Let $F$ be a non-archimedean local field and $\mathcal{O}_F$ the ring of integers in $F$. Let $G_F=GL_2(F)$.
Let $\pi_i$, $i=1,\ldots,n$,be non-equivalent finite dimensional irreducible representations of $GL_2(\mathcal{O}_F)$ and $$ \zeta(g):=\sum_i \dim(\pi_i) tr(\pi_i(g^{-1})). $$
I am trying to prove that $\zeta*\zeta=\zeta$. Here $*$ is the multiplication in the Hecke algebra associated to $G_F=GL_2(F)$. We have $$ (\zeta*\zeta)(h) \\ = \int_{G_F} \zeta(g) \zeta(g^{-1} h ) dg \\ = \int_{G_F} (\sum_i \dim(\pi_i) tr(\pi_i(g^{-1})) ) (\sum_i \dim(\pi_i) tr(\pi_i(h^{-1} g)) ) dg \\ = \sum_{i,j} \int_{G_F} \dim(\pi_i) \dim(\pi_j) tr(\pi_i(g^{-1})) tr(\pi_j(h^{-1} g)) dg. \quad (1) $$ But how could we show that $(1) = \zeta(h)$? Thank you very much.
Edit: this is an excise on the top of page 5 of the notes.
All these computations are happening just on $GL_2(\mathcal O_F)$. (If you like, just extend the functions by zero from $GL_2(\mathcal O_F)$ to $GL_2(F)$.)
Then, as the exercise hints, this is just the orthogonality relations for representations of the pro-finite group $GL_2(\mathcal O_F)$. (Note that since only finitely many reps. are involved, they are all actually reps. of some finite quotient $G$ of $GL_2(\mathcal O_F)$, and so this reduces to orthogonality for the finite group $G$.)