WE start by reviewing the classical theory of maps on orientable surfaces without boundary as presented in [4]. (For a survey see [3]). We consider an allowed ...
FOUNDATIONS OF THE THEORY OF MAPS ON SURFACES WITH BOUNDARY By ROBIN P. BRYANT and DAVID SINGERMAN [Received 16 August 1982; the Editors regret the delay to which this paper has been subject.]
1. Introduction
WE start by reviewing the classical theory of maps on orientable surfaces without boundary as presented in [4]. (For a survey see [3]). We consider an allowed graph (see [4] § 2), which consists of a graph all of whose vertices have finite valency and a set of edges each of which is homeomorphic to the closed interval [0,1] or the circle S 1 . As described in more detail in [4] edges homeomorphic to [0,1] have one or two vertices and edges homeomorphic to S 1 have one vertex. An edge homeomorphic to [0,1] with one vertex is called a free edge. A dart is an ordered pair consisting of a vertex and an edge directed towards that vertex so that non-free edges give two darts and free edges give one dart as illustrated in the following diagram.
(a)
(b)
(c)
Consider an imbedding of an allowed graph where we regard relations of the form (TR)°°=1 and (RL)°°= 1 as vacuous. We define the universal algebraic map of type (m, n) to be si = (F, |r|, T,L,R) where |F| denotes the underlying set of F and each g e F acts on |F| by right multiplication g: h -* hg, for all h e |F|. Let M be any subgroup of F. Then M acts as a group of automorphisms of s^ by m: h^*m~xh for all meM, he|F|; since m"1(hg) = (m~1h)g the action of M and F commute. We can form the quotient map M = siJM = (F/M*, F/M, M*T, M*L, M*R) where F/M = {Mh \ h e F}, M* is the core of M (the intersection of all conjugates of M in F) and F/M* acts on F/M by M*g: Mh —*• Mhg. THEOREM 6.1. Every algebraic map si of type (m, n) is isomorphic to a quotient of the universal algebraic map si of type (m, n).
Proof. Let si = (G, ft, T, A, p) and let F, si be as above. There is an epimorphism 6: F —> G given by T —» T, L —> A, R —> p. If G a = {gj e G | ag! = a} for some a e ft and if M = 0~\Ga) then ker 6 = M* the core of M. Defining cr: F/M*-»G by M*g-»-g0 and : M t h —> M2h and a=id: M*g —> M*g gives the required isomorphism (k) ^/M^/A* Now suppose that sllJM is isomorphic to A^jM. Then there exists a group isomorphism T/Mf mapping M t T - ^ M ^ T , M*L -»• M\L, M}i*->M£R. As M*tT, M^L, M*,R generate T/M*,, a: for all veT. We also have a bijection obeying and hence Now suppose that where e is the identity of T. Then (M1i))^ = (M1 If v € M l t then (MiU)^ = (M1e) = M2uo and so M2Uo = M2u0v.
Therefore for all ueM, and thus u o M 1 Uo 1 cM 2 From (*) we obtain and so if)'1: M2e •-* M t U o l .
Therefore we can exchange Mx with M 2 and u0 with UQ1 in the above calculation to obtain Uo 1 M 2 u o cM 1 . Therefore U O M 1 UQ 1 = M 2 . Automorphisms of Algebraic Maps. Let sd = J4/M = (T/M*, T/M, M*T, Af*L, M*R) be an algebraic map. Let Nr(M) = {ueT\ uMu~x = M] be the normalizer of M in T. Define u : T/M —*• T/M by for all g e F. As u e NjiM) this action is well-defined and it is easy to see that it defines an automorphism of si. We thus obtain a homomorphism from Nr(M) to Aut si, the automorphism group of si, defined by u >-» u. The kernel of this homomorphism is M so that NT(M)/M acts as a group
FOUNDATIONS OF THE THEORY OF MAPS
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of automorphisms of si. In fact an argument similar to that used in Theorem 6.2 shows that Aut si is isomorphic to Nr(M)/M. Now let N/M^Nr(M)/M. Then N/M acts as a group of automorphisms of si. We have the following result. THEOREM 6.3. The quotient of the algebraic map si by the group N/M is the algebraic map silN=(T/N*, T/N, N*T, N*L, N*R).
Proof. (See the paragraph entitled 'Quotient Maps'.) The group N/M acts on the F/M-cosets by Mu : Mg *-* Mu"1 g, for ueN. This divides the N/M-cosets into orbits and it is easy to see that these orbits are in one-to-one correspondence with the F/N-cosets. We now show that N*/M* is the subgroup of T/M* which fixes each orbit. Firstly F/M* acts on the F/M-cosets by
Thus M*7 fixes each N/M-orbit if given g e F there exists ueN such that
Hence gvg"1 e N for all g e F and thus -ye n g " 1 N g = N*. ger
Conversely, if yeN* we see that M*y fixes each N/M-orbit. Now F/M*/N*/M*sF/N* and the result follows.
7. Universal topological maps
We noted in the previous section that given a topological map M we can determine an algebraic map AlgAL We wish to show that for each algebraic map si we can find a topological map M such that Alg si = JL The method will be to first define a universal topological map M(m, n) which has the property that Alg M(m, n) is the universal algebraic map of type (m, n). The map M we require is then a quotient of M(m, n). For each pair of integers m, n ^ 2 , the universal map M =M{m, n) is described in § 5 of [4]. The map consists of the regular tessellation of one of the three simply-connected Riemann surfaces by regular n-gons at which m meet at each vertex. The simply-connected Riemann surface (which we always denote by °U) is the hyperbolic plane if 1/m + 1/n £. For example, (3,6) consists of a tessellation of the Euclidean plane by regular
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ROBIN P. BRYANT AND DAVID SINGERMAN
hexagons of which 3 meet at each vertex. Each n-gon in At(m, n) is the union of 2n triangles with angles TT/2, -n/m, TT/TI.
FIG.
20
Let ABC denote one of these triangles with A = ir/2, J§ = v/m, C = TT/TI and let t be the reflection in AB, I be the reflection in AC and r be the reflection in BC. The group generated by t, I, r has presentation gp(t, I,r\t2=l2
= r2 = (It)2 = (rr) m = (rl)" = 1>
(see [6]) so that this group is isomorphic to F[m, n]. We shall denote this group by T[m, n] too. It is a discontinuous group of conformal isometries of °U which maps At to itself. The triangle ABC is a fundamental region for this group. The edges of At are the images of the edge BB' under the group F[m, n] the vertices are the images of B and the face centres are the images of C. On each edge of At we can construct four blades in the usual way. Note that the triangle ABC contains precisely one of these blades and hence there is one blade in each F[m, n\ image of ABC. As ABC is a fundamental region for F[m, n] there is a one-to-one correspondence between the elements of T[m, n] and the blades of M. If b0 is the blade in ABC then every blade of M is of the form 60g, g € F[m, n]. We shall call b0 the fundamental blade of At. Let il denote the set of blades of M. Then as in § 3 we obtain three involutions f, A, p of A, the transverse, longitudinal and rotary reflections. THEOREM
7.1. The group G- generated by f, A, pis isomorphic to F[m, n].
fVoof. 6 is a group of permutations of Cl. We first observe that the elements of T[m, n] commute with the elements of (J. To see this it is enough to observe that the involutions f, A, p commute with elements of
FOUNDATIONS OF THE THEORY OF MAPS
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G. For example if g e F[m, n] then the effect of gA is
7
bgX
FIG.
21
while the effect of Ag is
7 bXg hi Fio. 22
giving gA = Ag. Similarly gr = fg and gp = pg. Now T[m, n] clearly acts transitively and freely on Cl by h: bog -* ^og^, f° r aU g,heT[m,n]. We now show that ( J acts freely on A. For suppose that ueG and bu = b for some b € & Then for some keT[m,n], b = bok. Hence boku = bok and thus bouk = bok. Therefore bou = b0 and so (bou)h = boh and thus (boh)u = b 0 ^ f° r all h 6 F[m, n]. As T[m, n] is transitive on ft, u fixes all blades of ft and so is the identity. Now suppose that uu u2eG and that
where fu g2 e IT"*, n]. Then
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ROBIN P. BRYANT AND DAVID SINGERMAN
If bou = bog we define a: G —*• F by cr(u) = g"1. This is well-defined as G acts freely on ft, and from above aiuiuj = (g2gi)"1 = g ^ g i 1 = criu^aiu-2) Thus cr is a homomorphism. As F acts freely on ft the kernel of a is trivial and as M is connected cr is surjective. Therefore a: G -* F is an isomorphism. THEOREM 7.2. Alg M (m, n) is isomorphic to the universal algebraic map of type (m, n).
Proof. Alg M{m, n) = (G, ft, f, A, p) and the universal map of type (m, n) is (F, |F|, T, L, i?) where F = F[m, n], which is obviously isomorphic to (r, |r|, t, I, r). It is now easy to verify that if we define : fl —* |r| by 4>: boh —> h" 1 and cr: G —* F as in the proof of Theorem 7.1, then (cr, ) an isomorphism of algebraic maps. Note. In § 5 of [4] the corresponding results to Theorems 7.1, 7.2 were accepted without proof. The proofs given here adapt straightforwardly to the situation of [4]. THEOREM 7.3. If si is an algebraic map then there is a topological map M such that Alg M = si. Proof. If si has type (m, n) let Ji = M(m, n) and F = T[m, n]. Suppose that si has map subgroup M < F . As F maps M to jtself so does the subgroup M. Hence we can form the quotient map M = MJM which is imbedded as a map in the surface
