Confusion with self-dual representations of $\mathrm{GL}_n$ over a $p$-adic field

Question

The following surely is kind of a trivial question, but it keeps me confused. It concerns a detail in Lust and Stevens' paper "On depth zero L-packets for classical groups" London Math. Soc. (3) 121 (2020), more precisely about the discussion in the first paragraphs of the 3rd part on depth zero supercuspidal representations. Let me give some more context.

Here, $F$ is an extension of degree at most $2$ of a $p$-adic field $F_0$ with $p$ odd, and $\sigma$ is the generator of the Galois group $\mathrm{Gal}(F/F_0)$. We use the letters $\mathfrak o$ and $k$ respectively for the rings of integers and residue fields. For $n\geq 1$, a smooth (complex) representation $\rho$ of $G = \mathrm{GL}_n(F)$ is said to be self-dual if we have an equivalence $\rho^{\sigma} \simeq \rho$ where $\rho^{\sigma}$ is defined by the formula $$\rho^{\sigma}(g) = \rho\left(\sigma(g^{-1})^T\right)$$ So far so good. Now, assume that $\rho$ is irreducible supercuspidal of depth zero. Thus, there is a well-determined (equivalence class of) representation $\tau$ of the maximal compact subgroup $K = \mathrm{GL}_n(\mathfrak o_F)$ such that, on the one hand $\tau$ is the inflation of a cuspidal irreducible representation of the finite group of Lie type $K/K^+ \simeq \mathrm{GL}_n(k_F)$, and on the other hand one may write $\rho \simeq \mathrm{c-Ind}_{F^{\times}K}^{G}\tilde{\tau}$ where $\tilde{\tau}$ is some extension of $\tau$ to a representation of the normalizer $N_{G}(K) = F^{\times}K$. Here, $F^{\times}$ is identified with the center of $G$. Such an extension $\tilde{\tau}$ is determined by the choice of a character $\chi$ of the center $F^{\times}$ which coincide with the central character of $\tau$ on $F^{\times}\cap K = \mathfrak{o}_F^{\times}$.
Then, it is claimed that $\rho$ is self-dual if and only if $\tau$, as a representation of $\mathrm{GL}_n(k_F)$, is self-dual (that is, it satisfies the same equation as above but with $\sigma$ denoting the generator of $\mathrm{Gal}(k_F/k_{F_0})$).

Alright, at this stage I wanted to check this in the special case $n=1$, and both $\rho$ and $\tau$ are identified with linear characters. I take $\tau$ as the trivial character of $K = \mathfrak o_F^{\times}$. Since the trivial representation of $\mathrm{GL}_1(k_F) = k_F^{\times}$ is cuspidal and because $\tau$ clearly is self-dual, I am in a special case of the situation described above. Thus, if I choose any character $\chi$ of the center $F^{\times} = G$ (the group $G$ is abelian) which is trivial on $\mathfrak o_F^{\times}$, then $\rho = \chi$ should be a self-dual (supercuspidal) character of $G$ - the compact induction being trivial in this case.

But now here's the thing. Any such $\chi$ is called an unramified character of $F^{\times}$, and it doesn't look like they are self-dual at all. Indeed, any such $\chi$ may be written $\chi = |\,\cdot\,|_F^{\alpha}$ for some complex number $\alpha$, where $|\,\cdot\,|$ is the normalized norm on $F$. Any two unramified characters are equivalent if and only if they share they have the same $\alpha$ (they are equal).
Then, since $\sigma$ acts on $F$ as an isometry, for $x\in F^{\times}$ we have $$\chi^{\sigma}(x) = \chi\left(\sigma(x^{-1})\right) = |\, \sigma(x^{-1})\,|_F^{\alpha} = |\, x \,|_F^{-\alpha}$$ So, $\chi$ seems to be self-dual if and only if $\alpha = - \alpha$, that is $\chi$ is trivial... Surely, something is wrong there.

I assume that I have misunderstood something somewhere, but I can't find what exactly. Could somebody point it out to me ?

Answer 1

You are right, there is an error in Lust-Stevens, and any unramified character provides a counter-example. Here is a way to fix their statement: $\rho=\mathrm{cInd}_{KF^\times}^G\widetilde\tau$ is self-dual if and only if $\tau=\widetilde\tau|_K$ is self-dual and the central character $\omega_{\widetilde\tau}\colon F^\times\to\mathbb C^\times$ of $\widetilde\tau$ is such that $\omega_{\widetilde\tau}(z\sigma(z))=1$ for all $z\in F^\times$ (when $F=F_0$, this means $\omega_{\widetilde\tau}^2=1$).

Indeed, $\rho$ being self-dual means $\mathrm{cInd}_{KF^\times}^F\widetilde\tau\cong\mathrm{cInd}_{KF^\times}^F\widetilde\tau^\sigma$, which is equivalent to $\widetilde\tau\cong\widetilde\tau^\sigma$ as representations of $KF^\times$.

Now the claim above follows from the observation that an irreducible representation $\widetilde\tau$ is uniquely determined by its central character $\omega_{\widetilde\tau}$, together with its restriction to $K$.