On the Grassmann modules for the symplectic groups

On the Grassmann modules for the symplectic groups Bart De Bruyn Ghent University, Department of Pure Mathematics and Computer Algebra, Krijgslaan 81 (S), B-9000 Gent, Belgium, E-mail: bdb@cage.ugent.be Abstract Let V be a n-dimensional vector space (n 1) over a field K equipped with a nondegenerate alternating bilinear form f, and let G = Sp(n, K) denote the group of isometries of (V, f). For every k {1,..., n}, there exists a natural representation of G on the subspace W k of k V generated by all vectors v 1 v k such that v 1,..., v k is totally isotropic with respect to f. With the aid of linear algebra, we prove some properties of this representation. In particular, we determine a necessary and sufficient condition for the representation to be irreducible and characterize the largest proper G-submodule. These facts allow us to determine when the Grassmann embedding of the symplectic dual polar space DW (n 1, K) is isomorphic to its minimal full polarized embedding. Keywords: Grassmann module, symplectic group, dual polar space, Grassmann embedding, minimal full polarized embedding MSC000: 15A75, 15A63, 0C33, 51A50 1 Notation and main results Let n 1 and let K be a field. Put K := K \ {0}. The characteristic of K is either 0 or a prime number p. If char(k) = p, then for every strictly positive integer m let h m be the largest nonnegative integer such that p hm m, and for every ɛ N let N ɛ,p denote the smallest multiple of p 1+h ɛ+1 greater than ɛ (and hence also greater than ɛ + 1). Let V be a n-dimensional vector space over K equipped with a nondegenerate alternating bilinear form f. An ordered basis (ē 1, f 1,..., ē n, f n ) of V 1

is called a hyperbolic basis of V if f(ē i, ē j ) = f( f i, f j ) = 0 and f(ē i, f j ) = δ ij for all i, j {1,..., n}. Let G denote the group of all isometries of (V, f), i.e. the set of all θ GL(V ) satisfying f(θ( x), θ(ȳ)) = f( x, ȳ) for all x, ȳ V. Then G = Sp(n, K). The elements of G are precisely the elements of GL(V ) which map hyperbolic bases of V to hyperbolic bases of V. For every k {0,..., n}, let k V be the k-th exterior power of V. Then 0 V = K and 1 V = V. If k {1,..., n}, then for every θ GL(V ), there exists a unique θ k GL( k V ) such that θ k ( v 1 v v k ) = θ( v 1 ) θ( v ) θ( v k ) for all vectors v 1, v,..., v k V. Now, let k {1,..., n} and let W k denote the subspace of k V generated by all vectors v 1 v v k such that v 1, v,..., v k is a k-dimensional subspace of V which is totally isotropic with respect to f. The dimension of W k is equal to ( ) ( n k n k ). For every θ G, θk stabilizes W k and hence the map θ θ k Wk defines a representation R k of the group G = Sp(n, K) on the ( ( ) ( n k n k ) )-dimensional vector space Wk. We call the corresponding KG-module a Grassmann module for Sp(n, K). Put G k := { θ k Wk θ G}. Let R k denote the set of all vectors α W k such that α v 1 v v n = 0 for any n linearly independent vectors v 1,..., v n of V such that v 1,..., v n is totally isotropic with respect to f. The subspace R k is stabilized by any element of G k. Hence, the representation R k induces a representation R (1) k of G on the quotient space W k /R k and a representation R () k of G on the subspace R k. From the theory of Lie algebras, it follows that W k, k {1,..., n}, has a largest proper G-submodule. In the case that K is a field of odd characteristic, a (complicated) recursive formula for the dimension of the largest proper G- submodule has been given in [6, Theorem (i)]. This result was generalized to the case char(k) = by Adamovich in her Ph.D. thesis, see [1] and []. With the aid of linear algebra, we will prove the following in Section 3.7: Theorem 1.1 If U is a proper subspace of W k stabilized by G k, then U R k. The following is an immediate corollary of Theorem 1.1. Corollary 1. (1) R k is the largest proper G-submodule of W k. () The representation R (1) k is irreducible. (3) The representation R k is irreducible if and only if R k = 0. Remark. In the special case that k = n, the conclusion mentioned in Theorem 1.1 was also obtained in [3] as part of a more general result regarding full polarized embeddings of dual polar spaces.

If char(k) = p is odd, then Premet and Suprunenko [6, Theorem (iv)] gave necessarily and sufficient conditions for the representation R k to be irreducible. They proved the following. Proposition 1.3 ([6]) If char(k) = p is odd, then the representation R k is irreducible if and only if p does not divide the number ( n j+k ) + 1, n k + 1 j where the product ranges over all j {0,..., k} having the same parity as k. The following is a corollary of Proposition 1.3. Corollary 1.4 Suppose char(k) = p is odd. Let ɛ N be fixed and n > ɛ be variable. Then there exists an integer n > ɛ such that R n ɛ is irreducible if and only if ɛ < n < n. Moreover, n = (N ɛ,p 1) ɛ. Proof. We will make use of Proposition 1.3. Notice that if k = n ɛ, then ( n j+k ) + 1 = ( ɛ + 1 + k j ), n k + 1 ɛ + 1 j j where the product ranges over all j {0,..., k} having the same parity as k. So, there exists a number n as in the statement of the corollary if and only if there exists a natural number m such that g(m) := ( ) ɛ+1+m ɛ+1 is divisible by p. Now, g(0) = 1, h g(0) = 0 and g(m) = ɛ+1+m for every m N \ {0}. We g(m 1) m have h g(m) > h g(m 1) if and only if h ɛ+1+m > h m, or equivalently, if and only if h ɛ+1+m > h ɛ+1. Hence, the smallest value of m for which h g(m) > 0 is equal to N ɛ,p ɛ 1. So, the values of k := n ɛ and n are respectively equal to (N ɛ,p ɛ 1) and n = k + ɛ = (N ɛ,p 1) ɛ. The proof of Proposition 1.3 given in [6] relies very much on the theory of Lie algebras and the representation theory for the symmetric groups (Specht modules). Using only linear algebra, we give a proof of the following facts in Section 3.8: Theorem 1.5 (1) If char(k) = 0, then R k = 0 for every k {1,..., n}. () Suppose char(k) = p and ɛ N. Then (for fixed ɛ and variable n > ɛ) R n ɛ = 0 if and only if ɛ < n < n := (N ɛ,p 1) ɛ. If n = n, then dim(r n ɛ ) = 1. If n = n + 1, then dim(r n ɛ ) = n. For all n > ɛ, dim(r n+1 ɛ ) dim(r n ɛ ). (3) Suppose char(k) = p and ɛ N. If n = n +1, then the representation R () n ɛ is isomorphic to the natural representation of G = Sp(n, K) on V. 3

By Corollary 1.(3) and Theorem 1.5, we have: Corollary 1.6 The representation R k is irreducible if and only if either char(k) = 0 or (char(k) = p and n < (N ɛ,p 1) ɛ), where ɛ = n k. Application to projective embeddings of symplectic dual polar spaces Let Π be a polar space (Tits [7]; Veldkamp [8]) of rank n. For every singular subspace ω of Π, let C ω denote the set of all maximal (i.e. (n 1)- dimensional) singular subspaces of Π containing ω. With Π, there is associated a point-line geometry = (P, L) which is called a dual polar space. The point-set P of consists of all maximal singular subspaces of Π, and the line-set L consists of all sets C ω, where ω is some (n )-dimensional singular subspace of Π. If ω 1 and ω are two points of, then d(ω 1, ω ) denotes the distance between ω 1 and ω in the collinearity graph of. The distance d(ω 1, ω ) is equal to n 1 dim(ω 1 ω ). So, the collinearity graph of has diameter n. For every point ω of and every i N, i (ω) denotes the set of points of at distance i from ω. A full (projective) embedding of a point-line geometry S is an injective mapping e from the point-set P of S to the point-set of a projective space Σ satisfying (i) e(p) Σ = Σ and (ii) e(l) is a line of Σ for every line L of S. Two full embeddings e 1 : S Σ 1 and e : S Σ of S are called isomorphic if there exists an isomorphism κ : Σ 1 Σ such that e = κ e 1. A full embedding e : Σ of a thick dual polar space = (P, L) is called polarized if e(p \ n (x)) Σ is a hyperplane of Σ for every point x of. The intersection N e of all subspaces e(p \ n (x)) Σ, x P, is called the nucleus of the polarized embedding e. The mapping x N e, e(x) Σ defines a full polarized embedding e/n e of into the quotient space Σ/N e. If e 1 and e are two full polarized embeddings of, then e 1 /N e1 and e /N e are isomorphic, see Cardinali, De Bruyn and Pasini [4]. The embedding e/n e, where e is an arbitrary full polarized embedding of, is called the minimal full polarized embedding of. As in Section 1, let V be a n-dimensional vector space over a field K equipped with a nondegenerate alternating bilinear form f. We suppose that n. With the pair (V, f), there is associated a symplectic polar space W (n 1, K) and a symplectic dual polar space DW (n 1, K). The singular subspaces of W (n 1, K) are the subspaces of PG(n 1, K) which are totally isotropic with respect to the symplectic polarity of P G(V ) defined by f. 4

Now, for every point ω = v 1, v,..., v n of DW (n 1, K), let e gr (ω) denote the point v 1 v v n of PG( n V ). Then e gr defines a full polarized embedding of DW (n 1, K) into PG(W n ), where W n is the subspace of n V as defined in Section 1. This embedding is called the Grassmann embedding of DW (n 1, K). The nucleus N egr of e gr is equal to PG(R n ), where R n is the subspace of W n as defined in Section 1. The following is a corollary of Theorem 1.5 (take ɛ = 0). Corollary.1 (1) If char(k) = 0, then the nucleus of the Grassmann embedding of DW (n 1, K) is empty. () If char(k) = p, then the nucleus of the Grassmann embedding of DW (n 1, K) is empty if and only if n < (p 1). If n = (p 1), then the nucleus is a point. If n = p 1, then the nucleus is a subspace of dimension n 1. (3) The Grassmann embedding of DW (n 1, K) is isomorphic to the minimal full polarized embedding of DW (n 1, K) if and only if either char(k) = 0 or (char(k) = p and n < (p 1)). 3 Proofs of Theorems 1.1 and 1.5 In this section, we will continue with the notation introduced in Section 1. Recall that V is a n-dimensional vector space (n 1) over K which is equipped with a nondegenerate alternating bilinear form f. 3.1 Hyperbolic bases of V If (ē 1, f 1,..., ē n, f n ) is a hyperbolic basis of V, then: (1) for every permutation σ of {1,..., n}, also (ē σ(1), f σ(1),..., ē σ(n), f σ(n) ) is a hyperbolic basis of V ; () for every λ K, also ( ē1 λ, λ f 1, ē, f,..., ē n, f n ) is a hyperbolic basis of V ; (3) for every λ K, also (ē 1 + λē, f 1, ē, λ f 1 + f, ē 3, f 3,..., ē n, f n ) is a hyperbolic basis of V ; (4) for every λ K, also (ē 1, f 1,..., ē n 1, f n 1, ē n, f n +λē n ) is a hyperbolic basis of V ; (5) for every λ K, also (ē 1, f 1,..., ē n 1, f n 1, ē n +λ f n, f n ) is a hyperbolic basis of V. For every i {1,, 3, 4, 5}, let Ω i denote the set of all ordered pairs (B 1, B ) of hyperbolic bases of V such that B can be obtained from B 1 as described in (i) above. The following lemma was proved in De Bruyn [5, Lemma.1]. 5

Lemma 3.1 ([5]) If B and B are two hyperbolic bases of V, then there exist hyperbolic bases B 0, B 1,..., B k (k 0) of V such that B 0 = B, B k = B and (B i 1, B i ) Ω 1 Ω 5 for every i {1,..., k}. Lemma 3. Let k, l {0,..., n} with 1 k n and 0 l min(n, k). Let v 1, v,..., v k be k linearly independent vectors of V such that v 1,..., v l is totally isotropic with respect to f. Then there exists a hyperbolic basis B = (ē 1, f 1,..., ē n, f n ) of V, a λ K, and an m {max(0, k n),..., min(k l, k )} such that v 1 v k = λ (ē 1 f 1 ) (ē m f m ) ē m+1 ē k m. Proof. Let R denote the radical of the alternating bilinear form of v 1,..., v k induced by f and let Z be a subspace of v 1,..., v k complementary to R. The alternating bilinear form f Z of Z induced by f is nondegenerate. Hence, dim(z) is even, say dim(z) = m 0. Let (ē 1, f 1,..., ē m, f m ) be a hyperbolic basis of Z with respect to f Z. Consider a basis {ē m+1,..., ē k m } of R. Then (ē 1, f 1,..., ē m, f m, ē m+1,..., ē k m ) can be extended to a hyperbolic basis (ē 1, f 1,..., ē n, f n ) of V. Obviously, v 1 v k = λ (ē 1 f 1 ) (ē m f m ) ē m+1 ē k m for some λ K. Clearly, dim(z) = m k, i.e. m k. Also, since the dimension of a maximal totally isotropic subspace of v 1,..., v k = R, Z is equal to k m, we have l k m n, i.e. k n m k l. 3. The linear maps θ k,l For every hyperbolic basis B = (ē 1, f 1,..., ē n, f n ) of V, for every k {1,..., n} and every l {0,..., k }, we now define a linear map θ k,l,b : k V k l V. If m N such that max(k n, 0) m k, if σ is a permutation of {1,..., n} such that σ(1) < σ() < < σ(m) and σ(m + 1) < < σ(k m), and if ḡ σ(i) {ē σ(i), f σ(i) } for every i {m + 1,..., k m}, then we put θ k,l,b [(ē σ(1) f σ(1) ) (ē σ(m) f σ(m) ) ḡ σ(m+1) ḡ σ(k m) ] equal to 0 if m < l and equal to (ēσ(i1 ) f σ(i1 )) (ē σ(im l) f σ(im l)) ḡ σ(m+1) ḡ σ(k m) if m l. Here, the summation ranges over all subsets {i 1, i,..., i m l } of size m l of {1,..., m} satisfying i 1 < i < < i m l. Since this defines θ k,l,b for every element of a basis of k V, we have defined θ k,l,b for all vectors of 6

k V. It is straightforward (but perhaps tedious) to verify that if B 1 and B are two hyperbolic bases of V such that (B 1, B ) Ω 1 Ω Ω 5, then θ k,l,b1 = θ k,l,b. Hence by Lemma 3.1, there exists a linear map θ k,l : k V k l V such that θ k,l,b = θ k,l for every hyperbolic basis B of V. 3.3 The subspaces W k,l of k V For every k {1,..., n} and every l {0,..., min(n, k)}, let W k,l be the subspace of k V generated by all vectors v 1 v v k, where v 1,..., v k are k linearly independent vectors of V such that v 1,..., v l is totally isotropic with respect to f. Here, we use the following convention: if l = 0, then v 1,..., v l = = {ō}. By definition, we put W 0,0 equal to 0 V = K. The following is the main result of De Bruyn [5]. Proposition 3.3 ([5]) For every k {0,..., n} and l {0,..., min(n, k)}, dim(w k,l ) = ( ) ( n k n l k ). In the previous proposition, we used the convention that ( a b) = 0 for every a N and every b Z \ {0,..., a}. The following proposition is precisely Lemma.5 of [5]. Proposition 3.4 If k {0,..., n} and 0 l k, then W k,l = k V. More generally, we can say the following. Theorem 3.5 For every k {0,..., n} and every l {0,..., min(n, k)}, W k,l = k i=k l+1 ker(θ k,i). (Here, we take the convention that if k l + 1 > k, then the intersection equals k V.) Proof. If k l+1 > k, or equivalently if l k, then k i=k l+1 ker(θ k,i) = k V = W k,l by Proposition 3.4. So, in the sequel we will suppose that l > k. We first prove that W k,l ker(θ k,i ) for every i {k l + 1,..., k }. It suffices to prove that θ k,i ( v 1 v v k ) = 0, where v 1,..., v k are k linearly independent vectors of V such that v 1,..., v l is totally isotropic with respect to f. By Lemma 3., there exists a hyperbolic basis B = (ē 1, f 1,..., ē n, f n ) of V, a λ K and an m {max(0, k n),..., min(k l, k )} such that v 1 v k = λ (ē 1 f 1 ) (ē m f m ) ē m+1 ē k m. Since m k l < k l + 1 i, θ k,i ( v 1 v k ) = 0. 7

By the previous paragraph, we know that W k,l k i=k l+1 ker(θ k,i). Suppose now that there exists a vector α k i=k l+1 ker(θ k,i) which is not contained in W k,l. Since W k,l W k,l 1 W k, k = k V, there exists an l {l 1, l,..., k } such that α W k,l and α W k,l +1. Now, let φ be the restriction of θ k,k l to the subspace W k,l of k V. (1) We prove that φ(w k,l ) W l k,l k. It suffices to prove that φ( v 1 v v k ) W l k,l k where v 1,..., v k are k linearly independent vectors of V such that v 1,..., v l is totally isotropic with respect to f. By Lemma 3., there exists a hyperbolic basis B = (ē 1, f 1,..., ē n, f n ) of V, a λ K and an m {max(0, k n),..., min(k l, k )} such that v 1 v k = λ (ē 1 f 1 ) (ē m f m ) ē m+1 ē k m. If m < k l, then φ( v 1 v v k ) = 0. If m = k l, then φ( v 1 v v k ) = λ ē m+1 ē k m W l k,l k. () We prove that φ(w k,l ) = W l k,l k. It suffices to prove that ē 1 ē l k Im(φ) for every l k linearly independent vectors ē 1,..., ē l k of V such that ē 1,..., ē l k is totally isotropic with respect to f. Now, extend (ē 1,..., ē l k) to a hyperbolic basis (ē 1, f 1,..., ē n, f n ) of V. Put α := (ē l k+1 f l k+1) (ē l f l ) ē 1 ē ē l k. Then α W k,l and φ(α) = ē 1 ē l k. (3) We prove that φ(w k,l +1) = 0. It suffices to prove that φ( v 1 v v k ) = 0 where v 1,..., v k are k linearly independent vectors of V such that v 1,..., v l +1 is totally isotropic with respect to f. By Lemma 3., there exists a hyperbolic basis B = (ē 1, f 1,..., ē n, f n ) of V, a λ K and an m {max(0, k n),..., min(k l 1, k )} such that v 1 v k = λ (ē 1 f 1 ) (ē m f m ) ē m+1 ē k m. Since m < k l, φ( v 1 v k ) = 0. (4) We prove that ker(φ) = W k,l +1. We have dim(ker(φ)) = dim(w k,l ) dim(im(φ)) = dim(w k,l ) dim(w l k,l k) = ( ) ( ) ( n k n l k n l k) + ( ( n l k ) = n ) ( k n l k) = dim(wk,l +1). Since W k,l +1 ker(φ), we necessarily have ker(φ) = W k,l +1. We are now ready to derive a contradiction. Since α k i=k l+1 ker(θ k,i), we necessarily have α ker(φ). On the other hand, since α W k,l \ W k,l +1, we necessarily have α ker(φ) by (4) above. 3.4 An invariant vector of k V, k {0,..., n} We can use Lemma 3.1 to prove the existence of some invariant vector of k V, k {0,..., n}. For every hyperbolic basis B = (ē 1, f 1,..., ē n, f n ) of V and every k {0,..., n}, we define α k (B) := (ē i1 f i1 ) (ē i f i ) (ē ik f ik ), 8

where the summation ranges over all ( n k) subsets {i1, i,..., i k } of size k of {1,..., n} satisfying i 1 < i < < i k. (By convention, α 0 (B) = 1 K.) One can readily verify that if B 1 and B are two hyperbolic bases of V such that (B 1, B ) Ω 1 Ω 5, then α k (B 1 ) = α k (B ) for every k {0,..., n}. Hence by Lemma 3.1, there exists a vector α k k V such that α k = α k(b) for every hyperbolic basis B of V. Lemma 3.6 Let p be a prime and ɛ N. Then the smallest k N \ {0} for which p ( ) ɛ+k+i i, i {1,..., k}, is equal to Nɛ,p ɛ 1. Proof. (1) Suppose k = N ɛ,p ɛ 1 and let i be an arbitrary element of {1,..., k}. Then ( ) ɛ + k + i = N ɛ,p Nɛ,p + 1 Nɛ,p + i 1. (1) i i 1 i 1 Recall that p h ɛ+1 is the largest power of p dividing ɛ + 1 and that N ɛ,p is the smallest multiple of p 1+h ɛ+1 bigger than ɛ+1. Hence, k = N ɛ,p ɛ 1 < p 1+h ɛ+1 and the largest power of p dividing a given element of {1,..., i} is at most p h ɛ+1. Now, for every j {1,..., i 1}, the larger power of p dividing N ɛ,p +j equals the largest power of p dividing j. The largest power of p dividing N ɛ,p is equal to p 1+h ɛ+1, while the largest power of p dividing i is at most p h ɛ+1. Hence, p divides ( ) ɛ+k+i i by equation (1). () Suppose k = N ɛ 1 where ɛ + 1 < N < N ɛ,p. Let p h be the largest power of p dividing N. Then h h ɛ+1 and hence N ɛ + 1 + p h, i.e. p h N ɛ 1 = k. We have ( ) ɛ + k + p h = N p N + 1 N + ph 1. () h 1 p h 1 p h For every j {1,..., p h 1}, the largest power of p dividing N +j equals the largest power of p dividing j. Recall that the largest power of p dividing N is equal to p h. So, p is not a divisor of ( ) ɛ+k+p h (recall ()) and p h {1,..., k}. Proposition 3.7 (1) If char(k) = 0, then α k W k for every k {1,..., n}. () Suppose char(k) = p. Then for fixed ɛ N and variable n > ɛ, n = (N ɛ,p 1) ɛ is the smallest value of n for which n ɛ > 0 is even and α n ɛ W n ɛ. Proof. (1) Suppose char(k) = 0. If α k would belong to W k = W k,k, then α k ker(θ k,1) by Theorem 3.5. Now, θ k,1 (α k ) = (n k + 1) α k 0 since n k + 1 0. 9 p h

() Suppose char(k) = p, ɛ N and k := n ɛ > 0 is even. By Theorem 3.5, W k = W k,k = k i=1 ker(θ k,i). Now, for every i {1,..., k}, θ k,i (αk ) = ( ) n k+i i α k i = ( ) ɛ+k+i i α k i. Claim () of the proposition now follows from Lemma 3.6. Corollary 3.8 Suppose char(k) = p, ɛ N and n = n = (N ɛ,p 1) ɛ. Then α n ɛ R n ɛ. Proof. By Proposition 3.7, αn ɛ W n ɛ. Now, let ē 1,..., ē n be n arbitrary linearly independent vectors of V such that ē 1,..., ē n is an n-dimensional totally isotropic subspace of V. Extend (ē 1,..., ē n ) to a hyperbolic basis (ē 1, f 1,..., ē n, f n ) of V. Then αn ɛ = (ēi1 f i1 ) (ē im f im ), where the summation ranges over all ( ) n subsets {i 1,..., i m } of size m := n ɛ of {1,..., n}. So, α n ɛ ē 1 ē n = (ēi1 f i1 ) (ē im f im ) ē 1 ē n = 0. 3.5 An invariant subspace of k+1 V, k {0,..., n 1} Let k {0,..., n 1} and let B = (ē 1, f 1,..., ē n, f n ) be a hyperbolic basis of V. For every i {1,..., n}, let α k,i (B) be the vector (ēi1 f i1 ) (ē i f i ) (ē ik f ik ), m where the summation ranges over all ( ) n 1 k subsets {i1, i,..., i k } of size k of {1,..., n} \ {i} satisfying i 1 < i < < i k. Let R k+1 (B) denote the ndimensional subspace of k+1 V generated by the n vectors ē i α k,i (B), f i α k,i (B) (i {1,..., n}). One readily verifies that if B 1 and B are two hyperbolic bases of V such that (B 1, B ) Ω 1 Ω 5, then R k+1 (B 1 ) = R k+1 (B ). Hence, by Lemma 3.1, there exists a n-dimensional subspace Rk+1 of k+1 V such that Rk+1 = R k+1(b) for any hyperbolic basis B of V. Lemma 3.9 If B = (ē 1, f 1,..., ē n, f n ) and B = (ē 1, f 1,..., ē n, f n) are two hyperbolic bases of V such that ē 1 = ē 1, then ē 1 α k,1 (B) = ē 1 α k,1 (B ). Proof. Let B = (ē 1, f 1, ē, f,..., ē n, f n) be the hyperbolic basis (ē 1, f 1, ē f(ē, f 1 ) ē 1, f f( f, f 1 ) ē 1,, ē n f(ē n, f 1 ) ē 1, f n f( f n, f 1 ) ē 1 ) of V. Then obviously, ē 1 α k,1 (B ) = ē 1 α k,1 (B ). The vectors ē, f,..., ē n, f n are f- orthogonal with ē 1, f 1 and hence belong to the subspace V := ē, f,..., ē n, f n. If f is the nondegenerate alternating bilinear form of V induced by f, then (ē, f,..., ē n, f n) and (ē, f,..., ē n, f n ) are two hyperbolic bases of V. So, by Section 3.4, α k,1 (B) = α k,1 (B ). Together with ē 1 α k,1 (B ) = ē 1 α k,1 (B ), this implies that ē 1 α k,1 (B) = ē 1 α k,1 (B ). 10

For every nonzero vector x of V, let η k ( x) denote the vector x α k,1 (B) of Rk+1, where B is an arbitrary hyperbolic basis of V having x as first component. We put η k (0) equal to the zero vector of k+1 V. Lemma 3.10 η k is a linear isomorphism from V to R k+1. Proof. (1) Obviously, η k ( x) = 0 if and only if x = 0. () We prove that η k (λ x) = λ η k ( x) for every λ K and every x V. Obviously, this holds if x = 0 or λ = 0. So, suppose x 0 and λ 0. Let B = ( x, f 1, ē, f,..., ē n, f n ) be an arbitrary hyperbolic basis of V having x as a first component. Then B = (λ x, f 1, ē λ, f,..., ē n, f n ) is also a hyperbolic basis of V. Now, η k (λ x) = λ x α k,1 (B ) = λ x α k,1 (B) = λ η k ( x). (3) We prove that η k ( x 1 + x ) = η k ( x 1 ) + η k ( x ) for any two vectors x 1 and x of V satisfying λ := f( x 1, x ) 0. Let B = ( x 1, x λ, ē, f,..., ē n, f n ) be a hyperbolic basis having x 1 and x λ as first two components. Then B = ( x 1 + x, x λ, ē, f,..., ē n, f n ) and B = ( x, x 1 λ, ē, f,..., ē n, f n ) are also hyperbolic bases of V. We have η k ( x 1 + x ) = ( x 1 + x ) α k,1 (B ) = x 1 α k,1 (B )+ x α k,1 (B ) = x 1 α k,1 (B)+ x α k,1 (B ) = η k ( x 1 )+η k ( x ). (4) We prove that η k ( x 1 + x ) = η k ( x 1 ) + η k ( x ) for any two linearly independent vectors x 1 and x of V satisfying f( x 1, x ) = 0. Let B = ( x 1, f 1, x, f, ē 3, f 3,..., ē n, f n ) be a hyperbolic basis of V having x 1 as first component and x as third component. Then B = ( x, f, x 1, f 1, ē 3, f 3,..., ē n, f n ) and B = ( x 1 + x, f 1, x, f 1 + f, ē 3, f 3,..., ē n, f n ) are also hyperbolic bases of V. One can readily verify that ( x 1 + x ) α k,1 (B ) = x 1 α k,1 (B) + x α k,1 (B ). Hence, η k ( x 1 + x ) = η k ( x 1 ) + η k ( x ). Lemma 3.11 The map θ θ k+1 induces a representation of G on the vector space Rk+1 which is isomorphic to the natural representation of G = Sp(n, K) on V. Proof. In view of Lemma 3.10, it suffices to show that η k (θ( x)) = θ k+1 (η k ( x)) for every θ G and every x V. Obviously, this holds if x = 0. So, suppose x 0 and consider a hyperbolic basis B = ( x, f 1, ē, f,..., ē n, f n ) of V having x as first component. Then θ k+1 (η k ( x)) = θ k+1 ( x ē i1 f i1 ē ik f ik ) = θ( x) θ(ē i1 ) θ( f i1 ) θ(ē ik ) θ( f ik ), and this is equal to η k (θ( x)) since (θ( x), θ( f 1 ), θ(ē ), θ( f ),..., θ(ē n ), θ( f n )) is a hyperbolic basis of V. 3.6 Preparation of an inductive approach Throughout this subsection we suppose that n k and that ē 1 and f 1 are two vectors of V such that f(ē 1, f 1 ) = 1. Let V denote the set of vectors of 11

V which are f-orthogonal with ē 1 and f 1 and let f denote the nondegenerate alternating bilinear form of V induced by f. Let W k 1 denote the subspace of k 1 V generated by all vectors of the form v 1 v v k 1 where v 1, v,..., v k 1 is a (k 1)-dimensional subspace of V which is totally isotropic with respect to f. Let R k 1 denote the set of all α W k 1 with the property that α v 1 v n 1 = 0 for all n 1 vectors v 1,..., v n 1 of V such that v 1,..., v n 1 is an (n 1)-dimensional subspace of V which is totally isotropic with respect to f. Let G be the group of isometries of (V, f ) and let G k 1 denote the subgroup of GL(W k 1 ) corresponding to G (see Section 1). For every vector α of k 1 V, let µ k (α) be the vector ē 1 α of k V. Then µ k defines an isomorphism between W k 1 and the subspace µ k (W k 1 ) of W k. In some of the proofs of this subsection, we will make use of the following obvious facts which hold for all vectors v 1, v,..., v k V, for all i {1,..., k} and all λ j K with j {1,..., k} \ {i}: ( v 1 v i v k = v 1 v i 1 v i + j i λ j v j ) v i+1 v k, (3) v 1 v i v i+1 v k = v 1 v i+1 ( v i ) v k (i k). (4) Now, every vector χ of k V can be written in a unique way as ē 1 f 1 α(χ) + ē 1 β(χ) + f 1 γ(χ) + δ(χ), where α(χ) k V, β(χ), γ(χ) k 1 V and δ(χ) k V. Notice that the maps α : k V k V, β : k V k 1 V, γ : k V k 1 V and δ : k V k V are linear. Lemma 3.1 If χ W k, then β(χ) and γ(χ) belong to W k 1. Proof. We will prove that β(χ) W k 1. In a completely similar way one can also prove that γ(χ) W k 1. By the linearity of β, we may restrict ourselves to the case where χ = v 1 v v k for some vectors v 1, v,..., v k V such that v 1, v,..., v k is a k-dimensional subspace of V which is totally isotropic with respect op f. By equations (3) and (4), we may suppose that v,..., v k f 1, V and v 3,..., v k V. Let λ 1, λ K such that v 1 λ 1 ē 1 f 1, V and v λ f1 V. Then β(χ) = λ 1 ( v λ f1 ) v 3 v k W k 1 since v λ f1, v 3,..., v k is a (k 1)- or (k )-dimensional subspace of V which is totally isotropic with respect to f. 1

Lemma 3.13 Suppose U is a subspace of W k which is stabilized by G k. Then µ 1 k (U µ k(w k 1 )) is a subspace of W k 1 which is stabilized by G k 1. Proof. Let α be an arbitrary vector of µ 1 k (U µ k(w k 1 )) and let θ be an arbitrary element of G k 1 corresponding to an element θ G. We need to show that θ(α) µ 1 k (U µ k(w k 1 )). We extend θ to an element θ of G by defining θ(ē 1 ) = ē 1 and θ( f 1 ) = f 1. Let θ be the element of G k corresponding to θ. Then for every vector α of W k 1, µ k θ(α ) = θ µ k (α ). Hence, θ stabilizes µ k (W k 1 ). Now, since µ k (α) U µ k (W k 1 ), also θ µ k (α) U µ k (W k 1 ). Hence, θ(α) = µ 1 k θ µ k (α) µ 1 k (U µ k(w k 1 )). Lemma 3.14 µ k (R k 1 ) = R k µ k (W k 1 ). Proof. (1) We prove that R k µ k (W k 1 ) µ k(r k 1 ). Let α be an arbitrary element of W k 1 such that ē 1 α R k. We need to show that α v v 3 v n = 0 for all vectors v,..., v n of V such that v,..., v n is an (n 1)-dimensional subspace of V which is totally isotropic with respect to f. Since ē 1 α R k, we have ē 1 α f 1 v v n = 0 since f 1, v,..., v n is an n-dimensional subspace of V which is totally isotropic with respect to f. This implies that α v v 3 v n = 0 as we needed to show. () We prove that µ k (R k 1 ) R k µ k (W k 1 ). Let α be an arbitrary element of R k 1. We need to show that ē 1 α R k, or equivalently, that ē 1 α v 1 v v n = 0 for all vectors v 1, v,..., v n of V such that v 1, v,..., v n is an n-dimensional subspace of V which is totally isotropic with respect to f. By equations (3) and (4), we may suppose that v,..., v n ē 1, V and v 3,..., v n V. Let λ K such that v λē 1 V. Then v λē 1, v 3,..., v n is an (n 1)- or (n )-dimensional subspace of V which is totally isotropic with respect to f. Now, ē 1 α v 1 v v n = ē 1 α v 1 ( v λē 1 ) v 3 v n = 0 since α R k 1. 3.7 Proof of Theorem 1.1 The following proposition is precisely Theorem 1.1. Proposition 3.15 If U is a proper subspace of W k which is stabilized by G k, then U R k. Proof. We will prove the proposition by induction on k. If k = 1, then W k = W 1 = 1 V = V and U = 0 since G acts transitively on the set of 13

1-spaces of V. Suppose therefore that k and that the proposition holds for smaller values of k. Suppose by way of contradiction that there exists a χ 0 U \ R k. Then there exists an n-dimensional subspace f 1, f,..., f n of V which is totally isotropic with respect to f such that χ 0 f 1 f f n 0. Now, extend ( f 1, f,..., f n ) to a hyperbolic basis (ē 1, f 1,..., ē n, f n ) of V. Let V, W k 1, R k 1 and µ k as in Subsection 3.6. We prove that µ 1 k (U µ k(w k 1 )) is a proper subspace of W k 1. If this would not be the case, then µ k (W k 1 ) U and hence U contains a vector of the form ē 1 v v k where ē 1, v,..., v k is a k-dimensional subspace of V which is totally isotropic with respect to f. Since G acts transitively on the set of all k-dimensional subspaces of V which are totally isotropic with respect to f, U must contain all vectors of the form v 1 v v k where v 1,..., v k is a k-dimensional subspace of V which is totally isotropic with respect to V. This however would imply that U = W k, which is impossible. By Lemma 3.13 and the induction hypothesis, we now have: Now, we can write χ 0 in a unique way as µ 1 k (U µ k(w k 1)) R k 1. (5) χ 0 = ē 1 f 1 α(χ 0 ) + ē 1 β(χ 0 ) + f 1 γ(χ 0 ) + δ(χ 0 ), where α(χ 0 ) k V, β(χ 0 ), γ(χ 0 ) k 1 V and δ(χ 0 ) k V. We prove that χ 1 := ē 1 f 1 α(χ 0 ) + δ(χ 0 ) belongs to U and satisfies χ 1 f 1 f f n 0. In order to achieve this goal, it suffices to prove that ē 1 β(χ 0 ) U, f 1 γ(χ 0 ) U and β(χ 0 ) f f n = 0. Let θ be the unique element of G mapping the hyperbolic basis (ē 1, f 1,..., ē n, f n ) of V to the hyperbolic basis (ē 1 + f 1, f 1, ē, f,..., ē n, f n ) of V. Then θ k (χ 0 ) = χ 0 + f 1 β(χ 0 ). Since χ 0 U, also θ k (χ 0 ) U and hence f 1 β(χ 0 ) U. Let θ be the element of G mapping the hyperbolic basis (ē 1, f 1,..., ē n, f n ) of V to the hyperbolic basis ( f 1, ē 1, ē, f,..., ē n, f n ) of V. Then since f 1 β(χ 0 ) U, also ē 1 β(χ 0 ) = θ k ( f 1 β(χ 0 )) U. In a similar way one proves that f 1 γ(χ 0 ) U. Now by Lemma 3.1, β(χ 0 ) W k 1. Hence, ē 1 β(χ 0 ) U µ k (W k 1 ). By (5), β(χ 0) R k 1. As a consequence, β(χ 0 ) f f n = 0. We now inductively define vectors χ i k V for every i {,..., n}. Let V i be the subspace of V which is f-orthogonal with ē i, f i. We can write χ i 1 in a unique way as χ i 1 = ē i f i α(χ i 1 ) + ē i β(χ i 1 ) + f i γ(χ i 1 ) + δ(χ i 1 ), 14

where α(χ i 1 ) k V i, β(χ i 1 ), γ(χ i 1 ) k 1 V i and δ(χ i 1 ) k V i. We define χ i := ē i f i α(χ i 1 ) + δ(χ i 1 ). With a completely similar reasoning as above, we can (inductively) prove that χ i U and χ i f 1 f f n 0 for every i {,..., n}. If k is odd, then χ n = 0. If k is even, then χ n is a sum of terms of the form λ (ē i1 f i1 ) (ē il f il ) where l = k. In any case, we have χ n f 1 f n = 0, which is a contradiction. Hence, U R k. 3.8 Proof of Theorem 1.5 (I) Let ɛ N be fixed and n > ɛ be variable. We suppose that there exists a value of n > ɛ for which R n ɛ 0. We moreover suppose that n is the smallest value for which this is the case. Then k := n ɛ. Let (ē 1, f 1,..., ē n, f n ) be a hyperbolic basis of V and let V, W k 1, R k 1 and µ k as in Subsection 3.6. By the minimality of n, we have R k 1 = 0. Let χ be an arbitrary vector of R k. Then χ = ē 1 f 1 α(χ) + ē 1 β(χ) + f 1 γ(χ) + δ(χ) where α(χ) k V, β(χ), γ(χ) k 1 V and γ(χ) k V. Consider the element θ of G mapping the hyperbolic basis (ē 1, f 1, ē, f,..., ē n, f n ) of V to the hyperbolic basis (ē 1, f 1 +ē 1, ē, f,..., ē n, f n ) of V. Then θ k (χ) = χ + ē 1 γ(χ). Since χ R k, also θ k (χ) R k and hence ē 1 γ(χ) R k. By Lemma 3.1, ē 1 γ(χ) µ k (W k 1 ). Hence, by Lemma 3.14, γ(χ) R k 1, i.e. γ(χ) = 0. In a completely similar way one can prove that β(χ) = 0. What we have just done, we can also do for any pair (ē i, f i ), i {1,..., n}. We can conclude: ( ) For every i {1,..., n} and every χ R k, χ can be written in the form ē i f i α i (χ) + δ i (χ) where α i (χ) k ē 1, f 1,..., ē i, fi,..., ē n, f n and δ i (χ) k ē 1, f 1,..., ē i, fi,..., ē n, f n. If k is odd, then ( ) implies that R k = 0, a contradiction. Hence, k = m is even. By ( ), every element χ of R k is of the form λ I ē i1 f i1 ē im f im, with the summation ranging over all subsets I = {i 1,..., i m } of size m of {1,..., n} satisfying i 1 < i < < i m. We will now show that all the coefficients λ I are equal to each other. Suppose first that I 1 and I are two subsets of size m of {1,,..., n} such that I 1 I = m 1. Without loss of generality, we may suppose that 15

I 1 \ I = {1} and I \ I 1 = {}. Write λ I ē i1 f i1 ē im f im in the form ē 1 f 1 ē f α + ē 1 f 1 β + ē f γ + δ, where α k 4 ē 3, f 3,..., ē n, f n, β, γ k ē 3, f 3,..., ē n, f n and δ k ē 3, f 3,..., ē n, f n. [If k =, then we omit the term ē 1 f 1 ē f α.] Let θ denote the element of G mapping the hyperbolic basis (ē 1, f 1, ē, f,..., ē n, f n ) of V to the hyperbolic basis (ē 1 + ē, f 1, ē, f 1 + f, ē 3, f 3,..., ē n, f n ) of V. Then θ k (χ) = χ + ē f 1 (β γ). Since χ R k, also θ k (χ) R k and hence ē f 1 (β γ) R k. By ( ), β = γ. Hence, λ I1 = λ I. Consider now the most general case and let I 1 and I be two arbitrary subsets of size m of {1,..., n}. Put I 1 I = m l. Then there exist l + 1 subsets J 0,..., J l of size m of {1,..., n} such that J 0 = I 1, J l = I and J i 1 J i = m 1 for every i {1,..., l}. By the previous paragraph, we know that λ I1 = λ J0 = λ J1 = = λ Jl = λ I. Hence, R k is 1-dimensional and equal to α m. By Proposition 3.7 and Corollary 3.8, we then know that char(k) = p and n = n := (N ɛ,p 1) ɛ. So, we have proved the following proposition: Proposition 3.16 (1) If char(k) = 0, then R k = 0 for every k {1,..., n}. () Suppose char(k) = p. Then for fixed ɛ and variable n > ɛ, R n ɛ = 0 if ɛ < n < n. If n = n, then R n ɛ = α n ɛ. (II) Let ɛ N be fixed and n > ɛ be variable. Let (ē 1, f 1,..., ē n, f n ) be a hyperbolic basis of V and let R k 1 and µ k as defined in Subsection 3.6. By Lemma 3.14, ē 1 R k 1 R k and f 1 R k 1 R k. This fact in combination with Proposition 3.16() gives us the following: Proposition 3.17 Suppose char(k) = p. Let ɛ N be fixed and n > ɛ be variable. Then for all n > ɛ, dim(r n+1 ɛ ) dim(r n ɛ ). As a consequence, R n ɛ = 0 if and only if ɛ < n < n := (N ɛ,p 1) ɛ. (III) Suppose char(k) = p and ɛ N. As before, put n = (N ɛ,p 1) ɛ. Put n = n + 1 and k + 1 = n ɛ. Proposition 3.18 We have R k+1 = R k+1. Proof. Let B = (ē 1, f 1, ē, f,..., ē n, f n ) be a hyperbolic basis of V. By Lemma 3.14 and the explicit description of the largest proper submodule in the case n = n (see Proposition 3.16), we know that ē i α k,i (B) R k+1 and f i α k,i (B) R k+1 for every i {1,..., n}. Hence, Rk+1 R k+1. We will now also prove that R k+1 Rk+1. 16

Let χ be an arbitrary vector of R k+1. For every i {1,..., n}, let α i k 1 ē 1, f 1,..., ē i, fi,..., ē n, f n, β i, γ i k ē 1, f 1,..., ē i, fi,..., ē n, f n and δ i k+1 ē 1, f 1,..., ē i, fi,..., ē n, f n such that χ = ē i f i α i +ē i β i + f i γ i +δ i. We will now prove that β i and γ i are multiples of α k,i (B). Since k+1 is odd and χ k+1 V, this fact then implies that χ = n i=1 (ē i β i + f i γ i ) R k+1. Consider the element θ G mapping the hyperbolic basis (ē 1, f 1,..., ē n, f n ) of V to the hyperbolic basis (ē 1, f 1,..., ē i + f i, f i,..., ē n, f n ) of V. Then θ k (χ) = χ + f i β i. Since χ R k+1, also θ k (χ) R k+1 and hence f i β i R k+1. In a similar way one proves that ē i γ i R k+1. By Lemmas 3.1 and 3.14 and the explicit description of the largest proper submodule in the case n = n, we then know that β i and γ i are multiples of α k,i (B). As mentioned before this fact implies that χ = n i=1 (ē i β i + f i γ i ) Rk+1. Hence, also R k+1 Rk+1. Lemma 3.11 and Proposition 3.18 then implies the following. Proposition 3.19 Suppose char(k) = p, ɛ N, n = (N ɛ,p 1) ɛ and n = n + 1. Then the representation R () n ɛ is isomorphic to the natural representation of G = Sp(n, K) on V. Acknowledgment The author would like to thank Ilaria Cardinali and Antonio Pasini whose question whether more geometric approaches can be given to some the problems considered in [6] was a motivation for the research carried out in this paper. References [1] A. M. Adamovich. Analog of the space of primitive forms over a field of positive characteristic. Moscow Univ. Math. Bull. 39 (1984), 53 56. [] A. M. Adamovich. The submodule lattice of Weyl modules for symplectic groups with fundamental highest weights. Moscow Univ. Math. Bull. 41 (1986), 6 9. [3] R. J. Blok, I. Cardinali, B. De Bruyn and A. Pasini. Polarized and homogeneous embeddings of dual polar spaces. J. Algebraic Combin. 30 (009), 381 399. 17

[4] I. Cardinali, B. De Bruyn and A. Pasini. Minimal full polarized embeddings of dual polar spaces. J. Algebraic Combin. 5 (007), 7 3. [5] B. De Bruyn. Some subspaces of the k-th exterior power of a symplectic vector space. Linear Algebra Appl. 430 (009), 3095 3104. [6] A. A. Premet and I. D. Suprunenko. The Weyl modules and the irreducible representations of the symplectic group with the fundamental highest weights. Comm. Algebra 11 (1983), 1309 134. [7] J. Tits. Buildings of spherical type and finite BN-pairs. Lecture Notes in Mathematics 386. Springer, Berlin, 1974. [8] F. D. Veldkamp. Polar Geometry. I-IV, V. Indag. Math. 1 and (1959), 51 551 and 07 1. 18