# The “argument” of a quaternion

My question is pretty simple. I've been trying to read a pretty introductory text on Clifford algebras, and I encountered how they define the "argument" of a quaternion as an ordered quadruple representing the components of the noramlized quaternion. Now, in $\mathbb{C}$, we have this map $\arg : \mathbb{C} \setminus \{ 0 \} \to [0, 2 \pi)$ such that \begin{align*} \arg (z_{1} z_{2}) & = \arg(z_{1}) + \arg(z_{2}) , \\ \arg(z_{1}) = \arg(z_{2}) & \iff \frac{z_{1}}{z_{2}} \in \mathbb{R}_{ > 0 } . \end{align*} So my question is, is there a map $$\arg' : \mathbb{H} \to [0, 2 \pi)^{2}$$ such that, again, $\arg'$ is additive under multiplication, $$\arg'(z_{1}) = \arg'(z_{2}) \iff \frac{z_{1}}{z_{2}} \in \mathbb{R}_{ > 0 }$$ and $$\arg' |_{\mathbb{C}} = (\arg, 0) ?$$

• You want the purely imaginary part of the logarithm of a quaternion. This has three components rather than two. – Qiaochu Yuan Jul 1 '15 at 5:29
• The "space of directions" in $\Bbb C$ is the ray projectivization of $\Bbb C - \{0\}$, which we can identify with the unit circle $\Bbb S^1$, which in particular is a Lie group. Now, we can identify $T_1 \Bbb S^1 \cong i \Bbb R$, and the exponential map is $T_1 \Bbb S^1 \cong i \Bbb R \to \Bbb S^1$. Then, we can regard a choice of argument as a right inverse for this map. We can proceed analogously for the quaternions and the Lie group $\Bbb S^3$, though the resulting "argument" function will not obey the same additivity, because this latter group is nonabelian. – Travis Jul 1 '15 at 5:40
• For quaternions $\arg(q_1q_2)$ may be different from $\arg(q_2q_1)$, so this cannot be done. The group of values of the quaternionic argument is topologically a 3-sphere as opposed to a direct product of copies of $\Bbb{R}/2\pi\Bbb{Z}$ that you seem to have hoped. – Jyrki Lahtonen Jul 1 '15 at 5:43
• @AJY Do you know about the quaternionic exponential map? That's really all one needs to write out the analogy. But again, as my previous comment mentions, and as Jyrki's makes usefully explicit, the short answer to your question is no, as the first equation cannot hold; there is a sort of replacement for it, though, see en.wikipedia.org/wiki/… . – Travis Jul 1 '15 at 5:50
• Well, Travis was the first to bring up $S^3$. His comment came while I was typing/editing. But anyway, as Travis also pointed out, the group operation is not commutative. You can try and "linearize" it by moving it to the Lie algebra of $S^3\cong SU(2)$ by taking the matrix logarithm (If I guessed correctly Qiaochu was suggesting that), but then the group operation becomes a mess. – Jyrki Lahtonen Jul 1 '15 at 6:04

First, note that the argument function $$\def\wtarg{\widetilde{\text{arg}}} \def\wtexp{\widetilde{\text{exp}}} \arg: \Bbb C - \{0\} \to [0, 2\pi)$$ as written does not actually satisfying the additive identity $$\arg (z w) = \arg z + \arg w:$$ if we take $z = w = -1$, then the l.h.s. is $0$ but the r.h.s. is $2 \pi$. The best we can ask for is that this identity hold true modulo $2 \pi$, that is, that $$\arg (z w) = \arg z + \arg w \bmod 2 \pi,$$ and this holds (more or less by definition) for any branch of the argument function.

The argument as a normalization map

Complex case

If, as OP specified in the comments, we compose this map with the canonical projection $\Bbb R \to \Bbb R / 2 \pi \Bbb Z$ and invoke the identification $\Bbb R / 2 \pi \Bbb Z \leftrightarrow \Bbb S^1$ given by $x + 2 \pi \Bbb Z \leftrightarrow e^{ix}$, we get the map $$\wtarg: \Bbb C - \{0\} \to \Bbb S^1$$ defined by $$\wtarg(z) := \frac{z}{|z|}.$$ When we view the argument this way, the additivity criterion becomes the familiar group homomorphism condition, $$\wtarg(z w) = \wtarg z \,\wtarg w.$$

Quaternionic case

Working instead with $\Bbb H$ leads to an analogous "argument" function $\wtarg': \Bbb H - \{0\} \to \Bbb S^3$ defined formally by the same rule as above. (Note that the codomain here is not $\Bbb S^1 \times \Bbb S^1$ but rather $\Bbb S^3$, which in particular has dimension $3$.) Applying this map gives exactly the "ordered quadruple representing the components of the normalized quaternion" in the text OP mentions. This map satisfies the desired conditions:

• $\wtarg'(q r) = \wtarg' q \,\wtarg' r$ (the group homomorphism condition)
• $\wtarg' q = \wtarg' r \Leftrightarrow \exists \lambda \in \Bbb R^+ : q = \lambda r$ (by definition, this is a description of the fibers of the homomorphism).

We can view $\Bbb C$ as a subset of $\Bbb H$ in the usual way, and since $\wtarg$ and $\wtarg'$ are given by the same rule, we have the desired identity

• $\wtarg'\vert_{\Bbb C - \{0\}} = \wtarg$.

In both cases, the argument map so viewed is perhaps not so interesting: It's simply the map that assigns to a nonzero vector the unit vector pointing in the same direction, and so it makes sense for any normed vector space.

The argument as a branch cut

Complex case

The situation where we do not pass to the appropriate quotient, and so work with what (at least in the complex setting) are called branch cuts of the argument function, e.g., $\arg$, is a good deal subtler. One way to view such functions that extends cleanly to the quaternionic setting is as follows: The restriction of the exponential map to the imaginary axis is a map $$\exp\vert_{i \Bbb R} : i \Bbb R \to \Bbb S^1$$ (via the identification $T_I \Bbb S^1 \leftrightarrow i \Bbb R$, this is precisely the exponential map of $\Bbb S^1$ regarded as a Lie group), and more or less by definition we can produce a branch cut of the argument function by taking (by convention, the imaginary part of) a right inverse for this function; this is essentially the restriction of the complex logarithm to the unit circle. If we restrict the map to $i \cdot [0, 2\pi)$, we get a bijection $\exp\vert_{i \cdot [0, 2 \pi)}: i \cdot [0, 2\pi) \to \Bbb S^1$, and by construction we have $$\arg z = \frac{1}{i} (\exp\vert_{i \cdot [0, 2\pi)})^{-1} (\wtarg z).$$

Quaternionic case

Similarly, the quaternionic exponential map $\wtexp: \Bbb H \to \Bbb H - \{0\}$ restricts to a map $\wtexp\vert_{\Bbb I} : \Bbb I \to \Bbb S^3$ (again, the exponential map of $\Bbb S^3$ regarded as a Lie group), and we can define a (branch $\arg'$ of the) (quaternionic) argument function to be a right inverse for this map. Here, $\Bbb I$ is the subspace $\langle i, j, k \rangle$ of the imaginary elements of $\Bbb H$.

Such a map $\arg'$, however, does not behave as well as its complex analogue $\arg$. We cannot expect an additivity property $$\arg' (q r) = \arg' q + \arg' r \bmod X$$ to hold for any lattice $X \subset \Bbb I$, in short because $\Bbb H$ is not commutative.

We can, however, produce an analogous condition that does hold for $q, r$ such that $\wtarg' q, \wtarg' r$ are close enough to $1$ (at least, provided that we choose our right inverse so that its domain includes an open neighborhood of $0$): Translating the Baker-Campell-Hausdorff Formula (and passing through the map $\wtarg'$) gives the identity $$\arg' (q r) = \arg' q + \arg' r + \frac{1}{2} [\arg' q, \arg' r] + \cdots,$$ where $$[a, b] := a b - b a$$ is the usual commutator (in this case, the quadratic correction term $\frac{1}{2} [\arg' q, \arg' r]$ in the expansion could be written as $\arg' q \times \arg' r$, where $\times$ is the usual cross product on $\Bbb I \cong \Bbb R^3$), and $\cdots$ denotes a particular remainder that depends on $\arg' q, \arg' r$ to third and higher order. (All of these higher-order terms are also brackets, so we can see the complex additivity identity as a special case of this general formula for which all of the bracket terms disappear, because $\Bbb S^1$ is commutative.)

On the other hand, a suitable map $\arg'$ so constructed does satisfy the other two conditions, namely that (a) $\arg' q = \arg' r$ iff $q$ and $r$ are positive multiples of one another, and (b) the restriction of $\arg'$ to $\Bbb C - \{0\}$ defines an argument function for $\Bbb C$ as defined above.