Combinatorial Problems Related to Automorphism Groups of Compact Riemann Surfaces Charles Camacho Oregon State University Rainwater Seminar University of Washington February 26, 2019 camachoc@math.oregonstate.edu Charles Camacho Oregon State University 1
Introduction Charles Camacho Oregon State University 2
Inspiration from the Platonic Solids The Platonic solids can be described by ramified coverings of the sphere branching over three points, 0 , 1 , ∞ , represented as white and black vertices, and face centers, respectively. The values below are the branch orders above 0 , 1 , ∞ . ∆(2 , 3 , 4) ∆(2 , 3 , 3) ∆(2 , 4 , 3) ∆(2 , 3 , 5) ∆(2 , 5 , 3) Charles Camacho Oregon State University 3
Klein Quartic of Genus Three (1879) Glue edges 1 ↔ 10 , 2 ↔ 7 , 3 ↔ 12 , 4 ↔ 9 , 5 ↔ 14 , 6 ↔ 11 , 8 ↔ 13. Rotation about the center of the hyperbolic 14-gon by 2 π/ 7 fixes three points. Charles Camacho Oregon State University 4
References for an Introduction to the Subject Broughton, S. A. (1990). The equisymmetric stratification of the moduli space and the Krull dimension of mapping class groups. Topology and its Applications, 37(2), 101-113. Broughton, S. A. (1991). Classifying finite group actions on surfaces of low genus. Journal of Pure and Applied Algebra, 69(3), 233-270. Girondo, E., Gonz´ alez-Diez, G. Introduction to Compact Riemann Surfaces and Dessins d’Enfants. Cambridge: Cambridge UP, 2012. Print. Jones, G. A., Wolfart, J. (2016). Dessins D’enfants on Riemann Surfaces. Springer. O’Sullivan, C., Weaver, A. A diophantine frobenius problem related to Riemann surfaces. Glasgow Mathematical Journal, 53(3), 501-522. (2011). Weaver, A. Genus spectra for split metacyclic groups. Glasgow Mathematical Journal, 43(2), 209-218. 2001. Charles Camacho Oregon State University 5
Main Talk Charles Camacho Oregon State University 6
Results - The Number of Quasiplatonic Cyclic Group Actions Theorem (C. 2018): Let QC ( n ) denote the number of distinct quasiplatonic topological C n -actions on surfaces of genus at least two. Write n = � r i =1 p a i i . Then � � QC ( n ) = 1 1 + 1 � − 1 + a · 2 r 6 n p p | n where 1 / 4 p 1 = 2 , a 1 = 1 1 / 2 p 1 = 2 , a 1 = 2 1 p 1 = 2 , a 1 ≥ 3 a = 5 / 6 p i ≡ 1 mod 6 for 1 ≤ i ≤ r . 2 / 3 p 1 = 3 , a 1 = 1 , p i ≡ 1 mod 6 for 2 ≤ i ≤ r 1 / 2 p 1 = 3 , a 1 ≥ 2 , p i ≡ 1 mod 6 for 2 ≤ i ≤ r 1 / 2 p i ≡ 5 mod 6 for some 1 ≤ i ≤ r Charles Camacho Oregon State University 7
Graph Charles Camacho Oregon State University 8
Preliminary Results - The Quasiplatonic Cyclic Genus Spectrum Let QCGS ( n ) be the set of genera two or greater admitting a quasiplatonic C n -action, called the quasiplatonic cyclic genus spectrum . Write QCGS ( n ) = { σ n , 1 , σ n , 2 , . . . , σ n , s n } , where 2 ≤ σ n , 1 < σ n , 2 < · · · < σ n , s n . Charles Camacho Oregon State University 9
Preliminary Results - The Quasiplatonic Cyclic Genus Spectrum Let QCGS ( n ) be the set of genera two or greater admitting a quasiplatonic C n -action, called the quasiplatonic cyclic genus spectrum . Write QCGS ( n ) = { σ n , 1 , σ n , 2 , . . . , σ n , s n } , where 2 ≤ σ n , 1 < σ n , 2 < · · · < σ n , s n . Theorem (C. 2019): If n = 2 p a , then σ n , 2 = ( p − 1) p a − 1 . � � � � p 2 − 1 If n = 2 p 2 p a 3 n 3 · · · p a r r , then σ n , 2 = · p 2 − 1 . 2 Charles Camacho Oregon State University 9
Conjecture for σ n , 2 � a − 1 � · n 2 a σ n , 2 = � b − 1 � � n � · b − 1 2 where p 1 p 1 ≥ 3 , a 1 = 1 p 1 = 2 and either p 2 a 1 = 1 , a 2 = 1 , r ≥ 3; p 2 p 1 = 2 and either or a 1 = 2 , p 2 = 3 , a 2 = 1 , r ≥ 3; a 1 = 1 , a 2 ≥ 2; or a 1 ≥ 3 , p 2 = 3 , a 2 ≥ 2 . b = or a 1 = 1 , r = 2; . Or p 1 ≥ 3 , a 1 ≥ 3 , a 2 = 1 , or a 1 = 2 , p 2 = 3 , a 2 = 1 , r = 2; p 2 < p 2 1 a = or a 1 ≥ 3 , p 2 = 3 , a 2 ≥ 2 . Or p 1 ≥ 3 , a 1 ≥ 3 , a 2 ≥ 2 , 4 p 1 = 2 , a 1 = 2 , p 2 > 3 . p 2 < p 2 1 4 p 1 = 2 , a 1 ≥ 3 , p 2 > 3 p 2 p 1 ≥ 3 , a 1 ≥ 3 , p 2 > p 2 1 . 1 Charles Camacho Oregon State University 10
Graph Charles Camacho Oregon State University 11
Part One: Counting Quasiplatonic Cyclic Actions Charles Camacho Oregon State University 12
∆(2 , 3 , 7) A tessellation of H by hyperbolic triangles, each with angles π/ 2 , π/ 3 and π/ 7. Charles Camacho Oregon State University 13
Example - C 7 Actions Consider the permutation ρ = (1 , 2 , 3 , 4 , 5 , 6 , 7) ∈ S 7 . Charles Camacho Oregon State University 14
Example - C 7 Actions Consider the permutation ρ = (1 , 2 , 3 , 4 , 5 , 6 , 7) ∈ S 7 . In terms of rotations about the white and black vertices, and the face, respectively: Charles Camacho Oregon State University 14
Example - C 7 Actions Consider the permutation ρ = (1 , 2 , 3 , 4 , 5 , 6 , 7) ∈ S 7 . In terms of rotations about the white and black vertices, and the face, respectively: 5 4 1 1 5 7 7 6 3 2 4 2 6 6 6 7 3 2 3 3 7 1 5 5 1 4 2 4 (3 , 3 , 1) is described by ρ 3 , ρ 3 , ρ ... ...and (2 , 4 , 1) is described by ρ 2 , ρ 4 , ρ . Charles Camacho Oregon State University 14
Counting C n -Actions with a Given Signature ( n 1 , n 2 , n 3 ) Theorem (Benim, Wootton 2014) Write n = � r i =1 p a i i . Suppose C n acts on X of signature ( n 1 , n 2 , n 3 ) . Let w ≥ 0 be the number of primes shared in common among n 1 , n 2 , n 3 which also share the same exponent. Relabel the primes to start with p 1 , . . . , p w . Then the number T of quasiplatonic topological actions of C n on X is given in the three possibilities below. 1. If each n i is distinct and w � = 0 , then w p i − 2 � T = φ (gcd( n 1 , n 2 , n 3 )) p i − 1 . i =1 If w = 0 , then T = φ (gcd( n 1 , n 2 , n 3 )) . Charles Camacho Oregon State University 15
Counting C n -Actions with a Given Signature ( n 1 , n 2 , n 3 ) Theorem (Benim, Wootton 2014) 2. If two n i are equal so that the signature is ( n 1 , n 2 , n 3 ) = ( n 1 , n , n ) with n 1 � = n and w � = 0 , then � w � p i − 2 T = 1 � τ 1 ( n , n 1 ) + φ ( n 1 ) . p i − 1 2 i =1 If w = 0 , then T = 1 2 ( τ 1 ( n , n 1 ) + φ ( n 1 )) . 3. If all n i are equal so that the signature is ( n 1 , n 2 , n 3 ) = ( n , n , n ) and w � = 0 , then � w � T = 1 p i − 2 � 3 + 2 τ 2 ( n ) + φ ( n ) . 6 p i − 1 i =1 If w = 0 , then T = 1 6 (3 + 2 τ 2 ( n ) + φ ( n )) . Charles Camacho Oregon State University 16
Example: Deriving QC ( n ) when p 1 = 2 , a 1 = 1 r Claim: QC ( n ) = 1 � � + 2 r − 2 − 1. � p a i − 1 ( p i + 1) i 2 i =2 Proof Sketch. First prove the following recursive formula by induction on r : r +1 ) = ( QC ( n ) + 1 − 2 r − 2 ) p a r +1 − 1 QC ( n · p a r +1 ( p r +1 + 1) − 1 + 2 r − 1 . r +1 Then prove the main claim by using induction again on r , with the base case being QC (2 p a ) = 1 2 p a − 1 ( p + 1) . Charles Camacho Oregon State University 17
Further Work Charles Camacho Oregon State University 18
Further Work Analyze the growth rate of QC ( n ). Find a geometric interpretation of the constant term in the formula for QC ( n ). Derive analogous formulas for quasiplatonic actions of other groups, e.g., abelian, elementary abelian C n p , semidirect products C m ⋊ C n . Does a count on the number of group actions have a physical interpretations (e.g., in string theory)? Visualization of these group actions? Charles Camacho Oregon State University 18
Part Two: The Quasiplatonic Cyclic Genus Spectrum Charles Camacho Oregon State University 19
Set-Up r � p a i Write n = i . Let i =1 A j = { 1 ≤ i ≤ r : k i = j , ℓ i = h i = a i } , B j = { 1 ≤ i ≤ r : ℓ i = j , k i = h i = a i } , C j = { 1 ≤ i ≤ r : h i = j , k i = ℓ i = a i } . Let t n = max 1 ≤ i ≤ r { a i } . Then 1 1 1 S ( n 1 , n 2 , n 3 ) = + + p k 1 1 · · · p k r p ℓ 1 1 · · · p ℓ r p h 1 1 · · · p h r r r r t n t n = 1 p a i − m p a i − m � � + � � i i n m =0 i ∈ A m m =0 i ∈ B m t n � � p a i − m . + i m =0 i ∈ C m Charles Camacho Oregon State University 20
Key Lemma Lemma (C. 2019) All signatures that achieve ˆ S 1 satisfy A j = B j = C j = ∅ for j ≥ 2 and | A 1 ∪ B 1 ∪ C 1 | ≤ 1 . If d ≥ 2 and n � = p a for prime p and a ≥ 1 , then all signatures that achieve ˆ S d satisfy A j = B j = C j = ∅ for j ≥ d and | A d − 1 ∪ B d − 1 ∪ C d − 1 | ≤ 1 . Charles Camacho Oregon State University 21
Example: C 12 -action with Signature (2 , 12 , 12) 6 1 12 7 5 2 6 11 4 11 8 9 1 2 4 8 3 7 3 12 5 6 10 10 9 8 115 1 2 12 3 4 7 3 9 10 4 9 2 5 10 8 11 1 6 12 7 (2 , 12 , 12) σ = 3 A 0 = { 2 } A 1 = { 1 } A 2 = ∅ B 0 = ∅ B 1 = ∅ B 2 = ∅ C 0 = ∅ C 1 = ∅ C 2 = ∅ Charles Camacho Oregon State University 22
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