I know that the only quaternion algebras over $\mathbb{R}$ are the quaternions and the split-quaternions. What is the characterization of the quaternion algebras over a particular field? Which of these be extended to octonion algebras?
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7$\begingroup$ For local fields see this post. For more classifications see this paper. $\endgroup$– Dietrich BurdeOct 27, 2022 at 18:17
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4$\begingroup$ This is the 80,000th abstract algebra question here (not including deleted questions) :) $\endgroup$– ShaunOct 27, 2022 at 20:05
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2 Answers
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In general the isomorphism classes of quaternion algebras over a field $K$ are in natural bijection with (among other things):
- the isometry classes of $2$-fold Pfister forms over $K$
- the isometry classes of $3$-dimensional quadratic forms over $K$ of determinant $1$
- the isomorphism classes of projective varieties $V$ over $K$ such that $V$ becomes isomorphic to $\mathbb{P}^2$ over an algebraic closure of $K$
- the cohomology classes in $H^1(K,PGL_2)$.
For a general field there is no neat classification, all those objects can be fairly complicated if the field is complex enough.
answered Oct 27, 2022 at 18:57
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$\begingroup$ Also the $2$-torsion in the Brauer group, right? $\endgroup$ Oct 27, 2022 at 21:47
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6$\begingroup$ @QiaochuYuan not quite. If the characteristic is not 2, then the 2-torsion in the Brauer group is generated by quaternion algebras, but not every element is represented by a quaternion algebra in general. $\endgroup$ Oct 27, 2022 at 23:42
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3$\begingroup$ As the question also asks about octonion algebras, they also have a cohomological description: $H^1(K,G_2)$, where $G_2$ is a certain exceptional algebraic group. $\endgroup$ Oct 28, 2022 at 2:42
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There's a cohomological description of certain algebras that is very general and covers in particular quaternion algebras and octonion algebras. I don't know if this is written down anywhere, but the proof is a standard Galois descent argument.
Let $k$ be a field and let $k^s$ be a fixed algebraic closure. Let $A$ be a finite-dimensional algebra over $k$ that is not required to be unital, commutative or associative. Then the group $G$ of $k$-algebra automorphism of $A$ is an algebraic group over $k$. An algebra $B$ (again without any assumptions) over $k$ is called a form of $A$ if $k^s \otimes_k A \cong k^s \otimes_k B$ as algebras over $k^s$. We have the following result:
Forms of $A$ are classified by the Galois cohomology group: $H^1(\operatorname{Gal}(k^s/k),G(k^s))$.
To recover something about quaternion and octonion algebras from this, one can show that forms of any fixed quaternion or octonion algebra are exactly all algebras of that type. Also for quaternion algebras the automorphism group is $PGL_2$, for octonions it's called $G_2$ and is an exceptional algebraic group.
But this result is more general than just these two cases, so one might try to inverstigate, e.g. what happens when one applies it to an algebra over a field $k$ with the same structure constants as the sedenions over $\Bbb R$ (this works as the structure constants are always $0$ or $\pm 1$).
For octonion algebras specifically, they also correspond to Pfister 3-forms, just like quaternion algebras correspond to Pfister 2-forms.
answered Nov 27, 2022 at 23:19
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$\begingroup$ There is no 16-dimensional composition algebra, so there don't exist any sedenion algebras. $\endgroup$ Nov 28, 2022 at 19:48
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1$\begingroup$ who says that a sedenion algebra has to be a composition algebra? anyway, if it makes you happy, I can remove any reference to sedenions in my answer, this does not take away from the general point. $\endgroup$ Nov 28, 2022 at 19:55
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$\begingroup$ If you look at the definition of quaternion and octonion algebras, you would see that they have to be composition algebras. $\endgroup$ Nov 28, 2022 at 19:56
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1$\begingroup$ @mathlander I have reworked the answer. Maybe now it's more accurate, because you're correct, there's no standard notion of a sedenion algebra (though in view of the theorem, one might try to define them as forms of the "standard" sedenion algebra over a field, then one gets a cohomological classification of those) $\endgroup$ Nov 28, 2022 at 20:00
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1$\begingroup$ @mathlander I know that the definition of quaternion and octonion algebras is in terms of composition algebras of a certain dimension. In the context of my answer, an alternative definition that leads to a cohomological classification is just as forms of a "standard" quaternion/octonion algebra and this definition carries over to sedenions. Perhaps I should have been more clear $\endgroup$ Nov 28, 2022 at 20:25