Notation B.1

Let π ∈ F with T(π) = 0 ≠ T(π^1+q). Use the nondegenerate symmetric K-bilinear form T(xy) on F to see that π^q ∉ {t ∈ F ∣ T(πt) = 0} = K + Kπ.

Lemma B.2

Ifp₁andp₂are distinct singular points not inΩ, then|p1⊥∩p2⊥∩Ω|≤5q−5.

Proof

By the transitivity of G we may assume that p₁ = 〈(0, 0, π, 0)〉 and p₂ = 〈(a, β, γ, d)〉 for some a, β, g, d. We need to estimate the number of solutions t to the equations

corresponding to points 〈(1, t, t^{q + q2}, t^1+q+q2)〉. By (B.1) we can write t = u + vπ with u, v∈ K. Then the second equation is

which expands to

For each u this is a K-polynomial in v of degree at most three, and hence has at most three roots v ∈ K if it is not the zero polynomial. Let B be the number of “bad” u for which this polynomial in v is the zero polynomial. Then |p1⊥∩p2⊥∩Ω|≤(q−B)3+Bq+1 (the last term occurs since 〈(0, 0, 0, 1)〉 may be in the intersection). We will show that B ≤ 2, which produces the bound in the lemma.

The coefficients of our polynomial show that, for a “bad” u, we must have aN(π) = 0, T(βπ^{q + q2}) = 0, uT(βπ) + T(γπ) = 0 and u²T(β) + uT(γ) + d = 0. If T(βπ) ≠ 0 then there is one “bad” u, while if T(βπ ) = T(γπ) = 0 then there are at most two “bad” u unless T(β) = T(γ) = d = 0.

Thus, we must show that T(βπ^{q + q2}) = T(βπ) = T(γπ) = T(β) = T(γ) = 0} cannot all occur. Since T(β) =T(βπ) = 0, by (B.1) we have β = xπ with x ∈ K. Then 0 = T(βπ^{q + q2})= xT(N(π)), so that x = 0. Similarly, since T(γ) = T(γπ) = 0 we have γ = yπ with y ∈ K. Now p₂=〈(0, 0, yπ, 0)〉 = p₁, which is not the case.□

Notation B.4

Let Ω₀ ⊂ Ω consist of 〈(0, 0, 0, 1) 〉 and 〈(1, t, t^{q + q2}, t^1+q+q2)〉, t ∈ K. There are (q + 1)² singular points in Ω0⊥, all having the form 〈(0, β, γ, 0)〉 with T(β) = T(γ) = T(βγ) = 0. The sets Ω₀ and Ω0⊥ are acted on by a naturally embedded subgroup G₀ = SL(2, q) of G containing the transformations

with s ∈ K. These act on each of the q + 1 lines 〈(0, β, 0, 0), (0, 0,β, 0) 〉 with T(β) = 0 ≠ β that partition the (q + 1)² singular points in Ω0⊥, sending

Definition B.6

An ordinary point is a singular point in Ω0⊥ of the form 〈(0, β, γ, 0)〉 such that either β = 0 and T(γ^1+q) ≠ 0, or T(β^1+q) ≠ 0 (recall that T(β) = T(γ) = T(βγ) = 0 ). Since any β ∈ F^* has characteristic polynomial x³ + T(β) x² + T(β^1+q)x + N(β), the ordinary requirement can fail for some β, γ if and only if q ≡ 1 (mod 3). Moreover, if β ∈ F-K then {β^q ∈ βK} ⇔ {β^{q − 1} ∈ K} ⇔ β^{(q − 1, q2 + q + 1)} ∈ K ⇔ β³ ∈ K ⇔ T(β^1+q) = 0.

For π in (B.1), since T((aπ + π^q)(aπ + π^q) ^q) = (a² + a + 1) T (π^1+q) all points of the line 〈(0, aπ + π^q, 0, 0), (0, 0, aπ + π^q, 0) 〉, a ∈ K, are ordinary if and only if a² + a + 1 ≠ 0, so that all points are ordinary if q ≡ 2 (mod 3), but there are two lines of this form all of whose points are not ordinary when q ≡ 1 (mod 3).

The significance of ordinary points is the following

Lemma B.7

Ifpis an ordinary point then

1. p has the form 〈(0, 0, γ, 0)〉 withT(γ) = 0 or 〈(0, β, aβ, 0)〉 withT(β) = 0 anda ∈ K, and

2. p^g = 〈(0, 0, π′, 0)〉 for some g ∈ G₀, whereπ′ behaves asπdoes in (B.1): T(π′) = 0 ≠ T(π^′1+q).

Proof

We may assume that p = 〈(0, β, γ, 0)〉 with β ≠ 0.

1. Since p is ordinary, we have seen that β^q ∉ Kβ, so that β and β^q span kerT. Write γ = kβ + bβ^q with k, b ∈ K. Then 0 = T(βγ) = bT(β^1+q) implies that b = 0.

2. By (B.5), p^u_kj = 〈(0, 0, β, 0)〉 behaves as stated.□

Lemma B.8

Ifp₁, p₂andp₃are pairwise non-perpendicular ordinary points, then

1. |Qp1⊥∩p2⊥∩Ω|=2qand

2. |p1⊥∩p2⊥∩p3⊥∩Ω|=q+2.

Proof

By Lemma B.7(ii) we may assume that p₁ has the form 〈(0, 0, π, 0)〉 and p₂ = 〈(0, β, γ, 0)〉, where T(β) = T(γ) = T(βγ) = 0. Also T(βπ) ≠ 0 since p₁ and p₂ are not perpendicular. All (0, 0, 0, 1) and (1, t, t^{q + q2}, N(t)), t ∈ K, are in each of the stated intersections, so we will focus on vectors (1, t, t^{q + q2}, N(t)) with t = u + vπ ∉ K that lie in an intersection.

1. Here (B.3) states that

   uvT(βπ)+v2T(βπq+q2)+vT(γπ)=0.(B.9)

   Since T(βπ) ≠ 0, each v ≠ 0 determines a unique u. This argument reverses: the intersection size is (q + 1) + (q − 1).

   Before continuing we massage (B.9). By Lemma B.7(i), γ = kβ for some k ∈ K. Since dim ker T = 2 we can write β = xπ + yπ^q with x, y ∈ K. Since 0 ≠ T(βπ) = yT(π^1+q) we have y ≠ 0 and β ∈ ((x/y)π + π^q)K. We may assume that β = aπ + π^q with a ∈ K. Then

   p2=〈(0,aπ+πq,k(aπ+πq),0)〉.(B.10)

   Also T(βπ) = T(π^1+q), so that (B.9) becomes

   uT(π1+q)+v[aN(π)+T(π2q+q2)]+kT(π1+q)=0.(B.11)

2. We may assume that p₃ = 〈(0, β′, γ′, 0〉) with γ′ = k′β′ and β′ = a′π + π^q for some k′, a′ ∈ K. Then (a + a′)(k + k′) T (ππ^q) = T(βγ′ + γβ′) ≠ 0 since p₂ and p₃ are not perpendicular. Then a ≠ a′, and the two versions of (B.11) imply that

   v=k+k′a+a′T(π1+q)N(π),u=k+k+k′a+a′(a+T(π2q+q2)N(π)),(B.12)

   which proves (ii). □

Example B.13

1. If 𝓢 ⊆ {〈(0, 0, π, 0)〉, 〈(0, aπ + π^q, a²π + aπ^q, 0)〉 ∣ {a ∈ K, a² + a + 1 ≠ 0}, then

   |⋂p∈Sp⊥∩Ω|=q2+1if|S|=12qif|S|=2q+2if|S|≥3.

2. If 𝓢 ⊆ {〈(0, 0, π, 0)〉, 〈(0, π^q, 0, 0)〉, 〈(0, π + π^q, π + π^q, 0)〉, 〈(0, aπ + π^q, a³π + a²π^q, 0)〉} for an arbitrary a ∈ K − {0, 1} such that a² + a + 1 ≠ 0, then

   |⋂p∈Sp⊥∩Ω|=q2+1if|S|=12qif|S|=2q+2if|S|=3q+1if|S|=4.

Proof

All of the stated points are ordinary. Since |p^⊥ ∩ Ω| = q² + 1 [15, p. 1204], we will assume that |𝓢| ≥ 2.

1. In (B.10), k = a for all listed points other than 〈(0, 0, π, 0)〉. By (B.12), t=T(π2q+q2)N(π)+T(π1+q)N(π)π is in every intersection (which is easily checked directly); so is Ω₀, so that every intersection has size ≥ q + 2. Since any intersection of three sets p^⊥ ∩ Ω has size q + 2 (by Lemma B.8(ii)), so does any intersection of at least four such sets.

2. The last three of these four ordinary points correspond to the pairs (a, k) = (0, 0), (1, 1), (a, a²) in (B.10). Then (B.12) and different 3-sets in 𝓢 produce different values of v, so that {|⋂_{p∈ 𝓢}p^⊥ ∩ Ω|} = q + 1 if |𝓢| = 4. The remaining sizes are given in Lemma B.8.

Lemma C.1

Every hyperplane meetsΩ. More precisely, there are exactly five G-orbits of hyperplanesHof U:

1. Tangent hyperplanes H = p^⊥for p ∈ Ω, withr ∈ Hand H ∩ Ω = {p};

2. Secant hyperplanes H = x^⊥containingr, where xis a singular point not inΩ, x^⊥ ∩ Ωis a circle of 𝓙(Ω) and 〈x^⊥ ∩ Ω〉 = x^⊥;

3. O⁻(4, q)-hyperplanes H for which H ∩ Ωis an orbit of a cyclic subgroup ofGof order |H ∩ Ω| = q ± 2q + 1 acting irreducibly on U/ r; and

4. O⁺(4, q)-hyperplanes Hfor whichH ∩ Ωcontains an orbit of a cyclic subgroup of G of order |H ∩ Ω| − 2 = q − 1 that fixes two points of H ∩ Ω. Moreover, H ∩ Ωis not one of the circles in (ii).

Proof

1. Projecting mod r shows that each point of Ω behaves as stated.

2. If x is a singular point not in Ω then each of the q + 1 t.s. lines on x meets Ω since Ω is an ovoid, so that |x^⊥ ∩ Ω| = q + 1. Also, dim〈x^⊥ ∩ Ω〉 = 4, as otherwise x^⊥ ∩ Ω would project into a plane of U/r, and hence be contained in a conic, which is not the case since q > 2 [30, pp. 51–52]. Since 〈x^⊥ ∩ Ω〉 lies in the 4-space x^⊥, these subspaces coincide.

3. This is [2, Theorem 1(a)].

4. The set of singular points of H is partitioned by q + 1 t.s. lines, and each t.s. line of U meets Ω since Ω is an ovoid. Thus, |H ∩ Ω| = q + 1.

We use the orbits of G to find G_H. There are exactly two point-orbits on U/r: Ω and the remaining q(q² + 1) points. There is a subgroup of G of order q − 1 that fixes four points of U/r and induces all scalars on each of these 1-spaces [14, P. 183]. Since each line containing r has a unique singular point, the two point-orbits on U/r produce four point-orbits on U − {r}.

Since G has five point-orbits it also has five hyperplane-orbits, so that all q²(q² + 1)/2 hyperplanes H in (iv) lie in an orbit. Then |G_H| = |G|/[q²(q² + 1)/2] = 2(q − 1), so that G_H is dihedral of order 2(q − 1), with orbits of size 2 and q − 1 on Ω [25, Theorem 9].

For the final assertion, if H ∩ Ω lies in two hyperplanes then it is in a plane, and hence is a conic, whereas in (ii) we already saw that 〈x^⊥ ∩ Ω〉 = x^⊥.□

Remarks C.2

Finally, we collect elementary properties of the group T appearing in Lemma C.1(iv). Consider the action of G = Sz(q) on Ω.

1. The stabilizer of a circle has order q(q − 1) [25, Theorem 9] and fixes a unique point c. Here G_c is a Frobenius group of order q²(q − 1).

2. A subgroup T of order q − 1 fixes two points c, d ∈ Ω and has q + 1 other orbits on Ω of size q − 1.

If 1 ≠ t ∈ T then t fixes exactly two circles: it lies in a unique subgroup of order q(q − 1) of the Frobenius group G_c (or G_d). If C is either of these circles then T is transitive on C − {c, d}.