Let $G$ be a split reductive group over a $p$-adic local field $k$. For $\pi$ an unramified representation of $G(k)$, and $r$ a finite dimensional representation of the L-group $^LG$, Langlands showed how to attach an L-function $L(s,\pi,r)$.
I have computed this L-function in the case of $\operatorname{GL}_1$ and am getting something that looks "wrong" to me. I was hoping to state it here and perhaps someone can explain where I am making a mistake.
Definition of $L(s,\pi,r)$:
Let me first say in my own words how to define $L(s,\pi,r)$. Let $\mathscr H_G = C_c^{\infty}(G(\mathcal O_k) \backslash G(k) / G(\mathcal O_k))$ be the spherical Hecke algebra of $G$, and let $\mathscr H_T = C_c^{\infty}(T(k)/T(\mathcal O_k))$ be the one of $T$. The Satake transform
$$S: \mathscr H_G \rightarrow \mathscr H_T$$
$$S(f)(t) = \delta_B(t)^{1/2} \int\limits_U f(tu) du,$$ for $B = TU$ a Borel subgroup of $G$, defines an isomorphism of $\mathscr H_G$ with a subalgebra $\mathscr H_T^W$ of Weyl group invariant functions on $\mathscr H_T$. If $\pi$ is an unramified representation of $G(k)$, then there is a nonzero vector $v$ in the space of $\pi$, and a character $\xi$ of $\mathscr H_G$, such that
$$\xi(f)v = \int\limits_{G(k)} f(g)\pi(g)vdg \tag{$f \in \mathscr H_G$}.$$
There is an unramified character $\chi$ of $T(k)$, unique up to Weyl group equivalence, such that
$$\xi(f) = \int\limits_{T(k)} S(f)(t) \chi(t)dt.$$
The Harish-Chandra map $H_T: T(k) \rightarrow \operatorname{Hom}_{\mathbb Z}(X(T),\mathbb Z) = X(^LT)$ defined by $H_T(t)(\eta) = -\operatorname{ord}_k\eta(t)$ is surjective with kernel $T(\mathcal O_k)$, and induces an isomorphism
$$T(k)/T(\mathcal O_k) \rightarrow X(^LT)$$
and therefore we get bijections
$$\operatorname{Hom}_{\textrm{grp}}(T(k)/T(\mathcal O_k), \mathbb C^{\ast}) \rightarrow \operatorname{Hom}_{\textrm{grp}}(X(^LT),\mathbb C^{\ast}) \rightarrow ^LT$$
through which $\chi$ corresponds to an element $A_{\pi} \in \space ^LT \subset \space ^LG$. Then Langlands defines
$$L(s,\pi,r) = \operatorname{det}(1 - q^{-s} r(A_{\pi}))^{-1}.$$
Computation in the case $\operatorname{GL}_1$:
Let $T = \operatorname{GL}_1$, so that $^LT = \mathbb C^{\ast}$. Let $r: \space ^LT \rightarrow \operatorname{GL}_1(\mathbb C) = \mathbb C^{\ast}$ be the identity map.
If $\chi$ is an unramified representation (unramified character) of $T(k)$, then by the way the Harish-Chandra map $H_T$ is defined, we actually get $A_{\pi} = \chi(\varpi)^{-1}$, where $\varpi$ is a uniformizer. Then Langlands' L-function is
$$L(s,\chi,r) = (1 - q^{-s} \chi(\varpi)^{-1})^{-1}.$$
This doesn't seem right. Taking $r = \textrm{identity}$ should yield the "standard L-function" which should be
$$L(s,\chi,r) = L(s,\chi) = (1 - q^{-s}\chi(\varpi))^{-1}.$$
In fact, if you identity $k^{\ast}$ with the abelianized Weil group via the local Artin map (the one that sends a uniformizer to a geometric Frobenius),then $(1 - q^{-s}\chi(\varpi))^{-1}$ is the Artin L-function attached to the one-dimensional Artin representation corresponding to $\chi$.
Have I made a mistake in computing these L-functions? If not, how should one reconcile these L-functions?
Of course, one very easy way to force things to match up is to use a different definition of $H_T$:
$$H_T(t)(\eta) = \textrm{ord}_k \eta(t)$$ and I have seen some authors define $H_T$ this way. Another thing one could do is to define standard L-functions differently:
$$L(s,\pi) = L(s,\pi, \tilde{r})$$ for $\pi$ a representation of $\operatorname{GL}_n(k)$ and $r$ the identity map on $^L\operatorname{GL}_n = \operatorname{GL}_n(\mathbb C)$.
