The Exceptional Tits Quadrangles Revisited

Tits polygons are generalizations of Moufang polygons in which the neighborhood of each vertex is endowed with an “opposition relation.” There is a standard construction that produces a Tits polygon from an arbitrary irreducible spherical building of rank at least 3 when paired with a suitable Tits index. In this note, we complete the proof of a characterization of the Tits quadrangles that arise in this way from the spherical building associated to an exceptional algebraic group.


Introduction
A Tits polygon is a bipartite graph in which the neighborhood of each vertex is endowed with an "opposition relation" satisfying certain axioms which reduce to the axioms of a Moufang polygon in the case that the opposition relations are all trivial.There is a standard construction (described in [3, §3]) that produces a Tits polygon from a pair ( , T ), where is an arbitrary irreducible spherical building of rank at least 3 and T is a suitable Tits index.We say that a Tits polygon is of index type if it arises in this way.We call a Tits polygon exceptional if it arises from such a pair ( , T ) for the spherical building associated to the group of rational points of an exceptional algebraic group.

Tits Polygons
In this section, we assemble the basic properties of Tits polygons, in addition to those listed in [3, 2.5-2.19], that we will require.Let X = ( , A, {≡ v } v∈V ) be a Tits n-gon for some n ≥ 3, let (γ , i → w i ) be a coordinate system of X, let i → U i be the corresponding root group labeling as defined in [3, 2.2 and 2.3], and let G = Aut(X).
We recall that vertices u, v both in w for some vertex w of are opposite at w if u ≡ w v.A path (v 0 , v 1 , . . ., v k ) is straight if v i−1 is opposite v i+1 at v i for all i in the interval [1, k − 1] and a root is a straight path of length n.

Proposition 2.1 Every straight path of length n + 1 is contained in a unique element of A.
Proof This holds by [5, 1.3.11].Definition 2.2 Two vertices, respectively edges, of are opposite if they are opposite vertices, respectively edges, of some element of A. Thus by Proposition 2.1, two vertices u, v are opposite if there is a root from u to v and two edges are opposite if and only if they are the first and last edges on a straight path of length n + 1. Proposition 2. 3 Let u and v be a pair of opposite vertices.Then for each w ∈ u , there exists a unique root (x 0 , x 1 , . . ., x n ) such that x 0 = u, x 1 = w and x n = v.
Proof This holds by [5, 1.3.18].Proposition 2. 4 Let G † denote the subgroup of G generated by all the root groups of X and let U [1,n] be as in [3, 2.5(i)].Then, the following hold: (i) G † acts transitively on the set of edges of .(ii) U [1,n] acts sharply transitive on the set of edges opposite the edge {w n , w n+1 }. (iii) G † acts transitively on the set of all ordered pairs of opposite edges and on all pairs of opposite vertices of .
Proposition 2.5 Let e = {u, v} and f = {x, y} be two edges of and suppose that u is opposite y and that v is opposite x.Then, e is opposite f .
Proof By Proposition 2.3, there exists a root α = (u, v, . . ., y) from u to y containing v and a root β = (v, . . ., y, x) from v to x containing y.By [3, 2.2(ii)], the subpath of α from v to y coincides with the subpath of β from v to y.Hence, the (n + 1)-path (u, v, . . ., y, x) is straight.As we observed in Definition 2.2, it follows that the edges e and f are opposite.

Definition 2.6
We recall that X is p-plump for some p ≥ 2 if for every vertex v and every subset S of v of cardinality at most p, there exists a vertex in v that is opposite every vertex in S at v. As we observed in [5, 1.4.22], the Tits polygon X ,T for a pair ( , T ) (see [3, §3] for the definition) is p-plump whenever the field of definition of the building contains at least p elements.
Proposition 2.7 Suppose that X is p-plump for some p ≥ 2. Then for each k ∈ [1, 2p − 2] and for each k-path (x 0 , . . ., x k ), there exists an edge that is opposite the edge Proof We proceed by induction with respect to k.By Proposition 2.4(i), the claim holds for k = 1.Suppose that (x 0 , . . ., x k ) is a k-path in for some k ∈ [2, 2p − 2] and that f is an edge that is opposite where m is the largest integer such that k − 1 − 2m ≥ 0, and let , the vertex y is opposite v i for all i ∈ [0, m].By Proposition 2.3, there exists for each i ∈ [0, m] a root α i = (v i , z i , . . ., y) from v i to y containing z i and if k is even, there exists a root β = (v m , z m+1 , . . ., y) from v m to y containing z m+1 .Let d i be the unique vertex in y on α i for each i ∈ [0, m] and if k is even, let b be the unique vertex in y on β.We have m ≤ (k − 1)/2 and thus m ≤ p − 2. Since X is p-plump, it follows that we can choose a vertex e ∈ y that is opposite d i at y for all i ∈ [0, m] and, if k is even, also opposite b at y. Since the (n+1)-path (α i , e) is straight, e is opposite z i for all i ∈ [0, m].If k is even, then the (n + 1)-path (β, e) is straight, so e is also opposite z m+1 .By Proposition 2.5, it follows that f is opposite c i for all i ∈ [0, m].
The following result refers to vertices w 1 , w 2 , etc., on γ in the coordinate system (γ , i → w i ) that we have chosen.Note, in particular, that {w 1 , w 2n } and {w n , w n+1 } are opposite edges of .Note, too, that by [3, 2.3 Then, {u, w 2n } is an edge opposite {w n , w n+1 }.By Proposition 2.8(i), therefore there exists [3, 2.3], on the other hand, U n fixes w 2n .Hence, the first claim holds.The second claim holds by a similar argument.Proposition 2.10 Suppose that X is 3-plump and let δ = (x, y, z) be a straight 2-path.Then, δ is the unique 2-path from x to z.
Proof Let ξ = (x, u, z) be an arbitrary 2-path from x to z.By Propositions 2.4(iii) and 2.7, we can assume that the coordinate system (γ , i → w i ) is chosen so that x = w 1 and y = w 2n and that the edges on δ and the edges on ξ are all opposite f := {w n , w n+1 }.By Proposition 2.8, therefore there exist a 1 ∈ U 1 and a n ∈ U n such that w 2n = u a 1 and z = w a n 1 .The element a 1 fixes f and {z a 1 , w 2n } = {z, u} a 1 .Hence, {z a 1 , w 2n } is an edge opposite f .Thus by one more application of Proposition 2.8, there exists b n ∈ U n such that z a 1 = w b n 1 .Therefore,

Tits Quadrangles
We continue with X, n, (γ , i → w i ), i → U i and G = Aut(X) as in the previous section, but now we assume that n = 4 and that X is sharp as defined in [3, 2.13].[3, 4.4 and 4.6], we let Y i denote the pointwise stabilizer in G of the set of vertices of at distance at most 2 from w i+2 .The group Y i is a normal subgroup of U i and it is normalized by the pointwise stabilizer G γ for all i.Since X is sharp, it follows that for each i, either

Notation 3.1 As in
Proof By Proposition 2.10, the distance from v to z is 3.It will suffice, therefore to show that δ is the unique path of length 3 from v to z.By [3, 4.8], we can assume that Y 1 = 1.Thus Y 1 = ∅ by Notation 3.1.Choose a ∈ Y 1 .Then, (w 0 , w 1 , w a 0 ) is a straight 2-path.By Proposition 2.4(i) and [3, 2.7], every straight path of length 3 is in the same G-orbit as (w 0 , w 1 , w 2 , w 3 ) or (w 3 , w 2 , w 1 , w 0 ).Replacing δ by (z, y, x, v) if necessary, we can assume, therefore that δ = (w 0 , w 1 , w 2 , w 3 ).Now let ξ = (v, x , y , z) be an arbitrary 3-path from v to z.Then, x is at distance 2 from z = w 3 and a ∈ Y 1 , so a fixes x .Thus {x , w a 0 } = {x , v} a is an edge and hence (w 0 , x , w a 0 ) is a path of length 2 from w 0 to w a 0 .Therefore, x = w 1 = x by Proposition 2.10 applied to (w 0 , w 1 , w a 0 ).By Proposition 2.10 applied to (x, y, z), it follows that y = w 2 = y and thus ξ = δ.Proposition 3.3 Suppose that X is 3-plump, let v and z be opposite vertices and let y ∈ z .Then, there exists a root α from v to z containing y and α is the unique path of length 4 from v to z containing y.
Proof The first claim holds by Proposition 2.3 and the second by Proposition 3.2.Proposition 3.4 Suppose that X is 3-plump and let u, v be opposite vertices.Then, the distance from u to v in is 4.
Proof It suffices to observe that by [3, 2.2(ii)], u = v and that by Proposition 3.2, there is no vertex z such that u, v ∈ z .
Since a 1 ∈ U 1 and a 4 , b 4 ∈ U 4 , α is a straight path of length 4 from w 1 to v by Proposition 2.8(ii).Therefore, w 1 and v are opposite.By Eq. 3.6, we have ) Proof By [3, 2.5(ii)], it suffices to observe that and

Linear Forms
We pause now to prove two elementary results about linear forms that we will require in the next section.
Notation 4.1 Let K be a field, let V be a vector space over K and let f, g : V → K be non-zero linear forms on V .Then, f and g are proportional if each is a scalar multiple of the other.
Proposition 4.2 Let c ∈ K and let f, g, h be linear forms on V := K ⊕ K such that f and g are non-zero and not proportional to each other.Then, there is an injection from Proof Since f and g are non-zero and not proportional to each other, there exist a, b ∈ K and u ∈ V such that h = af + bg, f (u) = −b and g(u) = −a.Let For each e ∈ K * , there is a unique v e ∈ V such that f (v e ) = e and g(v e ) = de −1 .We have q(v e ) = 0 for each e ∈ K * and hence e → v e + u is an injection from where := {v | f (v)g(v) + h(v) = c and j (v) = d}.Then, j is proportional to f or g.
Proof There exists z ∈ V such that j (z) = d.Replacing the variable v by v − z, we can assume that d = 0. Suppose that u, v are arbitrary distinct non-zero elements of .Since j = 0 and d = 0, we have u = λv for some λ ∈ K\{0, 1}.Hence, and thus (λ Suppose c = 0.Then, 0 ∈ and by Eq. 4.6, f (v)g(v) = 0. Hence by Eq. 4.5, λ is uniquely determined by f (v), g(v), and h(v).Since u and v are arbitrary, it follows that there are no other non-zero elements in .By Eq. 4.4, we conclude that c = 0. Thus by Eq. 4.6, λf (v)g(v) = 0. Since λ = 0, we have f (v)g(v) = 0 and thus f (v) = 0 or g(v) = 0. Since j (v) = 0 and v = 0, it follows that j is proportional to f or g.

The Exceptional Tits Quadrangles
Let X, n, (γ , i → w i ) and i → U i be as in Section 2. Again we assume that n = 4.
The main result of this section is Proposition 5.15.
Notation 5.1 Suppose that = (K, L, q, f, ε, X, •, h, θ) is an 8-tuple of algebraic data satisfying all the axioms of a quadrangular algebra in [3, 7.1] except possibly D1.Let g be as in [3, 7.1(C3)] and let S ,g denote the group defined in [3, 8.1].Thus the underlying set of S ,g is X × K and multiplication is given by the formula Let π(a) = θ(a, ε) for all a ∈ X as in [3, 7.1(D1)].
Hypothesis 5.3 Let = (K, L, q) be a quadratic space, let f be the associated bilinear form and suppose that X, (γ , i → w i ), , and ε satisfy all the hypotheses of [3, 8.2] except possibly [3, 8.2(b)].In particular, |K| > 4. In addition, we assume that X is 3-plump as defined in [3, 2.19].As indicated in [3, 8.59], there exist algebraic data X, •, h, and θ such that the 8-tuple satisfies all the axioms of a quadrangular algebra in [3, 7.1] except possibly D1 as well as isomorphisms x i from the group S ,g to U i for i odd (where g and S ,g are as in Notation 5.1) and from the additive group of L to U i for i even such that the following commutator relations hold: for all (a, t), (b, s) ∈ S ,g and all v, w ∈ L, where av = a • v and φ is as in [3, 7.1(C4)].
Notation 5.7 Suppose that the quadratic space = (K, L, q) is isotropic.By [3, 6.8], we can assume that the element ε in the hypothesis [3, 8.2(a)] lies in a hyperbolic plane P of .Since q(ε) = 1, we can choose a basis z 1 , z 2 of P such that q(z 1 ) = q(z 2 ) = 0, f (z 1 , z 2 ) = 1 and ε = z 1 + z 2 .Since |K| > 4, we can choose r 1 , r 2 ∈ K such that r 1 = r 2 , r 1 r 2 = 0, and r i = 1 for i = 1 and 2. Let u = r 1 z 1 + r 2 z 2 .Then, for all s, t ∈ K.In particular, q(u) = r 1 r 2 = 0 and f (u, ε) = r 1 + r 2 .Notation 5.8 Suppose that is isotropic and let u be as in Notation 5.7.We choose an arbitrary element a of X and set b = −au −1 , where u −1 is as in Notation 5.6.Remark 5.9 Let a, b, u be as in Notation 5.8 and let v = −ε+u.Then, a+b+bv = 0 (by [3, 7. The proof of the following result involves a few minor calculations in .These calculations require only the axioms A1-A3 (for the first assertion in Remark 5.9) and C1.In particular, they do not require D1.

Proposition 5.10 Suppose that
= (K, L, q) is isotropic and that |K| > 5. Let a, b be as in Notation 5.8, let v = −ε + u, and let a 1 = x 1 (0, 1).Then, a 1 ∈ U 1 and there exist s, t ∈ K and a 4 , b 4 ∈ U 4 such that a 4 b 4 = d 4 c 4 and Eq.3.8 holds with Proof By Proposition 5.5(i), a 1 ∈ U 1 .Let g be as in Hypothesis 5.3 and let π be as in Notation 5.1.Thus θ(b, v) a, t), and b 1 = x 1 (b, s) for some s, t ∈ K. Using Eq. 5.2, Eq. 5.4, and Remark 5.9, we calculate that with these choices, the expression on the right-hand side of Eq. 3.8 is x 2 (w)x 3 (0, p), where = h(b, a) + π(a) + θ(b, u) + tε + su (5.11) and p = t + q(u)s + β (5.12) for By Eq. 5.4, the expression on the left-hand side of Eq. 3.8 with our choices for a 1 and a 4 is Thus q(w) = p holds if and only if Eq. 3.8 holds.
The quantities w ∈ L and p ∈ K are functions of s and t.By Proposition 5.5(ii), a 4 = x 4 (w) ∈ U 4 and b 4 = x 4 (u−w) ∈ U 4 if and only if both q(w) and q(u−w) are non-zero.Our goal, therefore is to show that s and t can be chosen so that q(w) = p, q(w) = 0, and q(u − w) = 0.
By Eq. 5.11, we have where z = h(b, a)+π(a)+θ(b, u) and r 1 , r 2 are as in Notation 5.7.Let c = β −q(z), where β is as in Eq. 5.13.By Eq. 5.12, Hence, q(w) = p if and only if s, t is a solution to the equation where Let S denote the solution set of this equation.Since r 1 = r 2 , the linear forms G 1 and G 2 are not equivalent.By Proposition 4.2, therefore |S| ≥ 5 and for all (s, t) ∈ S, we have q(w) = p.Let j denote the linear form q(u)s + t.Thus j = r 1 r 2 s + t and by Eq. 5.12, p = j (s, t) + β.By the choice of r 1 and r 2 in Notation 5.7, j is equivalent to neither G 1 nor G 2 .By Proposition 4.3 with d = −β, therefore there are at most two elements (s, t) of S such that p = j (s, t) + β = 0. Next we note that by Eq. 5.11, q(w − u) = q(w) − f (w, u) + q(u) = q(w) − 2q(u)s − f (ε, u)t + δ (5.14) for all (s, t), where δ = q(u) − f (u, z).Let j 0 denote the linear form By the choice of r 1 and r 2 in Notation 5.7, we have and thus j 0 is equivalent to neither G 1 nor G 2 .By another application of Proposition 4.3, therefore there are at most two elements (s, t) of S such that j 0 (s, t) = −β − δ.Since |S| ≥ 5, we conclude that there exists (s, t) ∈ S such that p = j (s, t) + β = 0 and j 0 (s, t) + β + δ = 0.For all such (s, t), we have q(w) = p = 0, hence by Eq. 5.14 and thus q(w − u) = 0.
Proposition 5.15 Suppose that = (K, L, q) is isotropic and |K| > 5. Then for each a ∈ X, there exists t ∈ K such that x 1 (a, t) ∈ U 1 .
Proof Let a ∈ X.By Proposition 5.10, we can choose

Pseudo-quadratic Tits Quadrangles
The main result of this section is Proposition 6.10.Definition 6.1 An involutory ring (K, K 0 , σ ) is an associative ring K endowed with an involution σ and an additive subgroup K 0 such that σ acts trivially on K 0 and 1 ∈ K 0 as well as a + a σ ∈ K 0 and a σ K 0 a ⊂ K 0 for all a ∈ K. Definition 6.2 A pseudo-quadratic module is a 6-tuple = (K, K 0 , σ, L, q, f ) such that (K, K 0 , σ ) is an involutory ring, L is a right K-module, f is a skewhermitian form on L, and q is a map from L to K such that q(a + b) ≡ q(a) + q(b) + f (a, b) (mod K 0 ) and q(at) ≡ t σ q(a)t (mod K 0 ) (6.3) for all a, b ∈ L and all t ∈ K.We call non-degenerate if f is non-degenerate, i.e., if {a ∈ L | f (a, L) = 0} = {0}.Notation 6.4 Let = (K, K 0 , σ, L, q, f ) be a pseudo-quadratic module.Let T denote the subset We define an associative multiplication on T by setting for all (a, t), (b, s).Setting b = a and t = −1 and 2 in Eq. 6.3, we obtain q(−a) ≡ q(a) and 2q(a) ≡ f (a, a) (mod K 0 ) for all a ∈ X and hence q(−a) + t − f (a, a) ∈ K 0 for all (a, t) ∈ T .It follows that T is a group with identity (0, 0) and (a, t) −1 = (−a, −t + f (a, a)) for all (a, t) ∈ T .Note that T = K 0 if L = 0. Proposition 6.6 Let (a, t) ∈ T and suppose that t has a right inverse s ∈ K.Then, (as, −s) ∈ T .
Proof By Definition 6.1, s + s σ ∈ K 0 .Thus q(as) + s ≡ s σ q(a)s − s σ (mod K 0 ) by Eq. 6.3.Since ts = 1, it follows that s σ q(a)s − s σ = s σ (q(a) − t)s ∈ s σ K 0 s ⊂ K 0 by Definition 6.1 and Eq.6.5.Hence, q(as) + s ∈ K 0 .Definition 6.7 Let = (K, K 0 , σ, L, q, f ) be a pseudo-quadratic module, let T be as in Notation 6.5, and let X be a Tits quadrangle.Then, X is pseudo-quadratic of type if there exists a coordinate system (γ , i → w i ) and isomorphisms x i from T to U i for i = 1 and 3 and from the additive group of K to U i for i = 2 and 4 such that the following commutator relations hold: for all (a, t), (b, s) ∈ T and all v, w ∈ K.If L = 0, we say that X is involutory of type (K, K 0 , σ ).Proposition 6.9 Let X be a pseudo-quadratic Tits quadrangle of type for some pseudo-quadratic module = (K, K 0 , σ, L, q, f ) and let T , (γ , i → w i ) and x 1 , . . ., x 4 be as in Definition 6.7.Suppose that is non-degenerate and that X is sharp as defined in [3, 2.13].Let U 1 and Y 1 be as in [3, 2.8 and 4.6].Then, x 1 (0, K 0 ) = Y 1 and if x 1 (0, 1) ∈ U 1 , (6.10) then the following hold: Proposition 7.1 Suppose that X, (γ , i → w i ) and = (K, L, q) satisfy all the hypotheses of [3, 8.2] except possibly [3, 8.2(b)].Suppose, too, that X is 3-plump and that |K| > 5.Then, the hypothesis [3, 8.2(b)] holds as well.
Proof Let = (K, L, q, f, ε, X, •, h, θ) be as in Hypothesis 5.3.Our goal is to show that for each a ∈ X there exists t ∈ K such that x 1 (a, t) ∈ U 1 . (7.2) If X is a Moufang quadrangle, i.e., if all the local opposition relations are trivial, then U 1 = U * 1 , so x 1 (a, 1) ∈ U 1 and thus Eq. 7.2 holds.If X is of type E 7 as defined in [6, 6.15], then Eq. 7.2 holds by [3, 10.4(ii)].We can thus assume that X is neither a Moufang quadrangle nor of type E 7 .By Proposition 5.15, on the other hand, we can assume that is anisotropic.By [3, 5.1(iii) and 6.4(ii)], this assumption implies that for all even i.We now think of X as a Veldkamp quadrangle.Let P denote the vertices of at even distance from w 1 and let L denote the vertices at even distance from w 4 .By [5, 1.4.15] and Eq.7.3, the local opposition relation ≡ v is trivial for all v ∈ L. We declare the elements of P to be points and the elements of L to be lines.With this choice of point set and line set, X is green as defined in [6, 3.1].

Conclusions
We suppose one last time that X, n, (γ , i → w i ) and i → U i are as in Section 2, that n = 4 and that Hypothesis 5.3 holds and let = (K, L, q, f, ε, X, •, h, θ) be as in Hypothesis 5.3.We assume in this section that X is 4-plump and that |K| > 5.By Proposition 7.1, we know that X satisfies all the conclusions of [3, 8.2].In Theorems 8.4 and 8.6, we summarize the main conclusions in [3] that follow from [3, 8.2].

Proposition 3 . 7
8(ii) and 3.4, β is also a path of length 4 from w 1 to v, or equivalently, c 4 , c 1 , and d 4 are all non-trivial.By Proposition 3.3, β is also straight.By Proposition 2.8(ii), therefore c 4 , d 4 ∈ U 4 and c 1 ∈ U 1 .Let a i , b i , c i , d i ∈ U i for i = 1 and 4 and suppose that a 1 = d 1 c 1 b 1 and a 4 b 4 = d 4 c 4 .Then, Eq. 3.6 holds if and only if

Proposition 4 . 3
. (This map is a bijection if d = 0.If d = 0, there are also solutions w + u for w ∈ V such that f (w) = 0 and g(w) is arbitrary.)Let c, d ∈ K and let f, g, h, j be linear forms on V := K ⊕ K such that f and g are non-zero and not proportional to each other and j is non-zero.Suppose that | | ≥ 3, (4.4) and t ∈ K such that a 4 b 4 = d 4 c 4 and the elements a 1 , b 1 , c 1 , a 4 , b 4 , c 4 , d 4 with c 1 = x 1 (a, t) satisfy Eq. 3.8.Let d 1 = a 1 b −1 1 c −1 1 .Then by Proposition 3.7, the elements a 1 , b 1 , c 1 , d 1 , a 4 , b 4 , c 4 , d 4 satisfy Eq. 3.6.By Proposition 3.5, therefore c 1 = x 1 (a, t) ∈ U 1 .
], U i fixes {w n , w n+1 } for all i ∈ [1, n].Let u ∈ w 1 \{w 2 } and v ∈ w 2n \{w 2n−1 } and suppose that the edges {u, w 1 } and {v, w 2n } are both opposite {w n , w n+1 }.Then, the following hold: The stabilizer of w 2n in U [1,n] is U n and the stabilizer of w 1