Pure-cycle Hurwitz factorizations and multi-noded rooted trees

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1 Pure-cycle Hurwitz factorizations and multi-noded rooted trees University of California, Davis CMS winter meeting, Vancouver, British Columbia December 4, 2010 This is joint work with Rosena R.X. Du. 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) M := σ 1,...,σ r is 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) M := σ 1,...,σ r is 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) M := σ 1,...,σ r is 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 (L-Osserman). In the genus-0 pure-cycle case, whenr = 3, h(d,3,(e 1,e 2,e 3 )) = 1. 5

10 Previous results on the pure-cycle case Lemma 4 (L-Osserman). In the genus-0 pure-cycle case, whenr = 3, h(d,3,(e 1,e 2,e 3 )) = 1. Theorem 5 (L-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 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

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. Example 7. We have(23)(45)(135) = (12345). Hence,((23),(45),(135)) is a factorization ofτ of type(2,2,3). 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. We have(23)(45)(135) = (12345). Hence,((23),(45),(135)) is a factorization ofτ of type(2,2,3). Definition 8. The factorization numberfac(d,r,(e 1,...,e r 1 )) is the number of factorizations of a fixedd-cycleτ of type(e 1,...,e r 1 ). 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. We have(23)(45)(135) = (12345). Hence,((23),(45),(135)) is a factorization ofτ of type(2,2,3). Definition 8. The factorization numberfac(d,r,(e 1,...,e r 1 )) is the number of factorizations of a fixedd-cycleτ of type(e 1,...,e r 1 ). It is clear that h(d,r;(e 1,...,e r 1,e r = d)) = 1 d fac(d,r,(e 1,...,e r 1 )) 7

16 The Formula Goulden and Jackson considered more general factorizations of a d-cycle, where they allow σ 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. Specializing their formula to the pure-cycle case gives the following formula. Theorem 9. Suppose r 1 i=1 (e i 1) = d 1. Then fac(d,r,(e 1,...,e r 1 )) = d r 2 Hence, h(d,r,(e 1,...,e r 1,d)) = d r 3 8

17 The Formula Goulden and Jackson considered more general factorizations of a d-cycle, where they allow σ 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. Specializing their formula to the pure-cycle case gives the following formula. Theorem 9. Suppose r 1 i=1 (e i 1) = d 1. Then fac(d,r,(e 1,...,e r 1 )) = d r 2 Hence, h(d,r,(e 1,...,e r 1,d)) = d r 3 Their proof involves calculation of generating functions. 8

18 The Formula Goulden and Jackson considered more general factorizations of a d-cycle, where they allow σ 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. Specializing their formula to the pure-cycle case gives the following formula. Theorem 9. Suppose r 1 i=1 (e i 1) = d 1. Then fac(d,r,(e 1,...,e r 1 )) = d r 2 Hence, h(d,r,(e 1,...,e r 1,d)) = d r 3 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. 8

19 Question: Can we give a de-symmetrized direct bijective proof for Theorem 9? What are the right combinatorial objects to use? 9

20 Question: Can we give a de-symmetrized direct bijective proof for Theorem 9? What are the right combinatorial objects to use? We construct a class of combinatorial objects that are counted by d r 2, and then describe a bijection between factorizations and these objects. 9

21 PART II: Multi-noded Rooted Trees 10

22 Definition of Multi-noded Rooted Trees Definition 10. 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. 11

23 Definition of Multi-noded Rooted Trees Definition 10. 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 11. A multi-noded rooted tree of vertex data(1,1,2,1,2,2,3,3,1,4): 11

24 Definition of Multi-noded Rooted Trees Definition 10. 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 11. 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

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

26 Counting Multi-noded Rooted Trees Theorem 12. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 13. 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

27 Counting Multi-noded Rooted Trees Theorem 12. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 13. 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

28 Counting Multi-noded Rooted Trees Theorem 12. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 13. 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

29 Counting Multi-noded Rooted Trees Theorem 12. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 13. 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

30 Counting Multi-noded Rooted Trees Theorem 12. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 13. 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

31 Counting Multi-noded Rooted Trees Theorem 12. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 13. 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

32 Counting Multi-noded Rooted Trees Theorem 12. MR S (f 0,f 1,...,f n ) = f 0 ( n i=0 f i) n 1. Corollary 13. 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

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

34 Factorization Graphs 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)( ) 14

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 s 6 s 5 s 2 s 3 s s 9 s s

37 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 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

38 Characterization of factorization graphs Proposition 14. 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

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

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 s 6 s 5 s 2 s 3 s s 9 s s

41 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

Pure-cycle Hurwitz factorizations and multi-noded rooted trees

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 PART I: Definitions