Euclidean Buildings. Guyan Robertson. Newcastle University

Euclidean Buildings Guyan Robertson Newcastle University

Contents 1 Overview of Buildings 2 Lattices, Coset Spaces, Normal Form 3 The Ãn 1 building of PGL n (F) 4 The Boundary 5 Ã 2 groups

Part I Overview of Buildings

Examples Buildings are geometric objects, invented by J Tits (c. 1965). He combined his own work on geometries associated to matrix groups ; previous algebraic work of F Bruhat and C Chevalley. Reference: K Brown, Buildings, Springer 1996. Buildings are simplicial complexes satisfying certain axioms.

Example 1 If V is a vector space of dimension n + 1 over a field k, construct the flag complex of proper linear subspaces of V : 1 Vertices of are proper linear subspaces of V. 2 Two vertices V 1, V 2 of are incident if V 1 V 2. 3 Simplices are chains V 1 V 2 V q. Note that Maximal simplices (called chambers) have n vertices. Therefore has dimension n 1.

A projective plane k = Z/2Z, dim V = 3. Vertices of are lines and planes in V. contains 28 hexagonal apartments. Any two chambers lie in a common apartment. GL 3 (k) is a group of symmetries of of order 168.

Example 2 is a tree and every vertex has degree 3. Apartments are lines (infinite both ways). There are uncountably many apartments. Any two chambers lie in a common apartment. Is there a natural group of symmetries of?

Chamber Complexes Let be a simplicial complex such that all maximal simplices (chambers) have the same dimension d. Chambers C, D are adjacent if C D has codimension 1. is a chamber complex if any two chambers are connected by a sequence of adjacent chambers (a gallery). C D

Buildings A chamber complex is thin [thick] if each simplex of codimension 1 is a face of exactly 2 [at least 3] chambers. thin thick A building is a thick chamber complex which is a union of thin chamber complexes, called apartments, satisfying: 1 Apartments have the same dimension as. 2 Any two simplices of lie in a common apartment. 3 If apartments Σ, Σ contain a common chamber, then there is an isomorphism Σ Σ fixing Σ Σ pointwise.

Types of buildings All apartments are isomorphic. If apartments are spheres, is called spherical. If apartments are euclidean spaces, is called euclidean. A projective plane is a 1-dimensional spherical building. A 1-dimensional euclidean building is a tree and its apartments are lines (infinite both ways). Focus now on euclidean buildings...

Apartments in euclidean buildings of dimension 2 Ã 2 B 2 G 2

The euclidean building of SL 2 (Q p ) The group SL 2 (Q p ) acts on its euclidean building, which is a (p + 1)-regular tree. A brief description of : A vertex is a maximal compact subgroup K of SL 2 (Q p ). e.g. K = SL 2 (Z p ). An edge is (K, K ) where K K is a maximal proper subgroup of K and K. PGL 2 (Q p ) acts on via K g 1 Kg

Example (p = 3) In the building of SL 2 (Q 3 ), a vertex K has four neighbours where g = ( ) 3 0, 0 1 g 1 Kg ( ) 3 1, 0 1 ( ) 3 2, 0 1 ( ) 3 1 0. 0 1

The building of SL 3 (Q p ) The group SL 3 (Q p ) acts on its euclidean building. A vertex is a maximal compact subgroup K of SL 3 (Q p ). e.g. K = SL 3 (Z p ). An edge is (K, K ) where K K is a maximal proper subgroup of K and K. PGL 3 (Q p ) acts on via K g 1 Kg

has type Ã2: its apartments are Ã2 Coxeter complexes. Each edge lies on p + 1 triangles. p = 2 :

The neighbours of a vertex (p = 2) A ball of radius one. The link of a vertex is a projective plane.

Local Structure and Global Structure One can replace Q p by an algebraic extension K. If K 1 K 2, then the buildings of SL 3 (K 1 ) and SL 3 (K 2 ) are not isomorphic. However, spheres of radius 1 are isomorphic :

SL 2 is different : e.g. the buildings of SL 2 (Q 2 ), SL 2 (Q 2 ( 1)) are isomorphic. SL 2 is atypical. Semisimple algebraic groups of higher relative rank behave like SL 3.

Part II Lattices, Coset Spaces, Normal Form

Local Fields: Notation F: a local field with residue field k of order q. O: ring of integers. O = {x F : x 1}. p = {x F : x < 1} = πo the maximal ideal. Uniformizer π chosen in p \ p 2. Thus p = πo The Haar measure µ on (F, +) satisfies µ(xe) = x µ(e). Normalized: µ(o) = 1.

Lattices A F n : n-dimensional vector space over F. B A lattice L F n is an O-submodule of the form Ov 1 + + Ov n for some basis (v 1,..., v n ) of F n. Canonical basis (e 1,..., e n ) Write 1 0 0 0 5 0 0 0 25 C G = GL(F n ) = GL n (F). Oe 1 + + Oe n = O n F n. v 1,..., v n = Ov 1 + + Ov n. 1 0 = O 0 + O 5 + O 0 0 0 0 25

D G acts transitively on bases, hence on lattices. E Therefore, K : = {g G : g(o n ) = O n } L : = {lattices} = G/K gk g(o n ) g(o n ) O n g has entries in O g(o n ) O n g 1 (O n ) O n K = GL n (O) := {g GL n (F) : g, g 1 have integer entries} = {g GL n (F) : det g O and entries of g are in O}.

F Double Cosets in General: (1) Let G be any group; and let H, K be subgroups of G. (2) G/H : = {cosets gh} H\G/K : = {cosets HgK} = orbits in G of right H-action = orbits in G of (left, right) (H K)-action = orbits in G/K of left H-action = orbits in H\G of right K-action (3) G acts diagonally on G/H G/K g (g 1 H, g 2 K) = (gg 1 H, gg 2 K). Space of orbits: G\(G/H G/K) identified with H\G/K. G (g 1 H, g 2 K) Hg 1 1 g 2 K G (H, gk) HgK

G Proposition. If G = GL n (F), K = GL n (O), then K\G/K has coset representatives: π j 1 0... ; j 1 j 2 j n. 0 π jn Interpretation: If (L 1, L 2 ) L L (= G/K G/K), then g G such that 1 0 π j 1 0 (g(l 1 ), g(l 2 )) =...,.... 0 1 0 π jn That is, for some basis (v 1,..., v n ), L 1 = Ov 1 + + Ov n L 2 = Oπ j 1 v 1 +... + Oπ jn v n

Uniqueness of j 1,..., j n : Look at #[(L 1 + π j L 2 )/L 1 ] for varying j. As a function of j, this determines j 1,..., j n up to ordering. Example L 1 = Ov 1 + Ov 2 L 2 = π 3 Ov 1 + π 2 Ov 2 L 1 + L 2 = π 3 Ov 1 + Ov 2 #[(L 1 + L 2 )/L 1 ] = #(π 3 O/O) = q 3 Exercise: Check #[(L 1 + π j L 2 )/L 1 ] for other values of j.

Sufficiency of coset representatives: i.e. diagonalization of an element of G by pre- and post-multiplying by matrices in K is always possible. Note: Kg K = KgK if g is obtained from g by (a) transposing rows/columns; (b) multiplying row/columns by scalar in O ; (c) subtracting an O-multiple of one row (column) from another. Exercise: Fill in the details.

H 0 B =.... G. 0... 0 Proposition. B\G/K has only one point. Therefore BK = G. Note: Bg K = BgK if g is obtained from g by: (a) interchanging columns; (b) subtracting an O-multiple of one column from another; (c) multiplying any row by any constant; (d) subtracting a multiple of row k from row j if k > j. Exercise. Fill in the details, to show BgK = BK for all g G.

I a 1 B 0 =... : a α = 1 G. 0 a n Proposition B 0 \G/K has coset representatives π j 1 A 0 =... : j α Z. π jn Sufficiency: As in H, but can only multiply a row by a constant in O, because B 0 replaces B. Uniqueness: q j α+1 j n = max norm of a determinant of an (n α) (n α) minor chosen from rows α + 1 to n.

Part III The Ãn 1 building of PGL n (F)

A: Projectivization From now on G = PGL n (F) : = GL n (F)/F g ag if a F, K = PGL n (O) : = GL n (O)/O k ak if a G. GL n (O) GL n (F) K G : GL n (O)/GL n (O) F GL n (F)/F and GL n (O) F = O.

The vertex set Initially is just a set of vertices, which will later be organized as the vertices of a simplicial complex. := L/F L al if a F. i.e. L π j L if j Z. [L] will be a vertex of the building. Let O be [O n ]. G = PGL n (F) acts transitively on. Also stab G O = K = PGL n (O). Therefore G/K =.

B: n-volumes Normalize by vol n (O n ) = 1. and vol n (g(o n )) = det g F vol n (O n ) = q i, for some i, vol n (πl) = vol n ((πid n )L) = q n vol n (L). Therefore x = L/F has volume q i, determined up to factors of q n. Therefore i is determined mod n.

C: as a typed set Define τ : I = Z/nZ by vol n (x) = q τ(x). Example τ π j 1 π j 2 π j 3 vol 3 π j 1 π j 2 π j 3 (O) = j 1 j 2 j 3 + 3Z. (O 3 ) = q j 1 j 2 j 3,

D: as a graph Abuse of notation: x will be used to denote both an element of and a representative in L. Definition There exists an edge between x, y if for some representative lattices πy x y. Reflexivity: πy x y πx πy x, πy y.

Normal Form: For some basis (v 1,..., v n ) and 1 µ n 1, y = Ov 1 + + Ov n x = πov 1 + + πov µ + Ov µ+1 + + Ov n πy = πov 1 + + πov µ + πov µ+1 + + πoµ n (See Part II.G) τ(y) = τ(x) + µ

Nearest Neighbours in the building Fix y L (and use the same notation for a lattice and its equivalence class). y, πy and y/πy are O-modules. If y = Ov 1 + + Ov n then y/πy = (O/πO) v 1 + + (O/πO) v n = kv 1 + + kv n #(y/πy) = q n where k = O/πO, the residue field with q elements. Now y/πy is a k-vector space, with dim k (y/πy) = n and so y/πy = k n.

The star of a vertex y is defined to be star(y) = {x : d(x, y) = 1}. πy x y 0 x/πy y/πy = k n. So x/πy is a k-subspace of y/πy = k n. Therefore star(y) = {V k n : V a k-subspace, 1 dim k V n 1}.

star(y) for n = 2. A vertex y has q + 1 neighbours corresponding to 1-dimensional subspaces of k 2. That is, corresponding to points of P 1 (k). Example. G = PGL 2 (Q 5 ), y = O 2 = 1 0, π = 5. 0 1 π 1 0 1 π 0 0 1 π 2 0 1 1 0 0 π π 3 0 1 π 4 0 1

star(y) for n = 3. Assume that τ(y) = 0. type 1 neighbours = { 1-dim. subspaces of k 3 } = points of P 2 (k). type 2 neighbours = { 2-dim. subspaces of k 3 } = lines of P 2 (k). No. of neighbours = 2.(q 2 + q + 1).

A star in the building of PGL 3 (Q 2 )

E: as a graph: distance Let x 0,..., x m be a chain. x 0 x 2 x 1 x m Then there exist representative lattices such that πx 1 x 0 x 1 πx 2 x 1 x 2. πx m x m 1 x m Then Therefore x 0 πx 1 π 2 x 2 π m x m. π m x m x 0 x 1 x m.

FACT: For x, y, d(x, y) = m, where m is the smallest integer such that for some representative lattices, π m y x y. Example. We now show an example where d(x, y) m for m = 3. y = Ov 1 + Ov 2 + Ov 3 x = π 3 Ov 1 + πov 2 + Ov 3 π 3 y = π 3 Ov 1 + π 3 Ov 2 + π 3 Ov 3.

Ov 1 + Ov 2 + Ov 3 πov 1 + Ov 2 + Ov 3 πov 1 + πov 2 + Ov 3 π 2 Ov 1 + Ov 2 + Ov 3 π 2 Ov 1 + πov 2 + Ov 3 π 3 Ov 1 + πov 2 + Ov 3 Therefore d(x, y) 3.

F: as a simplicial complex Say {x 1,..., x J } is a simplex if d(x j, x k ) = 1 for 1 j, k J. Only possible if (in some ordering, for some representative lattices), πx J x 1 x 2 x J. [Proof omitted.] Note. 0 x 1 /πx J x 2 /πx J x J /πx J = k n, is a flag of k-subspaces of x J /{πx J }.

G: Chambers of. (1) A chamber is a maximal simplex C = {x 1,..., x n } with πx n x 1 x 2 x n, τ(x j ) = j + τ(x n ). (2) 0 x 1 /πx n x 2 /πx n x n /πx n = k n is a complete flag of k-subspaces. (3) rank( ) := rank(c) := n. n = 3 : However rank(g as algebraic group) = n 1.

H: Apartments of The apartment A associated to a fixed basis (v 1,..., v n ) of F n is A = {π j 1 Ov 1 + + π jn Ov n : j α Z} Recall that π j 1 Ov 1 + + π jn Ov n denote lattice classes. Notation: [j 1,..., j n ] = π j 1 Ov 1 + + π jn Ov n τ ([j 1,..., j n ]) = j 1 j n. Note that [j 1,..., j n ] = [j + j 1,..., j + j n ]. G acts transitively on bases, hence on apartments.

Example. n = 2. [0, 2] [0, 1] [0, 0] [0, 1] [0, 2] [2, 0] [1, 0] Example. n = 3. [ 1, 0, 0] [0, 0, 1] [0, 1, 0] [0, 1, 0] [0, 0, 1] [1, 0, 0]

0 2 2 0 1 2 0 0 2 0 1 1 0 0 1 0 1 0 0 0 0 τ = 0 τ = 1 τ = 2 0 0-1 0-1 -1 Any two vertices lie in a common apartment. Reason: If L 1, L 2 L then, for some basis (v 1,..., v n ), (c.f. Section 1.G.) L 1 = Ov 1 + + Ov n, L 2 = Oπ jn v 1 + + Oπ jn v n.

Part IV The Boundary

A: Sectors (Weyl Chambers) Let v 1,..., v n be a basis for F n. W = {π j 1 Ov 1 + + π jn Ov n : j 1 j n } is a sector with basepoint Ov 1 + + Ov n. n = 2 n = 3

An apartment is an union of n! sectors intersecting only at the walls, defined by the conditions j α = j α+1. If x, y, there exists a basis v 1,..., v n with x = Ov 1 + + Ov n, y = π j 1 Ov 1 + + π jn Ov n, j 1 j n. Therefore there exists a sector based at x which contains y. x y

Consider the sector W 0 = {π j 1 Oe 1 + + π jn Oe n : j 1 j n } based at O. Then O W 0 is a set of coset representatives for G\{ }. Any (x, y) can be written with g G, z W 0. (x, y) = (g(o), g(z)), y x O z

To each sector W we associate a boundary point ω, as follows. If W = {π j 1 Ov 1 + + π jn Ov n : j 1 j n } and k 1 k n, then W = {π j 1+k 1 Ov 1 + + π jn+kn Ov n : j 1 j n } is a subsector of W. W W W, W will represent the same boundary point ω.

Assertion: If j 1 < j 2 < < j n, (meaning [j 1,..., j n ] is an interior vertex of W ), then O stab G (O, [j 1,..., j n ]) = g PGL n(o) : g αβ O π jα j βo α < β α = β α > β This is proved by using the facts that stab G (O) = PGL n (O) and stab G (ho) = hpgl n (O)h 1, where π j 1 0 h =.... 0 π jn

Example (n = 2) [0, 0] [0, 4] O x x = [0, 4] = Oe 1 + π 4 Oe 2. Here stab G (O, x) = {( )} O O π 4 O O.

Consequence: If k 1 k n, j 1 j n, then stab G (O, [k 1,..., k n ]) stab G (O, [j 1 + k 1,..., j n + k n ]). Geometrically: [0, 0] [0, 3] fixed fixed also fixed

[0, 0, 0] fixed also fixed [0, 2, 3] fixed

The convex hull of [0,..., 0] and [l 1,..., l n ] W is {[j 1,..., j n ] : j 1 j n, l 1 j 1 l n j n }. It s points are fixed by G if [0,..., 0] and [l 1,..., l n ] are fixed. The definition of convex hull extends to any pair x, y, by G-action; also to any set of points.

The n 1 neighbours of O which lie in the sector W 0 are of the form [0,..., 0, 1,..., 1]. [0, 1, 1] [0, 0, 1] n = 3 C 0 [0, 0, 0] C 0 = base chamber of W 0.

I = stab G (all x C O ) (Iwahori subgroup) O O O = πo O O (n = 3). πo πo O Fact: G acts transitively on {basepointed chambers}. Therefore G/I = {basepointed chambers}. Theorem Any two chambers lie in a common apartment.

Outline of Proof Any chamber can be written x n = Ov 1 + + Ov n x 1 = Ov 1 + + Ov n 1 + πov n x 2 = Ov 1 + + Ov n 2 + πov n 1 + πov n...... x n 1 = Ov 1 + πov 2 + Ov n 2 + πov n 1 + πov n

I \G/I has coset representatives like 0 π 7 0 0 π 3 0 0 0 0 0 π 9 0 0 0 0 π 1 Given two basepointed chambers C 1, C 2, there exist bases giving C 1 and C 2 respectively and differing only by permutations and factors π j on each element. Exercise. Fill in the details.

Notation: [j 1,..., j n ] = B: Stabilizers π j 1 0... 0 π jn = π j 1 Oe 1 + + π jn Oe n π j 1 0 =... (O) 0 π jn Thus W 0 = {[j 1,..., j n ] : j 1 j n }. Also stab G (O) = PGL n (O). Therefore π j 1 0 π j 1 0 stab G ([j 1,..., j n ]) =... PGL n (O)... 0 π jn 0 π jn. = {g GL n (F) : g αβ, (g 1 ) αβ π jα j β O}/O.

Therefore stab G (O, [j 1,..., j n ]) = { } {g GL n (F) : g αβ, (g 1 O if α < β, ) αβ π jα j /O βo. if α > β. Hence stab G (W O ) = {PGL n (O) : g αβ = 0 if α > β}. Example:[n = 2] n 0 ( ) ( ) O O O O π n O O = 0 O.

Suppose: [0,..., 0] W 0 [k 1,..., k n ] [k 1,..., k n ]-subsector W

W = stab G (W ) = π k 1 π k 1... (W O ), π kn... stab G (W O ) π kn π k 1... π kn = 0 if α > β = g PGL n(o) : g αβ O if α = β. π kα k βo if α < β

C: Horocycles W = {Sectors}, W = {π j 1 Ov 1 + + π jn Ov n : j 1 j n }. For k 1 k n, the [k 1,..., k n ]-subsector is {π j 1+k 1 Ov 1 + + π jn+kn Ov n : j 1 j n }. Definition W 1 W 2 if, for some k 1 k n, the [k 1,..., k n ]-subsectors coincide. n = 2 W 1 W 2

n = 3 W 1 common W 2 common

Definition H = W/ Let h 0 = [W 0 ] H. Then, stab G (h 0 ) = stab G k 1 k n π k 1... (W 0 ) π kn { } = 0 if α > β = {g G : g αβ O if α = β := B 0 Example n = 2: B 0 = ( ) O F 0 O G/B 0 = H : gb0 g(h 0 ).

Recall: coset representatives for B 0 \G/K are π j 1 A 0 =... : j α Z. π jn Therefore A 0 parametrizes G\{G/K G/B 0 }. That is, G\ H. Any (x, h) H can be written g π j 1... (O), g(h 0 ) = (g([j 1,..., j n ]), g(h 0 )). π jn for unique [j 1,..., j n ].

The retraction map is defined by r h (x) := [j 1,..., j n ]. Note. r h (x) = r h (y) gx = y with g stab G (h). n = 2 W h = [W ] r h (x) = [1, 0] r h (x) = [0, 0]

n = 3 W r h (x) = [1, 0, 1]

The set is called a horocycle. r 1 h ([0,..., 0]) = {basepoint(w ): [W ] = h} r 1 h 1 ([0,..., 0]) = r 1 h 2 ([0,..., 0]) h 1 = h 2. Therefore identify H with the space of horocycles.

D: The boundary Ω Let W 1, W 2 W. Say W 1 = W2 if W 1, W 2 have a common subsector. W 1 common subsector n = 2 W 2

n = 3 W 1 common W 2 common

Definition Theorem Ω = W/ = Ω = G/B where B = {g G : g αβ = 0 if α > β}. Proof Let ω 0 = [W 0 ] Ω. Then stab G (ω 0 ) = π k 1 g : g... (W 0 ) = π kn k 1 kn l 1 l n π l 1... (W 0 ) π ln

Thus stab G (ω 0 ) = π k 1 k 1 kn l 1 l n π l1 k1... π ln kn... stab G (W 0 ) π kn π k 1... π kn = j 1,...,j n Z = j 1,...,j n Z = A 0 B 0 = B. π j 1 π j 1... stab G π jn k 1 k n... B 0 π jn π k 1... W 0 π kn Therefore, G/B = Ω : gb gω 0

Recall: #(B\G/K) = 1. i.e. G = BK; i.e. G\{Ω } has 1 orbit. x, ω Ω, g G such that x = g(o) and ω = g(ω 0 ). Consequence: There exists an unique sector W based at x representing ω. Uniqueness: Can assume x = 0, ω = ω 0. stab G (O) stab G (ω 0 ) = PGL n (O) B = stab G (W 0 ). Notation: W = conv(o, ω), (recalling that conv(o, ω) is the convex hull of O and ω).

E: Topology of Ω Fix a basepoint y 0. Say ω n ω if conv(y 0, ω n ) conv(y 0, ω) pointwise. (This topology is independent of choice of y 0.) ω 1, ω 2 are close if they have representative sectors based at y o with large common initial segment T. Ω is a totally disconnected compact Hausdorff space.

F: Ω as (the set of chambers of) a building The spherical building of type A n 1 associated with PGL n (F) has as simplices, flags in F n, The space of complete flags V 1 < V 2 < < V k. {0} < V 1 < V 2 < < V n 1 < F n, is identified with G/B = Ω, since B is the stabilizer of Example: n = 3 {0} < [e 1 ] < [e 1, e 2 ] < < [e 1,..., e n ]. An apartment in Ω is (6 chambers)

G: Geometric realization Let be a euclidean building. is a metric space in which each apartment has the euclidean metric. Any two vertices x, y lie in a common apartment E and d(x, y) = d E (x, y). x y is contractible.

A ray is a subset of isometric to [0, ). Two rays are equivalent if they are parallel, i.e. at finite Hausdorff distance from each other. The building at infinity is a spherical building. A point ξ is an equivalence class of rays. For each vertex x of and each ξ, there is a unique ray [x, ξ) with initial vertex x in the parallelism class of ξ. x [x, ξ)

Now let be of type Ã2. Each sector in determines a 1-simplex (chamber) of. Two sectors determine the same chamber of if and only if they contain a common subsector. If [x, ξ) is a sector wall then ξ is a vertex of. If the initial edge of this sector wall is [x, y], and if τ(y) = τ(x) + i then the vertex ξ is said to be of type i 1. Let P be the set of vertices ξ of type 0 and let L be the set of vertices η of type 1. Then (P, L) is a projective plane. A point ξ P and a line η L are incident if and only if they are the vertices of a common chamber in.

[x, η) η L x [x, ξ) ξ P

Part V Ã 2 groups

à 2 groups Let be a locally finite building of type Ã2. α Aut( ) is type rotating if there exists i Z/3Z such that for all vertices v τ(α(v)) = τ(v) + i. An Ã2 group Γ is a group of type rotating automorphisms, which acts freely and transitively on the vertex set of. q = 2 : 8 groups. All are subgroups of PGL 3 (F). q = 3 : 89 groups. Only 21 are subgroups of PGL 3 (F). à 2 groups exist for any prime power q.

Presentation Let (P, L) be a projective plane of order q: P = L = q 2 + q + 1. Let λ : P L be a bijection. A triangle presentation compatible with λ is a set T P P P such that (i) If x, y P, then (x, y, z) T for some z P if and only if y and λ(x) are incident (i.e. y λ(x)). (ii) (x, y, z) T (y, z, x) T. (iii) Given x, y P, then (x, y, z) T for at most one z P.

Form the group Γ = Γ T = P xyz = 1 for (x, y, z) T. (1) The Cayley graph of Γ with respect to the generating set P is the 1-skeleton of a building of type Ã2. Vertex set : Γ Edges : where x P x g gx Γ acts on (by left multiplication)

Example: Mumford s group (1979) D. Mumford constructed Γ M < PGL 3 (Q 2 ) : Γ M is a torsion free lattice in PGL 3 (Q 2 ) ; Γ M acts regularly on the vertex set of. Γ M has generators x 0, x 1,..., x 6, and relators { x 0 x 0 x 6, x 0 x 2 x 3, x 1 x 2 x 6, x 1 x 3 x 5, x 1 x 5 x 4, x 2 x 4 x 5, x 3 x 4 x 6. The 1-skeleton of is the Cayley graph of Γ M. x 0 x 2 x 3 = 1 x 0 x 3 x 2

The projective plane of neighbours of 1 x 1 2 x 3 x 1 1 x 2 x 4 x 1 3 x 2 x 1 5 x 1 x 0 x 5 x 3 x 1 6 x 1 4 x 6 x 1 0 x 0

à 2 groups of Tits type The projective plane (P, L) = PG(2, q) P and L are the 1- and 2-dimensional subspaces of a 3-dimensional vector space V over F q. F q is a subfield of F q 3. Consequence: F q 3 is a 3-dimensional vector space over F q. Identify P with F q 3 /F q. F q 3 = {1, ζ, ζ 2,..., ζ q3 1 }, a cyclic group. F q = ζ 1+q+q2, unique subgroup of order q 1. Therefore P is a cyclic group of order 1 + q + q 2 generated by the element [ζ] = F q ζ. P is a collineation group of order q 2 + q + 1 which acts regularly on P (and on L): a Singer group.

The F q -linear mapping Tr : F q 3 F q is defined by Tr(a) = a + σ(a) + σ 2 (a) where σ(a) = a q, a F q 3. [σ is an automorphism of F q 3 over F q.] The map (x 0, y 0 ) Tr(x 0 y 0 ) is a regular symmetric bilinear form on F q 3, and so the map S S = {y 0 F q 3 : Tr(x 0 y 0 ) = 0 for all x 0 S} is a bijection P L.

If x = F q x 0 P = F q 3 /F q, write Tr(x) = 0 if Tr(x 0 ) = 0; x := (F q x 0 ). Identify a line, i.e., 2-dimensional subspace, with the set of its 1-dimensional subspaces. Then the lines are the subsets of the group P. For x P, let λ 0 (x) = x = {y P : Tr(xy) = 0} ( ) 1 { ( y ) } = y P : Tr = 0. x x This defines a point-line correspondence λ 0 : P L.

The following is a triangle presentation compatible with λ 0 : T 0 = {(x, xξ, xξ q+1 ) : x, ξ P and Tr(ξ) = 0}. (2) The only nontrivial thing to check is that T 0 is invariant under cyclic permutations. If (x, xξ, xξ q+1 ) T 0 then (xξ, xξ q+1, x) = (xξ, (xξ)ξ q, (xξ)(ξ q ) q+1 ) T 0 since Tr(ξ q 0 ) = Tr(ξ 0) and ξ 1+q+q2 0 F q for each ξ 0 F q 3. The group Γ T associated with the presentation (2) is called a group of Tits type.

Groups of Tits type embed as lattice subgroups of PGL 3 (F q ((X ))). For q = 2, Γ T has generators x 0, x 1,..., x 6, and relators { x 0 x 1 x 3, x 1 x 2 x 4, x 2 x 3 x 5, x 3 x 4 x 6, x 4 x 5 x 0, x 5 x 6 x 1, x 6 x 0 x 2.

Appendix: Application to geometric analysis. The Baum-Connes conjecture proposes (for any group Γ) an isomorphism Geometric Group (Group Homology) = Analytic Group K (C r (Γ)) where C r (Γ) := CΓ B(l 2 (Γ)). V. Lafforgue: B-C is true if Γ acts (properly isometrically with compact quotient) on a space of nonpositive curvature and has the following property. (RD) Every rapidly decaying function ξ in l 2 (Γ) lies in C r (Γ).

rapidly decaying means : ξ(g) 2 (1 + g ) 2s <, s R. g Γ P. Jolissaint (1986): Hyperbolic groups (in the sense of Gromov) satisfy (RD). (RD) holds if Γ is a torsion free cocompact lattice in (A) PGL 3 (F), F nonarchimedean; (Ramagge, Robertson, Steger - 1998) (B) PGL 3 (R), PGL 3 (C). (Lafforgue - 2000) These are the first examples of higher rank groups satisfying (RD). Until 1998, B-C was not known for any higher rank group. Open problem for PGL 4.

Underlying obstruction to generalising the method: An apartment in the building of PGL 4 (Q p ) is an à 3 Coxeter complex: Fundamental simplex has dihedral angles: 3 π 3, π 2. 1 1 2 1