Properties of finite dimensional, real division algebras that only yield $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$ It is a classical result by Kervaire and Milnor that every finite-dimensional, real division algebra has dimension 1,2,4 or 8, with the most prominent examples being $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$.
However, as far as I am aware, a full classification of all such finite-dimensional division algebras is still unknown, although there are plenty of classification results for certain additional requirements. For brevity, let algebra always mean finite-dimensional, real division algebra.
Aiming at further requiring additional properties to reduce to the examples of $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$, one has the following results of

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*Hopf, asserting that every commutative, unitary such algebra is either (isomorphic to) $\mathbb{R}$ or $\mathbb{C}$.

*Zorn, stating that every alternative such algbera is $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$ (being a special case of Frobenius, who classified the associative algebras, where $\mathbb{O}$ drops out)

*Albert, stating that the only absolute valued algebras are $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$, this is mentioned in [1] and can be found in [2]

*In [3], Koecher and Remmert also link the existence of (n-1) dimensional vector product algebras to the existence of unitary, compositional algebras of dimension n, thus again somehow establishing that these 4 stand out from the rest.

It is, however, unsatisfying that the above actually don't use the result of Milnor and Kervaire, and thus feel like an understatement of how special these algebras are - you could say that it is not the fact of them being commutative / alternative / absolute valued that forces them to be 'special', but Milnor and Kervaire actually showed us that it is already the mere fact of being division algebras that force the algebras to be of dimensions 1,2,4,8, and only to further reduce one needs these additional properties.
So I am asking whether there are mild / weak additional assumptions to require from a finite-dimensional, real division algebra, that already restricts the possible examples to $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$

Some small remarks:

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*For the two-dimensional case, Hopf actually first showed that any commutative algebra has dimension 1 or 2 and just with the additional assumption of requiring the algebra to be unitary, one concludes that the only examples left are $\mathbb{R}$ and $\mathbb{C}$. In fact, all 2-dimensional algebras are isotopes of $\mathbb{C}$, as is described in [1]

*For four dimensions, this however does not suffice anymore. In [1], Darpö revises the classification of the four-dimensional, quadratic algebras (in particular, these are all unitarian), yielding many more than just $\mathbb{H}$. Thus, further assumptions as described above are neccessary to uniquely characterise $\mathbb{H}$.

*I am not specifically asking for further classifications, e.g. of all flexible algebras, since these will yield more examples than just $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$. Rather, i really want an assumption that is strong enough to rule out all other algebras, but definitely weaker than 'alternative' or 'absolute valued'. Of course, this is just a matter of taste, but since we are more or less taught the moral that even when dropping the alternativity assumption, one does not get any more 'nice' algebras, e.g. the sedenions fail to be a divisions algebra, which is covered in the 1-2-4-8 theorem by Kervaire and Milnor.


[1] Erik Darpö. "Some modern developments in the theory of real division algebras". In: (2010). DOI: 10.3176/proc.2010.1.09
[2] Albert, A. A. "Absolute valued real algebras". Ann. Math. (2), 1947, 48, 495–501
[3] Ebbinghaus et al. "Numbers", Springer 1991
 A: I spent a day looking into this myself and I didn't find anything.
Given a real unital composition algebra with a positive-definite form, Hurwitz's theorem will tell you that it must be one of those four examples, and it's evident that they are all normed division algebras.
Having encountered many statements like "there are exactly four normed division algebras," I was expecting some sort of converse: start with a normed division algebra -> it must be one one of the four, and in particular must be finite/real/unital/absolute-valued, etc.
However, it seems true only when "normed division algebra" means "real unital absolute-valued division algebra." And in that case, you can apply an updated version [0] of the Albert result to show that it's one of the four. But then the division algebra structure is never used! So it's not fully clear to me what we gain from division or a non-absolute-value norm. To go backwards, additional strong assumptions appear necessary, which, as you say, is unsatisfying.
[0] Kazimierz Urbanik and Fred B. Wright, Absolute-valued algebras, Proc. Amer. Math. Soc. 11 (1960), 861-866, doi:10.1090/S0002-9939-1960-0120264-6.
