Why is the Hecke algebra commutative I am following some course notes about local representation theory. It introduces the Hecke algebra of $G=GL_n(F)$ for a non-Archimedean local field F as the space of compactly supported, bi-$K$-invariant functions on $G$, considered as an algebra under convolution.
I would like to show that this Hecke algebra is commutative. It advises to "consider the transpose map and show that it induces a map on the Hecke algebra which is both an involution and an anti-involution". However I do not see what map to consider: is it $f(g) \mapsto f({}^t g)$? (which is still $K$-invariant)
 A: We follow Exe 5.14 of a GTM preprint by Getz and Hahn. What to be considered is indeed $f\mapsto f({}^t*)=:f^\dagger$. We show this is an anti-homomorphism. First, for $f\in C_c^\infty (G)$, we have
$$
\int f(g)dg = \int f({}^tg^{-1})dg.
$$
This is because sending $f$ to the RHS of the above is left invariant, and because the above is true for $f = \mathbf 1_{GL_n(\mathcal O_F)}$.
By the above, for $f_1, f_2\in C_c^\infty (G)$ and $g\in GL_n(F)$, we have
$$
(f_1*f_2)({}^tg)=\int f_1(h)f_2(h^{-1}{}^tg)dh = \int f_1({}^th^{-1})f_2({}^th{}^tg)dh = \int f_1({}^tg{}^th^{-1})f_2({}^th)dh = (f_2^\dagger*f_1^\dagger)(g).
$$
In place of proving that $\dagger$ is a homomorphism on $C_c^\infty(GL_n(F)//GL_n(\mathcal O_F))$, we show that $\dagger$ is the identity here. Fixing a uniformizer $\pi\in\mathcal O_F$, notice
$$
GL_n(F) = \bigcup_{a_1\geqq a_2\geqq\dots a_n} GL_n(\mathcal O_F)\mathrm{diag}(\pi^{a_1},\dots, \pi^{a_n})GL_n(\mathcal O_F).
$$
This is because for an arbitrary $A\in GL_n(F)$, we can arrange that $A_{11}$ has the largest absolute value by the multiplication of elementary matrices on both sides, conduct the Gaussian elimination with respect to $A_{11}$, and then continue.
Now, $\dagger$ is the identity on $C_c^\infty(GL_n(F)//GL_n(\mathcal O_F))$ since the same is true on the set of diagonal matrices.
