# Is a 8-dimensional quadratic form recognized by its Lie algebra, modulo equivalence and scalar multiplication?

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.
• 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).