# What can we say about the Local Langlands Correspondence for GL_n without using Bernstein-Zelevinski?

I have two specific questions regarding the LLC for $GL_n$, and in particular, what we can say about the conjecture if we don't have the ideas of Bernstein and Zelevinski, which reduce the problem to finding a correspondence between supercuspidal representations on the automorphic side, and irreducible representations on the Galois side.

Question 1: If we don't have B-Z, is there a way to see that supercuspidal representations MUST correspond to irreducible representations? Obviously, there is lots of evidence that these two things SHOULD correspond. First, supercuspidals are the "building blocks" of automorphic representations just as irreducibles are the building blocks of Weil-Deligne representations. Moreover, it's easy to prove that the $L$-functions match: in the degree-one case this is just match of characters under Class Field Theory, and in higher degree it's easy to prove you have $L = 1$ on both sides. But my question is: without having the explicit construction of Bernstein and Zelevinski, is there a proof that if a local Langlands Correspondence exists, that supercuspidals must correspond with irreducibles?

Question 2: This is built from my comment at the bottom of: Weil group, Weil-Deligne group scheme and conjectural Langlands group

Let's say we're explicitly trying to find two automorphic representations whose corresponding Weil representations are the same. Under the ideas of BZ, we can say that if two representations have the same supercuspidal support, then they correspond to two Weil-Deligne representations with the same underlying Weil representation. Is it easy to prove that this is so, without using the ideas of Bernstein and Zelevinski?

As an example, let's take the degree-$2$ representation induced from the characters $\chi |\cdot|^{-1/2},\, \chi|\cdot|^{1/2}$. This decomposes into a one-dimensional representation $\pi$ of $GL_2$, and then a Steinberg representation $\sigma$. In this simple case, can we see that $\rho(\sigma) \cong \rho(\pi)$, without using BZ? (Here $\rho$ denotes the underlying Weil representation).

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