Combinatorial Problems Related to AutomorphismGroups of Compact Riemann Surfaces
Charles CamachoOregon State University
Rainwater SeminarUniversity of Washington
February 26, [email protected]
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 spherebranching over three points, 0, 1,∞, represented as white and blackvertices, and face centers, respectively.
The values below are the branch orders above 0, 1,∞.
∆(2, 3, 5)∆(2, 3, 4)∆(2, 3, 3) ∆(2, 4, 3) ∆(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 thehyperbolic 14-gon by 2π/7 fixesthree points.
Charles Camacho Oregon State University 4
References for an Introduction to the Subject
Broughton, S. A. (1990). The equisymmetric stratification of themoduli 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 surfacesof low genus. Journal of Pure and Applied Algebra, 69(3), 233-270.
Girondo, E., Gonzalez-Diez, G. Introduction to Compact RiemannSurfaces and Dessins d’Enfants. Cambridge: Cambridge UP, 2012.Print.
Jones, G. A., Wolfart, J. (2016). Dessins D’enfants on RiemannSurfaces. Springer.
O’Sullivan, C., Weaver, A. A diophantine frobenius problem related toRiemann surfaces. Glasgow Mathematical Journal, 53(3), 501-522.(2011).
Weaver, A. Genus spectra for split metacyclic groups. GlasgowMathematical 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 GroupActions
Theorem (C. 2018): Let QC (n) denote the number of distinctquasiplatonic topological Cn-actions on surfaces of genus at least two.Write n =
∏ri=1 p
aii . Then
QC (n) =1
6n∏p|n
(1 +
1
p
)− 1 + a · 2r
where
a =
1/4 p1 = 2, a1 = 11/2 p1 = 2, a1 = 21 p1 = 2, a1 ≥ 35/6 pi ≡ 1 mod 6 for 1 ≤ i ≤ r2/3 p1 = 3, a1 = 1, pi ≡ 1 mod 6 for 2 ≤ i ≤ r1/2 p1 = 3, a1 ≥ 2, pi ≡ 1 mod 6 for 2 ≤ i ≤ r1/2 pi ≡ 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 GenusSpectrum
Let QCGS(n) be the set of genera two or greater admitting a quasiplatonicCn-action, called the quasiplatonic cyclic genus spectrum. Write
QCGS(n) = {σn,1, σn,2, . . . , σn,sn},
where 2 ≤ σn,1 < σn,2 < · · · < σn,sn .
Theorem (C. 2019):
If n = 2pa, then σn,2 = (p − 1)pa−1.
If n = 2p2pa33 · · · parr , then σn,2 =
(p2−12
)·(
np2− 1)
.
Charles Camacho Oregon State University 9
Preliminary Results - The Quasiplatonic Cyclic GenusSpectrum
Let QCGS(n) be the set of genera two or greater admitting a quasiplatonicCn-action, called the quasiplatonic cyclic genus spectrum. Write
QCGS(n) = {σn,1, σn,2, . . . , σn,sn},
where 2 ≤ σn,1 < σn,2 < · · · < σn,sn .
Theorem (C. 2019):
If n = 2pa, then σn,2 = (p − 1)pa−1.
If n = 2p2pa33 · · · parr , then σn,2 =
(p2−12
)·(
np2− 1)
.
Charles Camacho Oregon State University 9
Conjecture for σn,2
σn,2 =
(a− 1
2
)· na(
b − 1
2
)·(nb− 1)
where
a =
p1 p1 ≥ 3, a1 = 1
p2 p1 = 2 and eithera1 = 1, a2 ≥ 2;or a1 = 1, r = 2;
or a1 = 2, p2 = 3, a2 = 1, r = 2;or a1 ≥ 3, p2 = 3, a2 ≥ 2.Or p1 ≥ 3, a1 ≥ 3, a2 ≥ 2,
p2 < p21
4 p1 = 2, a1 ≥ 3, p2 > 3
p21 p1 ≥ 3, a1 ≥ 3, p2 > p2
1 .
b =
p2 p1 = 2 and eithera1 = 1, a2 = 1, r ≥ 3;
or a1 = 2, p2 = 3, a2 = 1, r ≥ 3;or a1 ≥ 3, p2 = 3, a2 ≥ 2.Or p1 ≥ 3, a1 ≥ 3, a2 = 1,
p2 < p21
4 p1 = 2, a1 = 2, p2 > 3.
.
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 andπ/7.
Charles Camacho Oregon State University 13
Example - C7 Actions
Consider the permutation ρ = (1, 2, 3, 4, 5, 6, 7) ∈ S7.
In terms of rotationsabout the white and black vertices, and the face, respectively:
1
2
3
4
5
6
7 5
6
7
1
2
3
4
(3, 3, 1) is described by ρ3, ρ3, ρ...
1
2
3
4
5
6
7 6
7
1
2
3
4
5
...and (2, 4, 1) is described byρ2, ρ4, ρ.
Charles Camacho Oregon State University 14
Example - C7 Actions
Consider the permutation ρ = (1, 2, 3, 4, 5, 6, 7) ∈ S7. In terms of rotationsabout the white and black vertices, and the face, respectively:
1
2
3
4
5
6
7 5
6
7
1
2
3
4
(3, 3, 1) is described by ρ3, ρ3, ρ...
1
2
3
4
5
6
7 6
7
1
2
3
4
5
...and (2, 4, 1) is described byρ2, ρ4, ρ.
Charles Camacho Oregon State University 14
Example - C7 Actions
Consider the permutation ρ = (1, 2, 3, 4, 5, 6, 7) ∈ S7. In terms of rotationsabout the white and black vertices, and the face, respectively:
1
2
3
4
5
6
7 5
6
7
1
2
3
4
(3, 3, 1) is described by ρ3, ρ3, ρ...
1
2
3
4
5
6
7 6
7
1
2
3
4
5
...and (2, 4, 1) is described byρ2, ρ4, ρ.
Charles Camacho Oregon State University 14
Counting Cn-Actions with a Given Signature (n1, n2, n3)
Theorem (Benim, Wootton 2014)
Write n =∏r
i=1 paii . Suppose Cn acts on X of signature (n1, n2, n3).
Let w ≥ 0 be the number of primes shared in common among n1, n2, n3which also share the same exponent. Relabel the primes to start withp1, . . . , pw .
Then the number T of quasiplatonic topological actions of Cn on X isgiven in the three possibilities below.
1. If each ni is distinct and w 6= 0, then
T = φ(gcd(n1, n2, n3))w∏i=1
pi − 2
pi − 1.
If w = 0, then T = φ(gcd(n1, n2, n3)).
Charles Camacho Oregon State University 15
Counting Cn-Actions with a Given Signature (n1, n2, n3)
Theorem (Benim, Wootton 2014)
2. If two ni are equal so that the signature is (n1, n2, n3) = (n1, n, n) withn1 6= n and w 6= 0, then
T =1
2
(τ1(n, n1) + φ(n1)
w∏i=1
pi − 2
pi − 1
).
If w = 0, then T = 12 (τ1(n, n1) + φ(n1)).
3. If all ni are equal so that the signature is (n1, n2, n3) = (n, n, n) andw 6= 0, then
T =1
6
(3 + 2τ2(n) + φ(n)
w∏i=1
pi − 2
pi − 1
).
If w = 0, then T = 16 (3 + 2τ2(n) + φ(n)).
Charles Camacho Oregon State University 16
Example: Deriving QC (n) when p1 = 2, a1 = 1
Claim: QC (n) =1
2
r∏i=2
{pai−1i (pi + 1)
}+ 2r−2 − 1.
Proof Sketch.
First prove the following recursive formula by induction on r :
QC (n · par+1
r+1 ) = (QC (n) + 1− 2r−2)par+1−1r+1 (pr+1 + 1)− 1 + 2r−1.
Then prove the main claim by using induction again on r , with the basecase being
QC (2pa) =1
2pa−1(p + 1).
Charles Camacho Oregon State University 17
Further Work
Analyze the growth rate of QC (n).
Find a geometric interpretation of the constant term in the formulafor QC (n).
Derive analogous formulas for quasiplatonic actions of other groups,e.g., abelian, elementary abelian Cn
p , semidirect products Cm o Cn.
Does a count on the number of group actions have a physicalinterpretations (e.g., in string theory)?
Visualization of these group actions?
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 formulafor QC (n).
Derive analogous formulas for quasiplatonic actions of other groups,e.g., abelian, elementary abelian Cn
p , semidirect products Cm o Cn.
Does a count on the number of group actions have a physicalinterpretations (e.g., in string theory)?
Visualization of these group actions?
Charles Camacho Oregon State University 18
Part Two: The Quasiplatonic Cyclic Genus Spectrum
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Set-Up
Write n =r∏
i=1
paii . Let
Aj = {1 ≤ i ≤ r : ki = j , `i = hi = ai},Bj = {1 ≤ i ≤ r : `i = j , ki = hi = ai},Cj = {1 ≤ i ≤ r : hi = j , ki = `i = ai}.
Let tn = max1≤i≤r{ai}. Then
S(n1, n2, n3) =1
pk11 · · · pkrr
+1
p`11 · · · p`rr
+1
ph11 · · · phrr
=1
n
tn∏m=0
∏i∈Am
pai−mi
+
tn∏m=0
∏i∈Bm
pai−mi
+
tn∏m=0
∏i∈Cm
pai−mi
.
Charles Camacho Oregon State University 20
Key Lemma
Lemma (C. 2019)
All signatures that achieve S1 satisfy Aj = Bj = Cj = ∅ for j ≥ 2 and|A1 ∪ B1 ∪ C1| ≤ 1.
If d ≥ 2 and n 6= pa for prime p and a ≥ 1, then all signatures thatachieve Sd satisfy Aj = Bj = Cj = ∅ for j ≥ d and|Ad−1 ∪ Bd−1 ∪ Cd−1| ≤ 1.
Charles Camacho Oregon State University 21
Example: C12-action with Signature (2, 12, 12)
1
2
3
4
5
67
8
9
10
11
12
1
2
3
4
5
67
8
9
10
11
12
(2, 12, 12)
1
72
8
9
3
10
4
115
6
12
8
1
6 11 4
927
125
10
3
σ = 3
A0 = {2} A1 = {1} A2 = ∅B0 = ∅ B1 = ∅ B2 = ∅C0 = ∅ C1 = ∅ C2 = ∅
Charles Camacho Oregon State University 22
Example: C12-action with Signature (3, 12, 12)
1
2
3
4
5
67
8
9
10
11
12
(3, 12, 12)
1
23
4
5
6
7
89
10
11
12
8
8
1
1
3
3
10
10
5
512
12
7
7
2
2
9
9
4
4
11
11
6
6
σ = 4
A0 = {1} A1 = {2} A2 = ∅B0 = ∅ B1 = ∅ B2 = ∅C0 = ∅ C1 = ∅ C2 = ∅
Charles Camacho Oregon State University 23
Example: C12-action with Signature (6, 12, 12)
1
2
3
4
5
67
8
9
10
11
12
(6, 12, 12)
7
89
10
11
12
1
23
4
5
6
1 122 3
3
4
455
6
67
788 9910
1011
1112
12
σ = 5
A0 = ∅ A1 = {1, 2} A2 = ∅B0 = ∅ B1 = ∅ B2 = ∅C0 = ∅ C1 = ∅ C2 = ∅
Charles Camacho Oregon State University 24
Preliminary Results
Theorem (C. 2019)
When n = 2pa, we have σn,2 = (p − 1)pa−1.
Proof Sketch.
The admissible signatures are of the form (2pk , p`, 2ph) (so assumeWLOG that 1 ∈ B0).
If |A1 ∪ B1 ∪ C1| = 0, these signatures achieve S1 = 3/(2pa) + 1/2.
When |A1 ∪ B1 ∪ C1| = 1, there are two cases: either A1 = {2} orB1 = {2} (omit C1 by symmetry). The value of S is verified to belarger in the case when B1 = {2}. This gives S2 = 1/pa + 1/p.
Therefore, B0 = {1} and B1 = {2}. The associated signature is(2pa, p, 2pa), so the Riemann-Hurwitz formula gives the result.
Charles Camacho Oregon State University 25
Preliminary Results
Theorem (C. 2019)
When n = 2pa, we have σn,2 = (p − 1)pa−1.
Proof Sketch.
The admissible signatures are of the form (2pk , p`, 2ph) (so assumeWLOG that 1 ∈ B0).
If |A1 ∪ B1 ∪ C1| = 0, these signatures achieve S1 = 3/(2pa) + 1/2.
When |A1 ∪ B1 ∪ C1| = 1, there are two cases: either A1 = {2} orB1 = {2} (omit C1 by symmetry). The value of S is verified to belarger in the case when B1 = {2}. This gives S2 = 1/pa + 1/p.
Therefore, B0 = {1} and B1 = {2}. The associated signature is(2pa, p, 2pa), so the Riemann-Hurwitz formula gives the result.
Charles Camacho Oregon State University 25
Preliminary Results
Theorem (C. 2019)
When n = 2pa, we have σn,2 = (p − 1)pa−1.
Proof Sketch.
The admissible signatures are of the form (2pk , p`, 2ph) (so assumeWLOG that 1 ∈ B0).
If |A1 ∪ B1 ∪ C1| = 0, these signatures achieve S1 = 3/(2pa) + 1/2.
When |A1 ∪ B1 ∪ C1| = 1, there are two cases: either A1 = {2} orB1 = {2} (omit C1 by symmetry). The value of S is verified to belarger in the case when B1 = {2}. This gives S2 = 1/pa + 1/p.
Therefore, B0 = {1} and B1 = {2}. The associated signature is(2pa, p, 2pa), so the Riemann-Hurwitz formula gives the result.
Charles Camacho Oregon State University 25
Preliminary Results
Theorem (C. 2019)
When n = 2pa, we have σn,2 = (p − 1)pa−1.
Proof Sketch.
The admissible signatures are of the form (2pk , p`, 2ph) (so assumeWLOG that 1 ∈ B0).
If |A1 ∪ B1 ∪ C1| = 0, these signatures achieve S1 = 3/(2pa) + 1/2.
When |A1 ∪ B1 ∪ C1| = 1, there are two cases: either A1 = {2} orB1 = {2} (omit C1 by symmetry). The value of S is verified to belarger in the case when B1 = {2}. This gives S2 = 1/pa + 1/p.
Therefore, B0 = {1} and B1 = {2}. The associated signature is(2pa, p, 2pa), so the Riemann-Hurwitz formula gives the result.
Charles Camacho Oregon State University 25
Preliminary Results
Theorem (C. 2019)
When n = 2pa, we have σn,2 = (p − 1)pa−1.
Proof Sketch.
The admissible signatures are of the form (2pk , p`, 2ph) (so assumeWLOG that 1 ∈ B0).
If |A1 ∪ B1 ∪ C1| = 0, these signatures achieve S1 = 3/(2pa) + 1/2.
When |A1 ∪ B1 ∪ C1| = 1, there are two cases: either A1 = {2} orB1 = {2} (omit C1 by symmetry). The value of S is verified to belarger in the case when B1 = {2}. This gives S2 = 1/pa + 1/p.
Therefore, B0 = {1} and B1 = {2}. The associated signature is(2pa, p, 2pa), so the Riemann-Hurwitz formula gives the result.
Charles Camacho Oregon State University 25
Preliminary Results (cont.)
Theorem (C. 2019)
When n = 2p2pa33 · · · parr for r ≥ 3,
σn,2 =
(p2 − 1
2
)·(
n
p2− 1
),
occurring with associated signature (p2, n/p2, n).
Charles Camacho Oregon State University 26
Proof Sketch.
Assume WLOG 1 ∈ B0.
If |A1 ∪ B1 ∪ C1| = 0, then S is of the form
S =1
n
∏i∈A0
paii + 2∏i∈B0
paii +∏i∈C0
paii
.
Write u =∏
i∈A0paii , v =
∏i∈B0
paii ,w =∏
i∈C0paii and note that
uvw = n/2.
Goal: find the second-largest maximum of the functionf (u, v) = u + 2v + n/(2uv).
This turns out to occur when B0 = {1, 3, 4, . . . , r} and A ∪ C = {2}.The associated signature is (p2, n/p2, n).
If |A1 ∪ B1 ∪ C1| = 1, S turns out to be smaller than in the previouscase.
Charles Camacho Oregon State University 27
Proof Sketch.
Assume WLOG 1 ∈ B0.
If |A1 ∪ B1 ∪ C1| = 0, then S is of the form
S =1
n
∏i∈A0
paii + 2∏i∈B0
paii +∏i∈C0
paii
.
Write u =∏
i∈A0paii , v =
∏i∈B0
paii ,w =∏
i∈C0paii and note that
uvw = n/2.
Goal: find the second-largest maximum of the functionf (u, v) = u + 2v + n/(2uv).
This turns out to occur when B0 = {1, 3, 4, . . . , r} and A ∪ C = {2}.The associated signature is (p2, n/p2, n).
If |A1 ∪ B1 ∪ C1| = 1, S turns out to be smaller than in the previouscase.
Charles Camacho Oregon State University 27
Proof Sketch.
Assume WLOG 1 ∈ B0.
If |A1 ∪ B1 ∪ C1| = 0, then S is of the form
S =1
n
∏i∈A0
paii + 2∏i∈B0
paii +∏i∈C0
paii
.
Write u =∏
i∈A0paii , v =
∏i∈B0
paii ,w =∏
i∈C0paii and note that
uvw = n/2.
Goal: find the second-largest maximum of the functionf (u, v) = u + 2v + n/(2uv).
This turns out to occur when B0 = {1, 3, 4, . . . , r} and A ∪ C = {2}.The associated signature is (p2, n/p2, n).
If |A1 ∪ B1 ∪ C1| = 1, S turns out to be smaller than in the previouscase.
Charles Camacho Oregon State University 27
Proof Sketch.
Assume WLOG 1 ∈ B0.
If |A1 ∪ B1 ∪ C1| = 0, then S is of the form
S =1
n
∏i∈A0
paii + 2∏i∈B0
paii +∏i∈C0
paii
.
Write u =∏
i∈A0paii , v =
∏i∈B0
paii ,w =∏
i∈C0paii and note that
uvw = n/2.
Goal: find the second-largest maximum of the functionf (u, v) = u + 2v + n/(2uv).
This turns out to occur when B0 = {1, 3, 4, . . . , r} and A ∪ C = {2}.The associated signature is (p2, n/p2, n).
If |A1 ∪ B1 ∪ C1| = 1, S turns out to be smaller than in the previouscase.
Charles Camacho Oregon State University 27
Proof Sketch.
Assume WLOG 1 ∈ B0.
If |A1 ∪ B1 ∪ C1| = 0, then S is of the form
S =1
n
∏i∈A0
paii + 2∏i∈B0
paii +∏i∈C0
paii
.
Write u =∏
i∈A0paii , v =
∏i∈B0
paii ,w =∏
i∈C0paii and note that
uvw = n/2.
Goal: find the second-largest maximum of the functionf (u, v) = u + 2v + n/(2uv).
This turns out to occur when B0 = {1, 3, 4, . . . , r} and A ∪ C = {2}.The associated signature is (p2, n/p2, n).
If |A1 ∪ B1 ∪ C1| = 1, S turns out to be smaller than in the previouscase.
Charles Camacho Oregon State University 27
Proof Sketch.
Assume WLOG 1 ∈ B0.
If |A1 ∪ B1 ∪ C1| = 0, then S is of the form
S =1
n
∏i∈A0
paii + 2∏i∈B0
paii +∏i∈C0
paii
.
Write u =∏
i∈A0paii , v =
∏i∈B0
paii ,w =∏
i∈C0paii and note that
uvw = n/2.
Goal: find the second-largest maximum of the functionf (u, v) = u + 2v + n/(2uv).
This turns out to occur when B0 = {1, 3, 4, . . . , r} and A ∪ C = {2}.The associated signature is (p2, n/p2, n).
If |A1 ∪ B1 ∪ C1| = 1, S turns out to be smaller than in the previouscase.
Charles Camacho Oregon State University 27
Further Work
Determine all the genera σn,d .
Analyze the sequence of gaps in the genus spectrum.
Extend to the non-quasiplatonic case.
Charles Camacho Oregon State University 28
Further Work
Determine all the genera σn,d .
Analyze the sequence of gaps in the genus spectrum.
Extend to the non-quasiplatonic case.
Charles Camacho Oregon State University 28
References
John Baez’s webpage: http://math.ucr.edu/home/baez/klein.html
Benim, R., Wootton, A. (2013). Enumerating quasiplatonic cyclic group actions.Rocky Mountain Journal of Mathematics, 43(5), 1459-1480.
Greg Egan’s webpage:http://www.gregegan.net/SCIENCE/KleinQuartic/KleinQuartic.html
Jones, G. A. (2014). Regular dessins with a given automorphism group.Contemporary Mathematics, 629, 245-260.
Jones, G. A., Singerman, D. (1987). Complex functions: an algebraic andgeometric viewpoint. Cambridge university press.
Kulkarni, R. S. (1991). Isolated Points in the Branch Locus of the Moduli Spaceof Compact Riemann Surfaces.
By Tamfang - Own work, Public Domain,https://commons.wikimedia.org/w/index.php?curid=12806647
Wootton, A. (2007). Extending topological group actions to conformal groupactions. Albanian Journal of Mathematics (ISNN: 1930-1235), 1(3), 133-143.
Charles Camacho Oregon State University 29
Thank you!
1
2
345
6
7
16
4 2
7
53 1
3
5
7
2
4
6
1
52
6
37
4
Charles Camacho Oregon State University 30
Appendix
τ1(m, n) denotes the number of nonzero, noncongruent solutions x tothe equation x2 + 2x ≡ 0 mod m where gcd(x ,m) = m/n.
τ2(n) is the number of nonzero, noncongruent solutions x to theequation x2 + x + 1 ≡ 0 mod n.
Charles Camacho Oregon State University 31
Appendix
Fuchsian groups ∆ having compact quotient space H/∆ of genus gcan be shown to have a presentation
∆ ∼=⟨a1, b1, . . . , ag , bg , c1, . . . , ck | cm1
1 = · · · = cmrr = 1,
g∏i=1
[ai , bi ] ·r∏
i=1
ci = 1⟩,
for mi positive integers, and [x , y ] := xyx−1y−1. We say ∆ hassignature (g ;m1, . . . ,mr ) with periods m1, . . . ,mr .
Theorem (e.g., Broughton (1991) and Wootton (2007)): Two(0;m1,m2,m3)-generating vectors of the finite group G define thesame equivalence class of G -actions if and only if the generatingvector lie in the same Aut(G )× Aut(Γ)-class.
(There is an analogous statement for general group actions, not necessarilyquasiplatonic.)Charles Camacho Oregon State University 32
RAINWATER SEMINAR NOTES
CHARLES CAMACHO
1. Introduction
1.1. Illustrative Example. [Genus-three surface with three-fold rotational sym-metry having genus one quotient, sphere with three-fold rotation related to dessinsand Belyı maps.]
We focus on two combinatorial problems inspired by the theory of automorphismgroups of compact Riemann surfaces:
(i) How many different ways are there to rotate a surface n times?(ii) What are the genera of surfaces having such an n-fold rotation?
The first problem is to enumerate the distinct topological actions of the cyclic groupCn on a surface, while the second problem is to determine the genus spectrum ofCn. General results are challenging, so we discuss the first nontrivial case: thequasiplatonic actions.
1.2. The General Set-up. Ramified covering map of degree d = |G| given as
X −→ X/G
where
• X: compact, connected surface of genus σ• G: finite group• X/G: quotient surface of genus g with k branch points of branch ordersn1, . . . , nk.
If such a map exists, we say X has signature (g;n1, . . . , nk). We generally assumeσ ≥ 2 so that ni ≥ 2 for each i. A quasiplatonic action is when g = 0 and k = 3.These quantities are related by the Riemann-Hurwitz formula:
σ = 1 +|G|2
(2g − 2 +
k∑i=1
(1− 1
ni
)).
2. Topological Group Actions
G acts topologically on X if there exists monomorphism ε : G ↪→ Homeo+(X).Another action ε′ : G′ ↪−→ Homeo+(Y ) will be equivalent to ε if there exists a groupisomorphism ω : G→ G′ and a homeomorphism h : X → Y such that the followingdiagram commutes:
G×X X
G′ × Y Y
εG
ω h h
εG′
1
2 CHARLES CAMACHO
That is, for any g ∈ G, x ∈ X,
h ◦ ε(g) ◦ h−1(x) = ε′(w(g))(x).
2.1. Facts.
• G acts topologically on a surface X of genus σ ≥ 2 iff G-action on Xis topologically equivalent to ∆/Γ-action on H/Γ, where ∆ is a Fuchsiangroup and ΓC∆ is torsion-free and Γ ∼= π1(X).
• G acts topologically on X iff there exists a surface kernel epimorphismϕ : ∆� G with kerϕ = Γ.
• ∆ has presentation
∆ ∼=
⟨a1, b1, . . . , ag, bg, c1, . . . , ck |
g∏i=1
[ai, bi]
k∏i=1
ci = 1, cn11 = · · · = cnk
k = 1
⟩.
• The images of the generators a1, b1, . . . , ag, bg, c1, . . . , ck of ∆ under ϕ formsa (g;n1, . . . , nk)-generating vector for G. Thus, G acts topologically on Xwith signature (g;n1, . . . , nk) iff G has a (g;n1, . . . , nk)-generating vectorand the Riemann-Hurwitz formula is satisfied.
• If G = Cn, Harvey’s Theorem gives necessary and sufficient conditionson the signatures (n1, . . . , nk) guaranteeing a Cn-action. The signaturessatisfying Harvey’s Theorem are called admissible signatures.
2.2. History.
(i) 1969: (Lloyd) Generating function for prime power cyclic group actions.(ii) 1979: (Belyı) Compact Riemann surfaces admitting a ramified covering of
the sphere over three points are in one-to-one correspondence with alge-braic curves over number fields.
(iii) 1983: (Kerckhoff) Groups of automorphisms of a surface of genus σ ≥ 2are subgroups of Γσ, and conversely, all subgroups of Γσ arise in this way.
(iv) 1999: (Jones) Enumerating subgroups of non-Euclidean crystallographicgroups.
(v) 2007: (Broughton, Wootton) Enumeration formulas for finite abelian groupactions.
(vi) 2007: (Wootton) Enumeration formulas for triangle group actions.(vii) 2013: (Jones) Enumerating regular dessins with a given automorphism
group.(viii) 2014: (Benim, Wootton) Explicit formulae enumerating the quasiplatonic
topological cyclic group actions with a given signature.
2.3. Motivations.
• Determining number of topological actions amounts to finding the conju-gacy classes of finite subgroups of the mapping class group Γσ for a fixedgenus σ ≥ 2. The formula for QC(n) explicitly enumerates the conjugacyclasses of cyclic subgroups of Γσ as σ ranges over the quasiplatonic genusspectrum.
• The moduli space Mσ of conformal equivalence classes of Riemann surfacesof genus σ is decomposed into finite, disjoint equisymmetric strata, eachof which corresponds to a unique equivalence class of actions of some finitegroup. See Broughton, 1990.
RAINWATER SEMINAR NOTES 3
• Many quasiplatonic groups are maximal in Γσ (e.g., the Hurwitz groups).Enumerating the number of quasiplatonic topological actions of the cyclicgroup of a given admissible signature yields a lower bound on the numberof conjugacy classes of maximal finite cyclic subgroups of Γσ. The formulafor QC(n) determines an explicit lower bound on the number of conjugacyclasses of maximal finite cyclic subgroups of Γσ for σ ∈ QCGS(n).
• Quasiplatonic G-actions are related to regular dessins d’enfants D withAut(D) ∼= G.
2.4. Enumerating Actions. Restrict to the quasiplatonic case. Let T (n1, n2, n3)denote the number of distinct topological Cn-actions with signature (n1, n2, n3). Wemay also refer to the number of topological actions of Cn of signature (n1, n2, n3)as the T -value of (n1, n2, n3). Let τ1(m,n) denote the number of nonzero, noncon-gruent solutions x to the equation
x2 + 2x ≡ 0 modm
where gcd(x,m) = m/n. Also, let τ2(n) be the number of nonzero, noncongruentsolutions x to the equation
x2 + x+ 1 ≡ 0 modn.
Theorem 2.1 (Benim and Wootton 2014). Let n be a positive integer and letp1, . . . , pr be the distinct primes dividing n. Suppose Cn acts on X of signature(n1, n2, n3). Let w ≥ 0 be the number of primes shared in common among n1, n2, n3which also share the same exponent and relabel these primes as p1, . . . , pw. Then thenumber T of distinct topological actions of Cn on X is given in the three possibilitiesbelow.
(i) If each ni is distinct and w 6= 0, then
T = φ(gcd(n1, n2, n3))
w∏i=1
pi − 2
pi − 1.
If w = 0, then
T = φ(gcd(n1, n2, n3)).
(ii) If two ni are equal so that the signature is (n1, n2, n3) = (n1, n, n) withn1 6= n and w 6= 0, then
T =1
2
(τ1(n, n1) + φ(n1)
w∏i=1
pi − 2
pi − 1
).
If w = 0, then
T =1
2(τ1(n, n1) + φ(n1)) .
(iii) If all ni are equal so that the signature is (n1, n2, n3) = (n, n, n) and w 6= 0,then
T =1
6
(3 + 2τ2(n) + φ(n)
w∏i=1
pi − 2
pi − 1
).
If w = 0, then
T =1
6(3 + 2τ2(n) + φ(n)) .
4 CHARLES CAMACHO
Let QC(n) denote the total number of quasiplatonic topological Cn-actions onsurfaces of genus two or greater. It follows that
QC(n) =∑
(n1,n2,n3)
T (n1, n2, n3),
the sum being taken over all admissible signature triples.
Theorem 2.2 (C. 2018).
QC(n) =1
6n∏p|n
(1 +
1
p
)− 1 + a · 2r
where
a =
14 p1 = 2, a1 = 1
12 p1 = 2, a1 = 2
1 p1 = 2, a1 ≥ 3
56 pi ≡ 1 mod 6 for 1 ≤ i ≤ r
23 p1 = 3, a1 = 1, pi ≡ 1 mod 6 for 2 ≤ i ≤ r
12 p1 = 3, a1 ≥ 2, pi ≡ 1 mod 6 for 2 ≤ i ≤ r
12 pi ≡ 5 mod 6 for some 1 ≤ i ≤ r
.
3. Genus Spectrum
The Riemann-Hurwitz formula applied to all admissible signatures (guaranteedby Harvey’s Theorem) yields a set QCGS(n) of genera admitting a quasiplatonicCn-action, called the (quasiplatonic) genus spectrum of Cn:
QCGS(n) = {σn,1, . . . , σn,sn},
where 2 ≤ σn,1 < σn,2 < · · · < σn,sn .
3.1. Facts.
• Harvey determined σn,1: The minimum genus σn,1 of a surface admittingan automorphism of order n = pa11 · · · parr is
σn,1 =
max{
2,(p1−1
2
)· np1}
a1 > 1 or n is prime
max{
2,(p1−1
2
)·(np1− 1)}
a1 = 1.
Moreover, the corresponding Cn-action on the surface of genus σn,1 isquasiplatonic.
• Without restricting the signature type, Kulkarni proved that all finitegroups will eventually act in all genera. That is, there exists a constantNG ∈ N called the stable genus increment such that if g is in the genusspectrum of G, then g ≡ 1 modNG and, expect for finitely many g whereg ≡ 1 modNG, g is in the genus spectrum.
RAINWATER SEMINAR NOTES 5
• The minimum genus µ(G) is known for cyclic groups, non-cyclic abeliangroups, metacyclic groups, alternating and symmetric groups, all sporadicsimple groups.
• The maximum genus of a surface admitting a quasiplatonic Cn-action isquickly computed from the Riemann-Hurwitz formula. This occurs forCn-actions resulting from Fuchsian triangle groups of signature (n, n, n)for n odd or (n/2, n, n) for n even. That is,
σn,sn =
n− 1
2n is odd
n− 2
2n is even
.
3.2. History.
(i) 1965: (Maclachlan) Minimum genus for noncyclic abelian.(ii) 1966: (Harvey) Minimum genus for cyclic group.(iii) 1985: (Conder) Minimum genus for alternating and symmetric groups.(iv) 1985: (Glover, Sjerve) Minimum genus for PSL2(p).(v) 1987: (Kulkarni) Stable genus and genus spectra for cyclic p-groups.
(vi) 1987: (Kulkarni) Determination of genus increment NG.(vii) 1998: (Machlachlan, Talu) Genus spectra for elementary abelian p-groups,
p-groups of cyclic p-deficiency at most 2, and some other p-groups.(viii) 1990s, 2001: (Conder, Wilson, Woldar) Minimum genus for all 26 sporadic
groups.(ix) 2001: (Weaver) Genus spectra for split metacyclic groups.
3.3. Motivations.
• The largest non-genus action is related to a Diophantine Frobenius prob-lem: the Frobenius number of a set {a1, . . . , ak} of relatively prime integersstrictly greater than 1 is the largest number which is not a linear combi-nation of the ai, which is NP-hard for k ≥ 3.
• There is a connection between gaps in the genus spectrum and “unstable”torsion in the cohomology of the mapping class group, i.e., no p-torsion inthe mapping class of the mapping class group of a particular genus κ(p),but there is p-torsion in H2(p−1)(Γκ(p);Z).
• There is interest in determining the stable genus increment NG.
3.4. How to Determine σn,d. Recall the Riemann-Hurwitz formula in the quasi-platonic case for an admissible signature (a, b, c) for Cn:
σ = 1 +|G|2
(1− 1
pk11 · · · pkrr
− 1
p`11 · · · p`rr
− 1
ph11 · · · p
hrr
).
(Here, we wrote a, b and c as factors of n =∏ri=1 p
aii .) Let
S(a, b, c) =1
pk11 · · · pkrr
+1
p`11 · · · p`rr
+1
ph11 · · · p
hrr
6 CHARLES CAMACHO
so that σ = 1 + (n/2)(1 − S(a, b, c)). We write S(a, b, c) = S for clarity. For each0 ≤ j ≤ max1≤i≤r{ai}, define the following subsets of the index set {1, . . . , r}:
Aj = {1 ≤ i ≤ r : ki = j, `i = hi = ai},Bj = {1 ≤ i ≤ r : `i = j, ki = hi = ai},Cj = {1 ≤ i ≤ r : hi = j, ki = `i = ai}.
Let tn = max1≤i≤r{ai}. It follows that
S =1
n
((tn∏m=0
∏i∈Am
pai−mi
)+
(tn∏m=0
∏i∈Bm
pai−mi
)+
(tn∏m=0
∏i∈Cm
pai−mi
)).
Define the following:
F = {(a, b, c) ∈ Z3 : (a, b, c) is an admissible triple for Cn},
Sd = the d-th largest maximum of S over F ,
Fd = {(a, b, c) ∈ F : S(a, b, c) = Sd}.
Then
S1 = max{S(a, b, c) : (a, b, c) ∈ F},and for d ≥ 1,
Sd+1 = max
S(a, b, c) : (a, b, c) ∈ F \
⋃1≤t≤d
Ft
.
The following lemma describes the sufficient conditions on the sets Ak, Bk andCk guaranteeing when a signature attains Sd.
Lemma 3.1. All signatures that achieve S1 satisfy Aj = Bj = Cj = ∅ for j ≥ 2 and|A1 ∪B1 ∪C1| ≤ 1. If d ≥ 2 and n 6= pa for prime p and a ≥ 1, then all signatures
that achieve Sd satisfy Aj = Bj = Cj = ∅ for j ≥ d and |Ad−1 ∪Bd−1 ∪Cd−1| ≤ 1.
3.5. Preliminary Results on σn,2.
Theorem 3.2. When n = 2pa, we have σn,2 = (p− 1)pa−1.
Theorem 3.3. When n = 2p2pa33 · · · parr for r ≥ 3,
σn,2 =
(p2 − 1
2
)·(n
p2− 1
),
occurring with associated signature (p2, n/p2, n).
Theorem 3.4 (CONJECTURE). The two types of signatures giving σn,2 are ofthe form (a, n, n) for a ∈ {p1, p2, 4, p21} and (b, n/b, n) for b ∈ {p2, 4}. Thus, thevalue of σn,2 is given by
σn,2 =
(a− 1
2
)· na(
b− 1
2
)·(nb− 1)
RAINWATER SEMINAR NOTES 7
where
a =
p1 p1 ≥ 3, a1 = 1
p2 p1 = 2 and eithera1 = 1, a2 ≥ 2;or a1 = 1, r = 2;
or a1 = 2, p2 = 3, a2 = 1, r = 2;or a1 ≥ 3, p2 = 3, a2 ≥ 2.
Or p1 ≥ 3, a1 ≥ 3, a2 ≥ 2, p2 < p21
4 p1 = 2, a1 ≥ 3, p2 > 3
p21 p1 ≥ 3, a1 ≥ 3, p2 > p21.
and
b =
p2 p1 = 2 and eithera1 = 1, a2 = 1, r ≥ 3;
or a1 = 2, p2 = 3, a2 = 1, r ≥ 3;or a1 ≥ 3, p2 = 3, a2 ≥ 2.
Or p1 ≥ 3, a1 ≥ 3, a2 = 1, p2 < p21
4 p1 = 2, a1 = 2, p2 > 3.
.
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