Timeline for When do two non-degenerate quadratic forms give rise to isomorphic Lie algebras?
Current License: CC BY-SA 3.0
20 events
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| Sep 11, 2016 at 23:54 | comment | added | Skip | Similar questions: mathoverflow.net/questions/87751/… and mathoverflow.net/questions/98003/… See also the related paper dx.doi.org/10.1016/j.jalgebra.2014.08.057 especially Proposition 7.2 | |
| May 11, 2015 at 10:06 | vote | accept | jmc | ||
| Mar 25, 2015 at 6:21 | comment | added | jmc | @user74230 — Thanks for your comment. However, it was not clear to me that the same statement is true for Lie algebras. Thanks for the reference! | |
| Mar 25, 2015 at 4:54 | comment | added | user74230 | @jmc: As I stated in my initial comments, over fields of any characteristic whatsoever, the isomorphism class of SO($q$) determines the conformal isometry class of $q$ in any dimension. This doesn't require any work with classification of representations, and works over any local ring. See C.3.14 and C.3.16 for generalizations over base schemes in the article on reductive group schemes in the recently published volume 1 of the Luminy summer school on SGA3. | |
| Mar 24, 2015 at 19:40 | answer | added | YCor | timeline score: 7 | |
| Mar 24, 2015 at 18:30 | history | edited | jmc | CC BY-SA 3.0 |
Formulate equivalent question
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| Mar 24, 2015 at 17:28 | comment | added | YCor | They're equivalent if $\phi$ and $t\psi$ are equivalent over $K$ for some $t\in K^*$ and I don't know if this is the only case. If the dimension is $n$, an attempt would be to classify the $n$-dimensional absolutely irreducible representations of $\mathfrak{so}(\phi)$ (we could also require "preserving at least one nonzero quadratic form"). In the most optimistic case, there would be only one such representation, hence this is the usual representation and all the invariant quadratic forms are proportional. | |
| Mar 24, 2015 at 17:13 | comment | added | jmc | @YCor — Ok, cool! I hadn't thought of that, but it makes sense. Do you have clue about when $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are isomorphic? | |
| Mar 24, 2015 at 16:49 | comment | added | YCor | 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 $K$). | |
| Mar 24, 2015 at 16:18 | comment | added | jmc | @YCor — Probably that is the question I am asking. But then you have to take Lie algebras over the ground field ($\mathbb{Q}$ in my case, not $\mathbb{C}$!), right? Over $\mathbb{C}$ the entire question is moot. | |
| Mar 24, 2015 at 15:58 | comment | added | YCor | 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? | |
| Mar 24, 2015 at 15:46 | comment | added | jmc | @user74230 — Ok, I solved the exercise. If I am correct, you can prove it when $\dim V$ is odd, because then the groups are adjoint. If $\dim V$ is even, then they are adjoint up to a $\mu_{2}$. Anyway, I was misguided (maybe by abelian varieties), that if $G \to H$ is an isogeny, then there exists an isogeny $H \to G$. This changes my question. Is it still true that the conformal isometry class determines everything? | |
| Mar 24, 2015 at 15:44 | history | edited | jmc | CC BY-SA 3.0 |
Adjust the question because it relied on the false assumption that all groups in the same isogeny class are isogenous.
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| Mar 19, 2015 at 9:05 | comment | added | jmc | @user74230 — Thanks for that comment! I'm fine with assuming that the forms are non-degenerate. I will look into Dieudonne's theorem, and try to work out the details of the proof myself (including the exercise). If you want, you can post the comment as an answer. | |
| Mar 19, 2015 at 9:02 | history | edited | jmc | CC BY-SA 3.0 |
Assume the qf's are non-degenerate
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| Mar 18, 2015 at 17:10 | history | edited | jmc | CC BY-SA 3.0 |
Switch from orthogonal to special orthogonal.
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| Mar 18, 2015 at 17:09 | comment | added | jmc | @user74230 — Ah, that is a good point. I'll switch to $\mathrm{SO}$. That actually suits my case even better. | |
| Mar 18, 2015 at 16:55 | comment | added | user74230 | 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, 2015 at 16:53 | comment | added | user74230 | 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, 2015 at 15:47 | history | asked | jmc | CC BY-SA 3.0 |