The Code of a Regular Generalized Quadrangle of Even Order

N. S. Narasimha Sastry Peter Sin To Professor John Thompson on the occasion of his 65th birthday

Introduction

Let V (q) ≅ _{q}^{4} be a 4-dimensional vector space over the field of q = 2^{n} elements, endowed with a nonsingular symplectic form ⟨−, −⟩. Our aim is to study the incidence between the set P of one-dimensional subspaces of V (q) (viewed as points of projective space) and the set L (lines) of isotropic 2-dimensional subspaces. This incidence system is the regular generalized quadrangle of order q. (See [8], p.37 and (5.2.1), p.77.) Let A be the incidence matrix, with rows labelled by the q^{3} + q^{2} + q + 1 points (in any order) and columns labelled by the q^{3} + q^{2} + q + 1 lines (in any order) and entries 0 or 1 according to whether or not the point lies on the line. The matrix A can be considered as a matrix over any commutative ring k and defines an k-linear map between the free k-modules k^{L} and k^{P} on L and P respectively, sending l ∈ L to ∑_{p∈l} p ∈ k^{P}. The most interesting case is when k = _{2}, when we have a natural identification of subsets of P with the elements of _{2}^{P}. The first theorem gives the rank of A in this case. Two fields are involved, the field _{q} defining the geometry and the coefficient field k = _{2}. The latter may be replaced by any extension without altering the rank, but the order q = 2^{n} of the first field is critical. Let N be the collection of those subsets of /2n which do not have two consecutive elements.

Theorem 1. Let k be any field of characteristic 2. Then

rank A = 1 + (\frac{1 + \sqrt{17}}{2})^{2n} + (\frac{1 − \sqrt{17}}{2})^{2n}

= 1 + ∑_{I∈N} 4^{|I|}.

We shall in fact prove that the rank of A is equal to the second expression. A simple combinatorial proof of the equality of the two expressions is given in the first lemma of [6, Appendix].

This work was done while one of us (NSN) was visiting the university of Florida during 1995-96. He wishes to thank Professor John Thompson for kindly arranging this visit and the department for its kind hospitality and the friendly atmosphere during his stay.

We shall not prove Theorem 1 directly but instead, we shall deduce it from a much stronger result (Theorem 2) for which we need some notation and background results. Let k be a field of characteristic 2. The group G = Aut(V (q), ⟨−, −⟩) ≅ Sp(4, q) acts on the sets P and L, preserving the incidence relation. Thus, the map from k^{L} to k^{P} defined by A is a map of kG-modules. Denote its image by C = C_{k}. This is the k-code of the generalized quadrangle. Theorem 1 gives its dimension. It is is well known that the words of minimum weight in the _{2}-code are precisely the columns of A ([3, Theorem 2.8,(b),p.3].]).

It is clear that the vector 1 = ∑_{p∈P} p spans a trivial kG-submodule of k^{P} and that the map sending every p ∈ P to 1 is a kG-projection onto this submodule. Thus

(1)  k^{P} = k1 ⊕ Y_{P} ,

where Y_{P} is the kernel of the projection. It is also immediate from the fact that each point lies on q + 1 lines that 1 ∈ C. Thus, we have a decomposition

(2)  C = k1 ⊕ \overline{C}.

We can think of \overline{C} as the k-span of the set of codewords of even weight in C_{_{2}} .

We shall take k to be an algebraically closed field of characteristic 2 from now on. Theorem 2 below describes the structure of the kG-module \overline{C} in detail.

We next recall some facts about the simple kG-modules. Let V = V (q)⊗_{_{q}} k ≅ k^{4} and extend the symplectic form to V. Then G is the subgroup of the algebraic group Sp(V) ≅ Sp(4, k) fixed by the n-th power of the Frobenius morphism F (which is the map squaring matrix entries of elements in SL(4, k)). It is well known [13, Theorem 28, p.146] that Sp(4, k) has an endomorphism τ with τ^{2} = F. Let S be any Sp(4, k)-module, given by some representation ϕ : Sp(4, k) ⟶ GL(S). Then for i ∈ , we denote by S_{i}, the vector space S with the “twisted” module structure given by the representation ϕ ∘ τ^{i}. Clearly, S ≅ S_{2n} as kG-modules, so for kG-modules, we may take the indices in /2n . For I ⊆ /2n , we set V_{I} = ⨂_{i∈I} V_{i} (with V_{∅} = k). Then by Steinberg’s Tensor Product Theorem [12,§11], the 4^{n} modules V_{I} are a complete set of nonisomorphic simple kG-modules. It makes sense to speak of even and odd elements of /2n , and so each I ⊆ /2n can be partitioned as I = I_{odd} ∪ I_{even}. We set h_{I} = |I_{even}| − |I_{odd}|.

Recall that the radical of a module is the smallest submodule with semisimple quotient (called the head). Iterating, we obtain the radical series. Dually, the maximal semisimple submodule is called the socle and we have the socle series. We are ready to state the main result, from which Theorem 1 follows immediately.

Theorem 2. The radical series of \overline{C} has length 2n + 1. The radical layers are

rad^{j}(\overline{C})/ rad^{j+1}(\overline{C}) ≅ ⨁_{\begin{matrix}I∈N \\ h_{I}+n=j \end{matrix}} V_{I} (0 ≤ j ≤ 2n).

Moreover, the socle series is the same in the sense that

soc^{j}(\overline{C}) = rad^{2n+1−j}(\overline{C}).

Remark. Since every composition factor of \overline{C} has multiplicity 1, it follows that every submodule of \overline{C} is determined by the isomorphism type of its head.

The proof of Theorem 2 is presented in §1. In §2 we discuss certain subcodes of C of geometric origin.

§1. The structure of C.

Let G_{α} and G_{β} be the stabilizers in G of a point and of a line respectively. We have the decomposition of G_{α} = U_{α}L_{α} as the semidirect product of its unipotent radical U_{α} with a Levi complement L_{α}. Further we have

(3)  L_{α} ≅ SL(2, q) × _{q}^{×}.

The facts we use about k SL(2, q)-modules can be found in [1]. The simple k SL(2, q)-modules are all obtained as twisted tensor products of the standard 2-dimensional module W; if W_{j} (j ∈ /n ) denotes the twist of this module by the j-th power of the Frobenius map and for K ⊆ /n , we set

W_{K} = ⨂_{j∈K} W_{j},

then the W_{K} are a complete set of nonisomorphic simple k SL(2, q)- modules. We fix an isomorphism in (3) and denote by [a] the character t ↦ t^{a} of _{q}^{×}, a ∈ /(q − 1) . Then, since U_{α} acts trivially on all simple kG_{α}-modules, these modules are of the form

W_{K} ⊗ [a] (outer tensor product).

For example, the module W_{/n} ⊗ [0] is the Steinberg module St_{L_{α}}; also we will write [a] for k ⊗ [a].

At several points later, we need facts about representations of G_{α}, which we collect in the next lemma.

Lemma 3.

(a) The restriction of V to G_{α} is a uniserial module with composition factors (in descending order) [−1], W ⊗[0], [1]. The module is semisimple for L_{α}.

(b) The restriction of V_{1} to G_{α} is a nonsplit extension of W ⊗ [−1] by W ⊗ [1]. The module is semisimple for L_{α}.

(c) The group U_{α} is elementary abelian. This _{2}L_{α}- module can be given a _{q}-vector space structure R(q) so that R = R(q)⊗_{_{q}} k is isomorphic as a kL_{α}-module to a nonsplit extension of W ⊗ [1] by [1].

(d) Ext_{kG_{α}}^{1}(M, [−2]) = 0 for M ≅ [0], W ⊗ [1] and W ⊗ [−1].

Proof: Parts (a), (b) and (c) are by direct calculations involving only the standard 4-dimensional matrix representation of G. In (c), to get the _{q}- structure, one must choose the isomorphisms between the root subgroups and the additive group of _{q} in a consistent way, so that Z_{α} acts with the same character (instead of different Galois conjugates of a character) on both composition factors. The 1-cohomology groups for G_{α} with values in simple modules can all be computed by means of the inflation-restriction sequence [10, Ch. VII.6, Prop. 5 and p.118, Remark]; since we have a semidirect product in which the kernel U_{α} acts trivially on any simple kG_{α}-module M, the inflation-restriction sequence takes the form of the short exact sequence: (4)

0 ⟶ H^{1}(L_{α}, M) ⟶^{inf} H^{1}(G_{α}, M) ⟶^{res} Hom_{kL_{α}}( ⨁_{σ∈Gal( _{q}/ _{2})} R^{(σ)}, M) ⟶ 0.

The first term of this sequence is easy to compute given the calculations of all extensions of simple k SL(2, q)-modules in [1], and the third term is straightforward to calculate. Thus we obtain the G_{α}- cohomology. Then (d) follows from the standard isomorphisms (cf. [10, p. 111]):

H^{1}(G_{α}, [−2]) ≅ Ext_{kG_{α}}^{1}(k, [−2]), etc.

The automorphism τ of G interchanges the two parabolic subgroups G_{α} and G_{β}, so that

(5)  k^{L} ≅ ind_{G_{β}}^{G}(k) ≅ [ind_{G_{α}}^{G}(k)]_{1} ≅ (k^{P})_{1}.

We set V_{odd} = V_{( /2n )_{odd}} and V_{even} = V_{( /2n )_{even}} .

The following lemma is a special case of Curtis’ theory of modular representations of groups with a split BN-pair [7, 6.6, 6.8].

Lemma 4.

(a) soc(Y_{P}) ≅ head(Y_{P}) ≅ V_{even}.

(b) soc(Y_{L}) ≅ head(Y_{L}) ≅ V_{odd}.

Proof: By (5), it suffices to prove (a). Now k^{P}, is self-dual, so Y^{P} is too. Further, all simple kG-modules are self-dual, since V is. Thus, we are reduced to showing that head(Y_{P}) ≅ V_{even}. By Frobenius reciprocity,

Hom_{kG}(ind_{G_{α}}^{G}(k), V_{I}) ≅ Hom_{kG_{α}}(k, V_{I}) ≅ (V_{I})^{G_{α}}.

Direct computation shows that the only simple modules V_{I} on which G_{α} has a nonzero fixed point are k and V_{even}, and for these the space of fixed points is 1-dimensional. The lemma now follows from (1).

Since \overline{C} is the image of the kG-map from Y_{L} to Y_{P} induced by A, Lemma 4 yields

(6)  head(\overline{C}) ≅ V_{odd} and soc(\overline{C}) ≅ V_{even}.

We denote by [Y : W]_{G} the composition multiplicity of the simple kGmodule W in the kG-module Y.

Lemma 5. [Y_{P} : V_{odd}]_{G} = 1 .

Proof: First, we observe that V_{/2n} is the Steinberg module St_{G}. As is true of all Steinberg modules for groups of Lie type, St_{G} is projective and its restriction to a parabolic subgroup G_{J} is the induction to G_{J} of the Steinberg module of a Levi complement L_{J} [5, Proposition 6.3.3]. Thus, using Frobenius reciprocity,

Hom_{kG}(St_{G} ⊗ V_{even}, ind_{G_{α}}^{G}(k)) ≅ Hom_{kG_{α}}(St_{G} ⊗ V_{even}, k)
≅ Hom_{kG_{α}}(St_{G}, V_{even})
≅ Hom_{kG_{α}}(ind_{L_{α}}^{G_{α}} St_{L_{α}}, V_{even})
≅ Hom_{kL_{α}}(St_{L_{α}}, V_{even}).

The last space has dimension [V_{even} : St_{L_{α}}]_{L_{α}} because St_{L_{α}} is projective (hence also injective).

We have L_{α} ≅ SL(2, q) × _{q}^{×} and the restriction of V to this subgroup is a direct sum of two one-dimensional modules and a two-dimensional simple module, by Lemma 3. Since St_{L_{α}} ≅ W_{/n} ⊗ [0], it follows that [V_{even} : St_{L_{α}}]_{L_{α}} ≤ 1. Next,

Hom_{kG}(St_{G} ⊗ V_{even}, V_{odd}) ≅ Hom_{kG}(St_{G}, V_{even} ⊗ V_{odd})

= Hom_{kG}(St_{G}, St_{G}) ≅ k, so the projective cover of V_{odd} is a direct summand of St_{G} ⊗ V_{even}. It now follows that [ind_{G_{α}}^{G}(k) : V_{odd}]_{G} ≤ 1 and equality holds by (6), which proves the lemma.

We can now give a module-theoretic characterization of \overline{C}.

Lemma 6. Suppose X is a kG-module satisfying the following conditions: (a) head(X) ≅ V_{odd};

(b) soc(X) ≅ V_{even} and [X : V_{even}]_{G} = 1;

(c) Hom_{kG_{α}}(X, k) ≠ 0. Then X ≅ \overline{C}.

Proof: By Frobenius reciprocity, (c) gives a nonzero kG-map from X to k^{P}. By (a), the image of this map is contained in Y_{P} and has head isomorphic to V_{odd}. This shows that the image is \overline{C}, for Lemma 5 shows that there is a unique submodule of Y_{P} with this property and (6) shows that \overline{C} is one such. Finally, by (b), this map is injective.

We will now construct a module satisfying the hypotheses of Lemma 6. The properties of \overline{C} stated in Theorem 2 will then be read off by closer examination of the construction itself.

To begin with, we assert that there exists a kG-module E which is uniserial with composition factors (in descending order) V_{1}, k, V_{2}. In fact (cf. [11, Lemma 2]) such a module exists as a subquotient of V ⊗ V. (It is even a module for the algebraic group; it’s dual admits a natural description as the quotient of the space of symmetric bilinear forms on V by the 1-dimensional span of the given symplectic form ⟨−, −⟩.) We then define X̃ to be the tensor product ⨂_{i=0}^{n−1} E_{2i} of all the Galois conjugates of E. We note the following properties of X̃ which follow from its definition. The descending composition series of the tensor factors E_{2i} induce a natural descending filtration on X̃ of length 2n + 1 as follows. If we designate the composition factors of each E_{2i} to be of level 0, 1 and 2 in descending order, then the t-th layer of filtration on X̃ is the direct sum of all the tensor products (involving one composition factor from each E_{2i}) such that the levels of the factors add up to t. We shall call this the descending tensor filtration. It is clear that the products formed by taking one composition factor from each E_{2i} are all simple modules. Therefore each layer of this filtration is semisimple. Moreover, the composition factors V_{J} of X̃ each occur with multiplicity one and have the property that J has no consecutive indices of the form “odd-even”. However, there are some in which consecutive indices of the form “even-odd” occur in J.

Lemma 7. X̃ has a submodule Z such that the composition factors of Z are precisely those composition factors V_{J} of X̃ for which J contains consecutive indices. In particular, the composition factors of X = X̃/Z are precisely the V_{J} with J in the collection N of subsets not having consecutive indices.

Proof: Suppose V_{J} is a composition of X̃ such that J contains consecutive indices r, r + 1. Note that r must be even and n > 1. Since V_{J} occurs with composition multiplicity one, there is a unique submodule A of X with head(A) ≅ V_{J}; A can be characterized as the smallest submodule having V_{J} as a composition factor. The Lemma will follow if we can prove that for every composition factor V_{K} of A, the set K also contains r and r+1. By induction, it is sufficient to consider the composition factors V_{K} of rad(A)/ rad^{2}(A). Suppose that V_{K} is such a factor. Then Ext_{kG}^{1}(V_{J}, V_{K}) ≠ 0. Thus J and K must satisfy the condition for this to hold given by the main theorem in [11], which states:

The space Ext_{kG}^{1}(V_{J}, V_{K}) is zero unless the symmetric difference of J and K is a singleton {i} with i − 1 ∉ J ∩ K, in which case it is one-dimensional.

If J ⊆ K, or i ≠ r, r + 1, we are done. Clearly i cannot be r + 1. Finally, suppose i = r. So J = K ∪{r}. But since r is even, V_{r} occurs at level 2 in the tensor factor E_{2(r−1)} of X̃, while k is at level 1. So the composition factor V_{J} appears in the layer of the tensor filtration on X̃ below that in which V_{K} appears. This contradicts the fact that V_{K} is a composition factor of A and the lemma is proved.

We shall next prove that X = X̃/Z has the structure claimed for \overline{C} in Theorem 2. Up to isomorphism there is a unique nonsplit extension D_{i} of V_{i} by the trivial module k (by [11]). The key fact is:

Lemma 8. If neither i nor i − 1 belongs to J ∈ N then D_{i} ⊗ V_{J} is a nonsplit extension of V_{J∪{i}} by V_{J}.

Proof: This follows from the fact that D_{i} is isomorphic to a submodule of V_{i−1} ⊗ V_{i−1}, so that D_{i} ⊗ V_{J} embeds in V_{i−1} ⊗ V_{J∪{i−1}}. Then this is a tensor factor of V_{i−1} ⊗ V_{/2n} , which by [11, Lemma 1] is a projective indecomposable module and so has a simple socle. Therefore D_{i} ⊗ V_{J} has simple socle and the lemma is proved.

Lemma 9. The filtration on X induced by the tensor filtration on X̃ is equal to the radical filtration, and the reversed (ascending) filtration is the socle filtration.

Proof: Let K ∈ N. Suppose that there is an odd index r in K. Then Lemma 8 applies with J = K \{r}, showing that D_{r} ⊗V_{K\{r}} is indecomposable of length 2. If r = 2j +1, then D_{r} is isomorphic to a quotient of E_{2j}. Let X̃_{j} be the tensor product of the other factors, so that X̃ = E_{2j} ⊗ X̃_{j}. Then V_{K\{r}} is a composition factor of X̃_{j}, since r+1 ∉ K. Therefore, D_{r} ⊗V_{K\{r}} is isomorphic to a subquotient of X̃. This shows that the unique composition factor V_{K} of X̃ occurs in a subquotient which is a nonsplit extension of V_{K} by V_{K\{r}}. Note that in the tensor filtration of X̃, V_{K} appears exactly one layer above V_{K\{r}}. Next suppose there is an even index s = 2j so that K ∪{s} is still in N. Then the dual module D_{r}^{∗} is isomorphic to a submodule of E_{2(j−1)} and a similar argument to the above shows that V_{K} occurs in X̃ in a subquotient which is a nonsplit extension of V_{K} by V_{K∪{r}}. Again, we note that V_{K∪{r}} is exactly one layer below V_{K} in the tensor filtration of X̃. None of what we have said is affected by passing to X. Thus we have shown that if V_{K} is any composition factor of X then all composition factors of X which can be obtained either by deleting an odd index or by adding an even index appear strictly lower in the radical filtration of X and exactly one layer lower in the filtration induced from the tensor fitration of X̃. It is easy to see that starting with any K ∈ N, we can reach the set ( /2n )_{even} by applying a suitable sequence of these two operations on indices.

Then since the induced filtration has semisimple layers, it follows that the two filtrations are equal. The dual argument shows that the ascending induced filtration is the socle filtration.

Proof of Theorem 2: It remains to show X ≅ \overline{C}. The hypotheses (a) and (b) of Lemma 6 are immediate consequences of Lemma 9. We now check condition (c). If we consider E as a kG_{α}-module, then by parts (a) and (b) of Lemma 3, E has a filtration with subquotients (in descending order) W ⊗[−1], W ⊗[1], k, [−2], W_{1}⊗[0], [2], where the last three factors come from a G_{α}-filtration of S = soc_{G}(E) ≅ V_{2}. Then by Lemma 3 (d), the nonzero kG_{α} homomorphism from S to [−2] extends to E. Therefore the induced homomorphism from ⨂_{i=0}^{n−1} S_{2i} ≅ V_{even} to [∑_{i=0}^{n−1} 2^{i}(−2)] = [0] extends to X̃ = ⨂_{i=0}^{n−1} E_{2i}. Since the submodule ⨂_{i=0}^{n−1} S_{2i} of X̃ maps isomorphically to its image in X, this proves Hom_{kG_{α}}(X, k) ≠ 0. So by Lemma 6, X ≅ \overline{C}.

Remarks.

(1) A closer look at the proof of Lemma 9 shows that the same argument, using lemma 8 will yield the entire submodule structure of \overline{C}. From J ∈ N we may obtain other elements of N by deleting an odd index or by adding an even one so that the resulting set still belongs to N. If K ∈ N is obtained from J by a sequence of these two types of operation, then we will write K ≺ J. This is clearly a partial order with unique maximal element ( /2n )_{odd} and unique minimal element ( /2n )_{even}. Consider the collection of those subsets X of N with the property that if J ∈ X and K ≺ J, then K ∈ X. This collection is partially ordered by inclusion. The argument of Lemma 9 shows that this partially ordered set is lattice-isomorphic to the collection of kG-submodules of \overline{C}.

(2) We can also obtain the submodule structure of the _{2}-code C_{_{2}} , identified with the set of fixed points in C = C_{_{2}} ⊗_{_{2}} k of the map σ : x⊗λ ↦ x⊗λ^{2^{−1}}. The _{2}G-submodules of C_{_{2}} correspond bijectively with the σ-invariant kGsubmodules of C, from which they may be recovered by taking σ-fixed points. Now the image under σ of a kG-submodule is isomorphic to its Frobenius twist, so the remark following Theorem 2 makes finding the σ-invariant kGsubmodules of C especially easy.

§2. Some geometric subcodes

Ovoids.

An ovoid in the 3-dimensional projective space (V (q)) ≅ P G(3, q) is a set O of q^{2} +1 points, of which no three are collinear. An ovoid in the symplectic geometry W(q) = (P, L) is a set O of q^{2} + 1 points, such that no two lie on a common (isotropic) line. By a well-known theorem of Segre [9, Theorem III, p.321] (see also [2, Prop. 1 and Remarks, p.138]), the classification of the ovoids of P G(3, q) and those of W(q) are equivalent problems.

The only known ovoids in W(q) are the (nondegenerate) elliptic quadrics and the Tits ovoids [14, n^{∘}.5]. An elliptic quadric is the set of one-dimension-al isotropic subspaces in V (q) for a quadratic form f of Witt index 1 which has the given symplectic form < −, − > as its associated bilinear form. The stabilizer in G of an elliptic quadric is conjugate to a subgroup Ω which has a subgroup of index 2 isomorphic to SL(2, q^{2}) ≅ SO(V (q), f). A Tits ovoid is the set of self-conjugate points of a polarity of W(q) which exists only when q is an odd power of 2. When q > 2, its stabilizer in G is conjugate to the Suzuki group Sz = Sp(4, k)^{τ^{n}}.

The codes Q and T .

Given a subset X of P, we can consider the sum of its elements as a vector [X] in k^{P}, and thereby obtain a bijection from the set of all subsets of P to the _{2}-span of P in k^{P}. In [2, Propositions 2 and 3] it was shown that the elements [O], where O is either an elliptic quadric or a Tits ovoid, belong to the code C. Let Q denote the subspace of C generated by the vectors [O] for elliptic quadrics, and let T denote the corresponding space for Tits ovoids. In this section we investigate these subcodes, which are clearly kG-submodules.

Up to isomorphism there is a unique nonsplit extension M_{i} of the trivial module by V_{i+2}. ( The dual of this module was denoted D_{i+2} earlier.) Thus, M is isomorphic to a submodule of the module E defined earlier in the construction of X. Thus in X̃ = ⨂_{i=1}^{n} E_{2i} there is a submodule Q̃ = ⨂_{i=1}^{n} M_{2i} of dimension 5^{n}. We also have [11, Lemma 2(b)]

(7)  M ≅ ∧^{2}(V_{1})/(∧^{2}(V_{1}))^{G}.

We will describe the structure of Q̃ and then show that it is isomorphic to Q as a kG-module. More precisely, we shall show that in the diagram

(8)  X̃ ⟶ X ≅ \overline{C} ; ↑ ; C

the modules Q̃ and Q map isomorphically to the same submodule of \overline{C}.

Lemma 10.

(a) The horizontal map in (8) restricts to an isomorphism of Q̃ with its image.

(b) The tensor filtration on X̃ induces the radical filtration (and socle filtration) of Q̃. Thus, Q̃ has radical length n + 1 with

rad^{j}(Q)/ rad^{j+1}(Q) ≅ ⨁_{\begin{matrix}I∈( /2n ) _{even} \\ |I|=j \end{matrix}} V_{I}  (0 ≤ j ≤ n).

Proof: Part (a) is clear since Q̃ has no composition factors in common with the kernel of the map. Part (b) is proved using Lemma 8, exactly as in the proof of Lemma 9.

In view of Lemma 10(a) we identify Q̃ with its image in \overline{C} and use the same symbol for both. Let \overline{Q} denote the image of Q in \overline{C}.

Lemma 11. Q̃ = \overline{Q}.

Proof: Let T be a cyclic subgroup of order q^{2} + 1 of the stabilizer Ω in G of an elliptic quadric O. First we have that \overline{C}^{T} is one-dimensional, either by direct calculation, knowing the composition factors by Theorem 2 or (more geometrically) by [2, Lemma 4]. Second, from Lemma 10 (b) we have head(Q̃) ≅ k. Since T acts semisimply, these two observations imply that T fixes some nonzero vector in Q̃, but none in rad Q̃. Thus, Q̃ is the kG-submodule of \overline{C} generated by the one-dimensional space \overline{C}^{T}. But it is clear that the image of [O] in \overline{C} is a nonzero vector fixed by T. Therefore, Q̃ is generated by the image of [O], hence equal to \overline{Q}.

Lemma 12. The vertical map in (8) restricts to an isomorphism of Q to \overline{Q}.

Proof: This amounts to checking that 1 does not lie in Q. Supose it did. Then we would have Q ≅ k ⊕ \overline{Q} ≅ k ⊕ Q̃, using Lemma 11 for the last isomorphism. Then since head(Q̃) ≅ k, by Lemma 10(b), it would follow that dim_{k} Hom_{kG}(Q, k) = 2. On the other hand Q is a quotient of the transitive permutation module ind_{Ω}^{G}(k) with basis consisiting of the set of elliptic quadrics, and by Frobenius reciprocity, dim_{k} Hom_{kG}(ind_{Ω}^{G}(k), k) = 1, so we have a contradiction.

We summarize our results on Q.

Theorem 13. Q is a 5^{n}-dimensional subcode of C isomorphic as a kGmodule to

⨂_{i=0}^{n−1}[∧^{2}(V_{1})/(∧^{2}(V_{1}))^{G}]_{2i}.

It has radical length n + 1 and the layers of its radical series are those stated for Q̃ in Lemma 10.

Remark. If we fix an isomorphism (k^{P})_{1} ≅ k^{L} and identify submodules of (k^{P})_{1} with their images, then the subcode Q_{1} of k^{L} can be described in terms of the incidences of points and lines in W(q) and the ambient space (V (q)). Let L̂ denote the set of all lines in (V (q)) and k_{L} the vector space having this set as basis. The same method used to prove Lemma 4 (see also [7], 6.6, 6.8) yields:

(9)  k_{L} ≅ ⟨1⟩ ⊕ Y_{L},  soc_{G}(Y_{L}) ≅ head_{G}(Y_{L}) ≅ ⨂_{i=0}^{n−1}[∧^{2}(V)]^{(i)},

where the superscript (i) indicates twisting by F^{i} and 1 is the sum of all lines.

Consider the following diagram:

Here the maps α̂ and α^{∗} are respectively the line-point and point-line incidence maps of (V (q)). These maps commute with the action of Ĝ = SL(V (q)). The map α^{∗} is the point-line incidence map in W(q) and π is the projection mapping isotropic lines to themselves and nonisotropic lines to 0. These are kG-maps. It is immediate from the definitions that the diagram commutes. We consider first the map α^{∗} ∘ α̂. Since Ĝ acts as a rank 3 permutation group on L̂, we have dim_{k} End_{k_{G}}(k_{L}) = 3, and from (9) one sees that the identity and the maps onto the two simple submodules form a basis for this space. From this it is not hard to see that the image of α^{∗} ∘ α̂ must be soc_{G}(Y_{L}). Since the diagram commutes, this shows that I := Im(α^{∗} ∘ α̂) = π(soc_{G}(Y_{L})). We claim

(10)  I = ⟨1⟩ ⊕ Q_{1}.

It is easy to see that ⟨1⟩ I. The image \overline{I} of I in \overline{C} is therefore a kGquotient of soc_{G}(Y_{L}) with simple socle equal to V_{even}. Since V_{even} occurs with composition multiplicity 1 in ⨂_{i=0}^{n−1}[∧^{2}(V)]^{(i)}, there is a unique such quotient, and Lemma 10 shows that it must be isomorphic to Q̃_{1}. Thus in particular head(\overline{I}) ≅ k, and we can conclude by the reasoning of Lemma 11 that \overline{I} is the submodule of \overline{C}_{1} generated by (\overline{C}_{1})^{τ(T)} = (\overline{C}^{T})_{1}, namely, \overline{Q}_{1}. Now (10) follows.

Assume now that n > 1 is odd. Let \overline{T} denote the image of T in \overline{C}.

Lemma 14. Let S be a Tits ovoid and let Ω be the stabilizer of an elliptic quadric O. Then there exists a cyclic subgroup T of Ω of order q^{2} + 1 for which

∑_{t∈T} t[S] = [O].

Proof: We may assume that S is the Tits ovoid stabilized by Sz. Denote the sum in the statement by f_{T} (S). Now Sz ∩ Ω contains a cyclic subgroup T′ of order q ± \sqrt{2q} + 1 and index 4 [4, Theorem 5(a)]^{1}

The centralizer in G of this subgroup is a cyclic subgroup of Ω of order q^{2} + 1. We take this to be our T. Let \{S_{j}\}_{j=1}^{q∓\sqrt{2q}+1} be the set of distinct images of S under T. Each S_{j} can be further decomposed into T′-orbits S′_{jl} (1 ≤ l ≤ q ∓ \sqrt{2q} + 1). Then, since k has characteristic 2 we have

1On p.147 of this paper appears the erroneous statement that there is a single conjugacy class of subgroups isomorphic to SL(2, q^{2}) in Sp(4, q). There are in fact two classes, interchanged by the automorphism τ. The results which we use from this paper remain true however, so long as they are understood as referring to that conjugacy class whose members stabilize elliptic quadrics.

f_{T} (S) = ∑_{j=1}^{q±\sqrt{2q}+1} ∑_{l=1}^{q∓\sqrt{2q}+1}[S′_{jl}].

We claim that for each j precisely one of the S′_{jl} is disjoint from all the other ovoids S_{j′}, while each of the other S′_{jl} is equal to precisely one S′_{j′l} for j′ ≠ j. If j′ ≠ j then S_{j′} ∩S_{j} is a T′-orbit (cf. [4, proof of Theorem 1(b), p.156-157]^{1}) and if j, j′ and j′′ are distinct, then the T′-orbits S_{j′} ∩ S_{j} and S_{j′′} ∩ S_{j} are distinct, since T acts semiregularly on (V (q)). This proves the claim. It follows the weight of f_{T} (S) is equal to (q + \sqrt{2q} + 1)(q − \sqrt{2q} + 1) = q^{2} + 1. In view of [2, Lemma 4] the only possibilty for f_{T} (S) is [O].

Theorem 15.

(a) Q T .

(b) \overline{T} ⊆ rad^{n}(\overline{C}) but \overline{T} rad^{n+1}(\overline{C}). Consequently, the radical length of \overline{T} is n + 1.

(c) T maps isomorphically to \overline{T} .

Proof: (a) Lemma 14 gives the inclusion. To prove that this inclusion is proper, it is enough to show that Sz does not fix any nonzero element of Q ≅ Q̃. Now each composition factor of Q̃ remains simple upon restriction to Sz, so [Q̃ : k]_{Sz} = 1. Now Q̃ has a kG quotient isomorphic to the nonsplit extension M of the trivial module by V_{2}. This extension does not split for Sz (e.g. using [11, Lemma 2]; here we use n > 1). We may therefore conclude that Q̃^{Sz} = 0. This establishes (a).

(b) We consider the possibilities for simple homomorphic images of \overline{T} . Now \overline{T} is a homomorphic image of the permutation module ind_{Sz}^{G}(k) and if V_{J} is a simple quotient of the latter then

(11) 0 ≠ Hom_{kG}(ind_{Sz}^{G}(k), V_{J}) ≅ Hom_{kSz}(k, V_{J}) ≅ Hom_{kSz}(V_{J_{even}}, V_{J_{odd}}).

Now V ≅ V_{n} for Sz, so both V_{J_{even}} and V_{J_{odd}} ≅ V_{J_{odd}+n} are simple kSzmodules, by the Tensor Product Theorem [12,§11]. Therefore, in order for (11) to hold, we must have h_{J} = |J_{even}| − |J_{odd}| = 0. The only layer of \overline{C} in which such V_{J} appear is rad^{n}(\overline{C})/ rad^{n+1}(\overline{C}). It follows that any nonzero image in \overline{C} of ind_{Sz}^{G}(k) is contained in rad^{n}(\overline{C}) but not in rad^{n+1}(\overline{C}). In particular this is true of \overline{T} . The radical length then follows from this and the fact that \overline{T} contains \overline{Q}.

(c) We must show that 1 ∉ T . We can argue exactly as in the proof of Lemma 12 (with T in place of Q and Sz in place of Ω) as long as we show Hom_{kG}(\overline{T} , k) ≠ 0. To see this, observe that by (a) and (b) \overline{Q} is contained in \overline{T} and has nonzero image in head(\overline{T} ), and that head(\overline{Q}) ≅ k.

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Math.-Stat. Division, Indian Statistical Institute, 8th mile, Mysore Road, R.V. College Post, Bangalore-560059, India e-mail: nsastry@isibang.ernet.in Department of Mathematics, University of Florida, Gainesville, FL 32611, USA e-mail: sin@math.ufl.edu