Yes. The point is that $Ind_P^G(\sigma)$ is by definition equal to the space of section 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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Yes. The point is that $Ind_P^G(\sigma)$ is by definition equal to the space of section 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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