$U(2,1)$ is quasi-split Let $$w = \begin{pmatrix} & & 1 \\ & -1 & \\ 1 & & \end{pmatrix}$$
Let $k = \mathbb{R}, K = \mathbb{C}$.  Let $G = \textrm{GL}_3$ as a group over $K$.  How can I show the existence of an algebraic group over $\mathbb{R}$ whose $\mathbb{R}$-points are
$$\{ x \in G(K) : w \space ^t \overline{x}^{-1}w = x \}$$
where the bar denotes entrywise complex conjugation?  How can I show the group is quasi-split, and compute a maximal split and anisotropic torus?  
 A: Let $\Gamma = \textrm{Gal}(K/k)$ with nontrivial element $\sigma$.  We have that $\Gamma$ acts on $\textrm{Aut}(G)$ by $\gamma.\phi(x) = \gamma \circ \phi \circ \gamma^{-1}$.  Define a map $c: \Gamma \rightarrow \textrm{Aut}(G)$ by setting $c(\gamma)(x) = w \space ^t x^{-1} w$.  Then $c$ is easily seen to be a one-cocycle, which is to say 
$$ c(\gamma_1 \gamma_2) = c(\gamma_1) \gamma_1. c(\gamma_2)$$
for all $\gamma_i \in \Gamma$.  General theory of forms tells us that there exists an algebraic group $H$ over $k$, such that $H \times_k \overline{k} = G$, and the difference between action of $\Gamma$ on $G$ coming from $H$, and the usual action of $\Gamma$ on $G$, produces $c$.  In other words,
$$c(\sigma)(x) = \overline{\sigma.x}$$
for all $x \in G$ and $\gamma \in \Gamma$.  Thus as an algebraic group coming from $H$, $\Gamma$ acts on $G$ by $\sigma.x = w \space ^t \overline{x}^{-1}w$.   This group $H$ is called $U(2,1)$.
$U(2,1)$ is quasi-split:
Let $B, T$ be the usual Borel subgroup and maximal torus of $G$.  To say that they are defined over $\mathbb{R}$ (for the algebraic group coming from $U(2,1)$) is to say that they are stabilized by $\Gamma$.  One checks that 
$$\sigma. \begin{pmatrix} x \\ & y \\ & & z \end{pmatrix} = \begin{pmatrix} \overline{z}^{-1} \\ & \overline{y}^{-1} \\ & & \overline{x}^{-1} \end{pmatrix}$$
and so $T$ is defined over $\mathbb{R}$, with $$T(\mathbb{R}) = \begin{pmatrix} re^{i \theta} \\ & e^{i \phi} \\ & & \frac{1}{r} e^{i \theta} \end{pmatrix}$$
Similarly, $B$ is defined over $\mathbb{R}$.  Its unipotent radical $U$ satisfies
$$U(\mathbb{R}) = \{ \begin{pmatrix} 1 & b & c \\ & 1 & \overline{b} \\ & & 1 \end{pmatrix} : |b|^2 = 2 \textrm{Re}(c) \}$$
Maximal split and anisotropic tori:
Let $e_1, e_2, e_3$ be the usual basis for $X(T)$.  Then $\Gamma$ acts as a group of automorphisms of $X(T)$ by $\gamma.\chi = \gamma \circ \chi \circ \gamma^{-1}$.  So it is easy to see that $\sigma.e_1 = -e_3, \sigma.e_3 = -e_1$, and $\sigma.e_2 = -e_2$.  
Define
$$X^{\Gamma} = \{ \chi \in X(T) : \sigma.\chi = \chi \}$$
which has basis $e_1 - e_3$, and 
$$X_0 = \{ \chi \in X(T) : \chi + \sigma.\chi = 0\}$$
which has basis $e_1 + e_3, e_2$.  Then the maximal split torus of $T$ is
$$A_0 = \bigcap\limits_{\chi \in X_0} \textrm{Ker } \chi = \begin{pmatrix} x \\ & 1 \\ & & x^{-1} \end{pmatrix}$$
and the maximal anisotropic torus is
$$T_0 =  \bigcap\limits_{\chi \in X^{\Gamma}} \textrm{Ker } \chi = \begin{pmatrix} 1 \\ & x \\ & & 1 \end{pmatrix} $$
Relative roots:
Let $\tilde{\Delta} = \{e_1 - e_2, e_2 - e_3\}$ be the base corresponding to $B$.  Let $\delta$ be the basis of $X(A_0)$ given by restriction of $e_1$ (or $-e_3$) to $A_0$.  Then the restriction of $\tilde{\Delta}$ to $X(A_0)$ is $\{\delta\}$.  In particular, no elements of $\tilde{\Delta}$ restrict to zero, so this gives another argument for why $U(2,1)$ is quasi-split.
Moreover, the root $e_1 - e_3$ restricts to $2 \delta$.  Thus $\Phi(A_0,G) = \{ \pm \delta, \pm 2 \delta \}$, and so the root system for $U(2,1)(\mathbb{R})$ is nonreduced.
