Finite field analogue of representations in same packet have equal central character - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:02:49Z http://mathoverflow.net/feeds/question/74763 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74763/finite-field-analogue-of-representations-in-same-packet-have-equal-central-charac Finite field analogue of representations in same packet have equal central character Dror Speiser 2011-09-07T15:39:38Z 2011-09-07T15:39:38Z <p>In Kevin Buzzard's <a href="http://mathoverflow.net/questions/74663/how-badly-can-strong-multiplicity-one-fail-in-the-theory-of-automorphic-represent" rel="nofollow">recent question</a>, a warm up question was: if two automorphic representations are nearly equivalent, then are the central characters of their local components equal?</p> <p>Working my way up to local and later global automorphic representations, I am currently studying the situation over finite fields.</p> <ol> <li>What is the analogue of Buzzard's question in the finite field case?</li> <li>Is it true?</li> </ol> <hr> <p>Here are some of my own thoughts on this.</p> <p>If $$(T_1,\theta_1)\sim (T_2,\theta_2)$$ are two geometrically conjugate pairs of torus+character, then I think the characters have to agree on the center of $G^F$ (elements in the center are norms [?], so the geometric conjugacy equation shows equality). Using 7.2 in [DL] we see that the value of $R_T^\theta$ on the center is $\theta$ (maybe up to sign), i.e. $$\frac{R_{T_1}^{\theta_1}(z)}{R_{T_1}^{\theta_1}(1)} = \frac{R_{T_2}^{\theta_2}(z)}{R_{T_2}^{\theta_2}(1)}=\theta(z)$$ for $z\in Z(G^F)$.</p> <p>This might be considered an analogue as requested, but a naive one at that. A deeper analogue should consider the irreducible representations of $G^F$, and not the Deligne-Lusztig virtual characters, which can be reducible.</p> <p>Consider the two cross sections $\rho_x$, $\rho_x^'$ from the set of geometric conjugacy classes to irreducible representations (10.7.1/2 in [DL]): $$\rho_x=\sum_{[(T,\theta)]=x} \frac{(-1)^{\sigma(G)-\sigma(T)}}{\lt R_T^\theta,R_T^\theta\gt }R_T^\theta$$ $$\rho_x^'=(-1)^{\sigma(G)-\delta_x} \sum_{[(T,\theta)]=x} \frac{1}{\lt R_T^\theta,R_T^\theta\gt }R_T^\theta$$</p> <blockquote> <p>Do these have the same central character?</p> </blockquote> <p>If the $\theta$'s are trivial on the center, then the answer is yes. Computations that I have done before show that this is the case for the $\theta$'s that are not in general position in $Sp_4$.</p> <p>We can divide in an obvious way the $[(T,\theta)]\in x$ into two sets, of "positive" and "negative", such that $\rho_x^'$ is a sum and $\rho_x$ is a difference. We see that for the two representations to have equal central character, either the sum of "negative" terms, or "positive" terms (depending on $\delta_x$), must be zero on the center (minus the identity).</p> <p>Note that $\rho_x$ appears in the Gelfand-Graev representation of $G^F$, whose character restricted to the center (minus the identity) is zero, so this supports in spirit the paragraph above. I'm not sure if it can be extended from spirit to an actual proof.</p> <p>[DL] - this is, of course, the original Deligne-Lusztig paper from 1976.</p>