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The Mackey restriction formula says, for an $H$-representation $V$,Frobenius reciprocity saysApplying both with an irreducible character $\chi$ givesThe identity double coset contributes one, and all terms are nonnegative integers. Therefore Mackey irreducibility criterion says that ${\mathrm{Ind}}_H^G\chi$ is irreducible exactly when every term belonging to a nonidentity double coset is zero.

Now take $G=S{L}_2(k)$. Writeso $B=T⋉U$. For $q\ge 4$, commutators of $T$ with $U$ generate $U$, because conjugation sends $u(b)$ to $u(a^{2}b)$ and one can choose $a^2\ne 1$. Every degree-one character of $B$ is therefore trivial on $U$ and has the formwhere $\theta :k^{\times }\to {\mathbb{C}}^{\times }$ is a multiplicative character.

The Bruhat decomposition of SL2 over a finite field has two double cosets,and $B\cap wB{w}^{-1}=T$. Conjugation by $w$ inverts $T$, soThe only nonidentity Mackey term vanishes exactly when these two one-dimensional characters differ. ConsequentlyThus the trivial character is excluded, as is the unique quadratic character when $q$ is odd. All other degree-one characters of $B$ induce irreducibly; the induced representations have degree $[G:B]=q+1$. This is the irreducible principal series of finite SL2.

The Mackey restriction formula says that for subgroups $H,K\le G$ and a $K$-representation $W$,where ${}^{x}W(xk{x}^{-1})=W(k)$.

For $H=K=B$, the Bruhat decomposition of SL2 over a finite field givesThus $B\backslash G/B$ has representatives $1,w$. The identity double coset contributes $\theta$, whilecontributes an induced representation from the diagonal subgroup. ThereforewhereFor $t(a)=\mathrm{diag}(a,a^{-1})$, one has $w^{-1}t(a)w=t(a^{-1})$.