Understanding Frobenius reciprocity I am stuck trying to understand the proof of the following proposition:

Let $\pi$ be an irreducible representation of $G=GL_2$. Then the following are equivalent:

*

*$\pi$ is equivalent to a subspace of $Ind_B^G \chi$ for a character $\chi$ of $T$ ;

*$\pi$ contains the trivial character of $N$.


Here, we use the usual notations: $B$ the standard Borel of upper triangular matrices, $N$ the unipotent upper triangular matrices and $T$ the torus of diagonal matrices.
It is mentioned that this is a simple consequence of Frobenius Reciprocity, but I don't get where it comes into play.
 A: I suppose you are talking about finite groups and their complex representations.
By Frobenius reciprocity, we know $Hom_N (1, \pi) \cong Hom_G (Ind^G _N 1, \pi)$ . We also know that $Ind^G _N 1 \cong \sum_\chi Ind^G _B \chi$ This proves the claim.
A: What you are describing is Harish-Chandra induction and restriction. If $\psi$ is a character of $T$, write $R_T^G(\psi)$ for $\psi$ inflated to $B$, and then induced to $G$. On the other hand, if $\chi$ is a character of $G$, write ${}^*R_T^G(\chi)$ for the character obtained first by restricting to $B$, and then taking the subspace of this space that is fixed by the unipotent subgroup $U$. This naturally becomes a character for $T$.
Frobenius recpirocity, applied to any character of $G$ and any character of $T$, yields $\langle R_T^G(\psi),\chi\rangle=\langle \psi,{}^*R_T^G(\chi)\rangle$. To see this notice that we are ignoring in HC-restriction all characters that are not inflated from the torus. Thus
$$\langle \psi,{}^*R_T^G(\chi)\rangle=\langle \psi,\chi{\downarrow_B}\rangle=\langle \psi,\chi{\downarrow_T}\rangle,$$ where $\downarrow$ is the standard restriction.
On the other hand, HC-induction is simply standard induction from a Borel, but only for certain characters. In this case $\langle R_T^G(\psi),\chi\rangle=\langle \psi\uparrow^G,\chi\rangle$. Thus Frobenius reciprocity completes the proof.
If $\pi$ contains the trivial character of $N$, then $\pi$ has (the inflation of) a character of $T$ in its restriction to $B$. Thus its Harish-Chandra restriction is non-zero. Let $\chi$ be one of the constituents of it. Then the HC-induction of $\chi$ must include $\pi$ by the above statement.
