On the Universal Embedding of the Sp2n(2) Dual Polar Space

Journal of Combinatorial Theory, Series A 94, 100117 (2001) doi:10.1006jcta.2000.3132, available online at http:www.idealibrary.com on

On the Universal Embedding of the Sp 2n(2) Dual Polar Space Paul Li Department of Mathematics, University of Chicago, 5734 South University Avenue, Chicago, Illinois 60637 E-mail: paullimath.uchicago.edu Communicated by the Managing Editors Received May 22, 2000

A. E. Brouwer has shown that the universal embedding of the Sp 2n(2) dual polar space has dimension at least (2 n +1)(2 n&1 +1)3 and has conjectured equality. The present paper settles this conjecture in the affirmative by proving a theorem about  2001 permutation modules for GL n(2) which implies the reverse inequality. Academic Press

Key Words: universal embedding; universal abelian representation; dual polar space.

1. INTRODUCTION Let G=(P, L) be a point-line geometry with exactly three points on each line. An embedding of G is a mapping % from P to the nonzero elements of a vector space E over the field F 2 of order 2, such that (a) E=( P % ) and (b) for every line [ p, q, r] # L, the vectors p %, q %, r % form a projective line in E, i.e., p % +q % +r % =0. The universal embedding U(G) can be defined as the quotient of the vector space of all formal F 2 -linear combinations of P by all relations of the form p+q+r as [ p, q, r] ranges over L. There is a natural sense in which all embeddings of G are quotients of U(G). The Sp 2n(2) dual polar space is the point-line geometry Gn constructed from a 2n-dimensional nondegenerate symplectic space over F 2 in the following way. The points are the maximal isotropic subspaces while the lines are the isotropic subspaces of dimension n&1, and incidence is given by inclusion. Note that each line in Gn is incident to exactly three points, so we may study its embeddings as defined above. One motivation for studying the embeddings of Gn is provided by Ivanov's work on classifying affine extended dual polar spaces [4]. 100 0097-316501 35.00 Copyright  2001 by Academic Press All rights of reproduction in any form reserved.

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A. E. Brouwer has shown that dim U(Gn )g(n)=(2 n +1)(2 n&1 +1)3 by producing an embedding of Gn of dimension exactly g(n). This follows from a short eigenvalue argument, outlined in [4]. Brouwer conjectured that his embedding is in fact isomorphic to U(Gn ), or equivalently, that dim U(Gn )= g(n). Note that, in order to prove his conjecture, it suffices to show that U(Gn ) is spanned by g(n) vectors. Yoshiara [6] proved the case n=3 as part of a more general investigation on the embeddings of rank 3 geometries. The cases n=4, 5 were settled by Cooperstein [3], using a ``geometric spanning'' argument. Computer calculations by Bardoe and Ivanov [1] covered the cases up to n=7. This paper presents a proof of the general case of Brouwer's conjecture. First we outline the BardoeIvanov reduction [1] of the conjecture to an assertion about the ranks of certain maps between permutation modules for GL n(2). The latter assertion is then proved in this paper. McClurg's ordering technique and ``condition N '' [5] are crucial to the argument. As a corollary of our result, we obtain an interesting isomorphism between certain GL n(2) modules which arise from studying the structure of the F 2 -permutation modules of GL n(2) on the subspaces of F n2 . Note. Several months after the author obtained his solution, Blokhuis and Brouwer [2] announced a different proof of the same result. Their arguments are completely independent of those of Bardoe and Ivanov and the author.

2. REDUCTION TO PERMUTATION MODULES In this section, let n be a fixed integer with n3. Let %: Pn Ä U(Gn ) be the universal embedding of the Sp 2n(2) dual polar space Gn =(Pn , Ln ). Let 1 be the collinearity graph of Gn . Thus the vertices of 1 are the points of Gn and two distinct vertices are adjacent if and only if they are collinear. Fix a vertex x 0 # 1, and let 1 k (0kn) denote the set of vertices at distance k from x 0 . Then y # 1 k if and only if dim( y & x 0 )=n&k (as a subspace of the symplectic space defining Gn ). The geometry Gn has the following special (``near n-gon'') property: every line of Gn contains two elements from 1 k and one from 1 k&1 , for some 1kn. Therefore we have a filtration 0( 1 %0 )  } } } ( 1 %n ) =E. It turns out that for two points p, q # 1 k , p and q lie in the same connected component of 1 k (viewed as an induced subgraph of 1) if and only if p & x 0 =q & x 0 . Thus the connected components of 1 k are in one-to-one correspondence with the n&k-subspaces of x 0 . The significance of the

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connected components of 1 k is that any two points in the same component are mapped to the same vector in the quotient (1 %k )(1 %k&1 ) in the filtration above. (This is because if p and q are distinct, collinear points in 1 k , then the third point on the line joining them must be in 1 k&1 .) In the case n=3, there exists a 7-cycle p 0 , ..., p 6 in 1 3 with the following property: if q i (i=0, ..., 6) denotes the third point on the line through p i and p i+1 (indices taken modulo 7), then the q i lie in the seven different connected components of 1 2 . Since  i q %i = i ( p %i + p %i+1 )=0, the images of the seven connected components of 1 2 in ( 1 %2 )( 1 %1 ) sum to zero. A similar situation persists for general n3. If 2kn&1 and L
dim E1+ : dim P k im . k +(2 n &1)+1. k=1

In Sections 4 and 5, we shall prove the bound 1+ n&2 k=1 dim P k im . k + (2 n &1)+1(2 n +1)(2 n&1 +1)3. This will give us the required upper bound for dim U(Gn ), and it will also follow that the natural surjections P n&k im . n&k Ä ( 1 %k )( 1 %k&1 ) are actually isomorphisms.

3. LEXICOGRAPHIC ORDER For this section, let us fix an integer n1. We write x 1 , ..., x n for the standard basis of F n2 . For any vector v=a 1 x 1 + } } } +a n x n # F n2 , define its

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support supp(v)=[i : a i {0] and its weight wt(v)= |supp(v)|; for any subspace LF n2 , define supp(L)= v # L supp(v). For a nonzero vector v # F n2 , set :(v)=min supp(v) and ;(v)=max supp(v). For LF n2 , set :(L)=[:(v) : v # L, v{0]. Then it is obvious that L$L implies :(L$) :(L). One easy consequence is that :(L) & :(L$)=< implies L & L$=[0]. It is an easy fact from linear algebra (``Gaussian elimination'') that dim(L)= |:(L)|. We define a total ordering on the vectors of F n2 as follows: a 1 x 1 + } } } + a n x n o b 1 x 1 + } } } +b n x n if there is some i # [1, ..., n] such that a j =b j for all ji. This gives the lexicographic order on the k-subspaces of F n2 . (Note the somewhat nonstandard use of the term ``lexicographic'': to compare two subspaces L, L$ of F n2 , we compare their reduced echelon bases starting with their last vectors and going backwards.)

Lemma 3.1. Let X, X $, and L 0 be subspaces of F n2 with dim X=dim X $ =k, X o X$, and supp(X ) & :(L 0 )=supp(X $) & :(L 0 )=<. Then X+L 0 and X$+L 0 have the same dimension, and X+L 0 oX $+L 0 . Proof. The sums X+L 0 and X$+L 0 are direct sums, so they have the same dimension. Let v 1 o } } } o v k and v$1 o } } } o v$k be the respective reduced echelon bases of X and X $. Suppose that v k =v$k . Set X =(v 1 , ..., v k&1 ), X $=( v$1 , ..., v$k&1 ), and L 0 =( v k ) +L 0 , so that X+L 0 =X +L 0 and X $+L 0 =X $+L 0 . We have dim X =dim X $ and X o X $ (the latter because v 1 o } } } ov k&1 and v$1 o } } } ov$k&1 are the reduced echelon bases of X and X$). Moreover, :(L 0 )= [:(v k )] _ :(L 0 ) is disjoint from supp(X ) and similarly :(L 0 ) & supp (X $)=<. Therefore if v k =v$k we can continue our argument with X, X $ and L 0 (i.e., use induction on dim X ) to deduce that X+L 0 =X +L 0 o X $+L 0 =X$+L 0 . Therefore from now on we assume that v k o v$k . If L 0 =[0] there is nothing to prove, so we assume otherwise. Let w 1 o } } } ow l be the reduced echelon basis of L 0 .

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Suppose that v$k is smaller than all the w i . (Since :(X $) & :(L 0 )=<, this is the same as supposing that :(v$k )>:(w l ).) Then v$k is the smallest nonzero vector in X $+L 0 (i.e., the last vector in its reduced echelon basis). Now either v k is smaller than all the w i , in which case v k is the smallest nonzero vector in X+L 0 , or if not then w l will be the smallest nonzero vector in X+L 0 . In either case, the smallest nonzero vector in X+L 0 will be greater than that in X $+L 0 , and we have X+L 0 o X$+L 0 . Now suppose w a$ is the largest possible such that v$k o w a$ . Then the reduced echelon basis of X $+L 0 ends with v$k , w a$ , w a$+1 , ... . Now define the index a analogously for v k in place of v$k . Clearly aa$. If a=a$ then the reduced echelon basis of X+L 0 ends with v k , w a$ , w a$+1 , ..., while if a
4. LARGEST ELEMENTS IN SYMMETRIC DIFFERENCES Let n be a fixed positive integer in this section. If X
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precisely equal to the cardinality T. The goal of this section is to identify an explicit subcollection of T (Theorem 4.1). This will give us an (implicit) lower bound for the rank of . k . Let N n denote the collection of all subspaces of F n2 whose reduced echelon basis v 1 o } } } ov k (where k=dim L) satisfies all of the following four conditions: (N1)

wt(v a )2 for every a # [1, ..., k].

(N2)

If v a ov b (i.e. a
(N3) If v a ov b o v c , wt(v a )=wt(v b )=wt(v c )=2, and ;(v a )=;(v b ) <;(v c ), then :(v c )>;(v a ). (N4) There do not exist v a o v b o v c o v d such that wt(v a )= wt(v b )=wt(v c )=wt(v d )=2 and ;(v a )=;(v b )=;(v c )<;(v d ). Furthermore, let N nk denote those subspaces in N n of dimension k. This definition first appeared in McClurg's paper [5]. (We shall describe his work in a remark at the end of this section.) For example, the following subspaces satisfy conditions (N1)(N4):

[0], 10001 10010 1000010 01000, 01001, 0100001 00011 00101 0010001 0001000 0000101

The following subspaces violate conditions (N1)(N4) (in that order):

10110, 10001, 1000000, 01001 01100 0100010 00011 0010010 0001000 0000101

100001000. 010000100 001000000 000100100 000010100 000000011

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Here each array of zeros and ones represents the reduced echelon basis of a subspace of F n2 (for the appropriate value of n). A row abcd } } } represents the vector ax 1 +bx 2 +cx 3 +dx 4 + } } } . Each of the four conditions (N1)(N4) stipulates that a subspace L of F n2 does not contain a certain ``forbidden configuration'' of 0's and 1's in its reduced echelon basis. These configurations are highlighted in bold in the examples above. We now proceed in Lemmas 4.14.4 to show that if a subspace L of F n2 possesses one of these forbidden configurations, then L arises as the lexicographically largest element in the symmetric difference of some collection of sets of the type S(X, Y ). Note that the proofs of these lemmas follow the same basic outline. Lemma 4.1. Let L be a subspace of F n2 , with reduced echelon basis v 1 o } } } o v k . Suppose that some v i has weight 3. Then there exist subspaces X;(v b ) hold. Then there exist a (k+2)-subspace Y of F n2 and (k&1)-subspaces X 1 , X 2 of Y, such that L is (lexicographically) the largest member of the symmetric difference of S(X 1 , Y ) and S(X 2 , Y ). Proof. Let p=:(v a ), q=:(v b ), r=;(v b ), s=;(v a ), so that p
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argument is provided in the Appendix to this paper.) So if we set X i =X i$+ L 0 and Y=R+L 0 , it follows from Lemma 3.2 that ( v a , v b ) +L 0 =L is the largest element of the symmetric difference of S(X 1 , Y ) and S(X 2 , Y ). K Lemma 4.3. Let L be a subspace of F n2 , with reduced echelon basis v 1 o } } } o v k . Suppose, for some v a o v b ov c , that wt(v a )=wt(v b )= wt(v c )=2 and :(v c )<;(v a )=;(v b )<;(v c ) hold. Then there exist a (k+2)-subspace Y of F n2 and (k&1)-subspaces X 1 , ..., X 4 of Y, such that L is the largest member of the symmetric difference of the S(X i , Y ). Proof. Let p=:(v a ), q=:(v b ), r=:(v c ), s=;(v a )=;(v b ), and t=;(v c ), so that p
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Lemma 3.2 that L$+L 0 =L is the largest element of the symmetric difference of S(X 1 , Y ), ..., S(X 21 , Y ). K Theorem 4.1. Let L be a k-subspace of F n2 , with 1kn&2. If L Â N n, then there exist a finite number of flags X i
.

t

k k P n&1&k, n&1 wÄ P k wÄ P k&1, k+2

(-)

of F 2 GL n(F 2 )-module homomorphisms is zero, where the maps c k and . tk are defined by c k(e X
()

Since Brouwer's result (stated in the Introduction) gives a lower bound for  k dim P k im . k , inequality () can be thought of as a refinement of Brouwer's bound. (This conjectured refinement lead to the author's

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discovery of Theorem 4.1.) The results of this paper together with Brouwer's result imply that dim P k im . k =nullity . tk = |N nk |. So if McClurg's construction works, the sequence (-) would be exact.

5. COUNTING THE ELEMENTS OF N The goal of this section is to prove the following: Theorem 5.1. elements.

For every n1, N n has exactly (2 n +1)(2 n&1 +1)3

The combination of Corollary 4.1 with Theorem 5.1 implies that the the universal embedding of the Sp 2n(2) dual polar space has at most the conjectured dimension. Proof of Theorem 5.1. The proof proceeds by induction on n. Let g(n)=(2 n +1)(2 n&1 +1)3. It is easy to verify the theorem directly in dimensions 1 and 2. So we assume n3. It will be convenient to abuse notation slightly by regarding linear combinations of x 1 , ..., x n&1 as and F n2 . elements of both F n&1 2 Let L be an element of N n. Write k=dim L, and let v 1 o } } } o v k be the reduced echelon basis of L. We find that L falls into exactly one of the following seven cases. Case 1. n  supp(L). It is easy to see that such subspaces L can be identified with the elements of N n&1. In terms of reduced echelon bases, this identification is given by ``deleting the rightmost column'': 100100 100100 10010 10010 wÄ 10010 = 1001 1010 1010 101 Therefore, by induction, the number of such subspaces is g(n&1). Case 2. x n # L. Clearly v k =x n , and v 1 , ..., v k&1 can be regarded as the reduced echelon basis for a subspace L$ # N n&1. The rule which sends L to L$ gives a one-to-one correspondence between the subspaces L # N n containing x n and the members of N n&1. For example:

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100100 100100 10010 10010 10010 1001 wÄ = 1000 1000 100 1 1 So again the count is g(n&1), by induction. Now for use in the remaining cases, let S n&1 denote the set of all subspaces L$ # N n&1 which do not fall into Cases 1 or 2. By induction, we have |S n&1 | = g(n&1)&2g(n&2). Case 3. There is some v i of weight 2 such that ;(v i )=n, and n&1 Â be the linear map sending x s [ x s for 1s supp(L). Let +: F n2 Ä F n&1 2 n&2, x n&1 [ 0, and x n [ x n&1 . Then +(v 1 ), ..., +(v k ) is the reduced . In fact, this rule L [ L$= echelon basis for some subspace L$ of F n&1 2 +(L) gives a bijection between the elements L # N n which fall into the present case and the elements of S n&1. For example: 10001 10001 1001 1001 wÄ 1001 = 101 100 100 10 (Informally, the rule is given in terms of reduced echelon bases by ``deleting the (n&1)st column.'') Hence the number of subspaces L # N n which fall into the present case equals g(n&1)&2g(n&2). Case 4. There is some v i of weight 2 such that ;(v i )=n, and x n&1 # L. Here we have v k =x n&1 . Let + be as in Case 3. Then +(v 1 ), ..., +(v k&1 ) is . In fact, this rule the reduced echelon basis for some subspace L$ of F n&1 2 L [ L$ gives a bijection between the elements L # N n which fall into the present case and the elements of S n&1. For example, 11000 11000 1100 101 wÄ 101 = 11 10 10 (Informally, the rule is given in terms of reduced echelon bases by ``deleting the last row and the (n&1)st column.'') Hence, as in the previous case, the count is g(n&1)&2g(n&2).

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Now if L # N n does not fall into Cases 14, then either v k =x n&1 +x n (Cases 5 and 7 below), or ;(v i )=n and ;(v j )=n&1 for some v i and v j of weight 2 (Case 6 below). Let us partition S n&1 into a disjoint union S n&1 1 _ S n&1 as follows. If L$ # S n&1 has reduced echelon basis v$1 o } } } o v$l , 2 if and only if ;(v$s )=n&1 for exactly one v$s , and then L$ # S n&1 1 if and only if ;(v$s )=n&1 for two or more v$s . L$ # S n&1 2 Case 5. v k =x n&1 +x n , and there is at least one v j other than v k such be the same map as in that wt(v j )=2 and ;(v j )=n. Let +: F n2 Ä F n&1 2 Case 3. It is easy to see that +(v 1 ), ..., +(v k&1 ) is the reduced echelon basis for some subspace L$ # N n&1. The rule sending L to this subspace L$ gives a bijection from the set of all L # N n falling into the present case onto S n&1. Informally, this rule can be described in terms of reduced echelon bases as ``deleting the bottom row and the (n&1)st column.'' For instance,

1000100 1000100 100010 100001 100001 10001 10001 wÄ 10001 = 1001 1000 1000 100 11 11

Again the count equals g(n&1)&2g(n&2). Case 6. There are v i and v j of weight 2 such that ;(v i )=n and ;(v j )=n&1. Note that v j is unique, as L satisfies condition (N3). Let be the linear map which sends x s [ x s for 1sn&1 and &: F n2 Ä F n&1 2 x n [ x n&1 . Then it is easy to see that &(v 1 ), ..., &(v k ) is the reduced echelon . Here is an example of the mapping basis for some subspace L$ of F n&1 2 L [ L$:

1000010 1000010 100001 100001 100010 10001 10000 wÄ 10000 = 1000 1001 1010 101 101 110 11

Thus, informally, the rule L [ L$ is given by ``sliding the 1's in the n th column (shown in bold in the above example) one place to the left and

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then deleting the n th column.'' Since the reduced echelon basis of L does not possess any of the four types of configurations forbidden by conditions (N1)(N4), it is not difficult to see that the process of ``sliding and deleting'' we have just described will not create any of those forbidden configura. tions. In other words, L$ # N n&1. It is then clear that in fact L$ # S n&1 2 The inverse to the rule L [ L$ may be informally described as follows:

100001 1000010 10001 100001 1000 wÄ 10000 101 1001 11 101

All the 1's in the (n&1)st (i.e., rightmost) column of the reduced echelon basis of L$, except the very top one, are pushed one place to the right, and then the necessarily zeros are added. (The 1's to be moved are shown in , there is at least bold in the above example. Since L$ is taken from S n&1 2 one such 1.) This shows that there is a bijection between those L # N n which fall into . Case 6 and the elements of S n&1 2 Case 7. v k =x n&1 +x n , and supp(v j ) & [n&1, n]=< for all j{k. The one-dimensional subspace ( x n&1 +x n ) falls into this case. Our goal now is to describe a bijection between those L # N n other than ( x n&1 +x n ) . which fall into the present case and the elements of S n&1 1 Let a denote the largest index in [1, ..., n&2] which lies in supp(L). There are three possibilities: (a) x a # L. In this case we must have v k&1 =x a . It is easy to see that the sequence of (k&1) vectors v 1 , ..., v k&2 , x a +x n&1 (call them w 1 , ..., w k&1 ) can be regarded as the reduced echelon basis of some sub. Our bijection sends L to this subspace L$. For example, space L$ # S n&1 1

1010000 101000 110000 wÄ 11000 1000 101 Â x a +x n&1 v k&1 Ä 11 A a

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Notice that L$ must have the following property: if we define a$ to be the largest index in [1, ..., n&2] & supp(L$), then in fact a$=a and w k&1 = x a$ +x n&1 . (b) wt(v j )=2 and ;(v j )=a for a unique j. Let w 1 , ..., w k be the vectors v 1 , ..., v j&1 , v j &x a +x n&1 , v j+1 , ..., v k&1 , x a . Then w 1 , ..., w k can be . Our bijection viewed as the reduced echelon basis for a subspace L$ # S n&1 1 sends L to this subspace L$. For example:

vj Ä

1010000 101000 100100 wÄ 10001 Â v j &x a +x n&1 1000 100 11 10 Â x a A a

Notice that L$ must have the following property: if we define a$ to be the largest index in [1, ..., n&2] & supp(L$), then in fact a$=a and x a$ # L$. (c) wt(v s )=2 and ;(v s )=a for (exactly) two different values of s. By condition (N4), there cannot be more than two such values of s. Let j be the larger of the two possible values of s. Let w 1 , ..., w k&1 be the vectors v 1 , ..., v j&1 , v j &x a +x n&1 , v j+1 , ..., v k&1 . Then w 1 , ..., w k&1 can be regarded . Our bijection sends as the reduced echelon basis for a subspace L$ # S n&1 1 L to this subspace L$. For example:

vj Ä

1001000 100100 101000 wÄ 10001 Â v j &x a +x n&1 10000 1000 11 A a

Notice that L$ must have the following property: if we define a$ to be the largest index in [1, ..., n&2] & supp(L$), then in fact a$=a and a$=;(w t ) for some w t of weight 2.

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This completes the description of the required bijection with S n&1 . Now 1 with reduced we describe the inverse of this correspondence. Let L$ # S n&1 1 , there is a unique v$s echelon basis v$1 o } } } o v$l . By definition of S n&1 1 such that ;(v$s )=n&1, and moreover v$s has weight 2. Let a be the largest index in [1, ..., n&2] such that a # supp(L$). There are again three possibilities: (a$) a=:(v$s ) (where v$s is as above). We must have v$s =v$l = x a +x n&1 . Send L$ to the subspace of F n2 spanned by v$1 , ..., v$l&1 , x a , x n&1 +x n . (This is a reduced echelon basis.) (b$) x a # L$. We must have v$l =x a . Send L$ to the subspace of F n2 spanned by v$1 , ..., v$s&1 , v$s +x a &x n&1 , v$s+1 , ..., v$l&1 , x n&1 +x n . (This is a reduced echelon basis.) (c$) a=;(v$t ) for some v$t of weight 2. By condition (N3), v$t is unique. Send L$ to the subspace of F n2 spanned by v$1 , ..., v$s&1 , v$s + x a &x n&1 , v$s+1 , ..., v$l , x n&1 +x n . (This is a reduced echelon basis.) This completes the description of the inverse bijection. The mappings L [ L$ and L$ [ L exchange cases (a), (b), and (c) with (a$), (b$), and (c$) respectively. It is not too difficult to see that they are in fact inverse bijections between the elements L # N n in Case 7 other than ( x n&1 +x n ) and . the elements of S n&1 1 So the number of elements L # N n that fall into either Case 6 or Case | +1+ |S n&1 | = |S n&1 | +1= g(n&1)&2g(n&2)+1. 7 is exactly |S n&1 2 1 Now the seven cases above are mutually exclusive and cover all possibilities for L # N n. So we have |N n | = g(n&1)+ g(n&1)+( g(n&1) &2g(n&2))+(g(n&1)&2g(n&2))+( g(n&1)&2g(n&2))+(g(n&1)& 2g(n&2)+1)= g(n). This completes the induction step of the proof of Theorem 5.1. K Theorem 5.1, Corollary 4.6 and Brouwer's result together imply our main result: Theorem 5.2. The dimension of the universal embedding of the Sp 2n(2) (2n +1)(2n&1 +1) . K dual polar space is equal to 3

6. ANOTHER DESCRIPTION OF ( 1 %K )( 1 %K&1 ) Recall the notation %, P k , 1 k , etc., from Section 2. The results of the preceding sections imply that ( 1 %k )( 1 %k&1 ) & P k im . k . Here we mention a different description of the F 2 GL n(2)-module (1 %k )( 1 %k&1 ).

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The following commutative diagram was shown to the author by Peter Sin: .

k P k&1, k+2 wwÄ Pk

?

\k

P k&1

\

k&1 wwÄ P n&2

Here 1kn&2, ? is the linear map which sends the basis element of P k&1, k+2 corresponding to a flag X
dim P n&2 &1= : dim A k A k&1 k=1 n&2

 : dim P k im . k k=1

=

(2 n &1)(2 n&1 &1) &1. 3

Since the first and last quantities are equal, we have Theorem 6.1. For 1kn&2, the natural maps from the quotients P k im . k onto the successive quotients A k A k&1 of the filtration for P n&2 above are isomorphisms. K Remark. An alternate approach to proving Brouwer's conjecture is to somehow prove directly (perhaps using a geometric argument) that the maps P k im . k Ä A k A k&1 ,

1kn&2

are injective. So far the author has not found a way to do this.

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PAUL LI

APPENDIX: SOME TECHNICAL ARGUMENTS Here we provide the arguments for some of the more technical claims we have made earlier in Section 4. Note to Lemma 4.2. Here we indicate why L$=( x p +x s , x q +x r ) is the largest element in the symmetric difference 2 of S(X$1 , R) and S(X $2 , R). Suppose L" is any element of 2. Let v be its smallest nonzero v) exactly when vector. Then supp(v)[q, r, s], and L" equals X $+( i (There is exactly one i # [1, 2] for which this happens.) Recall X $L". i that to compare two subspaces of F n of the same dimension, one starts by comparing their respective smallest nonzero vectors. Therefore, we now examine the possibilities for v in descending lexicographic order. Suppose v=x q +x r +x s ; then X$1 +( v) =X$2 +( v), contradicting the fact that L" contains exactly one of X $1 and X $2 . If v=x q +x r , then X$1 +(v) { X$2 +( v), and the larger of them is equal to L$. This shows not only that L$ # 2, but that in fact L$ is the largest element of 2. Note to Lemma 4.3. Here we indicate why L$=( x p +x s , x q +x s , x r +x t ) is the largest element in the symmetric difference 2 of S(X $1 , R), ..., S(X$4 , R). Let L" any element of 2, and consider its smallest nonzero v) exactly when X $L. vector v. Then supp(v)[r, s, t], and L"=X $+( i i As in the previous paragraph, it suffices to examine the possibilities for v in descending lexicographic order. Suppose v=x r +x s +x t ; then we find v) there are two equal pairs, that among the four subspaces X $+( i contradicting the fact that L" contains X $i for exactly an odd number of choices of i # [1, 2, 3, 4]. The same contradiction occurs if v=x r +x s . If v=x r +x t , then L$ is the largest subspace which equals X i$+( v) for exactly an odd number of values of i # [1, 2, 3, 4]. This shows not only that L$ # 2, but that in fact L$ is the largest element of 2. Note to Lemma 4.4. To verify the assertion, one may as well assume that ( p, q, r, s, t, u)=(1, 2, 3, 4, 5, 6), R=F 62 , and L$=( x 1 +x 4 , x 2 +x 4 , x 3 +x 4 , x 5 +x 6 ). With the help of a computer, the author has obtained the following 3-spaces X$1 , ..., X $21 of F 62 : 100100 100010 100000 100110 100000 101000 101000 010011 001010 010011 010100 010011 011000 010000 001111 000101 000100 001010 001010 000101 000101 100000 010011 001100

010011 001010 000101

100011 011001 000101

100011 100001 010011 001001 000111 000101

100100 010011 001001

110011 001001 000101

100011 001011 000101

100111 010100 001100

010010 001010 000110

100011 100011 010001 010100 000101 001100

100010 010100 001010

010011 001010 000111

Sp 2n(2) DUAL POLAR SPACE

117

Here each X i$ is specified by giving its reduced echelon basis. It can be checked that L$ is the largest 4-subspace of F 62 which contains exactly an odd number of the X $i .

ACKNOWLEDGMENTS The author thanks Alexander Ivanov for suggesting this problem, and Barbara Baumeister and George Glauberman for directing attention to McClurg's work. Suggestions from Professors Glauberman and Ivanov and from Larry Wilson have led to many improvements in the present paper. The author is also grateful to Matt Bardoe and Peter Sin for helpful suggestions, and to the referee for pointing out several errors in a previous version of this paper.

REFERENCES 1. M. K. Bardoe and A. A. Ivanov, Draft report: Natural representations of dual polar spaces, unpublished. 2. A. Blokhuis and A. E. Brouwer, The universal embedding dimension of the binary symplectic dual polar space, preprint. 3. B. Cooperstein, On the generation of some dual polar spaces of symplectic type over GF(2), European J. Combin. 18 (1997), 741749, doi:10.1006eujc.1997.0142. 4. A. A. Ivanov, Affine extended dual polar spaces, in ``Trends in Math.,'' pp. 107121, Birkhauser, Basel, 1998. 5. P. McClurg, On the universal embedding of dual polar spaces of type Sp 2n(2), J. Combin. Theory Ser. A 90 (2000), 104122, doi:10.1006jcta.1999.3020. 6. S. Yoshiara, Embeddings of flag-transitive classical locally polar geometries of rank 3, Geom. Dedicata 43 (1992), 121165.