Timeline for Conjugation of root subgroups by the Weyl group
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2022 at 14:56 | comment | added | LSpice | @JoshuaRuiter, a smooth map of manifolds has a derivative that is a function between tangent bundles. In particular, a smooth homomorphism of Lie groups has a derivative that is a map between identity sections of tangent bundles, i.e., Lie algebras. (This much is in any book on Lie groups.) The Lie algebra of the root group $U_b$ is the root subspace $\mathfrak g_b$ (more or less by definition); and the derivative of $U_b \to G$ is the map $\mathfrak g_b \to \mathfrak g$. (For this you'd want to read about reductive groups; e.g., in Borel's book.) | |
| Apr 7, 2022 at 23:53 | comment | added | Joshua Ruiter | I know this was a very long time ago, but I have only recently had time to look more closely at this. I do not understand what you mean in the first paragraph by "whose derivative at the identity is the inclusion of the $b$-root subspace of $\operatorname{Lie}(G)$." In particular, I don't know how to interpret "derivative" in this context. Is there somewhere I can learn more about this? | |
| Sep 11, 2021 at 23:19 | vote | accept | Joshua Ruiter | ||
| Sep 10, 2021 at 22:15 | comment | added | LSpice | $\alpha^\vee(u_\alpha) \in T$ is the value of the coroot $\alpha^\vee \in \operatorname{Hom}(\operatorname{GL}_1, T)$ at the scalar $u_\alpha$ (where $T$ is the maximal torus containing $S$ that we use to define absolute roots). $\operatorname{Int}(g) \in \operatorname{Aut}(G)$ is the interior automorphism $h \mapsto g h g^{-1}$. | |
| Sep 10, 2021 at 21:21 | comment | added | Joshua Ruiter | I don't know exactly what you mean by $\alpha^{\vee}(u_\alpha)$ or by $\operatorname{Int}$. | |
| Sep 9, 2021 at 22:03 | comment | added | LSpice | Yes, and that's all we need for my comment, since we're only dealing with absolute roots whose restriction is $\pm$ a given relative root. (Then we use quasisplitness further to note that, if $\alpha$ restricts to $a$ and $\alpha'$ restricts to $-a$ and $\alpha + \alpha' \ne 0$, then the $\alpha$ and $\alpha'$ root groups commute.) | |
| Sep 9, 2021 at 12:44 | comment | added | Joshua Ruiter | It's not true that all root groups commute in the non-multipliable case. Did you just mean that the root groups associated to the absolute roots lying above a given relative root commute? | |
| Sep 9, 2021 at 1:05 | comment | added | LSpice | In fact, the way I was writing it, with $v \in V_b$ (rather than $v \in V_{w_a b}$), I should have said $\operatorname{Int}(w_a(1))\bigl(X_b(v)\bigr) = X_{w_a b}\bigl(\operatorname{Ad}(w_a(1))v\bigr)$. I fixed that, too. | |
| Sep 9, 2021 at 1:04 | history | edited | LSpice | CC BY-SA 4.0 |
Fixing the conjugation
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| Sep 9, 2021 at 0:57 | comment | added | LSpice | By $w_a(1)X_b(w_a(1)^{-1}v)$ I meant $\operatorname{Int}(w_a(1))\bigl(X_b\bigl(\operatorname{Ad}(w_a(1))^{-1}v\bigr)\bigr)$; there are two conjugations, one on the group and one on the Lie algebra (identifying $V_a$ with the $a$-root space). So your proposed version misses one conjugation. But I apologise; it was definitely written as if a multiplication, which would make sense in the group but wasn't what I meant. (Edited accordingly, and fixed typo $\beta \to w_a\beta$.) | |
| Sep 9, 2021 at 0:56 | comment | added | LSpice | Re again, special orthogonal groups are of type $B_n$ (if ($2n + 1$)-dimensional orthogonal space) or $D_n$ (if $2n$-dimensional). Unitary groups are of type $A_n$. | |
| Sep 9, 2021 at 0:55 | comment | added | LSpice | Re, in both cases, because root groups commute (since we're in the non-multipliable case). $X_b$ is the unique homomorphism whose derivative is the inclusion, and $v \mapsto \prod X_\beta(v_\beta)$ has that derivative. $w_a(u)$ is the unique element of $U_{-a}u U_{-a} \cap N$. Since $U_{-a}u U_{-a}$ equals $\prod U_{-\alpha}u_\alpha U_{-\alpha}$ (now also using quasisplit), it contains $\prod w_\alpha(u_\alpha)$. | |
| Sep 9, 2021 at 0:32 | comment | added | Joshua Ruiter | Can you provide any reference on the fact that $X_b(v) = \prod_\beta X_\beta(v_\beta)$, or that $w_\alpha(u) = \prod w_\alpha(u_\alpha)$? I believe the first equation, but I am skeptical of the second. In particular, in concrete calculations in special orthogonal groups, I have found that the elements $w_\alpha(u)$ associated with multidimensional root groups are not monomial matrices. But if what you say is true, it is a product of the absolute versions $w_\alpha(u_\alpha)$ which are monomial matrices (at least for absolute type $A_n$, which I think these special orthogonal groups are). | |
| Sep 9, 2021 at 0:28 | comment | added | Joshua Ruiter | I am confused by your placement of parentheses in the expression $w_a(1) X_b(w_a(1)^{-1} v)$ both times. Why have you not written this as $w_a(1) X_b(v) w_a(1)^{-1}$? | |
| Sep 8, 2021 at 0:45 | comment | added | LSpice | The usual definition of the $w_\alpha(t)$ requires that $t \ne 0$, so in our setting we must restrict to an open subset of $V_a$. I tried to check whether this is consistent with your pointer to the definition of $w_a$, but I cannot find it in Deodhar. However, this is how it "should" be defined, so I will proceed that way and correct if necessary once I know where to look. | |
| Sep 7, 2021 at 22:58 | history | edited | LSpice | CC BY-SA 4.0 |
Maybe qs comes earlier
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| Sep 7, 2021 at 15:31 | history | answered | LSpice | CC BY-SA 4.0 |