Non-supercuspidal representations are principal series

I am reading Godement Jacquet's notes on Jacquet Langlands. I have an issue justifying a certain point. I want to prove

If $$\pi$$ is an irreducible admissible representation of $$GL_2(\mathbb{Q}_p)$$ that is not supercuspidal, then it is isomorphic to a "principal series" $$\rho_{\mu_1, \mu_2}$$ for certain characters $$\mu_1, \mu_2$$.

The proof is pretty quick. We already know that, letting $$K$$ the Kirillov model of $$\pi$$ and $$S(\mathbb{Q}_p^\times)$$ the space of locally constant compactly supported functions on $$\mathbb{Q}_p$$, we have $$K/S(\mathbb{Q}_p^\times)$$ is of finite dimension. Supposing $$\pi$$ non supercuspidal means that we suppose this dimension $$\geq 2$$.

Now here is the point where I am missing something: Godement claims that there is necessarily a non-zero linear form $$B$$ and characters $$\mu_1, \mu_2$$ of $$\mathbb{Q}_p^\times$$ such that $$B\left(\pi \pmatrix{a & \star \\ & b}\xi\right) = \mu_1(a) \mu_2(b) |a/b|^{1/2} B(\xi)$$

This is pretty enough to settle the proof since that already gives functions in the principal series. However, why can we conclude to the existence of this linear functional? Why specifically of this form?

• Where does this appear in Godement's notes? If $\pi$ isn't supercuspidal, it could still be a special representation (twist of a Steinberg representation). A special representation is isomorphic to a subrepresentation or a subquotient of a reducible principal series representation. Nov 12, 2020 at 23:31
• @PeterHumphries It's Theorem 4, page 1.24. I agree with your precision, this is why I chose to put quotes around "principal series" Nov 13, 2020 at 9:46

Since $$\pi$$ is non-supercuspidal, its Jacquet module $$\pi_N$$, a representation of $$T\cong\mathbb Q_p^\times\times\mathbb Q_p^\times$$, is nonzero. So, there is some homomorphism $$\pi_N\to\chi_1\otimes\chi_2$$, where $$\chi_i$$ are characters of $$\mathbb Q_p^\times$$. The composition $$\pi\to\pi_N\to\chi_1\otimes\chi_2$$ is the desired linear form.