Hesselink normal forms of unipotent elements in some representations of classical groups in characteristic two
aa r X i v : . [ m a t h . G R ] M a y HESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS INSOME REPRESENTATIONS OF CLASSICAL GROUPS INCHARACTERISTIC TWO
MIKKO KORHONEN
Abstract.
Let G be a simple linear algebraic group over an algebraicallyclosed field K of characteristic two. Any non-trivial self-dual irreducible K [ G ]-module W admits a non-degenerate G -invariant alternating bilinear form, thusgiving a representation f : G → Sp( W ). In the case where G = SL n ( K ) and W has highest weight ̟ + ̟ n − , and in the case where G = Sp n ( K ) and W has highest weight ̟ , we determine for every unipotent element u ∈ G the conjugacy class of f ( u ) in Sp( W ). As a part of this result, we describe theconjugacy classes of unipotent elements of Sp( V ) ⊗ Sp( V ) in Sp( V ⊗ V ). Introduction
Let G be a simple algebraic group over an algebraically closed field K of charac-teristic p >
0, and let f : G → SL( W ) be a non-trivial finite-dimensional rationalirreducible representation. Recall that an element u ∈ G is unipotent , if its imageunder every rational representation of G is a unipotent linear map. Equivalently u is unipotent if it has order p α for some α ≥ Problem 1.1.
Let u ∈ G be a unipotent element. What is the Jordan normal formof f ( u ) ? There are relatively few cases where a complete answer to Problem 1.1 is known.Computations done by Lawther [Law95, Law98] give an answer in most cases where G is simple of exceptional type and W is either minimal-dimensional or the adjointmodule. Consider the case where G is a simple classical group (SL( V ), Sp( V ), orSO( V )). For almost all irreducible representations f with dim W ≤ (rank G ) / p is good for G . (For a simple classical group G , the prime p is good for G if G = SL( V ) or p >
2. Otherwise p is bad for G .)In this paper we will extend the results of [Kor19] to the case where p is bad for G , but our main concern will be a somewhat more general problem in characteristictwo. Suppose from now on that p = 2, and suppose that W is self-dual. By Fong’slemma [Fon74], there exists a non-degenerate G -invariant alternating bilinear form b on W , which is unique up to scalar multiples. Thus we may consider f as arepresentation f : G → Sp(
W, b ). In the main results of this paper, we give asolution to the following problem in some special cases. Date : May 15, 2020.2010
Mathematics Subject Classification.
Primary 20G05.The author was supported by a postdoctoral fellowship from the Swiss National Science Foun-dation (grant number P2ELP2 181902).
Problem 1.2.
Let u ∈ G be a unipotent element. What is conjugacy class of f ( u ) in Sp(
W, b ) ?Remark . We note here that although in odd characteristic it is true that the Jor-dan normal form of u ∈ Sp(
W, b ) determines the conjugacy class of u in Sp( W, b )[Ger61, Proposition 2 of Chapter II], this no longer holds in characteristic two.Knowing the Jordan normal form of f ( u ) is essential in the solution of Problem1.2, but one also needs specific information about the action of u on W with respectto the bilinear form b . Motivation.
One basic motivation for considering Problem 1.1 and Problem 1.2 isin the problem of determining the fusion of unipotent classes in maximal subgroupsof simple algebraic groups. That is, for a simple algebraic group Y and a maximalsubgroup X < Y , what is the Y -conjugacy class of each unipotent element u ∈ X ?Here solutions to Problem 1.1 and Problem 1.2 provide answers in the case where Y is of classical type and X is an irreducible simple subgroup.Solutions of Problem 1.1 in specific cases have found applications in variouscontexts, see for example [Law95], [Law09, Section 3], and [TZ02]. It seems thatso far there are very few results on Problem 1.2 in the literature, although somecomputations are contained in the PhD thesis of the present author. In this paper,we solve Problem 1.2 in the smallest cases where the answer is not known. Asan application of our results, in the final section of this paper we classify somesimple subgroups of Sp( V, b ) that contain distinguished unipotent elements. (Aunipotent element in a simple algebraic group is distinguished , if its centralizerdoes not contain a non-trivial torus.)Let λ be the highest weight in W . As the main result of this paper, we will solveProblem 1.2 in the following cases: • G = SL( V ) and λ = ̟ + ̟ n − , where n = dim V (Theorem A). • G = Sp( V, b ) and λ = ̟ (Theorem B).In order to describe our main results in more detail, we will first have to de-scribe how the unipotent conjugacy classes are classified in the symplectic groups.Throughout we will describe the conjugacy class of u in Sp( W, b ) using the
Hes-selink normal form described in [Hes79]. For the purposes of this introduction wewill give a brief description, a more detailed exposition of the relevant results andconcepts is given in Sections 5 and 6.Let g be a generator of a cyclic 2-group of order q , and denote the group algebraof h g i by K [ g ]. Then there exist a total of q indecomposable K [ g ]-modules V , . . . , V q up to isomorphism, where dim V i = i and g acts on V i as a unipotent i × i Jordanblock. For a K [ g ]-module V we denote V = 0 and V n = V ⊕ · · · ⊕ V ( n copies)for all n >
0. Then notation V ∼ = ⊕ d ≥ V r d d can be used to say that in the action of g on V , a Jordan block of size d occurs with multiplicity r d .For any finite-dimensional K [ g ]-module V equipped with a g -invariant alternat-ing bilinear form b (not necessarily non-degenerate), we define a map ε V,b : Z ≥ →{ , } by ε V,b ( d ) = ( , if b (( g − d − v, v ) = 0 for all v ∈ V such that ( g − d v = 0 . , otherwise.We note that ε V,b ( d ) = 1 is only possible for d even (Lemma 6.10 (i)). ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 3
Let b be a non-degenerate alternating bilinear form on a finite-dimensional K -vector space V . For a unipotent element u ∈ Sp(
V, b ), as K [ u ]-modules we have V ∼ = V n d ⊕ · · · ⊕ V n t d t , where 0 < d < · · · < d t and n i > ≤ i ≤ t (Jordan normal form). It turns out (Theorem 6.7) that the Sp( V, b )-conjugacyclass of u ∈ Sp(
V, b ) is uniquely determined by the integers d i , n i , and ε V,b ( d i ).In our main results, we will describe these integers for f ( u ), thus describing theconjugacy class of f ( u ) in Sp( W, b ).For our first main result, let G = SL( V ) and λ = ̟ + ̟ n − . To setup thestatement, we need the following facts which are proven in Section 8. One can showthat V ⊗ V ∗ admits a non-zero alternating G -invariant bilinear form b V which isunique up to scalar multiples, and furthermore b V is non-degenerate if and onlyif dim V is even. We can identify W = L G ( ̟ + ̟ n − ) = Z ⊥ /Z , where Z is theunique 1-dimensional G -submodule of V ⊗ V ∗ . In all cases, the bilinear form b V induces a non-degenerate G -invariant alternating bilinear form on Z ⊥ /Z .Given a unipotent element u ∈ G , we can write V ∼ = V n d ⊕ · · · ⊕ V n t d t as K [ u ]-modules, where 0 < d < · · · < d t and n i > ≤ i ≤ t . There exists arecursive algorithm — involving only calculations with the integers d i and n i —for computing the indecomposable summands of V ⊗ V ∗ and their multiplicities(Theorem 4.1 and Remark 4.2). That is, we can assume V ⊗ V ∗ ∼ = ⊕ d ≥ V λ ( d ) d ,where λ ( d ) ≥ f ( u ) in Sp( W, b V ), and whendim V is even, the conjugacy class of the image of u in Sp( V ⊗ V ∗ , b V ). The resultis given in terms of the integers d i and λ ( d ). The theorem also includes the Jordannormal form of f ( u ), which was described before in [Kor19, Theorem 6.1]. Theorem A.
Let G = SL( V ) , where dim V = n for some n ≥ . Let u ∈ G beunipotent and V ∼ = V d ⊕ · · · ⊕ V d t as K [ u ] -modules, where t ≥ and d r ≥ forall ≤ r ≤ t . Set α = ν (gcd( d , . . . , d t )) . Suppose that V ⊗ V ∗ ∼ = ⊕ d ≥ V λ ( d ) d and L G ( ̟ + ̟ n − ) ∼ = ⊕ d ≥ V λ ′ ( d ) d as K [ u ] -modules, where λ ( d ) , λ ′ ( d ) ≥ for all d ≥ .Set ε := ε V ⊗ V ∗ ,b V and ε ′ := ε L G ( ̟ + ̟ n − ) ,b V .Then the values of λ ′ are given in terms of λ as follows: (i) If ∤ n , then λ ′ (1) = λ (1) − and λ ′ ( d ) = λ ( d ) for all d > . (ii) If | n and α = 0 , then λ ′ (1) = λ (1) − and λ ′ ( d ) = λ ( d ) for all d > . (iii) If | n and α > : (a) If | n α , then λ ′ (2 α ) = λ (2 α ) − , λ ′ (2 α −
1) = 2 , and λ ′ ( d ) = λ ( d ) forall d = 2 α , α − . (b) If α > and ∤ n α , then λ ′ (2 α ) = λ (2 α ) − , λ ′ (2 α −
2) = 1 , and λ ′ ( d ) = λ ( d ) for all d = 2 α , α − . (c) If α = 1 and ∤ n , then λ ′ (2) = λ (2) − and λ ′ ( d ) = λ ( d ) for all d = 2 .Furthermore, the values of ε and ε ′ are given as follows: (iv) ε ( d ) = 1 if and only if d = 2 β for some β > occurring in the consecutive-ones binary expansion (Definition 4.6) of d r for some ≤ r ≤ t . (v) If (iii)(b) holds, then ε (2 α −
2) = 0 , ε ′ (2 α −
2) = 1 , and ε ′ ( d ) = ε ( d ) for all d = 2 α − . (vi) If (iii)(b) does not hold, then ε ′ ( d ) = ε ( d ) for all d ≥ . For our other main result, let G = Sp( V, b ) and λ = ̟ . To set up the statement,we need the following facts from Section 9 — these are very much analogous to the MIKKO KORHONEN case with SL( V ) above. One can show that ∧ ( V ) admits a non-zero G -invariantalternating bilinear form a V which is unique up to scalar multiples, and furthermore a V is non-degenerate if and only if dim V / W = L G ( ̟ ) = Q ⊥ /Q , where Q is the unique 1-dimensional G -submodule of ∧ ( V ). In all cases,the bilinear form a V induces a non-degenerate G -invariant alternating bilinear formon Q ⊥ /Q .Let u ∈ G be a unipotent element. Let d , . . . , d t be the Jordan block sizesof u , and let n i > d i . To compute the Jordanblock sizes of u on ∧ ( V ), there exists a recursive algorithm which only involvescomputations with the integers d i and n i (Theorem 4.8 and Remark 4.11). Hencewe can assume that ∧ ( V ) ∼ = ⊕ d ≥ V λ ( d ) d , where the integers λ ( d ) ≥ f ( u ) in Sp( W, a V ), and whendim V / u in Sp( ∧ ( V ) , a V ). The resultis given in terms of the integers d i , n i , and λ ( d ). Theorem B.
Let G = Sp( V, b ) , where dim V = 2 n for some n ≥ . Let u ∈ G be unipotent. For t ≥ , let d , . . . , d t be the Jordan block sizes d of u such that ε V,b ( d ) = 0 , and for s ≥ let d t +1 , . . . , d t + s be the Jordan block sizes d of u suchthat ε V,b ( d ) = 1 . Write V ∼ = V n d ⊕ · · ·⊕ V n t d t ⊕ V n t +1 d t +1 ⊕ · · ·⊕ V n t + s d t + s as K [ u ] -modules,where n r > for all ≤ r ≤ t + s .Set α = ν (gcd( d , . . . , d t + s )) . Suppose that ∧ ( V ) ∼ = ⊕ d ≥ V λ ( d ) d and L G ( ̟ ) ∼ = ⊕ d ≥ V λ ′ ( d ) d as K [ u ] -modules, where λ ( d ) , λ ′ ( d ) ≥ for all d ≥ . Set ε := ε ∧ ( V ) ,a V and ε ′ := ε L G ( ̟ ) ,a V .Then the values of λ ′ are given in terms of λ by the rules (i) – (iii) of TheoremA. Furthermore, the values of ε and ε ′ are given as follows: (iv) We have ε ( d ) = 1 if and only if one of the following conditions holds: (a) d = 2 β for some β > occurring in the consecutive-ones binary expan-sion (Definition 4.6) of d r for some ≤ r ≤ t . (b) d > occurs as a Jordan block size of ∧ ( V d r ) for some t + 1 ≤ r ≤ t + s . (c) d = d ′ β +1 , where: • β = ν ( d r ) = ν ( d r ′ ) for some t + 1 ≤ r ≤ r ′ ≤ t + s ; • n r > if r = r ′ ; • d ′ is the unique odd Jordan block size in V d r / β ⊗ V d r ′ / β (Lemma4.3). (v) If ∤ n or α = 0 , then ε ( d ) = ε ′ ( d ) for all d ≥ . (vi) If | n and α > , then ε ( d ) = ε ′ ( d ) for all d = 2 α , α − , and: (a) ε (2 α ) = 1 . (b) ε ′ (2 α ) = 1 if and only if ν ( d r ) = α for some ≤ r ≤ t . (c) If α > , then ε (2 α −
2) = 0 . (d) If α > , then ε ′ (2 α −
2) = 1 if and only if ∤ n α . As part of our main result for G = Sp( V, b ), we will also have to resolve thefollowing problem.
Problem 1.4.
Let u ∈ Sp( V , b ) and u ∈ Sp( V , b ) be unipotent. What is theconjugacy class of u ⊗ u in Sp( V ⊗ V , b ⊗ b ) ? Here b ⊗ b is the usual product form on V ⊗ V given by b and b , see Definition5.7. We will give a complete solution to Problem 1.4 in Section 7. ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 5
Remark . Let G = Sp( V, b ), where dim
V >
4. In characteristic two, we alwayshave SO(
V, q ) < G , where q is a quadratic form on V such that q ( v + w ) + q ( v ) + q ( w ) = b ( v, w ) for all v, w ∈ V . It follows for example from [Sei87, Theorem4.1] that the restriction of the irreducible K [ G ]-module with highest weight ̟ to SO( V, q ) remains irreducible. Furthermore, the unipotent conjugacy classes ofSO(
V, q ) can be described in terms of unipotent conjugacy classes of G [LS12,Proposition 6.22]. Thus from our main result for G = Sp( V, b ) (Theorem B), it isstraightforward to deduce the corresponding result for SO(
V, q ). Acknowledgements.
The author would like to acknowledge the anonymous refereefor their useful comments and suggestions.2.
Notation
We fix the following notation and terminology, some of which was already men-tioned in the introduction. Throughout the text, let K be an algebraically closedfield. We will always assume that K has characteristic two. For an integer n ∈ Z ,we will denote the element n · K of K by n , and it will be clear from the contextwhen n is considered as an element of K .For a K -vector space V and non-negative integer n , we use the notation V n forthe direct sum V ⊕ · · · ⊕ V , where V occurs n times. Note that V = 0.Let u be a generator of a cyclic 2-group of order q . We will denote the groupalgebra of h u i over K by K [ u ]. Recall that K [ u ] has exactly q indecomposablemodules V , . . . , V q up to isomorphism, where dim V i = i and u acts on V i asa full i × i Jordan block. For convenience of notation, we denote V = 0. Anynon-zero K [ u ]-module V has a decomposition V ∼ = V n d ⊕ · · · ⊕ V n t d t , where t ≥ < d < · · · < d t , and n i > i (Jordan normal form). We call the d i the Jordan block sizes of u on V , and n i is the multiplicity of d i in V .When considering K [ u ]-modules, we will denote by X the element u − K [ u ].Let Y ∈ K [ u ]. If a K [ u ]-module V has a K [ u ]-submodule W , we will usually usethe notation Y W for the linear map Y W : W → W induced by the action of Y on W , and similarly Y V/W for the linear map Y V/W : V /W → V /W induced by theaction of Y on V /W .Throughout the text G will always denote a group. Any K [ G ]-module that weconsider will be finite-dimensional. If a K [ G ]-module V has a filtration V = W ⊃ W ⊃ · · · ⊃ W t ⊃ W t +1 = 0 with soc( V /W i +1 ) = W i /W i +1 ∼ = Z i for all 1 ≤ i ≤ t ,we will denote this by V = Z | Z | · · · | Z t . Let G be a group and H < G a subgroup.We denote the restriction of a K [ G ]-module V to H by Res GH ( V ). For a K [ H ]-module W , the induced module of W from H to G is Ind GH ( W ) := K [ G ] ⊗ K [ H ] W .A bilinear form b on a vector space V is non-degenerate , if its radical rad b = { v ∈ V : b ( v, w ) = 0 for all w ∈ V } is zero. For a subspace W of V , we call W totally singular with respect to b if b ( w, w ′ ) = 0 for all w, w ′ ∈ W . We say that b is alternating , if b ( v, v ) = 0 for all v ∈ V , and symmetric if b ( v, w ) = b ( w, v )for all v, w ∈ V . Note that since we are working over a field of characteristic two,any alternating bilinear form is also symmetric. If V is a K [ G ]-module, then b is G -invariant if b ( gv, gw ) = b ( v, w ) for all g ∈ G and v, w ∈ V . For a non-degeneratealternating bilinear form b on V , we denote Sp( V, b ) = { g ∈ GL( V ) : b ( gv, gw ) = b ( v, w ) for all v, w ∈ V } . MIKKO KORHONEN
Suppose that G is a simple linear algebraic group over K . In the context ofalgebraic groups, the notation that we use will be as in [Jan03]. For basic termi-nology and results on algebraic groups, see [Hum75]. We note however that notmuch will be needed from the theory of algebraic groups. For the most part, theonly algebraic groups that appear in this paper are G = SL( V ) or G = Sp( V, b ).When G is an algebraic group, by a K [ G ]-module we will always mean a finite-dimensional rational K [ G ]-module. We fix a maximal torus T of G with charactergroup X ( T ), and a base ∆ = { α , . . . , α ℓ } for the root system of G , where ℓ =rank G . Here we use the standard Bourbaki labeling of the simple roots α i , asgiven in [Hum72, 11.4, p. 58]. We denote the dominant weights with respect to ∆by X ( T ) + , and the fundamental dominant weight corresponding to α i is denotedby ̟ i . For a dominant weight λ ∈ X ( T ) + , we denote the rational irreducible K [ G ]-module with highest weight λ by L G ( λ ).For a simple linear algebraic group G ≤ GL( V ), an element u ∈ G is unipotent ,if it is unipotent as a linear transformation on V . That is, if ( u − V ) n = 0 forsome n >
0. Since char K = 2, an equivalent definition is that u ∈ G is unipotentif and only if it has order 2 k for some k ≥ a and b we denote by (cid:0) ab (cid:1) the usual binomial coefficient,using the convention that (cid:0) ab (cid:1) = 0 if a < b . We denote by ν the 2-adic valuationon the integers, so ν ( a ) is the largest integer k ≥ k divides a .3. Preliminaries
In this section, we list some preliminary results needed in the paper. All of theresults in this section are well known, and furthermore the results and their proofsgeneralize to arbitrary characteristic p >
0. We begin with some basic results aboutunipotent linear maps.
Lemma 3.1.
Let u be a generator of a cyclic -group of order q , and suppose that α ≤ q . For an integer < n ≤ q , write n = a α + r for ≤ r < α and a ∈ Z .Then Res h u ih u α i ( V n ) ∼ = V ra +1 ⊕ V α − ra . Proof.
Let e , . . . , e n be a basis of V n such that ue = e and ue i = e i + e i − forall 1 < i ≤ n . Set e j = 0 for j ≤ j > n . Now ( u − k e i = e i − k for all k ≥ i >
0. Since ( u − α = u α −
1, it follows that(3.1) u α e i = e i + e i − α for all 1 ≤ i ≤ n . For all 1 ≤ i ≤ α , define W i to be the subspace spanned by { e i + j α } j ≥ . Then V = W ⊕ · · · ⊕ W α . Furthermore, from (3.1) we find that each W i is u α -invariant and as K [ u α ]-modules W i ∼ = V a +1 for 1 ≤ i ≤ r and W i ∼ = V a for r < i ≤ α . From this the lemma follows. (cid:3) Lemma 3.2.
Let u be a generator of a cyclic -group of order q , and suppose that α ≤ q . Then Ind h u ih u α i ( V n ) ∼ = V α n for all < n ≤ q/ α .Proof. The lemma is an immediate consequence of Green’s indecomposability the-orem [Gre59, Theorem 8]. For an elementary proof, let 0 < n ≤ q/ α and set ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 7 V = Ind h u ih u α i ( V n ). To prove the lemma, it will suffice to show that the u -fixedpoint space of V is one-dimensional. Let W = 1 ⊗ V n , so V = M ≤ i ≤ α − u i W, where as K [ u α ]-modules u i W ∼ = V n for all 0 ≤ i ≤ α −
1. The u α -fixed pointspace of W is one-dimensional, spanned by some w ∈ W . Then for all 0 ≤ i ≤ α − u α -fixed point space of u i W is spanned by u i w . From this it easily followsthat the u -fixed point space of V is spanned by P ≤ i ≤ α − u i w . (cid:3) Lemma 3.3 ([Kor19, Lemma 3.3]) . Let u ∈ GL( V ) be unipotent and denote X = u − . Suppose that W ⊆ V is a subspace invariant under u such that dim V /W = 1 .Write V ∼ = ⊕ d ≥ V λ ( d ) d and W ∼ = ⊕ d ≥ V λ ′ ( d ) d as K [ u ] -modules, where λ ( d ) , λ ′ ( d ) ≥ for all d ≥ .Let m ≥ be such that Ker X m − ⊆ W and Ker X m W . Then: (i) if m = 1 , we have λ ′ (1) = λ (1) − and λ ′ ( d ) = λ ( d ) for all d > . (ii) if m > , we have λ ′ ( m ) = λ ( m ) − , λ ′ ( m −
1) = λ ( m − , and λ ′ ( d ) = λ ( d ) for all d = m, m − . Lemma 3.4.
Let u ∈ GL( V ) be unipotent and denote X = u − . Suppose that W ⊆ V is a subspace invariant under u such that dim W = 1 . Write V ∼ = ⊕ d ≥ V λ ( d ) d and V /W ∼ = ⊕ d ≥ V λ ′ ( d ) d as K [ u ] -modules, where λ ( d ) , λ ′ ( d ) ≥ for all d ≥ .Let m ≥ be such that Im X m − ⊇ W and Im X m W . Then: (i) if m = 1 , we have λ ′ (1) = λ (1) − and λ ′ ( d ) = λ ( d ) for all d > . (ii) if m > , we have λ ′ ( m ) = λ ( m ) − , λ ′ ( m −
1) = λ ( m − , and λ ′ ( d ) = λ ( d ) for all d = m, m − .Proof. It is clear that Im X iV ⊆ W for all 0 ≤ i ≤ m − X iV ∩ W = 0for all i ≥ m . For all i ≥
0, we have Im X iV/W ∼ = Im X i / Im X i ∩ W as vectorspaces, so we conclude that rank X iV/W = rank X iV − ≤ i ≤ m − X iV/W = rank X iV for all i ≥ m . Now the claim follows from [Kor19, Lemma3.2]. (cid:3) The following results are used to construct the irreducible representations thatwe consider in our main results.
Lemma 3.5 ([McN98, Proposition 4.6.10]) . Let G = SL( V ) , where dim V = n forsome n ≥ . Then as K [ G ] -modules, we have V ⊗ V ∗ ∼ = ( L G ( ̟ + ̟ n − ) ⊕ L G (0) , if ∤ n,L G (0) | L G ( ̟ + ̟ n − ) | L G (0) , if | n. Lemma 3.6 ([Sei87, 1.14, 8.1 (c)], [McN98, Lemma 4.8.2]) . Let G = Sp( V, b ) ,where dim V = 2 n for some n ≥ . Then as K [ G ] -modules, we have ∧ ( V ) ∼ = ( L G ( ̟ ) ⊕ L G (0) , if ∤ n,L G (0) | L G ( ̟ ) | L G (0) , if | n. MIKKO KORHONEN Decomposition of tensor products and exterior squares
In this section, we give results on the decomposition of tensor products andexterior squares of unipotent linear maps. Throughout, we let u be a generatorof a cyclic 2-group of order q >
1, and denote the indecomposable K [ u ]-modulesby V , . . . , V q as defined in Section 2. A recursive algorithm for calculating thedecomposition of V m ⊗ V n into indecomposable summands is given by the followingtheorem, see for example [Gre62, (2.5a)] and [GL06, Lemma 1] for a proof. Theorem 4.1.
Let < m ≤ n ≤ q and α ≤ n < α +1 . Then the followingstatements hold: (i) If m + n > α +1 , then V m ⊗ V n ∼ = V m + n − α +1 α +1 ⊕ ( V α +1 − n ⊗ V α +1 − m ) . (ii) If n = 2 α , then V m ⊗ V n = V m α . (iii) If n > α and m + n ≤ α +1 , then V m ⊗ V n ∼ = V α +1 − d m ⊕ · · · ⊕ V α +1 − d ,where V m ⊗ V α +1 − n ∼ = V d ⊕ · · · ⊕ V d m .Remark . Taking tensor products of K [ u ]-modules is an additive functor, sousing Theorem 4.1 one can decompose any tensor product of two K [ u ]-modulesinto indecomposable summands. Lemma 4.3.
Let m and n be odd integers such that < m ≤ n ≤ q . Suppose that V m ⊗ V n ∼ = V d ⊕ · · · ⊕ V d t , where d i > for all i . There exists a unique i such that d i is odd.Proof. We prove the lemma by induction on n . The case n = 1 is obvious. Supposethen that 0 < m ≤ n are odd integers and n >
1. Let α > α < n < α +1 . If m + n > α +1 , by Theorem 4.1 (i) V m ⊗ V n ∼ = V m + n − α +1 α +1 ⊕ ( V α +1 − n ⊗ V α +1 − m )so the claim follows by applying induction on the tensor product V α +1 − n ⊗ V α +1 − m .The other possibility is that m + n ≤ α +1 , in which case by Theorem 4.1 (iii) wehave V m ⊗ V n ∼ = V α +1 − d m ⊕ · · · ⊕ V α +1 − d , where V m ⊗ V α +1 − n ∼ = V d ⊕ · · · ⊕ V d m .Thus the claim follows by applying induction on V m ⊗ V α +1 − n . (cid:3) Remark . One can also describe the unique odd Jordan block size of Lemma4.3 explicitly. Let 0 < m ≤ n ≤ q be odd integers. Write m = P ti =0 a i i and n = P ti =0 b i i , where a i , b i ∈ { , } for all 0 ≤ i ≤ t . We shall omit the proof fromthis paper, but one can show that the unique odd Jordan block size in V m ⊗ V n isequal to n + P ti =1 a i ( − b i i . Example . To give an example of Theorem 4.1 in a small case, consider V m ⊗ V n for m = 3. This particular example will also be useful later (Example 7.7). In anycase, with Theorem 4.1 it is easy to show that for all n ≥ V ⊗ V n ∼ = V n , if n ≡ .V n − ⊕ V n +2 , if n ≡ .V n − ⊕ V n ⊕ V n +2 , if n ≡ .V n − ⊕ V n +1 , if n ≡ . It also clear from this decomposition that the conclusion of Lemma 4.3 holds inthis case.Following [GPX15, p. 231], we make the following definition.
ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 9
Definition 4.6.
The consecutive-ones binary expansion of an integer n > n = P ki =1 ( − i +1 e i such that e > · · · > e k ≥ k is minimal.The consecutive-ones binary expansion can be calculated as follows. Groupingtogether the blocks of consecutive ones in the binary expansion of n , we write n = P ℓi =1 P a i − j = b i j , where ℓ ≥ a > b > · · · > a ℓ > b ℓ ≥
0. Now P a i − j = b i j = 2 a i − b i , and the consecutive-ones binary expansion of n is given by n = P ℓi =1 (2 a i − b i ) if a ℓ > b ℓ + 1 and n = P ℓ − i =1 (2 a i − b i ) + 2 b ℓ if a ℓ = b ℓ + 1.For example, we have consecutive-ones binary expansions 3 = 2 − , 4 = 2 ,5 = 2 − + 2 , and 6 = 2 − . Note that e k − > e k + 1 for any consecutive-onesbinary expansion with k > < n ≤ q thedecomposition of V n ⊗ V n was described explicitly in terms of the consecutive-onesbinary expansion of n . Theorem 4.7 ([GPX15, Theorem 15]) . Suppose that < n ≤ q , and let n = P ki =1 ( − i +1 e i be the consecutive-ones binary expansion of n , where e > · · · >e k ≥ . Then V n ⊗ V n ∼ = M ≤ i ≤ k V d i ei , where d k = 2 e k , and d i = 2 e i − P kj = i +1 ( − i + j +1 e j +1 for all ≤ i < k . We finish this section by discussing some results on the decomposition of ∧ ( V n ).The following recursive description of ∧ ( V n ) is due to Gow and Laffey [GL06]. Theorem 4.8 ([GL06, Theorem 2]) . Suppose that q/ < n ≤ q . Then ∧ ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ − q ⊕ V q/ − n . Example . Applying Theorem 4.8 with n = q = 2 α , it is immediate that ∧ ( V α ) ∼ = V α − ⊕ V α − − α for all α > ∧ ( V ⊕ W ) = ∧ ( V ) ⊕ ∧ ( W ) ⊕ ( V ∧ W )for all K [ u ]-modules V and W . Since V ∧ W ∼ = V ⊗ W as K [ u ]-modules, thisdecomposition gives the following result. Lemma 4.10.
Let V be a K [ u ] -module such that V ∼ = V d ⊕ · · · ⊕ V d t , where d i > are integers. Then ∧ ( V ) ∼ = M ≤ i ≤ t ∧ ( V d i ) ⊕ M ≤ i Let < n ≤ q/ and set α = ν ( n ) . Then the smallest Jordan blocksize in ∧ ( V n ) is α , occurring with multiplicity one. Proof. By induction on n . In the case n = 1, the claim holds since ∧ ( V n ) = ∧ ( V ) = V . Suppose then n > < n ′ < n .Without loss of generality, we can assume that q/ < n ≤ q . If 2 n = q , then theclaim follows from Example 4.9. Suppose that q/ < n < q . Then(4.1) ∧ ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ − q ⊕ V q/ − n by Theorem 4.8. Now ν (( q − n ) / 2) = ν ( n ) = α since q > α +1 , so by inductionthe smallest Jordan block size in ∧ ( V q − n ) is 2 α , occurring with multiplicity one.Furthermore, we have q > q/ − n > q/ ≥ α , so the result follows from (4.1). (cid:3) Lemma 4.13. Let < n ≤ q/ . Then every Jordan block size in ∧ ( V n ) has oddmultiplicity.Proof. By induction on n . The steps of the proof are essentially the same as in theproof of Lemma 4.12, so we omit the details. (cid:3) Lemma 4.14. Let n > and suppose that all Jordan block sizes in ∧ ( V n ) havemultiplicity at most . Then n ∈ { , , , } .Proof. For n = 4, an easy calculation with Theorem 4.8 shows that ∧ ( V ) ∼ = V ⊕ V . Thus we may assume n > q be a power of 2 suchthat q/ < n ≤ q . Suppose that all Jordan block sizes in ∧ ( V n ) have multiplicityat most 2. Then by Lemma 4.13 each Jordan block size in ∧ ( V n ) has multiplicityone. By Theorem 4.8, we have(4.2) ∧ ( V n ) ∼ = ∧ ( V q − n ) ⊕ V n − q/ − q ⊕ V q/ − n so 2 n − q/ − ≤ 1, which forces 2 n = q/ ∧ ( V n ) ∼ = ∧ ( V q/ − ) ⊕ V q ⊕ V q − . Now q/ < q/ − < q/ 2, so applying Theorem 4.8 we get ∧ ( V q/ − ) ∼ = V ⊕ V q/ ⊕ V q/ − q/ , and therefore q/ − ≤ 1, giving q ≤ 16. Since n > q/ < n ≤ q , it followsthat q = 16. In this case 2 n = q/ n = 5. (cid:3) Modules equipped with a bilinear form Let G be a group. It is an elementary fact in representation theory that theGL( V )-conjugacy classes of homomorphisms G → GL( V ) are in bijection with theisomorphism classes of K [ G ]-module structures on V . Similarly, it is convenient tostudy the conjugacy classes of subgroups of Sp( V, b ) in terms of modules equippedwith a non-degenerate alternating bilinear form. In later sections of this paper, thiswill be useful for us when describing the conjugacy class of a unipotent element u ∈ Sp( V, b ).For some generalities on modules equipped with a bilinear form, see for example[Wil77], [QSSS76], and [Mur16]. We give the basic definitions and results neededin this paper in what follows. Definition 5.1. A bilinear K [ G ] -module ( V, b ) is a K [ G ]-module V with a G -invariant bilinear form b : V × V → K . A bilinear K [ G ]-module ( V, b ) is said to be non-degenerate if b is non-degenerate, symmetric if b is symmetric, and alternating if b is alternating. ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 11 Definition 5.2. An isomorphism of bilinear K [ G ]-modules ( V, b ) and ( V ′ , b ′ ) is anisomorphism ϕ : V → V ′ of K [ G ]-modules such that b ′ ( ϕ ( v ) , ϕ ( w )) = b ( v, w ) forall v, w ∈ V . Definition 5.3. Let ( V, b ) be a bilinear K [ G ]-module and W a K [ G ]-submoduleof V . We denote ( W, b ) := ( W, b | W × W ). Furthermore, if W is totally singular withrespect to b , then we set ( W ⊥ /W, b ) := ( W ⊥ /W, b ′ ), where b ′ ( v + W, v + W ) = b ( v , v ) for all v , v ∈ W ⊥ . Definition 5.4. The orthogonal direct sum of two bilinear K [ G ]-modules ( V, b )and ( V ′ , b ′ ) is the bilinear K [ G ]-module ( V ⊕ V ′ , b ⊥ b ′ ), where( b ⊥ b ′ )( v + v ′ , v + v ′ ) = b ( v , v ) + b ′ ( v ′ , v ′ )for all v , v ∈ V and v ′ , v ′ ∈ V ′ . We denote ( V ⊕ V ′ , b ⊥ b ′ ) := ( V, b ) ⊥ ( V ′ , b ′ ).In the context of bilinear K [ G ]-modules, for n ≥ V, b ) n to denote( V, b ) ⊥ · · · ⊥ ( V, b ), where ( V, b ) occurs n times in the orthogonal direct sum. Notethat ( V, b ) = 0. Definition 5.5. We call a bilinear K [ G ]-module ( V, b ) orthogonally indecomposable ,if V = 0 and whenever V = V ⊥ V for two K [ G ]-submodules V and V , we have V = 0 or V = 0. Remark . It is clear that any bilinear K [ G ]-module decomposes into an orthog-onal direct sum of orthogonally indecomposable bilinear K [ G ]-modules. However,there is no analogue of the Krull-Schmidt theorem in this setting, as noted in [Wil76,3.13]. In fact, even the number of orthogonally indecomposable summands is notunique, see for example Lemma 5.15 below. Definition 5.7. The tensor product of bilinear K [ G ]-modules ( V, b ) and ( V ′ , b ′ ) isthe bilinear K [ G ]-module ( V ⊗ V ′ , b ⊗ b ′ ), where b ⊗ b ′ is defined by( b ⊗ b ′ )( v ⊗ v ′ , v ⊗ v ′ ) = b ( v , v ) b ′ ( v ′ , v ′ )for all v , v ∈ V and v ′ , v ′ ∈ V ′ . We denote ( V ⊗ V ′ , b ⊗ b ′ ) := ( V, b ) ⊗ ( V ′ , b ′ ).We will also need to consider induction and restriction of bilinear K [ G ]-modules,as defined for example in [GW95, Lemma, p. 1242], see also [Mur16, Section 4]. Definition 5.8. Let H < G be a subgroup. For a bilinear K [ G ]-module ( V, b ), its restriction to H is Res GH ( V, b ) := (Res GH ( V ) , b ). Definition 5.9. Let H < G . For a bilinear K [ H ]-module ( L, b ), the bilinear K [ G ] -module induced by ( L, b ) is Ind GH ( L, b ) := (Ind GH ( L ) , a ), where Ind GH ( L ) = K [ G ] ⊗ K [ H ] L is the K [ G ]-module induced by L and a ( g ⊗ ℓ , g ⊗ ℓ ) = ( b ( g − g ℓ , ℓ ) , if g H = g H. , if g H = g H. for all g , g ∈ G and ℓ , ℓ ∈ L . Lemma 5.10. Let H < G . Let ( L, b ) be a bilinear K [ H ] -module and ( W, b ′ ) abilinear K [ G ] -module. Then Ind GH ( L, b ) ⊗ ( W, b ′ ) ∼ = Ind GH ( L ⊗ Res GH ( W ) , b ⊗ b ′ ) as bilinear K [ G ] -modules. Proof. The corresponding result for K [ G ]-modules is a basic result [Alp86, Lemma5 (5), p. 57], and one can see that the map θ : Ind GH ( L ) ⊗ W → Ind GH ( L ⊗ Res GH ( W )) defined by ( g ⊗ ℓ ) ⊗ w g ⊗ ( ℓ ⊗ g − w ) for all g ∈ G , ℓ ∈ L , and w ∈ W , is an isomorphism of K [ G ]-modules. A straightforward check shows that θ is also an isometry with respect to the bilinear forms on Ind GH ( L, b ) ⊗ ( W, b ′ ) andInd GH (( L, b ) ⊗ Res GH ( W, b ′ )). (cid:3) Definition 5.11. Let M be a K [ G ]-module. The paired module associated with M is the bilinear K [ G ]-module ( M ⊕ M ∗ , a ), where a ( v + f, v ′ + f ′ ) = f ( v ′ ) + f ′ ( v )for all v, v ′ ∈ M and f, f ′ ∈ M ∗ .Note that the paired module associated with a K [ G ]-module M is always anon-degenerate alternating bilinear K [ G ]-module. Lemma 5.12. Let ( V, b ) be a non-degenerate alternating bilinear K [ G ] -module.Then ( V, b ) is a paired module if and only if there exists a totally singular decom-position V = W ⊕ W ′ , where W and W ′ are K [ G ] -submodules of V . Furthermore,in this case ( V, b ) is the paired module associated with W .Proof. If ( V, b ) = ( M ⊕ M ∗ , a ) is a paired module as in Definition 5.11, then V = M ⊕ M ∗ is a totally singular decomposition with respect to b . Conversely, supposethat V admits a totally singular decomposition V = W ⊕ W ′ into K [ G ]-submodules W and W ′ . For w ′ ∈ W ′ , define ϕ w ′ ∈ W ∗ by ϕ w ′ ( w ) = b ( w ′ , w ) for all w ∈ W .Then it is straightforward to see that the map w + w ′ w + ϕ w ′ is an isomorphism( V, b ) → ( W ⊕ W ∗ , a ) of bilinear K [ G ]-modules, where ( W ⊕ W ∗ , a ) is the pairedmodule associated with W . (cid:3) The following two lemmas are easy consequences of Lemma 5.12. Lemma 5.13. Let ( V, b ) be a paired K [ G ] -module. Then for any bilinear K [ G ] -module ( W, b ′ ) , the tensor product ( V, b ) ⊗ ( W, b ′ ) is a paired K [ G ] -module. Lemma 5.14. Let H < G and let ( W, b ) be a paired K [ H ] -module. Then Ind GH ( W, b ) is a paired K [ G ] -module. We finish this section with a proof of the following lemma from [Mur16, Example2.1]. Lemma 5.15. Let ( W, b ) be a bilinear K [ G ] -module. Then as bilinear K [ G ] -modules ( W, b ) ⊥ ( W, b ) ⊥ ( W, b ) ∼ = ( W, b ) ⊥ ( W ⊕ W ∗ , a ) , where ( W ⊕ W ∗ , a ) is the paired module associated with W .Proof. Let V = ( W, b ) ⊥ ( W, b ) ⊥ ( W, b ). It is straightforward to see that thediagonal subspace Z = { ( w, w, w ) : w ∈ W } is non-degenerate, and that Z ∼ = ( W, b )as bilinear K [ G ]-modules. The orthogonal complement of Z in V is Z ⊥ = Z ⊕ Z ,where Z = { ( w, w, 0) : w ∈ W } and Z = { ( w, , w ) : w ∈ W } . Both Z and Z are totally singular and Z ∼ = W ∼ = Z and K [ G ]-modules. Thus by Lemma5.12, as a bilinear K [ G ]-module Z ⊥ is the paired module associated with W . Since V = Z ⊥ Z ⊥ , the lemma follows. (cid:3) ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 13 Unipotent classes in Sp( V )Throughout this section, we denote by u a generator of a cyclic 2-group of order q > 1, and denote by X the element u − K [ u ]. Recall (Section 2) that wedenote the indecomposable K [ u ]-modules by V , . . . , V q , where dim V i = i and u acts on V i as a full i × i Jordan block.For the symplectic groups Sp( V, b ), the conjugacy classes of unipotent elementsof order at most q correspond to the isomorphism classes of non-degenerate alter-nating bilinear K [ u ]-modules. This is the basic approach taken in [Hes79], whereHesselink classifies the unipotent conjugacy classes of Sp( V, b ) in terms of orthog-onally indecomposable bilinear K [ u ]-modules. We give an explicit construction ofthese modules in the following definitions. Definition 6.1. Let d > 0. We define W ( d ) to be the paired module (Definition5.11) ( V d ⊕ V ∗ d , a ) associated with V d . Definition 6.2. Let d > d = 2 k . Fix a basis e , . . . , e d of the K [ u ]-module V d such that ue = e ,ue i = e i + e i − + · · · + e for all 2 ≤ i ≤ k + 1 ,ue i = e i + e i − for all k + 1 < i ≤ d. We define V ( d ) to be the bilinear K [ u ]-module ( V d , b ) where b ( e i , e j ) = 1 if i + j = d + 1 and 0 otherwise.Here W ( d ) is orthogonally indecomposable by [PM18, Section 2.3], while V ( d )in Definition 6.2 is orthogonally indecomposable since it is indecomposable as a K [ u ]-module.We note that Definition 6.2 is the same as [LS12, Section 6.1, p. 91], anddescribes the action of a regular unipotent element of Sp( V, b ) on the basis ( e i ) of V .More specifically, Definition 6.2 describes the action of the product x α (1) · · · x α k (1)of fundamental root elements of Sp( V, b ).To describe the conjugacy classes in Sp( V, b ), we will first need the followingresult from [Hes79]. Theorem 6.3 ([Hes79, Proposition 3.5]) . Up to isomorphism, the orthogonallyindecomposable non-degenerate alternating bilinear K [ u ] -modules are V ( d ) ( d even)and W ( d ) . As a consequence of Theorem 6.3, each non-degenerate alternating bilinear K [ u ]-module has a certain normal form which is described in the next theorem. We willcall this the Hesselink normal form . Theorem 6.4. Let ( V, b ) a non-degenerate alternating bilinear K [ u ] -module. Let < d < · · · < d t be the Jordan block sizes of u on V , and for ≤ i ≤ t let n i > be the multiplicity of d i in V .There exists a unique sequence W , . . . , W t of non-degenerate alternating bilinear K [ u ] -modules such that V ∼ = W ⊥ · · · ⊥ W t and the following hold for all ≤ i ≤ t : (i) If d i is odd, then W i = W ( d i ) n i / . (ii) If d i is even, then W i = W ( d i ) n i / or W i = V ( d i ) n i . The normal form in Theorem 6.4 is the same as that described by Hesselink in[Hes79, 3.7]. One can see this using [Hes79, 3.7 – 3.9], but we will give a proof laterin this section to keep this paper more self-contained.Note that the conjugacy class of a unipotent element u ∈ Sp( V, b ) is determinedby the Hesselink normal form of u on ( V, b ). Remark . There is also a distinguished normal form defined in [LS12, p. 61],which is different from the Hesselink normal form and useful for describing cen-tralizers of unipotent elements in Sp( V, b ). Translating between these two normalforms is straightforward, using the fact that V (2 d ) ∼ = W (2 d ) ⊥ V (2 d ) as bilinear K [ u ]-modules (Lemma 5.15). In this paper, we will only use the Hesselink normalform.Following [Spa82, 2.6, p. 20], we make the following definition. Definition 6.6. Let ( V, b ) be a bilinear K [ u ]-module. We define ε V,b : Z ≥ → { , } by ε V,b ( d ) = 0 if b ( X d − v, v ) = 0 for all v ∈ V such that X d v = 0, and ε V,b ( d ) = 1otherwise.It turns out that the Hesselink normal form of u ∈ Sp( V, b ) (and hence itsconjugacy class in Sp( V, b )) is determined by the Jordan normal form of u and thevalues of ε V,b on the Jordan block sizes of u . This is a well known result which isstated in [Spa82, 2.6, p. 20]. Theorem 6.7. Suppose that u ∈ Sp( V, b ) , and set ε := ε V,b . Let < d < · · · < d t be the Jordan block sizes of u on V , with block size d i having multiplicity n i > .Let ( V, b ) ∼ = W ⊥ · · · ⊥ W t be the Hesselink normal form of u on ( V, b ) as inTheorem 6.4. Then for all ≤ i ≤ t , we have W i ∼ = W ( d i ) n i / if ε V,b ( d i ) = 0 and W i ∼ = V ( d i ) n i if ε V,b ( d i ) = 1 .In particular, the Hesselink normal form of u on ( V, b ) is uniquely determinedby the tuple ( d n ε ( d ) , . . . , d tn t ε ( d t ) ) . Since our main results rely on Theorem 6.7, we will give a proof in what follows.First we need a few lemmas which will also be useful later for the computation of ε V,b for various bilinear K [ u ]-modules ( V, b ). Lemma 6.8. Suppose that u ∈ SL( V ) , let b be a u -invariant alternating bilinearform on V , not necessarily non-degenerate. Let d > be an integer. Then: (i) For all v, w ∈ Ker X d and ≤ i, j ≤ d − with i + j = d , we have b ( X i − v, X j w ) = b ( X i v, X j − w ) . (ii) For all v, w ∈ Ker X d and ≤ i, j ≤ d with i + j ≥ d , we have b ( X i v, X j w ) =0 . (iii) For all v, w ∈ Ker X d , we have b ( X d − v, w ) = b ( v, X d − w ) . (iv) The map v b ( X d − v, v ) is additive on Ker X d .Proof. Set ( e i ) ≤ i ≤ d = ( X d − i v ) ≤ i ≤ d and ( f j ) ≤ j ≤ d = ( X d − j w ) ≤ j ≤ d . For all1 ≤ i, j ≤ d − i + j = d , it follows from [Spa82, Lemme II.6.10 b), pg.99] that b ( e d − i +1 , f d − j ) + b ( e d − i , f d − j +1 ) = 0, which gives (i). For all 1 ≤ i, j ≤ d with i + j ≥ d , we have b ( e d − i , f d − j ) = 0 by [Spa82, Lemme II.6.10 a), pg. 99],which gives (ii).For claim (iii), using (i) repeatedly we find that b ( X d − v, w ) = b ( X d − v, Xw ) = · · · = b ( v, X d − w )for all v, w ∈ Ker X d . Claim (iv) is an easy consequence of (iii). (cid:3) ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 15 Lemma 6.9. Let ( V, b ) be an alternating bilinear K [ u ] -module, not necessarilynon-degenerate. The following statements are equivalent: (i) b ( X d − v, v ) = 0 for some v ∈ V such that X d v = 0 . (ii) d is even, and V ( d ) occurs as an orthogonal direct summand of V . (iii) d is even, and for any decomposition V = W ⊥ · · · ⊥ W t into orthogonallyindecomposable K [ u ] -submodules, we have ( W i , b ) ∼ = V ( d ) for some i .Proof. We first show that (i) and (ii) are equivalent. Suppose that b ( X d − v, v ) = 0for some v ∈ V such that X d v = 0. Let W be the subspace of V spanned by v, Xv, . . . , X d − v , so now W ∼ = V d as K [ u ]-modules. It follows from Lemma 6.8 (i)– (ii) that the matrix of b | W × W with respect to the basis v, Xv, . . . , X d − v is of theform(6.1) ∗ λ ... λ where λ = b ( X d − v, v ). Since λ = 0, it follows that b | W × W is non-degenerate, so d must be even since b is alternating. Furthermore, we have V = W ⊥ W ⊥ and W ∼ = V ( d ) since W is non-degenerate, so (ii) holds.Conversely, suppose that d is even and V = W ⊥ W ′ with W ∼ = V ( d ). Choosesome v ∈ W such that X d − v = 0. Then v, Xv, . . . , X d − v is a basis of W , andthe matrix of b | W × W with respect to this basis is as in (6.1), with λ = b ( X d − v, v ).Thus we must have b ( X d − v, v ) = 0 since W is a non-degenerate subspace. Weconclude then that (i) and (ii) are equivalent.It is obvious that (iii) implies (ii). Next we will show that (i) implies (iii),which will complete the proof. Suppose that (i) holds, and let v ∈ V be such that b ( X d − v, v ) = 0 and X d v = 0. Let V = W ⊥ · · · ⊥ W t be a decomposition intoorthogonally indecomposable K [ u ]-submodules. We can write v = w + · · · + w t with w i ∈ W i for all 1 ≤ i ≤ t . Now X d w i = 0 for all 1 ≤ i ≤ t , so by Lemma 6.8(iv) b ( X d − v, v ) = b ( X d − w , w ) + · · · + b ( X d − w t , w t ) . Thus b ( X d − w i , w i ) = 0 for some 1 ≤ i ≤ t . From the equivalence of (i) and (ii),it follows that d is even and ( W i , b ) has V ( d ) as an orthogonal direct summand.Since ( W i , b ) is orthogonally indecomposable, this proves that ( W i , b ) ∼ = V ( d ). (cid:3) We can now prove Theorem 6.4 and Theorem 6.7. Proof of Theorem 6.4. Let ( V, b ) be a non-degenerate alternating bilinear K [ u ]-module. One can write V as an orthogonal direct sum V = Z ⊥ · · · ⊥ Z t oforthogonally indecomposable bilinear K [ u ]-modules, and by Theorem 6.3 each Z i is isomorphic to V ( d i ) ( d i even) or W ( d i ) for some d i > V (2 d ) ∼ = W (2 d ) ⊥ V (2 d ) by Lemma 5.15. From this it follows thatfor a, b ≥ 0, we have V (2 d ) a ⊥ W (2 d ) b ∼ = ( V (2 d ) a +2 b , if a > .W (2 d ) b , if a = 0 . as bilinear K [ u ]-modules. Thus by collecting the orthogonal direct summands Z i with equal Jordan block sizes, we get the Hesselink normal form on V . For uniqueness, let 0 < d < · · · < d t be the Jordan block sizes of u on V , withblock size d i having multiplicity n i > 0. Write V = W ⊥ · · · ⊥ W t , where for all i we have W i ∼ = W ( d i ) n i / or W i ∼ = V ( d i ) n i . By the equivalence of (i) and (iii)Lemma 6.9, we see that W i ∼ = V ( d i ) n i if and only if d i is even and b ( X d i − v, v ) = 0for some v ∈ V such that X d i v = 0. From this we get the uniqueness of theHesselink normal form. (cid:3) Proof of Theorem 6.7. The proof follows from the argument at the end of the pre-vious proof. Indeed, by Lemma 6.9, in Theorem 6.4 we have W i ∼ = W ( d i ) n i / if ε V,b ( d i ) = 0 and W i ∼ = V ( d i ) n i if ε V,b ( d i ) = 1. (cid:3) Lemma 6.10. Let ( V, b ) be a non-degenerate alternating bilinear K [ u ] -module.Then the following statements hold: (i) If d is odd, then ε V,b ( d ) = 0 . (ii) If V d occurs with odd multiplicity in V , then d is even and ε V,b ( d ) = 1 .Proof. Claim (i) is immediate from Lemma 6.9. For (ii), suppose that V d occurswith odd multiplicity in V . Since V d has multiplicity 2 in W ( d ), it follows that d iseven and ( V, b ) must have V ( d ) as an orthogonal direct summand. Thus ε V,b ( d ) = 1by Lemma 6.9. (cid:3) Let G = SL( V ) and set n = dim V . In one of our main results, Theorem A,we describe the Hesselink normal form of any unipotent element u ∈ G on theirreducible K [ G ]-module L G ( ̟ + ̟ n − ). In the proof, we make use of the factthat up to scalar multiples there is a unique non-zero alternating bilinear form on V ⊗ V ∗ , and an isomorphism L G ( ̟ + ̟ n − ) ∼ = h v i ⊥ / h v i where v ∈ V ⊗ V ∗ isa G -fixed point — see Section 8. A natural approach then is to first consider theaction of u on V ⊗ V ∗ and use it to deduce information about the action of u on h v i ⊥ / h v i . For this we need the following general lemma, which will also be usefulin the proof of our main result concerning Hesselink normal forms on L G ( ̟ ) for G = Sp( V, b ) (Theorem B). Lemma 6.11. Let ( V, b ) be a non-degenerate alternating bilinear K [ u ] -module andlet v ∈ V be a non-zero vector fixed by u . Write V ∼ = ⊕ d ≥ V λ ( d ) d and h v i ⊥ / h v i ∼ = ⊕ d ≥ V λ ′ ( d ) d as K [ u ] -modules, where λ ( d ) , λ ′ ( d ) ≥ for all d ≥ . Set ε := ε V,b and ε ′ := ε h v i ⊥ / h v i ,b .Let m ≥ be such that Ker X m − V ⊆ h v i ⊥ and Ker X mV 6⊆ h v i ⊥ . The followingstatements hold: (i) For all d ≥ , we have Ker X dV ⊆ h v i ⊥ if and only if h v i ⊆ Im X dV . Inparticular, there exists δ ∈ V such that X m − δ = v , and v Im X dV for d ≥ m . (ii) If m = 1 , then: (a) λ ′ (1) = λ (1) − , and λ ′ ( d ) = λ ( d ) for all d > . (b) ε ′ ( d ) = ε ( d ) for all d ≥ . (iii) If m > and δ ∈ h v i ⊥ , then: (a) λ ′ ( m ) = λ ( m ) − , λ ′ ( m − 1) = λ ( m − 1) + 2 , and λ ′ ( d ) = λ ( d ) for all d = m − , m . (b) ε ′ ( d ) = ε ( d ) for all d = m . (iv) If m > and δ 6∈ h v i ⊥ , then: ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 17 (a) λ ′ ( m ) = λ ( m ) − , λ ′ ( m − 2) = λ ( m − 2) + 1 (if m > ), and λ ′ ( d ) = λ ( d ) for all d = m − , m . (b) ε ( d ) = ε ′ ( d ) for all d = m − , m . (c) ε ( m ) = 1 . (d) ε ′ ( m − 2) = 1 (if m > ).Proof. Since b is non-degenerate, we have h v i ⊥ ∼ = ( V / h v i ) ∗ as K [ u ]-modules. Every K [ u ]-module is self-dual, so in fact h v i ⊥ ∼ = V / h v i as K [ u ]-modules. Then withLemma 3.3 and Lemma 3.4, we conclude that for all d ≥ 0, we have Ker X d ⊆ h v i ⊥ if and only if h v i ⊆ Im X d , which proves (i). Let δ ∈ V be such that X m − δ = v .For claims (ii) – (iv), we first consider the description of λ ′ . By (i) and Lemma3.4, we find that V / h v i ∼ = ⊕ d ≥ V µ ( d ) d as K [ u ]-modules, where µ ( d ) ≥ • If m = 1, then µ (1) = λ (1) − µ ( d ) = λ ( d ) for all d > • If m > 1, then µ ( m ) = λ ( m ) − µ ( m − 1) = λ ( m − 1) + 1, and µ ( d ) = λ ( d )for all d = m, m − V / h v i and h v i ⊥ / h v i in order to describe λ ′ . Firstnote that v Im X d for all d ≥ m by (i), so(6.2) Ker X dV/ h v i = Ker X dV / h v i for all d ≥ m . If 0 ≤ d ≤ m − 1, then X d ( X m − − d δ ) = v and any solution to X d v ′ = v is unique modulo Ker X d , so(6.3) Ker X dV/ h v i = (cid:0) Ker X dV ⊕ h X m − − d δ i (cid:1) / h v i for all 0 ≤ d ≤ m − u acts trivially on the 1-dimensional K [ u ]-module V / h v i ⊥ , so Im X ⊆h v i ⊥ . Thus (6.3) implies thatKer X dV/ h v i ⊆ h v i ⊥ / h v i for all 0 ≤ d < m − . (6.4)Furthermore, by (6.3) and (6.2) the following hold:Ker X m − V/ h v i ⊆ h v i ⊥ / h v i if and only if δ ∈ h v i ⊥ . (6.5) Ker X mV/ h v i 6⊆ h v i ⊥ / h v i . (6.6)Now combining Lemma 3.3, statements (6.4) – (6.6), and the description of µ ( d )above, it follows easily that λ ′ is given as described in (ii) – (iv). This completesthe proof of the claims for λ ′ .For the rest of the proof we will consider the claims about ε and ε ′ in (ii) – (iv).Let ˆ δ be such that X m ˆ δ = 0 and ˆ δ 6∈ h v i ⊥ . Then(6.7) Ker X dV = Ker X d h v i ⊥ ⊕ h ˆ δ i for all d ≥ m . If d > m , then b ( X d − ˆ δ, ˆ δ ) = 0 since X m ˆ δ = 0. Thus it followsfrom (6.7) and Lemma 6.8 (iv) that ε ′ ( d ) = ε ( d ) for all d > m .We always have ε ′ (1) = ε (1) = 0, so if m = 1, then ε ′ ( d ) = ε ( d ) for all d ≥ m > X ⊆ h v i ⊥ , and Ker X dV ⊆ h v i ⊥ for all 1 ≤ d ≤ m − 1, so(6.8) Ker X d h v i ⊥ / h v i = (cid:0) Ker X dV ⊕ h X m − d − δ i (cid:1) / h v i for all 1 ≤ d ≤ m − Note that X m δ = 0 since v is fixed by u . Thus if 1 ≤ d < m − 2, then b ( X d − ( X m − d − δ ) , X m − d − δ ) = b ( X m − δ, X m − d − δ ) = 0by Lemma 6.8 (ii). It follows then from (6.8) and Lemma 6.8 (iv) that ε ′ ( d ) = ε ( d )for all 1 ≤ d < m − ε ′ ( d ) = ε ( d ) for all d = m − , m − , m , as claimedby (iii) and (iv). For d = m − , m − , m , we will consider the two cases (iii) and(iv) separately. Case (iii): δ ∈ h v i ⊥ . In this case b ( X m − δ, δ ) = b ( v, δ ) = 0. SinceKer X m − h v i ⊥ / h v i = (cid:0) Ker X m − V ⊕ h δ i (cid:1) / h v i by (6.8), it follows from Lemma 6.8 (iv) that ε ′ ( m − 1) = ε ( m − m > 2, with Lemma 6.8 (i) we get(6.9) b ( X m − ( Xδ ) , Xδ ) = b ( X m − δ, Xδ ) = b ( X m − δ, δ ) = b ( v, δ ) . Thus b ( X m − ( Xδ ) , Xδ ) = 0. SinceKer X m − h v i ⊥ / h v i = (cid:0) Ker X m − V ⊕ h Xδ i (cid:1) / h v i by (6.8), it follows from Lemma 6.8 (iv) that ε ′ ( m − 2) = ε ( m − ε ′ ( d ) = ε ( d ) for all d = m , as claimed in (iii). Case (iv): δ 6∈ h v i ⊥ . In this case b ( X m − δ, δ ) = b ( v, δ ) = 0, so ε ( m ) = 1. Thus m must be even and ε ′ ( m − 1) = ε ( m − 1) = 0 by Lemma 6.10.If m > 2, then we see from (6.9) that b ( X m − ( Xδ ) , Xδ ) = 0. Thus ε ′ ( m − 2) = 1,since Xδ + h v i ∈ Ker X m − h v i ⊥ / h v i . This completes the proof of (iv) and the lemma. (cid:3) We finish this section by describing the induction and restriction of orthogonallyindecomposable bilinear K [ u ]-modules. First we need a small lemma, which willalso be useful later. Lemma 6.12. Let ( Z, b ) be a non-degenerate alternating bilinear K [ u ] -module.Then the following statements are equivalent: (i) ( Z, b ) is a paired module. (ii) There exists a totally singular decomposition Z = W ⊕ W ′ where W and W ′ are K [ u ] -submodules of Z . (iii) For all d > and v ∈ Z such that X d v = 0 , we have b ( X d − v, v ) = 0 .Furthermore, if Z = W ⊕ W ′ as in (ii) and W ∼ = V d ⊕ · · ·⊕ V d t as K [ u ] -modules,then ( Z, b ) ∼ = W ( d ) ⊥ · · · ⊥ W ( d t ) as bilinear K [ u ] -modules.Proof. The equivalence of (i) and (ii) is given by Lemma 5.12. We show that (ii)implies (iii). Let Z = W ⊕ W ′ be a totally singular decomposition, where W and W ′ are K [ u ]-submodules of Z . For any v ∈ Ker X d , we can write v = w + w ′ where w ∈ Ker X dW and w ′ ∈ Ker X dW ′ . Then b ( X d − v, v ) = b ( X d − w, w ) + b ( X d − w ′ , w ′ ) by ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 19 Lemma 6.8 (iv). Since W and W ′ are totally singular, it follows that b ( X d − v, v ) =0. Next we show that (iii) implies (i). If (iii) holds, then ( Z, b ) does not have anyorthogonal direct summands of the form V ( m ) by Lemma 6.9. It follows fromTheorem 6.3 that ( Z, b ) ∼ = W ( d ) ⊥ · · · ⊥ W ( d t ) for some integers d i > 0, andconsequently ( Z, b ) is a paired module since each W ( d i ) is.For the last statement of the lemma, suppose that Z = W ⊕ W ′ as in (ii) and W ∼ = V d ⊕ · · · ⊕ V d t as K [ u ]-modules. Then W = W ⊥ since Z = W ⊕ W ′ is atotally singular decomposition, so W ′ ∼ = V /W = V /W ⊥ ∼ = W ∗ . Hence Z ∼ = W ⊕ W ∗ ∼ = V d ⊕ · · · ⊕ V d t . As in the previous paragraph, as a bilinear K [ u ]-module ( Z, b ) decomposes into anorthogonal direct sum involving only summands of the form W ( d ), so we must have( Z, b ) ∼ = W ( d ) ⊥ · · · ⊥ W ( d t )as bilinear K [ u ]-modules. (cid:3) Lemma 6.13. Let α > be such that α ≤ q and let < d ≤ q/ α . Then: (i) If d is even, then Ind h u ih u α i ( V ( d )) ∼ = V (2 α d ) . (ii) Ind h u ih u α i ( W ( d )) ∼ = W (2 α d ) . (iii) Res h u ih u i ( V (2 d )) ∼ = ( V ( d ) , if d is even .W ( d ) , if d is odd . (iv) Write d = a α − + r for ≤ r < α − . Then Res h u ih u α i ( V (2 d )) ∼ = ( V ( d/ α − ) α , if α | d.W ( a + 1) r ⊥ W ( a ) α − − r , if α ∤ d. where we define W (0) = 0 .Proof. For (i), note that by Lemma 3.2 we have Ind h u ih u α i ( V d ) ∼ = V α d as K [ u ]-modules. Thus from Theorem 6.3 it is clear that Ind h u ih u α i ( V ( d )) ∼ = V (2 α d ).By Lemma 5.14 the induced module Ind h u ih u α i ( W ( d )) is a paired K [ u ]-module,so (ii) follows from Lemma 6.12 and the fact that Ind h u ih u α i ( V d ) ∼ = V α d as K [ u ]-modules.Claim (iii) is [LLS14, Lemma 4.1]. For claim (iv), note that the case α = 1 isthe same as (iii). For α > 1, we prove by induction on α that Res h u ih u α i ( V (2 d )) ∼ = V ( d/ α − ) α if 2 α | d . If α > α | d , then Res h u ih u α − i ( V (2 d )) ∼ = V ( d/ α − ) α − by induction. On the other hand Res h u α − ih u α i ( V ( d/ α − )) ∼ = V ( d/ α − ) by (iii), sowe conclude that Res h u ih u α i ( V (2 d )) ∼ = V ( d/ α − ) α .Next consider the case where 2 α ∤ d . We show first that Res h u ih u α i ( V (2 d )) is apaired module. To this end, let 0 ≤ β < α be such that 2 β | d and 2 β +1 ∤ d . Thenwe have already shown that Res h u ih u β i ( V (2 d )) ∼ = V ( d/ β − ) β . Since d/ β is odd, itfollows from (iii) that Res h u β ih u β +1 i V ( d/ β − ) ∼ = W ( d/ β ) and so Res h u ih u β +1 i ( V (2 d )) ∼ = W ( d/ β ) β . Thus Res h u ih u α i ( V (2 d )) is a paired module since Res h u ih u β +1 i ( V (2 d )) is.Now 2 d = a α + 2 r , so by Lemma 3.1Res h u ih u α i ( V d ) ∼ = V ra +1 ⊕ V α − ra and (iv) follows from Lemma 6.12. (cid:3) Tensor products of bilinear K [ u ] -modules We keep the setup of the previous section, so let u be a generator of a cyclic2-group of order q > X = u − K [ u ]-modules into orthogonally indecomposable summands.Clearly, it suffices to do this for the orthogonally indecomposable bilinear K [ u ]-modules, which (up to isomorphism) are of the form V (2 d ) or W ( d ) for some integer d > W ( d ), the following proposition isan easy consequence of Lemma 5.13 and Lemma 6.12. Proposition 7.1. Let < d, d ′ ≤ q . Suppose that V d ⊗ V d ′ ∼ = V d ⊕ · · · ⊕ V d t . Then: (i) If d ′ is even, then W ( d ) ⊗ V ( d ′ ) ∼ = W ( d ) ⊥ · · · ⊥ W ( d t ) . (ii) W ( d ) ⊗ W ( d ′ ) ∼ = W ( d ) ⊥ · · · ⊥ W ( d t ) . Here Proposition 7.1 describes the tensor products W ( d ) ⊗ V ( d ′ ) and W ( d ) ⊗ W ( d ′ ) in terms of indecomposable summands of V d ⊗ V d ′ , which can be calculatedwith Theorem 4.1.For the rest of this section, we will consider the decomposition of V (2 d ) ⊗ V (2 d ′ )into orthogonally indecomposable summands, and Theorem 7.4 gives a completeanswer in terms of the decomposition of V d ⊗ V d ′ . We begin with a series oflemmas that deal with the case where d and d ′ are odd. Lemma 7.2. Let < ℓ ≤ q be an odd integer. Then there exists a non-degeneratealternating u -invariant bilinear form a on Ind h u ih u i ( V ℓ ) such that (Ind h u ih u i ( V ℓ ) , a ) ∼ = V (2 ℓ ) and Ind h u ih u i ( V ℓ ) = (1 ⊗ V ℓ ) ⊕ ( u ⊗ V ℓ ) is a totally singular decomposition with respect to a .Proof. Consider first V = ( V (2 ℓ ) , b ) with a basis e , . . . , e ℓ such that b ( e i , e j ) = 1if i + j = 2 ℓ + 1 and 0 otherwise, and with the action of u on the e i as in Definition6.2.Let W = h e , e , . . . , e ℓ i . Then W is u -invariant, since u e i = e i + e i − + · · · + e for 2 ≤ i ≤ ℓ + 1 and u e i = e i + e i − for ℓ + 3 ≤ i ≤ ℓ . Furthermore,we claim that W ∩ u ( W ) = 0. To this end, note that W ∩ u ( W ) is u -invariant since u ( W ) ⊆ W . On the other hand, there are no non-zero u -fixed points in W ∩ u ( W )since there are none in W , so we must have W ∩ u ( W ) = 0.Therefore V = W ⊕ u ( W ) and this is a totally singular decomposition. Wehave W ∼ = V ℓ as h u i -modules, so it follows from a basic property of inducedmodules [Alp86, Proof of Lemma 4, pp. 56-57] that there exists an isomorphism ϕ : Ind h u ih u i ( V ℓ ) → V of K [ u ]-modules with ϕ (1 ⊗ V ℓ ) = W and ϕ ( u ⊗ V ℓ ) = u ( W ).Now we can define a non-degenerate u -invariant alternating bilinear form a onInd h u ih u i ( V ℓ ) via a ( x, y ) = b ( ϕ ( x ) , ϕ ( y )) for all x, y ∈ Ind h u ih u i ( V ℓ ). It is clear that ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 21 (Ind h u ih u i ( V ℓ ) , a ) ∼ = V (2 ℓ ), and furthermore Ind h u ih u i ( V ℓ ) = (1 ⊗ V ℓ ) ⊕ ( u ⊗ V ℓ ) is atotally singular decomposition since V = W ⊕ u ( W ) is. (cid:3) Lemma 7.3. Let < ℓ, k ≤ q/ be odd integers. Then we have an orthogonaldecomposition V (2 ℓ ) ⊗ V (2 k ) = W ⊥ W ′ , where W and W ′ are K [ u ] -submodulesof V (2 ℓ ) ⊗ V (2 k ) such that W ∼ = Ind h u ih u i ( V ℓ ⊗ V k ) ∼ = W ′ as K [ u ] -modules.Proof. It will suffice to prove the lemma for (Ind h u ih u i ( V ℓ ) , a ) ⊗ (Ind h u ih u i ( V k ) , a ′ ) where a and a ′ are as in Lemma 7.2. Now Ind h u ih u i ( V ℓ ) = (1 ⊗ V ℓ ) ⊕ ( u ⊗ V ℓ ) is a totallysingular decomposition, so there exist bases e , . . . , e ℓ and f , . . . , f ℓ of V ℓ suchthat a (1 ⊗ e i , u ⊗ f j ) = δ i,j for all 1 ≤ i, j ≤ ℓ . Similarly, one finds bases e ′ , . . . , e ′ k and f ′ , . . . , f ′ k of V k such that a (1 ⊗ e ′ i , u ⊗ f ′ j ) = δ i,j for all 1 ≤ i, j ≤ k .Consider the map θ : Ind h u ih u i ( V ℓ ⊗ V k ) → Ind h u ih u i ( V ℓ ) ⊗ Ind h u ih u i ( V k ) defined by θ ( g ⊗ ( x ⊗ y )) = ( g ⊗ x ) ⊗ ( g ⊗ y ) for all g ∈ h u i , x ∈ V ℓ , and y ∈ V k . It isstraightforward to see that θ is an injective map of K [ u ]-modules.We claim that W = Im θ is a non-degenerate subspace of Ind h u ih u i ( V ℓ ) ⊗ Ind h u ih u i ( V k )with respect to the tensor product form b = a ⊗ a ′ . For this, first note that W hasas a basis the elements v i,i = (1 ⊗ e i ) ⊗ (1 ⊗ e ′ i ) and w j,j = ( u ⊗ f j ) ⊗ ( u ⊗ f ′ j ) for1 ≤ i, j ≤ ℓ and 1 ≤ i , j ≤ k . We have b ( v i,i , w j,j ) = δ i,j δ i ,j for all 1 ≤ i, j ≤ ℓ and 1 ≤ i , j ≤ k , so it follows that W is non-degenerate. ThereforeInd h u ih u i ( V ℓ ) ⊗ Ind h u ih u i ( V k ) = W ⊥ W ′ , where W ′ is the orthogonal complement to W with respect to b .Now W ∼ = Ind h u ih u i ( V ℓ ⊗ V k ), and by [Alp86, Lemma 5 (5), p. 57]Ind h u ih u i ( V ℓ ) ⊗ Ind h u ih u i ( V k ) ∼ = Ind h u ih u i ( V ℓ ⊗ Res h u ih u i Ind h u ih u i ( V k )) ∼ = Ind h u ih u i ( V ℓ ⊗ V k ) ∼ = W ⊕ W as K [ u ]-modules. From the Krull-Schmidt theorem for K [ u ]-modules, we concludethat W ′ ∼ = W . (cid:3) With the lemmas above, we can now prove the main result of this section. Theorem 7.4. Let < ℓ, k ≤ q/ . Then: (i) V ℓ ⊗ V k ∼ = V k d ⊕ · · · ⊕ V k t d t for some integers < d < · · · < d t and k i > . (ii) If ν ( ℓ ) = ν ( k ) , then as a bilinear K [ u ] -module V (2 ℓ ) ⊗ V (2 k ) is isomorphicto ⊥ ≤ i ≤ t W (2 d i ) k i . (iii) If ν ( ℓ ) = ν ( k ) = α , then there a unique j such that ν ( d j ) = α , and d j / α is the unique odd Jordan block size in V ℓ/ α ⊗ V k/ α . Furthermore k j = 2 α ,and as a bilinear K [ u ] -module V (2 ℓ ) ⊗ V (2 k ) is isomorphic to V (2 d j ) α +1 ⊥ ⊥ ≤ i ≤ ti = j W (2 d i ) k i . Proof. We have V ℓ ⊗ V k ∼ = Ind h u ih u i ( V ℓ ⊗ V k ) ∼ = Ind h u ih u i ( V ℓ ⊗ Res h u ih u i ( V k )) by[Alp86, Lemma 5 (5), p. 57], so(7.1) V ℓ ⊗ V k ∼ = Ind h u ih u i ( V ℓ ⊗ V k ) as K [ u ]-modules. Thus V ℓ ⊗ V k ∼ = V k d ⊕· · ·⊕ V k t d t , where V ℓ ⊗ V k ∼ = V k d ⊕· · ·⊕ V k t d t for some integers 0 < d < · · · < d t and k i > 0, which proves (i). We note herethat (i) follows also from [GPX16, Theorem 5].For (ii), we assume without loss of generality that α = ν ( ℓ ) > ν ( k ). Write ℓ = 2 α ℓ ′ . We have V (2 ℓ ) ∼ = Ind h u ih u α i ( V (2 ℓ ′ )) by Lemma 6.13 (i), so it follows withLemma 5.10 that V (2 ℓ ) ⊗ V (2 k ) ∼ = Ind h u ih u α i ( V (2 ℓ ′ )) ⊗ V (2 k ) ∼ = Ind h u ih u α i ( V (2 ℓ ′ ) ⊗ Res h u ih u α i ( V (2 k )))(7.2)as bilinear K [ u ]-modules.Write k = 2 α − k ′ + r for 0 ≤ r < α − . Since 2 α ∤ k , by Lemma 6.13 (iv)Res h u ih u α i ( V (2 k ))) ∼ = W (2 k ′ + 1) r ⊥ W (2 k ′ ) α − − r as bilinear K [ u ]-modules. Thus (7.2) is a paired module by Lemma 5.13 and Lemma5.14, which combined with Lemma 6.12 gives (ii).Next we consider (iii), so suppose that α = ν ( ℓ ) = ν ( k ), and write ℓ = 2 α ℓ ′ , k = 2 α k ′ , where ℓ ′ , k ′ are odd integers. Similarly to (7.1), we see that(7.3) V ℓ ⊗ V k ∼ = Ind h u ih u α +1 i ( V ℓ ′ ⊗ V k ′ ) α +1 as K [ u ]-modules. By Lemma 4.3 the tensor product V ℓ ′ ⊗ V k ′ has a unique Jordanblock of odd size d ′ , occurring with multiplicity 1. Hence we conclude from (7.3)that ν ( d j ) = α for a unique j , and k j = 2 α . Note that d j = 2 α d ′ and ν ( d i ) > α for all i = j .We will now proceed to show that V (2 d i ) occurs as an orthogonal direct sum-mand of V (2 ℓ ) ⊗ V (2 k ) if and only if i = j , which will complete the proof of (iii)and the theorem. First note thatRes h u ih u α +1 i ( V (2 ℓ ) ⊗ V (2 k )) ∼ = W ( ℓ ′ ) ⊗ W ( k ′ )is paired module by Lemma 6.13 (iv) and Lemma 5.13. On the other hand, byLemma 6.13 (iv) we have Res h u ih u α +1 i V (2 d i ) ∼ = V ( d i / α ) for i = j (since ν ( d i ) >α ). Thus we conclude that if i = j , then V (2 d i ) cannot be an orthogonal directsummand of V (2 ℓ ) ⊗ V (2 k ).What remains is to show that V (2 d j ) occurs as an orthogonal direct summandof V (2 ℓ ) ⊗ V (2 k ). For this, first note that Res h u ih u α i ( V (2 k ))) ∼ = V (2 k ′ ) α by Lemma6.13 (iv). Thus by Lemma 5.10(7.4) V (2 ℓ ) ⊗ V (2 k ) ∼ = Ind h u ih u α i ( V (2 ℓ ′ ) ⊗ V (2 k ′ )) α as bilinear K [ u ]-modules, as in (7.2).By Lemma 7.3, we have V (2 ℓ ′ ) ⊗ V (2 k ′ ) = W ⊥ W ′ , where W ∼ = Ind h u ih u i ( V ℓ ′ ⊗ V k ′ ) ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 23 as K [ u ]-modules. Since V d ′ occurs with multiplicity 1 in V ℓ ′ ⊗ V k ′ , we concludethat V d ′ occurs with multiplicity 1 in W . In this case V (2 d ′ ) must occur as anorthogonal direct summand of W by Lemma 6.10 (ii) and Lemma 6.9. Now it followsfrom (7.4) and Lemma 6.13 that V (2 α +1 d ′ ) = V (2 d j ) occurs as an orthogonal directsummand of V (2 ℓ ) ⊗ V (2 k ). (cid:3) We finish this section by giving some examples that illustrate Theorem 7.4. Example . For any 0 < k ≤ q/ 2, it follows from (7.1) that V ⊗ V k ∼ = V k . Thuswe conclude from Theorem 7.4 that V (2) ⊗ V (2 k ) ∼ = ( W (2 k ) , if k ≡ .V (2 k ) , if k ≡ . Example . For 1 < k ≤ q/ 2, it is well known that V ⊗ V k ∼ = V k if k ≡ V ⊗ V k ∼ = V k − ⊕ V k +1 if k ≡ V ⊗ V k ∼ = V k if k ≡ V ⊗ V k ∼ = V k − ⊕ V k +2 if k ≡ V (4) ⊗ V (2 k ) ∼ = W (2 k ) , if k ≡ .W (2 k − ⊥ W (2 k + 2) , if k ≡ .V (2 k ) , if k ≡ .W (2 k − ⊥ W (2 k + 2) , if k ≡ . by Theorem 7.4. Example . For 2 < k ≤ q/ 2, similarly to Examples 7.5 and Example 7.6,from (7.1) and the decomposition of V ⊗ V k (Example 4.5) one finds using Theorem7.4 that V (6) ⊗ V (2 k ) ∼ = W (2 k ) , if k ≡ .W (2 k − ⊥ V (2 k + 4) , if k ≡ .W (2 k − ⊥ W (2 k ) ⊥ W (2 k + 4) , if k ≡ .V (2 k − ⊥ W (2 k + 2) , if k ≡ . An alternating bilinear form on V ⊗ V ∗ Let V be a finite-dimensional vector space over K with n = dim V and set G = SL( V ). The purpose of this section is to describe a non-zero alternating G -invariant bilinear form on V ⊗ V ∗ explicitly, and to give some of its basic properties.Fix a basis e , . . . , e n of V and the corresponding dual basis e ∗ , . . . , e ∗ n of V ∗ ,so e ∗ i ( e j ) = δ i,j for all 1 ≤ i, j ≤ n .There is a natural bilinear form ˆ b V on V ⊗ V ∗ defined byˆ b V ( v ⊗ f, v ′ ⊗ f ′ ) = f ( v ′ ) f ′ ( v )for all v, v ′ ∈ V and f, f ′ ∈ V ∗ . A straightforward calculation shows that ˆ b V is anon-degenerate G -invariant symmetric bilinear form. Note that ˆ b V ( e i ⊗ e ∗ i , e i ⊗ e ∗ i ) =1 for all 1 ≤ i ≤ n , so ˆ b V is not alternating. However, as in [Kor18, Lemma 4.1],one can use ˆ b V to define an alternating bilinear form on V ⊗ V ∗ .Let γ V = X ≤ i ≤ n e i ⊗ e ∗ i . It is well known that the choice of γ V does not depend on the choice of the basis ( e i ),and furthermore γ V spans the fixed point space of G on V ⊗ V ∗ , see for example[Kor19, Lemma 3.7].We have a morphism of G -modules ψ V : V ⊗ V ∗ → K defined by ψ V ( v ⊗ f ) = f ( v )for all v ∈ V and f ∈ V ∗ . By calculating ψ V ( x ) on basis elements e i ⊗ e ∗ j of V ⊗ V ∗ ,one finds that ψ V ( x ) = ˆ b V ( x, γ V ) for all x ∈ V ⊗ V ∗ .We define τ : V ⊗ V ∗ → V ⊗ V ∗ by τ ( x ) = x + ψ V ( x ) γ V for all x ∈ V ⊗ V ∗ .Then τ is a morphism of G -modules, so b V defined by(8.1) b V ( x, y ) = ˆ b V ( τ ( x ) , y ) = ˆ b V ( x, y ) + ψ V ( x ) ψ V ( y )for all x, y ∈ V ⊗ V ∗ is a G -invariant bilinear form on V ⊗ V ∗ . For calculations, itis useful to note that b V ( e i ⊗ e ∗ i , e j ⊗ e ∗ j ) = ( , if i = j. , if i = j. for all 1 ≤ i, j ≤ n . Furthermore, for i = jb V ( e i ⊗ e ∗ j , e i ⊗ e ∗ j ) = ( , if i = j and j = i . , otherwise.We will now make some basic observations about b V . Lemma 8.1. The following statements hold: (i) The bilinear form b V is alternating. (ii) If n is even, then rad b V = 0 and Ker ψ V = h γ V i ⊥ . (iii) If n is odd, then rad b V = h γ V i and V ⊗ V ∗ = Ker ψ V ⊕ h γ V i . (iv) We have h γ V i ⊥ / h γ V i ∼ = L G ( ̟ + ̟ n − ) as G -modules.Proof. For (i), first note that b V is symmetric since ˆ b V is. It is easy to verify that b V ( x, x ) = 0 for all basis elements x = e i ⊗ e ∗ j , so b V is alternating.The bilinear form ˆ b V is non-degenerate, so rad b V = Ker τ . Clearly Ker τ ⊆ h γ V i ,and τ ( γ V ) = γ V + ψ V ( γ V ) γ V = ( n + 1) γ V . Thus γ V ∈ Ker τ if and only if n isodd, from which the claims about rad b V in (ii) and (iii) follow. For other claim in(ii), note that Ker ψ V ⊆ h γ V i ⊥ . If n is even, then h γ V i ⊥ = V ⊗ V ∗ and so equalityholds since both subspaces have codimension one. The other claim in (iii) followssince γ V Ker ψ V when n is odd.Since γ V spans the unique 1-dimensional G -submodule of V ⊗ V ∗ , claim (iv)follows easily from (ii), (iii), and Lemma 3.5. (cid:3) Lemma 8.2. Every G -invariant alternating bilinear form on V ⊗ V ∗ is a scalarmultiple of b V .Proof. It will suffice to show that V ⊗ V ∗ has a unique G -invariant alternatingbilinear form up to a scalar multiple. If n is even, this follows from [Kor18, Lemma4.2]. If n is odd, it follows from Lemma 3.5 that V ⊗ V ∗ = W ⊕ h γ V i , where W ∼ = L G ( ̟ + ̟ n − ).Let b be a G -invariant alternating bilinear form on V ⊗ V ∗ . Then the map f : V ⊗ V ∗ → K defined by f ( v ) = b ( v, γ V ) is a morphism of G -modules, where G acts trivially on K . The map f must vanish on W since W is a non-trivial irreducible K [ G ]-module, and furthermore f vanishes on γ V since b is alternating. Thus f is ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 25 zero, which means that γ V ∈ rad b . Now the claim follows, since W is irreducibleand thus has a unique G -invariant bilinear form up to a scalar multiple. (cid:3) Remark . An alternative point of view that could have been used in this sectionis the following. Recall that there is a natural isomorphism V ⊗ V ∗ → End( V )of G -modules, where for v ∈ V and f ∈ V ∗ the image of v ⊗ f is the linear map V → V defined by w f ( w ) v . Here G acts on End( V ) by conjugation.Under this isomorphism, the element γ V corresponds to the identity map Id V on V , and the map ψ V corresponds to the trace map End( V ) → K . The bilinearform ˆ b V corresponds to the bilinear form on End( V ) defined by ( A, B ) Tr( AB ).The map τ corresponds to A A + Tr( A ) Id V , so the alternating bilinear form b V corresponds to the bilinear form defined by ( A, B ) Tr( AB ) − Tr( A ) Tr( B ).9. An alternating bilinear form on ∧ ( V )Let b be a non-degenerate alternating bilinear form on a vector space V over K with dim V = 2 n . Set G = Sp( V, b ). This section is analogous to the previous one,and we will be concerned with a non-zero alternating G -invariant bilinear form on ∧ ( V ) and its basic properties.The G -invariant bilinear form b on V induces a G -invariant bilinear form ˆ a V on ∧ ( V ) via ˆ a V ( v ∧ v , w ∧ w ) = det( b ( v i , w j )) ≤ i,j ≤ for all v i , w j ∈ V . That is,ˆ a V ( v ∧ v , w ∧ w ) = b ( v , w ) b ( v , w ) + b ( v , w ) b ( v , w )for all v , v , w , w ∈ V . The bilinear form ˆ a V is a non-degenerate G -invariantsymmetric bilinear form. Now ˆ a V is not alternating, but as in Section 8, with asmall modification we can construct a G -invariant alternating bilinear form.Fix a basis e , . . . , e n of V such that b ( e i , e j ) = 1 if i + j = 2 n +1 and b ( e i , e j ) = 0otherwise. Define β V = X ≤ i ≤ n e i ∧ e n +1 − i . It follows from [DB10, 3.4] that β V does not depend on the choice of the basis ( e i ),and thus it is fixed by the action of Sp( V, b ) on ∧ ( V ). Furthermore, it is clearfrom Lemma 3.6 that β V is the unique Sp( V, b )-fixed point in ∧ ( V ), up to scalarmultiples.We have a morphism of G -modules ϕ V : ∧ ( V ) → K defined by ϕ V ( v ∧ v ′ ) = b ( v, v ′ ) for all v, v ′ ∈ V . Similarly to ψ V in Section 8, we see that ϕ V ( x ) = ˆ a V ( x, β V )for all x ∈ ∧ ( V ).Define σ : ∧ ( V ) → ∧ ( V ) by σ ( x ) = x + ϕ V ( x ) β V for all x ∈ ∧ ( V ). Then σ isa morphism of G -modules, and so a V defined by a V ( x, y ) = ˆ a V ( σ ( x ) , y ) = ˆ a V ( x, y ) + ϕ V ( x ) ϕ V ( y )for all x, y ∈ ∧ ( V ) is a G -invariant bilinear form on ∧ ( V ). Lemma 9.1. The following statements hold: (i) The bilinear form a V is alternating. (ii) If n is even, then rad a V = 0 and Ker ϕ V = h β V i ⊥ . (iii) If n is odd, then rad a V = h β V i and ∧ ( V ) = Ker ϕ V ⊕ h β V i . (iv) We have h β V i ⊥ / h β V i ∼ = L G ( ̟ ) as G -modules.Proof. Same as Lemma 8.1. (cid:3) Lemma 9.2. Every G -invariant alternating bilinear form on ∧ ( V ) is a scalarmultiple of a V .Proof. If n is even, the claim follows from [Kor18, Lemma 4.2]. If n is odd, thelemma follows with the same proof as Lemma 8.2. (cid:3) Lemma 9.3. Let H < G , and let ( V, b ) = ( W ⊕ W ∗ , b ) be the paired moduleassociated with some K [ H ] -module W . Then ∧ ( W ⊕ W ∗ ) = ∧ ( W ) ⊕ ∧ ( W ∗ ) ⊕ ( W ∧ W ∗ ) , where ( ∧ ( W ) ⊕ ∧ ( W ∗ ) , a V ) is the paired module associated with ∧ ( W ) , and ( W ∧ W ∗ , a V ) ∼ = ( W ⊗ W ∗ , b W ) as bilinear K [ H ] -modules.Proof. The restriction of a V to ∧ ( W ) ⊕∧ ( W ∗ ) is non-degenerate, and furthermore ∧ ( W ) ⊕ ∧ ( W ∗ ) is a totally singular decomposition with respect to a V . Thus( ∧ ( W ) ⊕ ∧ ( W ∗ ) , a V ) is the paired module associated with ∧ ( W ) by Lemma5.12.For W ∧ W ∗ , a straightforward verification shows that w ∧ f w ⊗ f defines anisomorphism ( W ∧ W ∗ , a V ) → ( W ⊗ W ∗ , b W ) of bilinear K [ H ]-modules. (cid:3) Hesselink normal forms on V ⊗ V ∗ In this section, we will prove Theorem A, one of the main results of this paper.At the end of this section, we will also give some examples which illustrate howTheorem A can be applied.Let V be a vector space over K with n = dim V . Set G = SL( V ). Recall (Lemma8.2) that we have an alternating G -invariant bilinear form b V on V ⊗ V ∗ which isunique up to scalar multiples. By Lemma 8.1, the bilinear form b V induces a non-degenerate G -invariant bilinear form on h γ V i ⊥ / h γ V i ∼ = L G ( ̟ + ̟ n − ), giving usa representation f : G → Sp( L G ( ̟ + ̟ n − ) , b V ). Furthermore, the bilinear form b V is non-degenerate if and only if n is even (Lemma 8.1), in which case we alsoget a representation f ′ : G → Sp( V ⊗ V ∗ , b V ).For each unipotent element u ∈ G , Theorem A describes the Hesselink normalform of f ( u ). Furthermore when n is even, Theorem A also gives the Hesselinknormal form of f ′ ( u ). We state the Hesselink normal forms in terms of the Jordannormal form of u on V ⊗ V ∗ , which one can calculate using Theorem 4.1.We will first need two lemmas, and to setup their statements we fix a basis e , . . . , e n of V and the corresponding dual basis e ∗ , . . . , e ∗ n of V ∗ . For convenienceof notation, we set e i = 0 and e ∗ i = 0 for all i ≤ i > n . Let u ∈ G be aunipotent Jordan block with respect to the basis ( e i ), that is, ue i = e i + e i − for all 1 ≤ i ≤ n . As usual, we denote by X the element u − K [ u ].Let α > α ≤ n < α +1 . For all 1 ≤ β ≤ α + 1 and 2 β − + 1 ≤ i ≤ β , we define(10.1) v ( β ) i = X j ≥ e i + j β ⊗ e ∗ i − β − + j β . The key lemma in this section is the following. Lemma 10.1. Let ≤ β ≤ α + 1 and write n = k β + r , where ≤ r < β . Thenfor all β − + 1 ≤ i ≤ β : ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 27 (i) We have X β · v ( β ) i = 0 if and only if ≤ r < i − β − or r ≥ i . (ii) If r < i , then b V ( X β − v ( β ) i , v ( β ) i ) = k . (iii) If r ≥ i , then b V ( X β − v ( β ) i , v ( β ) i ) = k + 1 .Proof. For (i), note first that X β = u β − 1, so X β · ( e i + j β ⊗ e ∗ i − β − + j β ) equals(10.2) ( X β e i + j β ⊗ u β e ∗ i − β − + j β ) + ( e i + j β ⊗ X β e ∗ i − β − + j β )for all j ≥ 0. Now by [Kor19, (5.1)] we have X β e i ′ = e i ′ − β and X β e ∗ i ′ = P j ′ > e ∗ i ′ + j β for all i ′ , so (10.2) equals q j − + q j , where we define q j = e i + j β ⊗ X j ′ >j e ∗ i − β − + j ′ β for all j ≥ − 1. Note that q − = 0 since e i − β = 0. Therefore(10.3) X β · v ( β ) i = X j ≥ ( q j − + q j ) = q − + q m = q m , where m ≥ i + m β ≤ n .It is clear that q m = 0 if and only if i − β − + ( m + 1)2 β > n . If r < i , then m = k − q m = 0 if and only if i − β − + k β > n , which is equivalentto r < i − β − . Similarly if r ≥ i , then m = k and so q m = 0 if and only if i − β − + ( k + 1)2 β > n , which is equivalent to r < i + 2 β − , and this always holdssince i > β − . We have shown that q m = 0 if and only if 0 ≤ r < i − β − or r ≥ i ,which together with (10.3) completes the proof of (i).For (ii) and (iii), we proceed to calculate b V ( X β − v ( β ) i , v ( β ) i ). By [Kor19, Lemma5.2], for all j ≥ X β − · ( e i + j β ⊗ e ∗ i − β − + j β ) equal to X ≤ t ≤ β − ≤ s ≤ t (cid:18) β − t (cid:19)(cid:18) ts (cid:19) X t e i + j β ⊗ X β − − s e ∗ i − β − + j β = X ≤ t ≤ β − ≤ s ≤ t (cid:18) ts (cid:19) e i − t + j β ⊗ X β − − s e ∗ i − β − + j β (10.4)where (10.4) holds since (cid:0) β − t (cid:1) ≡ ≤ t ≤ β − v ( β ) i with respect to b V occur only for 0 ≤ t ≤ β − i − t + j β = i − β − + j ′ β , which isonly possible for t = 2 β − . Furthermore, by Lucas’ theorem for 0 ≤ s ≤ β − wehave (cid:0) β − s (cid:1) ≡ s = 0 and s = 2 β − . Thus (10.4) equals(10.5) e i − β − + j β ⊗ ( X β − e ∗ i − β − + j β + X β − − e ∗ i − β − + j β ) mod h v ( β ) i i ⊥ . We show next that (10.5) equals(10.6) e i − β − + j β ⊗ e ∗ i + j β mod h v ( β ) i i ⊥ . We divide the proof of (10.6) into two cases. Case 1: Suppose that β = 1. In this case (10.5) equals e i − j ⊗ ( Xe ∗ i − j + e ∗ i − j )= e i − j ⊗ X j ′ ≥ i − j e ∗ j ′ (10.7) ≡ e i − j ⊗ e ∗ i +2 j mod h v ( β ) i i ⊥ where (10.7) is given by [Kor19, Lemma 5.1 (ii)]. Case 2: Suppose that β > e i − β − + j β ⊗ X β − e ∗ i − β − + j β equal to a sum of some basis elements e i − β − + j β ⊗ e ∗ j ′ such that j ′ ≥ i + 2 β − − j β . Here j ′ > i + j β since β > 1, so we conclude that e i − β − + j β ⊗ X β − e ∗ i − β − + j β has zero product with v ( β ) i with respect to b V . Thus (10.5)equals e i − β − + j β ⊗ X β − − e ∗ i − β − + j β mod h v ( β ) i i ⊥ = e i − β − + j β ⊗ X j ′ ≥ i − j β (cid:18) j ′ − i + 2 β − − j β − β − − (cid:19) e ∗ j ′ mod h v ( β ) i i ⊥ (10.8) = e i − β − + j β ⊗ (cid:18) β − − β − − (cid:19) e ∗ i + j β mod h v ( β ) i i ⊥ = e i − β − + j β ⊗ e ∗ i + j β mod h v ( β ) i i ⊥ where (10.8) is given by [Kor19, Lemma 5.1 (ii)]. This completes the proof of (10.6).We have shown that X β − · ( e i + j β ⊗ e ∗ i − β − + j β ) equals (10.6) modulo h v ( β ) i i ⊥ .From this we conclude that b V ( X β − v ( β ) i , v ( β ) i ) = X j ≥ b V ( e i − β − + j β ⊗ e ∗ i + j β , v ( β ) i )= X j ≥ i + j β ≤ n k if r < i and k + 1 if r ≥ i , proving (ii) and (iii). (cid:3) Lemma 10.2. Let ≤ β ≤ α + 1 . If V n ⊗ V n has a Jordan block of size β , then X β · v ( β ) i = 0 and b V ( X β − v ( β ) i , v ( β ) i ) = 0 for some β − + 1 ≤ i ≤ β .Proof. Suppose that V n ⊗ V n has a Jordan block of size 2 β . By Theorem 4.7, thismeans that 2 β occurs in the consecutive-ones binary-expansion of n . Equivalently,either (a) 2 β occurs in the binary expansion of n and 2 β − does not; or (b) 2 β − occurs in the binary expansion of n and 2 β does not.If (a) holds, then n = k β + r , where 0 ≤ r < β − and k is odd. By Lemma10.1 (i) and (ii), for any r + 2 β − + 1 ≤ i ≤ β we have X β · v ( β ) i = 0 and b V ( X β − v ( β ) i , v ( β ) i ) = k = 0. For example, one can choose i = 2 β . ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 29 If (b) holds, then n = k β + r where 2 β − ≤ r < β and k is even. ByLemma 10.1 (i) and (ii), for any 2 β − + 1 ≤ i ≤ r we have X β · v ( β ) i = 0 and b V ( X β − v ( β ) i , v ( β ) i ) = k + 1 = 0. For example, we could choose i = r . (cid:3) With Lemma 10.2, we will be able to prove Theorem A, our first main result.We refer the reader to the introduction for the statement of the theorem. Proof of Theorem A. We first recall the setup of the theorem. Let G = SL( V ),where dim V = n for some n ≥ 2. Let u ∈ G be unipotent and V ∼ = V d ⊕· · ·⊕ V d t as K [ u ]-modules, where t ≥ d r ≥ ≤ r ≤ t . Set α = ν (gcd( d , . . . , d t )).Suppose that V ⊗ V ∗ ∼ = ⊕ d ≥ V λ ( d ) d and L G ( ̟ + ̟ n − ) ∼ = ⊕ d ≥ V λ ′ ( d ) d as K [ u ]-modules, where λ ( d ) , λ ′ ( d ) ≥ d ≥ 1. We identify L G ( ̟ + ̟ n − ) asthe subquotient ( h β V i ⊥ / h β V i , b V ) of ( V ⊗ V ∗ , b V ) — see Lemma 8.1 (iv). Set ε := ε V ⊗ V ∗ ,b V and ε ′ := ε L G ( ̟ + ̟ n − ) ,b V .For statements (i) – (iii) of the theorem, the description of λ ′ is just [Kor19,Theorem 6.1] in characteristic p = 2.For the proof of claims (iv) – (vi) concerning ε and ε ′ , we first setup somemore notation. Let V = W ⊕ · · · ⊕ W t , where W r are u -invariant subspaces and W r ∼ = V d r for all 1 ≤ r ≤ t . For each r , choose a basis ( e ( r ) j ) ≤ j ≤ d r of W r such that ue ( r ) j = e ( r ) j + e ( r ) j − for all 1 ≤ j ≤ d r , where we set e ( r ) j = 0 for all j ≤ 0. For thebasis ( e ( r ) j ) of V , we let ( e ( r ) ∗ j ) be the corresponding dual basis of V ∗ .We consider (iv). Suppose first that ε ( d ) = 1, so now d is even by Lemma 6.9.For each 1 ≤ r ≤ t , we identify W ∗ r with the subspace spanned by the ( e ( r ) ∗ j ). Then V ⊗ V ∗ = M ≤ r ≤ t ( W r ⊗ W ∗ r ) ⊕ M ≤ r 1, as claimed.We assume next that 2 | n , so now b V is non-degenerate and h γ V i ⊥ = Ker ψ V byLemma 8.1 (ii). By [Kor19, (6.2)], there exists a ˆ δ ∈ V ⊗ V ∗ such that X α ˆ δ = 0 and ˆ δ 6∈ h γ V i ⊥ . In the case where α = 0, it follows from Lemma 6.11 (ii) that ε ′ ( d ) = ε ( d ) for all d ≥ 1, proving the theorem in this case. Assume then for therest of the proof that α > ε ′ ( d ) = ε ( d ) for all d = 2 α − , α by Lemma 6.11 (iii) – (iv). We showthat ε (2 α ) = ε ′ (2 α ). To this end, pick some 1 ≤ r ≤ t such that ν ( d r ) = α . Then2 α occurs in the consecutive-ones binary expansion of d r , so ε (2 α ) = ε ′ (2 α ) = 1as shown at the end of the proof of (iv). In the case where α = 1, it follows that ε ( d ) = ε ′ ( d ) for all d ≥ 1, as claimed. Thus we can assume α > ε ( d ) = ε ′ ( d ) for all d = 2 α − 2, as is claimed by(v) and (vi). For ε (2 α − 2) and ε ′ (2 α − u in V ⊗ V ∗ is 2 α . Thus u has no Jordan blocks ofsize 2 α − V ⊗ V ∗ , and so ε (2 α − 2) = 0 by Lemma 6.9. If (iii)(b) holds, then2 α − L G ( ̟ + ̟ n − ) with multiplicity one, so ε ′ (2 α − 2) = 1 by Lemma6.10 (ii). If (iii)(b) does not hold, then u has no Jordan blocks of size 2 α − L G ( ̟ + ̟ n − ), and so ε ′ (2 α − 2) = 0. (cid:3) In the following we show with small examples how Theorem A is applied. Let G = SL( V ) and n = dim V , and let u ∈ G be a unipotent element. Let α = ν (gcd( d , . . . , d t )) as in Theorem A, so 2 α is the largest power of two dividing everyJordan block size of u on V . Set ε := ε V ⊗ V ∗ ,b V and ε ′ := ε L G ( ̟ + ̟ n − ) ,b V . In whatfollows we shall use Theorem 6.7 to describe the decomposition of ( V ⊗ V ∗ , b V ) and( L G ( ̟ + ̟ n − ) , b V ) into bilinear K [ u ]-modules. Example . When n = 2 and V ∼ = V as K [ u ]-modules, we have α = 1 and byTheorem 4.1 we get V ⊗ V ∗ ∼ = V as K [ u ]-modules. We have a consecutive-onesbinary expansion n = 2 , so ε (2) = 1 and ε ( d ) = 0 for all d = 2 by Theorem A(iv). Thus ( V ⊗ V ∗ , b V ) ∼ = V (2) as bilinear K [ u ]-modules. In this case Theorem A(iii)(c) and (vi) apply, giving L G ( ̟ + ̟ n − ) ∼ = V and ε ′ ( d ) = ε ( d ) for all d ≥ L G ( ̟ + ̟ n − ) , b V ) ∼ = V (2) as bilinear K [ u ]-modules. Example . When n = 4 and V ∼ = V as K [ u ]-modules, we have α = 2 and byTheorem 4.1 we get V ⊗ V ∗ ∼ = V as K [ u ]-modules. We have a consecutive-onesbinary expansion n = 2 , so ε (4) = 1 and ε ( d ) = 0 for all d = 4 by Theorem A(iv). Thus ( V ⊗ V ∗ , b V ) ∼ = V (4) as bilinear K [ u ]-modules. In this case Theorem A(iii)(b) and (vi) apply, giving L G ( ̟ + ̟ n − ) ∼ = V ⊕ V , ε ′ (2) = 1, and ε ′ ( d ) = ε ( d )for all d = 2. Hence ( L G ( ̟ + ̟ n − ) , b V ) ∼ = V (2) ⊥ V (4) as bilinear K [ u ]-modules. Example . For n = 6 and V ∼ = V ⊕ V as K [ u ]-modules, we have α = 0 andby Theorem 4.1 we get V ⊗ V ∗ ∼ = V ⊕ V ⊕ V ⊕ V . We have a consecutive-onesbinary expansion 5 = 2 − + 2 , so ε (8) = 1, ε (4) = 1 and ε ( d ) = 0 for all d = 4 , V ⊗ V ∗ , b V ) ∼ = W (1) ⊥ V (4) ⊥ W (5) ⊥ V (8) as bilinear K [ u ]-modules. In this case Theorem A (ii) and (vi) apply, giving( L G ( ̟ + ̟ n − ) , b V ) ∼ = V (4) ⊥ W (5) ⊥ V (8) as bilinear K [ u ]-modules. Example . In Table 1, we illustrate Theorem A for all 2 ≤ n ≤ 7. In the firstcolumn we use notation ( d n , . . . , d n t t ) to denote that V ∼ = V n d ⊕ · · · ⊕ V n t d t as K [ u ]-modules. In the third and second columns, we use notation as in Theorem 6.7.That is, for an alternating bilinear K [ u ]-module ( W, b ), we use ( d n ε , . . . , d tn t ε t ) todenote that W ∼ = V n d ⊕ · · · ⊕ V n t d t as K [ u ]-modules and ε W,b ( d i ) = ε i for 1 ≤ i ≤ t . ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 31 Table 1. n V ↓ K [ u ] ( V ⊗ V ∗ , b V ) ( L G ( ̟ + ̟ n − ) , b V ) n = 2 (2) (2 ) (2 ) n = 3 (3) (1 , ) (4 )(1 , 2) (1 , ) (2 ) n = 4 (4) (4 ) (2 , )(1 , 3) (1 , , ) (3 , )(2 ) (2 ) (1 , )(1 , 2) (1 , ) (1 , ) n = 5 (5) (1 , , ) (4 , )(1 , 4) (1 , ) (4 )(2 , 3) (1 , , ) (2 , )(1 , 3) (1 , , ) (1 , , )(1 , ) (1 , ) (2 )(1 , 2) (1 , ) (1 , ) n = 6 (6) (2 , ) (2 , )(1 , 5) (1 , , , ) (4 , , )(2 , 4) (2 , ) (2 , )(1 , 4) (1 , ) (1 , )(3 ) (1 , ) (1 , )(1 , , 3) (1 , , , ) (2 , , )(1 , 3) (1 , , ) (1 , , )(2 ) (2 ) (2 )(1 , ) (1 , ) (1 , )(1 , 2) (1 , ) (1 , ) n = 7 (7) (1 , ) (8 )(1 , 6) (1 , , , ) (2 , , )(2 , 5) (1 , , , , ) (2 , , , )(1 , 5) (1 , , , ) (1 , , , )(3 , 4) (1 , ) (4 )(1 , , 4) (1 , , ) (2 , )(1 , 4) (1 , ) (1 , )(1 , ) (1 , , ) (1 , , )(2 , 3) (1 , , ) (2 , )(1 , , 3) (1 , , , ) (1 , , , )(1 , 3) (1 , , ) (1 , , )(1 , ) (1 , ) (2 )(1 , ) (1 , ) (1 , )(1 , 2) (1 , ) (1 , ) Hesselink normal forms on ∧ ( V )In this section, we will give a proof of Theorem B. At the end of this section wehave included a number of examples to illustrate how Theorem B can be used.Let G = Sp( V, b ), where V is a K -vector space of dimension 2 n and b is anon-degenerate alternating bilinear form on V . We recall the following which isanalogous to the setup of the previous section. By Lemma 9.2, we have an al-ternating G -invariant bilinear form a V on ∧ ( V ) which is unique up to scalar multiples. By Lemma 9.1, the bilinear form a V induces a non-degenerate G -invariant bilinear form on h β V i ⊥ / h β V i ∼ = L G ( ̟ ), which gives us a representa-tion f : G → Sp( L G ( ̟ ) , a V ). Furthermore, by Lemma 9.1 the bilinear form a V is non-degenerate if and only if n is even, in which case we get a representation f ′ : G → Sp( ∧ ( V ) , a V ).In Theorem B, we describe the Hesselink normal form of f ( u ) for each unipotentelement u ∈ G . When n is even, Theorem B also describes the Hesselink normalform of f ′ ( u ).We begin with some observations in case where u ∈ Sp( V, b ) and ( V, b ) ∼ = V (2 n )or ( V, b ) ∼ = W ( n ) as bilinear K [ u ]-modules. Let n ≥ V, b ) = V (2 n )with a basis e , . . . , e n as in Definition 6.2. That is, we define b ( e i , e j ) = 1 if i + j = 2 n + 1 and 0 otherwise. Furthermore, the action of u on the e i is given by ue = e ,ue i = e i + e i − + · · · + e for all 2 ≤ i ≤ n + 1 ,ue i = e i + e i − for all n + 1 < i ≤ n. Throughout we will denote e j = 0 for all j ≤ j > n .Suppose that 2 α | n , where α ≥ 0. We define(11.1) δ α = X ≤ i ≤ n e i ∧ e n − α t ( α ) i where t ( α ) i = (cid:4) i − α (cid:5) for all 1 ≤ i ≤ n , cf. [Kor19, (5.6)]. Note that δ = β V . Proposition 11.1. Suppose that α | n , where α > . Then X α − δ α = δ α − .Proof. We first note that X α − = u α − − 1, so it follows that(11.2) X α − · ( v ∧ w ) = ( X α − v ) ∧ ( X α − w ) + v ∧ ( X α − w ) + ( X α − v ) ∧ w for all v, w ∈ V .It is clear from the definition that ( u − e i = ue i − for all 1 ≤ i ≤ n + 1. Thus( u − k e i = u k e i − k for all k ≥ ≤ i ≤ n + 1. With k = 2 α − , we see that X α − e i = e i − α − + X α − e i − α − for all 1 ≤ i ≤ n + 1. Consequently(11.3) X α − e i = X j ≥ e i − α − j for all 1 ≤ i ≤ n + 1.It is clear that X α − e i = e i − α − for all i > n + 2 α − . Since 2 n − α t ( α ) i >n + 2 α − , it follows that(11.4) X α − e n − α t ( α ) i = e n − α t ( α ) i − α − for all 1 ≤ i ≤ n .Applying (11.3) and (11.4) on (11.2), we see that X α − δ α equals X ≤ i ≤ n e i ∧ e n − α t ( α ) i − α − + X ≤ i ≤ nj ≥ e i − α − j ∧ (cid:16) e n − α t ( α ) i + e n − α t ( α ) i − α − (cid:17) , where collecting the terms of the form e i ∧ v , we get(11.5) X ≤ i ≤ n e i ∧ e n − α t ( α ) i − α − + X j ≥ s j , ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 33 with s j = e n − α t ( α ) i +2 α − j + e n − α t ( α ) i +2 α − j − α − for all j ≥ ≤ i ≤ n , we have i − t ( α ) i α + r i , for some 0 ≤ r i < α . Considerfirst the case where 0 ≤ r i < α − . Then t ( α ) i +2 α − j = t ( α ) i + j if j is even, and t ( α ) i +2 α − j = t ( α ) i + j − if j is odd. Hence if j > s j = s j +1 , whichimplies that P j ≥ s j = s . Now t ( α ) i +2 α − = t ( α ) i , so e n − α t ( α ) i − α − + P j ≥ s j = e n − α t ( α ) i . Because 0 ≤ r i < α − , we have t ( α − i = 2 t ( α ) i , so the summandin (11.5) corresponding to i is e i ∧ e n − α − t ( α − i .If 2 α − ≤ r i < α , a similar calculation shows that s j = s j +1 for all j ≥ P j ≥ s j = 0. In this case the summand in (11.5) correspondingto i is e i ∧ e n − α t ( α ) i − α − = e i ∧ e n − α − t ( α − i , since t ( α − i = 2 t ( α ) i + 1 when2 α − ≤ r i < α . Thus we conclude that (11.5) equals X ≤ i ≤ n e i ∧ e n − α − t ( α − i = δ α − as claimed. (cid:3) Corollary 11.2. Suppose that α | n , where α ≥ . Then X α − δ α = δ = β V .Proof. If α = 0, the claim follows since δ α = δ = β V . If α > 0, then 2 α − P ≤ β ≤ α − β , so the claim follows using Proposition 11.1. (cid:3) Lemma 11.3. Let ( V, b ) be a bilinear K [ u ] -module such that ( V, b ) ∼ = V (2 n ) or ( V, b ) ∼ = W ( n ) , where n > . Assume that α | n , where α ≥ . Then: (i) There exists δ ∈ ∧ ( V ) such that X α − δ = β V and ϕ V ( δ ) = n/ α . (ii) We have Ker X α − ∧ ( V ) ⊆ Ker ϕ V . (iii) If α = ν ( n ) , then Ker X α ∧ ( V ) Ker ϕ V . (iv) If α < ν ( n ) , then a V ( X α − v, v ) = 0 for all v ∈ Ker X α ∧ ( V ) . (v) If α = ν ( n ) and ( V, b ) ∼ = V (2 n ) , then a V ( X α − v, v ) = 0 for all v ∈ Ker X α Ker ϕ V . (vi) If α = ν ( n ) > and ( V, b ) ∼ = W ( n ) , then a V ( X α − v, v ) = 0 for some v ∈ Ker X α Ker ϕ V .Proof. For (i), we first consider the case where ( V, b ) ∼ = V (2 n ). Take a basis e , . . . , e n as above and δ α as in (11.1). Then X α − δ α = β V by Corollary 11.2.For the summands in (11.1), for 1 ≤ i ≤ n we have ϕ V ( e i ∧ e n − α t ( α ) i ) = 1 if2 n − α t ( α ) i + i = 2 n + 1 and 0 otherwise. Now 2 n − α t ( α ) i + i = 2 n + 1 if and onlyif 2 α | i − 1, so we deduce that ϕ V ( δ α ) = n/ α .Next consider (i) in the case where ( V, b ) ∼ = W ( n ). Fix a basis e , . . . , e n , f , . . . , f n of V such that the subspaces A = h e , . . . , e n i and B = h f , . . . , f n i are u -invariant and totally singular, and b ( e i , f j ) = δ i,j . With a suitable choice ofbasis, we also arrange ue = e and ue i = e i + e i − for all 1 < i ≤ n . Then B ∼ = A ∗ , with an isomorphism of K [ u ]-modules defined by f i e ∗ i . Recall that ∧ ( V ) decomposes as ∧ ( V ) = ∧ ( A ) ⊕ ∧ ( B ) ⊕ A ∧ B. By the proof of Lemma 9.3, we have ( A ∧ B, a V ) ∼ = ( A ⊗ A ∗ , b A ), with an isomorphismof bilinear K [ u ]-modules given by e i ∧ f j e i ⊗ e ∗ j . Thus by [Kor19, Corollary 5.5],with δ = X ≤ i ≤ α X ≤ j ≤ n/ α − e j α + i ∧ f j α +1 we get X α − δ = P ≤ i ≤ n e i ∧ f i = β V . Since ϕ V ( δ ) = n/ α , this completes theproof of (i).Claim (ii) is obvious in the case where α = 0. If α > 0, then n is even andKer ϕ V = h β V i ⊥ . In this case (ii) follows from (i) and Lemma 6.11 (i).For (iii), suppose that α = ν ( n ). If δ is as in (i), then δ ∈ Ker X α ∧ ( V ) since β V is a fixed point. Since ϕ V ( δ ) = n/ α = 0, claim (iii) follows.In the case where V ∼ = V (2 n ), claim (iv) follows from Lemma 6.9, since thesmallest Jordan block size of u on ∧ ( V ) is 2 ν ( n ) (Lemma 4.12). Consider then(iv) in the case where V ∼ = W ( n ), and let V = A ⊕ B as in the proof of (i) above.For v ∈ Ker X α ∧ ( V ) , we can write v = v ′ + v ′′ , where v ′ ∈ Ker X α ∧ ( A ) ⊕∧ ( B ) and v ′′ ∈ Ker X α A ∧ B . We have a V ( X α − v ′ , v ′ ) = 0 since ( ∧ ( A ) ⊕∧ ( B ) , a V ) is a pairedmodule (Lemma 9.3). Furthermore, by Lemma 6.9 we get a V ( X α − v ′′ , v ′′ ) = 0,since the smallest Jordan block size of u in A ∧ B ∼ = A ⊗ A ∗ is 2 ν ( n ) > α [Kor19,Lemma 4.3]. Thus a V ( X α − v, v ) = 0 by Lemma 6.8 (iv).Claim (v) is clear if α = 0, so suppose that α > δ ∈ ∧ ( V ) be as in(i). Now 2 ∤ n α , so δ 6∈ h β V i ⊥ and from Lemma 6.11 (iv) and Lemma 4.12 weconclude that u has no Jordan blocks of size 2 α on h β V i ⊥ / h β V i = Ker ϕ V / h β V i .Thus a V ( X α − v, v ) = 0 for all v ∈ Ker X α Ker ϕ V by Lemma 6.9.For (vi), suppose that ( V, b ) ∼ = W ( n ) and let V = A ⊕ B as in the proof of (i)above. For 2 α − + 1 ≤ i ≤ α , similarly to (10.1) we define w i = X j ≥ e i + j α ∧ f i − α − + j α . Now 2 α occurs as a Jordan block size in V n ⊗ V n for example by [Kor19, Lemma 4.3],so from ( A ∧ B, a V ) ∼ = ( A ⊗ A ∗ , b A ) and Lemma 10.2 we conclude that X α · w i = 0and a V ( X α − w i , w i ) = 0 for a suitable choice of i . Since w i ∈ Ker ϕ V , this proves(vi). (cid:3) We will now be able to prove Theorem B, our second main result. We refer thereader to the introduction for the statement of the theorem. Proof of Theorem B. We first recall the setup of the theorem, as stated in theintroduction. Let G = Sp( V, b ), where dim V = 2 n for some n ≥ 2. Let u ∈ G be unipotent. For t ≥ 0, let d , . . . , d t be the Jordan block sizes d of u such that ε V,b ( d ) = 0, and for s ≥ d t +1 , . . . , 2 d t + s be the Jordan block sizes d of u suchthat ε V,b ( d ) = 1. Write V ∼ = V n d ⊕ · · ·⊕ V n t d t ⊕ V n t +1 d t +1 ⊕ · · ·⊕ V n t + s d t + s as K [ u ]-modules,where n r > ≤ r ≤ t + s .Set α = ν (gcd( d , . . . , d t + s )). Suppose that ∧ ( V ) ∼ = ⊕ d ≥ V λ ( d ) d and L G ( ̟ ) ∼ = ⊕ d ≥ V λ ′ ( d ) d as K [ u ]-modules, where λ ( d ) , λ ′ ( d ) ≥ d ≥ 1. We identify L G ( ̟ ) as the subquotient ( h β V i ⊥ / h β V i , a V ) of ( ∧ ( V ) , a V ), which is justified byLemma 9.1 (iv). Set ε := ε ∧ ( V ) ,a V and ε ′ := ε L G ( ̟ ) ,a V . ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 35 Before the beginning the actual proof, we still need some additional notation.By Theorem 6.7, we have( V, b ) ∼ = W ( d ) n / ⊥ · · · ⊥ W ( d t ) n t / ⊥ V (2 d t +1 ) n t +1 ⊥ · · · ⊥ V (2 d t + s ) n t + s as bilinear K [ u ]-modules. Then writing V as an orthogonal direct sum of indecom-posables, we have V = W ⊥ · · · ⊥ W t ⊥ W t +1 ⊥ · · · ⊥ W t + s where W r is u -invariant for all 1 ≤ r ≤ t + s , and furthermore for all 1 ≤ r ≤ t we have ( W r , b ) ∼ = W ( d π ( r ) ) for some 1 ≤ π ( r ) ≤ t , and for all t + 1 ≤ r ≤ t + s we have ( W r , b ) ∼ = V (2 d π ( r ) ) for some t + 1 ≤ π ( r ) ≤ t + s . For 1 ≤ r ≤ t ,let W r = A r ⊕ B r be a totally singular decomposition into two K [ u ]-submodules A r ∼ = B r ∼ = V d π ( r ) .It is easy to see that(11.6) ∧ ( V ) = M ≤ r ≤ t + s ∧ ( W r ) ⊕ M ≤ r 1, as in Theorem A (i).We will next describe the values of λ ′ in the case where 2 | n , using Lemma 6.11.Note that in this case h β V i ⊥ = Ker ϕ V (Lemma 9.1 (ii)). We will first show that(11.7) Ker X α − ∧ ( V ) ⊆ h β V i ⊥ and(11.8) Ker X α ∧ ( V ) 6⊆ h β V i ⊥ . Clearly W r ∧ W r ′ ⊆ h β V i ⊥ for all 1 ≤ r < r ′ ≤ t + s , so Ker X α − W r ∧ W r ′ ⊆ h β V i ⊥ .Furthermore Ker X α − ∧ ( W r ) ⊆ Ker ϕ W r = ∧ ( W r ) ∩ Ker ϕ V for all 1 ≤ r ≤ t + s byLemma 11.3 (ii), so (11.7) follows.Let 1 ≤ r ≤ t + s be such that ν ( d π ( r ) ) = α . Then Ker X α ∧ ( W r ) Ker ϕ W r = ∧ ( W r ) ∩ Ker ϕ V by Lemma 11.3 (iii), which proves (11.8).For all 1 ≤ r ≤ t + s , let δ r ∈ ∧ ( W r ) be as in Lemma 11.3 (i), so X α − δ r = β W r and ϕ W r ( δ r ) = ϕ V ( δ r ) = d π ( r ) / α . Set δ = δ + · · · + δ t + s . Then X α − δ = β W + · · · + β W t s = β V and furthermore ϕ V ( δ ) = d π (1) / α + · · · + d π ( t + s ) / α = n/ α . Thus(11.9) δ ∈ h β V i ⊥ if and only if 2 | n α . With (11.7) – (11.9) and Lemma 6.11, we conclude that the values of λ ′ are givenin terms of λ as in (i) – (iii) of Theorem A.Next we consider the values of ε and prove (iv). Let d ≥ ε ( d ) =1. Then d > 1, and there exists v ∈ Ker X d ∧ ( V ) such that a V ( X d − v, v ) = 0.Furthermore by (11.6) and Lemma 6.8 (iv), we can choose v such that v ∈ ∧ ( W r )for some 1 ≤ r ≤ t + s , or v ∈ W r ∧ W r ′ for some 1 ≤ r < r ′ ≤ t + s . Suppose first that v ∈ ∧ ( W r ) for some 1 ≤ r ≤ t . Then ∧ ( W r ) = ∧ ( A r ) ⊕ ∧ ( B r ) ⊕ A r ∧ B r , so v = v ′ + v ′′ , where v ′ ∈ Ker X d ∧ ( A r ) ⊕∧ ( B r ) and v ′′ ∈ Ker X dA r ∧ B r . The bilin-ear K [ u ]-module ( ∧ ( A r ) ⊕ ∧ ( B r ) , a V ) is a paired module (Lemma 9.3). Hence a V ( X d − v ′ , v ′ ) = 0 by Lemma 6.12 and then a V ( X d − v, v ) = a V ( X d − v ′′ , v ′′ ) = 0by Lemma 6.8 (iv). Thus by Lemma 6.9, there is a Jordan block of size d in A r ∧ B r ∼ = V d π ( r ) ⊗ V d π ( r ) . Now it follows from Theorem 4.7 that d = 2 β for some2 β > d π ( r ) , so (iv)(a) holds.If v ∈ ∧ ( W r ) for some t + 1 ≤ r ≤ t + s , then by Lemma 6.9 there is a Jordanblock of size d in ∧ ( W r ) ∼ = ∧ ( V d π ( r ) ). In other words, case (iv)(b) holds.Suppose that v ∈ W r ∧ W r ′ for some 1 ≤ r < r ′ ≤ t + s . If 1 ≤ r ≤ t , then W r ∼ = W ( d π ( r ) ) is a paired module, and thus so is ( W r ∧ W r ′ , a V ) ∼ = W ( d π ( r ) ) ⊗ ( W r ′ , a V )(Lemma 5.13). But in that case a V ( X d − v, v ) = 0 by Lemma 6.12, contradiction.Thus we must have t + 1 ≤ r ≤ t + s , and so ( W r ∧ W r ′ , a V ) ∼ = V (2 d π ( r ) ) ⊗ V (2 d π ( r ′ ) ). By Theorem 7.4 and Lemma 6.9, the fact that a V ( X d − v, v ) = 0 impliesthat β = ν ( d π ( r ) ) = ν ( d π ( r ′ ) ), and furthermore d = 2 β +1 d ′ , where d ′ is the uniqueodd Jordan block size in V d π ( r ) / β ⊗ V d π ( r ′ ) / β . In other words, we are in case (iv)(c).For the converse of (iv), we consider (iv)(a) – (iv)(c). In case (iv)(a), we have d = 2 β for some 2 β > d π ( r ) for some 1 ≤ r ≤ t . Since ( A r ∧ B r , a V ) ∼ = ( A r ⊗ A ∗ r , b A r ) as bilinear K [ u ]-modules by Lemma 9.3, it follows from Theorem A that a V ( X d − v, v ) = 0 for some v ∈ Ker X dA r ∧ B r . Thus ε ( d ) = 1 in this case.For statement (iv)(b), let d > u in ∧ ( V d π ( r ) ) forsome t +1 ≤ r ≤ t + s . Note that d occurs in ∧ ( V d π ( r ) ) with odd multiplicity byLemma 4.13. If d π ( r ) is even, then ( ∧ ( W r ) , a V ) is non-degenerate (Lemma 9.1 (ii)),and thus a V ( X d − v, v ) = 0 for some v ∈ Ker X d ∧ ( W r ) by Lemma 6.10 (i). In thecase where d π ( r ) is odd, it follows from Lemma 9.1 that ∧ ( W r ) = Ker ϕ W r ⊕ h β W r i ,where (Ker ϕ W r , a V ) is non-degenerate. The multiplicity of d in Ker ϕ W r is the sameas in ∧ ( W r ), in particular the multiplicity is odd. Hence a V ( X d − v, v ) = 0 forsome v ∈ Ker X d Ker ϕ Wr by Lemma 6.10 (i). We conclude then that ε ( d ) = 1.In case (iv)(c), we have d = d ′ β +1 , where β = ν ( d π ( r ) ) = ν ( d π ( r ′ ) ) for some t + 1 ≤ r < r ′ ≤ t + s and d ′ is the unique odd Jordan block size in V d π ( r ) / β ⊗ V d π ( r ′ ) / β . We have ( W r ∧ W r ′ , a V ) ∼ = ( W r , b ) ⊗ ( W r ′ , b ) ∼ = V (2 d π ( r ) ) ⊗ V (2 d π ( r ′ ) )as bilinear K [ u ]-modules. Hence V (2 d ) is an orthogonal direct summand of ( W r ∧ W r ′ , a V ) by Theorem 7.4, and so it follows from Lemma 6.9 that a V ( X d − v, v ) = 0for some v ∈ Ker X dW r ∧ W r ′ . Thus ε ( d ) = 1, which completes the proof of (iv).Next we calculate ε ′ and prove claims (v) and (vi). If 2 ∤ n , then ∧ ( V ) =Ker ϕ V ⊕ h β V i , where Ker ϕ V ∼ = L G ( ̟ ) and β V ∈ rad a V (Lemma 9.1). In thiscase it is clear that ε ′ ( d ) = ε ( d ) for all d ≥ 1. If 2 | n and α = 0, then ε ′ ( d ) = ε ( d )for all d ≥ | n and α > ε ′ ( d ) = ε ( d )for all d = 2 α , α − 2, as is claimed by (vi). We prove (vi)(a) next, that is, we showthat ε (2 α ) = 1. Let 1 ≤ r ≤ t + s be such that ν ( d π ( r ) ) = α . If 1 ≤ r ≤ t , then2 α > d π ( r ) , and thus ε (2 α ) = 1 ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 37 by (iv)(a). If t + 1 ≤ r ≤ t + s , then 2 α > ∧ ( V d π ( r ) ) (Lemma 4.12), and so ε (2 α ) = 1 by (iv)(b).For (vi)(b), suppose first that ν ( d π ( r ) ) = α for some 1 ≤ r ≤ t . By Lemma11.3, there exists v ∈ Ker ϕ W r such that X α v = 0 and a V ( X α − v, v ) = 0. Then v ∈ Ker ϕ V = h β V i ⊥ , so we conclude that ε ′ (2 α ) = 1.For the other direction of (vi)(b), suppose that ν ( d π ( r ) ) > α for all 1 ≤ r ≤ t .We will show that ε ′ (2 α ) = 0, which is equivalent to the claim that a V ( X α − v, v ) =0 for all v ∈ Ker X α Ker ϕ V . First we describe Ker X α Ker ϕ V . If ν ( d π ( r ) ) > α , thenKer X α ∧ ( W r ) ⊆ Ker ϕ W r ⊆ Ker ϕ V by Lemma 11.3 (ii). If ν ( d π ( r ) ) = α , then ϕ V ( δ r ) = d π ( r ) / α = 1, so Ker X α ∧ ( W r ) = Ker X α Ker ϕ Wr ⊕ h δ r i . Furthermore, forall 1 ≤ r < r ′ ≤ t + s we have W r ∧ W r ′ ⊆ Ker ϕ V .Thus any v ∈ Ker X α Ker ϕ V can be written in the form v = X ≤ r ≤ t + s ν ( d π ( r ) ) >α z r + X ≤ r ≤ t + s ν ( d π ( r ) )= α ( z r + µ r δ r ) + X ≤ r 1. For (vi)(d),if 2 | n α , then δ ∈ h β V i ⊥ by (11.9) and thus ε ′ (2 α − 2) = ε (2 α − 2) = 0 by Lemma6.11 (iii). Similarly, if 2 ∤ n α , then δ 6∈ h β V i ⊥ by (11.9) and thus ε ′ (2 α − 2) = 1 byLemma 6.11 (iv). This completes the proof of (vi) and the theorem. (cid:3) In the following we show with small examples how Theorem B is applied. Let G = Sp( V, b ), where dim V = 2 n for some n ≥ 2. Let u ∈ G be a unipotent element and V ∼ = V n d ⊕· · ·⊕ V n t d t ⊕ V n t +1 d t +1 ⊕· · ·⊕ V n t + s d t + s as K [ u ]-modules, where d i and n i areas in Theorem B. Equivalently by Theorem 6.7, we have V ∼ = W ( d ) n / ⊥ · · · ⊥ W ( d t ) n t / ⊥ V (2 d t +1 ) n t +1 ⊥ · · · ⊥ V (2 d t + s ) n t + s as bilinear K [ u ]-modules. Let α = ν (gcd( d , . . . , d t + s )) as in Theorem B. Set ε := ε ∧ ( V ) ,a V and ε ′ := ε L G ( ̟ ) ,a V . Example . If n = 2 and V ∼ = V (4), then ∧ ( V ) ∼ = V ⊕ V by Theorem 4.8. Wehave ε (2) = 1, ε (4) = 1 by Theorem 4.8 (iv)(b), so ( ∧ ( V ) , a V ) ∼ = V (2) ⊥ V (4) asbilinear K [ u ]-modules. Now α = 1, so L G ( ̟ ) ∼ = V by Theorem B (applying rule(iii)(c) from Theorem A). Hence ( L G ( ̟ ) , a V ) ∼ = V (4) as bilinear K [ u ]-modules. Example . If n = 4 and V ∼ = W (4), then V ∼ = V ⊕ V and so ∧ ( V ) ∼ = V ⊕ V byTheorem 4.8. In this case we have a consecutive-ones binary expansion 4 = 2 , so byTheorem B we find that ε (4) = 1 and ε (2) = 0. Hence ( ∧ ( V ) , a V ) ∼ = W (2) ⊥ V (4) as bilinear K [ u ]-modules. Now α = 1, so L G ( ̟ ) ∼ = V ⊕ V by Theorem B(applying rule (iii)(a) from Theorem A). From Theorem B (vi) we conclude that ε ′ (2) = 1 and ε ′ (4) = 1, so ( L G ( ̟ ) , a V ) ∼ = V (2) ⊥ V (4) as bilinear K [ u ]-modules. Example . If n = 6 and V ∼ = V (4) , then ∧ ( V ) ∼ = V ⊕ V by Theorem 4.8. Wehave ε (2) = 1, ε (4) = 1 by Theorem 4.8 (iv)(b), so ( ∧ ( V ) , a V ) ∼ = V (2) ⊥ V (4) asbilinear K [ u ]-modules. Now α = 1, so L G ( ̟ ) ∼ = V ⊕ V by Theorem B (applyingrule (iii)(c) from Theorem A). From Theorem B (vi), we conclude that ε ′ (2) = 0and ε ′ (4) = 1, so ( L G ( ̟ ) , a V ) ∼ = W (2) ⊥ V (4) as bilinear K [ u ]-modules. Example . In Table 2, we illustrate Theorem B for 2 ≤ n ≤ 8. For n > α > 0. As in Example 10.6, for a bilinear K [ u ]-module ( W, b ), we use ( d n ε , . . . , d tn t ε t ) to denote that W ∼ = V n d ⊕ · · · ⊕ V n t d t as K [ u ]-modules and ε W,b ( d i ) = ε i for 1 ≤ i ≤ t . It is straightforward to seefrom the results of Section 6 that if ( V, b ) corresponds to ( d n ε , . . . , d tn t ε t ), then α = ν (gcd( d ′ , . . . , d ′ t )), where d ′ i = d i if ε i = 0 and d ′ i = d i / ε i = 1. Note thatthe examples in Table 2 illustrate all possible cases of (iv) – (vi) in Theorem B. ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 39 Table 2. Example cases of Theorem B for 2 ≤ n ≤ 8, see Example 11.7. n ( V, b ) ( ∧ ( V ) , a V ) ( L G ( ̟ ) , a V ) αn = 2 (4 ) (2 , ) (4 ) 1(2 ) (1 , ) (2 ) 0(2 ) (1 , ) (1 , ) 1(1 , ) (1 , ) (2 ) 0 n = 3 (6 ) (1 , , ) (6 , ) 0(2 , ) (1 , , ) (2 , ) 0(1 , ) (1 , , ) (2 , ) 0(3 ) (1 , , ) (3 , ) 0(2 ) (1 , ) (1 , ) 0(1 , ) (1 , ) (1 , ) 0(1 , ) (1 , ) (1 , ) 0(1 , ) (1 , ) (1 , ) 0 n = 4 (8 ) (4 , ) (2 , ) 2(4 ) (2 , ) (1 , ) 1(4 ) (2 , ) (2 , ) 2(2 , ) (1 , , ) (1 , , ) 1(2 ) (1 , ) (1 , ) 1 n = 6 (12 ) (2 , , , ) (4 , , ) 1(4 , ) (2 , , ) (4 , ) 1(2 , ) (1 , , , ) (1 , , , ) 1(6 ) (1 , , , ) (1 , , , ) 1(4 ) (2 , ) (2 , ) 1(2 , ) (1 , , ) (1 , , ) 1(2 , ) (1 , , ) (1 , , ) 1(2 , ) (1 , , ) (1 , , ) 1(2 ) (1 , ) (1 , ) 1 n = 8 (16 ) (8 , ) (6 , ) 3(4 , ) (2 , , , ) (1 , , , ) 1(2 , ) (1 , , , , ) (1 , , , , ) 1(8 ) (4 , ) (3 , ) 2(8 ) (4 , ) (4 , , ) 3(4 , ) (2 , , ) (1 , , ) 1(4 , ) (2 , , ) (2 , , , ) 2(2 , , ) (1 , , , ) (1 , , , ) 1(2 , ) (1 , , , ) (1 , , , ) 1(4 , ) (1 , , , , ) (1 , , , , ) 1(2 , ) (1 , , , ) (1 , , , ) 1(4 ) (2 , ) (1 , , ) 1(4 ) (2 , ) (2 , , ) 2(2 , ) (1 , , ) (1 , , ) 1(2 , ) (1 , , ) (1 , , ) 1(2 , ) (1 , , ) (1 , , ) 1(2 , ) (1 , , ) (1 , , ) 1(2 ) (1 , ) (1 , ) 1 Overgroups of distinguished unipotent elements Let G be a simple algebraic group over K . One approach towards understandingthe subgroup structure of G is to classify subgroups by the elements that theycontain. See for example the survey [Sax98] for some results in this direction and their applications. To give a specific example, all connected reductive subgroupscontaining a regular unipotent element of G are known by the results in [SS97,TZ13]. Overgroups of regular unipotent elements were studied further in [GM14,Section 3], motivated by an application to the inverse Galois problem.In the PhD thesis of the present author, the main result classifies all maximalclosed connected subgroups G that contain a distinguished unipotent element , inany characteristic p > 0. Recall that a unipotent element of G is distinguished , ifits centralizer in G does not contain a non-trivial torus.In this section, we keep our assumption that char K = 2, and apply our mainresults to classify some subgroups of Sp( V, b ) that contain distinguished unipotentelements. The results of Proposition 12.4, Proposition 12.5, and Proposition 12.7below appeared first in the PhD thesis of the present author. However, using ourresults we are able to give proofs which are shorter and do not rely on many case-by-case calculations.The following definition is convenient for describing distinguished unipotent el-ements in Sp( V, b ). Definition 12.1. Let u be a generator of a cyclic 2-group and let ( V, b ) be a bilinear K [ u ]-module. We say that u acts on ( V, b ) as a distinguished unipotent element , ifone of the following equivalent conditions hold:(i) The image of u in Sp( V, b ) is a distinguished unipotent element of Sp( V, b ).(ii) ( V, b ) ∼ = V (2 k ) d ⊥ · · · ⊥ V (2 k t ) d t as bilinear K [ u ]-modules where 0 < k < · · · < k t and d i ≤ ≤ i ≤ t .(iii) Every Jordan block size d of u on V is even, has multiplicity at most two, and ε V,b ( d ) = 1.(iv) The bilinear K [ u ]-module ( V, b ) does not have any orthogonal direct sum-mands of the form W ( m ) for m > V, b ) ∼ = V (2 k ) d ⊥ · · · ⊥ V (2 k t ) d t as bilinear K [ u ]-modules forsome integers 0 < k < · · · < k t and d i > 0. If d i > 2, then from the isomorphism V (2 k i ) ∼ = W (2 k i ) ⊥ V (2 k i ) of bilinear K [ u ]-modules (Lemma 5.15) we see that W (2 k i ) occurs as an orthogonal direct summand of ( V, b ), contradiction. Thus d i ≤ ≤ i ≤ t , which proves that (iv) implies (ii).We shall need the following easy lemma, after which we will be able to prove themain results of this section. Lemma 12.2. Let G be a simple algebraic group and let f : G → Sp( V, b ) be anon-trivial representation of G . If u ∈ G is a unipotent element that acts on ( V, b ) as a distinguished unipotent element, then u is a distinguished unipotent element of G .Proof. If u is not a distinguished unipotent element of G , then u is centralized bysome non-trivial torus S < G . In this case f ( S ) is a non-trivial torus centralizing f ( u ), so f ( u ) is not a distinguished unipotent element of Sp( V, b ). (cid:3) ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 41 Proposition 12.3. Let G = SL( V ) , where n = dim V is even. A unipotent element u ∈ G acts on ( V ⊗ V ∗ , b V ) as a distinguished unipotent element if and only if n = 2 and V ∼ = V as K [ u ] -modules. Proposition 12.4. Let G = SL( V ) and set n = dim V , where n > . A unipotentelement u ∈ G acts on ( L G ( ̟ + ̟ n − ) , b V ) as a distinguished unipotent elementif and only if V ∼ = V n as K [ u ] -modules and n ∈ { , , } .Proof of Proposition 12.3 and Proposition 12.4. The only distinguished unipotentelements in G are the regular unipotent elements [LS12, Proposition 3.5], so byLemma 12.2 we may assume that V ∼ = V n as K [ u ]-modules for some n > 1. Aneasy calculation with Theorem 4.7 and Theorem A shows the following: • If n = 2, then ( V ⊗ V ∗ , b V ) ∼ = V (2) and ( L G ( ̟ + ̟ n − ) , b V ) ∼ = V (2). • If n = 3, then ( L G ( ̟ + ̟ n − ) , b V ) ∼ = V (4) . • If n = 5, then ( L G ( ̟ + ̟ n − ) , b V ) ∼ = V (4) ⊥ V (8) .This proves sufficiency in Proposition 12.3 and Proposition 12.4. We show nextthat these are the only cases where u acts on ( V ⊗ V ∗ , b V ) or ( L G ( ̟ + ̟ n − ) , b V )as a distinguished unipotent element.Let n = P ki =1 ( − i +1 e i be the consecutive-ones binary expansion of n , where e > · · · > e k ≥ 0. Note that e k − > e k + 1 if k ≥ 2. By Theorem 4.7 V ⊗ V ∗ ∼ = M ≤ i ≤ k V d i ei as K [ u ]-modules, where d i = 2 e i − P kj = i +1 ( − i + j e j +1 for all 1 ≤ i ≤ k .For the other direction of Proposition 12.3, suppose that n is even and that u actson ( V ⊗ V ∗ , b V ) as a distinguished unipotent element. Then d i ≤ ≤ i ≤ k .We have e k > n is even. Thus if k > 1, then d k − = 2 e k − − e k +1 ≥ e k +1 > 2, contradiction. Hence k = 1, so n = 2 e . Since d = 2 e , we must have n = 2, asclaimed by Proposition 12.3.For Proposition 12.4, suppose that u acts on ( L G ( ̟ + ̟ n − ) , b V ) as a distin-guished unipotent element. Since ν ( n ) = e k , by Theorem A we have d k ≤ d i ≤ ≤ i < k . If e k > 1, then d k = 2 e k ≥ 4, contradiction. Thus e k ≤ e k = 1. If k > 1, then it follows from e k − > e k + 1 that d k − =2 e k − − e k +1 ≥ 4, contradiction. Thus k = 1, and so n = 2 e = 2.Consider next e k = 0. Then d k − = 2 e k − − ≤ e k − ≤ 2. But e k − > e k + 1, which forces e k − = 2. If k = 2, then n = 2 − = 3. Suppose thenthat k > 2. In this case d k − = 2 e k − − ≤ 2, so we must have e k − = 3. If k = 3,then this gives n = 2 − + 2 = 5. Finally if k > 3, then d k − = 2 e k − − ≥ (cid:3) Proposition 12.5. Let ( V , b ) and ( V , b ) be non-degenerate alternating bilinear K [ u ] -modules, where < dim V ≤ dim V . Then u acts on ( V , b ) ⊗ ( V , b ) =( V ⊗ V , b ⊗ b ) as a distinguished unipotent element if and only if we have thefollowing isomorphisms of bilinear K [ u ] -modules: (i) ( V , b ) ∼ = V (2) , (ii) ( V , b ) ∼ = V (2 k ) ⊥ · · · ⊥ V (2 k t ) , where < k < · · · < k t are odd integers.Furthermore, if (i) and (ii) hold, then ( V , b ) ⊗ ( V , b ) ∼ = V (2 k ) ⊥ · · · ⊥ V (2 k t ) as bilinear K [ u ] -modules. Proof. Suppose that u acts on ( V , b ) ⊗ ( V , b ) as a distinguished unipotent ele-ment. Then u must act on ( V i , b i ) as a distinguished unipotent element for i = 1 , V i , b i ) had any orthogonal direct summands of the form W ( m ), then sowould ( V , b ) ⊗ ( V , b ) by Proposition 7.1.We consider first the case where ( V i , b i ) are orthogonally indecomposable, sosuppose that ( V , b ) ∼ = V (2 l ) and ( V , b ) ∼ = V (2 k ) for some 1 ≤ l ≤ k . Then byTheorem 7.4 there are at most two indecomposable summands in V (2 l ) ⊗ V (2 k ),as otherwise some summand would have multiplicity > W ( m ) wouldoccur as an orthogonal direct summand. Thus ( V , b ) ∼ = V (2), since the numberof indecomposable summands in the K [ u ]-module V l ⊗ V k is 2 l . By Example 7.5,we must have k odd and ( V , b ) ⊗ ( V , b ) ∼ = V (2 k ) .For the general case, let V = W ⊥ · · · ⊥ W s and V = W ′ ⊥ · · · ⊥ W ′ t , where W i and W ′ j are orthogonally indecomposable K [ u ]-modules for all 1 ≤ i ≤ s and1 ≤ j ≤ t . Clearly u acts as a distinguished unipotent element on ( W i ⊗ W ′ j , b ⊗ b )for all 1 ≤ i ≤ s and 1 ≤ j ≤ t , so it follows from the indecomposable casethat ( W i ⊗ W ′ j , b ⊗ b ) ∼ = V (2 k ) for some k odd, where ( W i , b ) ∼ = V (2) and( W ′ j , b ) ∼ = V (2 k ), or ( W i , b ) ∼ = V (2 k ) and ( W ′ j , b ) ∼ = V (2). From this and thefact that dim V ≤ dim V , it is straightforward to see that ( V , b ) ∼ = V (2) and( V , b ) ∼ = V (2 k ) ⊥ · · · ⊥ V (2 k t ), where 0 < k < · · · < k t are odd integers, asclaimed by the proposition.The other direction of the proposition is immediate from Example 7.5, whichshows that ( V , b ) ⊗ ( V , b ) ∼ = V (2 k ) ⊥ · · · ⊥ V (2 k t ) as bilinear K [ u ]-modules. (cid:3) Proposition 12.6. Let G = Sp( V, b ) , where dim V = 2 n and n is even. A unipo-tent element u ∈ G acts on ( ∧ ( V ) , a V ) as a distinguished unipotent element if andonly if n = 2 and ( V, b ) ∼ = V (4) as bilinear K [ u ] -modules. Proposition 12.7. Let G = Sp( V, b ) , where dim V = 2 n and n ≥ . A unipotentelement u ∈ G acts on ( L G ( ̟ ) , a V ) as a distinguished unipotent element if andonly if one of the following conditions hold: (i) ( V, b ) ∼ = V (2 n ) as bilinear K [ u ] -modules and n ∈ { , , } . (ii) ( V, b ) ∼ = V (2) ⊥ V (2 n − as bilinear K [ u ] -modules and n ∈ { , } .Proof of Proposition 12.6 and Proposition 12.7. Let u ∈ G be unipotent. We canassume that u is a distinguished unipotent element (Lemma 12.2), in which case( V, b ) = W ⊥ · · · ⊥ W t , where ( W i , b ) ∼ = V (2 d i ) for all 1 ≤ i ≤ t , for some integers d i > t > 1, then ( W i ∧ W j , a V ) ∼ = ( W i , b ) ⊗ ( W j , b ) is a non-degenerate subspace of( ∧ ( V ) , a V ) for all 1 ≤ i < j ≤ t . In fact, since W i ∧ W j is contained in h β V i ⊥ and( W i ∧ W j ) ∩ h β V i = 0, it follows that ( W i ∧ W j , a V ) embeds into ( L G ( ̟ ) , a V ) asa non-degenerate subspace.Thus if u acts on ( ∧ ( V ) , a V ) or ( L G ( ̟ ) , a V ) as a distinguished unipotentelement, then u acts on ( W i ∧ W j , a V ) ∼ = ( W i , b ) ⊗ ( W j , b ) as a distinguishedunipotent element. In this case, by Theorem 12.5 the integers d i and d j are odd,and furthermore d i = 1 or d j = 1. Consequently if u acts on ( ∧ ( V ) , a V ) or( L G ( ̟ ) , a V ) as a distinguished unipotent element, then ( V, b ) ∼ = V (2 n ) or n iseven and ( V, b ) ∼ = V (2) ⊥ V (2 n − V, b ) ∼ = V (2 n ). If n ≥ V occurs in ∧ ( V n ) withmultiplicity one by Lemma 4.12. Thus if u acts on ( L G ( ̟ ) , a V ) as a distinguished ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 43 unipotent element, then each Jordan block size in ∧ ( V n ) has multiplicity at most2, and thus n ∈ { , , } by Lemma 4.14. Suppose next that n is even, and let α = ν ( n ). By Lemma 4.12 and Theorem B, as K [ u ]-modules ∧ ( V ) ∼ = V α ⊕ W and L G ( ̟ ) ∼ = V α − ⊕ W , where W has no Jordan blocks of size 2 α . Therefore if u acts on ( ∧ ( V ) , a V ) or ( L G ( ̟ ) , a V ) as a distinguished unipotent element, theneach Jordan block size in ∧ ( V n ) has multiplicity at most 2, and so n = 2 byLemma 4.14.Next we consider the other possibility, which is that n is even and ( V, b ) ∼ = V (2) ⊥ V (2 n − K [ u ]-modules ∧ ( V n − ) ∼ = V ⊕ W , where W has no Jordan blocks of size 1. Hence(12.1) ∧ ( V ) ∼ = ∧ ( V ) ⊕ ∧ ( V n − ) ⊕ ( V ⊗ V n − ) ∼ = V ⊕ W ⊕ V n − as K [ u ]-modules. Thus u does not act on ( ∧ ( V ) , a V ) as a distinguished unipotentelement, since it has Jordan blocks of size 1 in ∧ ( V ).Note that L G ( ̟ ) ∼ = W ⊕ V n − as K [ u ]-modules by (12.1) and Theorem B. Thusif u acts on ( L G ( ̟ ) , a V ) as a distinguished unipotent element, each Jordan blocksize in ∧ ( V n − ) ∼ = V ⊕ W has multiplicity at most two, and so n ∈ { , , , } by Lemma 4.14. Here n = 3 is ruled out since we are assuming that n is even.For n = 4, a calculation with Theorem 4.8 shows that L G ( ̟ ) ∼ = V ⊕ V as K [ u ]-modules, so u does not act on ( L G ( ̟ ) , a V ) as a distinguished unipotent element.We still need to check that in the cases listed u does indeed act as a distinguishedunipotent element. To this end, a straightforward computation with Theorem 4.8and Theorem B shows the following. • If n = 2 and ( V, b ) ∼ = V (4), then ( ∧ ( V ) , a V ) ∼ = V (2) ⊥ V (4). • If n = 2 and ( V, b ) ∼ = V (4), then ( L G ( ̟ ) , a V ) ∼ = V (4). • If n = 2 and ( V, b ) ∼ = V (2) , then ( L G ( ̟ ) , a V ) ∼ = V (2) . • If n = 3 and ( V, b ) ∼ = V (6), then ( L G ( ̟ ) , a V ) ∼ = V (6) ⊥ V (8). • If n = 5 and ( V, b ) ∼ = V (10), then ( L G ( ̟ ) , a V ) ∼ = V (6) ⊥ V (8) ⊥ V (14) ⊥ V (16). • If n = 6 and ( V, b ) ∼ = V (2) ⊥ V (10), then ( L G ( ̟ ) , a V ) ∼ = V (6) ⊥ V (8) ⊥ V (10) ⊥ V (14) ⊥ V (16).This completes the proof of Proposition 12.6 and Proposition 12.7. (cid:3) References [Alp86] J. L. Alperin. Local representation theory , Volume 11 of Cambridge Studies in AdvancedMathematics . Cambridge University Press, Cambridge, 1986.[DB10] B. De Bruyn. On the Grassmann modules for the symplectic groups. J. Algebra ,324(2):218–230, 2010.[Fon74] P. Fong. 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ESSELINK NORMAL FORMS OF UNIPOTENT ELEMENTS 45 School of Mathematics, The University of Manchester, Manchester M13 9PL, UnitedKingdom E-mail address :: d r .Conversely, suppose that d = 2 β > d r for some 1 ≤ r ≤ t . By Lemma 10.2 and Theorem 4.7, for asuitable choice of 2 β − + 1 ≤ i ≤ β the element(10.9) v ( β ) i = X j ≥ e ( r ) i + j β ⊗ e ( r ) ∗ i − β − + j β is such that X β v ( β ) i = 0 and b V ( X β − v ( β ) i , v ( β ) i ) = 0. Thus ε ( d ) = 1, whichcompletes the proof of (iv). Note that here v ( β ) i ∈ h γ V i ⊥ , so this also shows that ε ′ ( d ) = 1 in this case.For (v) and (vi), we proceed to calculate ε ′ . Suppose first that 2 ∤ n . Then γ V ∈ rad b V by Lemma 8.1 (iii), so h γ V i ⊥ / h γ V i = V / h γ V i . In this case it is clearthat ε ′ ( d ) = ε ( d ) for all d ≥