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 $Ind\_P^G(\sigma)$.


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 $\sigma$? 



what's the contragredient of induced representation
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.


