# A proposition on square-integrable representations as Langlands quotient

In Stephen Kudla's paper "The local Langlands correspondence: the non-Archimedean case", we find the following result about square-integrable (mod center) representations and Langlands quotients $$Q(\Delta)$$.

Proposition 1.2.3 (I. N. Bernstein, [48, Proposition 11]). Every square integrable representation $$\pi$$ of $$G$$ has the form $$Q(\Delta)$$, where $$\Delta=(\sigma, \sigma(1), \sigma(2), \ldots, \sigma(r-1))$$ with $$\sigma\left(\frac{r-1}{2}\right)$$ unitary.

The reference [48] is Rodier's Bourbaki seminar talk, which states

PROPOSITION 11.- Toute représentation de carré intégrable de $$\mathrm{G}_{\mathrm{n}}$$ est de la forme $$\mathrm{L}(\Delta)$$ pour un segment $$\Delta=[\rho, \rho(\mathrm{r})]$$$$\rho$$ est une représentation cuspidale irréductible telle que $$\rho\left(\frac{r}{2}\right)$$ soit unitaire.

Notice the difference between $$r-1$$ and $$r$$, and between $$Q(\Delta)$$ and $$\mathrm{L}(\Delta)$$, in their notations.

I find this proposition false in the simplest case where $$G=\mathrm{GL}_1(\mathbb{Q}_p)$$ in which case we must have $$r-1=0$$ in Kudla's notation (or $$r=0$$ in Rodier's notation). But then every character of $$G$$ is square-integrable (modulo center) as $$G$$ is abelian, but not every smooth character $$\chi: \mathbb{Q}_p^\times \to \mathbb{C}^\times$$ is unitary (i.e., with values in $$S^1$$) : we are free to send $$p$$ to any nonzero complex number.

Question. I wonder what the correct statement of this proposition is. Could any expert give a correction or a idea of the proof? Is it fine if we restrict ourselves to $$G = \mathrm{GL}_n$$ for $$n>1$$? In fact, Rodier's usage of $$r$$, as opposed to $$r-1$$, seems to suggest this.