most recent 30 from http://mathoverflow.net 2013-05-25T15:09:09Z http://mathoverflow.net/feeds/question/75994 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75994/whats-the-contragredient-of-induced-representation unknown (google) 2011-09-20T21:03:37Z 2011-09-20T23:25:12Z

<p>Let $G$ be a real reductive Lie group, $P=MN$ its parabolic subgroup with Levi decomposition. Suppose $\sigma$ is a smooth admissible irreducible representation of $M$, extend this to $P$ by letting $N$ act trivially. Form the unitarily induced representation $Ind_P^G(\sigma)$.</p> <p>My question is what is the contragredient representation (smooth admissible dual) of $Ind_P^G(\sigma)$ in terms of $\sigma$ ? In particular, is it equal to $Ind_P^G(\sigma')$, where $\sigma'$ is the smooth admissible dual of $\sigma$? </p>

http://mathoverflow.net/questions/75994/whats-the-contragredient-of-induced-representation/76003#76003 Alexander Braverman 2011-09-20T22:22:04Z 2011-09-20T23:25:12Z

<p>Yes. The point is that $Ind_P^G(\sigma)$ is by definition equal to the space of sections of a certain $G$-equivariant vector bundle $E_{\sigma}$ on $G/P$ and $Ind_P^G(\sigma')$ is equal to the sections of the corresponding bundle $E_{\sigma'}$. Now the point is that because you are using unitary induction there is a natural map $E_{\sigma}\otimes E_{\sigma'}\to \Omega_{G/P}$ where $\Omega$ is the bundle on differential forms of top degree (more precisely, it has to be tensored with the corresponding orientation sheaf which we can trivialize if we choose a $G$-equivaruiant orientation of $G/P$ - let me for simplicity assume that we can do that). This gives a map $Ind_P^G(\sigma)\otimes Ind_P^G(\sigma')$ to differential forms which we can integrate (since I assumed that we have chosen an orientation on $G/P$). This gives a pairing between the two induced representations and the fact that it is a perfect pairing is easy.</p>