Decomposition of exterior and symmetric squares in characteristic two
aa r X i v : . [ m a t h . R T ] J a n DECOMPOSITION OF EXTERIOR AND SYMMETRIC SQUARESIN CHARACTERISTIC TWO
MIKKO KORHONEN
Abstract.
Let V be a finite-dimensional vector space over a field of char-acteristic two. As the main result of this paper, for every nilpotent element e ∈ sl ( V ), we describe the Jordan normal form of e on the sl ( V )-modules ∧ ( V ) and S ( V ). In the case where e is a regular nilpotent element, we areable to give a closed formula.We also consider the closely related problem of describing, for every unipo-tent element u ∈ SL( V ), the Jordan normal form of u on ∧ ( V ) and S ( V ). Arecursive formula for the Jordan block sizes of u on ∧ ( V ) was given by Gowand Laffey (J. Group Theory 9 (2006), 659–672). We show that their proofcan be adapted to give a similar formula for the Jordan block sizes of u on S ( V ). Introduction
Let V be a finite-dimensional vector space over a field. Let u ∈ SL( V ) be aunipotent linear map and let e ∈ sl ( V ) be a nilpotent linear map. We consider thefollowing two basic questions in representation theory. Q1.
What are the Jordan block sizes of u in its SL( V )-action on the exteriorsquare ∧ ( V ) and the symmetric square S ( V )? Q2.
What are the Jordan block sizes of e in its sl ( V )-action on ∧ ( V ) and S ( V )?As we will see later in this introduction, results from the literature quickly reduceboth problems to the case where u and e act on V with a single Jordan block, sowe assume that this is the case. Then good answers to both questions are knownin odd characteristic [Bar11, Theorem 2] [McN02, Theorem 24].In this paper we will consider the characteristic two case, where the previouslyknown results are as follows. A formula for the Jordan block sizes of u on ∧ ( V )has been given by Gow and Laffey [GL06, Theorem 2]. With [GL06, Theorem 2]and [Sym07, Corollary 3.11], one can calculate the Jordan decomposition of u on S ( V ), modulo Jordan blocks of even size. Then [Sym07, Proposition 2.2] providesa recursive algorithm for computing the Jordan block sizes of u on S ( V ).The main purpose of this paper is to provide explicit formulae in characteristictwo for the Jordan block sizes of u on S ( V ) (Theorem 1.3) and the Jordan blocksizes of e on ∧ ( V ) and S ( V ) (Theorems 1.6 - 1.9).For the Jordan block sizes of u on S ( V ), we give a recursive formula which isanalogous to [GL06, Theorem 2]. In the nilpotent case, we will compute a Jordan Date : January 19, 2021.Partially supported by NSFC grants 11771200 and 11931005. basis for the action of e on V ⊗ V (Theorem 3.6), and use it to find a closed formulafor the Jordan block sizes of e on ∧ ( V ) and S ( V ) (Theorems 1.6 - 1.7).For the rest of this paper, we fix a field K and make the following assumption. Assume that char K = 2 . To describe our results, it will be convenient to do so in terms of representationtheory. Let q = 2 α , where α is a positive integer. Let C q be a cyclic group oforder q . Recall that there are a total of q indecomposable K [ C q ]-modules V , . . . , V q , where dim V i = i and a generator of C q acts on V i as a single i × i unipotentJordan block. Denote V = 0. For a K -vector space W , we denote W = 0 and W d = W ⊕ · · · ⊕ W ( d copies) for an integer d > Q1 is equivalent to the problem of decomposing ∧ ( V ) and S ( V )into indecomposable summands for every K [ C q ]-module V . Note that we haveisomorphisms ∧ ( V ⊕ W ) ∼ = ∧ ( V ) ⊕ ( V ⊗ W ) ⊕ ∧ ( W ) , (1.1) S ( V ⊕ W ) ∼ = S ( V ) ⊕ ( V ⊗ W ) ⊕ S ( W )(1.2)of K [ C q ]-modules. Thus Q1 is reduced to the problem of decomposing V m ⊗ V n , ∧ ( V n ), and S ( V n ) into indecomposable summands for integers 0 < n, m ≤ q .The decomposition of V m ⊗ V n has been extensively studied in all characteristics,see for example [Sri64], [Ral66], [McF79], [Ren79], [Nor95], [Nor08], [Hou03], and[Bar11]. In our setting of characteristic two, we can use the following result, whichgives a recursive description for the decomposition of V m ⊗ V n . Theorem 1.1 ([Gre62, (2.5a)], [GL06, Lemma 1, Corollary 3]) . Let < m ≤ n ≤ q and suppose that q/ < n ≤ q . Then the following statements hold: (i) If n = q , then V m ⊗ V n ∼ = V mq as K [ C q ] -modules. (ii) If m + n > q , then V m ⊗ V n ∼ = V n + m − qq ⊕ ( V q − n ⊗ V q − m ) as K [ C q ] -modules. (iii) If m + n ≤ q , then V m ⊗ V n ∼ = V q − d t ⊕ · · · ⊕ V q − d as K [ C q ] -modules, where V m ⊗ V q − n ∼ = V d ⊕ · · · ⊕ V d t . Note that with Theorem 1.1, we are able to calculate V m ⊗ V n for any given0 < m ≤ n ≤ q . We either get an explicit decomposition (case (i)), or an expressionof V m ⊗ V n in terms of a tensor product V m ′ ⊗ V n ′ for some 0 < m ′ ≤ n ′ with n ′ < n .In the latter case we can consider V m ′ ⊗ V n ′ as a K [ C q ′ ]-module, where q ′ is a powerof 2 such that q ′ / < n ′ ≤ q ′ . Thus by applying Theorem 1.1 repeatedly, we canquickly calculate the decomposition of V m ⊗ V n into indecomposable summands.For the decomposition of ∧ ( V n ), a recursive formula in similar vein as Theorem1.1 was found by Gow and Laffey [GL06]. Theorem 1.2 ([GL06, Theorem 2]) . Suppose that q/ < n ≤ q . Then we have ∧ ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ − q ⊕ V q/ − n as K [ C q ] -modules. It turns out that there is a similar recurrence for the decomposition of S ( V n ),which we will prove in the next section. Our proof will follow along the same linesas the proof of Theorem 1.2 in [GL06]. Theorem 1.3.
Suppose that q/ < n ≤ q . Then we have S ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ q ⊕ V q/ as K [ C q ] -modules. Note that in Theorem 1.2 we have q − n < q/
2, so the result can be appliedrepeatedly to find efficiently the decomposition of ∧ ( V n ) for any given n . Similarly S ( V n ) can be decomposed by applying Theorem 1.3 together with Theorem 1.2.The main part of this paper will be concerned with problem Q2 about nilpotentlinear maps. Here the most natural way to describe our results will be in termsof representations of Lie algebras. Let w q be the abelian p -Lie algebra over K generated by a single nilpotent element e ∈ w q such that e [ q ] = 0, so as a K -vectorspace w q = M ≤ i<α h e [2 i ] i . There are a total of q indecomposable restricted w q -modules W , . . . , W q , wheredim W i = i and e acts on W i as a single i × i nilpotent Jordan block. Denote W = 0.In analogue with the unipotent case, question Q2 is equivalent to the problem ofdecomposing ∧ ( V ) and S ( V ) into indecomposable summands for every restricted w q -module V .The isomorphisms (1.1) and (1.2) hold for w q -modules as well, so we are reducedthe problem of decomposing W m ⊗ W n , ∧ ( W n ), and S ( W n ) into indecomposablesummands.By the following result, one can calculate the decomposition of W m ⊗ W n usingTheorem 1.1. This is a special case of a result of Fossum [Fos89] on formal grouplaws, alternatively a short proof can be found in [Nor93, Corollary 5 (a)]. Proposition 1.4 ([Fos89, Section III]) . Let < n, m ≤ q and suppose that wehave V m ⊗ V n ∼ = V r ⊕ · · · ⊕ V r t as K [ C q ] -modules for some r , . . . , r t > . Then W m ⊗ W n ∼ = W r ⊕ · · · ⊕ W r t as w q -modules. The analogue of Proposition 1.4 fails for ∧ ( W n ) and S ( W n ). The followingexample was noted in [Fos89, p. 286]: we have ∧ ( V ) ∼ = V ⊕ V , but ∧ ( W ) ∼ = W .Furthermore, we have S ( V ) ∼ = V ⊕ V , but S ( W ) ∼ = W ⊕ W .In our main results for ∧ ( W n ) and S ( W n ), we will give a closed formula fortheir decomposition into indecomposable summands. For this, we will need thefollowing definition from [GPX15, p. 231]. Definition 1.5.
The consecutive-ones binary expansion of an integer n > n = P ≤ i ≤ r ( − i +1 β i such that β > · · · > β r ≥ r isminimal.For example, we have consecutive-ones binary expansions 3 = 2 − , 4 = 2 ,5 = 2 − + 2 , 6 = 2 − , and 7 = 2 − . Note that for any consecutive-onesbinary expansion, we have β r − > β r + 1 if r > n , Glasby, Prager and Xia havegiven an explicit expression for the indecomposable summands of V n ⊗ V n and theirmultiplicities [GPX15, Theorem 15]. By Proposition 1.4, this also gives us thedecomposition of W n ⊗ W n into indecomposable summands.In Section 3, we give a different proof of [GPX15, Theorem 15] by constructinga Jordan basis for the action of e on W n ⊗ W n . This Jordan basis can be used tofind Jordan bases for the action of e on ∧ ( W n ) and S ( W n ) as well, allowing usto compute the indecomposable summands of ∧ ( W n ) and S ( W n ) explicitly. Thisleads to the following results, which will be proven in Section 4. MIKKO KORHONEN
Theorem 1.6.
Let n > be an integer, with consecutive-ones binary expansion n = P ≤ i ≤ r ( − i +1 β i , where β > · · · > β r ≥ . For ≤ k ≤ r with β k > ,define d k := 2 β k − + P k W d k βk − as w q -modules. Theorem 1.7.
Let n > be an integer, and let β > · · · > β r ≥ and d k be as inTheorem 1.6. Then S ( W n ) ∼ = W ⌈ n/ ⌉ ⊕ M ≤ k ≤ rβ k > W d k βk as w q -modules. As a corollary of Theorem 1.6 and Theorem 1.7, we also get reciprocity theoremsfor the decomposition of S ( W n ) and ∧ ( W n ), analogously to Theorem 1.2 andTheorem 1.3 above. The proofs will be given in Section 4. Theorem 1.8.
Suppose that q/ < n ≤ q . Then we have ∧ ( W n ) ∼ = ∧ ( W q − n ) ⊕ W n − q/ q − as w q -modules. Theorem 1.9.
Suppose that q/ < n ≤ q . Then we have S ( W n ) ∼ = S ( W q − n ) ⊕ W n − q/ q ⊕ W n − q/ as w q -modules.Remark . As a corollary of Theorems 1.2 - 1.3, one can also give explicit expres-sions for the decompositions of ∧ ( V n ) and S ( V n ) in terms of the consecutive-onesbinary expansion of n . We omit the details, but the main observation to make isthat if q = 2 β is the first term in the consecutive-ones binary expansion of n , then q/ < n ≤ q .We end this introduction with the following table of examples, which illustratesTheorems 1.2 - 1.3 and 1.6 - 1.9. Table 1.
Exterior and symmetric squares of V n and W n . n ∧ ( V n ) S ( V n ) ∧ ( W n ) S ( W n )1 0 V W V V ⊕ V W W ⊕ W V V ⊕ V W W ⊕ W V ⊕ V V ⊕ V W W ⊕ W V ⊕ V V ⊕ V ⊕ V W ⊕ W W ⊕ W ⊕ W V ⊕ V ⊕ V V ⊕ V ⊕ V W ⊕ W W ⊕ W ⊕ W V ⊕ V V ⊕ V W W ⊕ W V ⊕ V V ⊕ V W W ⊕ W V ⊕ V ⊕ V V ⊕ V ⊕ V W ⊕ W W ⊕ W ⊕ W Decomposition of S ( V n )In this section, we will prove Theorem 1.3, which gives a recursive description forthe decomposition of S ( V n ) into indecomposable summands. As mentioned in theintroduction, the proof follows essentially the same steps as the proof of Theorem1.2 in [GL06].Let G be a cyclic 2-group of order q > g , and let H = h g i be the unique subgroup of index 2 in G . As in the introduction, we set V = 0and denote the indecomposable K [ G ]-modules by V , . . . , V q . Similarly we willset U = 0 and denote the indecomposable K [ H ]-modules by U , . . . , U q/ , wheredim U i = i for all 1 ≤ i ≤ q/ K [ G ]-module V to H will be denoted by V H . For a K [ H ]-module U , we denote the induced module of U from H to G by U G := K [ G ] ⊗ K [ H ] U .A basic fact we will use in this section without mention is that U Gs ∼ = V s for all 1 ≤ s ≤ q/
2. This follows either by a direct calculation or by Green’sindecomposability theorem [Gre59, Theorem 8].We will denote the tensor induced module of U from H to G by U ⊗ G [CR90, § U ⊗ G = U ⊗ U as a K [ H ]-module, and the action of G on U ⊗ G is defined by g · ( v ⊗ w ) = g w ⊗ v for all v, w ∈ U .We begin with a series of lemmas which are similar (or the same) as those in[GL06]. After this we will proceed with the proof of Theorem 1.3. Lemma 2.1 ([GL06, Lemma 5]) . Let U be a K [ H ] -module. Then ∧ ( U G ) ∼ = ∧ ( U ) G ⊕ U ⊗ G as K [ G ] -modules. Lemma 2.2.
Let U be a K [ H ] -module. Then S ( U G ) ∼ = S ( U ) G ⊕ U ⊗ G as K [ G ] -modules.Proof. Let Z be the subspace of S ( U G ) spanned by (1 ⊗ v )(1 ⊗ w ) and ( g ⊗ v )( g ⊗ w )for v, w ∈ U , and let Z be the subspace spanned by (1 ⊗ v )( g ⊗ w ) for v, w ∈ U .We have S ( U G ) = Z ⊕ Z . Arguing as in [GL06, proof of Lemma 5], we see that Z and Z are G -submodules, with Z ∼ = S ( U ) G and Z ∼ = U ⊗ G . (cid:3) Lemma 2.3 ([GL06, Lemma 6]) . Let V be a K [ G ] -module with V H ∼ = L ≤ j ≤ q/ U r j j .Suppose that for all odd ≤ j < q/ , we have r j ∈ { , } . Then the isomorphismtype of V is uniquely determined by V H . Lemma 2.4.
Let n > and suppose that Theorem 1.3 holds for n . Write S ( V n ) ∼ = L j ≥ V r j j , where r j ≥ . Then the following statements hold: (i) r = 1 if n ≡ , and r = 0 otherwise. (ii) Let j > be an odd integer. Then r j = 0 if n is even, and r j ∈ { , } if n isodd. MIKKO KORHONEN
Proof.
For n = 2 we have S ( V n ) ∼ = V ⊕ V and clearly the claim holds. Supposethen that n >
2. We have S ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ q ⊕ V q/ as K [ G ]-modules,so the claim is immediate from the observation in [GL06, p. 670]. (cid:3) Proof of Theorem 1.3.
By induction on n . For the base case n = 2 an easy calcu-lation shows that S ( V n ) ∼ = V ⊕ V , so the claim holds. Suppose then that n > Case 1: n is even. Write n = 2 s . We have V n ∼ = U Gs , so by Lemma 2.2(2.1) S ( V n ) ∼ = S ( U s ) G ⊕ U ⊗ Gs . By the induction assumption, we have S ( U s ) ∼ = ∧ ( U q/ − s ) ⊕ U s − q/ q/ ⊕ U q/ as K [ H ]-modules, and by [GL06, Corollary 4] we have U ⊗ Gs ∼ = U ⊗ Gq/ − s ⊕ V s − q/ q as K [ G ]-modules. Plugging these isomorphisms into (2.1), we get S ( V n ) ∼ = ∧ ( U q/ − s ) G ⊕ U ⊗ Gs ⊕ V n − q/ q ⊕ V q/ as K [ G ]-modules. Thus S ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ q ⊕ V q/ by Lemma 2.1. Case 2: n is odd. Write n = 2 s + 1. We have ( V n ) H = U s ⊕ U s +1 , so by (1.2) S ( V n ) H ∼ = S ( U s ) ⊕ S ( U s +1 ) ⊕ ( U s ⊗ U s +1 )as K [ H ]-modules. Exactly one of s and s + 1 is odd, so by Lemma 2.4 and [GL06,Corollary 2] we conclude that for all odd 1 ≤ j < q/
2, the multiplicity of U j in S ( V n ) H is either 0 or 1. Thus Lemma 2.3 applies and S ( V n ) is uniquelydetermined up to isomorphism by the restriction S ( V n ) H .Applying the induction assumption and Theorem 1.1 (ii), we get S ( U s ) ∼ = ∧ ( U q/ − s ) ⊕ U s − q/ q/ ⊕ U q/ S ( U s +1 ) ∼ = ∧ ( U q/ − s − ) ⊕ U s +1 − q/ q/ ⊕ U q/ U s ⊗ U s +1 ∼ = ( U q/ − s − ⊗ U q/ − s ) ⊕ U n − q/ q/ as K [ H ]-modules. Hence S ( V n ) H ∼ = ∧ ( U q/ − s ⊕ U q/ − s − ) ⊕ U n − qq/ ⊕ U q/ , and so S ( V n ) has the same restriction to H as ∧ ( V q − n ) ⊕ V n − q/ q ⊕ V q/ . Thus S ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ q ⊕ V q/ as K [ G ]-modules. (cid:3) A Jordan basis for W n ⊗ W n For this section, fix an integer n >
0, and let q > q ≥ n . Recall that for a nilpotent linear map e : V → V , a Jordan chain is a set ofnon-zero vectors { w, ew, . . . , e k w } , where k ≥ e k +1 w = 0. A Jordan basis forthe action of e on V is a basis of V which is a disjoint union of such Jordan chains.We denote V e := { v ∈ V : ev = 0 } In this section, we give an explicit description of the indecomposable summandsof W n ⊗ W n , in terms of the consecutive-ones binary expansion of n . This isessentially due to Glasby, Praeger, and Xia [GPX15, Theorem 15] — see Proposition1.4. We give a different proof by constructing a Jordan basis for the action of agenerator e of w q on W n ⊗ W n (Theorem 3.6). XTERIOR AND SYMMETRIC SQUARES IN CHARACTERISTIC TWO 7
Our construction of the Jordan basis is based on the following elementary lemmaconcerning Jordan chains of nilpotent linear maps.
Lemma 3.1.
Let e : V → V be a nilpotent linear map. Let { z , . . . , z t } be aset of linearly independent vectors from V e . Let k , . . . , k t ≥ be integers and w , . . . , w t ∈ V such that e k i w i = z i for all ≤ i ≤ t . Then { e j w i : 1 ≤ i ≤ t and ≤ j ≤ k i } is a set of linearly independent vectors.Proof. By induction on dim V . There is nothing to prove when dim V = 0, since inthis case t = 0. Suppose then that dim V > e k i w i = z i for all 1 ≤ i ≤ t , the image of { e k i − w i : 1 ≤ i ≤ t and k i > } in V /V e is linearly independent and lies in ( V /V e ) e . Thus by applying inductionon V /V e , it follows that the image of S = { e j w i : 1 ≤ i ≤ t and 0 ≤ j < k i } in V /V e is linearly independent. From this we conclude that S ∪ { z , . . . , z t } = { e j w i : 1 ≤ i ≤ t and 0 ≤ j ≤ k i } is a set of linearly independent vectors. (cid:3) For all that follows, we fix a generator e of w q and let v , . . . , v n be a basis of W n such that ev = 0 and ev i = v i − for all 1 < i ≤ n . For convenience of notation,we define v j = 0 for all j ≤ j > n .The action of e on W n ⊗ W n is given by f ⊗ id + id ⊗ f , where f is the action of e on W n . Thus an application of the binomial theorem shows that for all integers i, j ≤ n and k ≥
0, we have(3.1) e k · ( v i ⊗ v j ) = X ≤ t ≤ k (cid:18) kt (cid:19) v i − t ⊗ v j − k + t . For 1 ≤ s ≤ n , define z s := X ≤ i ≤ s v i ⊗ v s +1 − i . It is clear that e · z s = 0 for all 1 ≤ s ≤ n , and in fact we have the following. Lemma 3.2.
The set { z , . . . , z n } is a basis of ( W n ⊗ W n ) e .Proof. A straightforward calculation — see for example [Nor95, Lemma 2]. (cid:3)
Let n = P ≤ i ≤ r ( − i +1 β i be the consecutive-ones binary expansion of n , where β > · · · > β r ≥
0, and β r − > β r + 1 if r >
1. Define n k := X k ≤ i ≤ r ( − i + k β i for all 1 ≤ k ≤ r , and set n r +1 := 0. Note that n = n > n > · · · > n r > n r +1 = 0.The rest of this section proceeds as follows. Consider 1 ≤ s ≤ n and let 1 ≤ k ≤ r be the unique integer such that n k > n − s ≥ n k +1 . Using the next two lemmas, wewill construct w s ∈ W n ⊗ W n such that e βk − w s = z s . From this, an applicationof Lemma 3.1 will give us a Jordan basis for the action of e on W n ⊗ W n . Lemma 3.3.
Let β = β k > and ≤ s ≤ n with n − s ≥ n k +1 . Then there existsan integer s ≤ x ≤ n − s such that x ≡ β mod 2 β +1 . MIKKO KORHONEN
Proof.
First note that the claim holds if n − s ≥ β , since in this case the interval[ s, n − s ] contains a complete set of representatives modulo 2 β +1 . This fact willbe used throughout the proof, which we split into two cases: Case 1: k ≡ . In this case n = n k +1 + 2 β + n ′ β +1 for some n ′ ≥ β occurs in the binary expansion of s , so s = s ′′ + 2 β + s ′ β +1 for some 0 ≤ s ′ ≤ n ′ and 0 ≤ s ′′ < β . If n ′ > s ′ , then n − s > β and the claimholds. If n ′ = s ′ , then n − s = n k +1 − s ′′ , so s ′′ = 0 since n − s ≥ n k +1 . Thus wecan choose x = s ≡ β mod 2 β +1 .If 2 β does not occur in the binary expansion of s , then s = s ′′ + s ′ β +1 for some0 ≤ s ′ ≤ n ′ and 0 ≤ s ′′ < β . We have n − s = n k +1 − s ′′ + 2 β ≥ β − s ′′ . Thus wecan choose x = s + (2 β − s ′′ ) ≡ β mod 2 β +1 . Case 2: k . We have n = − n k +1 + 2 β + n ′ β +1 for some n ′ ≥ β occurs in the binary expansion of s . Then s = s ′′ + 2 β + s ′ β +1 for some 0 ≤ s ′ < n ′ and 0 ≤ s ′′ < β . If n ′ > s ′ + 1,then n − s > β and the claim holds. If n ′ = s ′ + 1, we have n − s = 2 β +1 − n k +1 − s ′′ ≥ β +1 − s ′′ − β − since n k +1 ≤ β − . It follows then that n − s ≥ β − , so2( n − s ) ≥ (2 β +1 − s ′′ − β − ) + 2 β − = 2 β +1 − s ′′ . Hence we can pick x = s + (2 β +1 − s ′′ ) ≡ β mod 2 β +1 .Consider then the case where 2 β does not occur in the binary expansion of s ,so s = s ′′ + s ′ β +1 for some 0 ≤ s ′ ≤ n ′ and 0 ≤ s ′′ < β . If n ′ > s ′ , we have n − s > β , so assume that n ′ = s ′ . In this case n − s = 2 β − s ′′ − n k +1 . Since n − s ≥ n k +1 , it follows that 2 β − s ′′ ≥ n k +1 . Thus2( n − s ) = 2(2 β − s ′′ ) − n k +1 ≥ β − s ′′ , so we can choose x = s + (2 β − s ′′ ) ≡ β mod 2 β +1 . (cid:3) Lemma 3.4.
Let β = β k > and ≤ s ≤ n with n − s ≥ n k +1 . Then there existsan integer j ≥ such that the following hold: (i) s ≤ ⌊ s/ ⌋ + 2 β − + j β ≤ n , (ii) s ≤ ⌈ s/ ⌉ + 2 β − + j β ≤ n .Proof. If s is even, both (i) and (ii) are equivalent to s ≤ β + j β +1 ≤ n − s ,so the existence of such a j ≥ s is odd, then both(i) and (ii) hold if and only if s + 1 ≤ β + j β +1 ≤ n − s −
1. In this case, theexistence of such a j ≥ s and 2 n − s are odd. (cid:3) For the next lemma, we fix 1 ≤ k ≤ r and set β := β k . For 1 ≤ s ≤ n with n k > n − s ≥ n k +1 , we define a vector w s ∈ W n ⊗ W n as follows. If β = 0, we set w s = z s . If β >
0, we define w s := X − j ≤ j ≤ j v ⌊ s/ ⌋ +2 β − + j β ⊗ v ⌈ s/ ⌉ +2 β − − j β , where j ≥ Lemma 3.5.
Let ≤ s ≤ n with n k > n − s ≥ n k +1 . Then e β − w s = z s . XTERIOR AND SYMMETRIC SQUARES IN CHARACTERISTIC TWO 9
Proof. If β = 0, there is nothing to prove since w s = z s . Suppose then that β > (cid:0) β − t (cid:1) ≡ ≤ t ≤ β −
1. Thuswith (3.1), we get e β − · ( v i ⊗ v j ) = X ≤ t ≤ β − v i − t ⊗ v j − β +1+ t = X i − β +1 ≤ t ≤ i v t ⊗ v i + j − β +1 − t (3.2)for all integers i, j ≤ n . By Lemma 3.4 each summand in the definition of w s is ofthe form v i ⊗ v j for some i, j ≤ n , so by (3.2) we have e β − · w s = X ℓ ≤ t ≤ ℓ ′ v t ⊗ v s +1 − t , where ℓ = ⌊ s/ ⌋ + 2 β − − ( j + 1)2 β + 1 and ℓ ′ = ⌊ s/ ⌋ + 2 β − + j β .Thus in order to prove that e β − · w s = z s , it will suffice to show that ℓ ≤ ℓ ′ ≥ s . First note that the inequality ℓ ′ ≥ s is just Lemma 3.4 (i). Next, byLemma 3.4 (ii), we have ⌊ s/ ⌋ = s − ⌈ s/ ⌉ ≤ β − + j β . Thus ℓ ≤ (2 β − + j β ) + 2 β − − ( j + 1)2 β + 1 = 1 , which completes the proof of the lemma. (cid:3) We are now ready to prove the main result of this section.
Theorem 3.6.
For ≤ k ≤ r , define B k := { e j w s : n k > n − s ≥ n k +1 and ≤ j ≤ β k − } . Then B := ∪ ≤ k ≤ r B k is a Jordan basis for the action of e on W n ⊗ W n .Proof. By Lemma 3.5 and Lemma 3.1, the vectors in B are linearly independent. Toprove that B is a Jordan basis, it will suffice to show that | B | = dim W n ⊗ W n = n .To this end, note first that | B k | = 2 β k ( n k − n k +1 ) for all 1 ≤ k ≤ r . Furthermore,we have(3.3) n k − n k +1 = 2 β k + 2 X k
Theorem 3.7 ([GPX15, Theorem 15]) . For ≤ k ≤ r , define d k := 2 β k + P k
Let B ′ := { π ( w s ) : 1 ≤ s ≤ n even } and B ′ k := { e j π ( w s ) : s odd, n k > n − s ≥ n k +1 , and ≤ j ≤ β k − } for all ≤ k ≤ r . Then B ′ := S ≤ k ≤ r B ′ k is a Jordan basis for the action of e on S ( W n ) .Proof. Let B be the Jordan basis of W n ⊗ W n as in Theorem 3.6. We will beginby showing that(4.1) B ′ = π ( B ) \ { } , which implies that B ′ spans S ( W n ). To this end, first note that for 1 ≤ s ≤ n odd,we have π ( z s ) = π ( v ( s +1) / ⊗ v ( s +1) / ) = 0. Furthermore, the map π is invariantunder the action of e and e βk w s = z s by Lemma 3.5. We conclude then that for odd s with n k > n − s ≥ n k +1 , we have π ( e j w s ) = e j π ( w s ) = 0 for all 0 ≤ j ≤ β k − ≤ s ≤ n even, we have π ( w s ) = π ( v s/ β − ⊗ v s/ β − ) ∈ S ( W n ) e forsome β >
0, and thus π ( e j w s ) = 0 for all j >
0. This completes the proof of (4.1).Now to show that B ′ is a Jordan basis for the action of e on S ( W n ), it willsuffice to show that | B ′ | = dim S ( W n ) = n ( n + 1) /
2. Consider first the case where n is even. Then X ≤ k ≤ r | B ′ k | = X ≤ k ≤ r β k ( n k − n k +1 )2 = n / | B ′ | = n/
2, so | B | = n ( n + 1) /
2. Suppose nextthat n is odd. In this case β r = 0 and | B r | = 1, so X ≤ k ≤ r | B ′ k | = 1 + X ≤ k ≤ r − β k ( n k − n k +1 )2 = ( n + 1) / | B ′ | = ( n − /
2, we again get | B | = n ( n + 1) /
2, as claimed. (cid:3)
We can now prove our main result for S ( W n ). Proof of Theorem 1.7.
This is just a matter of counting Jordan chains in the basis B ′ described in Theorem 4.1. The Jordan block sizes that occur are 2 β k for 1 ≤ k ≤ r . For β k >
0, the multiplicity of 2 β k is | B ′ k | / β k = ( n k − n k +1 ) / d k .Now what remains is to count the number of blocks of size 1. If n is even, this isgiven by | B ′ | = n/
2. If n is odd, we have β r = 0 and the multiplicity is given by | B ′ | + | B ′ r | = ( n + 1) / (cid:3) As a corollary of Theorem 4.1, we also get the following.
Corollary 4.2.
The action of e has n Jordan blocks on S ( W n ) . XTERIOR AND SYMMETRIC SQUARES IN CHARACTERISTIC TWO 11
Proof.
Let B and B ′ be the Jordan bases described in Theorem 3.6 and Theorem4.1, respectively. Each Jordan chain in B is mapped by π to a Jordan chain in B ′ .Therefore e has the same number of Jordan blocks on W n ⊗ W n and S ( W n ), andthe claim follows from Lemma 3.2. (cid:3) We are now ready to prove the rest of our main results: the decompositiontheorem for ∧ ( W n ) (Theorem 1.6) and the recurrence relations for ∧ ( W n ) and S ( W n ) (Theorems 1.8 and 1.9). Proof of Theorem 1.6.
Let π ′ : S ( W n ) → ∧ ( W n ) be the map defined by π ′ ( vw ) = v ∧ w for all v, w ∈ W n . Then we have a short exact sequence0 → W [2] n → S ( W n ) π ′ −→ ∧ ( W n ) → w q -modules, where W [2] n is the subspace of S ( W n ) spanned by elements of theform v for v ∈ V . Now dim W [2] n = n and W [2] n is annihilated by the action of e ,so by Corollary 4.2 we have W [2] n = S ( W n ) e . Therefore(4.2) ∧ ( W n ) ∼ = S ( W n ) /S ( W n ) e as w q -modules.For any w q -module V , it is easy to see that if V ∼ = W r ⊕ · · · ⊕ W r t for someintegers r , . . . , r t >
0, then
V /V e ∼ = W r − ⊕ · · · ⊕ W r t − . Thus the claim followsfrom (4.2) and Theorem 1.7. (cid:3) Proof of Theorem 1.8.
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