Counting factorizations in complex reflection groups Joel Brewster - PowerPoint PPT Presentation

Counting factorizations in complex reflection groups Joel Brewster Lewis (George Washington University) joint work with Alejandro Morales (UMass Amherst) arXiv:1906.11961 FPSAC Ljubljana, July 5, 2019 Joel Lewis (GWU) Factorizations in CRGs July 5, 2019 1 / 16

Intro: counting factorizations in the symmetric group Question How many ways are there to write the n-cycle ς = (1 · · · n ) in S n as a product ς = π 1 · · · π k of permutations π 1 , . . . , π k ? Answer: | S n | k − 1 = ( n !) k − 1 (any group, nothing special about permutations)

Intro: counting factorizations in the symmetric group Question How many ways are there to write the n-cycle ς = (1 · · · n ) in S n as a product ς = π 1 · · · π k of permutations π 1 , . . . , π k so that π i has exactly r i cycles? Since # cycles is conjugacy invariant, the following general tool works: Lemma (Frobenius, 1898) Let G be a finite group, g ∈ G, and A 1 , . . . , A k subsets of G that are closed under conjugacy by G. Then the number of factorizations of g as a product g = t 1 · · · t k such that t i ∈ A i for each i is � 1 λ ∈ Irr( G ) dim( λ ) 1 − k χ λ ( g − 1 ) χ λ ( z 1 ) · · · χ λ ( z k ) , | G | where Irr( G ) is the set of irreducible representations of G, dim( λ ) is the dimension of the representation λ , χ λ is the character associated to λ , and z i is the formal sum in the group algebra of elements in A i .

Jackson’s theorem Theorem (Jackson (1988) as formulated by Schaeffer–Vassilieva (2008)) Let ς be a fixed n-cycle in S n , and let a r 1 ,..., r k be the number of tuples ( π 1 , . . . , π k ) of elements in S n such that π i has r i cycles for all i and π 1 · · · π k = ς . Then ( x 1 ) p 1 · · · ( x k ) p k � � a r 1 ,..., r k x r 1 1 · · · x r k k = ( n !) k − 1 · M n − 1 p 1 − 1 ,..., p k − 1 p 1 ! p k ! r 1 ,..., r k ≥ 1 p 1 ,..., p k ≥ 1 where ( x ) p denotes the falling factorial ( x ) p := x ( x − 1) · · · ( x − p + 1) and � � n p 1 ,..., p k := [ x p 1 1 · · · x p k M n k ] (1 + x 1 ) · · · (1 + x k ) − x 1 · · · x k . � x � Observation: ( x ) p . Interpretation: ( n !) k − 1 M n − 1 p ! = p 1 − 1 ,..., p k − 1 counts p cycle-colored factorizations . Recent combinatorial proof: Bernardi–Morales (2016), via maps on surfaces & sign-reversing involutions

What is this talk about? Often, interesting questions about S n have nice generalizations or analogues in other contexts E.g.: Jackson (1988) also enumerated factorizations of an n -cycle as a product of transpositions This was extended to well generated complex reflection groups by Chapuy–Stump (2014) I’m going to describe some initial attempts to do the same thing for the more general result First step: translate S n Joel Lewis (GWU) Factorizations in CRGs July 5, 2019 4 / 16

S n is a reflection group V a C -vector space; a transformation T : V → V is a reflection if it fixes a hyperplane pointwise a complex reflection group is a finite subgroup of GL( V ) that is generated by its subset of reflections E.g.: S n is a CRG S n : n × n permutation matrices, act on C n       0 0 0 1 w z 0 1 0 0 x x  ·  =  fixes w = z Transpositions are the ref’ns:    0 0 1 0 y y 1 0 0 0 z w S n generated by transpositions In general, # cycles = dimension of fixed space:   0 0 0 0 1 0 0 0 0 0 0 1   1 0 0 0 0 0   (153)(26)(4) =   0 0 0 1 0 0   0 0 1 0 0 0 0 1 0 0 0 0 fixes vectors like ( a , b , a , c , a , b )

S n is a reflection group V a C -vector space; a transformation T : V → V is a reflection if it fixes a hyperplane pointwise a complex reflection group is a finite subgroup of GL( V ) that is generated by its subset of reflections E.g.: S n is a CRG E.g.: group of n × n signed permutation matrices is a CRG E.g.: every finite Coxeter group is a CRG E.g.: wreath product ( Z / m Z ) ≀ S n is a CRG (definition on next slide!)

Generalized permutations Think of Z / m Z as m -th complex roots of unity; wreath product is � ( Z / m Z ) ≀ S n ∼ n × n monomial matrices = � whose nonzero entries are m -th roots of 1 m = 1: S n , Coxeter type A n − 1 m = 2: signed permutations, Coxeter type B n   0 1 0 m ≥ 3: things like − 1 0 0  , not Coxeter groups  0 0 exp(4 π i / 5) The weight of an element is the Z / m Z value corresponding to the product of the nonzero entries: matrix above is m = 10, weight = 9 because − 1 · 1 · exp(4 π i / 5) = exp(2 π i · 9 10 ). Another example: the subgroup G ( m , m , n ) of ( Z / m Z ) ≀ S n consisting of elements of weight 0 (when m = 2, Coxeter type D n )

n -cycles are Coxeter elements In S n , we factor n -cycle Analogue in a well generated CRG is a Coxeter element In ( Z / m Z ) ≀ S n , with ω = exp(2 π i / m ), these are things like   ω 1     ...       ...     1 In G ( m , m , n ), things like   ω 1     ...       1   ω − 1

Our question Jackson counted factorizations of the n -cycle in the symmetric group S n as an arbitrary product, keeping track of the number of cycles of each factor, using a change of basis to binomial coefficients. We count factorizations of a Coxeter element in ( Z / m Z ) ≀ S n (or other CRG) as an arbitrary product, keeping track of the fixed space dimension of each factor, using a change of basis to ???????. Joel Lewis (GWU) Factorizations in CRGs July 5, 2019 8 / 16

Reminder for comparison: Jackson Theorem (Jackson (1988)) Let ς be a fixed n-cycle in S n , and let a r 1 ,..., r k be the number of tuples ( π 1 , . . . , π k ) of elements in S n such that π i has r i cycles for all i and π 1 · · · π k = ς . Then ( x 1 ) p 1 · · · ( x k ) p k � � 1 · · · x r k a r 1 ,..., r k x r 1 k = ( n !) k − 1 · M n − 1 p 1 − 1 ,..., p k − 1 p 1 ! p k ! r 1 ,..., r k ≥ 1 p 1 ,..., p k ≥ 1 where ( x ) p denotes the falling factorial ( x ) p := x ( x − 1) · · · ( x − p + 1) and � � n p 1 ,..., p k := [ x p 1 1 · · · x p k M n k ] (1 + x 1 ) · · · (1 + x k ) − x 1 · · · x k . Joel Lewis (GWU) Factorizations in CRGs July 5, 2019 9 / 16

Our answer (wreath product) Theorem (L–Morales (2019)) For m > 1 , let c be a fixed Coxeter element in G = ( Z / m Z ) ≀ S n , and let a r 1 ,..., r k be the number of factorizations of c as a product of k elements of G with fixed space dimensions r 1 , . . . , r k , respectively. Then ( x 1 − 1) ( m ) · · · ( x k − 1) ( m ) � k = | G | k − 1 � a r 1 ,..., r k x r 1 1 · · · x r k p 1 p k M n , p 1 ,..., p k m p 1 p 1 ! m p k p k ! r 1 ,..., r k p 1 ,..., p k r i ≥ 0 p i ≥ 0 ( x − 1) ( m ) := ( x − 1)( x − m − 1) · · · ( x − m ( p − 1) − 1) and M n p 1 ,..., p k is p exactly the same coefficient as before . Observation: | ( Z / m Z ) ≀ S p | = m p p ! Joel Lewis (GWU) Factorizations in CRGs July 5, 2019 10 / 16

Proof idea Proof with character theory is technical, straightforward, not illuminating We think it is more interesting to give a combinatorial proof Fixed space dimension counts cycles of weight 0 Rewrite the desired result as � x 1 � � x k � � a r 1 ,..., r k ( mx 1 +1) r 1 · · · ( mx k +1) r k = | G | k − 1 � M n · · · p 1 ,..., p k p 1 p k r 1 ,..., r k p 1 ,..., p k r i ≥ 0 p i ≥ 0 Interpret left side as an elaborate cycle-coloring scheme Colored factorizations in ( Z / m Z ) ≀ S n project to colored factorizations of an n -cycle in S n Carefully count pre-images to get result Recover Chapuy–Stump result (reflection factorizations) for this group as corollary Joel Lewis (GWU) Factorizations in CRGs July 5, 2019 11 / 16

What about other CRGs? � G ( m , m , n ) = n × n monomial matrices whose nonzero � entries are m -th roots of 1 and have product 1 Coxeter elements look like   ω 1     ...       1   ω − 1 Under projection, gives an ( n − 1)-cycle in S n . Joel Lewis (GWU) Factorizations in CRGs July 5, 2019 12 / 16

Aside: factoring an ( n − 1)-cycle Two very different ways to factor an ( n − 1)-cycle: (13)(24)(5) · (1432)(5) = (1234)(5) vs. (15)(24)(3) · (152)(34) = (1234)(5) Say ( π 1 , . . . , π k ) is a transitive factorization of its product if group � π 1 , . . . , π k � acts transitively on { 1 , . . . , n } Every factorization of an n -cycle in S n is transitive A factorization of an ( n − 1)-cycle in S n is nontransitive if and only if all factors share a fixed point.

Aside: factoring an ( n − 1)-cycle Two very different ways to factor an ( n − 1)-cycle: (13)(24)(5) · (1432)(5) = (1234)(5) vs. (15)(24)(3) · (152)(34) = (1234)(5) Theorem (L–Morales (2019) (??!!??)) Let ς n − 1 be a fixed ( n − 1) -cycle in S n . For integers r 1 , . . . , r k let b r 1 ,..., r k be the number of k-tuples ( π 1 , . . . , π k ) of elements in S n such that π i has r i cycles for i = 1 , . . . , k, π 1 · · · π k = ς n − 1 , and these tuples are a transitive factorization. Then k = n ! k − 1 ( x 1 ) p 1 ( x k ) p k � � b r 1 ,..., r k x r 1 1 · · · x r k M n ( p 1 − 1)! · · · ( p k − 1)! , p 1 ,..., p k n k r 1 ,..., r k ≥ 1 p 1 ,..., p k ≥ 1 where M n p 1 ,..., p k still the same as ever. Proof. Character theory. (Is there a combinatorial proof?) �