Timeline for Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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May 24, 2019 at 11:37 | comment | added | user140765 | @Marty can you comment on the concerns of Marc Palm? | |
Mar 24, 2014 at 13:27 | comment | added | Marc Palm | @Marty I thought more about your claim about the intertwiner. You can see that JL are using the one that I describe as does Bump, etc. By Schur's lemma, there can be only one intertwiner up to a constant. It is the standard one for the adelic Eisenstein series as well. Yours does not match up with the standard one times a constant, or does it? | |
Mar 24, 2014 at 7:16 | comment | added | eric | In fact I'm silly -- this phenomenon happens for GL(2,R) too and in this case I already knew it well :-/ | |
Mar 23, 2014 at 21:27 | comment | added | eric | Example: if $\chi_1=z^p\overline{z}^q$ and $\chi_2=1$ then $\chi_3=z^p$ and $\chi_4=\overline{z}^q$. In both cases the inf chars are (with obvious notation) $\{p,0\}$ and $\{q,0\}$ but the $q$ has switched from the character-with-$z^p$-in to the other one. | |
Mar 23, 2014 at 21:25 | comment | added | eric | Thanks a lot Marty. Although I didn't put it in the question, I half-wanted to say "surely we can't have $\{\chi_3,\chi_4\}\not=\{\chi_1,\chi_2\}$ because one can look at the infinitesimal character". But there's a subtlety. Very briefly: the point is that the inf char is obtained by regarding $GL(2,C)$ as a real Lie group and then complexifying, and there are two ways to match up the two inf chars you get in this way; if $\chi_1,\chi_2$ correspond to one way then $\chi_3,\chi_4$ correspond to the other. This subtlety is what I'd missed. Many thanks! | |
Mar 23, 2014 at 21:19 | vote | accept | eric | ||
Mar 23, 2014 at 13:18 | comment | added | Marc Palm | Are you sure? I can't see the invariance. The integral transfer needs to map $I(\mu)$ to $I(\mu^w)$. | |
Mar 23, 2014 at 1:01 | comment | added | Marty | That one should work too, with U the (say) upper-triangular unipotent radical. It's equivalent, I think after substitution, to the one I wrote down with the lower-triangular unipotent radical. The important thing is to use whatever unipotent radical is opposite of the one used for parabolic induction in $I(\chi_1, \chi_2)$. | |
Mar 22, 2014 at 15:57 | comment | added | Marc Palm | The intertwiner seems wrong. You need $\int\limits_{U} f(wux) du$ for $w$ the Weyl element, or not? | |
Mar 22, 2014 at 1:36 | history | answered | Marty | CC BY-SA 3.0 |