For any field $K$, I denote by $\Gamma_K = \mathrm{Gal}(\overline K/K)$ its absolute Galois group. Let $p$ be a prime number and let $\mathbb Q_{p^2}$ denote a quadratic unramified extension of the field $\mathbb Q_p$ of $p$-adic numbers. Let $\sigma_0$ be the non-trivial element of the Galois group $\mathrm{Gal}(\mathbb Q_{p^2}/\mathbb Q_{p})$. Let $n$ be a positive integer, and define on the vector space $\mathbb Q_{p^2}^{n}$ the $\sigma$-hermitian form which, in the canonical basis, is represented by the following matrix
$$J = \begin{bmatrix}
& & & & 1\\
& & & -1 & \\
& & 1 & & \\
& \unicode{x22F0} & & & \\
(-1)^{n-1} & & & &
\end{bmatrix}$$
With such data, one can first define the associated unitary group denoted by $\mathrm U_n(\mathbb Q_p)$. It is the group consisting of all $\mathbb Q_{p^2}$-linear automorphisms of $\mathbb Q_{p^2}^n$ which preserve the hermitian form above. We can put it into a geometrico-algebraic setting, by recognizing that $U_n(\mathbb Q_p)$ is no other than the group of $\mathbb Q_p$-rational points of the algebraic (reductive) group $\mathbb G = \mathrm{GL}_n$ over $\overline{\mathbb Q_p}$, equipped with the following action of the absolute Galois group $\Gamma = \Gamma_{\mathbb Q_p}$ :
$$\forall \sigma \in \Gamma, \forall g \in \mathbb G, \quad\quad \sigma\cdot g :=
\left\{ \begin{matrix}
\sigma(g) & \text{ if } & g \in \Gamma_{\mathbb Q_{p^2}}\\
J\sigma(g)^{-T}J^{-1} & \text{ if } & g \in \Gamma\setminus \Gamma_{\mathbb Q_{p^2}}
\end{matrix} \right.$$
where $\sigma(g)$ is the matrix $g$ with $\sigma$ applied to all its coefficients, and where $g^{-T}$ is the inverse of the transpose of $g$.
In essence, it means that we have a natural identification (equality)
$$\mathrm U_n(\mathbb Q_p) \simeq \mathbb G(\mathbb Q_p) = \mathbb G(\overline{\mathbb Q_p})^{\Gamma} = \mathrm{GL}_n(\overline{\mathbb Q_p})^{\Gamma}$$
In the terminology of classical groups, I think that $\mathrm U_n(\mathbb Q_p)$ is called the $p$-adic quasi-split unitary group of dimension $n$. It's nothing but $\mathrm{GL}_n$ with a twisted $\mathbb Q_p$-structure.
Now, what if I want to replace the unitary group with the group of unitary similitudes $\mathrm{GU}_n(\mathbb Q_p)$ ? This is defined as the group of $\mathbb Q_{p^2}$-linear automorphisms of $\mathbb Q_{p^2}^{n}$ which preserve the hermitian form up to some unit scalar in $\mathbb Q_{p}^{\times}$. In a similar fashion as above, I would like to find a reductive group $\mathbb H$ over $\overline{\mathbb Q_p}$ with some Galois action, so that its group of $\mathbb Q_p$-rational points is naturally $\mathrm{GU}_n(\mathbb Q_p)$. Of course, just $\mathrm{GL}_n$ won't do as we need to keep track of the unit scalar somehow. Could $\mathrm{GL}_n \times \mathbb G_m$ do the work instead ?
Best Answer
Your guess is correct. Let $J$ be a hermitian matrix (yours is only if $n$ is odd) and write $K=\mathbb{Q}_{p^2}$. Then in fact $$GU_K\cong GL_{n,K}\times \mathbb{G}_{m,K} $$ and we have the short exact sequence $$1\to GL_{n,K}\to GL_{n,K}\times \mathbb{G}_{m,K}\to \mathbb{G}_{m,K}\to 1$$ which is the base change to $K$ of the short exact sequence (over $\mathbb{Q}_p$) $$1\to U\to GU\to \mathbb{G}_{m,\mathbb{Q}_p}\to 1.$$ The above isomorphism is given as follows: Let $R$ be a $K$-algebra, then $R\otimes_{\mathbb{Q}_p} K\cong R\oplus R$ and thus $$ GU(R)=\{(g,h)\in GL_n(R\oplus R)\mid (g,h)(J,J^t)(h^t,g^t)=\lambda\cdot(J,J^t)\}.$$ Writing out these conditions, it means you need $$ gJh^t=\lambda\cdot J$$ and $$ hJ^tg^t=\lambda\cdot J^t.$$ But then you can choose any $g\in GL_n(R)$ and set $h=\lambda J^tg^{-t}J^{-t}$ . You can see here for more details.