The non-canonical gluings of two affine spaces.

Barbara Baumeister, Gernot Stroth

April 9, 1997

1 Introduction.

Given two geometries each equipped with a suitable parallelism giving rise to the same geometry at infinity, a new geometry can be constructed by gluing the two geometries together along their geometry at infinity. This method of constructing new geometries was introduced by Buekenhout, Huybrechts and Pasini [BHP]. Suppose that the obtained geometry is of rank 3. Then each point is incident to each plane, i.e. Γ is flat. Conversely, many flat geometries of rank three are gluings – see for instance the classification of flat geometries of type Af.c^{∗}, Af.Af^{∗}, c.c^{∗} [BDMP, DP, BP]. In [Pas2] Pasini examines the special case of gluing two affine spaces.

Let (A_{0}, ||_{0}) and (A_{2}, ||_{2}) be two affine spaces isomorphic to AG(n, q) and let Γ^{∞} = Γ_{i}^{∞} be the set of parallel classes of ||_{i} (i = 0, 2). Then the gluing AS(n, q) ∘_{α} AS(n, q) of the points and lines of A_{0} and of A_{2} via the permutation α of Γ^{∞} is a geometry of rank three consisting of points, lines and planes defined as follows (see [Pas2]): The points are the points of A_{0}, the planes the points of A_{2} and the lines the pairs (l_{0}, l_{2}) with l_{i} a line in A_{i} such that l_{0}^{∞α} = l_{2}^{∞}. Any point is incident with any plane and a point p (resp. plane u) is incident with a line (l_{0}, l_{2}) iff p (resp. u) is incident to l_{0} (resp. l_{2}) in A_{0} (resp. A_{2}).

The gluing is said to be canonical, if α ∈ P ΓL_{n}(q), see [Pas2, 1.3.4]. In this paper we determine all the flag-transitive non-canonical gluings of two isomorphic desarguesian affine spaces AS(n, q) ∘_{α} AS(n, q).

Let Γ = AS(n, q) ∘_{α} AS(n, q), let A_{i} = Aut(A_{i}, ||_{i})) and let K_{i}^{∞} be the kernel of the action of A_{i} on ||_{i}, for i = 0, 2. Then A_{i}^{∞} = A_{i}/K_{i}^{∞} is the group induced by A_{i} on Γ_{i}^{∞} (i = 0, 2). Set X = A_{0}^{∞} ∩ A_{2}^{∞α}. Then Aut(Γ) = (K_{0}^{∞} × K_{2}^{∞})X, see Lemma 3.1.

There are sixteen sporadic examples (nine with n = 2, three with n = 4, two with n = 6) and two infinite series. More precisely we prove the following.

Theorem 1.1 Let Γ be a non-canonical gluing AS(n, q) ∘_{α} AS(n, q) where q = p^{r}. Then one of the following holds.

(1) (i) n = 2, q ∈ {5, 7, 11, 23} and X ≅ S_{4};

(ii) n = 2, q = 11 and X ≅ A_{4}; (2) n = 2, q ∈ {11, 19, 29, 59} and X ≅ A_{5};

(3) n = 4, q = 3, O_{2}(X) ≅ E_{2^{4}} and ZZ_{5} ≤ X/O_{2}(X) ≤ F rob(5 : 2);

(4) n = 6, q = 3 and X ≅ L_{2}(13);

(5) n > 2 or n = 2 and q ∉ {2, 3, 4, 5, 8, 9}; X = N_{ΓL_{n}(q)}(T), with T a Singer cycle of ΓL_{n}(q);

(6) gcd(r · n, (q^{n} − 1)/(q − 1)) ≠ 1, X is naturally embedded into ΓL_{1}(q^{n}), but not of type (5).

In cases (1), (2) and (3) we may choose α such that [X, α] = 1. In case (4) there exists just two examples: one with [X, α] = 1 and the other with X⟨α⟩ ≅ P GL_{2}(13).

Moreover, we determine the universal covers of the 15 exceptional gluings of type (1) – (4) as well as of the canonical gluings AS(n, q) ∘_{α} AS(n, q). We prove that the gluings in (1), (2) and (3) and in (4) with X⟨α⟩ ≅ P GL_{2}(13) are simply connected (see Propositions (5.4) – (5.6)). In case (4), if [X, α] = 1, then the universal cover of Γ is a triple cover (see Proposition (5.6)). Let Γ be the canonical gluing Γ = AS(n, q) ∘_{α} AS(n, q), then the universal cover is a q^{\binom{n}{2}} (resp. q^{2^{n}−n−1}) cover, if q is odd (resp. q is even) (see Theorem 6.4).

Theorem1.1 and the determination of certain universal covers solve some of the problems posed by Pasini in [Pas2, (5.4)].

Let Aut(Γ) = (K_{0}^{∞} × K_{2}^{∞})X. Then the group X acts faithfully on the set, say Γ_{0}^{∞}, of parallel classes of lines of A_{0}. Note, that Γ_{0}^{∞} is just the set of points of the projective space P G(n − 1, q). Moreover, Aut(Γ) acts flag-transitively on Γ if and only if X acts transitively on Γ_{0}^{∞}. In [Pas2, (4.3)] Pasini posed the question, if a group X ≤ Sym(Γ_{0}^{∞}) = Sym((q^{n}−1)/(q−1)) is transitive on the set of points of P G(n−1, q) precisely, if X contains a Singer-cycle. Since (1) - (4) and (6) of the main theorem do not contain a Singer-cycle, our answer is negative. Pasini also asked, whether, if X contains a Singer-cycle T and if q is large enough, X is contained in the normalizer of T in P ΓL_{n}(q). Here our answer is affirmative.

Our notation is fairly standard, see for instance [Pas1]. For G = Aut(Γ) and {a_{0}, a_{1}, a_{2}} a maximal flag of Γ we denote by G_{i} the stabilizer of a_{i} in G.

The proof of the Main Theorem runs as follows. Let (Γ, Aut(Γ)) be a flag-transitive non-canonical gluing AS(n, q) ∘_{α} AS(n, q). Then G_{0}/Z(G_{0}) is doubly transitive of degree q^{n}. Moreover, if X is not isomorphic to a subgroup of ΓL_{1}(q^{n}), then we prove that we may suppose that α normalizes X. By definition Γ is a non-canonical gluing if and only if α is in Sym((q^{n} − 1)/(q − 1)) \ A_{0}^{∞}. Hence we have just to determine those doubly transitive permutation groups of degree q^{n} for which the normalizer of X in Sym((q^{n} − 1)/(q − 1)) is not contained in A_{0}^{∞}.

2 Preliminaries.

Lemma 2.1 Let G be a group which acts faithfully and transitively on a set Ω. Then C_{Sym(Ω)}(G) acts regularly on the set of fixed points, say Λ, of G_{ω} on Ω for any ω ∈ Ω and C_{Sym(Ω)}(G) ≅ N_{G}(G_{ω})/G_{ω}.

Proof: The representations of G on Ω̃ = {G_{ω}g | g ∈ G} and on Ω are equivalent: for ψ : Ω̃ → Ω, G_{ω}g ↦ ω^{g} holds x^{ψg} = x^{gψ} for all x ∈ Ω̃ and all g ∈ G. Therefore C_{Sym(Ω)}(G) acts regularly on Λ.

Suppose that G acts regularly on Ω. Then the action of G on Ω and the action of G on G via right multiplication are equivalent. Moreover,

ψ : G → C_{Sym(Ω)}(G), g ↦ c_{g} with ω^{c_{g}} = ω^{g^{−1}}

for all ω ∈ Ω and all g ∈ G is an isomorphism. As C_{Sym(Ω)}(G) andN_{G}(G_{ω})/G_{ω} act regularly on Λ, we may embed C_{Sym(Ω)}(G) into C_{Sym(Λ)}(N_{G}(G_{ω})/G_{ω}). Application of the above reasonning to N_{G}(G_{ω})/G_{ω} and Λ shows C_{Sym(Λ)}(N_{G}(G_{ω})/G_{ω}) ≅ N_{G}(G_{ω})/G_{ω}. Since C_{Sym(Ω)}(G) and N_{G}(G_{ω})/G_{ω} are of the same order, also the last assertion of the lemma holds.

Let G be an affine doubly transitive permutation group of degree p^{n}, p prime. Then G = O_{p}(G) : Y, V = O_{p}(G) is elementary abelian and Y acts transitively and faithfully on V^{#}.

In the following we will use the notation introduced by Hering [He]. Let L ⊆ Hom(V, V) be maximal in respect with the conditions that

(i) L is normalized by Y;

(ii) L contains the identity and L is a field with the addition and multiplication of Hom(V, V).

Then L is a field extension of GF (p). Let |L| = p^{r} = q and n such that |V | = q^{n}. We say that Y is naturally embedded into ΓL_{n}(q). The affine doubly transitive permutation groups are classifyed by Hering and Liebeck [He,Li]. There are four infinite series: n = 1 or H ≤ Y ≤ ΓL_{n}(q), q = p^{r} with H ∈ {SL_{n}(q), Sp_{n}(q), n ≥ 4, G_{2}(q)} (see [He] or [Li]). We will call the remaining affine doubly transitive permutation groups the exceptional doubly transitive permutation groups.

Lemma 2.2 Let H = O_{p}(H) : Y be an affine doubly transitive permutation group and suppose that Y is naturally embedded into G = ΓL_{n}(q), n ≥ 2. Let Z be a subgroup of G which is isomorphic to Y and which acts transitively on O_{p}(H)^{#}. Then Y and Z are conjugate in ΓL_{n}(q).

Proof: Set V = O_{p}(H). As Z acts transitively on V^{#}, also V : Z is a doubly transitive affine permutation groups of degree q^{n}. Hence Y and Z are one of the groups listed by Hering (see [He] or [Li]).

Let Z be naturally embedded into ΓL_{n_{z}}(p^{r_{z}}). Set L_{y} = GF (q) and L_{z} = GF (p^{r_{z}}).

Case 1. Let Y be non solvable. Then E(Y) ∈ {SL_{n}(q); Sp_{n}(q), n ≥ 4; G_{2}(q), n = 6; SL_{2}(5), n = 2, q = 9, 11, 19, 29, 59; n = 4, A_{6}, q = 2; A_{7}, n = 4, q = 2; SL_{2}(13), n = 6, q = 3; G_{2}(2)′ ≅ U_{3}(3), n = 6, q = 2} or Y ≅ 2^{1+4}A_{5} or 2^{1+4}Sym(5) and n = 4, q = 3.

Suppose E(Y) ≅ SL_{n}(q). Then q = p^{r_{z}} and n = n_{z}. Therefore L_{y}^{#} and L_{z}^{#} are cyclic of order q − 1. Moreover, L_{y}^{#} (and also L_{z}^{#}) decompose V into GF (p)-subspaces of dimension r such that L_{y}^{#} (resp. L_{z}^{#}) acts as field multiplication on these subspaces. Since Y (resp. Z) acts irreducibly on V, these subspaces are isomorphic GF (q)L_{y}^{#} (resp. GF (q)L_{z}^{#})-modules. Therefore L_{y}^{#} and L_{z}^{#} are conjugate in G. This implies that E(C_{G}(L_{y}^{#})) = E(Y) and E(C_{G}(L_{z}^{#})) = E(Z) are conjugate in G. Let E = E(Y)^{g} = E(Z). As Y ≅ Z, there is a ϕ ∈ Aut(E) such that Y^{gϕ} = Z. Since we may assume ϕ ∈ ΓL_{n}(q), the groups Y and Z are conjugate in G.

Now let E(Y) ≅ Sp_{n}(q), (n, q) ≠ (4, 2). Again q = p^{r_{z}} and n = n_{z}. E(Y) and E(Z) are contained in subgroups S, T of G isomorphic to SL_{2n}(q). Since E(Y) and E(Z) act transitively on V, also S and T act transitively on V and therefore S and T are conjugate in G and we may assume E(Y), E(Z) ≤ S. As there exist up to equivalence only one symplectic scalar product on the L-vectorspace V (see for instance [As]), it follows that E(Y) and E(Z) are conjugate in G. As above, we obtain that Y and Z are conjugate in G.

Let E(Y) ≅ G_{2}(q) or G_{2}(2)′. Then again q = p^{r_{z}} and we may assume E(Y), E(Z) ≤ H ≤ G, H ≅ Sp_{6}(q). According to [Kl] E(Y) and E(Z) are conjugate in N_{G}(H). Since the outer automorphism group of E(Y) is cyclic, the assertion also holds in this case.

So let E(Y) ≅ SL_{2}(5) and q ∈ {9, 11, 19, 29, 59}. Again q = p^{r_{z}}. According to [Kl] there are two conjugacy classes of subgroups isomorphic to SL_{2}(5) in SL_{2}(q). Since P GL_{2}(q) does not contain S_{5} or A_{5} × ZZ_{2}, E(Y) and E(Z) are conjugate in G. The fact that Y ≤ E(Y)Z(G) and Z ≤ E(Z)Z(G) yields again the assertion.

In L_{4}(2) ≅ A_{8} there is just one conjugacy class of subgroups isomorphic to A_{6}. Therefore the assertion holds for Y ≅ A_{6} and Y ≅ A_{7}.

If Y ≅ SL_{2}(13), q = 3 and n = 6, then Y ≤ H ≤ G, H ≅ Sp_{6}(3). We may assume Z ≤ H, as well. Now according to [CCNPW] all subgroups of Aut(Sp_{6}(3)) isomorphic to SL_{2}(13) are conjugate.

Finally, if (n, q) = (4, 3) and Y ≅ 2^{1+4}A_{5} or 2^{1+4}S_{5}, then again according to [CCNPW] Y and Z are conjugate in G. Thus the assertion holds in case 1.

Case 2. Y is solvable. Let n = 2. Then Y contains a subgroup isomorphic to SL_{2}(3)

[Hu] and q ∈ {3, 5, 7, 11, 23}. We have ΓL_{1}(q^{2}) ≅ ZZ_{8} : ZZ_{2}, ZZ_{24} : ZZ_{2}, ZZ_{48} : ZZ_{2}, ZZ_{120} : ZZ_{2}, ZZ_{528} : ZZ_{2}, respectively. The facts that |X| |̸|P ΓL_{1}(q^{2})| or that O_{3}(ΓL_{1}(q^{2})) ≠ 1 imply n_{z} = 2. Therefore Z is of the same exceptional type as Y and by [Hu] Y and Z are conjugate in G.

If (n, q) = (4, 3) and Y is containing a subgroup isomorphic to 2^{1+4}ZZ_{5}, then according to [Hu] the assertion holds for Y and Z. This proves the lemma.

Lemma 2.3 Let G = O_{p}(G) : Y be an affine doubly transitive permutation group naturally embedded into ΓL_{n}(q), q = p^{r}, n ≥ 2 such that H ≤ Y with H ∈ {SL_{n}(q), Sp_{n}(q), n ≥ 4, G_{2}(q)}. Let m and t be natural numbers with n | m and q^{n} = t^{m} and let Γ^{∞} be the set of 1-dimensional GF (t)-subspaces of O_{p}(G). Moreover, let X be the group induced by Y on Γ^{∞}. Then C_{Sym(Γ^{∞})}(X) = C_{P ΓL_{m}(t)}(X) and N_{Sym(Γ^{∞})}(X) = N_{P ΓL_{m}(t)}(X).

Proof: Let ω ∈ Γ^{∞} and let M = stab_{X}(ω). Then N_{X}(M)/M ≅ C_{ΓL_{m}(t)}(E(X))/Z(GL_{n}(q)). Therefore, Lemma 2.1 yields the first part of the assertion.

Since ΓL_{n}(q) embeds into ΓL_{m}(t) also the second part of the assertion follows.

Lemma 2.4 Let G = O_{p}(G) : Y be an affine exceptional doubly transitive permutation group with Y ≤ ΓL_{n}(q), |O_{p}(G)| = q^{n}. Let Γ^{∞} be the set of 1-spaces of V = O_{p}(G) considering V as a natural GF (q)-module for Y. Then Y induces a group X on Γ^{∞} as listed in Table 1. Moreover, X_{ω}, ω ∈ Γ^{∞}, and N_{X}(X_{ω})/X_{ω} are as listed in Table 1.

Proof: In order to determine N_{Sym(Γ^{∞})}(X) we are using the fact that by Lemma 2.1 C_{Sym(Γ^{∞})}(X) ≅ N_{X}(X_{ω})/X_{ω}.

If n = 2, then it is easy to calculate N_{X}(X_{ω})/X_{ω} and N_{Sym(Γ^{∞})}(X).

If n = 4 and q = 2, then |Γ^{∞}| = 15, X ≅ A_{6} or A_{7} and in both cases C_{Sym(Γ^{∞})}(X) = 1. Since Sym(7) has no permutation representation of degree 15, we obtain N_{Sym(Γ^{∞})}(X) ≅ Sym(6) or A_{7}, as claimed.

Now let n = 4 and q = 3. If Y ≅ SL_{2}(5) ∗ ZZ_{8} (resp. SL_{2}(5)2 ∗ ZZ_{8}), then X ≅ A_{5} × ZZ_{4} (resp. Sym(5) × ZZ_{4}), X_{ω} ≅ Sym(3) (resp. Sym(3) × ZZ_{2}) and therefore C_{Sym(Γ^{∞})}(X) = Z(X) ≅ ZZ_{4} and N_{Sym(Γ^{∞})}(X) = Sym(5) × ZZ_{4}.

If X = O_{2}(X)N_{X}(S), S ∈ Syl_{5}(X), then according to the Theorem of Schur-Zassenhaus N_{Sym(Γ^{∞})}(X) = O_{2}(X)N_{M}(N_{X}(S)) with M = N_{Sym(Γ^{∞})}(O_{2}(X)). If N_{GL_{4}(3)}(N_{X}(S)) induces the full automorphism group on N_{X}(S), then N_{M}(N_{X}(S)) = C_{Sym(Γ^{∞})}(X) N_{GL_{4}(3)}(N_{X}(S)). In all cases it is straightforward to determine N_{X}(X_{ω})/X_{ω}. Therefore, the assertion holds, if X = O_{2}(X)N_{X}(S). If X/O_{2}(X) ≅ A_{5}, then X_{ω} ≅ ZZ_{2}A_{4} and therefore N_{X}(X_{ω}) = X_{ω}. This proves the assertion for X/O_{2}(X) ≅ A_{5} or Sym(5). Finally, let n = 6, q = 3 and X ≅ L_{2}(13). Then X_{ω} ≅ ZZ_{3} which implies the assertion. This proves the lemma.

Corollary 2.5 Let G = O_{p}(G) : Y be an affine exceptional doubly transitive permutation group with Y ≤ ΓL_{n}(q), |O_{p}(G)| = q^{n}. Let Γ^{∞} be the set of 1-spaces of V = O_{p}(G) considering V as a natural GF (q)-module for Y. Let X be the group induced by Y on Γ^{∞}. If N_{Sym(Γ^{∞})}(X) > N_{ΓL_{n}(q)}(X), then one of the following holds.

(1) n = 2, q ∈ {5, 7, 11, 23} and A_{4} ≤ X;

(2) n = 2, q ∈ {11, 19, 29, 59} and X ≅ A_{5};

(3) n = 4, q = 3, O_{2}(X) ≅ 2^{4} and ZZ_{5} ≤ X/O_{2}(X) ≤ F rob(5 : 2);

Figure 1: Table 1.

n q X  X_{ω}  N_{X}(X_{ω})/X_{ω} N_{Sym(Γ^{∞})}(X)
2 3 A_{4}  ZZ_{3}  1  Sym(4)
Sym(4)  Sym(3)  1  Sym(4)
5 A_{4}  ZZ_{2}  ZZ_{2}  Sym(4) × ZZ_{2}
Sym(4)  ZZ_{4}  ZZ_{2}  Sym(4) × ZZ_{2}
7 Sym(4)  ZZ_{3}  ZZ_{2}  Sym(4) × ZZ_{2}
11 A_{4}  1  A_{4}  (A_{4} × A_{4})2
Sym(4)  ZZ_{2}  ZZ_{2}  Sym(4) × ZZ_{2}
23 Sym(4)  1  Sym(4)  Sym(4) × Sym(4)
9 A_{5}  Sym(3)  1  Sym(5)
Sym(5)  Sym(3) × ZZ_{2} 1  Sym(5)
11 A_{5}  ZZ_{5}  ZZ_{2}  Sym(5) × ZZ_{2}
19 A_{5}  ZZ_{3}  ZZ_{2}  Sym(5) × ZZ_{2}
29 A_{5}  ZZ_{2}  ZZ_{2}  Sym(5) × ZZ_{2}
59 A_{5}  1  A_{5}  (A_{5} × A_{5})2
4 2 A_{6}  Sym(4)  1  Sym(6)
2 A_{7}  L_{3}(2)  1  A_{7}
3 A_{5} × ZZ_{4}  Sym(3)  ZZ_{4}  Sym(5) × ZZ_{4}
Sym(5) × ZZ_{4} Sym(3) × ZZ_{2} ZZ_{4}  Sym(5) × ZZ_{4}
E_{2^{4}} : ZZ_{5}  ZZ_{2}  E_{2^{3}}  ((E_{2^{4}} : ZZ_{5}) × E_{2^{3}})ZZ_{3} : ZZ_{4}
E_{2^{4}} : ZZ_{5} : ZZ_{2} ZZ_{2}ZZ_{2}  ZZ_{2}  ((E_{2^{4}} : (ZZ_{15} : ZZ_{2})) × ZZ_{2})ZZ_{2}
E_{2^{4}} : ZZ_{5} : ZZ_{4} ZZ_{2}ZZ_{4}  1  E_{2^{4}} : (ZZ_{5} : ZZ_{4})
E_{2^{4}} : A_{5}  ZZ_{2}A_{4}  1  E_{2^{4}} : Sym(5)
E_{2^{4}} : Sym(5) ZZ_{2}Sym(4)  1  E_{2^{4}} : Sym(5)
6 3 L_{2}(13)  ZZ_{3}  E_{4}  (L_{2}(13) × E_{4})2

(4) n = 6, q = 3 and X ≅ L_{2}(13);

Lemma 2.6 Let q = p^{r} be a power of a prime p. Let n > 2 or n = 2 and q ≠ 2, 3, 4, 5, 8, 9. Then φ(q^{n} − 1/q − 1) > rn.

Proof. Let r = 1, n = 2. Then we claim φ(p + 1) > 2, which is obviously true besides p = 2, 3, or 5. Assume next q^{n} = 64. Then q = 2, 4, or 8. Hence q^{n} − 1/q − 1 = 63, 21, or 7 and φ(q^{n} − 1/q − 1) = 36, 12, or 6. So the conclusion holds in all these cases. Hence we may assume that there is some Zsigmondy prime t dividing q^{n}−1. In particular φ(q^{n}−1/q−1) ≥ t − 1.

Assume by way of contradiction φ(q^{n} − 1/q − 1) ≤ rn. Then we have t − 1 ≤ rn. But by the choice of t we have rn | t − 1. This now shows t − 1 = rn and φ(q^{n} − 1/q − 1) = t − 1. Then we get q^{n} − 1/q − 1 = t or 2t. As t is odd, we have an integer u = (t − 1)/2 and t | p^{u} + 1. As p^{t−1} − 1 = t(q − 1) or 2t(q − 1), we see that p^{u} − 1 divides q − 1 or 2(q − 1). So in any case we get p^{u} − 1 | q − 1, as 2 | p^{u} + 1 in case p odd. Then u | r. But 2u = nr. This shows n = 2 and u = r. We now have p^{(t−1)/2} + 1 = t, (p = 2) or 2t, p odd. Let p > 3. Then p^{(t−1)/2} > 2^{t−1}. Hence we get t ≥ 2^{t−2}. This shows t = 3. But then u = f = 1, and we are done. So we are left with p = 2 and p = 3.

Let p = 2. Then 2^{(t−1)/2} + 1 = t. So t ≤ 5. But then q = 2 or 4, which are exceptional cases in the assertion. Let p = 3, then 3^{(t−1)/2} + 1 = 2t and we get t ≤ 5. Then q = 3 or q = 9 and again we have the exceptional cases. This proves the lemma.

3 On gluings.

3.1 The construction.

Let (A_{0}, ||_{0}) and (A_{2}, ||_{2}) be two affine spaces isomorphic to A(n, q) and let Γ^{∞} = Γ_{i}^{∞} be the set of parallel classes of ||_{i} (i = 0, 2). Then Pasini introduced in [Pas2] the gluing AS(n, q) ∘_{α} AS(n, q) of the points and lines of A_{0} and of A_{2} via the permutation α of Γ^{∞} as follows: The points are the points of A_{0}, the planes the points of A_{2} and the lines the pair of lines (l_{0}, l_{2}) with l_{i} a line in A_{i} such that l_{0}^{∞α} = l_{2}^{∞}. Any point is incident with any plane and a point p (resp. plane u) is incident with a line (l_{0}, l_{2}) iff p (resp. u) is incident to l_{0} (resp. l_{2}) in A_{0} (resp. A_{2}).

Let Γ = AS(n, q) ∘_{α} AS(n, q) be the gluing of two desargues affine spaces and let A_{i} = Aut(A_{i}, ||_{i})), K_{i}^{∞} the kernel of the action of A_{i} on ||_{i} and A_{i}^{∞} the group induced by A_{i} on Γ_{i}^{∞} (i = 0, 2). Then A_{0}^{∞} = A_{2}^{∞} = A^{∞}. Moreover, let K_{i} = O_{p}(K_{i}^{∞}) and Z_{i} be the stabilizer in K_{i}^{∞} of the zero-point of A_{i}, for i = 0, 2. Then K_{i} ≅ E_{q^{n}} and Z_{i} ≅ ZZ_{q−1}(i = 0, 2).

Lemma 3.1 Let Γ = AS(n, q) ∘_{α} AS(n, q). Then G = Aut(Γ) = (K_{0}^{∞} × K_{2}^{∞})X, where X = A^{∞} ∩(A^{∞})^{α}, G_{0} = (Z_{0} ×K_{2}^{∞})X, G_{2} = (Z_{2} ×K_{0}^{∞})X and G_{1} = ((⟨x_{1}⟩×(⟨α(x_{1})⟩)(Z_{0} × Z_{2})N_{X}(⟨x_{1}⟩) with ⟨x_{1}⟩ ≤ K_{2}, |⟨x_{1}⟩| = q. In particular, G acts flag-transitively on Γ iff G_{0}/Z_{0} and G_{2}/Z_{2} are doubly transitive permutation groups of degree q^{n}.

Proof: See [BHP, Theorem (3.6)].

In the following we will use the notation just introduced. According to [BHP, Theorem (3.9)] we have

Lemma 3.2 Let Γ = AS(n, q) ∘_{α} AS(n, q) and Γ′ = AS(n, q) ∘_{α′} AS(n, q). Then Γ and Γ′ are isomorphic iff α′ ∈ A^{∞}αA^{∞}.

Remark Lemma 3.2 implies that, if Γ is the canonical gluing AS(n, q) ∘_{α} AS(n, q) we may assume that α is the identity on Γ^{∞} and that α induces a GF (q)-linear mapping from K_{2} onto K_{0}.

3.2 Examples.

Let H = O_{p}(H) : Y ≤ AΓL_{n}(q) such that O_{p}(H) is the translation group of A = AΓL_{n}(q) and such that Y acts transitively on O_{p}(H)^{#}. Let Γ^{∞} be the set of 1-spaces of O_{p}(H), A^{∞} the subgroup of Sym(Γ^{∞}) induced by A on Γ^{∞} and let N be the normalizer of Y^{∞} in Sym(Γ^{∞}). Suppose N > N_{A^{∞}}(Y^{∞}). Then Γ = AS(n, q) ∘_{α} AS(n, q) is a non-canonical gluing of the two affine spaces isomorphic to AS(n, q) via the bijection α ∈ N \ N_{A^{∞}}(Y^{∞}). Moreover,

Aut(Γ) = (K_{0}^{∞} × K_{2}^{∞})X, where K_{i}^{∞} ≅ E_{q^{n}}ZZ_{q−1} and Y^{∞} ≤ X < A^{∞},
with X = A^{∞} ∩ (A^{∞})^{α}.

If C_{A^{∞}}(Y^{∞}) = 1, but C_{Sym(Γ^{∞})}(Y^{∞}) ≠ 1, then we may chose in particular α ∈ C_{Sym(Γ^{∞})}(Y^{∞})^{#}. The following examples are constructed in this way.

(a) Let n = 2, q ∈ {5, 7, 11, 23} and let Y be a group listed in [Hu]. By the Lemma of Schur C_{A^{∞}}(Y^{∞}) = 1, but by Lemma 2.1 and Table 1 C_{Sym(Γ^{∞})}(Y^{∞}) ≠ 1. Let α ∈ C_{Sym(Γ^{∞})}(Y^{∞})^{#}. According to Table 1 either X ≅ Sym(4) or q = 11 and X ≅ A_{4}.

(b) Let n = 2, q ∈ {11, 19, 29, 59} and let Y be a group listed in [He, Paragraph 5, E1]. Then Y^{∞} ≅ A_{5} and C_{A^{∞}}(Y^{∞}) = 1. By Lemma 2.1 and Table 1 C_{Sym(Γ^{∞})}(Y^{∞}) ≠ 1 and for α ∈ C_{Sym(Γ^{∞})}(Y^{∞})^{#} we have X ≅ A_{5}.

(c) Let n = 4, q = 3 and let Y ≅ 2^{1+4}ZZ_{5}. Again C_{P ΓL_{4}(3)}(Y^{∞}) = 1. By Table 1 C_{Sym(Γ^{∞})}(Y^{∞}) ≅ 2^{3} and X/O_{2}(X) ≅ ZZ_{5} or F rob(5 : 2).

(d) Let n = 6, q = 3 and let Y ≅ SL_{2}(13). Then C_{A^{∞}}(Y^{∞}) = 1 and according to Table 1 N_{Sym(Γ^{∞})}(Y^{∞}) ≅ (L_{2}(13) × 2^{2})2. So we may chose α either such that α commutes with Y^{∞} or such that Y^{∞}⟨α⟩ ≅ P GL_{2}(13). In both cases X ≅ L_{2}(13).

(e) Let Y be naturally embedded into ΓL_{1}(q^{n}) and assume that Y^{∞} contains a Singercycle T of A^{∞}. Moreover, let n > 2 or n = 2 and q ∉ {2, 3, 4, 5, 8, 9}. Recall that T is cyclic of order (q^{n} − 1)/(q − 1). Let q = p^{r}. By Lemma 2.6 |Aut(T)| > r · n. Therefore N_{A^{∞}}(T) ≅ T : ZZ_{r·n} is a proper subgroup of N_{Sym(Γ^{∞})}(T) ≅ T : Aut(T) (see Lemma 2.1). Let α ∈ N_{Sym(Γ^{∞})}(T) \ A^{∞}. Since N_{Sym(Γ^{∞})}(T)/T is abelian, X is the normalizer of T in A^{∞}.

(f) Let T be a Singer-cycle of A^{∞}. Let q = p^{r} and set m = r · n. Suppose gcd(m, (q^{n} − 1)/(q − 1)) ≠ 1. Let s be a prime which divides gcd(m, (q^{n} − 1)/(q − 1)) and let U be a subgroup of T of order |T |/s. Let t be an element of T of order |T |_{s} = s^{b} and let g be an element of N_{A^{∞}}(T) of order s which fixes a point in Γ^{∞}. Set Y^{∞} = U⟨tg⟩. Then |Y^{∞}| = |T |.

We claim that Y^{∞} acts regularly on Γ^{∞} (and thereby the preimage Y of Y^{∞} in ΓL_{n}(q) acts transitively on O_{p}(AΓL_{n}(q))^{#} ). If ⟨t, g⟩ is an abelian group, then tg is of order o(t) = s^{b}. If ⟨t, g⟩ is not an abelian group and if s is odd, then ⟨t, g⟩ is isomorphic to Mod_{s^{b+1}} (for the definition of the modular group Mod_{s^{b+1}} see [As, p. 107]) and therefore tg is again of order s^{b}. Hence any Sylow s-subgroup of Y^{∞} is cyclic. Therefore, any element of order s of Y^{∞} is contained in U which implies that Y^{∞} acts regularly on Γ^{∞}. If ⟨t, g⟩ is not an abelian group and s = 2, then g and tg are not conjugate. Thus also in this case Y^{∞} acts regularly, as claimed.

By Lemma 2.1 C = C_{Sym(Γ^{∞})}(Y^{∞}) ≅ Y^{∞} and N_{Sym(Γ^{∞})}(Y^{∞}) ≤ CAut(T). We claim that C_{A^{∞}}(Y^{∞}) is a proper subgroup of C. If there exists a Zsigmondy prime t for p^{m} − 1 = q^{n}−1, then according to [HuI, II. (7.3)] C_{A^{∞}}(U) = T which implies C_{A^{∞}}(Y^{∞}) = C_{T} (g) < C. If there exists no Zsigmondy prime for p^{m} − 1, then (m, p) = (6, 2) or m = 2 and p is a Mersenne prime [Zsig]. If (m, p) = (6, 2), then s = 3, U ≅ ZZ_{21} and C_{A^{∞}}(Y^{∞}) = Z(Y^{∞}) < C. Now let m = 2 and p be a Mersenne prime. Then m = n = s = 2, U ≅ ZZ_{p+1/2} and C_{A^{∞}}(Y^{∞}) = Z(Y^{∞}) < C. Thus our claim turns out to be true in all cases and therefore, there exists a non-canonical gluing via α ∈ C \ A^{∞} such that Y^{∞} ≤ X.

Clearly, the above construction can be generalized in such a way that [Y^{∞} : T ∩ Y^{∞}] will not be a prime.

4 The Main Theorem.

Theorem 4.1 Let Γ be a non-canonical gluing AS(n, q) ∘_{α} AS(n, q), where q = p^{r} . Then Γ is one of the examples listed above, i.e. one of the following holds.

(1) (i) n = 2, q ∈ {5, 7, 11, 23} and X ≅ S_{4}; (ii) n = 2, q = 11 and X ≅ A_{4};

(2) n = 2, q ∈ {11, 19, 29, 59} and X ≅ A_{5};

(3) n = 4, q = 3, O_{2}(X) ≅ 2^{4} and ZZ_{5} ≤ X/O_{2}(X) ≤ F rob(5 : 2);

(4) n = 6, q = 3 and X ≅ L_{2}(13); (5) n > 2 or n = 2 and q ∉ {2, 3, 4, 5, 8, 9}; X = N_{ΓL_{n}(q)}(T), with T a Singer cycle of

ΓL_{n}(q);

(6) gcd(r · n, (q^{n} − 1)/(q − 1)) ≠ 1, X is naturally embedded into ΓL_{1}(q^{n}), but not of type (5).

In cases (1), (2) and (3) we may choose α such that [X, α] = 1. In case (4) there exists just two examples: one with [X, α] = 1 and the other with X⟨α⟩ ≅ P GL_{2}(13).

Proof: By Lemma 3.1 G_{0}/Z_{0} ≅ K_{2}^{∞}X is a doubly transitive permutation group of degree q^{n}. Therefore X and X^{α} are subgroups of A^{∞} ≅ P ΓL_{n}(q) which act both transitively on Γ^{∞} and the preimages X̂ and X̂^{α} of X and X^{α} in ΓL_{n}(q) act transitively on O_{p}(G_{2})^{#}.

Notice, since Γ is a non-canonical gluing n is at least 2. If moreover, α normalizes X, then N_{Sym(Γ^{∞})}(X) is not contained in N_{A^{∞}}(X). Suppose that X̂ is naturally embedded into ΓL_{m}(t). Then m divides n and q^{n} = t^{m}.

Suppose first m ≥ 2. According to Lemma 2.2 X̂ and X̂^{α} are conjugate in ΓL_{n}(q) which implies X = X^{αg} for some g ∈ A^{∞}. Hence by Lemma 3.2 we may assume α ∈ N_{Sym(Γ^{∞})}(X). Since Γ is a non-canonical gluing Lemma 3.2 yields N_{A^{∞}}(X) is a proper subgroup of N_{Sym(Γ^{∞})}(X). Now Lemma 2.3 and Corollary 2.5 imply (1) – (4) of the Theorem.

Now suppose that m = 1. We distinguish the cases that X contains a Singer-cycle T or not. First suppose T ≤ X. Then X ≤ N_{A^{∞}}(T). As X^{α} has also a Singer-cycle T̃ and is also contained in N_{A^{∞}}(T̃), X and X^{α} are conjugate in A^{∞}. Thus we may assume α ∈ N_{Sym(Γ^{∞})}(X). Since N_{Sym(Γ^{∞})}(T) is not contained in N_{A^{∞}}(T) Lemma 2.6 implies n > 2 or n = 2 and q ∉ {2, 3, 4, 5, 8, 9}. Hence (5) of the theorem holds. Finally suppose that X does not contain a Singer-cycle. Since X ≤ ΓL_{1}(q^{n}) acts transitively on Γ^{∞}, we obtain that gcd(an, (q^{n} − 1)/(q − 1)) ≠ 1 and that Γ is as in (6).

Let Γ be an example of type (1), (2) or (3). In all these examples we have N_{Sym(Γ^{∞})}(X) = C_{Sym(Γ^{∞})}(X)N_{A^{∞}}(X). Therefore the last assertion follows with Lemma 3.2. This proves the theorem.

Corollary 4.2 Let Γ = AS(n, q) ∘_{α} AS(n, q) be a non-canonical gluing not of type (4),(5) or (6) of Theorem 1.1. Then we may choose α such that [X, α] = 1.

5 Exceptional coverings

In this chapter we are going to describe the universal covers of the non-canonical gluings obtained in the previous chapter. Here we just determine those which are not related to the affine operating group. The question for these groups is still open.

The basic idea reads as follows. Let G be the group of the universal covering. Then G is generated by the three parabolic subgroups G_{0}, G_{1}, G_{2}, where G_{0} = (Z_{0} × K_{2}^{∞})X and G_{2} = (Z_{2} × K_{0}^{∞})X^{α}. As has been seen in (4.2) we may choose α as a permutation inside of Σ_{n} which centralizes X besides of the case X ≅ SL_{2}(13). From the structure of G_{1} = ⟨⟨x_{1}⟩, α(⟨x_{1}⟩)⟩X_{1}, where ⟨x_{1}⟩ is a 1-space in K_{2}, we see that [⟨x_{1}⟩, α(⟨x_{1}⟩)] = 1. Now G = ⟨K_{0}^{∞}, K_{2}^{∞}, X⟩, where the action of X on K_{0}^{∞} and K_{2}^{∞} are known. The only additional relation we have is that there is some bijective mapping α from the 1-spaces of K_{2} onto the 1-spaces of K_{0} which commutes with the action of X and satisfies [U, α(U)] = 1 for any 1-space U in K_{2}. Hence we are going to determine precisely those groups.

Lemma 5.1 Let p be a prime, K_{0} ≅ K_{2} ≅ E_{q^{2}}, q = p^{r}, be subgroups of a group G, G = ⟨K_{0}, K_{2}⟩. Suppose that K_{0} and K_{2} admit GF (q)-vector space structures such that there is an GF (q)-linear isomorphism f : K_{2} → K_{0}. Moreover, suppose that [U, f(U)] = 1 for any 1-dimensional subspace U of K_{2}. Then G′ ≤ Z(G) and |G′| ≤ q.

Proof: Let U and V be two different subspaces of K_{2}. Then K_{2} = U × V. Observe that [xy, f(x)f(y)] = 1 (x, y ∈ K_{2}) implies [x, f(y)] = [y, f(x)]^{−1}. Let u_{1}, u_{2} ∈ U and v_{1}, v_{2} ∈ V. As [v_{1}, f(v_{2})] = 1, we see [u_{1}, f(v_{1}), v_{2}] = 1. As [u_{1}, f(u_{2})] = 1, we see [v_{1}, f(u_{1}), u_{2}] = 1. So [u_{1}, f(v_{1}), K_{2}] = 1. The same argument also shows [u_{1}, f(v_{1}), K_{0}] = 1. So [u_{1}, f(v_{1})] ∈ Z(G) and G′ ≤ Z(G). Furthermore [u_{1}, f(v_{2})]^{p} = [u_{1}^{p}, f(v_{2})] = 1.

Let λ ∈ GF (q) such that ⟨λ⟩ = GF (q)^{#} and let x_{1} ∈ U^{#} and y_{1} ∈ V^{#}. Set x_{i+1} = x_{1}^{λ^{i}} and y_{i+1} = y_{1}^{λ^{i}} for i = 1, …, r − 1. Then {x_{1}, …, x_{r}} and {y_{1}, …, y_{r}} are a basis of the GF (p)-vector spaces U and V, respectively.

Set D = ⟨[x_{1}, f(y_{i})] | i = 1, …, r⟩. We claim that G′ = D. Since G′ ≤ Z(G), we have G′ = [U, f(V)]. The fact that x_{i}y_{i} = (x_{1}y_{1})^{λ^{i−1}} implies [x_{1}y_{1}, f(x_{i})f(y_{i})] = 1 (i = 1, …, r). Therefore [y_{1}, f(x_{i})] = [x_{1}, f(y_{i})]^{−1} and, as [x_{i}, f(y_{1})] = [y_{1}, f(x_{i})]^{−1}, it follows [x_{i}, f(y_{1})] ∈ D (i = 1, …, r). Now suppose that [x_{i}, f(y_{j})] ∈ D for i = 1, …, r and j ≤ k and [x_{i}, f(y_{k+1})] ∈ D for i = 1, …, l. Since (x_{l}y_{k})^{λ} = x_{l}^{λ}y_{k+1} we obtain [x_{l}y_{k}, f(x_{l}^{λ})f(y_{k+1})] = 1 and [x_{l}, f(y_{k+1})] = [y_{k}, f(x_{l}^{λ})]^{−1}.

If l < r, then x_{l}^{λ} = x_{l+1} and [x_{l}, f(y_{k+1})] = [y_{k}, f(x_{l+1})]^{−1} = [x_{l+1}, f(y_{k})] ∈ D.

If l = r, then x_{r}^{λ} = ∏_{i=1}^{r} x_{i}^{a_{i}}, a_{i} ∈ GF (p), and [x_{r}, f(y_{k+1})] = ∏_{i=1}^{r}([y_{k}, f(x_{i})]^{a_{i}})^{−1} = ∏_{i=1}^{r}[x_{i}, f(y_{k})]^{a_{i}} ∈ D. This proves G′ = D. So |G′| ≤ q.

Using the same method as in the first paragraph of the previous proof we obtain the following more general result in the case q = p.

Lemma 5.2 Let p be a prime and K_{0} ≅ K_{2} ≅ E_{p^{2}} be subgroups of a group G, G = ⟨K_{0}, K_{2}⟩ and f be an isomorphism from K_{2} onto K_{0}. Let K_{2} = ⟨x_{1}, x_{2}⟩ and suppose [x_{1}, f(x_{1})] = [x_{2}, f(x_{2})] = [x_{1}x_{2}, f(x_{1})f(x_{2})] = 1, then |G′| | p.

Lemma 5.3 Let p be a prime and K_{0} ≅ K_{2} ≅ E_{p^{2}} be subgroups of a group G, G = ⟨K_{0}, K_{2}⟩, where G is extraspecial of order p^{5}. Then there is some isomorphism f : K_{2} → K_{0} with [x, f(x)] = 1 for all x ∈ K_{2}.

Proof: Let K_{2} = ⟨x_{1}, x_{2}⟩ and K_{2} = ⟨y_{1}, y_{2}⟩ with [x_{1}, y_{1}] = [x_{2}, y_{2}] = z and [x_{1}, y_{2}] = [x_{2}, y_{1}] = 1, ⟨z⟩ = Z(G). Now define f as follows f(x_{1}^{i}x_{2}^{j}) = y_{2}^{i}y_{1}^{−j}, i, j ∈ {0, …, p − 1}. Obviously f is an isomorphism. Using the fact G′ ≤ Z(G) we now have

[x_{1}^{i}x_{2}^{j}, f(x_{1}^{i}x_{2}^{j})] = [x_{1}^{i}, f(x_{1}^{i}x_{2}^{j})][[x_{2}^{j}, f(x_{1}^{i}x_{2}^{j})]
= [x_{1}, f(x_{1}^{i}x_{2}^{j})]^{i}[x_{2}, f(x_{1}^{i}x_{2}^{j})]^{j}
= [x_{1}, f(x_{1})]^{i^{2}}[x_{1}, f(x_{2})]^{ji}[x_{2}, f(x_{1})]^{ij}[x_{2}, f(x_{2})]^{j^{2}}
= [x_{1}, y_{2}]^{i^{2}}[x_{1}, y_{1}]^{−ij}[x_{2}, y_{2}]^{ij}[x_{2}, y_{1}]^{−j^{2}} = 1.

Lemma 5.4 Let p be a prime and K_{0} ≅ K_{2} ≅ E_{p^{2}} be subgroups of a group G, G = ⟨K_{0}, K_{2}⟩, where G is extraspecial of order p^{5}. Let α be a bijective mapping from the 1-spaces of K_{2} onto the 1-spaces of K_{0} such that [U, α(U)] = 1 for each 1-space U of K_{2}. Then there is an isomorphism f : K_{2} → K_{0} with [x, f(x)] = 1 for x ∈ K_{2}, such that f(U) = α(U) for all 1-spaces U of K_{2}.

Proof: Let K_{2} = ⟨x_{1}, x_{2}⟩ and K_{0} = ⟨y_{1}, y_{2}⟩ with [x_{1}, y_{1}] = [x_{2}, y_{2}] = z and [x_{1}, y_{2}] = [x_{2}, y_{1}] = 1, ⟨z⟩ = Z(G). Then for each x ∈ K_{2} we see that α(⟨x⟩) is uniquely determined. So α(⟨x⟩) = f(⟨x⟩), where f is the mapping from 5.3.

Proposition 5.5 Let Γ be a non-canonical gluing AS(2, p) ∘_{α} AS(2, p), p an odd prime. Then Γ is simply connected.

Proof: By (3.1) we have G_{0} = Z_{0}K_{2}^{∞}X, G_{2} = Z_{2}K_{0}^{∞}X and G_{1} = Y (Z_{0} × Z_{2})X_{1}, where X_{1} ≤ X and stabilizes in the elementary abelian group K_{2} a point ⟨x_{1}⟩ and in K_{0} a point α(⟨x_{1}⟩). Hence Y = ⟨⟨x_{1}⟩, α(⟨x_{1}⟩)⟩ is elementary abelian of order p^{2}. So we get [⟨x⟩, α(⟨x⟩)] = 1 for all x ∈ K_{2}.

Let K_{2} = ⟨x_{1}, x_{2}⟩ and y_{1}, y_{2} ∈ K_{2} with α(⟨x_{i}⟩) = ⟨y_{i}⟩, i = 1, 2. Let α(⟨x_{1}x_{2}⟩) = ⟨y_{1}^{a}y_{2}^{b}⟩. Define the linear mapping f : K_{2} → K_{0} by f(x_{1}) = y_{1}^{a}, f(x_{2}) = y_{2}^{b}. Then by 5.2 we have |⟨K_{0}, K_{2}⟩′| | p. If ⟨K_{0}, K_{2}⟩ is nonabelian, then it is extraspecial of order p^{5}. Now by 5.4 there is some linear mapping g : K_{2} → K_{0} with g(⟨x⟩) = α(⟨x⟩) for all x ∈ K_{2}. But this means that g = α is in ΓL_{2}(p) ≤ Σ_{p+1}, which contradicts (4.2) and example (3.2)(a),(b). So we have [K_{0}, K_{2}] = 1 and Γ is simply connected.

Proposition 5.6 Let Γ be a non-canonical gluing AS(4, 3) ∘_{α} AS(4, 3), then Γ is simply connected.

Proof: By (3.1) G_{0} contains a subgroup H_{0} = K_{2}Y, where K_{2} is elementary abelian of order 81 and Y is an extention of an extraspecial group of order 32 by Z_{5}. Furthermore G_{2} contains H_{2} = K_{0}Y, where K_{0} is elementary abelian of order 81 too. The action of Y on K_{2} is described in [Hu]. Let {x_{1}, x_{2}, x_{3}, x_{4}} be the basis used there and {y_{1}, y_{2}, y_{3}, y_{4}} the corresponding basis of K_{0}. We have [⟨x_{1}⟩, α(⟨x_{1}⟩)] = 1. As α(⟨x_{1}⟩) has to be fixed by the one point stabilizer of Y on K_{0} and α centralizes Y by (4.2) we see that α(⟨x_{1}⟩) ⊆ ⟨y_{1}, y_{3}⟩. As α is not the identity we see that α(⟨x_{1}⟩) ∈ {⟨y_{3}⟩, ⟨y_{1}y_{3}⟩, ⟨y_{1}^{−1}y_{3}⟩}. Set Y_{1} = G_{1} ∩ Y and H_{1} = ⟨⟨x_{1}⟩, α(⟨x_{1}⟩)⟩Y_{1}. Let H = ⟨H_{1}, H_{2}, H_{3}⟩. Using the relations from H_{1} and H_{3} and the additional relation [x_{1}, y] = 1, for some y ∈ {y_{3}, y_{1}y_{3}, y_{1}y_{3}^{−1}}, we get in any case by the Todd-Coxeter algorithm that H = (K_{0} × K_{2})Y. This shows that G = K_{0}^{∞}K_{2}^{∞}X and so Γ is simply connected.

Proposition 5.7 Let Γ be a non-canonical gluing AS(6, 3) ∘_{α} AS(6, 3). Then one of the following holds

(i) If α induces an outer automorphism on X, then Γ is simply connected.

(ii) If [X, α] = 1, then the universal cover of Γ is a 3-fold covering.

Proof: Let H_{0} = K_{2}X, where K_{2} is elementary abelian of order 729 and X ≅ SL_{2}(13). We first determine the structure of H_{0}. According to [HPl] the following is a presentation of X :

⟨a, x, y, r | x^{13} = r^{3} = 1, x^{r} = x^{3}, y^{4} = 1, x^{y} = x^{−1}, [y, r] = 1, a^{2} = y^{2}, y^{a} = y^{−1}, r^{a} = r^{−1}, (ax)^{3} = 1, x^{6}ax^{−1}axax^{7}a = (ry)^{−1}⟩

We are going to calculate the action of X on K_{2}. First of all, ⟨x, r⟩ decomposes K_{2} into two 3-dimensional modules. As GL(3, 3) does not contain a subgroup of order 13 · 12, we see that y has to interchange these two modules. So we get the following matrices according to a basis {x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}} of K_{2}

x → ( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & −1 & −1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} ) , r → ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 1 & −1 & −1 & 0 & 0 & 0 \\ −1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & −1 & −1 \\ 0 & 0 & 0 & −1 & 1 & 0 \end{matrix} )

y → ( \begin{matrix}0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ −1 & 0 & 0 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 & 0 & 0 \\ 0 & 0 & −1 & 0 & 0 & 0 \end{matrix} )

Now we have that a has to act on the fixed points of r i.e. ⟨x_{1}, x_{4}⟩. So ⟨y, a⟩ acts as a quaternion group on ⟨x_{1}, x_{4}⟩ and so it is uniquely determined. Using the relations above we see x_{1}^{a} = x_{1}^{−1}x_{4} and x_{4}^{a} = x_{1}x_{4}. This then yields

a → ( \begin{matrix}−1 & 0 & 0 & 1 & 0 & 0 \\ 1 & −1 & 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & −1 & −1 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & −1 & 1 & 0 \\ 1 & −1 & −1 & −1 & −1 & −1 \end{matrix} )

So let now K_{0} = ⟨y_{1}, y_{2}, y_{3}, y_{4}, y_{5}, y_{6}⟩. If [X, α] = 1 then X acts by the same matrices on K_{2} as on K_{0}. Let X_{1} be the stabilizer of ⟨x_{1}⟩ in X. Then α(⟨x_{1}⟩) is contained in the set of fixed points of X_{1} on K_{0} and so is some 1-space in ⟨y_{1}, y_{4}⟩. Hence we get α(⟨x_{1}⟩) ∈ {⟨y_{4}⟩, ⟨y_{1}^{−1}y_{4}⟩, ⟨y_{1}y_{4}⟩}. Let H = ⟨K_{0}X, K_{2}X⟩ with the defining relations of K_{0}X and K_{2}X and the additional relation [x_{1}, t] = 1 for some t ∈ {y_{4}, y_{1}^{−1}y_{4}, y_{1}y_{4}}. Then application of the Todd-Coxeter algorithm in all three cases shows that H = ⟨K_{0}, K_{2}⟩X where ⟨K_{0}, K_{2}⟩ is extraspecial of order 3^{13}, the assertion.

So assume now that α induces an outer automorphism on X. Then x acts as x^{8} and a as ay on K_{0}, while the action of r, y is unchanged. Hence the fixed points of X_{1} on K_{0} are the same as before. So we now choose t ∈ {y_{1}, y_{4}, y_{1}^{−1}y_{4}, y_{1}y_{4}} and add the relation [x_{1}, t]. Then we apply the Todd-Coxeter algorithm to the group ⟨K_{0}, K_{2}, X⟩ again. This then shows that ⟨K_{0}, K_{2}⟩ is elementary abelian in all four cases, hence Γ is simply connected.

6 Canonical coverings

In this section we are going to determine the universal coverings of the canonical gluings.

Lemma 6.1 Let G = ⟨K_{0}, K_{2}⟩ with elementary abelian p-groups K_{0}, K_{2} of order at most p^{3}, p an odd prime. Let f : K_{2} → K_{0} be an isomorphism with [x, f(x)] = 1 for all x ∈ K_{2}. Then G′ ≤ Z(G) and |G′| ≤ |K_{2}|.

Proof: The assertion is trivial for |K_{2}| = p. Let |K_{2}| = p^{2}. Then the assertion follows from 5.1. So we may assume that |K_{2}| = p^{3}. Set K_{2} = ⟨x_{1}, x_{2}, x_{3}⟩ and K_{0} = ⟨y_{1}, y_{2}, y_{3}⟩, where f(x_{i}) = y_{i}, i = 1, 2, 3. Let X be the free group in the generators x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}. We extend f to a mapping g of {x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}} by setting g(x_{i}) = y_{i} and g(y_{i}) = x_{i}, i = 1, 2, 3. As X is free g can be extended to an automorphism of X which we will denote by g too. Let N = ⟨x_{i}^{p}, y_{i}^{p}, [x_{i}, x_{j}], [y_{i}, y_{j}], i, j = 1, 2, 3, [x, f(x)], [f(x), x], x ∈ K_{2}⟩. Then we see that N is invariant under g. So g is an automorphism of H = X/N. As G is a factorgroup of H it is enough to prove the assertion for H. Hence we may assume that g is an automorphism of G. By definition we have g^{2} = 1.

As [ab, g(a)g(b)] = 1, we see [a, g(b)] = [b, g(a)]^{−1} for all a, b ∈ K_{2}. Now g([b, g(a)]) = [g(b), a] = [b, g(a)] and so [G′, g] = 1. Set L = ⟨x_{1}g(x_{1})^{−1}, x_{2}g(x_{2})^{−1}, x_{3}g(x_{3})^{−1}⟩. Then L ≤ [G, g]. Let a ∈ G with g(a) = a^{−1}. Let x ∈ G′. Then [x, a] = g([x, a]) = [x, g(a)] = [x, a^{−1}]. This implies x^{−1}a^{−1}xa = x^{−1}axa^{−1} or [a^{2}, x] = 1. If a is of odd order we get [a, x] = 1. Now choose for a one of the elements x_{i}g(x_{i})^{−1}. Then o(a) = p is odd and so we have [a, G′] = 1. In particular [L, G′] = 1. This shows

(1) L′ ≤ Z(L).

We have

(2) [x_{1}g(x_{1})^{−1}, x_{2}g(x_{2})^{−1}] = [x_{1}, x_{2}g(x_{2})^{−1}]^{g(x_{1})^{−1}}[g(x_{1})^{−1}, x_{2}g(x_{2})^{−1}] = [x_{1}, g(x_{2})^{−1}][g(x_{1})^{−1}, x_{2}].

By 5.1 we have [g(x_{i}), x_{j}]^{g(x_{i})} = [g(x_{i}), x_{j}] for all i, j. So we get

(3) [g(x_{i})^{n}, x_{j}] = [g(x_{i}), x_{j}]^{n} for all n ∈ \overline{Z} and all i, j.

This now shows

[x_{1}, g(x_{2})^{−1}][g(x_{1})^{−1}, x_{2}] = [x_{1}, g(x_{2})]^{−1}[g(x_{1}), x_{2}]^{−1} = [x_{1}, g(x_{2})]^{−2}.

So (1) - (3) imply

(4) L′ = ⟨[x_{i}, g(x_{j})]^{−2} | i, j = 1, 2, 3⟩

By 5.1 we have that o([x_{i}, g(x_{j})]) | p. Hence we get

(5) L′ = ⟨[g(x_{1}), x_{2}], [g(x_{1}), x_{3}], [g(x_{2}), x_{3}]⟩.

In particular, we have |L′| ≤ p^{3}.

It remains to show that L′ = G′ ≤ Z(G). Obviously we have ⟨x_{j}g(x_{j})^{−1}, x_{k}⟩ ≤ ⟨x_{j}, x_{k}, g(x_{j}), g(x_{k})⟩ = Y. By 5.1 we see that Y′ = ⟨[x_{j}, g(x_{k})]⟩. So by (5) L is normal in G. We have that ⟨x_{1}, x_{3}⟩ acts on ⟨[g(x_{1}), x_{2}], [g(x_{3}), x_{2}]⟩ = C_{L′}(x_{2}). Hence there are i, j with x_{1}^{i}x_{3}^{j} ≠ 1 and [x_{1}^{i}x_{3}^{j}, C_{L′}(x_{2})] = 1. This shows that [g(x_{1}), x_{2}, x_{3}^{j}] = 1 = [g(x_{3}), x_{2}, x_{1}^{i}] by 5.1. We may assume [g(x_{1}), x_{2}, x_{3}] = 1. By the Witt identity we have 1 = [g(x_{1}), x_{2}, x_{3}]^{x_{2}^{−1}}[x_{3}, g(x_{1}), x_{2}]^{g(x_{1})^{−1}} and so [g(x_{1}), x_{3}, x_{2}] = 1. Also we get 1 = [g(x_{2}), x_{3}, x_{1}]^{x_{3}^{−1}}[x_{1}, g(x_{2}), x_{3}]^{g(x_{2})^{−1}}. As [x_{1}, g(x_{2})] = [x_{2}, g(x_{1})]^{−1} and [[x_{2}, g(x_{1})]^{−1}, x_{3}] = 1, we get [g(x_{2}), x_{3}, x_{1}] = 1. Hence we have [K_{2}, L′] = 1. As [g, L′] = 1, we now get [G, L′] = 1. This implies G′ = L′ ≤ Z(G).

Lemma 6.2 Let Γ be a canonical gluing AS(n, q) ∘_{α} AS(n, q), q = p^{r}, p an odd prime, with group G. Let G̃ be the group belonging to the universal covering. Then G = G̃/N, where N is an elementary abelian p-group of order at most q^{\binom{n}{2}}.

Proof: We have G̃_{0} = (Z_{0} × K_{2}^{∞})X, G̃_{2} = (Z_{2} × K_{0}^{∞})X and G̃_{1} = Y (Z_{0} × Z_{2})X_{1}, where K_{0} = O_{p}(K_{0}^{∞}) and K_{2} = O_{p}(K_{2}^{∞}) are elementary abelian of order q^{n}. Further |Y ∩ K_{0}| = |K_{2} ∩ Y | = q, and Y = (K_{0} ∩ Y)(K_{2} ∩ Y) is elementary abelian of order q^{2}.

Finally there is an GF (q)-linear isomorphism α : K_{2} → K_{0} which commutes with the action of X and maps K_{0} ∩ Y onto K_{2} ∩ Y. Hence we have [U, α(U)] = 1 for every 1-dimensional GF (q)-subspace U of K_{2}.

Set H = ⟨K_{0}, K_{2}⟩. Let u, v, w ∈ K_{2}. Then [u, α(v), w] ∈ ⟨u, v, w, α(u), α(v), α(w)⟩. Therefore Lemma 6.1 implies [u, α(v), w] = 1, so [u, α(v), K_{2}] = 1 for all u, v ∈ K_{2}. Using the same argument we also obtain [u, α(v), K_{0}] = 1 for all u, v ∈ K_{2}. This shows H′ ≤ Z(H). In particular, H′ is elementary abelian and H′ = ⟨[u, α(v)] | u, v ∈ K_{2}⟩.

Let U_{1}, …, U_{n} be 1-dimensional GF (q)-subspaces of K_{2} with K_{2} = U_{1} × … × U_{n}. Since [U_{i}, α(U_{j})] ≤ ⟨U_{i}, U_{j}, α(U_{i}), α(U_{j})⟩ (1 ≤ i, j ≤ n), Lemma 5.1 implies [U_{i}, α(U_{j})] = [U_{j}, α(U_{i})] is a group of order at most q (1 ≤ i, j ≤ n). Hence H′ = ⟨[U_{i}, α(U_{j})] | i < j ≤ n⟩ is of order at most q^{\binom{n}{2}}. As G̃ = H(Z_{0} × Z_{2})X, the assertion follows.

Proposition 6.3 Let Γ be a canonical gluing AS(n, q) ∘_{α} AS(n, q), q = p^{r}, p an odd prime, with group G. Let G̃ be the group belonging to the universal covering. Then G = G̃/N, where N is an elementary abelian p-group of order q^{\binom{n}{2}}.

Proof: Let W be an (n + 1)-dimensional vectorspace over GF (q) with basis {e_{1}, …, e_{n+1}} and let V = ⟨e_{1}, …, e_{n}⟩. Let S = SL(W) ≅ SL_{n+1}(q). Then N_{S}(V) = C_{S}(V)N_{S}(⟨e_{n+1}⟩) with K_{2} = C_{S}(V) an elementary abelian group of order q^{n} and X = N_{S}(⟨e_{n+1}⟩) ≅ GL_{n}(q). Set U = W ∧ W and N = V ∧ V. Then N is elementary abelian of order q^{\binom{n}{2}} and U/N ≅ ⟨e_{i} ∧ e_{n+1} | i = 1, …, n⟩ elementary abelian of order q^{n}.

Set K_{0} = ⟨e_{i} ∧ e_{n+1} | i = 1, …, n⟩. Let H = U : K_{2}X. Then H is generated by H_{0} = K_{2}X and H_{2} = K_{0}X.

Let α be the isomorphism from K_{2} onto K_{0} with α(a) = [a, e_{n+1}] ∧ e_{n+1} for all a ∈ K_{2}. Then α is an GF (q)Xisomorphism. Observe that [U, α(U)] = 1 for all 1-dimensional subspaces U of K_{2}. Set H_{1} = (T × α(T))N_{X}(T), where T is a GF (q)-subspace of K_{2}. Then {H_{0}, H_{1}, H_{2}} generates a cover of Γ. By 6.2 this is the universal cover of Γ.

Now we want to determine the universal cover of the canonical gluing assuming that q is even.

Lemma 6.4 Let Γ be a canonical gluing AS(n, q) ∘_{α} AS(n, q), q = 2^{r}. Then the universal cover \overline{Γ} of Γ has at most q^{2^{n}−1} points.

Proof: Since Γ is a canonical gluing, the subgroup G = (K_{2} × K_{0})X of Aut(Γ) acts flagtransitively on Γ, where X ≅ GL_{n}(q), K_{0} ≅ K_{2} ≅ E_{q^{n}} are isomorphic GF (q)X-modules with isomorphism α : K_{2} → K_{0}. We may choose notation such that G_{0} = K_{2}X, G_{2} = K_{0}X and G_{1} = (⟨u⟩_{GF (q)} × ⟨α(u)⟩_{GF (q)})N_{X}(⟨u⟩_{GF (q)}), where u ∈ K_{2} and where ⟨u⟩_{GF (q)} is the GF (q)-span of u in K_{2}. Then G_{0} ∩ G_{1} = ⟨u⟩_{GF (q)}N_{X}(⟨u⟩_{GF (q)}), G_{0} ∩ G_{2} = X, G_{1} ∩ G_{2} = ⟨α(u)⟩_{GF (q)}N_{X}(⟨u⟩_{GF (q)}) and B = N_{X}(⟨u⟩_{GF (q)}).

The universal cover of Γ is the group geometry \overline{Γ} = Γ(\overline{G}, (G_{0}, G_{1}, G_{2})), with \overline{G} the universal completion of the amalgam (G_{i}, G_{i} ∩ G_{j}, B | i, j = 0, 1, 2).

Since K_{0} and K_{2} are GF (q)X-modules, we may also read them as GF (2)X-modules. Consider the following amalgam:

H_{0} = G_{0}, H_{2} = G_{2}, H_{1} = (⟨u⟩_{GF (2)} × ⟨α(u)⟩_{GF (2)})N_{X}(⟨u⟩_{GF (2)})

with

H_{0} ∩ H_{1} = ⟨u⟩_{GF (2)}N_{X}(⟨u⟩_{GF (2)}), H_{0} ∩ H_{2} = X, H_{1} ∩ H_{2} = ⟨α(u)⟩_{GF (2)}N_{X}(⟨u⟩_{GF (2)})

and

B = N_{X}(⟨u⟩_{GF (2)}).

For H = ⟨H_{0}, H_{1}, H_{2}⟩ = G the group geometry Γ̃ = Γ(H, (H_{0}, H_{1}, H_{2})) is a canonical gluing AS(nr, 2) ∘_{α} AS(nr, 2) with flag-transitive group H = G.

We may lift H to a flag-transitive group of automorphism \overline{H} of the universal cover of Γ̃. According to [BMP, Corollary 9] \overline{H} contains a normal subgroup N, N ≅ ∑_{i=1}^{rn} Λ^{i}(K_{2}), and \overline{H} = N : H_{2} = N : G_{2}.

Any relation which holds in the amalgam (H_{i}, H_{i} ∩ H_{j}, H_{0} ∩ H_{1} ∩ H_{2} | i, j = 0, 1, 2) also holds in the amalgam (G_{i}, G_{i} ∩ G_{j}, B | i, j = 0, 1, 2). Therefore \overline{G} is a factor group of \overline{H}, so \overline{G} = (N : G_{2})/M for a normal subgroup M of \overline{H}.

Moreover, according to [BMP] [K_{2}, K_{0}] = N′ and N/N′ ≅ Λ(K_{2}) ≅ K_{2}. Hence, for v, w ∈ K_{2} we have [v, α(w)] = v∧w mod M′′, i.e. [v, α(w)]+v∧w ∈ M′′. In \overline{G} for any v ∈ K_{2} we have [v, α(λv)] = 1. This implies v∧λv = 0 for all v ∈ K_{2}. Therefore (v+w)∧λ(v+w) = 0 for all v, w ∈ K_{2}. But this implies 0 = v ∧ λw + w ∧ λv, so v ∧ λw = w ∧ λv = λv ∧ w. Thus NM/M becomes an GF (q)-module for G_{2}. In particular, ∧_{i=1}^{k}v_{i} = 0, v_{i} ∈ K_{2}, whenever v_{j} and v_{l} are linear dependant over GF (q) for some 1 ≤ j < l ≤ k. Notice, if v_{k} lies in the GF (q)-span of v_{1}, …, v_{k−1}, then ∧_{i=1}^{k}v_{i} = 0, as well.

In order to determine the order of NM/M we count for any k, 1 ≤ k ≤ rn, the elements of the subspace Λ^{k}(K_{2})_{q} of NM/M which is generated by ∧_{i=1}^{k}v_{i}, v_{i} ∈ K_{2}. Let {w_{1}, …, w_{n}} be a GF (q)-basis of K_{2}. Then {w_{i_{1}} ∧…∧w_{i_{k}} | 1 ≤ i_{1} < i_{2} < … < i_{k} ≤ n} is a GF (q)-basis of Λ^{k}(K_{2})_{q}, so |Λ^{k}(K_{2})_{q}| = q^{\binom{n}{k}}. Therefore

|NM/M| = ∏_{k=1}^{n} q^{\binom{n}{k}} = q∑{n \atop k=1} ^{\binom{n}{k}} = q^{2^{n}−1}.

This implies that \overline{Γ} has at most q^{2^{n}−1} points, as asserted.

Let Γ be the canonical gluing AS(n, q) ∘_{α} AS(n, q), q = 2^{r}. Let K_{2} : X ≤ Aut(Γ)_{0} ( K_{0} : X ≤ Aut(Γ)_{2}), where K_{2} ≅ E_{q^{n}} (K_{0} ≅ E_{q^{n}}) and X ≅ GL_{n}(q). Then G = (K_{0} × K_{2})X acts flag-transitively on Γ. According to Lemma 3.2 we may assume that α is a GF (q)Xisomorphism between K_{2} and K_{0}. Let N = ∑_{i=1}^{n} Λ^{i}(K_{0}). Then there is a natural action of X on N. We may also define an action of K_{2} on N. Let {v_{1}, …, v_{n}} be a GF (q)-basis of K_{0}. Let u ∈ K_{2}. Then u = α(v) for some v ∈ K_{0}. Let a_{1}, …, a_{n} ∈ GF (q) with v = ∑_{i=1}^{n} a_{i}v_{i}. For w ∈ ⋀^{k}(K_{0}) set

w^{u} = ∑_{J⊆{1,...,n}}(w ∧_{i∈J} a_{i}v_{i}).

Since q is even, this defines an action of K_{2} on N.

Proposition 6.5 Let \overline{Γ} be the universal cover of the canonical gluing Γ = AS(n, q) ∘_{α} AS(n, q), q = 2^{r}. Then \overline{Γ} is a q^{2^{n}−n−1}-fold cover of Γ and \overline{G} = N : K_{2} : X acts flagtransitively on \overline{Γ}.

Proof: Let V ≤ \overline{G} be the diagonal V = \{uα(u) | u ∈ K_{2}\}. Then V is invariant under the action of X. Set \overline{G}_{0} = V : X, \overline{G}_{2} = K_{2} : X and \overline{G}_{1} = (⟨u⟩_{GF (q)}×⟨α(u)_{GF (q)}⟩)N_{X}(⟨u⟩_{GF (q)}), where u ∈ K_{2}^{#}. Clearly, \overline{G} = ⟨\overline{G}_{0}, \overline{G}_{1}, \overline{G}_{2}⟩ and Γ(\overline{G}, (\overline{G}_{0}, \overline{G}_{1}, \overline{G}_{2})) is a cover of Γ. By 6.4 this is the universal cover of Γ.

Theorem 6.6 Let Γ be a canonical gluing AS(n, q) ∘_{α} AS(n, q), q = p^{r} with group G and Γ̃ the universal covering with group G̃. Then G̃/N = G, where N is an elementary abelian p-groups and

(1) If q is odd |N| = q^{\binom{n}{2}}

(2) If q is even, then |N| = q^{2^{n}−n−1}

Proof: The assertion (1) follows from 6.3 and (2) follows from 6.5.

Remark The covers of AS(n, q) ∘_{α} AS(n, q), q odd or q = 2, which turned out to be universal (Theorem 6.6), were already constructed in [Pas2]. For the convenience of the reader we include the constructions of the universal covers. Moreover we give a geometric description of the universal cover for q = 2^{r}, r > 1.

First let q = 2. Let Γ_{m} be the Coxeter complex of type D_{m}, m = 2^{n} and let T r(Γ_{m}) be the {+, 0, −}-truncation of Γ_{m} formed by the elements of type +, 0 and −, where +, 0 and − are as follows:

[figure: •; •; •; 0; •; .....; •; •; −; +]

Then T r(Γ_{m}) is the universal cover of the canonical gluing AS(n, 2) ∘_{α} AS(n, 2).

This universal cover is isomorphic to the GF (2)-expansion D(A), where A is the affine space AG(n, 2), see [BPM, Corollary 9]. Similiar, we may define the GF (q)-expansion D(A), where A is the affine space AG(n, q). Then D(A) is a cover of the canonical gluing which is by Lemma 6.4 the universal cover.

Now let q be odd. Let Δ be the building of type D_{n+1}, n ≥ 2. As above take the types as follows:

[figure: •; •; •; 0; •; 1; .....; •; n − 3; •; n − 2; −; +]

Let us write ε to denote any of the two types + or −. For ε ∈ {+, −} let Δ^{ε} be the half-spin geometry relative to the type ε, that is Δ^{ε} is the geometry of rank 2 having the elements of Δ of type ε as points and those of type 0 as lines, with the incidence inherited from Δ. The diameter of Δ^{ε} is n/2 (resp. (n + 1)/2) for n even (resp. n odd). Let a^{+} and a^{−} be elements of type + and − respectively which are incident. Let H^{ε} be the set of those points of Δ^{ε} which are not at maximal distance from a^{ε} for ε ∈ {+, −}. Then H^{ε} is a hyperplane in Δ^{ε}.

Let \overline{Δ} be the following geometry: The elements of \overline{Δ} are the elements of type ε which are not in H^{ε}, ε ∈ {+, −}, and the elements of type 0 which meet H^{ε} in exactly one element for ε ∈ {+, −}; the incidence is inherited from Δ.

Then \overline{Δ} is the universal cover of the canonical gluing AS(n, q) ∘_{α} AS(n, q), q odd.

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Fachbereich Mathematik und Informatik Martin-Luther Universität Halle Wittenberg 06099 Halle Germany

baumeis@coxeter.mathematik.uni-halle.de stroth@coxeter.mathematik.uni-halle.de