Timeline for Parabolics and simple roots for a special unitary group: reference request
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2021 at 10:41 | vote | accept | Mikhail Borovoi | ||
| Mar 10, 2021 at 20:45 | history | edited | LSpice | CC BY-SA 4.0 |
Editing in content from comments
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| Mar 9, 2021 at 20:35 | comment | added | Mikhail Borovoi | Thank you! But I need some time to digest your answer before I accept it... | |
| Mar 9, 2021 at 14:21 | comment | added | LSpice | In the quasi-split case ($\dim W \le 1$), the root system we get is the one that comes from folding $\mathsf A_n$ using its order-2 diagram automorphism. See the lovely Stembridge - Folding by automorphisms. (That paper doesn't apply literally to $\dim W = 1$ since the "middle roots" are not orthogonal but belong to an orbit; but I think it is still reasonable to say that $\mathsf{BC}_m$ arises by folding in that case.) | |
| Mar 9, 2021 at 14:20 | comment | added | LSpice | If you identify $I$ with $\{1, \dotsc, m\}$, then the root system is $\mathsf C_m$ if $W = 0$ and $\mathsf{BC}_m$ if $W \ne 0$. One set of simple roots is $\{a_{i(i + 1)} \mathrel: 1 \le i < m\} \cup \{a_{m m^*}\}$ if $W = 0$ and $\{a_{i(i + 1)} \mathrel: 1 \le i < m\} \cup \{a_m\}$ if $W \ne 0$. | |
| Mar 9, 2021 at 14:17 | comment | added | LSpice | This approach also picks out a Levi (the centraliser of $\lambda$) and identifies the unipotent radical (those $g$ for which $\lim_{t \to 0} \lambda(t)g\lambda(t)^{-1} = 1$). It is classical, but it was Conrad, Gabber, and Prasad - Pseudo-reductive groups that drove home its importance for me. | |
| Mar 9, 2021 at 14:15 | comment | added | LSpice | @MikhailBorovoi, standard parabolics (containing the Borel used to define 'simple') can be parameterised as you say, but all parabolics can be parameterised by cocharacters: the parabolic associated to $\lambda$ is the variety of all $g \in G$ such that $\lim_{t \to 0} \lambda(t)g\lambda(t)^{-1}$ exists; it is generated by the torus and those root subgroups corresponding to roots $\alpha$ for which $\langle\alpha, \lambda\rangle \ge 0$. In terms of simple roots, if $\lambda$ lies in the dominant chamber, then this corresponds to the set of simple roots satisfying the same condition. | |
| Mar 9, 2021 at 9:49 | comment | added | Mikhail Borovoi | I do not quite understand how you describe all parabolics. Their conjugacy classes should correspond to the subsets of the basis (the system of simple roots). | |
| Mar 9, 2021 at 9:43 | comment | added | Mikhail Borovoi | How can one choose a system of simple roots (a basis) and what is the type of the root system? $\rm BC_\ell$ ? | |
| Mar 8, 2021 at 23:32 | comment | added | LSpice | @FrancoisZiegler, right you are. Fixed, thanks! | |
| Mar 8, 2021 at 23:32 | history | edited | LSpice | CC BY-SA 4.0 |
Fixing typos pointed out by @FrancoisZiegler
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| Mar 8, 2021 at 23:23 | comment | added | Francois Ziegler | Shouldn’t the displayed formula have $v_j$? Also there is an “$i\in i$”. | |
| Mar 8, 2021 at 22:46 | history | edited | LSpice | CC BY-SA 4.0 |
First change (\ge to \le) was wrong; should have been \ge after all
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| Mar 8, 2021 at 22:43 | comment | added | LSpice | I am used to thinking of the stabiliser of a self-dual flag, but of course every self-dual flag has an isotropic flag as its "bottom half", and every isotropic flag can be completed to a self-dual flag by tossing the duals "on top". | |
| Mar 8, 2021 at 22:38 | history | edited | LSpice | CC BY-SA 4.0 |
No need to name the elements of the root spaces
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| Mar 8, 2021 at 22:18 | history | edited | LSpice | CC BY-SA 4.0 |
No need separately to specify \pair{a_i}\lambda \ge 0 (since a_i = 2a_{i i^*})
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| Mar 8, 2021 at 22:13 | history | edited | LSpice | CC BY-SA 4.0 |
No need separately to specify \pair{a_i}\lambda \ge 0 (since a_i = 2a_{i i^*})
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| Mar 8, 2021 at 22:06 | history | edited | LSpice | CC BY-SA 4.0 |
We only need j \ne i, not j \ne i^*; and, oh hey, we handle the mixed-anisotropic case too; no need to order I
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| Mar 8, 2021 at 22:01 | history | edited | LSpice | CC BY-SA 4.0 |
We only need j \ne i, not j \ne i^*; and, oh hey, we handle the mixed-anisotropic case too
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| Mar 8, 2021 at 20:07 | history | answered | LSpice | CC BY-SA 4.0 |