Let's consider the following number:
$n = n_0 + n_1\textbf{i} + n_2\textbf{j} + n_3\textbf{k}$
Where $\{1,\textbf{i},\textbf{j},\textbf{k}\}\subset\mathbb{H}$ and in turn each component can be written as following:
$n_i = n_{i0} + n_{i1}\textbf{e}_1 + n_{i2}\textbf{e}_2 + n_{i3}\textbf{e}_2$
Where $\{n_{i0},n_{i1},n_{i2},n_{i3}\}\subset\mathbb{R}$ and $\textbf{e}_1$,$\textbf{e}_2$,$\textbf{e}_3$ form the following algebra:
${\textbf{e}_1}^2={\textbf{e}_2}^2={\textbf{e}_3}^2={\textbf{e}_1}{\textbf{e}_2}{\textbf{e}_3}=+1$
${\textbf{e}_1}{\textbf{e}_2}=-{\textbf{e}_2}{\textbf{e}_1}={\textbf{e}_3}$
${\textbf{e}_2}{\textbf{e}_3}=-{\textbf{e}_3}{\textbf{e}_2}={\textbf{e}_1}$
${\textbf{e}_3}{\textbf{e}_1}=-{\textbf{e}_1}{\textbf{e}_3}={\textbf{e}_2}$
Basically, n is a quaternion whose each component is hyperbolic quaternion. Alternatively it can be rewritten as hyperbolic quaternion whose each component is a regular quaternion. Do any literature deal with this construction, and if so, does this number have any special name? Quadrotessarine? Tetraquaternion? Split-sedenion?
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1$\begingroup$ My brain might not function correctly because of the heat wave, but doesn't the tensor product of any two of the quaternions and the split-quaternions simply give the real $4×4$ matrices? (But if you really want a term, I think “quaterquaternions” is somewhat standard.) $\endgroup$– Gro-TsenCommented Jun 25, 2019 at 16:34
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1$\begingroup$ Ah no, it can also give $2×2$ matrices over the quaternions, as Bugs Bunny points out. It remains to see exactly what goes by the name “quaterquaternions”. $\endgroup$– Gro-TsenCommented Jun 25, 2019 at 16:44
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2 Answers
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Ther are not sedenions because sedenions are not associative: https://en.wikipedia.org/wiki/Sedenion
I doubt they have a name but this algebra is well understood. The hyperbolic quaternions is just $M_2({\mathbb R})$. You take a tensor product of them with usual quaternions ${\mathbb H}$. As a result you get $M_2({\mathbb H})$ that deserves no special name.
answered Jun 25, 2019 at 12:19
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3$\begingroup$ The sedenions are non-associative. The algebra described in the question, if I understand it correctly, is a tensor product of two associative algebras, so it is associative. $\endgroup$– Gro-TsenCommented Jun 25, 2019 at 16:26
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$\begingroup$ Yes, you are correct. I change my answer. $\endgroup$ Commented Jun 25, 2019 at 16:30
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$\begingroup$ "As a result you get $M_2(\mathbb{H})$ that deserves no special name."<br\>But such algebra would allow to simultaneously simplify both space rotation and Lorentz transformation in Minkowski's space-time. It deserves it's own name. $\endgroup$ Commented Jun 25, 2019 at 19:14
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3$\begingroup$ @Gro-Tsen hyperbolic quaternions ARE NOT associative, so the algebra in question is also non-associative. en.wikipedia.org/wiki/Hyperbolic_quaternion $\endgroup$– AnixxCommented Sep 24, 2021 at 17:08
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Isn't this an example of a Clifford algebra attached to a quadratic form?
It seems so to me, though maybe the intention has some quirks... but, if so, it is certainly associative, and its structure as a semi-simple algebra is well understood, at least over real, complex, or p-adic scalars.
answered Sep 25, 2021 at 0:43