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 sameisogenyclass?

Remarks:

- 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.)***[Edit]**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]** - 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.

mustbe 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 $\dim V = 2$. $\endgroup$ – user74230 Mar 18 '15 at 16:538more comments