Question. Let $K$ be a field of characteristic zero (large characteristic should be fine too). Let $q,q'$ be two non-degenerate quadratic forms on $K^n$ with $n=8$. Suppose that the Lie algebras $\mathfrak{so}(q,K)$ and $\mathfrak{so}(q',K)$ are isomorphic (these are simple of type $D_4$, 28-dimensional). Does it follow that $q$ is equivalent to some nonzero scalar multiple of $q'$?

A restatement of the question is whether $\mathrm{SO}(q)$ and $\mathrm{SO}(q')$ being isogeneous over $K$ implies the same conclusion.

This is asking the converse of an obvious fact (since $\mathfrak{so}(q,K)$ and $\mathfrak{so}(tq,K)$ are equal for every nonzero scalar $t$. By an elementary argument (see this MO answer), the converse holds for $n\ge 3$ with the possible exception $n=8$ (while it fails for $n=2$ as soon as $K$ has a non-square). The difficulty comes from the existence of triality, namely automorphisms $\mathfrak{so}(q,K)$ not induced by $\mathrm{O}(q,K)$.

The argument can be used to give a positive answer if "the" absolute Galois group of $K$ does not admit as quotient a group of order 3 or 6. This applies to the reals, in which case we can also argue using the signature of the Killing form. An ad-hoc argument can also probably be done for $p$-adic fields.

(In the comments to the linked answer, some hints were given towards a positive answer for $n=8$. I don't know if they're enough to conclude but obviously if so they should be promoted to a full answer.)


Yes: Proposition C.3.14 in Brian Conrad's article Reductive group schemes is that $SO(q)$ determines $q$ up to similarity for all $q$ of dimension $> 2$. (This was pointed out by @user74230 in a comment somewhere.)

This could be viewed as a special case of a more general phenomenon where there is a simple algebraic group $G$ acting on a vector space $V$ and a $G$-invariant homogeneous polynomial $f$ on $V$ so that the twisted forms of $G$ are in bijection with twisted forms of $f$ up to similarity, see Bermudez and Ruozzi Classifying forms of simple groups via their invariant polynomials, where the fact about quadratic forms is stated as Proposition 7.2.

  • $\begingroup$ Great, thank you. In addition from the Bermudez-Ruozzi statement, the algebraic group version is true for all $n$ (including $n=2$) and for all fields (including characteristic 2). $\endgroup$
    – YCor
    May 3 '20 at 14:44

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