Timeline for Weight of the defining function of a Bruhat cell in a simply connected semisimple group
Current License: CC BY-SA 4.0
4 events
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| May 4, 2023 at 13:29 | comment | added | LSpice | @AllenLee, re, since this function transforms by the weight $(-w_0\omega, \omega)$ under $B\times B$, it suffices to check what elements of the Weyl group have a representative on which $f$ doesn't vanish. For $n \in N_G(T)(k)$, we have that $f(n)$ is non-$0$ if and only if $n\omega - w_0\omega$ equals $0$, i.e., if and only if $\omega$ is fixed by $n^{-1}w_0$. It's not obvious to me that's equivalent to $s_i \not\le n^{-1}w_0$, but, if so, then that's equivalent to $s_i \not\le w_0 n$, and so to $n \not\le w_0 s_i$. | |
| May 4, 2023 at 13:15 | comment | added | LSpice | What does it mean to be dual to a lowest weight vector? That it pairs non-$0$-ly with a lowest-weight vector, and $0$-ly with all other weight vectors? Is that different from just being a highest-weight vector in $L(\omega)^*$? \\ Also, my computation gave a weight $(-w_0\omega, \omega)$ (or at least a multiple of that), and it seems that yours does too, not $(\omega, -w_0\omega)$. Is that correct? | |
| May 4, 2023 at 9:36 | comment | added | Allen Lee | >Thanks for your answer, could you be more specific about why this $f$ defines that Bruhat cell? | |
| May 4, 2023 at 7:37 | history | answered | Peter McNamara | CC BY-SA 4.0 |