Pure-cycle Hurwitz factorizations and multi-noded rooted trees

Transcription

1 Pure-cycle Hurwitz factorizations and multi-noded rooted trees by Rosena Ruoxia Du East China Normal University Combinatorics Seminar, SJTU August 29, 2013 This is joint work with Fu Liu. 1

2 PART I: Definitions and Backgrounds 2

3 Hurwitz s problem Definition 1. Given integersdandr, andr partitionsλ 1,...,λ r d, a Hurwitz factorization of type (d,r,(λ 1,...,λ r )) is an r-tuple (σ 1,...,σ r ) satisfying the following conditions: (i) σ i S d has cycle type (or is in the conjugacy class)λ i, for everyi; (ii) σ 1 σ r = 1; (iii) σ 1,...,σ r generate a transitive subgroup ofs d. 3

4 Hurwitz s problem Definition 1. Given integersdandr, andr partitionsλ 1,...,λ r d, a Hurwitz factorization of type (d,r,(λ 1,...,λ r )) is an r-tuple (σ 1,...,σ r ) satisfying the following conditions: (i) σ i S d has cycle type (or is in the conjugacy class)λ i, for everyi; (ii) σ 1 σ r = 1; (iii) σ 1,...,σ r generate a transitive subgroup ofs d. Definition 2. The Hurwitz numberh(d,r,(λ 1,...,λ r )) is the number of Hurwitz factorizations of type(d,r,(λ 1,...,λ r )) divided byd!. 3

5 Hurwitz s problem Definition 1. Given integersdandr, andr partitionsλ 1,...,λ r d, a Hurwitz factorization of type (d,r,(λ 1,...,λ r )) is an r-tuple (σ 1,...,σ r ) satisfying the following conditions: (i) σ i S d has cycle type (or is in the conjugacy class)λ i, for everyi; (ii) σ 1 σ r = 1; (iii) σ 1,...,σ r generate a transitive subgroup ofs d. Definition 2. The Hurwitz numberh(d,r,(λ 1,...,λ r )) is the number of Hurwitz factorizations of type(d,r,(λ 1,...,λ r )) divided byd!. Question: What is the Hurwitz numberh(d,r,(λ 1,...,λ r ))? This question originally arises from geometry: Hurwitz number counts the number of degree-d covers of the projective line with r branch points where the monodromy over theith branch point has cycle typeλ i. 3

6 The pure-cycle case A number of people (Hurwitz, Goulden, Jackson, Vakil...) have studied Hurwitz numbers. However, they often restricted their attention to the case where all but one or twoσ i s are transpositions. 4

7 The pure-cycle case A number of people (Hurwitz, Goulden, Jackson, Vakil...) have studied Hurwitz numbers. However, they often restricted their attention to the case where all but one or twoσ i s are transpositions. We consider instead the pure-cycle case. This means each λ i has the form (e i,1,..., 1), for some e i 2, or equivalently, each σ i is an e i cycle. In this case, we use the notationh(d,r,(e 1,...,e r )) for the Hurwitz number. We also focus on the genus-0 case, which simply means that r 2d 2 = (e i 1). i=1 4

8 The pure-cycle case A number of people (Hurwitz, Goulden, Jackson, Vakil...) have studied Hurwitz numbers. However, they often restricted their attention to the case where all but one or twoσ i s are transpositions. We consider instead the pure-cycle case. This means each λ i has the form (e i,1,..., 1), for some e i 2, or equivalently, each σ i is an e i cycle. In this case, we use the notationh(d,r,(e 1,...,e r )) for the Hurwitz number. We also focus on the genus-0 case, which simply means that 2d 2 = r (e i 1). Example 3. Letd = 5,r = 4,(e 1,e 2,e 3,e 4 ) = (2,2,3,5). One can check that i=1 ((23),(45),(135),(54321)) is a genus-0 pure-cycle Hurwitz factorization. (Genus-0: 2d 2 = 8 = 4 i=1 (e i 1) = ) 4

9 Previous results on the pure-cycle case Lemma 4 (Liu-Osserman). In the genus-0 pure-cycle case, when r = 3, h(d,3,(e 1,e 2,e 3 )) = 1. 5

10 Previous results on the pure-cycle case Lemma 4 (Liu-Osserman). In the genus-0 pure-cycle case, when r = 3, h(d,3,(e 1,e 2,e 3 )) = 1. Theorem 5 (Liu-Osserman). In the genus-0 pure-cycle case, when r = 4, h(d,4,(e 1,e 2,e 3,e 4 )) = min{e i (d+1 e i )} 5

11 Hurwitz factorizations with a d-cycle We study a special case of genus-0 pure-cycle Hurwitz factorizations: when one of thee i isd. W.L.O.G, we assumee r = d. Then the genus-0 condition becomes: 2d 2 = r (e i 1) i=1 r 1 i=1 (e i 1) = d 1. Sinceσ r is ad-cycle, σ 1,...,σ r is automatically transitive ins d. Moreover, σ 1...σ r = 1 σ 1...σ r 1 = σ 1 r. 6

12 Factorizations of a d-cycle 7

13 Factorizations of a d-cycle Definition 6. Assumed,r 1,e 1,...,e r 1 2 are integers satisfying r 1 i=1 (e i 1) = d 1. Fix a d-cycle τ S d, We say (σ 1,...,σ r 1 ) is a factorization of τ of type(e 1,...,e r 1 ) if the followings are satisfied: i. For eachi,σ i is ane i -cycle ins d. ii. σ 1 σ r 1 = τ. 7

14 Factorizations of a d-cycle Definition 6. Assumed,r 1,e 1,...,e r 1 2 are integers satisfying r 1 i=1 (e i 1) = d 1. Fix a d-cycle τ S d, We say (σ 1,...,σ r 1 ) is a factorization of τ of type(e 1,...,e r 1 ) if the followings are satisfied: i. For eachi,σ i is ane i -cycle ins d. ii. σ 1 σ r 1 = τ. Example 7. Letd = 5,r = 4,(e 1,e 2,e 3 ) = (2,2,3),τ = (12345),σ 1 = (23), σ 2 = (45), σ 3 = (135). It is easy to check that (σ 1,σ 2,σ 3 ) is a factorization of τ of type(2,2,3): (23)(45)(135) = (12345). 7

15 Factorizations of a d-cycle Definition 6. Assumed,r 1,e 1,...,e r 1 2 are integers satisfying r 1 i=1 (e i 1) = d 1. Fix a d-cycle τ S d, We say (σ 1,...,σ r 1 ) is a factorization of τ of type(e 1,...,e r 1 ) if the followings are satisfied: i. For eachi,σ i is ane i -cycle ins d. ii. σ 1 σ r 1 = τ. Example 7. Letd = 5,r = 4,(e 1,e 2,e 3 ) = (2,2,3),τ = (12345),σ 1 = (23), σ 2 = (45), σ 3 = (135). It is easy to check that (σ 1,σ 2,σ 3 ) is a factorization of τ of type(2,2,3): (23)(45)(135) = (12345). Question 8. Given a d-cycle τ and integers e 1,...,e r 1 2, how many factorizations are there ofτ of type(e 1,...,e r 1 )? 7

16 Main Result Theorem 9. Suppose r 1 i=1 (e i 1) = d 1. Then the number of factorizations of a d-cycle of type(e 1,...,e r 1 ) is fac(d,r;e 1,...,e r 1 ) = d r 2. 8

17 Main Result Theorem 9. Suppose r 1 i=1 (e i 1) = d 1. Then the number of factorizations of a d-cycle of type(e 1,...,e r 1 ) is fac(d,r;e 1,...,e r 1 ) = d r 2. Example 10. There are3 = 3 1 factorizations of(123) of type(2,2): (12)(23) (23)(13) (13)(12) 8

18 Main Result Theorem 9. Suppose r 1 i=1 (e i 1) = d 1. Then the number of factorizations of a d-cycle of type(e 1,...,e r 1 ) is fac(d,r;e 1,...,e r 1 ) = d r 2. Example 10. There are3 = 3 1 factorizations of(123) of type(2,2): (12)(23) (23)(13) (13)(12) Example 11. There are25 = 5 2 factorizations of(12345) of type(2,2,3): (12)(23)(345) (23)(34)(451) (34)(45)(512) (45)(51)(123) (51)(12)(234) (23)(13)(345) (34)(24)(451) (45)(35)(512) (51)(41)(123) (12)(52)(234) (13)(12)(345) (24)(23)(451) (35)(34)(512) (41)(45)(123) (52)(51)(234) (12)(34)(245) (23)(45)(351) (34)(51)(412) (45)(12)(523) (51)(23)(134) (34)(12)(245) (45)(23)(351) (51)(34)(412) (12)(45)(523) (23)(51)(134) 8

19 Special Case Whene 1 = = e r 1 = 2, from r 1 i=1 (e i 1) = d 1we haved = r. Then Theorem 9 gives the following well-known result: Corollary 12. The number of factorizations of a d-cycle into d 1 transpositions is d d 2. 9

20 Special Case Whene 1 = = e r 1 = 2, from r 1 i=1 (e i 1) = d 1we haved = r. Then Theorem 9 gives the following well-known result: Corollary 12. The number of factorizations of a d-cycle into d 1 transpositions is d d 2. Note that d d 2 is also the number of trees on d vertices. Different bijective proofs of this result were given by Dénes (1959), Moszkowski (1989), Goulden-Pepper (1993) and Goulden-Yong (2002). 9

21 Special Case Whene 1 = = e r 1 = 2, from r 1 i=1 (e i 1) = d 1we haved = r. Then Theorem 9 gives the following well-known result: Corollary 12. The number of factorizations of a d-cycle into d 1 transpositions is d d 2. Note that d d 2 is also the number of trees on d vertices. Different bijective proofs of this result were given by Dénes (1959), Moszkowski (1989), Goulden-Pepper (1993) and Goulden-Yong (2002). Main Idea to prove Theorem 9: Construct a class of combinatorial objects that are counted byd r 2, and then find a bijection between factorizations and them. 9

22 PART II: Multi-noded Rooted Trees 10

23 Definition of Multi-noded Rooted Trees Definition 13. Suppose f 0,f 1,...,f n are positive integers and S = {s 1,...,s n }. We say G is a multi-noded rooted tree on S {0} of vertex data (f 0,f 1,..., f n ) if we have the followings: (i) The vertex set ofgiss {0}. (ii) For each vertexs i, it includesf i ordered nodes (by convention,s 0 := 0). (iii) Considering only vertices and edges, G is a rooted tree with root 0, but in addition each edge is connected to a particular node of the parent vertex. We denote bymr S (f 0,f 1,...,f n ) the set of multi-noded rooted trees. Example 14. A multi-noded rooted tree of vertex data(1,1,2,1,2,2,3,3,1,4): 0 s 3 s 5 s 8 s 2 s 9 s 6 s 4 s 1 s 7 11

24 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. 12

25 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 16. Suppose r 1 j=1 (e j 1) = d 1. Then MR S (1,e 1 1,...,e r 1 1) = d r 2. 12

26 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 16. Suppose r 1 j=1 (e j 1) = d 1. Then MR S (1,e 1 1,...,e r 1 1) = d r 2. 0 s 3 s 5 s 8 s 2 s 9 s 6 s 4 s 1 s 7 12

27 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 16. Suppose r 1 j=1 (e j 1) = d 1. Then MR S (1,e 1 1,...,e r 1 1) = d r 2. 0 s 3 s 5 s 8 s 2 s 9 s 6 s 4 s 1 s 7 s

28 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 16. Suppose r 1 j=1 (e j 1) = d 1. Then MR S (1,e 1 1,...,e r 1 1) = d r 2. 0 s 3 s 5 s 2 s 9 s 6 s 4 s 1 s 7 s 3 s

29 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 16. Suppose r 1 j=1 (e j 1) = d 1. Then MR S (1,e 1 1,...,e r 1 1) = d r 2. 0 s 3 s 5 s 2 s 9 s 6 s 4 s 1 s 3 s 9 s

30 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 16. Suppose r 1 j=1 (e j 1) = d 1. Then MR S (1,e 1 1,...,e r 1 1) = d r 2. 0 s 3 s 5 s 2 s 9 s 4 s 1 s 3 s 9 s 2 s

31 Counting Multi-noded Rooted Trees Theorem 15. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 16. Suppose r 1 j=1 (e j 1) = d 1. Then MR S (1,e 1 1,...,e r 1 1) = d r 2. 0 s 3 s 5 s 8 s 2 s 9 s 6 s 4 s 1 s 7 s 3 s 9 s 2 s 9 s 3 0 s 9 s

32 PART III: Bijection between Factorizations and Multi-noded Rooted Trees 13

33 Factorization Graphs 14

34 Factorization Graphs A factorization ofτ = (12 20) of type(2,3,2,3,3,4,4,2,5): (1011)(141519)(119)(345)(1213)( )(78911)(1920)( ) 14

35 Factorization Graphs A factorization ofτ = (12 20) of type(2,3,2,3,3,4,4,2,5): (1011)(141519)(119)(345)(1213)( )(78911)(1920)( ) The factorization graph associated to this factorization is: s s 6 s 5 s 2 s 3 s s 9 s s

36 Factorization Graphs A factorization ofτ = (12 20) of type(2,3,2,3,3,4,4,2,5): (1011)(141519)(119)(345)(1213)( )(78911)(1920)( ) The factorization graph associated to this factorization is: s 8 4 Facts: 1. G is a bipartite graph ons [d] s 6 s 5 s 2 s 3 s Any vertexs i has degreee i. 15 s 9 s s

37 Characterization of factorization graphs Proposition 17. Suppose r 1 j=1 (e j 1) = d 1, and G is a bipartite graph on S [d] such that vertexs i has degreee i. Then G is a factorization graph associated to a factorization of τ of type (e 1,..., e r 1 ) if and only ifgsatisfies the following conditions: i. G is a tree. ii. For each [d]-vertex ν of G, suppose {s j1 < s j2 < < s jt } are the vertices adjacent to ν in G. We get t subtrees after removing ν and all its incident edges. Then (a) The[d]-vertices of the t subtrees partition[d]\{ν} into contiguous pieces. (b) If we order the pieces in counterclockwise order on τ starting from ν, then the m-th piece is exactly the subtree that contains vertexs jm for any1 m t. 15

38 Factorization Graphs to Labeled Multi-noded Rooted Trees A factorization ofτ = (12 20): (1011)(141519)(119)(345)(1213)( )(78911)(1920)( ) 16

39 Factorization Graphs to Labeled Multi-noded Rooted Trees A factorization ofτ = (12 20): (1011)(141519)(119)(345)(1213)( )(78911)(1920)( ) The factorization graph associated to a factorization of type(2,3,2,3,3,4,4,2,5) s s 6 s 5 s 2 s 3 s s 9 s s

40 Factorization Graphs to Labeled Multi-noded Rooted Trees A factorization ofτ = (12 20): (1011)(141519)(119)(345)(1213)( )(78911)(1920)( ) The factorization graph associated to a factorization of type(2,3,2,3,3,4,4,2,5) s 6 20 s s 5 s 2 s s 9 s 7 s 1 10 s s 8 20 A labelled multi-noded rooted tree of vertex data(1,1,2,1,2,2,3,3,1,4) s 3 19 s s s s s s 1 10 s

41 Remark Goulden and Jackson considered more general factorizations of a d-cycle, where they allowσ i to be any cycle type, that is,σ i does not have to be a cycle. They gave a formula for the factorization number in this situation. 17

42 Remark Goulden and Jackson considered more general factorizations of a d-cycle, where they allowσ i to be any cycle type, that is,σ i does not have to be a cycle. They gave a formula for the factorization number in this situation. But their proof involves calculation of generating functions. 17

43 Remark Goulden and Jackson considered more general factorizations of a d-cycle, where they allowσ i to be any cycle type, that is,σ i does not have to be a cycle. They gave a formula for the factorization number in this situation. But their proof involves calculation of generating functions. An equivalent symmetrized version of Theorem 9 was proved by Springer and Irving separately: e.g., when (e 1,e 2,e 3 ) = (2,2,3), we only allow factorizations where the first and second cycles have length 2 and the third cycle has length 3. They included all factorizations with one 3-cycle and two 2-cycles. 17

44 Remark Goulden and Jackson considered more general factorizations of a d-cycle, where they allowσ i to be any cycle type, that is,σ i does not have to be a cycle. They gave a formula for the factorization number in this situation. But their proof involves calculation of generating functions. An equivalent symmetrized version of Theorem 9 was proved by Springer and Irving separately: e.g., when (e 1,e 2,e 3 ) = (2,2,3), we only allow factorizations where the first and second cycles have length 2 and the third cycle has length 3. They included all factorizations with one 3-cycle and two 2-cycles. We believe that our proof is the first de-symmetrized direct bijective proof. 17