The cyclic Bruhat decomposition of flag manifolds

Transcription

1 The cyclic Bruhat decomposition of flag manifolds Allen Knutson (Cornell) Joint Meetings, Atlanta, January 207 Abstract If we intersect the n cyclic translates of the Bruhat decomposition of the Grassmannian Gr(k, n), we get the celebrated positroid stratification studied by Lusztig, Postnikov, Williams, Rietsch, Knutson-Lam-Speyer... It is also the stratification by projected Richardson varieties [KLS], and its most natural flagmanifoldversionisjust the Richardsonvarieties{X π σ}. Nonetheless, I ll look at the intersection of the n cyclic translates of the Bruhat decomposition, and index the strata with cyclic flag pipe dreams. Alas: unlike ongr(k,n), these strata canbeempty, orofbaddimension. They are determined by ( dim(f k C [i,j] ) ) where [i,j] varies over cyclic intervals; unfortunately their closures are not given by inequalities on those dimensions, and(relatedly) this decomposition is not a stratification. Taking [i, j] only from (non-cyclic) intervals, as was useful for Schubert calculus [K], I do get ness, smoothness, irreducibility, and dimension. These transparencies are available at

2 Arrays of dimension jumps, with pipes. IdentifyFl(n) := B \GL(n), using F k := span oftopkrowsofm GL(n). For eachi j i+n, consider columns [i,j] mod n ofm, andrecord J ij := {k [n] : rank(top k rowsincols[i,j]) > rank(top k rowsincols[i,j])} e.g. M = 0 a b 0 0 a b Theorem. J ij increasesbyone elementasyougo NorthorEast, i.e. {(i,j) : J ij k} is anorderidealabovesome k-pipe. These transparencies are available at

3 Tiles for CF pipe dreams (CF = cyclic flag ). Wecan thereforethink ofthese CFpipedreamsasbeingbuiltout oftiles: Sa Sab Sb Sab Sa Sab a<b S Sb S Sa S Sa It s easyto show the elbowstile (the third type) doesn toccur witha > b. The pipelabelsonthe verticaledgesareweakly increasing ineachrowofj: a a, b b, a b inthe three tiles. For J an n-periodic assemblage of these tiles into a CF pipe dream, letx(j) Fl(n) denotethe correspondinglocally closed subsetoffl(n). Basedonthe exampleofgr(k,n), Iwasmovedto Conjecture. X(J) issmoothand irreducible,withcodimensiongiven by the number of horizontal tiles, of the left type. (Which equals the number of vertical tiles, by the Jordan curve theorem.) But this turns out to be false, much like most conjectures about matroid strata! These transparencies are available at 2

4 Counterexamples: an empty stratum, and stratification failure Which a b c d e f g h ? The 2,,2, downthe i = j diagonaltellus a = c = 0, b,d 0. The 24 at [,3] says that det a b c e f = det 0 b 0 e f = b(eh fg) = 0. g h g h Butthe23at[3,5]saysthatdet c d a f e = d(fg eh) 0. Sotherearenone. h g The same phenomenon,b 0 = eh fg = 0, leadsto a stratum whose closure isnotaunionofstrata. Moral: When an inequality helps prove an equality, watch out! These transparencies are available at 3

5 A flag of positroids. The data definingj tells us whichpositroidstratum eachk-planef k isin, i.e. we get an n-tuple of bounded affine permutations. In this interpretation, the k-pipe says which dots move when going from the k-ball affine permutation to the (k + )-ball affine permutation Question. For which J is X(J) determined as a set by intersecting the flag manifold with the preimages of those positroid varieties? These transparencies are available at 4

6 Partial flag manifolds, and the loop amplituhedron. The same recipes work on Fl(n < n 2 <... < n m = n), except now we have n i n i many n i -pipes,and theymustn t crosseachother. In the case Gr(k,n), in row i Z the pipe labels on the n vertical edges go k,k,...,k,n,n,...,n. If we let π(i) := i + #ks in that row, we get the corresponding bounded affine permutation defining the positroid stratum. The l-loop amplituhedrona k,l isthe Gr(2,4) l -bundleovergr(k,k+4). SoA k, = Fl(k,k+2,k+4),andAk,l Fl(k,k+2,k+4) l,fromwhichitinherits a cyclic Bruhat decomposition. Now there are k pipes labeled V, and 2 pipes labeled each of,...,l, that can lie along one another (but not along V-pipes). Ifwe don tbotherdrawingthe remainingpipes, weget pipedreamslike this: M V = [ ] L = L 2 = [ ] [ ] V2 These transparencies are available at 5 V V2 V V22 V2 V V22 V22 V2 2 V22 V V22 V2

7 Interval rank flag strata and IF pipe dreams. If we only study intervals [i,j] [n] of columns, rather than cyclic intervals, we get a coarser decomposition into IF strata, indexed by triangular(rather than periodic) IF pipe dreams. It still is finer than the Richardson stratification. Associated to an IF pipe dream J are two permutations π and σ, from the lists ofpipescrossedacrossthe Northside and thendownthe Eastside. Theorems. LetJ be anifpipedream,and πand σ as above. X(J) isnonempty, smooth, and irreducible. X(J) X σ π codim ( X(J) X σ ) = #vertical tiles. codim ( X(J) X π ) = #horizontaltiles(equivalentto the previous). Not everything is great: the same counterexample still works to show that this coarser decomposition is not a stratification. David Speyer and I are trying to relate this decomposition to Deodhar s. On the Grassmannian, this was the stratification I used in arxiv: to extend Vakil s geometric Littlewood-Richardson rule to equivariant K-theory. On there, though, it was coarser than the projected Richardson stratification. These transparencies are available at 6

8 Towards a geometric L-R rule for IF pipe dreams. I defined the geometric shift α of X P\G as lim t exp(te α ) X, connecting a construction from Vakil with Erdős-Ko-Rado combinatorial shifting. Here,exp(te α ) meansaddingttimescolumnitocolumnj,andtakingthelimit; the rank conditions on column j thereby move backwards to column i. Vakil gave a list of shifts to apply to (initially Richardson, eventually Schubert) varieties in Gr(k,n). His list rasters the rows of the pipe dream bottom to top, and right to leftwithin rows;weindicate his{(i,j)} belowat yellowtiles. If the corners of a yellow tile have NW < SE, the shift switches those sets (and otherwise does nothing). The resulting array of subsets may be combinatorially illegal, reflectingthe geometrythat i j X(J) hasbecomereducible = >3 3 >3 2 2 = 2 < 3 3 > U 0 0 * = 2 This is the calculation [X 23 ][X 32 ] = [X 23 ]+[X 32 ] (don t forget the inverting!). The mainholdup: what does i j do to the equations-from-inequations? These transparencies are available at 7