# The induced representations of $SL(2, F)$.

Let $G=SL(2,F)$ and $I=J_{0}\cap J_{1}$ be the Iwahori subgroup of $SL(2, F)$, where $J_{0}=\left( \begin{array}{cc} \mathcal{O}_{\mathbb{F}} & \mathcal{O}_{\mathbb{F}} \\ \mathcal{O}_{\mathbb{F}} & \mathcal{O}_{\mathbb{F}} \\ \end{array} \right)\cap SL(2)$ and $J_{1}=\left( \begin{array}{cc} \mathcal{O}_{\mathbb{F}}& \varpi_{\mathbb{F}}^{-1}\mathcal{O}_{\mathbb{F}} \\ \varpi_{\mathbb{F}} \mathcal{O}_{\mathbb{F}}& \mathcal{O}_{\mathbb{F}} \\ \end{array} \right)\cap SL(2)$ are the maximal compact subgroups of $SL(2, F)$. Here $F$ is a local non-Archimedean $p$-adic field, $\mathcal{O}_{F}$ is the valuation ring and $\varpi$ is the uniformizer.

Now, we know that if $\lambda^{2}= 1$, then $Ind_{I}^{G}\lambda=\lambda^{-}\oplus\lambda^{+}$. Also , we can write $Ind_{I}^{G}\lambda=Ind_{J_{0}}^{G}(Ind_{I}^{J_{0}}\lambda)$.

Does the following equality hold?

$Ind_{I}^{G}\lambda=Ind_{J_{0}}^{G}\lambda\oplus Ind_{J_{1}}^{G}\lambda$.

I do apologize for my bad English and my bad temper to create a question.

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I assume that $\lambda$ is a character!? How is your $\lambda$ a representation of $J_0$ or $J_1$, do you mean the induction $Ind_I^{J_k} \lambda$ instead, but the induction is never irreducible.

On some related matters on $GL(2)$:

The question whether $Ind_{I}^{J_i} \lambda$ splits has been answered by Silberger for $PGL(2)$ for $J_0$ and $J_1$ in odd residue characteristic, and by Casselman for $PGL(2)$ for $J_0$ in all residue characteristics.

Silberger, Allan J. Irreducible representations of a maximal compact subgroup of ${\rm pgl}_{2}$pgl2 over the $p$p-adics. Math. Ann. 229 (1977), no. 1, 1–12.

Silberger, Allan J. ${\rm PGL}_{2}$PGL2 over the $p$p-adics: its representations, spherical functions, and Fourier analysis. Lecture Notes in Mathematics, Vol. 166

Casselman, William The restriction of a representation of ${\rm GL}_{2}(k)$GL2(k) to ${\rm GL}_{2}({\germ o})$GL2(o). Math. Ann. 206 (1973), 311–318.

In these cases, the restriction induction formula $Res_{SL(2,F)} Ind^{GL(2,F)}_{J_i} \lambda =Ind_{J_i}^{SL(2,F)} \lambda$! Induction by steps imply the results, if it $Ind_{I}^{J_i} \lambda$ splits.

I am not sure, if that is necessary, but you can look at the induction restriction formula for $Res_{J_i} Ind^{SL(2,F)}_{J_i} \lambda$ and see what happens.

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