Quaternion exponential

Question

Given an imaginary quaternion $ \mathbf{v}=\alpha \mathbf{i}+ \beta \mathbf{j}+\gamma \mathbf{k} $ its exponential is: $ e^\mathbf{v}=\cos \theta +\mathbf{v}\dfrac {\sin \theta}{\theta} $ where $\theta=|\mathbf{v}|=\sqrt{\alpha^2+\beta^2+\gamma^2}$. It's easy to show that a sufficient condition to obtain: $$ e^\mathbf{v}=e^{\alpha \mathbf{i}+ \beta \mathbf{j}+\gamma \mathbf{k}} = e^{\alpha \mathbf{i}} e^{\beta \mathbf{j}} e^{\gamma \mathbf{k}} $$ is that $ \alpha=h\pi$, $\beta=k\pi $ and $\gamma=j \pi$ , where $h,k,j$ are the fist three numbers of a Pythagorean quadruple $h^2+k^2+j^2=l^2 \quad h,k,j,l \in \mathbb{N}/0$. But I can not verify whether this condition is also necessary. In other words: there is an imaginary quaternion such that $$e^\mathbf{v}=e^{\alpha \mathbf{i}+ \beta \mathbf{j}+\gamma \mathbf{k}} = e^{\alpha \mathbf{i}} e^{\beta \mathbf{j}} e^{\gamma \mathbf{k}}$$ and $\alpha/\pi, \beta/\pi,\gamma/\pi$ are not the fist three numbers of a Pythagorean quadruple?

Answer 1

Thanks to the result of @Achille Hui I can sketch an answer. The equation $$ e^\mathbf{v}=e^{\alpha \mathbf{i}+ \beta \mathbf{j}+\gamma \mathbf{k}} = e^{\alpha \mathbf{i}} e^{\beta \mathbf{j}} e^{\gamma \mathbf{k}} \qquad (1) $$ is equivalent to the system

(2)\begin{cases} \theta\left( \sin \alpha \cos \beta \cos \gamma + \cos \alpha \sin \beta \sin\gamma\right)=\alpha \sin \theta\\ \theta\left(\cos \alpha \sin \beta \cos \gamma - \sin \alpha \cos \beta \sin\gamma\right)= \beta\sin \theta\\ \theta\left(\cos \alpha \cos \beta \sin \gamma + \sin \alpha \sin \beta \cos\gamma \right)=\gamma \sin \theta\\ \cos \alpha \cos \beta \cos \gamma - \sin \alpha \sin \beta \sin\gamma= \cos \theta\\ \theta^2=\alpha^2+\beta^2+\gamma^2 \end{cases}

that has a trivial solution for $$ (3)\quad \alpha=h\pi \;\land \; \beta=k\pi \;\land \; \gamma=j \pi \;\land \; \theta=l \pi \;\land \; \;h,k,,j,l \in \mathbb{N}^* \land \; h^2+k^2+j^2=l^2 $$ Substituting in the system we easily show that, if two of the exponent coefficents $\alpha,\beta,\gamma$ are integer multiples of $\pi$, then also the last one must be an integer multple of $\pi$. Because Hui finds that ``there is a solution of the equation (1) near $\alpha=\gamma \thickapprox 1,573919413377416 \pi$ and $\beta \thickapprox 2,322863603545524 \pi$.'' condition (3) is not, in general, necessary. So we can only say that the condition (3) is sufficient to verify (1) but it's necessary only if at least two of the exponet coefficents are integer multiples of $\pi$. It remains an open problem how many solutions there are when conditions (3) is not verified.

We can add some other related claims.

It's not difficult to show that $e^{\alpha\mathbf{i}}e^{\beta\mathbf{j}}=e^{\beta\mathbf{j}}e^{\alpha\mathbf{i}} \iff \alpha=h \pi \lor \beta=k \pi$ and the same is for the other couples . So we have: $$ e^{\alpha\mathbf{i}}e^{\beta\mathbf{i}}e^{\gamma\mathbf{k}}=e^{\beta\mathbf{i}}e^{\alpha\mathbf{i}}e^{\gamma\mathbf{k}}=e^{\beta\mathbf{i}}e^{\gamma\mathbf{k}}e^{\alpha\mathbf{i}} $$ if and only if at least two exponent coefficients are integer multiples of $\pi$. And in such a case the six permutations of the triple product are all equal to $\pm1$. So we have: $$ e^{\alpha \mathbf{i}+ \beta \mathbf{j}+\gamma \mathbf{k}} = e^{\alpha\mathbf{i}}e^{\beta\mathbf{i}}e^{\gamma\mathbf{k}}=e^{\beta\mathbf{i}}e^{\alpha\mathbf{i}}e^{\gamma\mathbf{k}}=e^{\beta\mathbf{i}}e^{\gamma\mathbf{k}}e^{\alpha\mathbf{i}}= \cdots $$ if and only if the condition (3) is verified.

Moreover we have: $$ e^\mathbf{v}=e^{\alpha \mathbf{i}+ \beta \mathbf{j}+\gamma \mathbf{k}} = e^{(\alpha \mathbf{i}+\beta \mathbf{j})} e^{\gamma \mathbf{k}}=e^{\gamma \mathbf{k}}e^{(\alpha \mathbf{i}+\beta \mathbf{j})} =e^{\beta \mathbf{j}}e^{(\alpha \mathbf{i}+\gamma \mathbf{k})} $$ if and only if: $$ \gamma=j\pi \;\land\; \alpha^2+\beta^2=n^2\pi^2 \;\land\; \alpha=h\pi \;\land\; \beta^2+\gamma^2=m^2\pi^2 \;\land\; \beta=k\pi \;\land\; \alpha^2+\gamma^2=p^2\pi^2 $$ So $\{h,k,j,l\}$ must be e perfect Pythagorean quadruple, such that $l^2=h^2+m^2=n^2+j^2=k^2+p^2=h^2+k^2+j^2$ with $h^2+k^2=n^2$, $h^2+j^2=p^2$ e $k^2+j^2=m^2$.

As far as I know the existence of such quadruples is an open question.