A specific $2$-dimensional Galois representation of $G_{\mathbb{Q}_2}$ and its Langlands correspondenceI am interested in understanding a situation in (classical, not $p$-adic) local Langlands for $\mathrm{GL}_p(\mathbb{Q}_p)$. An example of it is as follows: Let $F=\mathbb{Q}_2$ and $E$ be the splitting field of $x^4-2$ over $F$. It is a totally wildly ramified extension. The Galois group $G:=\operatorname{Gal}(E/F)$ is isomorphic to $D_4$, the dihedral group of order $8$. We have its upper numbering filtration (see comment for a code) $$ G^v=\left\{\begin{array}{ll} D_4,&\text{when }v\in [-1,1]\\ C_4,&\text{when }v\in (1,2]\\ (D_4,D_4)=Z(D_4),&\text{when }v\in (2,3]\\ \{1\},&\text{when }v\in (3,+\infty) \end{array}\right. $$ Let $\tilde{G}:=W_F$ be the Weil group of $F$ and $\rho$ be the pullback of the $2$-dimensional standard representation of $D_4$ under $\tilde{G}\twoheadrightarrow G$. By properties of local Langlands for $\mathrm{GL}_2(\mathbb{Q}_2)$ (see below), there exists a character $\chi:\tilde{G}\rightarrow\mathbb{C}^{\times}$ such that $\rho\otimes\chi$ is trivial on $\tilde{G}^3$ (the upper numbering subgroup of $\tilde{G}=W_F$ with index $3$). What makes this non-trivial is that $\chi$ cannot factor through $G=\operatorname{Gal}(E/F)$, as $\rho$ is non-trivial on $(D_4,D_4)$. My question is
How do we find, or understand, such $\chi$?
Thank you for any comment!
Here is a proof of the existence of $\chi$ using local Langlands: suppose such $\chi$ does not exist. Then $\rho$ has minimal normalized level, or minimal depth (which is $3$), among its central twists. Note also that $\rho$ is "totally ramified," namely is not isomorphic to $\rho\otimes\chi_2$ where $\chi_2$ is the unramified quadratic character, because $\operatorname{ker}(\rho)\not\subset\operatorname{ker}(\chi_2)$.
By [BH06, Cor. 13.3, Thm. 14.5 and Lemma 20.3], since our $\rho$ is totally ramified and has minimal normalized level among central twists. The corresponding supercuspidal representation $\pi\in\mathrm{Irr}(\mathrm{GL}_2(\mathbb{Q}_2))$ has a ramified simple stratum, and in particular has normalized level = depth $\in\frac{1}{2}+\mathbb{Z}$. By [ABPS16, Prop. 4.2], $\pi$ and $\rho$ has the same depth, but $3\not\in\frac{1}{2}+\mathbb{Z}$, a contradiction.
Aubert, Anne-Marie; Baum, Paul; Plymen, Roger; Solleveld, Maarten, The local Langlands correspondence for inner forms of (\mathrm{SL}_{n}) [1], Res. Math. Sci. 3, Paper No. 32, 34 p. (2016). ZBL1394.22015 [2].
Bushnell, Colin J.; Henniart, Guy, The local Langlands conjecture for GL(2). Grundlehren der Mathematischen Wissenschaften 335. Berlin: Springer (ISBN 3-540-31486-5/hbk). xii, 347 p. (2006). ZBL1100.11041 [3].
Turns out that the answer is very simple as J.-K. Yu kindly reminds me. Such a character has to factor through $\tilde{G}_{ab}:=\tilde{G}/(\tilde{G},\tilde{G})\cong\mathbb{Q}_2^{\times}$. By the condition posed by the problem, we need $\chi$ to be trivial on $(\tilde{G}_{ab})^{3+}=1+16\mathbb{Z}_2$, non-trivial on $(\tilde{G}_{ab})^{3}=1+8\mathbb{Z}_2$, and we actually only care about its restriction to $\tilde{G}^{3}$. Essentially $\chi$ is the unique non-trivial character on $(1+8\mathbb{Z}_2)/(1+16\mathbb{Z}_2)$. Local langlands for $\mathrm{GL}_2(\mathbb{Q}_2)$ then guarantees that $\chi$ has the asserted property of my problem.
In elementary terms, what happens is the following: Let $F':=F(\mu_{16})$ be the splitting field of $x^{16}-1$ over $F=\mathbb{Q}_2$ and $E':=F'E$. We have $H':=\operatorname{Gal}(F'/F)\cong(\mathbb{Z}/16)^{\times}$ and $G':=\operatorname{Gal}(E'/F)$ has order $16$. What one may check (I do with Magma) is:
The upper numbering subgroup $(G')^3$ has order $2$.
The natural maps $(G')^3\rightarrow G^3$ and $(G')^3\rightarrow (H')^3$ have to be surjective thanks to properties of the upper numbering filtration. Let $\chi:H'\rightarrow\mathbb{C}^{\times}$ be any character that is non-trivial on $(H')^3=\langle 9+16\mathbb{Z}\rangle$. Then (the pullbacks to $G'$ of) both $\rho$ and $\chi$ are non-trivial on $(G')^3$, which forces $\rho\otimes\chi$ to be trivial on $(G')^3$, as desired.
