Order two unramified character of $\mathrm{SO}_{2n+1}(F)$ Let $F$ be a $p$-adic field. Then $\mathrm{SO}_{2n+1}(F)$ has an order-two unramified character "corresponding to" the nontrivial central element of its dual group $\mathrm{Sp}_{2n}(\mathbb{C})$. What is this character explicitly, say when $n=2$? I thought I understood the relationship between the real Lie algebra and unramified characters, but I guess I still don't.
For example, when $n=1$, the character I am confused about must be the character of $\mathrm{PGL}_2(F)\simeq \mathrm{SO}_3(F)$ given by composing the determinant with the nontrivial order two unramified character.
 A: Unless I have made a mistake (which is possible, so please check it), I can explain how to 'canonically' construct this order $2$ unramified character. Of course, 'canonical' and 'explicit' are not the same thing, and I still think it would be nice if someone could point out more explicitly what this character is (EDIT: see the edit at the bottom of the post). Maybe this will help you (or someone else) find such a description. For simplicitly, let me assume that the residue characteristic of our fields is greater than $2$.

Proposition: Let $F$ be a finite extension of $\mathbb{Q}_p$, and let $G$ be a semisimple algebraic group over $F$. Then, there is a
canonical isomorphism
$$\left\{\text{continuous }\chi \colon G(F)\to
 \mathbf{C}^\times\right\} \xrightarrow{\approx} \left\{\nu\colon
 H^1(F,\pi_1(G))\to \mathbf{C}^\times\right\}.$$

Let me clarify the notation here. Let $G^\mathrm{sc}$ be the universal cover of $G$, and $\pi_1(G)=\ker(G^\mathrm{sc}\to G)$ the fundamental group of $G$, considered as a finite abelian group $F$-scheme. As $\pi_1(G)$ is abelian the Galois cohomology group $H^1(F,\pi_1(G))$ is an abelian group, and thus it makes sense to consider characters of $H^1(F,\pi_1(G))$.
To prove this claim, let us first observe that every continuous character $G^\mathrm{sc}(F)\to\mathbf{C}^\times$ is trivial. The reason for this is that by Chevalley's theorem (see Theorem 10.3.4 of this) ${G}^\mathrm{sc}(F)$ is generated by subgroups isomorphic to $\mathrm{SL}_2(F)$. This reduces us to proving the claim for $G=\mathrm{SL}_{2,F}$ which can be easily done by hand as $\mathrm{SL}_2(F)^\mathrm{ab}$ is trivial.
From this, we see that for any continuous character $\chi\colon G(F)\to \mathbf{C}^\times$, the composition
$$G^\mathrm{sc}(F)\to G(F)\xrightarrow{\chi}\mathbf{C}^\times$$
is trivial. That said, from the short exact sequence of group $F$-schemes
$$1\to \pi_1(G)\to G^\mathrm{sc}\to G\to 1,$$
(where note that $\pi_1(G)$ is central in $G^\mathrm{sc}$) we get the exact sequence of groups
$$1\to \pi_1(G)(F)\to G^\mathrm{sc}(F)\to G(F)\to H^1(F,\pi_1(G))\to 1.$$
Here I have used Kneser's theorem that $H^1(F,X)$ vanishes for any simply connected group $X$. From this we see that we have a canonical isomorphism
$$G(F)/G^\mathrm{sc}(F)\xrightarrow{\approx}H^1(F,\pi_1(G)).$$
As this latter group is abelian, the proposition quickly follows.
Now, for the case $G=\mathrm{SO}_{2n+1,F}$ we can explicitly compute things. Namely, $G^\mathrm{sc}=\mathrm{Spin}_{2n+1,F}$ and $\pi_1(G)=\mu_{2,F}$. Thus, from the above proposition we have a bijection
$$\{\text{continuous }\chi\colon\mathrm{SO}_{2n+1}(F)\to\mathbf{C}^\times\}\xrightarrow{\approx}\{\nu\colon F^\times/(F^\times)^2\to\mathbf{C}^\times\},$$
where I have applied Kummer theory to form the canonical isomorphism between $H^1(F,\mu_{2,F})$ and $F^\times/(F^\times)^2$.
I claim that there is a unique order 2 unramified character of $\mathrm{SO}_{2n+1}(F)$ and that it corresponds, under this bijection, to the unique order $2$ unramifed character of $F^\times$, which matches your example for $n=1$. Indeed, let us observe that we have a commuting diagram
$$\begin{matrix}\mathrm{SO}_{2n+1}(\mathcal{O}_F) & \to & H^1(\mathcal{O}_F,\mu_{2,\mathcal{O}_F}) & \to & 1\\ \downarrow & & \downarrow & &\\ \mathrm{SO}_{2n+1}(F) & \to & H^1(F,\mu_{2,F}) & \to & 1\end{matrix}$$
where the vertical maps are injections, and the top row is exact. To justify this exactness it suffices to note that $H^1(\mathcal{O}_F,\mathrm{Spin}_{2n+1,\mathcal{O}_F})=0$ by Lang's theorem and the fact that $H^1$ of a reductive group scheme is insensitive to passing to the closed point of a Henselian local ring.
After identifying $H^1(\mathcal{O}_F,\mu_{2,\mathcal{O}_F})=\mathcal{O}_F^\times/(\mathcal{O}_F^\times)^2$ we deduce from this that under the bijection of the proposition, $\chi\colon \mathrm{SO}_{2n+1}(F)\to \mathbf{C}^\times$ is unramified if and only if $\nu\colon F^\times/(F^\times)^2\to \mathbf{C}^\times$ kills $\mathcal{O}_F^\times$ as desired.
To identify this character more explicitly, or to identify the characters of $\mathrm{SO}_{2n+1}(F)$ in general, all you need from the above analysis is a character $\chi\colon G(F)\to \mathbf{C}^\times$ such that
$$\ker\chi=\mathrm{im}(\mathrm{Spin}_{2n+1}(F)\to \mathrm{SO}_{2n+1}(F)).$$
For $n=1$ you apparently found this by taking the character
$$\omega\circ \det\colon \mathrm{PGL}_2(F)\to \mathbf{C}^\times$$
where $\omega$ is the unique order $2$ unramified character of $F^\times$, and transferring it to $\mathrm{SO}_3(F)$. Note that this character is well-defined as $\omega(\det(\alpha I_2))=1$. This works because $\mathrm{SL}_{2,F}=(\mathrm{PGL}_{2,F})^\mathrm{sc}$ and $\ker(\omega\circ \det)=\mathrm{im}(\mathrm{SL}_2(F)\to\mathrm{PGL}_2(F))$.
EDIT: Despite the above being purely abstract, it turns out that the map $\mathrm{SO}_{2n+1}(F)\to F^\times/(F^\times)^2$ has a name, and a more explicit description! It's the so-called 'spinor norm'. See Section 5.1 here, and here. So, if we denote this spinor norm by $\varsigma$, the above discussion shows that the characters of $\mathrm{SO}_{2n+1}(F)$ are precisely those of the form $\chi\circ\varsigma$ where $\chi$ is an order $2$ character of $F^\times$, and that this unramified precisely when $\chi$ is.
