TITLE: Explicit description of SU(2,2)/U
QUESTION [7 upvotes]: Consider the real diagonal $4\times 4$ - matrix
$$I_{2,2}={\rm diag}(1,1,-1,-1)$$ 
and  the corresponding special unitary group
$$ G={\rm SU}(2,2)=\{g\in {\rm SL}(4,{\mathbb{C}})\ |\ g\cdot I_{2,2}\cdot \bar g ^{\rm tr}=I_{2,2}\}.$$
We regard $G$ as an algebraic group over ${\mathbb{R}}$.
It is known that $G$ is quasi-split, that is, $G$ contains a Borel subgroup $B$ defined over ${\mathbb{R}}$.
Let $U$ denote the unipotent radical of $B$, it is defined over ${\mathbb{R}}$. 
Set $X=G/U$.

Question. How can one describe the homogeneous space  $X$ explicitly (by equations and "inequalities") as a quasi-projective variety over ${\mathbb{R}}$ on which ${\rm SU}(2,2)$ naturally acts?

EDIT: I explain why I need explicit equations for $G/U$. I want to twist the desired variety by the 1-cocycle $c=I_{2,2}\in G(\mathbb{R})_2$ and to obtain explicit equations for the twisted variety $Y={}_c X$. 
This new variety $Y$ over $\mathbb{R}$ is a homogeneous space of the twisted group $_cG={\rm SU}(4)$ for which $U$ is the stabilizer of a $\mathbb C$-point. 
Of course, $Y$ has no $\mathbb R$-points, because the stabilizer of an $\mathbb R$-point would be a compact form of $U$, but we know that the unipotent group $U$ has no compact forms.

REPLY [4 votes]: Although the question has been answered in comments (by Victor Petrov), I prefer to post an answer.
I assume that $G={\rm U}(2,2)$ rather than $G={\rm SU}(2,2)$.

My variety $G/U$ is the variety $X$ whose real points are  the triples
$$(W,w,b),$$
where $W\subset \mathbb C^4$ is an isotropic 2-dimensional subspace, $w\in W$ a nonzero vector (which is automatically isotropic),
and $b$ is a  nonzero element of $\Lambda^2W$.
This variety is a $R_{\mathbb C/\mathbb R}\mathbb G_{m,\mathbb C}^2$-torsor over the variety $\mathcal F$ of isotropic flags: the map is
$$X\to\mathcal F\colon\quad (W,w,b)\mapsto (W,\langle w\rangle),$$
and the action of $(\mathbb C^\times)^2$ on $X$ is
$$ (\lambda,\mu)*(W,w,b)=(W,\lambda w,\lambda\mu b)\quad \text{for } \lambda,\mu\in\mathbb C^\times.$$
By Witt's theorem for Hermitian forms, $G(\mathbb R)$ transitively acts on $X(\mathbb R)$, and my calculations show
that the stabilizer of the point  $$(\langle e_1,e_2\rangle, e_1, e_1\wedge e_2)\in X(\mathbb R)$$
is a maximal unipotent subgroup of $G$.
Thus $X\simeq G/U$.
The twisted form of $X$ is the same variety, but for another Hermitian form (and if the form is not hyperbolic, there is no such  triples $(W,w,b)$ over $\mathbb R$, as Victor has mentioned).
The real points of the  variety $\mathcal V$ of Victor's answer are pairs of non-proportional isotropic vectors
$(w_1,w_2)$ in $\mathbb C^4$.
This variety  is a $R_{\mathbb C/\mathbb R}\mathbb G_{a,\mathbb C}$-torsor over  $X$:  the map is
$$\mathcal V\to X\colon\quad (w_1,w_2)\mapsto
(\,\langle w_1,w_2\rangle,\, w_1,\, w_1\wedge w_2)$$
and the action of  $\mathbb C$ on $\mathcal V$ is
$$a*(w_1,w_2)=(w_1, w_2+aw_1)\quad\text{for } a\in\mathbb C.$$
The real points of the variety $\mathcal V'$ of Victor's comment is the set of pairs $(v,u)$, where $v$ is a nonzero isotropic vector in $\mathbb C^4$,
and $u$ is a nonzero vector in $v^\perp/\langle v \rangle$ that is isotropic with respect to the induced Hermitian form on $v^\perp$.
We construct a $G$-equivariant  isomorphism $$\varphi\colon\mathcal V'\to X.$$
Let $(v,u)\in \mathcal V'(\mathbb R)$. We lift $u$ to an isotropic vector $\tilde u\in \mathbb C^4$ and set
$$\varphi(v,u)=(\langle v, \tilde u\rangle, v, v\wedge\tilde u)\in X(\mathbb R).$$
In the opposite direction, if we have  $(W,w,b)\in X(\mathbb R)$, we choose $y\in W$ such that $b=w\wedge y$, and we set
$$\psi(W,w,b)=(w, y+\langle w\rangle)\in\mathcal V'(\mathbb R).$$
Since $\varphi$ and $\psi$ are mutually inverse, we see that $\varphi$ is an isomorphism.