# When do two non-degenerate quadratic forms give rise to isomorphic Lie algebras?

Let $V$ be a vector space over some number field $k$. (I'm fine with $\mathbb{Q}$.)

Let $\phi \colon V \to k$ be a non-degenerate quadratic form. Associated with $\phi$ is the orthogonal group $\mathrm{O}(V,\phi) \subset \mathrm{GL}(V)$ of linear automorphisms preserving $\phi$. The connected component of the identity is $\mathrm{SO}(V,\phi)$.

Suppose $\psi \colon V \to k$ is another non-degenerate quadratic form. If $\phi \sim \psi$ (equivalent as quadratic forms), then definitely $\mathrm{SO}(V,\phi) \cong \mathrm{SO}(V,\psi)$. The same is true if $\phi \sim \lambda \cdot \psi$, for some scalar $\lambda \in k^{*}$.

Question: Is there a general statement about when $\phi$ and $\psi$ have special orthogonal groups in the same isogeny class?

Remarks:

1. I am asking about isogeny classes, not isomorphism classes of groups. I don't know if this makes the question harder or easier. (With isogeny, I mean a homomorphism between algebraic groups of the same dimension (trivial in this case) such that the kernel is finite.)  I changed the question, so that it is no longer about isogenous groups, but about isogeny classes. In particular, I would like to know what the conditions on $\phi$ and $\psi$ are, so that there exists a group $G$, with isogenies $G \to \mathrm{SO}(V,\phi)$ and $G \to \mathrm{SO}(V,\psi)$. [/Edit]
2. I would prefer a statement similar to the classification of quadratic forms. So in terms of data similar to local Hasse invariants or such. (But maybe this is optimistic, because, for example, my above remark shows that the discriminant can be changed to anything, by twisting with a scalar $\lambda$.)

[Edit2] As YCor points out in the comments below, the current version of the question is equivalent to

When do $\phi$ and $\psi$ induce isomorphic Lie algebras $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ over $k$?

[/Edit2]

Somehow the literature (which most likely exists) cannot be found easily via Google and the likes.

• Are your quadratic forms non-degenerate? Assuming so, in general if those identity components are isogenous then they must be isomorphic (exercise!), so the question is whether for a non-degenerate quadratic space $(V,q)$ over a field $k$, the isomorphism class of ${\rm{SO}}(q)$ determines the conformal isometry class of $(V,q)$. The answer is affirmative because of Dieudonne's theorem ${\rm{Aut}}_{{\rm{SO}}(q)/k} = {\rm{PGO}}(q)$ if $\dim V \ge 3$ (needs extra care to give characteristic-free proof valid in characteristic 2) and consideration of splitting fields when $\dim V = 2$. Mar 18 '15 at 16:53
• Just as an aside, beyond the connected case it isn't clear that "isogeny" is a good notion with smooth groups of finite type, or at least one should demand surjectivity too (automatic in the connected case). Mar 18 '15 at 16:55
• @user74230 — Ah, that is a good point. I'll switch to $\mathrm{SO}$. That actually suits my case even better.
– jmc
Mar 18 '15 at 17:09
• If you don't like isogenies and if you're fine with characteristic zero, the question is: when are the Lie algebras $\mathfrak{so}(\psi)$ and $\mathfrak{so}(\phi)$ isomorphic?
– YCor
Mar 24 '15 at 15:58
• Yes of course, the Lie algebras over the ground field. In characteristic zero, two semisimple Lie groups (over $K$) are isogenous iff their Lie algebras are isomorphic (over $K$).
– YCor
Mar 24 '15 at 16:49

Here's a proof assuming $n\ge 3,n\neq 8$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$).

Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are $K$-defined subalgebras of $\mathfrak{sl}_n$ preserving a nondegenerate quadratic form on the $n$-dimensional space. Then both are $C$-conjugate to $\mathfrak{so}(n)$. Since I assume $n\neq 8$ odd, over $C$, all automorphisms of $\mathfrak{so}(n)$ can be realized by some element of $\mathrm{GL}_n(C)$ (actually, of $\mathrm{O}_n(C)$). It follows that $f$ can be realized by a conjugation, namely there exists $A\in\mathrm{GL}_n(C)$ satisfying: $f(g)A=Ag$ for all $g\in \mathfrak{so}(\phi)$. The set of $A$ satisfying this condition is a $K$-defined linear subspace on which the determinant map does not vanish; hence it contains a $K$-point with nonzero determinant. That is, $A$ can be found in $GL_n(K)$. Hence $\mathfrak{so}(\psi)$ preserves the $K$-defined quadratic form $x\mapsto \phi'(x):=\phi(A^{-1}x)$. Since the set of $K$-defined invariant forms is 1-dimensional (because the standard representation of $\mathfrak{so}(\psi)$ is absolutely irreducible), it follows that $\phi'$ and $\psi$ are collinear.

I don't know what's going on for $n=8$.

For $n=2$, while there's only one 1-dimensional Lie algebra over $K$ so it's not enough to classify. Nevertheless it's still true that two quadratic forms are $K$-isomorphic up to rescaling [equivalently, have same determinant in $K^*/(K^*)^2$] iff they have isogenous SO(-). The point is that for 1-dimensional $K$-tori, $K$-isogenous is the same as $K$-isomorphic, and we can run the same proof as the above Lie-algebra-theoretic one, where we need to use the fact that every automorphism of $\mathrm{SO}_2(C)$ can be realized by some element of $\mathrm{GL}_2$. Actually this latter proof works for all $n\ge 2$ to show that if $\mathrm{SO}(\phi)$ and $\mathrm{SO}(\psi)$ are $K$-isomorphic then $\phi$ and $\psi$ are $K$-equivalent up to rescaling.

• For $n = 8$, the answer is the same: if the special orthogonal groups of $\phi$ and $\psi$ are isogenous then they are isomorphic (done) or related by triality. More than 1 SO quotient of Spin($\phi$) is SO of a q.form iff $\phi$ is similar to a 3-Pfister quadratic form, in which case all three quotients are isomorphic to SO($\phi$).
– Skip
Sep 11 '16 at 23:46
• @Skip How do you prove this last statement (or what's a reference)? It's equivalent to the claim that some outer automorphism of order 3 of $\mathrm{Spin}(\phi)$ is defined over the ground field $K$.
– YCor
Sep 12 '16 at 7:47
• If $G$ is split simply connected of type $D_4$ then it has 3 inequivalent 8-dimensional irreducible representations that are interchanged by an automorphism of $G$ of order 3. The same is true if $G$ is Spin($q$) for a 3-Pfister quadratic form $q$ by Galois descent or using the explicit description of triality from the Book of Involutions. If $G$ is a $K$-form of one of these such that at least 2 of the 8-dimensional representations are $K$-defined, then $G$ has inner type $D_4$ (because outer types have at most 1) ...
– Skip
Sep 12 '16 at 23:02
• ... and moreover all 3 are $K$-defined because the Tits algebras of these three representations sum to zero in the Brauer group and you assume that 2 of them are already zero. Therefore, $G$ is Spin($\phi$) for an 8-dimensional quadratic form $\phi$ with trivial discriminant and trivial Clifford invariant, i.e., is an 8-dimensional form in $I^3$, so it is a scalar multiple of a 3-Pfister form.
– Skip
Sep 12 '16 at 23:03
• @Skip I think you could consider writing a separate answer for the case $n=8$
– YCor
Sep 28 '16 at 23:46