$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}$Setup: Let $G$ be the group $\GL(4, \mathbb{R})$, $B$ denotes the Borel subgroup consisting of upper triangular matrices and $P_{(2,2)}$ be the standard parabolic whose Levi is $\GL(2)\times \GL(2)$. We write $D_i$ for the unitary $(\mathfrak{g}, K)$-module in the discrete series representations of $\GL(2, \mathbb{R})$ with lowest non-negative $K$-type being $$ \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \mapsto e^{-\iota(i+1)\theta} $$
Let $\mathrm{Ind}^G_P()$ denotes the normalized parabolic induction from $P$ to $G$. Recall that $$ D_i \subset \mathrm{Ind}(|\cdot|^{i/2}, |\cdot|^{-i/2}), $$ where $|\cdot|$ denotes the modulus character of $\mathbb{R}^{\times}$. Thus we can get the following map of induced representations $$ \mathrm{Ind}^G_{P_{(2,2)}}(D_1 \otimes D_3) \xrightarrow{\phi} \mathrm{Ind}(|\cdot|^{1/2}, |\cdot|^{-1/2}, |\cdot|^{3/2}, |\cdot|^{-3/2}) $$ Consider the map induced by $\phi$ in the $(\mathfrak{g}, \SO(2)\mathbb{R}^{> 0})$-cohomology $$ \mathrm{H}^i(\mathfrak{g}, \SO(2)\mathbb{R}^{> 0}, \mathrm{Ind}^G_{P_{(2,2)}}(D_1 \otimes D_3) \xrightarrow{\phi^*} \mathrm{H}^i( \mathfrak{g}, \SO(2)\mathbb{R}^{\geq 0}, \mathrm{Ind}(|\cdot|^{1/2}, |\cdot|^{-1/2}, |\cdot|^{3/2}, |\cdot|^{-3/2}) $$ It is known that the left hand side has cohomology in degree 4 and 5 and moreover it is one dimensional. I have the following questions
Question : Is the map $\phi^*$ nonzero in degree $i=4$?
We can also ask more generally an analogue of this question for $G = \GL(2n, \mathbb{R})$ and the parabolic being $P_{(2, \dots, 2)}$. The representations in question being modified in the most natural way $\mathrm{Ind}(D_1 \otimes D_3 \otimes \dots D_{n+1}) \subset \mathrm{Ind}^G_B(|\cdot|^{1/2}, |\cdot|^{-1/2}, |\cdot|^{3/2}, |\cdot|^{-3/2}, \dots, |\cdot|^{n/2}, |\cdot|^{-n/2})$.
Question : Is the map $\phi^*$ induced in Lie algebra cohomology nonzero in degree $n^2$?
Any suggestions, comments or references are welcome. Thank you.
