Configuration space of flags Let $U\subset \mathbb{P}^1\times\mathbb{P}^1\times\mathbb{P}^1\times\mathbb{P}^1$ be the Zariksi open set of ordered quadruple of distinct points in the projective line. The quotient of $U$ by the projective transformation group $PSL(2)$ can be identified to $\mathbb{P}^1$ by cross-ratio.
Motivated by a paper of Fock and Goncharov (arXiv:math/0311149), I want to understand the configuration space of flags. 
Question: Let $\mathscr{F}$ be the variety of flags in $\mathbb{P}^2$.Let $n\geq 3$. Is there a Zariski open set $U\subset\mathscr{F}^{(n)}$ like above, consisting of $n$ flags "in general position" in some sense, such that the quotient of $PSL(3)$-action on $U$ is a projective variety? And what is the quotient? 
For those who know a little geometric invariant theory, I could have just asked "How to describle (semi-)stable points of the $PSL(3)$-action on  $\mathscr{F}^{(n)}$, and what is the quotient?"
 A: (This should be a comment but I don't have enough reputation to leave one.)  The quotient of $U$ by $PGL(2)$ is $\mathbb{P}^1 \backslash \{ 0, 1, \infty \}$.  To get something compact, you need to allow the four points to lie on a stable curve -- in this case, you need to allow the union of two $\mathbb{P}^1$'s with two of your points on each component (none of the points coinciding with the node of intersection).
You might want to look at Multiple Flag Varieties of Finite Type by Magyar, Weyman, and Zelevinsky.  It characterizes the conditions under which you get finitely many $PGL(m)$ orbits on products of (partial) flag varieties and may help give a sense of the complexities involved.
A: There are many GIT quotients, since to define one requires a choice of $G$-line bundle, so a pair of naturals for each $F$. 
There's an obvious democratic choice -- $(a,b) = (1,1)$ for every $F$ -- but I think this will in general lead to properly semistable reduction, which is a little icky. (Instead of just taking the quotient of an open set, further identifications are required.)
I will be lazy and put all the burden on the first three: $(2n,2n), (4n-1,3n), (n,1)$, and $(1,1)$ for each of the rest. I think the effect on the stable set is the following. The first two flags must be in general position w.r.t. one another, so we can use $PGL(3)$ to move them to the base and anti-base flag, respectively. This leaves over the diagonal matrices $T$ acting on $F^{n-2}$. The line in the third flag shouldn't be in the $xy$, $xz$, or $yz$ coordinate planes. That defines an open set in $F$ whose $T$-quotient turns out to be ${\mathbb P}^1$. So the final quotient space is ${\mathbb P}^1 \times F^{n-3}$.
For more about these undemocratic GIT quotients, see the purple book [Guillemin-Lerman-Sternberg].
