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Given complex numbers, we can calculate e.g. $i^i$.

Given quaternions, how can we calculate something like $i^j$? Wolfram Mathematica choked on that and googling did not produce any useful results. My guess is that this could be something ill defined, similar to quaternion derivative or, perhaps, even worse.

Konstantin Konstantinov
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    You'll probably need a good look at what the exponential map does in quaternion-land. Similarly to matrices, start with its power series and look at what the extension does. Actually, pass through to one of the 4x4 representations and look at its matrix exponential. – obscurans Jun 17 '20 at 00:26
  • Related. See also other threads linked to that one. – Jyrki Lahtonen Jun 20 '20 at 20:20
  • @JyrkiLahtonen - the answer by md2perpe is simple, precise, and surprisingly exact in comparison to the discussion in the "related" answer. The comments by RobertLewis are also great as they show that $i^i$ is not well defined. It is then interesting that $i^j$ is well defined under some very reasonable assumptions. – Konstantin Konstantinov Jun 22 '20 at 00:31
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    Konstantin, md2perpe even refused to give a precise definition of a quaternion power. I agree that we can define $e^q$, but the same ambiguity is present. Admittedly there is the surprise that for this power the phase ambiguity disappears (that is a bonus). So I did not downvote their answer. – Jyrki Lahtonen Jun 22 '20 at 03:50
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    Anyway, the whole concept of quaternion power is fuzzy. md2perpe apparently defines $q_1^{q_2}$ as follows. Write $q_1=a+b\vec{u}$ for some unit vector $\vec{u}$. Then we can write $q_1=e^{\ell}$ where $\ell=\ln(\sqrt{a^2+b^2}+\vec{u}\phi$, where $\phi$ is the phase of the 2D point $(a,b)$. So far so good (apart from the multiple of $2\pi$ ambiguity). But, can we really define $q_1^{q_2}:=e^{\ell q_2}$? Why not $q_1^{q_2}:=e^{q_2\ell}$? That would also extend the complex power. What criteria are we using to call this definition the correct one? – Jyrki Lahtonen Jun 22 '20 at 03:57
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    Observe that, unlike in the complex case, we may have $e^{q_1+q_2}\neq e^{q_1}\cdot e^{q_2}$. What are the justifications for the particular choices made? – Jyrki Lahtonen Jun 22 '20 at 04:01
  • Sorry about the missing parens in $\ln(\sqrt{a^2+b^2})$. Of course, $a,b$ are real numbers. – Jyrki Lahtonen Jun 22 '20 at 04:13

3 Answers3

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$$ i^j = (e^{i\pi/2})^j = e^{ij\pi/2} = e^{k\pi/2} = k $$

md2perpe
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I know I'm late to this, but there is one thing I'm not seeing in the other responses. For some reason $$ i=e^{\frac{\pi}{2}i+2\pi ni }, n\in\mathbb{Z} $$ but since we're dealing with quaternions, it shouldn't be unreasonable to say that $$ i=e^{\frac{\pi}{2} i + 2 \pi x i + 2 \pi y j + 2 \pi z k }, \text{ for } x,y,z\in\mathbb{Z}\\ \text{since } e^{2 \pi x i} = e^{2 \pi y j} = e^{2 \pi z k} = 1 $$ In this case $$ i^{j} = (e^{\frac{\pi}{2} i + 2 \pi x i + 2 \pi y j + 2 \pi z k })^{j}\\ = e^{\frac{\pi}{2} ij + 2 \pi x ij + 2 \pi y jj + 2 \pi z kj }\\ = e^{\frac{\pi}{2} k + 2 \pi x k - 2 \pi y - 2 \pi z i }\\ = e^{\frac{\pi}{2} k} e^{ 2 \pi x k} e^{ - 2 \pi y} e^{ - 2 \pi z i }\\ = k \cdot 1 \cdot e^{ - 2 \pi y} \cdot 1\\ = e^{ - 2 \pi y} k$$ which is not well-defined since the choice for $y$ (and less importantly $x$ and $z$) is arbitrary.

Note that even with all of this, it shouldn't work out so nicely in general due to quaternions not being commutative as $(q_{1}q_{2})^{n}\neq q_{1}^{n} q_{2}^{n}$ unless $q_{1}q_{2} = q_{2}q_{1}$, but a lot of the cases here are ultimately multiplying things by 1 (or some other real number), which might commute well, but I'm unsure.

Similarly, the order of addition of exponents isn't changed so that it can be used more as shorthand since $e^{q_{1}+q_{2}}$ isn't exactly well-defined as $e^{q_{1}}e^{q_{2}}\neq e^{q_{2}}e^{q_{1}}$ in general.

A little bit of a tangent, but another property that may not necessarily hold, especially if the previous line did without additive commutativity in the exponent (except where $e^{q}=1$) is that $e^{q_{1}}=e^{q_{2}}\Rightarrow q_{1} = q_{2}$. Assuming it did hold, then using $i$ as above and a similar $j$ and $k$, you could show that $ij=k \Rightarrow k = ni + mj \text{ for some } n,m\in\mathbb{Q}$, which is clearly false.

Jordan Miller
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How do we meaningfully calculate $i^i$? One way which comes to mind is to write

$i = e^{\pi i \ 2}, \tag 1$

and thus

$i^i = (e^{\pi i / 2})^i; \tag 2$

if we now assume the rule

$(a^b)^c = a^{bc}, \; a, b, c \in \Bbb C, \tag 3$

is valid, then (2) becomes

$i^i = (e^{\pi i / 2})^i = e^{\pi i^2 / 2} = e^{-\pi / 2}. \tag 4$

So far, so good; however, the equation (1) does not uniquely define $i$ as an exponential; indeed, we have

$i = e^{(2n + 1/ 2)\pi i}, \; n \in \Bbb Z, \tag 5$

whence

$i^i = (e^{(2n + 1/2)\pi i})^i = e^{(2n + 1/2)\pi i^2} = e^{-(2n + 1/2)\pi}, n \in \Bbb Z. \tag 6$

If we accept these results, we may move on to attempting to calculate $i^j$; from (5), assuming (3) holds for quaternions,

$i^j = (e^{(2n + 1/2)\pi i})^j = e^{(2n + 1/2)\pi ij} = e^{(2n + 1/2)\pi k}$ $= e^{2n\pi k} e^{\pi k / 2} = e^{\pi k / 2} = k, \; n \in \Bbb Z. \tag 7$

It is perhaps worth noting that

$i^j = k = ij, \tag 8$

and that via cyclic permutation $i \to j \to k$ we find

$j^k = i = jk, \tag 9$

and

$k^i = j = ki. \tag{10}$

Note Added in Edit, Friday 19 June 2020 12:51 PM PST: A few words in response to the comment of Konstantin Konstantinov on this answer. The defining relations of the quaternion algebra,

$i^2 = j^2 = k^2 = -1, \tag{11}$

$ij = k, jk = i, ki = j, \tag{12}$

found in this wikipedia entry, lead naturally to the formulas

$e^{2n\pi i} = e^{2n \pi j} = e^{2n\pi k} = 1, n \in \Bbb Z, \tag{13}$

from which we see that for

$t \in [0, 2\pi) \tag{14}$

$e^{(2n\pi + t)i} = e^{2n\pi i}e^{t\pi i} = e^{t\pi i}, n \in \Bbb Z, \tag{15}$

and corresponding formulas with $j$ and $k$ replacing $i$. From this we see that there are a countably infinite number of distinct solutions to (15), viz.

$2n\pi + t, \; n \in \Bbb Z, \tag{16}$

with corresponding solution sets for the $j$ and $k$ variants of (15). We further observe that

$e^{i \theta} = \cos \theta + i \sin \theta, \tag{17}$

again with the corresponding relations for $j$ and $k$; of course (17) gives rise to (5) in the usual manner:

$e^{(2n + 1/ 2)\pi i} = \cos ((2n + 1/2)\pi) + i \sin ((2n + 1/2)\pi)$ $= i\sin((2n + 1/2)\pi) = i\sin((1/2) \pi) = i, \; n \in \Bbb Z, \tag{18}$

and again, the corresponding statements hold for $j$ and $k$. End of Note.

Robert Lewis
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    It is then interesting that it is branch cut discontinuity of $log$ on a complex plane that results in that ambiguity of $i^i$, whereas for quaternions the value of $i^j$ is well defined and does not have such a discontinuity. – Konstantin Konstantinov Jun 18 '20 at 21:44
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    I know nothing about quaternions. Is the rule $(a^b)^c=a^{bc}$ rather than $(a^b)^c=a^{cb}$ based on anything more principled than the notational happenstance that exponents are placed on the right of the base rather than on the left or directly above? – bof Jun 21 '20 at 06:13
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    In addition to the criticism from @bof I want to point out that in general the rule $e^{q_1+q_2}=e^{q_1}\cdot e^{q_2}$ is only valid when $q_1q_2=q_2q_1$. In the cases where you apply it that holds. But the whole concept is just very fuzzy. – Jyrki Lahtonen Jun 22 '20 at 04:08