 3$ is odd, meaning $\ecnj\gamma = \egrd\gamma$, the action of $\gamma \in \Gamma^+$ on pseudoparavectors is identical to the action on vectors; that is to say that imaginary quaternions transform exactly the same way as vectors.

Because of the unintuitive form $a - bk + cj$ of paravectors, it is perhaps best to think of the polar decomposition in terms of pseudoparavectors. Note that $\ei^{-1} = -\ei$ and the quaternion paravectors are complex. The unit pseudoparavector $\pseud z = \ei e^{-\theta\ei(bk + cj)}$ with $b^2 + c^2 = 1$ represents the vector $v = I\pseud z$ such that, upon choosing a direction $be_2 + ce_3$, our chosen axis $r = e_1$ is rotated in this direction by an angle $\theta$, resulting in $v$.

The Internal Perspective

I will not go in depth on this, but the real power of paravectors comes from viewing them internally. Rather than looking at paravectors of $\Cl^+_{p,q}$ from the perspective of $\Cl_{p,q}$, we look at the paravectors $\R\oplus\R^n$ of $\Cl_{p,q}$ itself. Then the the transformations which fix this set of paravectors correspond to the even Lipschitz group of $\Cl_{p,q+1}$ or of $\Cl_{q+1,p}$ (these groups being isomorphic). This means that we can derive (without any reference to $\Cl_{2,0}$) how complex numbers $\Cl_{0,1}$ must rotate if they are to represent the plane; and derive (without any reference to $\Cl_{3,0}$) how imaginary quaternions $\Cl^2_{0,2}\oplus\Cl^1_{0,2}$ must rotate if they are to represent space. As another example, $\Cl_{3,0}$ is isomorphic to the even space-time algebra $\Cl^+_{1,3}$, so we can derive the proper Lorentz transformations simply by considering transformations of paravectors $\R\oplus\R^4 \subset \Cl_{3,0}$.

  • $\begingroup$ Wow, this is an incredible answer. I've only read the first few sections so far but already it's exactly what I was hoping for and much more. Now I understand the need to find an element of the Clifford algebra that maps vectors to their representation ($-I$ for the quaternions, and $e_1$ for the complex numbers, which are both really neat operations I never would have thought of), and that the rotation of the representation (w.r.t. representation of the rotation) is then determined by how that element commutes with rotors. I look forward to reading the rest of it! $\endgroup$ Sep 29, 2022 at 16:38
  • $\begingroup$ @AdamHerbst It took me a long time to write, so I'm happy you're happy! $\endgroup$ Sep 29, 2022 at 16:58

Usually, in this context "vectors" refer to elements of the base inner product space $V$. This canonically embeds into $\text{Cl}(V)$ as $\text{Cl}^1(V)\cong V$. This explains what you're seeing in dimension $d=3$, since there is an isomorphism (canonical up to sign) between $\text{Cl}^1(V)$ and $\text{Cl}^{d-1}(V)$.

The $d=2$ case, on the other hand, is more exceptional, and the two notions of rotations don't really coincide. While it is true that $\text{Cl}^{\text{even}}(\mathbb{R}^2)\cong\mathbb{C}$ as algebras, this is not an isomorphism of representations.

In the context of Clifford algebras, "rotations" usually refers to the conjugation actions by the pin group $\text{Pin}(V)\subset\text{Cl}(V)$. On $\text{Cl}^1(V)\cong V$ this action factors through $O(V)$ and gives the standard double covering (likewise for the subgroups $\text{Spin}$ and $\text{SO}$). The isomorphisms $\text{Cl}^1(V)\cong\text{Cl}^{d-1}(V)$ is equivariant w.r.t. this action, but the action of $\text{Pin}(2)$ on $\text{Cl}^{\text{even}}(\mathbb{R}^2)$ is trivial, and so not isomorphic to the standard action of $\text{Pin}(2)$ on $\mathbb{C}$.


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