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A quaternion $q \in \mathbb{H}$ can be write like

$$q = a_0 + a_1 i + a_2j + a_3 k; ~ a_\iota \in \mathbb{R}$$

, where:

$$i^2 = j^2 = k^2 = ijk = -1$$ $$ij = -ji = k$$ $$jk = -kj = i$$ $$ki= -ik = j$$

But, when we calculate

$$i^j = \left( e^{\frac{\pi}{2}i} \right)^j = e^{\frac{\pi}{2}ij} = e^{\frac{\pi}{2}k} = k$$

My question: is this right? If yes, why $i^j = ij = k$? Is there a geometric/algebraic meaning of this?

Gabriel Fanini
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    Why would $e^{\frac{\pi}{2}k}=k$? What $i^j$ is even supposed to mean? Moreover the exp rule you use does not even hold for quaternions. This is a good example of careless generalizations. – freakish Dec 25 '23 at 19:25
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    In your last displayed equation, you write $i^j$ without having defining it first. Tell us what $q_1^{q_2}$ means for general (nonzero?) quaternions $q_1$ and $q_2$? You also write $(e^{(\pi/2)i})^j = e^{(\pi/2)ij}$. Did you ever prove that $(e^{q})^{q'} = e^{qq'}$? It looks to me like you're symbol pushing without first checking whether the expressions you use have any actual meaning and desired properties first. – KCd Dec 25 '23 at 19:26
  • It is better to avoid words like "crazy" for a serious question. – Dietrich Burde Dec 25 '23 at 19:27
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    Without going to quaternions you can deduce equally crazy results by just manipulating symbols without knowing what you are doing. – Kurt G. Dec 25 '23 at 19:34
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    It is at least possible to define the exponential map in terms of a power series like $e^q = \sum\limits_{n=0}^\infty \frac{q^n}{n!}$. In a domain where this map is bijective, one could also define the logarithm and then define $p^q = e^{q\ln p}$. I haven't checked if these definitions make sense for $i^j$. – Tob Ernack Dec 25 '23 at 19:37
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    @freakish. Just as $e^{\frac\pi 2 i} = i$ for complex numbers. – md2perpe Dec 25 '23 at 19:44
  • There is no function algebra of quaternions, simply because order of arguments matters. In naive $i^j$ notation you have to find a convergent series like $$e^{j\cdot \log i}$$ if i,j commute as for complex numbers. The rest of function algebra remaining for quaternions, Clifford and Grassman algebras is functions of a single variable, while functions of several variables degenerate into multilinear algebra of the different symmetry types . – Roland F Dec 26 '23 at 11:08

2 Answers2

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There is no definition of ${q_1}^{q_2}$ for arbitrary quaternions $q_1, q_2$. Even on $\Bbb{C}$ you have problems of whether ${c_1}^{c_2}$ is well-defined and single-valued. For the quaternions there is an additional complication that they don't commute so a formula like $(a^b)^c=a^{bc}$ is unlikely.

There is a definition of the exponential map on imaginary quaternions $\Bbb{R}i\oplus\Bbb{R}j\oplus\Bbb{R}k$ which is consistent with the exponential map in Lie groups (from their Lie algebra).

If $x=ai+bj+ck$ then $x^2=-a^2-b^2-c^2$. So take $a,b,c$ with $a^2+b^2+c^2=1$. Then $x^2=-1$. Then $$\exp(tx)=\sum_{n=0}^\infty\frac{1}{n!}(tx)^n=\sum_{n\text{ even}} \frac{1}{n!}(-1)^{n/2}t^n\mathbf{1}+\sum_{n\text{ odd}} \frac{1}{n!}(-1)^{(n-1)/2}t^n x = \cos(t)\mathbf{1}+\sin(t)x$$

The Lie group generated by the exponential map is the sphere of the quaternions which is isomorphic to $SU(2)$.

Chad K
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According to this document, quaternion powers can be defined.

Using the notation $\mathbf q = s + \mathbf v$ where $s\in\mathbb R$ and $\mathbf v\in\operatorname{span}\{i,j,k\}$ we define $$ \exp(\mathbf q) = \exp(s) \left( \cos|\mathbf v| + \frac{\mathbf v}{|\mathbf v|}\sin|\mathbf v| \right) \\ \ln(\mathbf q) = \ln|\mathbf q| + \frac{\mathbf v}{|\mathbf v|} \arccos\frac{s}{|\mathbf q|} \\ $$ and powers by $$ \mathbf q^{\mathbf p} = \exp(\ln(\mathbf q) \mathbf p). $$

For $\mathbf q=i$ we get $$ \ln(\mathbf q) = \ln|\mathbf q| + \frac{\mathbf v}{|\mathbf v|} \arccos\frac{s}{|\mathbf q|} = \ln|i| + \frac{i}{|i|} \arccos\frac{0}{|i|} = i\frac{\pi}{2} $$ which is not surprising since we know from the complex numbers that $e^{i\frac{\pi}{2}} = i.$

Thus, with $\mathbf p=j,$ we get $$ \ln(\mathbf q)\mathbf p = i\frac{\pi}{2} j = \frac{\pi}{2} k $$ which then gives us $$ \mathbf q^{\mathbf p} = \exp(\ln(\mathbf q)\mathbf p) = \exp(\frac{\pi}{2}k) = \exp(0) \left( \cos|\frac{\pi}{2}k| + \frac{\frac{\pi}{2}k}{|\frac{\pi}{2}k|} \sin|\frac{\pi}{2}k| \right) = 1 \cdot (0 + k\cdot 1) = k. $$

So your calculation is correct, but I can unfortunately not give you any interpretation of this.

md2perpe
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