Finer rook equivalence: Classifying Dings Schubert varieties Mike - PDF document

Finer rook equivalence: Classifying Ding’s Schubert varieties Mike Develin (AIM) Jeremy Martin (University of Minnesota) Victor Reiner (University of Minnesota Preprint: arXiv:math.AG/0403530 math.umn.edu/ ∼ martin/math/pubs.html

Rook theory Let λ = (0 < λ 1 ≤ λ 2 ≤ · · · ≤ λ n ) be a partition. A k -rook placement on λ consists of k squares of the Ferrers Defn diagram (or “Ferrers board”) of λ , no two in the same row or column. λ = (4 , 4 , 6 , 6 , 8 , 9) R k ( λ ) = number of k -rook placements on λ Defn λ, µ are rook-equivalent iff R k ( λ ) = R k ( µ ) ∀ k . Defn λ = µ = Example R 1 ( λ ) = R 1 ( µ ) = 4 R 2 ( λ ) = R 2 ( µ ) = 2 R k ( λ ) = R k ( µ ) = 0 for k > 2

Rook equivalence (Foata–Sch¨ utzenberger 1970) Theorem Each rook-equivalence class contains a unique partition with distinct parts. (Goldman–Joichi–White 1975) Theorem Two partitions λ = (0 < λ 1 ≤ · · · ≤ λ n ) µ = (0 < µ 1 ≤ · · · ≤ µ n ) are rook-equivalent iff { λ i − i } n i =1 = { µ i − i } n i =1 as multisets. GJW ( λ ) = { 0 , 1 , 1 , 2 } Example 2 1 1 1 2 2 1 1 0 0 0 1

q -counting maximal rook placements Enumerate rook placements by an “inversion” statistic (generalizing inversions of permutations): � q inv( σ ) R k ( λ, q ) = k -rook placements σ (Garsia–Remmel 1986) Theorem (1) λ, µ are rook-equivalent iff they are q -rook equivalent. If λ = ( λ 1 ≤ · · · ≤ λ n ), then up to a factor of q , (2) n � R n ( λ, q ) = [ λ i − i + 1] q i =1 = 1 + q + q 2 + · · · + q m − 1 . q m − 1 where [ m ] q = q − 1 Observations (1) If λ i < i for some i (that is, λ does not contain a staircase), then R n ( λ, q ) = 0. (2) If λ n = n , then λ is rook-equivalent to ( λ 1 , . . . , λ n − 1 ).

Ding’s Schubert varieties • λ = ( λ 1 ≤ · · · ≤ λ n = m ), λ i ≥ i ( λ contains a staircase) C 0 ⊂ C 1 ⊂ · · · ⊂ C m : • standard flag � flags 0 ⊂ V 1 ⊂ V 2 ⊂ · · · ⊂ V n ⊂ C m : � X λ = . Defn ∀ i : dim C V i = i, V i ⊂ C λ i • X λ is a Schubert variety X w in a type-A partial flag manifold Y w = 43521 ∈ S 5 Example λ = (4 , 4 , 5 , 5 , 5) 1 2 3 4 5 5 4 3 2 1 • w is 312-avoiding; in particular X w is smooth • [ X w ] ∈ H ∗ ( Y ) is a Schubert polynomial indexed by the dominant permutation w 0 ww 0

The cohomology ring of X λ R λ := H ∗ ( X λ ; Z ) = � H 2 i ( X λ ; Z ) Defn i (because X λ has no torsion or odd-dimensional cohomology) (Ding) Theorem q i rank Z H 2 i ( X λ ) � = R n ( λ, q ) . i (Gasharov–Reiner) Theorem ∼ H ∗ ( X λ ) = Z [ x 1 , . . . , x n ] /I λ where I λ = � h λ i − i +1 ( x 1 , . . . , x i ) : 1 ≤ i ≤ n � . If λ i < i for some i (that is, λ does not contain a Observation staircase), then X λ = ∅ .

Trivial isomorphisms among the X λ ’s Suppose that λ i = i for some i : Observation 0 0 1 1 0 2 1 0 2 1 X λ = { V • : V 1 ⊂ V 2 ⊂ V 3 = C 3 ⊂ V 4 ⊂ C 5 } ∼ = Fl 3 × Fl 2 X µ = { V • : V 1 ⊂ V 2 = C 2 ⊂ V 3 ⊂ V 4 ⊂ C 5 } ∼ = Fl 2 × Fl 3 R λ = Z [ x 1 , . . . , x 5 ] / � h 3 (1) , h 2 (2) , h 1 (3) , h 2 (4) , h 1 (5) � = Z [ x 1 , x 2 , x 3 ] / � e 1 , e 2 , e 3 � ⊗ Z Z [ x 4 , x 5 ] / � e 4 , e 5 � R µ = Z [ x 1 , x 2 ] / � e 1 , e 2 � ⊗ Z Z [ x 3 , x 4 , x 5 ] / � e 3 , e 4 , e 5 � In general, R λ ∼ R λ ( j ) X λ ∼ � � X λ ( j ) , = = j j where λ ( j ) are the indecomposable components of λ .

Fine rook equivalence 1 2 0 1 0 1 2 2 0 1 2 0 0 1 0 1 2 0 1 2 0 0 1 2 2

Rook equivalence is not enough λ = (2 , 2 , 4) µ = (2 , 3 , 3) 1 0 0 1 1 1 R λ ∼ R µ ∼ x 2 , y 2 � s 2 , st + t 2 � � � = Z [ x, y ] / = Z [ s, t ] / λ and µ are rook-equivalent, and both cohomology rings have Poincar´ e series 1 + 2 q + q 2 . But consider 1 : f 2 = 0 } { primitive f ∈ R λ { x, y } , = 1 : f 2 = 0 } { primitive f ∈ R µ { s, s + 2 t } . = The former is a Z -basis for H 1 ( X λ ), while the latter is not a Z -basis for H 1 ( X µ ). Therefore X λ �∼ = X µ . In fact, R λ ∼ = Z [ x ] / � x � ⊗ Z [ y ] / � y � , while R µ does not decompose as a tensor product of smaller rings.

The main classification theorem (D–M–R) For partitions λ and µ with indecomposable Theorem components λ (1) , . . . , λ ( r ) , µ (1) , . . . , µ ( s ) , the following are equivalent: The multisets { λ ( i ) } r i =1 and { µ ( i ) } s (1) i =1 are identical. X λ ∼ (2) = X µ as algebraic varieties. H ∗ ( X λ ; Z ) ∼ (3) = H ∗ ( X µ ; Z ) as graded rings. (1) = ⇒ (2): Follows from trivial isomorphisms. (2) = ⇒ (3): Immediate. • ⇒ (1). The hard part is (3) =

Overview of the proof Main idea: In order to recover λ 1 , . . . , λ n from the structure of R λ = H ∗ ( X λ ) as a graded Z -algebra . . . . . . study nilpotence orders of linear forms. The nilpotence order of a homogeneous element f ∈ R λ is Defn nilpo( f ) = min { n ∈ N : f n = 0 } . Proposition If λ is indecomposable, then nilpo( f ) : f ∈ R λ � � min = λ 1 . 1 R λ / � x 1 � ∼ = R µ , where µ is the partition obtained Proposition by “peeling off” the leftmost column and bottom row of λ : → So we can just read off λ from the structure of R λ by taking successive quotients by linear forms of appropriate nilpotence order, right? Well. . .

Good and bad nilpotents Identify a λ 1 -nilpotent linear form f with Problem H ∗ ( X λ ) / � f � ∼ = H ∗ ( X λ ) / � x 1 � (for instance, f = x 1 ), independently of the presentation H ∗ ( X λ ) ∼ = R λ /I λ . For λ indecomposable and Theorem k = λ 1 = λ 2 = · · · = λ m < λ m +1 , the λ 1 -nilpotents in R λ 1 are exactly the following: x 1 , x 2 , . . . , x m (in all cases) x 1 + . . . + x m (iff m = k − 1) x 1 + . . . + x m + 2 x m +1 (iff m = k − 1, λ k = k + 1, and k is even) • The “good” nilpotents x 1 , . . . , x m can be distinguished intrinsi- cally from the “bad” ones. Necessary to show that R λ has a unique maximal tensor product • decomposition into the R λ ( i ) ’s. (This is probably not true for standard graded Z -algebras in general!)

λ 1 -nilpotents in R λ Partitions λ 1 k = 4, m = 2 x 1 , x 2 , x 3 k = 4, m = 3 x 1 , x 2 , x 3 , x 1 + x 2 + x 3 k = 4, m = 3, λ 4 = 5 x 1 , x 2 , x 3 , x 1 + x 2 + x 3 , x 1 + x 2 + x 3 + 2 x 4

Gr¨ obner bases, cores and stickiness → X λ and R λ ։ R µ . If µ ⊂ λ , then X µ ֒ Fact ⊂ ⊂ (4 , 4 , 4 , 5 , 6 , 7) λ = (4 , 4 , 6 , 6 , 7 , 8) (8 , 8 , 8 , 8 , 8 , 8) core of λ rectangle If you want to prove that f = 0 in R λ . . . • . . . replace λ with a larger rectangle. If you want to prove that f � = 0 in R λ . . . • . . . replace λ with its core. If λ is indecomposable and its own core, then the Proposition generators of I λ can be manipulated to produce a Gr¨ obner basis in which the variables x λ 1 , . . . , x n are “sticky”. I.e., if λ 1 ≤ j ≤ n and f ∈ R λ involves x j , then all partial Gr¨ obner reductions of f involve x j .

Questions for further study 1. Poset rook equivalence When are two rook-placement posets RP λ , RP µ isomorphic? • Strictly stronger than rook equivalence Strictly weaker than X λ ∼ • = X µ 2. Nilpotence and the Schubert variety • What do all these (Gr¨ obner) calculations say about the (enumer- ative) geometry of X λ ? • Nilpotence ⇐ ⇒ self-intersection numbers? 3. Other Schubert varieties • Find a presentation for H ∗ ( X w ; Z ), where X w ⊂ GL n /B • Can these be used to classify arbitrary X w up to isomorphism?