Configuration space of flags - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:20:09Z http://mathoverflow.net/feeds/question/76797 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76797/configuration-space-of-flags Configuration space of flags Xin Nie 2011-09-29T20:11:38Z 2011-10-06T09:09:52Z <p>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.</p> <p>Motivated by a paper of Fock and Goncharov (arXiv:math/0311149), I want to understand the configuration space of flags. </p> <p><strong>Question:</strong> 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? </p> <p>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?"</p> http://mathoverflow.net/questions/76797/configuration-space-of-flags/76808#76808 Answer by Michael Joyce for Configuration space of flags Michael Joyce 2011-09-29T21:49:37Z 2011-09-29T21:49:37Z <p>(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).</p> <p>You might want to look at <a href="http://arxiv.org/abs/math/9805067" rel="nofollow">Multiple Flag Varieties of Finite Type</a> 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.</p> http://mathoverflow.net/questions/76797/configuration-space-of-flags/77344#77344 Answer by Allen Knutson for Configuration space of flags Allen Knutson 2011-10-06T09:09:52Z 2011-10-06T09:09:52Z <p>There are many GIT quotients, since to define one requires a choice of $G$-line bundle, so a pair of naturals for each $F$. </p> <p>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.)</p> <p>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 ${\mathbb P}^1 \times F^{n-3}$.</p> <p>For more about these undemocratic GIT quotients, see the purple book [Guillemin-Lerman-Sternberg].</p>