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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 $\sigma$?

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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.

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Thanks a lot,Alex. How about the following argument? Note any element in $Ind_P^G(\sigma)$ is uniquely determined by its restriction to max. compact subgroup $K$. For any $ϕ∈Ind(\sigma)′$, $f∈Ind(\sigma′)$, let $$ϕ(f)=∫_K<f(k),ϕ(k)>dk$$ where $<,>$ denotes the pairing between $\sigma$ and $\sigma′$. I haven't check all the necessary properties, but it seems fine. By the way, is there some reference containing this? –  user1832 Sep 23 '11 at 12:24
    
Well, in this argument you have to explain why it is important that you are using unitary induction - without that you formula makes sense but it is not $G$-invariant. You have to integrate over $G/P$ rather than over $K$. –  Alexander Braverman Sep 24 '11 at 18:35
    
Why does one have to integrate over $G/P$? –  user1832 Sep 25 '11 at 23:55
    
To see that the procedure is $G$-invariant –  Alexander Braverman Sep 26 '11 at 16:32
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