Timeline for $\mathbb{Q}$-forms of $\operatorname{SL}_4(\mathbb{R})$ inside $\operatorname{SL}_8(\mathbb{R})$
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| Nov 6, 2023 at 14:11 | vote | accept | CommunityBot | ||
| Nov 6, 2023 at 14:13 | |||||
| Oct 28, 2023 at 7:18 | answer | added | Mikhail Borovoi | timeline score: 1 | |
| Oct 27, 2023 at 16:40 | comment | added | YCor | @MikhailBorovoi Oops, I computed the normalizer of the wrong group. Actually, the normalizer is bigger than that. Indeed, the centralizer in the matrix algebra is a copy of $M_2(\mathbf{C})$, so in $\mathrm{GL}_8$ it can be identified to $\mathrm{GL}_2(\mathbf{C})$, and in $\mathrm{SL}_8$ this corresponds to matrices in $A\in\mathrm{GL}_2$ with $\det(A)^4=1$. The normalizer is thus generated by this and the given copy of $\mathrm{SL}_4$ (no element of the normalizer acts as non-inner automorphism) — they intersect in scalar matrices $\lambda I_8$, $\lambda^4=1$. | |
| Oct 27, 2023 at 13:51 | comment | added | Mikhail Borovoi | @YCor: I think we need $B=\lambda A$ for some scalar $\lambda$. | |
| Oct 27, 2023 at 10:54 | comment | added | YCor | @MikhailBorovoi the normalizer has two components: one is the group of block-diagonal matrices $(A,B)$ with $\det(A,B)=1$, $A,B\in\mathrm{GL}_4$. The other one contains the flip. The action is the obvious one by conjugation. | |
| Oct 27, 2023 at 9:11 | history | edited | Mikhail Borovoi |
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| Oct 27, 2023 at 9:10 | comment | added | Mikhail Borovoi | Can you compute the normalizer of ${\bf G}(\Bbb C)$ in ${\bf H}(\Bbb C)$ and the action of this normalizer on ${\bf G}(\Bbb C)$ ? Then I will answer your question. | |
| Oct 27, 2023 at 9:06 | comment | added | Mikhail Borovoi | The question is about Galois cohomology. We have a subgroup ${\bf G}={\rm SL}_{4, {\Bbb Q}}$ and an embedding $\iota\colon {\bf G}\hookrightarrow {\bf H}$ where ${\bf H}={\rm SL}_{8, {\Bbb Q}}$. | |
| Oct 24, 2023 at 20:36 | comment | added | Vincent | Ah wait you want them to be conjugate to embeddings into $\mathfrak{gl}(n, \mathbb{Q})$! I thought you wanted them to be conjugate to eachother. Now I get it, thank you! | |
| Oct 24, 2023 at 14:21 | comment | added | user515519 | Continuation of the previous comment: comparing to our situation $\mathfrak{g}_{\mathbb{Q}}=\mathfrak{sl}(4,\mathbb{Q})$ and $\mathfrak{h}_{\mathbb{Q}}$ should be some anisotropic $\mathbb{Q}$-form of $\operatorname{SL}_4(\mathbb{R})$. The embeddings of $\mathfrak{g}_{\mathbb{R}}$ and $\mathfrak{h}_{\mathbb{R}}$ are into $\mathfrak{sl}(8,\mathbb{R})$.... (I hope this makes sense) | |
| Oct 24, 2023 at 14:12 | comment | added | user515519 | @Vincent Thanks for pointing this out. The general situation should be as follows (If I understand correctly your argument): we have two algebras $\mathfrak{g}_{\mathbb{Q}}, \mathfrak{h}_{\mathbb{Q}}$ over $\mathbb{Q}$ and embeddings of $\mathfrak{g}_{\mathbb{R}}$ and $\mathfrak{h}_{\mathbb{R}}$ into some $\mathfrak{gl}(n,\mathbb{R})$ which are conjugate over $\mathbb{R}$. But we want also the embeddings of $\mathfrak{g}_{\mathbb{Q}}, \mathfrak{h}_{\mathbb{Q}}$ to be conjugate over $\mathbb{R}$ to embeddings into $\mathfrak{gl}(n,\mathbb{Q})$ (this is the whole point). | |
| Oct 24, 2023 at 13:15 | comment | added | Vincent | I'm sorry you came here to ask a question and now I am asking you questions, but I really find this whole set-up fascinating and hope to learn something here | |
| Oct 24, 2023 at 13:14 | comment | added | Vincent | I.e. for most $X \in \mathfrak{h}$ wouldn't $g^{-1}Xg$ be some linear combination of elements of $\mathfrak{g}$ that is not in $\mathfrak{g}$ itself? Or is this different for groups than for Lie-algebras? | |
| Oct 24, 2023 at 13:12 | comment | added | Vincent | Now if you have very concrete embeddings of $\mathfrak{g}$ and $\mathfrak{h}$ into $\mathfrak{gl}(n, \mathbb{R})$, then we can realize $\mathfrak{g}_\mathbb{R}$ and $\mathfrak{h}_\mathbb{R}$ also in a very concrete way: as the set of all $\mathbb{R}$-linear combinations of the elements of $\mathfrak{g}, \mathfrak{h}$ respectively. I also see how there must be some $g \in GL(n, \mathbb{R})$ such that for these concrete versions $\mathfrak{h}_\mathbb{R} = g^{-1}\mathfrak{g}_\mathbb{R}g$. But what I don't get is how that would imply $\mathfrak{h} = g^{-1}\mathfrak{g}g$ for the same $g$. | |
| Oct 24, 2023 at 13:05 | comment | added | Vincent | I find it a bit easier to think about Lie algebras than groups. I can see how you have two Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ over $\mathbb{Q}$ that are non-isomorphic while their 'realifications' $\mathfrak{g}_\mathbb{R} := \mathfrak{g} \otimes_\mathbb{Q} \mathbb{R}$ and $\mathfrak{h}_\mathbb{R} := \mathfrak{h} \otimes_\mathbb{Q} \mathbb{R}$$ are isomorphic. (Ctd in next comment) | |
| Oct 24, 2023 at 10:18 | comment | added | user515519 | @Vincent Consider rational quadratic forms on $\mathbb{R}^3$ of signature $2+1$. Their orthogonal groups are $\mathbb{Q}$-algebraic and conjugate over $\mathbb{R}$ but not over $\mathbb{Q}$ in general as some of them are $\mathbb{Q}$-isotropic and some are not. This also happens for $\operatorname{SL}_2$ acting on $2\times 2$ matrices i.e there are $\mathbb{Q}$-isotropic $\mathbb{Q}$-forms conjugate in $\operatorname{SL}_4(\mathbb{R})$ to anisotropic ones. | |
| Oct 24, 2023 at 9:20 | comment | added | Vincent | I think I have to see an example in order to understand it, so I really hope someone can type an answer. So far I cannot phantom how $x \mapsto g^{-1}xg$ can not be a bijective group homomorphism, no matter what bigger group $g$ comes from. | |
| Oct 23, 2023 at 16:46 | comment | added | user515519 | @Vincent I think that the difference is that we allow conjugacy by $\operatorname{SL}_8(\mathbb{R})$, instead of just $\operatorname{SL}_8(\mathbb{Q})$, such that the resulting group is $\mathbb{Q}$-algebraic. This yields, in principle, different isomorphism classes over $\mathbb{Q}$. | |
| Oct 23, 2023 at 16:37 | comment | added | Vincent | Intruiging. I knew there was a bijection between $\mathbb{Q}$-forms of $SL_2$ and quaternion-algebras, but I thought that conjugating left everything isomorphic, so you wouldn't get to a truly different algebra/group. What am I missing here? | |
| Oct 23, 2023 at 16:28 | comment | added | user515519 | @Vincent Yes for $\operatorname{SL}_2$ this happens. It can be obtained using a quaternionic central division algebra over $\mathbb{Q}$. | |
| Oct 23, 2023 at 16:21 | comment | added | user515519 | @YCor the second one $A\to \begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix}$. | |
| Oct 23, 2023 at 16:17 | history | edited | YCor | CC BY-SA 4.0 |
changed crooked math mode to standard boldface
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| Oct 23, 2023 at 16:15 | comment | added | YCor | The "natural embedding" is $A\mapsto \begin{pmatrix} A & 0\\ 0 & I_4\end{pmatrix}$ or $A\mapsto \begin{pmatrix} A & 0\\ 0 & A\end{pmatrix}$? | |
| Oct 23, 2023 at 16:03 | comment | added | Vincent | Do you know of an example of this happening, e.g. for $SL_2$? | |
| S Oct 23, 2023 at 15:09 | review | First questions | |||
| Oct 23, 2023 at 15:20 | |||||
| S Oct 23, 2023 at 15:09 | history | asked | user515519 | CC BY-SA 4.0 |