Timeline for Reference Request: Derived group of $\mathscr R_u(B)$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2017 at 19:33 | answer | added | Jim Humphreys | timeline score: 3 | |
| Jun 21, 2017 at 12:35 | vote | accept | D_S | ||
| Jun 21, 2017 at 11:11 | answer | added | Mikko Korhonen | timeline score: 4 | |
| Jun 21, 2017 at 1:26 | comment | added | D_S | But now I see what I'm doing is a special case of what you said, since the choice of a "$k$-splitting" gives the quotient group $U/U_{\Phi^+ - \Delta}$ the structure of a finite dimensional $k$-vector space. However, it's fine for me (at least for the near future) that this appears to depend on $T$. | |
| Jun 21, 2017 at 1:20 | comment | added | D_S | $$U\rightarrow \prod\limits_{\alpha \in \Delta} U_{\alpha} \rightarrow \mathbf G_a$$ where the right most map is $(x_{\alpha}) \mapsto \sum\limits_{\alpha} x_{\alpha}$. In order to define the left map as a homomorphism of algebraic groups, one also needs a choice of root vectors $x_{\alpha} : \alpha \in \Phi^+ - \Delta$, and project. Although neither homomorphism is defined over $k$, the composition is.The existence of such a homomorphism with kernel $U_{\Phi^+ - \Delta}$ implies that $U_{\Phi^+ - \Delta}$ must contain the derived group of $U$, so this is what naturally led to my question. | |
| Jun 21, 2017 at 1:17 | comment | added | D_S | Thank you as always for your informative answers and references. Actually, my question was more motivated by representation theory: for $G$ quasisplit over a $p$-adic field $k$ and a choice of splitting $x_{\alpha}: \mathbf G_a \rightarrow U_{\alpha} : \alpha \in \Delta(B,T)$ such that $\sigma.x_{\alpha} = x_{\sigma.\alpha}$ for all $\sigma \in \textrm{Gal}(\overline{k}/k)$, the choice of an additive unitary character $\psi$ of $k$ allows you to define a unitary character $\chi: U(k) \rightarrow S^1$ via a composition of homomorphisms | |
| Jun 21, 2017 at 0:45 | comment | added | nfdc23 | You gave no motivation, but I conjecture that what you seek (for which the above is unnecessary) is that $U$ has a canonical $B$-equivariant composition series (so independent of $T$!) with each successive quotient a vector group having a unique $B$-equivariant linear structure. This is true even for parabolic subgroup schemes of any reductive group scheme over any base scheme: see SGA3 XXVI 2.1, or Theorem 5.4.3 of the article Reductive Group Schemes in smf4.emath.fr/Publications/PanoramasSyntheses/2014/42-43/html/… (for a simpler proof via dynamic arguments). | |
| Jun 20, 2017 at 23:45 | comment | added | nfdc23 | There are exotic commutations among root groups in characteristics 2 and 3 for types B$_2$=C$_2$ and G$_2$, so probably one cannot expect a purely combinatorial recipe for computing the derived group of $U$ that works in all characteristics. In particular, you should check ${\rm{Sp}}_4$ and ${\rm{G}}_2$ directly (using the Chevalley commutation relations, which are given in Humphreys' book on linear algebraic groups, as well as in SGA3) with special attention to characteristics 2 and 3 before contemplating the general case. The rank-2 case is the essential case to understand. | |
| Jun 20, 2017 at 23:18 | history | asked | D_S | CC BY-SA 3.0 |