(Co)isotropic triples and poset representations
aa r X i v : . [ m a t h . S G ] J un (Co)isotropic triples and poset representations Christian Herrmann ∗ Jonathan Lorand ◦ Alan Weinstein † ∗ Fachbereich Mathematik, Technische Universit¨at Darmstadt, 64289 Darmstadt, Germany, [email protected] ◦ Institute of Mathematics, University of Zurich, CH-8057 Zurich, Switzerland, [email protected] † Department of Mathematics, University of California Berkeley, CA 94720-3840 USA. [email protected]
Abstract
We study triples of coisotropic or isotropic subspaces in symplectic vector spaces;in particular, we classify indecomposable structures of this kind. The classificationdepends on the ground field, which we only assume to be perfect and not of charac-teristic 2. Our work uses the theory of representations of partially ordered sets with(order reversing) involution; for (co)isotropic triples, the relevant poset is “2+2+2”consisting of three independent ordered pairs, with the involution exchanging themembers of each pair.A key feature of the classification is that any indecomposable (co)isotropic tripleis either “split” or “non-split”. The latter is the case when the poset represen-tation underlying an indecomposable (co)isotropic triple is itself indecomposable.Otherwise, in the “split” case, the underlying representation is decomposable andnecessarily the direct sum of a dual pair of indecomposable poset representations;the (co)isotropic triple is a “symplectification”.If one ignores the involution, representations of the poset “2 + 2 + 2” are es-sentially the same as representations of a certain quiver of extended Dynkin type˜ E , and our work relies on the known classification of indecomposable represen-tations of such quivers. In particular from these results it follows that the posetrepresentations underlying (co)isotropic triples come in two types: there are fami-lies of indecomposables which depend on a parameter taking a continuum of values(“continuous-type”), and there are those indecomposables which are characterizedonly by the dimensions of the spaces involved (“discrete-type”). This pattern isreflected in the classification of (co)isotropic triples; in particular, there are familiesof triples which depend on a parameter. Also, indecomposable triples exist in everyeven dimension.In the course of the paper we develop the framework of “symplectic poset rep-resentations”, which can be applied to a range of problems of symplectic linearalgebra. The classification of linear hamiltonian vector fields, up to conjugation,is an example; we briefly explain the connection between these and (co)isotropictriples. The framework lends itself equally well to studying poset representationson spaces carrying a non-degenerate symmetric bilinear form; we mainly keep ourfocus, however, on the symplectic side. ontents + + . . . 294.6 Triples in dimension 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.7 Higher dimensions: A preview . . . . . . . . . . . . . . . . . . . . . . . . 344.8 Hamiltonian vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . 35 A (3 k + 1 ,
0) . . . . . . . . . . . . . . . . . . . . . . . . 395.3 Implementing the A (3 k + 2 ,
0) . . . . . . . . . . . . . . . . . . . . . . . . 415.4 Existence of compatible forms . . . . . . . . . . . . . . . . . . . . . . . . 435.5 Uniqueness of compatible forms . . . . . . . . . . . . . . . . . . . . . . . 49 C and R R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707.3 Hamiltonian vector fields over R from non-split framed sextuples . . . . 732 Continuous non-split isotropic triples over perfect fields 76 S η . . . . . . . . . . . . . . . . . . . . . . . . . 88 In a previous paper [21], the second two authors have given a complete classificationof pairs of coisotropic subspaces in Poisson vector spaces and, equivalently by duality,pairs of isotropic subspaces in presymplectic vector spaces. Each such pair is uniquely(up to isomorphism and ordering) decomposable as a direct sum of multiples of tenindecomposable pairs, for which there are simple normal forms in ambient spaces ofdimension 1, 2, or 3. These decomposition problems are special cases of the problem ofclassifying triples of coisotropic or isotropic subspaces in symplectic vector spaces, withan extra condition relating the third subspace to the first two.In this paper, we will deal with all (co)isotropic triples in symplectic vector spaces.The decomposition into indecomposables is still possible with summands essentiallyunique, but there are many more indecomposables. In dimensions up to 4, there are stillonly finitely many isomorphism classes of indecomposables, while in higher dimensionsthe moduli space of such classes includes parametrized families as well as single pointswhich may or may not be in the closure of such families.The classification of pairs in [21] was done by elementary arguments in linear alge-bra, but the results there (as well as those in [22] on the classification of (co)isotropicrelations) suggested links with the representation theory of quivers, particularly of thoseassociated with (extended) Dynkin diagrams and the closely related representation the-ory of partially ordered sets (posets). (See e.g. Gabriel and Roiter [12].) We rely onthese to carry out the classification of triples. In fact, we largely reduce our problemto that of studying representations which are maps from a certain 6-element poset withinvolution to the poset of subspaces of a symplectic vector space with the involutiongiven by symplectic orthogonality. The classification of these representations, withoutthe involution, is essentially that of certain representations of a quiver associated to theextended Dynkin diagram ˜ E , which is a tree consisting of a central vertex attached tothree “branches” containing two vertices each. The quiver in question is obtained from˜ E by orienting all of its edges toward the central vertex. Here are two depictions ofthis quiver; the first is the most common, while we will use the latter, to emphasize theposet structure: 3or our results, we rely on the classification of representations of extended Dynkinquivers given in [7] and [8], for the case of ˜ E . Representations of this particular quiverhave also been studied in quite some detail by Stekolchshik, see e.g. [31] and [32].The study of poset representations in spaces equipped with an (anti-)symmetric innerproduct was first developed, to our knowledge, by Scharlau and collaborators; see [28]for a concise and enlightening overview.For (co)isotropic triples, the connection with the ˜ E -type quiver described above isthis: the central vertex corresponds to the ambient symplectic vector space, while eachbranch corresponds to an isotropic subspace and (adjacent to the central vertex) thecoisotropic orthogonal in which it is contained.The associated six-element poset consists of the vertices in the branches, with partialordering given by the arrows connecting them, and (order-reversing) involution givenby exchanging the elements of each pair. We will use one of the standard notations, + + , for this poset.Since the operation of “taking the symplectic orthogonal” induces a one-to-one cor-respondence between isotropic subspaces and coisotropic ones, in the following we focuson and refer simply to isotropic triples. The concomitant results for coisotropic triplesare implicit.A crucial part of our analysis is a result due to Quebbemann, Scharlau, and Schulte[25] and Sergeichuk [30], which, in our setting, says that every indecomposable isotropictriple in dimension 2 n is either already indecomposable as a linear representation ofthe poset + + or is obtained from an indecomposable linear representation ofthe same poset in dimension n by a “doubling construction” known as hyperbolization([25, p. 267]), and which in our context we will call symplectification . (It is closelyconnected to the cotangent bundle construction in symplectic geometry, though thelatter always produces lagrangian subspaces.) This dichotomy reduces our problem todeciding which such indecomposable linear representations actually come from inde-composable isotropic triples, and finding the nonisomorphic isotropic triples which maygive rise to the same indecomposable linear representation. Acknowledgements
J. L. would like to thank Alberto Cattaneo, Wilhelm Karlsson, Marcel Wild, andThomas Willwacher; he acknowledges support from the NCCR SwissMAP and SNF The simplest triple illustrating this possibility consists of three distinct lines in a symplectic plane.These are lagrangian, so each one corresponds to both the isotropic and coisotropic subspaces in anested pair. If we forget the symplectic structure, there is no further invariant, but in the case of realcoefficients, there is a symplectic invariant given by the cyclic order of the three lines with respect tothe symplectic orientation. This is sometimes called the Maslov index or Kashiwara-Vergne index ofthe lagrangian triple.
In this section we give a summary of our results. We hope that placing this summaryhere, rather than as a final section, will give the reader an initial rough overview and aplace to refer back to while reading the paper. Some of the terminology and notationused below is only defined later in the course of the paper; we include here, however, ashort list of our most essential terms and notions: • A linear poset representation (or just “poset representation”) of a poset P in a vector space V (always assumed finite-dimensional) is an order-preservingmap from P to the poset of subspaces of V . Poset representations will usuallybe denoted by the letter ψ . Representations of the poset + + are calledsextuples. • If a poset P is equipped with an order-reversing involution, then to each repre-sentation ψ of P in V , there is a dual representation ψ ∗ of P in V ∗ . (For thedefinition, see (4) in Section 3.5.) In this paper, we always assume the poset + + to be equipped with the order-reversing involution that exchanges thetwo elements of each of the three ordered pairs. • An isotropic triple ϕ is an ordered triple of isotropic subspaces of a symplecticvector space. The three isotropics, together with their corresponding symplecticorthogonals, form a sextuple of subspaces which define a linear representation ofthe poset + + , the underlying linear representation ˆ ϕ of the triple. Theunderlying poset representation ˆ ϕ of an isotropic triple is necessarily isomorphicto its dual; i.e ˆ ϕ is “self-dual”. • Given a sextuple ψ in V , a compatible symplectic form is a symplectic formon V with respect to which ψ becomes the underlying poset representation of anisotropic triple. (See Section 3.5.) • Isotropic triples are examples of symplectic poset representations (see Defini-tion 3.2). There are distinct notions of decomposition (and of “indecomposable”)for linear poset representations and for symplectic poset representations, respec-tively (the latter are orthogonal decompositions.) In particular, an indecompos-able isotropic triple may be decomposable as a linear poset representation. • Any representation ψ of a poset P has an associated dimension vector : this isthe function which assigns to each element x ∈ P the dimension of the associatedsubspace ψ ( x ). We usually write the dimension vector not as a function, but as atuple (i.e. a “vector”). 5 Throughout we assume that the ground field k is perfect and does not have char-acteristic 2. The condition of “perfectness” is not a strong one: perfect fieldsinclude fields which are algebraically closed, fields of characteristic zero, and finitefields. Resum´e :1. Symplectically indecomposable isotropic triples ϕ come in two kinds: • Split: the underlying sextuple ˆ ϕ of the isotropic triple is decomposable as alinear poset representation. By Lemma 3.24, ϕ is isomorphic to the symplec-tification of some indecomposable poset representation ψ of + + whichis not self-dual; in particular ˆ ϕ ∼ = ψ ⊕ ψ ∗ . • Non-split: ϕ is such that ˆ ϕ indecomposable and self-dualOur main work is the identification of the duals of indecomposable sextuples andthe classification of the indecomposable non-split isotropic triples. Once this isdone, by Lemma 3.24, the classification of the split-type indecomposable isotropictriples is essentially automatic. Throughout, it is understood that when we speakof classification of indecomposables, we mean the classification of isomorphismclasses of indecomposables.2. In order to classify the non-split indecomposable isotropic triples, we first identifywhich sextuples ψ are self-dual. We have the following subcases (we use labels (a)through (e), which are also referred to further below): • Discrete-type: in these cases, ψ is based on an indecomposable nilpotent en-domorphism. Up to isomorphism, ψ is uniquely determined by its dimensionvector d . There are the following types, with k ∈ Z ≥ :(a) A (3 k + 1 ,
0) with d = (3 k + 1; 2 k + 1 , k, k + 1 , k, k + 1 , k ) , (b) A (3 k + 2 ,
0) with d = (3 k + 2; 2 k + 1 , k + 1 , k + 1 , k + 1 , k + 1 , k + 1) . • Continuous-type: in these cases, ψ has a dimension vector of the form(3 k, k, k, k, k, k, k ), with k ∈ Z > , and is based on an indecomposableendomorphism η (this is a generalization of the ∆( k, λ ) of Donovan-Freislich[8], and we also use normal forms based on the homogeneous representationsof Dlab-Ringel [7]). In our encoding, the underlying endomorphism η of aself-dual ψ is necessarily such that η is similar to ( id − η ) ∗ . The followingtypes of endomorphism η (up to similarity) account for all the indecompos-able self-dual continuous-type sextuples:If η has an eigenvalue λ in the base field:(c) λ = . If η has has no eigenvalue in the base field:6d) Over the reals: the complex eigenvalues of η are ± b η √− b η > η = id + ζ where ζ = 0 is similar to − ζ ∗ .The characteristic (= minimal) polynomial of ζ is of the form r ( x ) m foran irreducible polynomial r which is of the form r ( x ) = p ( x ) for somepolynomial p .3. With the indecomposable self-dual sextuples classified, we then determine whichof these admit compatible symplectic forms (and we give such forms explicitly viacoordinate matrices). We find that compatible symplectic forms exists as follows:(a) for A (3 k + 1 ,
0) if and only if k is odd.(b) for A (3 k + 2 ,
0) if and only if k is even.(c) for η having an eigenvalue in k : if and only if k is even.(d-e) for η having no eigenvalue in k : for all k .4. Following the question of existence of compatible symplectic forms for indecom-posable self-dual sextuples, we then address the question of uniqueness. We findthat compatible symplectic forms for a sextuple ψ are unique up to isomorphism(i.e. up to an isometry which is an automorphism of ψ ) and, in(a-c), multiplication by a scalar. There are no compatible symmetric forms.(d), multiplication by a scalar. There are also compatible symmetric forms.(e), multiplication by a ‘scalar’ Z ∈ F + H . Here F = k [ x ] /q ( x ) is considered as asubring of k k × k and F + H consists of all Z ∈ F such that Z t H = HZ , where H is the coordinate matrix of a particular compatible form. There are alsocompatible symmetric forms.5. The following provides a complete list of isomorphism types of isotropic triplesin the non-split case. Given a non-split isotropic triple ϕ with symplectic form ω (and associated coordinate matrix H ) there is an automorphism of ˆ ϕ which is(a-d) an isometry from ω to cω , for some 0 = c ∈ k , if and only if c is a squarein k . Thus, compatible symplectic forms for a given sextuple ψ = ˆ ϕ areparametrized by the square class group k × / ( k × ) .(e) an isometry from H to HZ , for some 0 = Z ∈ F , if and only if Z = X − Y where X ∈ F + H and Y ∈ F − H . Here, F ± H = { Z | Z t H = ± HZ } .6. For a fixed indecomposable sextuple ψ , the set (if non-empty) of all compatiblesymplectic forms is given via linear expressions involving(a) k parameters,(b) k + 1 parameters, Here we mean all compatible forms, i.e. not only up to isometry. k + 1 parameters. Proof of the Resum´e
1. By Lemma 3.24.2. For the cases (a-b), in Proposition 4.4 it is stated which discrete-type indecom-posable sextuples are self-dual, and in Theorem 5.3 it is shown which ones admitcompatible symplectic forms.For continuous-type indecomposable sextuples, it is established in Section 6.5that self-duality only occurs for framed sextuples, and from Proposition 6.17 itfollows that a framed sextuple S η is self-dual if and only if η is similar to id − η ∗ .Furthermore, those S η which underly non-split isotropic triples are identified:(c) in Theorem 7.1 for the case when the underlying endomorphism has an eigen-value in the ground field.(d) in Theorem 7.3 for the case when k = R and the underlying endomorphismhas no eigenvalue in R .(e) in Theorem 8.9 for the general case when k is perfect and the underlyingendomorphism has no eigenvalue in k . See also Corrollary 8.10.3. For (a-b), see Theorem 5.3; for (c), see Theorem 7.1; (d-e), see Theorem 8.9.4. For (a-b), see Theorem 5.9; for (c), see Theorem 7.1; for (d), see Theorem 7.3;for (e), see Theorem 8.17. The uniqueness statements in the theorems above for(a-d) make use of Lemma 3.26, and in the case (e), Theorem 8.17 makes use ofLemma 8.15.5. As in the previous point: for (a-b), see Theorem 5.9; for (c), see Theorem 7.1; for(d), see Theorem 7.3; for (e), see Theorem 8.17.6. For (a-b), see Remark 5.8; for (c-d), see Theorem 7.1 and Theorem 7.3. Corollary 2.1.
For an indecomposable representation ψ of the poset P = + + ina vector space V the following are equivalent1. There is a symplectic representation ϕ such that ˆ ϕ = ψ .2. ψ is self-dual and dim V is even. Remark 2.2.
The following topics are deferred to possible future work. • For those representations which do not admit compatible symplectic structures(since they are not self-dual or admit only symmetric structure), a detailed de-scription of the isotropic triples resulting from their symplectifications.8
Analysis of how the classification of isotropic triples changes when the groundfield is changed. • A description of the isometry groups of indecomposable isotropic triples. • A discussion of the classification of isotropic triples within the category-theoreticframework of Quebbemann, Scharlau, Schulte [25] and Sergeichuk [30]. • Analysis of the relation between poset representations which are sextuples andthose which involve four subspaces. • The study of how a general (not necessarily indecomposable) isotropic tripledecomposes into indecomposable summands; in particular the question of howunique such a decomposition is. • The question of defining invariants for isometry types of isotropic triples (in par-ticular in relation to the perfect elements established by Stekolchshik [31]). • An explanation in further detail of the relation between isotropic triples and linearhamiltonian vector fields.
Here we provide a review of some aspects of symplectic linear algebra and the repre-sentation theory of partially ordered sets. We have tried to do this in such as way asto be accessible to readers who might be familiar with only one of these two areas. Inparticular, we give an elementary presentation of the relevant results from the generaltheory of Quebbemann, Scharlau, and Schulte [25] and Sergeichuk [30].We work with a fixed ground field, which we assume to be perfect (e.g. of character-istic zero, finite, or algebraically closed) and not of characteristic 2; otherwise we usuallyleave the field unspecified. All vector spaces are assumed to be finite-dimensional.
A pair of subspaces (
A, B ) in a vector space V (without further structure) is completelydetermined up to isomorphism by four invariants: the dimensions of V , A , B , and A ∩ B . For a triple ( A, B, C ) ⊆ V of subspaces, one needs to know, in addition to thedimensions of V , A , B , C , all pairwise intersections, and the triple intersection, thedimension of one more space, such as ( A + B ) ∩ C , giving a total of nine invariants. (Forinstance, this ninth invariant is needed to distinguish the two arrangements of threedistinct lines in 3-space, which may or may not be coplanar.) At this point, we haveeffectively introduced the lattice h A, B, C i generated by A , B , and C as a sublattice ofthe lattice Σ( V ) of subspaces in V , i.e. the subspaces generated by A , B , and C underiterations of the operations of sum and intersection. The study of such structures datesback to Dedekind and Thrall, immanent in early results of representation theory andleading into abstract lattice theory. The present work is mainly oriented around posets,9ith Σ( V ) being a poset with respect to inclusion, and we will not deal with abstractlattice theory. Nevertheless, the lattice structure of Σ( V ) will play an important role,in particular in calculations.A non-degenerate antisymmetric bilinear, i.e. symplectic , form ω on V producesadditional structure.First of all, there is a naturally associated linear map ˜ ω from V to V ∗ defined by˜ ω ( v )( w ) = ω ( v, w ). Non-degeneracy means that ˜ ω is an isomorphism.Next, there is a natural order-reversing involution A A ⊥ on Σ( V ), where A ⊥ , the symplectic orthogonal of A , is the subspace { v ∈ V | ∀ w ∈ A, ω ( v, w ) = 0 } . Thisinvolution is related to another order-reversing operation, namely the one that mapsa subspace A ⊆ V to its annihilator A ◦ = { ξ ∈ V ∗ | ξ ( v ) = 0 ∀ v ∈ A } in V ∗ . Thefollowing result is easy to verify; we state it as a lemma to refer to later. Lemma 3.1.
For any subspace A in a symplectic space ( V, ω ) , ˜ ω ( A ⊥ ) ⊆ V ∗ is theannihilator A ◦ of A . Using the involution ⊥ , we may define several special types of subspaces. There arethe isotropic subspaces, for which A ⊆ A ⊥ and the coisotropic subspaces, for which A ⊥ ⊆ A . Subspaces which are both isotropic and coisotropic, i.e. fixed points of theorthogonality involution, are called lagrangian . Subspaces for which A ∩ A ⊥ = { } are called symplectic ; the restriction of the symplectic form to such a subspace is non-degenerate, and hence again a symplectic form. For such spaces, the order-reversingproperty implies that this condition is equivalent to A + A ⊥ = V , so that we can alsowrite the condition as V = A ⊕ A ⊥ . Involutivity implies that A ⊥ is symplectic as well,giving a symplectic direct sum decomposition of V . Such symplectic direct sumscan also be built from the external direct sum of symplectic spaces by equipping thesum with the direct sum form, which is again symplectic. The aim of this paper is tostudy the decomposition of (co)isotropic triples with respect to symplectic direct sums(not just of two summands).Given a symplectic direct sum decomposition V = A ⊕ A , we can define in a purelylattice-theoretic way the involution on Σ( A ) associated with the restricted symplecticstructure. Let C be any subspace of A . Then C ⊥ contains A , and the modularlaw implies that C ⊥ = ( C ⊥ ∩ A ) ⊕ A . So we may define the operation ⊥ by setting C ⊥ := C ⊥ ∩ A . It is clearly order-reversing, and it is easy to check that it is involutive.Of course, we can do the same thing in A . We see from here that for subspaces C j ⊆ A j ,the direct sum C ⊕ C is (co)isotropic or symplectic if and only if C and C are. Similarstatements to all of those above hold for orthogonal decompositions with any numberof summands.A linear representation of a partially ordered set, or poset, P in a vector space V is an order preserving map ψ from P to the poset Σ( V ). The dimension vector of ψ is the function which assigns to each p ∈ P the dimension of ψ ( p ). We will also add The modular law states that, for subspaces A , B , C of a given vector space, if B ⊆ A , then A ∩ ( B + C ) = B + A ∩ C . We use the word “linear” here simply to distinguish this notion of representation from the notionof symplectic poset representation, which we define below. V as a first component to the dimension vector. A morphism f : ψ → ψ ′ betweenrepresentations ψ in V and ψ ′ in V ′ of the same poset P is a linear map f : V → V ′ such that f ( ψ ( p )) ⊆ ψ ′ ( p ) for all p ∈ V ; it is an isomorphism if f is bijective and f ( ψ ( p )) = ψ ′ ( p ) for all p ∈ P .The term poset with involution will denote a poset P equipped with an order-reversing involution ⊥ : P → P . The relevance of poset representations and involutionsarises when we observe that each isotropic subspace A in a symplectic space V is natu-rally associated with the subspace A ⊥ which contains it and hence with the pair ( A, A ⊥ ).Conversely, if ( A, B ) is any pair of subspaces for which A ⊆ B and A ⊥ = B , then A isisotropic and B is its coisotropic orthogonal. Thus we see that isotropic (or equivalently,coisotropic) subspaces in V may be identified with involution-preserving representationsinto Σ( V ) of a poset consisting of two elements in linear order, equipped with the order-reversing involution which exchanges the two elements. We will denote this poset by . Definition 3.2.
Let ( P, ⊥ ) be a poset with involution, ( V, ω ) a symplectic vector space,and (Σ( V ) , ⊥ ) the poset of its subspaces equipped with the involution induced by ω . A symplectic representation of ( P, ⊥ ) in V is an order-preserving map ϕ : P → Σ( V ) such that ϕ ( p ⊥ ) = ϕ ( p ) ⊥ ∀ p ∈ P. If ϕ is such a representation, forgetting the involutions gives us simply a linear repre-sentation of P in the vector space V , which we denote by ˆ ϕ and call the underlying linear representation.Given another symplectic representation ϕ ′ in ( V ′ , ω ′ ) of the same poset with invo-lution ( P, ⊥ ) , an isomorphism from ϕ to ϕ ′ is an isomorphism of linear poset repre-sentations ˆ ϕ → ˆ ϕ ′ which is also an isometry ( V, ω ) → ( V ′ , ω ′ ) . Now, for any k , the k -tuples of isotropic-coisotropic pairs in ( V, ω ) are simply thesymplectic representations in (
V, ω ) of the partially ordered set k · := + + · · · + ( k times) consisting of k copies of which are independent in the sense that a ≤ b only when a and b belong to the same copy, with the involution interchanging the elements of eachcopy of . In the present paper, we derive for k = 0 , , ,
3, up to direct decomposition,the classification of isotropic k -tuples from the classification of linear representations ofthe poset k · ; these essentially correspond to the representations of a quiver associatedwith the Dynkin diagrams A , A , and A , and the extended Dynkin diagram ˜ E ,respectively. The first three cases are easy, with only finitely many indecomposables upto isomorphism, while the latter case is more involved – in particular there are infinitelymany indecomposables, some appearing in 1-parameter families.Our general strategy for classifying isotropic triples will involve three basic opera-tions related to symplectic representations. First of all, given a symplectic representa-tion ϕ of ( P, ⊥ ) in ( V, ω ), one may ignore the involution and symplectic structure toobtain a linear representation ˆ ϕ of P in V . Second, starting with a linear representation ψ of ( P, ⊥ ) in V , one can ask whether there exists a symplectic form on V such that We use the same symbol to denote two different involutions – the one on P and the one on Σ( V ). is in fact a symplectic representation in ( V, ω ); in this case ω is a compatible sym-plectic form. Third, one can build a symplectic representation out of each linear one bya “doubling” construction called symplectification , which we define in subsection 3.6below. Remark 3.3.
These operations are close to ones in symplectic geometry. The first onecorresponds to forgetting the symplectic structure on a symplectic manifold, the secondis analogous to asking whether a given manifold admits a symplectic structure, whilethe third is similar to the cotangent bundle construction.
We fix a poset P . Given linear poset representations ψ and ψ ′ , on V and V ′ respectively,their (external) direct sum is the poset representation on V ⊕ V ′ defined by( ψ ⊕ ψ ′ )( x ) = ψ ( x ) ⊕ ψ ′ ( x ) ∀ x ∈ P. A subrepresentation of a linear representation ψ on V is a representation ψ ′ ona subspace U ⊆ V such that ψ ′ ( x ) ⊆ ψ ( x ) ∀ x ∈ P . Given subrepresentations ψ ′ and ψ ′′ of ψ , on U ′ and U ′′ respectively, we say they form an (internal) direct sumdecomposition of ψ if V = U ′ ⊕ U ′′ and ψ ( x ) = ψ ′ ( x ) ⊕ ψ ′′ ( x ) ∀ x ∈ P. In this case, ψ ′ ( x ) = ψ ( x ) ∩ U ′ for all x ∈ P , and similarly for ψ ′′ .We note that, given a subrepresentation ψ ′ of ψ , there may not exist a subrepresen-tation ψ ′′ such that ψ = ψ ′ ⊕ ψ ′′ . For an example, consider P = { x , x , x } endowedwith the empty partial order, i.e. there are no order relations between the elements. Let ψ be a representation of P on a two-dimensional space V such that the subspaces ψ ( x i )are three independent lines, and let ψ ′ be the subrepresentation on U ′ := ψ ( x ) suchthat ψ ′ ( x ) = ψ ( x ) and ψ ′ ( x ) = ψ ′ ( x ) = 0. Suppose there existed a subrepresenta-tion ψ ′′ such that ψ = ψ ′ ⊕ ψ ′′ . Then ψ ′′ would be need to be defined on a subspace U ′′ such that V = U ′ ⊕ U ′′ ; thus U ′′ would be a line in V . By the requirement that ψ ( x i ) = ψ ′ ( x i ) ⊕ ψ ′′ ( x i ), it follows that necessarily ψ ( x ) = 0 ⊕ ψ ′′ ( x ) and ψ ( x ) = 0 ⊕ ψ ′′ ( x ) . But since the subspaces ψ ′′ ( x ) and ψ ′′ ( x ) must be contained in the line U ′′ , this wouldimply that ψ ′′ ( x ) = ψ ′′ ( x ) = U ′′ , a contradiction to the fact that ψ ′′ ( x ) and ψ ′′ ( x )are independent.For any vector space V , idempotents π in the algebra End( V ) of its endomorphismscorrespond to direct sum decompositions V = A ⊕ B of V , where A = Im π and B = Ker π = Im (1 − π ). (The zero and the identity endomorphisms are denoted by 0 and1, respectively.) The same is true for the endomorphism algebra End( V, ψ ) of a linearrepresentation ψ in V of a fixed partially ordered set P . Recall that an endomorphismof a linear representation ψ is a linear map f : V → V such that f ( ψ ( p )) ⊆ ψ ( p ) for12ll p ∈ P . It is easy to check that End( V, ψ ) is a unital subalgebra of End( V ). Sincethe ground field embeds into End( V, ψ ) via its action on the algebra unit, we sometimesrefer simply to the ring of endomorphisms of ψ .The basic theory of poset representations parallels that of modules of finite compo-sition length as presented, for example, in [17, § E of some End( V ) local if each of its elements is either invertibleor nilpotent. It follows that, for any f ∈ E , either f or 1 − f is invertible; namely, if f is not invertible, then f n = 0 for some n and 1 − f has inverse P ni =0 f i . More generally,if g = P mi =1 f i is invertible then so is at least one of the f i . The nilpotent elements forman ideal RadE, the radical of E . (Namely, given f and g , if one of them is nilpotent andif f g were to be invertible, then f would be surjective and g injective, so actually bothwould be invertible. And if both f and g are nilpotent then f + g cannot be invertible.)It follows that E/ RadE is a division ring and RadE the unique maximal ideal.The following result is a version of Fitting’s Lemma.
Lemma 3.4.
For a linear poset representation ψ in V the following are equivalent1. ψ is indecomposable.2. End(
V, ψ ) is local.3. End(
V, ψ ) has only the trivial idempotents and . Proof
Clearly, 2) ⇒ ⇒ ψ indecomposable. The rank of f j is anon-increasing and non-negative function of j = 1 , , , · · · , so it stabilizes after finitelymany steps, say d steps. Then f j (Im f d ) = Im f d + j = Im f d for all negative j ; Call thisimage space I . It follows that the nondecreasing sequence of kernels of these powers alsosatisfies N := Ker f d + j = Ker f d ; call this null space N . If v ∈ I ∩ N , then v = f d ( w )for some w , and 0 = f d v . Hence, f d ( w ) = 0, so v = f d ( w ) must be zero as well. So I ∩ N = { } . By dimension counting, we have V = I ⊕ N , where I and N are both f -invariant, f | I is invertible (since it is surjective), and f N is nilpotent.If K is any f -invariant subspace, then f k ( K ) ⊆ K is the projection π I ( K ) of K on I . Then the other projection π N ( K ) = (1 − π I )( K ) is contained in K as well. Thisshows that any f -invariant subspace is the direct sum of its components in I and N . Itfollows that, if f leaves invariant an indecomposable family of subspaces, either I or N must be zero, and f is either nilpotent or invertible. Remark 3.5.
Even without the indecomposability assumption, the decomposition intoinvertible and nilpotent parts is unique: for any such decomposition f I ⊕ f N , I and N must be the image and kernel respectively of all sufficiently large powers of f .We now state the Krull-Remak-Schmidt theorem in the form that we will need. Theorem 3.6.
Let ψ = ψ ⊕ ψ ⊕ · · · ⊕ ψ n and let ψ ′ = ψ ′ ⊕ ψ ′ ⊕ · · · ⊕ ψ ′ n ′ be directsum decompositions of isomorphic linear poset representations, e.g., of the same repre-sentation, into indecomposable summands. (Such decompositions always exist in finite imensions.) Then n = n ′ , and the two decompositions are the same up to isomorphismand permutation of the summands. Proof
We may assume n ′ ≥ n , and we denote the underlying vector spaces as V ⊕ · · · ⊕ V n and V ′ ⊕ · · · ⊕ V ′ n ′ . First, consider n = n ′ = 2 and an isomorphism f : ψ → ψ ′ such that f ( v ,
0) = ( g ( v ) , h ( v )) with morphisms g, h : ψ → ψ ′ where g is an isomorphism. We claim that ψ and ψ ′ are isomorphic, too. To prove this,we may assume h = 0: replace f by f ′ where f ′ ( v , v ) = ( w , w − hg − ( w )) if f ( v , v ) = ( w , w ). Then ψ ∼ = ψ/ ( ψ ⊕ ∼ = ( f ψ ) / ( gψ ⊕
0) = ψ ′ / ( ψ ′ ⊕ ∼ = ψ ′ .In general, let f : ψ → ψ ′ be a given isomorphism and consider the canonicalembeddings ε i , ε ′ i and projections π i , π ′ i given by the decompositions. Put g i = π ′ ◦ f ◦ ε i and h i = π i ◦ f − ◦ ε ′ . Then P ni =1 g i ◦ h i = 1 ∈ End( V ′ , ψ ′ ). Since this ring islocal, one of the summands is invertible, say the first. Then h ◦ g ∈ End( V , ψ )is not nilpotent, whence invertible. Thus, g : ψ → ψ ′ is an isomorphism. Clearly, f ( v , , . . . ,
0) = ( g ( v ) , w , . . . ) and it follows, by the special case, that ψ ⊕ · · · ⊕ ψ n isisomorphic to ψ ′ ⊕ · · · ⊕ ψ ′ n ′ . We repeat the argument until only ψ n is left on one side.Since ψ n is indecomposable one has n = n ′ whence ψ n ∼ = ψ ′ n . Poset representations of particular interest are ones which can be built from a vectorspace U and a linear map η : U → U . We will will refer to such a couple ( U, η ) simplyas an endomorphism when no conflation with other notions is to be feared. Fromsuch an endomorphism one can build a poset representation ( V ; U , U , U , U ), i.e. arepresentation of the poset + + + : define V = U × UU = U × U = 0 × UU = { ( x, − x ) | x ∈ V } U = { ( x, − η ( x )) | x ∈ V } . (1)We will call this the quadruple associated to ( U, η ). Note that U and U are thenegative graphs, respectively, of η and the identity map on U . It is straightforward tocheck that any endomorphism of the poset representation (1) is of the form f × f , where f : U → U is such that η ◦ f = f ◦ η. (2)The collection of linear maps f : U → U satisfying (2) form an algebra End( U, η ), the endomorphism algebra of ( U, η ). More generally, a morphism of endomorphisms(
U, η ) → ( U ′ , η ′ ) is a linear map f : U → U ′ satisfying η ′ ◦ f = f ◦ η ; such a map is an isomorphism if, additionally, f is invertible as a linear map.14 emma 3.7. A poset representation ( V ; U , U , U , U ) is isomorphic to one of the form (1) for some ( U, η ) if and only if U i ⊕ U j = V for i < j ≤ and U ⊕ U = V. (3) Proof
It is easily seen that, given (
U, η ), the associated quadruple satisfies (3). Forthe converse, suppose ( V ; U , U , U , U ) is a quadruple satisfying (3). First, note thatthe conditions (3) imply that all of the subspaces U i have the same dimension. Chooseany vector space U having the same dimension as the U i and choose linear isomorphisms ϕ : U → U and ϕ : U → U . Next, note that since U is independent of both U and U , there exists an isomorphism ϕ : U → U such that U = { x + ϕ x | x ∈ U } ⊆ U ⊕ U . Finally, using this data we construct the following map ϕ : V = U ⊕ U −→ U × U, x + y ( ϕ x, − ϕ ϕ − y ) . Clearly ϕ is a linear isomorphism and maps U to U × U to 0 × U . Furthermore,given x + ϕ x ∈ U , its image under ϕ is ( ϕ x, − ϕ x ), as desired. Since U is independentof U , it is the graph of a linear map ϕ : U → U (though this map may not be anisomorphism). Now given x + ϕ x ∈ U ⊆ U ⊕ U , we have ϕ ( x + ϕ x ) = ( ϕ x, − ϕ ϕ − ϕ x ) = ( ϕ x, − ϕ ϕ − ϕ ϕ − ϕ x ) . Setting η = ϕ ϕ − ϕ ϕ − , we have a quadruple of the form (1) which is isomorphic to( V ; U , U , U , U ).An endomorphism ( U, η ) is decomposable if there exist non-zero subspaces U , U ⊆ U which are invariant under η and form a decomposition U = U ⊕ U . As in the caseof poset representations, such decompositions correspond to idempotents in the endo-morphism algebra of ( U, η ). For future reference we state:
Lemma 3.8.
The endomorphism algebra of ( U, η ) is isomorphic to the endomorphismalgebra of the associated quadruple (1) . In particular, ( U, η ) is indecomposable if andonly if its endomorphism algebra is local. Proof
An isomorphism is given by mapping an endomorphism f of ( U, η ) to theendomorphism f × f of (1). Its inverse is “restriction to U ”.The characterization of indecomposabilty follows from Lemma 3.4.Another point of view is that an endomorphism ( U, η ) defines a k [ x ]-module U k [ η ] ,where the action of x on U k [ η ] is defined by the action of η on U , i.e. ( P i λ i x i ) · u :=( P i λ i η i )( u ) for u ∈ U , λ i ∈ k . In this case, morphisms f : ( U, η ) → ( U ′ , η ′ ) arethe same as module homomorphisms f : U k [ η ] → U ′ k [ η ′ ] and decompositions of ( U, η )correspond to direct sum decompositions of U k [ η ] .15 emark 3.9. If η is an indecomposable endomorphism of U with an eigenvalue λ inthe base field k , then there is a basis such that η is described by a single λ -Jordan block.With respect to such a basis, the members of the endomorphism algebra E = End( U, η )are given by the matrices P d − i =0 a i N i where d = dim U and N is the nilpotent matrixwith n i,i +1 = 1 for i = 1 , . . . , d −
1, and 0 otherwise. In particular, E is local (namelyisomorphic to k [ x ] / ( x d )) and E = k id ⊕ RadE.
Proof
Consider the endomorphism ζ = η − λ id and apply the Fitting Lemma. Since ζ is non-invertible, there is n with ζ n = 0, that is η admits Jordan normal form withunique eigenvalue λ and there is a single block J = λI + N only. Now, E is given bythe A such that AJ = J A , which is equivalent to the condition AN = N A . Since anyinvariant subspace of N is one of A also, A must be upper triangular and with only asingle scalar on each upper diagonal. In other words, from AN = B = N A one has that a i,j − = b ij = a i +1 ,j for i < j , and 0 as entry otherwise.If an indecomposable endomorphism ( U, η ) does not have an eigenvalue in the groundfield k , a generalization of Remark 3.9 holds; we recall this in Sections 8.2.We note that, for ( U, η ) indecomposable, the endomorphism algebra End(
U, η ) isgenerated, as a unital k -subalgebra of End( U ), by the single element η (see Proposition8.11). In other words, it consists simply of “polynomials in η ”. In particular, it is acommuatative algebra, and any subspace of U is invariant under End( U, η ) if and only ifit is invariant under η . Further details about these endomorphism algebras are discussedin Section 8.4. If V carries a symplectic form ω , we define a transpose operation t on its endomorphismsby the usual formula ω ( f ( x ) , y ) = ω ( x, f t ( y )). Then the condition π t π = 0 on anidempotent π means that the image of π is an isotropic subspace. In fact, for any x and y in V , we have ω ( π ( x ) , π ( y )) = ω ( x, π t π ( y )).Similarly, if ππ t = 0, then the image of π t is isotropic. It follows that, if π + π t = 1,then the images of π and π t give a decomposition of V as the direct sum of two isotropicsubspaces which must be lagrangian and hence in duality by the bilinear form. We statethis result in the form of a lemma for use below. Lemma 3.10.
Decompositions of a symplectic vector space V as a direct sum of (two)lagrangian subspaces are in 1-1 correspondence with idempotent endomorphisms π : V → V such that π t π = 0 = ππ t and π + π t = 1 . Similarly, for symplectic direct sum decompositions, we have the following charac-terization in terms of idempotents.
Lemma 3.11.
Symplectic direct sum decompositions of V into two subspaces are in 1-1correspondence with idempotent endomorphisms π : V → V which are self-adjoint, i.e. π t = π . roof It is a standard fact that Ker ( π ) = (Im π t ) ⊥ , and Ker π is also the image of1 − π . It follows that π = π t if and only if the images of π and 1 − π are orthogonal,which means that the corresponding direct sum decomposition is symplectic.Now suppose that ϕ is a symplectic representation in V of an involutive poset P . Lemma 3.12. If f is an endomorphism of ϕ , then f t is an endomorphism of ϕ as well. Proof
We must show that, for each p ∈ P , f t ( ϕ ( p )) ⊆ ϕ ( p ). This is equivalent toshowing that, for a ∈ ϕ ( p ) and b ∈ ϕ ( p ) ⊥ , we have ω ( f t ( a ) , b ) = 0, or, equivalently, ω ( a, f ( b )) = 0. Since ϕ is symplectic, we have b ∈ ϕ ( p ⊥ ); since f is an endomorphismof ϕ , f ( b ) ∈ ϕ ( p ⊥ ) = ϕ ( p ) ⊥ as well. Lemma 3.13.
Symplectic direct sum decompositions of ϕ into two representations arein - correspondence with self-adjoint idempotents π = π t which are endomorphismsof the underlying linear representation, i.e. π ∈ End ( ˆ ϕ ) . Proof
A direct sum decomposition V = V ⊕ V of ϕ is nothing else than a decom-position of ˆ ϕ where the summands V and V are also mutual orthogonals. Such adecomposition corresponds to an idempotent π ∈ End( ˆ ϕ ) such that π t = π (c.f. Lemma3.11). Lemma 3.14. If ϕ is an indecomposable symplectic representation and if f ∈ End ( ˆ ϕ ) isan endomorphism of ˆ ϕ such that f t = ± f , then f is either nilpotent or an isomorphism. Proof
From the proof of Lemma 3.4 we know that there exists a non-negative integer d such that V = Ker f d ⊕ Im f d is a decomposition of ˆ ϕ . Since ( f d ) t = ( f t ) d = ± f d , thisdecomposition of V is one into orthogonal summands; hence it is a symplectic directsum decomposition of ϕ . Since ϕ is assumed to be indecomposable as a symplecticrepresentation, either Ker f d or Im f d must be zero. Let ψ : P → Σ( V ) be any linear representation in V of a poset P equipped with anorder-reversing involution ⊥ . We can think of ψ as a linear representation of ( P, ⊥ )which doesn’t “see” the involution. Define the dual representation of ψ in V ∗ by ψ ∗ : ( P, ⊥ ) −→ Σ( V ∗ ) , ψ ∗ ( p ) = ψ ( p ⊥ ) ◦ ∀ p ∈ P. (4)Note that this definition makes use of the involution on P , i.e. ψ ∗ is the dual of ψ withrespect to the involution ⊥ on P . In particular, the definition only makes sense viewing ψ as a linear representation of ( P, ⊥ ) (rather than only of P ). The combination oftwo order inversions - once due to the poset involution and once due to the annihilatoroperation - leads to ψ ∗ being order preserving.17y Lemma 3.1, if ϕ is a symplectic representation of ( P, ⊥ ) in ( V, ω ), then ˜ ω is anisomorphism from ˆ ϕ to ˆ ϕ ∗ . This shows that a representation of P can be compatible witha symplectic structure only if it is self-dual, i.e. isomorphic to its dual. In particular,it has a dimension vector which is self-dual in the sense that dim ϕ ( p ) + dim ϕ ( p ⊥ ) =dim V for any p ∈ P .The results above lead to the natural question of determining the relation betweenisomorphisms of a linear representation to its dual, the self-duality of its dimensionvector, and the existence (and uniqueness) of compatible symplectic structures.We will see later that many representations of interest to us are characterized up toisomorphism by their dimension vectors. Hence we record the following simple obser-vation. Lemma 3.15.
If a representation ψ is isomorphic to its dual, then it has a self-dualdimension vector. If a representation ψ is characterized up to isomorphism by its di-mension vector, and this dimension vector is self-dual, then ψ is isomorphic to its dual. Proof
The first statement has already been noted. For the second, we observe that, ifthe dimension vector of ψ is self-dual, then ψ and ψ ∗ have the same dimension vector,and so by assumption they must be isomorphic.In studying symplectic structures, it will sometimes be important to consider non-degenerate symmetric bilinear forms, as well. For these, there is an analogous notion oforthogonality and, therefore, an analogous notion of representation of a partially orderedset with involution. To capture both kinds of forms, we’ll speak of ε - symmetric bilinearforms, where ε = 1 for symmetric forms and ε = − ε - symmetric representation of a poset with involution is the generalization of thedefinition of symplectic representation to ε - symmetric forms. We will often identifya non-degenerate bilinear form B on V with the linear isomorphism B : V → V ∗ , v B ( v, · ), setting B ∗ : V → V ∗ , v B ( · , v ). With this notation, B is ε -symmetric if andonly if B ∗ = εB . In referring to the parity ε of B we sometimes write ε ( B ).For a fixed linear representation ψ on V of a poset ( P, ⊥ ) with involution, one canask how many different non-degenerate ε -symmetric forms B exist (if any) which arecompatible with ψ in the sense that ψ ( p ⊥ ) = ψ ( p ) ⊥ ∀ p ∈ P, where the involution ⊥ on V is the one induced by B . A non-degenerate ε -symmetricform which is compatible with ψ in this sense will be called a compatible form . Lemma 3.16.
Let ψ be a linear representation of ( P, ⊥ ) in V , and B : V → V ∗ anon-degenerate ε -symmetric form. Then B is a compatible form (for ψ ) if and only if B is an isomorphism ψ → ψ ∗ . Note that if B : V → V ∗ is a non-degenerate bilinear form such that B ∗ = λB for some λ ∈ k , thenin general, for a given subspace A ⊆ V , one no longer has “the” orthogonal “ A ⊥ ”, but rather one mustconsider the right- and left-orthogonal of A , which in general will not coincide. roof That B is compatible means that ψ ( p ⊥ ) = ψ ( p ) ⊥ for all p ∈ P . This isequivalent with B ( ψ ( p ⊥ )) = ψ ( p ) ◦ , and since ψ ( p ) ◦ = ψ ∗ ( p ⊥ ), this is the same as B ( ψ ( p ⊥ )) = ψ ∗ ( p ⊥ ) for all p ∈ P .The following is Proposition 2.5 (2) in [25]. Lemma 3.17.
Let ψ be an indecomposable linear representation in V of a poset withinvolution ( P, ⊥ ) . Then ψ is isomorphic to its dual if and only if there exists a compatibleform. Proof
If there exists a compatible form, then by Lemma 3.16, such a form defines anisomorphism between ψ and ψ ∗ .Conversely, suppose that B : ψ → ψ ∗ is an isomorphism. Then so is B ∗ , andhence the symmetric and antisymmetric parts ( B + B ∗ ) / B − B ∗ ) / B arealso morphisms of representations, as are the endomorphisms B − ( B + B ∗ ) / B − ( B − B ∗ ) /
2, whose sum is the identity morphism. By Lemma 3.4, the ring ofendomorphisms of ψ is local, so the two summands cannot both be degenerate. Itfollows that either the symmetric or antisymmetric part of B gives a compatible non-degenerate bilinear form. Definition 3.18.
The symplectification ψ − of a linear representation ψ is the rep-resentation ψ − : ( P, ⊥ ) −→ Σ( V ∗ ⊕ V, Ω) , ψ − ( x ) = ψ ∗ ( x ) ⊕ ψ ( x ) , where V ∗ ⊕ V is endowed with the canonical symplectic structure Ω(( ξ, v ) , ( η, w )) := ξ ( w ) − η ( v ) ξ, η ∈ V ∗ v, w ∈ V. Proposition 3.19. ψ − is a symplectic representation. Proof
For any p ∈ P , we have ψ − ( p ⊥ ) = ψ ∗ ( p ⊥ ) ⊕ ψ ( p ⊥ ) = ψ ( p ) ◦ ⊕ ψ ( p ⊥ ) . This isthe symplectic orthogonal of ψ ( p ⊥ ) ◦ ⊕ ψ ( p ) = ψ ∗ ( p ) ⊕ ψ ( p ) = ψ − ( p ) . Here are some fundamental properties of the symplectification operation.First of all, given a symplectic vector space V = ( V, ω ), we denote by V the sym-plectic vector space ( V, ω ), where we define ω := − ω . Given a symplectic poset repre-sentation ϕ on ( V, ω ), we define a symplectic poset representation ϕ on V by setting ϕ ( x ) = ϕ ( x ) for all p ∈ P , i.e. as morphisms of posets, ϕ and ϕ are the same, only theform on V has been changed. This construction is sometimes known in a more general setting as hyperbolization , see [25], be-cause the analogue for symmetric bilinear forms leads to isotropic subspaces in spaces with forms ofsignature zero, sometimes called “hyperbolic”. We use “symplectification” rather than “symplectiza-tion” because the latter term already refers to the construction of symplectic manifolds from contactmanifolds by adding one dimension. roposition 3.20. For any symplectic representation ϕ on V , the symplectic represen-tations ˆ ϕ − = ˆ ϕ ∗ ⊕ ˆ ϕ on V ∗ ⊕ V and ϕ ⊕ ϕ on V ⊕ V are isomorphic. Proof
An isomorphism of symplectic representations is given by τ : ϕ ⊕ ϕ → ˆ ϕ ∗ ⊕ ˆ ϕ, ( v, w ) ( ˜ ω ( v + w ) , v − w ) . Indeed, τ is a morphism of representations, since when ( v, w ) ∈ ϕ ( x ) ⊕ ϕ ( x ) = ϕ ( x ) ⊕ ϕ ( x ), then τ ( v, w )) ∈ ˜ ω ( ϕ ( x )) ⊕ ϕ ( x ) = ϕ ( x ⊥ ) ◦ ⊕ ϕ ( x ) = ˆ ϕ − ( x ). And τ is a symplecticisomorphism, sinceΩ( τ ( v, w ) , τ ( v ′ , w ′ )) = ˜ ω ( v + w )( v ′ − w ′ ) − ˜ ω ( v ′ + w ′ )( v − w )= [ ω ( v, v ′ ) + ω ( v, − w ′ ) + ω ( w, v ′ ) + ω ( w, − w ′ ) − ω ( v ′ , v ) − ω ( v ′ , − w ) − ω ( w ′ , v ) − ω ( w ′ , − w )]= [2 ω ( v, v ′ ) − ω ( w, w ′ )]= ω ⊕ ω (( v, v ′ ) , ( w, w ′ )) . Remark 3.21.
The symplectic isomorphism used in the proposition above is the sameas the one behind the Weyl symbol calculus for pseudodifferential operators (c.f. forinstance [6], formula (7) and Theorem 8) and the definition of “Poincare’s generatingfunction” in hamiltonian mechanics (c.f. [34]).
Proposition 3.22.
The symplectification ψ − of an indecomposable linear representa-tion ψ is symplectically decomposable if and only if ψ admits a compatible symplecticstructure. Proof ϕ = ψ − is by definition decomposed linearly into the indecomposables ψ ∗ and ψ .Suppose that it is also symplectically decomposable into two symplectic representations, ϕ and ϕ . The latter decomposition is also a linear decomposition of ˆ ϕ , and so byTheorem 3.6 the linear representations ϕ and ϕ must be isomorphic to ψ ∗ and ψ insome order. In particular ψ (as does ψ ∗ ) admits a compatible symplectic structure.Conversely, suppose that ϕ admits a compatible symplectic structure. Then, byProposition 3.20, ( ˆ ϕ ) + = ( ˆ ϕ ) ∗ ⊕ ˆ ϕ and ϕ ⊕ ϕ are isomorphic symplectic representations.Since the latter is decomposable, so is the former. Proposition 3.23. If ψ and ψ are indecomposable linear representations, then ψ isisomorphic to ψ or to ψ ∗ if and only if the symplectifications ψ − and ψ − are isomorphicas symplectic representations.In particular, two symplectifications of indecomposable linear representations areisomorphic as symplectic representations if and only if they are isomorphic as linearrepresentations. roof If ψ is isomorphic to ψ or to ψ ∗ , then ψ − is isomorphic to ψ − or to ( ψ ∗ ) − . Ifthe former holds, we are done. For the latter, we must show that ψ − is isomorphic to( ψ ∗ ) − . They are clearly isomorphic as linear representations, but under the isomorphismwhich exchanges the summands, the symplectic structures differ by a factor of −
1. Tocorrect for this factor, we compose with the antisymplectic isomorphism ( ξ, v ) ( − ξ, v )from ψ − to itself.Conversely, if ψ ∗ ( x ) ⊕ ψ ( x ) and ψ ∗ ( x ) ⊕ ψ ( x ) are isomorphic as symplectic repre-sentations, then they are in particular also isomorphic as linear representations. Thisimplies that their indecomposable summands are isomorphic in some order, so either ψ ≃ ψ or ψ ≃ ψ ∗ Example 3.1.
The following are the symplectifications of the indecomposable repre-sentations of the poset , i.e. nested pairs of subspaces. Each pair is contained in k , sothe symplectification is contained in k ∗ ⊕ k and is symplectically indecomposable. • The symplectification of k ⊇ k ∗ ⊕ k ⊇ ⊕ • The symplectification of 0 ⊇ k ∗ ⊕ ⊇ k ∗ ⊕ • The symplectification of k ⊇ k is 0 ⊕ k ⊇ ⊕ k .The first example is self-dual, while the latter two examples are dual to one another,and their symplectifications are isomorphic by a “90-degree rotation”.Given a pair of linear poset representations, we say that they are mutually dual or,synonymously, a dual pair if each representation is isomorphic to the dual of the other.On the level of isomorphism classes, symplectification builds symplectic representationsby taking the direct sum of dual pairs of linear representations. The following lemma, due to Quebbemann et al. [25, Thm.3.3] and Sergeichuk [30,Lemma 2] will be an essential tool in this paper. It shows that symplectically indecom-posable but linearly decomposable representations arise only through symplectification.The analogous result holds in the symmetric setting.
Lemma 3.24.
Suppose that ϕ : ( P, ⊥ ) → Σ( V ) is an indecomposable symplectic repre-sentation such that ˆ ϕ is (linearly) decomposable. Then there exists an indecomposablelinear representation ψ such that ϕ ≃ ψ − . Proof
Because ˆ ϕ is linearly decomposable, there exists a non-trivial idempotent π ∈ End( ˆ ϕ ). After two modifications, π will be conjugated into an idempotent endo-morphism π satisfying the hypotheses π t π = 0 = ππ t and π + π t = 1 of Lemma 3.10,giving the required decomposition.By Lemma 3.12, the idempotent π t is also an endomorphism. By Lemma 3.13, π t = π , since otherwise ϕ would be decomposable as a symplectic representation. Set21 = π π t . Note that ρ is self-adjoint and lies in End( ˆ ϕ ) . By Lemma 3.14, ρ must beeither nilpotent or an isomorphism. But ρ cannot be an isomorphism, since π and π t have nontrivial kernels and cokernels. So ρ must be nilpotent.Now set h := s ( ρ ), where s ( X ) is the binomial series for (1 − X ) / ; s ( ρ ) is well-defined because ρ is nilpotent, which implies that the power series is just a polynomial in ρ . Note that h ∈ End( ˆ ϕ ) , and that h is also self-adjoint. Furthermore, h is invertible,its inverse being defined by substituting ρ in the binomial series for (1 − X ) − / . Define π := h π h − , and note that π lies in End( ˆ ϕ ) and is again a non-trivialidempotent. Furthermore, π t π = h − π h π t h − = h − π (1 − π π t ) π t h − = h − ( π π t − π π t ) h − = 0 . We are half-way there. Now ρ := π π t is a nilpotent, self-adjoint element of End( ˆ ϕ ),and h := s ( ρ ) is again an invertible, self-adjoint endomorphism of ˆ ϕ . Then π := h − π h ∈ End( ˆ ϕ ) is a non-trivial idempotent such that ππ t = h − π h π t h − = h − π (1 − π π t ) π t ˜ h − = h − ( π π t − π π t ) h − = 0and π t π = h π t ( h − ) π h = h π t (1 − π π t ) − π h = h π t (1 + π π t ) π h = h ( π t π + π t π π t π ) h = 0 , since π t π = 0. Furthermore, π + π t ∈ End( ˆ ϕ ) is idempotent: ( π + π t ) = π + π t π + ππ t + ( π t ) = π + π t . But π + π t is also self-adjoint, so, by Lemma 3.13, π + π t mustbe a trivial idempotent. It cannot be that π + π t = 0, since this would imply π t = − π ,whence 0 = π t π = − π = π , a contradiction to π = 0. Thus π + π t = 1. Corollary 3.25.
If two symplectically indecomposable but linearly decomposable repre-sentations are isomorphic as linear representations, then they are isomorphic as sym-plectic representations.
Proof
By Lemma 3.24, each of the two symplectic representations is the symplectifi-cation of an irreducible linear representation. Since the two representations are linearlyisomorphic, by Lemma 3.23, they are symplectically isomorphic.Lemma 3.24 tells us that every indecomposable symplectic representation is eitherlinearly indecomposable or the symplectification of a linearly indecomposable represen-tation, but not both. In the former case we say that ϕ is of non-split type; in thelatter case we say that ϕ is of split type. We briefly discuss the question of uniqueness of compatible bilinear forms and formulatea lemma which will be the basis of our analysis of uniqueness of compatible forms. Later22e will verify the hypothesis of the lemma for the representations which concern us,and one particular type will require a generalization. The proof uses ideas from thatof Lemma 5 in [30]; it is interesting that the proof uses a version of the “square root”construction of Lemma 2 of that paper (which is our Lemma 3.24). In addition, we usean idea (simplified for our context) from Proposition 2.5 in [25] when showing that anytwo compatible forms must both be symmetric or antisymmetric.
Lemma 3.26.
Let ψ be a linearly indecomposable representation in V of an involutiveposet for which the endomorphism algebra E (which is local by Corollary 3.4) and hasthe property that E = k id ⊕ Rad E . If ψ admits two compatible bilinear forms, thenthese forms are equal up to a constant scalar multiple and an automorphism of ψ . Inparticular, the forms must both be symmetric or antisymmetric. Proof If B and B are the isomorphisms from ψ to ψ ∗ corresponding to two com-patible forms, then C = B − B is an automorphism of ψ .Let † denote the antiautomorphism of E given by the operation of adjoint withrespect to B , i.e. B ( A † x )( y ) = B ( x )( Ay ) , which is equal to A ∗ ( B ( x ))( y ), so we have A † = B − A ∗ B . Define the signs ε and ε by B ∗ i = ε i B i , and let ǫ = ε ε . Then we have C † = ǫC .In fact, C † = ( B − B ) † = B − B ∗ ( B − ) ∗ B = ǫB − B B − B = ǫB − B = ǫC. By our hypothesis, we may write C as c id − r , where r ∈ R and c is a scalar and c = 0 since C is invertible. By replacing B by c − B and repeating the argument upto this point, we may assume that c = 1 , so that C = 1 − r for an r ∈ R .Now if ǫ were equal to −
1, we would have1 − r † = (1 − r ) † = − (1 − r ) , which would imply that 1 + 1 = r † − r , which is impossible since R is closed underaddition and contains no invertible elements (and, by assumption, 2 = 0 in our groundfield). So ǫ = 1, and both forms are either symmetric or antisymmetric. Furthermore, r = r † .By Lemma 3.4, we know that r is nilpotent, and so we can use the formal powerseries for √ − r to construct an automorphism h such that h † = h and h = C. Now we have h ∗ B h = B h † B − B h = B h † h = B h = B C = B , which shows that h is an isomorphism between the bilinear forms B and B . Remark 3.27.
A form ω on V and its scalar multiple aω are equivalent by a homothetyof V (which is automatically an automorphism of any poset representation) if and onlyif a is a square in the coefficient field. This means that the set of equivalence classes23f compatible forms under homothety is a principal homogeneous space of the squareclass group of k , defined as the quotient k × / k × of the multiplicative group of nonzeroelements of k by the perfect squares.Even if a is not a square, ω and aω might, a priori, still be isomorphic by a linearisomorphism which preserves a particular poset representation. This may also be thecase if the representation is linearly decomposable, though in the present paper we willnot consider the question of compatible forms for decomposable linear representations. Remark 3.28.
In Section 5.5 we will see that certain self-dual poset representationsfulfill the hypotheses of Lemma 3.26. In these cases it turns out that compatible formswhich differ by a scalar c ∈ k are in fact equivalent if and only if c is a square.For certain other self-dual poset representations, however, only a generalization ofLemma 3.26 applies and both symmetric and antisymmetric compatible forms exist fora given such representation; see in particular Sections 8.5 and 8.6. By Theorem 3.6 and Lemma 3.25 one obtains the following.
Corollary 4.1.
For a given symplectic representation, consider orthogonal decomposi-tions into symplectically indecomposable summands. Then any summand is either splitor non-split and the following are uniquely determined:1. The isomorphism types and multiplicities of split summands.2. The linear isomorphism types and multiplicities of non-split summands.
At this point, we have finished our discussion of the general theory of symplecticposet representations and are ready to move on to specific cases related to isotropictriples. We will do the following things. • In section 4.2 we review some essentials of the theory of quiver representations– as background for the classification results to be used. We explain how theclassification, obtained in [7, 8], of indecomposable representations of a quiverrelated to the Dynkin diagram ˜ E gives the classification of indecomposable linearrepresentations of the poset + + . • As a warmup, we derive, in sections 4.3 and 4.4, the classification of isotropic k -tuples, for k = 0 , ,
2, from the classification of representations of quivers asso-ciated with the Dynkin-diagrams A , A , and A . Then in section 4.5 we give anoverview of the quiver representation classification results for the Dynkin diagram˜ E , which we will use for classifying isotropic triples. In section 4.6, again as awarmup, we discuss isotropic triples in ambient dimension 2, and give, in section4.7, a preview of the situation in higher dimensions.24 Using the results of [7, 8], we identify which indecomposable linear representationsof P = + + are dual to one another when we equip + + with the involution ⊥ which exchanges the respective elements of the three nested pairs in + + .From now on we will use the term sextuples to refer to linear representations ofthis poset with involution. • From the general theory we know that self-dual sextuples admit compatible sym-metric or symplectic forms (or both). We determine which self-dual sextuplesadmit compatible symplectic forms (thus giving non-split isotropic triples), andwe give explicit constructions of such forms. • For the self-dual sextuples, we reduce the classification of compatible forms toa field-theoretic description. When k = R or when k is algebraically closed,compatible forms are parametrized by the square class group k × / ( k × ) . Forgeneral perfect fields, a similar, though slightly more complicated, description isobtained; see Theorem 8.17. We recall here some basic definitions and results in quiver representation theory, refer-ring to the literature for more details. A quiver Q is simply a directed graph, i.e. aset V of vertices and a set A of arrows, with source and target maps s and t from A to V . We allow multiple arrows with a given source and target, but assume the sets A and V to be finite. With a chosen ground field k , a representation ρ of Q is simplyan assignment to each vertex v a (finite dimensional, for our purposes) k -vector space ρ ( v ) and to each arrow a a linear map ρ ( a ) : ρ ( s ( a )) → ρ ( t ( a )). A morphism µ from ρ to ρ is a family of linear maps µ v : ρ ( v ) → ρ ( v ) making the obvious diagramscommute. When the family of linear maps consists of isomorphisms, then µ is called anisomorphism. The collection of representations of a fixed quiver with their morphismsform a category.A fundamental problem in the theory of quiver representations (as is the case forrepresentations of just about anything) is to describe the structure of the set of isomor-phism classes of representations, and, among these, the indecomposable representations,which are those not decomposable into nontrivial direct sums.The first basic result is the Krull-Schmidt theorem, which states that each repre-sentation of a quiver is isomorphic to a direct sum of indecomposables, and that thesummands in this decomposition, with their multiplicities, are unique up to isomorphismand reordering. This reduces the classification of representations to the enumeration ofthose which are indecomposable. When the set of isomorphism classes of indecompos-ables is finite, the quiver is of finite type .As mentioned above, we will be studying isotropic triples by considering them aslinear representations of the poset + + . These poset representations can be identified For example { [1], [2], [5], [10], [14], [29] } is a small sample subset of the available references, to givethe reader a starting point. E : v c c c i i i (5)We’ll refer to this quiver also as ˜ E . The labels on the vertices are in principle arbitrary(and sometimes unnecessary); we fix this choice since it is suggestive for our applicationto (co)isotropic triples. We identify the vertices other than v with the elements of theposet + + , with the partial order indicated by the arrows, i.e. i ≤ c , i ≤ c and i ≤ c .When considering representations of this quiver, we will denote the space associatedto the vertex v by V , and the spaces associated to the vertices c , c , c and i , i , i willbe denoted by C , C , C and I , I , I , respectively. When it is clear what the maps are,a representation of (5) will be denoted by the 7-tuple of spaces ( V ; C , I ; C , I ; C , I )or just ( V ; C i , I i ). We will also call such representations sextuples , just as we do posetrepresentations of + + . Abusing notation slightly, dimension vectors of sextupleswill be denoted ( v, c , c , c , i , i , i ), where the entries denote the dimensions of the(sub)spaces of a representation ( V ; C , I ; C , I ; C , I ).It is straightforward to see that, on the level of isomorphism classes, linear posetrepresentations of + + are in one-to-one correspondence with quiver representationsof the quiver ˜ E where all arrows are represented by injective linear maps (we will callthese injective representations). This correspondence is compatible with the notionsof direct sum for poset and quiver representations, respectively. Furthermore, one canprove that an indecomposable quiver representation of ˜ E is an injective representationif and only if the space assigned to the central vertex is non-zero (see [32], PropositionA.7.1).Using this, one can read off the indecomposable representations of + + fromthose of ˜ E as given in [7, 8]. We will mainly use the explicit normal forms given in [7].Since that reference (in contrast to [8]) only treats the case when the ground field isalgebraically closed, we will use a straightforward generalization of their normal formsfor more general fields. By inspection of the proofs [7], it is only the normal formsfor continuous-type representations that must be generalized. These are discussed inSection 6.An essential tool for the study of quiver representations is the Tits form of a quiver Q . This is the quadratic form q on the Z -module generated by the vertices defined by q ( n ) = P V n v − P V×V a v,w n v n w , where n v is the coefficient of v in n , and a v,w is thenumber of arrows from v to w . Note that this form does not depend on the directionof the arrows.The idea behind the Tits form is that, if the coefficients of n are the dimensionsof vector spaces assigned to the vertices, then the second term is (the negative of) thedimension of the linear space of all representations of Q in this family of vector spaces,while the first term is the dimension of the group, acting on the space of representations,26hose orbits are the isomorphism classes. In fact, the scalar multiples of the identity acttrivially, so we may say that the “virtual dimension” of the moduli space of isomorphismclasses of representations with dimension vector n is 1 − q ( n ). The actual dimension isat least this large (and larger if more of the group acts trivially), so if our ground fieldis, e.g., the real or complex numbers, the only way in which there can be finitely manyisomorphism classes with dimension vector n is if q ( n ) is at least 1.This suggests (but does not prove, for various reasons), that a quiver is of finitetype if and only if its Tits form is positive definite. In fact, this is true (for any groundfield!) and is part of what is known as Gabriel’s Theorem . The other part of thetheorem states that the connected quivers of finite type are exactly those for which theassociated undirected graph is a Dynkin diagram of type A , D , or E [11]. For thesequivers, it turns out that the nonnegative solutions of q ( n ) = 1, known as the positiveroots , are precisely the dimension vectors of indecomposable representations, and thereis exactly one isomorphism class corresponding to each positive root.The quiver (5) relevant to the classification of isotropic triples is not of finite type.It does belong, however, to the “next best” class, that of the so-called tame quivers.For these, the Tits form is positive semi definite, with one-dimensional null space whichwe will denote by N . N has a smallest positive element, which is the dimension vectorof a family of representations whose isomorphism classes are parametrized, in the caseof k algebraically closed, by a 1-dimensional variety. The positive roots thus fall intolines parallel to N . They still correspond to indecomposable representations, which nowbelong to families indexed by the positive integers.An extension of Gabriel’s theorem (c.f. [7], [8]) tells us that a quiver is of (infinite)tame type if and only if the corresponding undirected graph is an extended Dynkindiagram ; these are obtained from certain Dynkin diagrams by the addition of an edgeattached to an extremal vertex. Among these, for instance, is f D , consisting of fouredges attached to a central vertex. If the edges are all oriented to point toward thevertex, the representations of the quiver are closely related to those of the partiallyordered set + + + consisting of four incomparable elements. Representations ofthis poset are just quadruples of subspaces in vector spaces. These were classifiedby Gel’fand and Ponomarev [13], who showed that many classification problems inlinear algebra reduce to the classification of certain subspace quadruples. For instance,endomorphisms of a vector space V in dimension n correspond to certain kinds ofquadruples of subspaces of dimension n in V ⊕ V (the “axes,” the diagonal, and thegraph of the endomorphism). Indecomposable representations of this kind correspondto indecomposable endomorphisms which, in the case of an algebraically closed field,are those given by a single Jordan block. Since the diagonal element and the size ofsuch a block is arbitrary, one sees immediately the presence of 1-parameter families witharbitrarily large dimension vectors. This example also shows the possible complicationsfor fields which are not algebraically closed, where indecomposable endomorphisms areparametrized by irreducible polynomials which are no longer necessarily linear. We willsee later that there are close connections between the representations of f D and thoseof f E , the latter also being connected to the classification of endomorphisms.27n the next two sections, we reinterpret, via the theory of quiver and poset repre-sentations, the easy classifications of symplectic vector spaces and isotropic subspaces,followed by the classification of isotropic pairs, c.f. [21]. The relevant Dynkin diagramsare A , A , and A . We start with the classification of symplectic vector spaces with no distinguished isotropicsubspaces. We may think of these as the symplectic representations of the empty poset,or of the Dynkin diagram A , whose quiver consists of a single vertex with no arrows,and whose Tits form in terms of the one-entry dimension vector ( v ) is the positive def-inite form v . It is well known that any symplectic vector space admits a basis (in factmany such bases) of the form ( e , f , . . . e n , f n ), with the symplectic form determined bythe conditions that ω ( e j , f j ) = 1 for all j and ω ( a, b ) = 0 for all other pairs ( a, b ) of basiselements. We call this a symplectic basis . As a consequence, any symplectic vectorspace can be decomposed as a direct sum of copies of the space k with symplectic basis( e, f ). The only invariant of a symplectic space is its dimension, which must be an evennonnegative integer. This is consistent with the fact that the Tits form has a singlepositive root, which is (1).Another viewpoint here is that there is one indecomposable representation of theempty poset, the 1-dimensional space k . The symplectification of this representationis k ∗ ⊕ k . In fact, we will use this description of the 2-dimensional symplectic space,rather than k .We now move on to the example of individual isotropic subspaces, which correspondto certain symplectic representations of the poset , i.e. nested pairs of subspaces, asnoted above, where isotropic I in V corresponds to the representation V ⊇ I ⊥ ⊇ I .The quiver associated to this poset iswith underlying Dynkin diagram A . For a dimension vector of the form ( v, c, i ), corre-sponding to a representation V ← C ← I , the Tits form is q ( v, c, i ) = v + c + i − vc − ci ,which can be rewritten as 12 ( v + ( v − c ) + ( c − i ) + i ) , which is clearly positive definite. The dimension vectors of indecomposable repre-sentations of the quiver are the positive roots, i.e. nonnegative integer solutions of q ( v, c, i ) = 1. These must be vectors such that exactly two of the four squared sum-mands are equal to 1. Of the six such solutions, those for which the arrows are repre-sented by injective maps are (1 , , , , , , . Over a ground field k , thecorresponding representations are (1; 0 ,
0) := k ⊇ ⊇
0, (1; 1 ,
0) := k ⊇ k ⊇
0, and(1; 1 ,
1) := k ⊇ k ⊇ k . (1; 1 ,
0) is self-dual, while (1; 0 ,
0) and (1; 1 ,
1) are (isomorphicto) the duals of one another. 28n (1; 1 , As we have seen in Example 3.1, the symplectifications of these representations,contained in k ∗ ⊕ k , are (1; 1 , − = k ∗ ⊕ k ⊃ ⊕
0, (1; 0 , − = k ∗ ⊕ ⊇ k ∗ ⊕
0, and(1; 1 , − = 0 ⊕ k ⊇ ⊕ k .The isotropic subspaces in k ∗ ⊕ k are the zero subspace in the first case, and (la-grangian) lines in the latter two cases. The latter two symplectic representations areisomorphic, but we will use both of them in the classification of multiple isotropic sub-spaces. The relevant poset here is + , to which we associate the quiverwith underlying Dynkin diagram A .For a representation of this quiver given by maps I → C → V ← C ← I , we willwrite the dimension vector in the form ( v ; c , i ; c , i ).The Tits form may be written as q ( v ; c , i ; c , i ) = 12 [( v − c ) + ( c − i ) + i + ( v − c ) + ( c − i ) + i ]Again, this is positive definite, and the positive roots are those vectors making exactlytwo of the squared terms equal to 1. Those giving representations by injective maps(which can be found by consulting a table of the positive roots of A ) are as follows. Theonly self-dual root is (1; 1 ,
0; 1 , ,
0; 0 ,
0) and (1; 1 ,
1; 1 , ,
0; 0 ,
0) and (1; 1 ,
0; 1 , ,
0; 1 ,
0) and (1; 1 ,
1; 1 , ,
1; 0 ,
0) and (1; 0 ,
0; 1 , k ∗ ⊕ k ;they are (0 ⊕ k , ⊕ k ) (two equal lines), (0 ⊕ , ⊕ k ) (zero and a line), (0 ⊕ k , ⊕
0) (aline and zero), ( k ∗ ⊕ , ⊕ k ) (two distinct lines), and (0 ⊕ , ⊕
0) (two zero subspaces).They correspond exactly (in a different order) to the five symplectic indecomposablesnumbered 6 through 10 in Theorem 2 of [21]. + + We come now to the central object of this paper. As noted earlier, the quiver which weassociate with the poset + + governing isotropic triples is The compatible nondegenerate bilinear forms on (1; 1 ,
0) are symmetric; the group k ∗ / ( k ∗ ) , where k ∗ is the multiplicative group of k , acts simply and transitively on the isomorphism classes of suchforms. This quotient group is Z in the case of a finite field or the real numbers, and the trivial groupfor an algebraically closed field. It can be much larger, for instance in the case of the rational numbers.In any case, the isotropic subspace is the zero subspace. f E . An explicit description of theindecomposable f E representations has been given by Donovan and Freislich in [8],organized into families described as follows: The dimension vectors of indecomposablesare arranged in lines parallel to N = N (3; 2 ,
1; 2 ,
1; 2 , N . Thesegive sequences in increasing dimensions. The dimension vectors in N are referred toas continuous , the others as discrete ; we also use this terminology for the associatedindecomposable representations. Remark 4.2.
The classification of discrete-type indecomposables is in fact independentof the ground field k , while the classification of continuous-type indecomposables does(partially) depend on k . This follows from DR [7], or by inspection of the proofs in DF[8]. Remark 4.3.
In view of Corollary 3.4 and the fact that indecomposable discrete-typesextuples are uniquely determined up to isomorphism by their dimension vectors, toshow that a sextuple is of a given isomorphism type of discrete type it is sufficient toshow that it has the required dimension vector and that its endomorphism ring is local.An important step in classifying symplectic representations of the poset P = + + with our chosen involution is to identify which linear representations are self-dual, sincethese may admit compatible symplectic forms. For the discrete-type dimension vectors,self-duality of the dimension-vector implies self-duality of the (uniquely) correspondingindecomposable linear representation, c.f. Lemma 3.15. In view of Remark 4.2, theself-dual discrete-type sextuples may be read off from the classification in [8]: Proposition 4.4.
The self-dual discrete dimension vectors of indecomposable repre-sentations of + + are of the form (3 k + 1; 2 k + 1 , k, k + 1 , k, k + 1 , k ) and (3 k + 2; 2 k + 1 , k + 1 , k + 1 , k + 1 , k + 1 , k + 1) . For each of these there is, up to isomor-phism, a unique indecomposable, named A (3 k + 1 , resp. A (3 k + 2 , . In particular,these are the only self-dual discrete sextuples. Explicit descriptions of the isotropic triples associated to the self-dual sextuples A (3 k + 1 ,
0) and A (3 k + 2 ,
0) are given in Sections 5.2 and 5.3. In Section 5.4 com-patible symplectic forms are constructed explicitly, and in Section 5.9 we discuss theiruniqueness. The (more difficult) question of duality for continuous-type indecomposablesextuples is discussed in Sections 6, 7, and 8.In the present section we give some further explanations of the DF-classification [8].The discrete indecomposable sextuples are labeled in the form L s (3 k + i, d ), where k can be any non-negative integer and i ∈ { , , , } (but does not always run over thatwhole set). 3 k + i gives the dimension of the ambient space V , and d is the defect P c j + P i j − v . L is a letter in { A, B, C, D } which encodes the degree of symmetry30f the dimension vector, with A = fully symmetric, i.e. all “arms” equal, B or C =exactly two arms equal, D = no arms equal. The subscript s is either empty (when L = A ), an integer in { , , } when L = B or C , or a pair of unequal integers in { , , } (when L = D ). This subscript encodes “where the asymmetry is”. In cases B or C , it tells which arm has a different dimension vector. In case D , it tells how twoasymmetric dimension vectors can be related via permutation of the arms. For example, D (3 k + 1 ,
0) is related to D (3 k + 1 ,
0) via the permutation 1 →
3, 2 →
1, 3 → L = B and L = C . The reason forusing two different letters seems to simply be that, in the defect − k , i , d , and s are fixed, then there are still two distinct dimension vectorsof indecomposables.The lowest dimensional members of each family of discrete-type indecomposablesextuples are listed in the following table (with i = 1 , , − A (1 , − A (2 , − − C i (1 , − C i (2 , − C i (3 , − − B i (1 , − B i (2 , − B i (3 , − − C i (1 , − C i (2 , − C i (3 , − D i +2 ,i +1 (1 , D i +2 ,i +1 (2 , A (1 , A (2 , D i,i +1 (1 , D i,i +1 (2 , C i (1 , C i (2 , C i (3 , B i (1 , B i (2 , B i (3 , C i (1 , C i (2 , C i (3 , A (1 , A (2 , A (3 k + 1 ,
0) and A (3 k + 2 , A (1 , −
3) and A (1 ,
3) are mutually dual, C i (1 , −
2) and C i (1 ,
2) are mutually dual, etc..In contrast to the discrete-type indecomposable sextuples, the classification of continuous-type indecomposable sextuples is dependent on the ground field (again, this follows fromthe classification in DF [8] and DR [7]). Although we will ultimately work in the set-ting where the ground field is only assumed to be perfect, for illustrative purposes, weassume for the moment that the ground field is the complex numbers.The indecomposable continuous-type sextuples can be arranged into families whoselowest dimensional members are listed in the following table. Here, i = 1 , , , and theparameter λ is understood as ranging in the disjoint union of the sets C \{ , } and { i , , , ∞ i } , where the latter 8 (formal) elements are labels for certain “exceptional”indecomposables. 31efect dim 30 ∆(1 , i )0 ∆(1 , λ ) , < | λ | < , or | λ | = 1 with Im λ >
00 ∆(1 , )0 ∆(1 , − , )0 ∆(1 , λ − ) , < | λ | < , or | λ | = 1 with Im λ >
00 ∆(1 , ∞ i +1 )As with the previous table, this one is also arranged so that dual sextuples are placedsymmetrically to each other with respect to reflection around the central horizontal row(and leaving this row fixed); in particular, the only self-dual element of the table is∆(1 , − .Finally, we review these results in the context of Section 4.2. The quiver we workwith, ˜ E , is tame but not of finite type, and its indecomposable representations, andhence those of the underlying poset, are infinite in number and of arbitrarily highdimension, with some of them of discrete type and others of continuous type. Fol-lowing the pattern in the preceding subsections, we will denote a dimension vector by( v ; c , i ; c , i ; c , i ). The Tits form q ( v ; c , i ; c , i ; c , i ) = 12 [ − v + ( v − c ) + ( c − i ) + i + ( v − c ) + ( c − i ) + i + ( v − c ) + ( c − i ) + i ]is now positive semi definite; its null space N is 1-dimensional, generated over the inte-gers by ν := (3; 2 ,
1; 2 ,
1; 2 , We will enumerate here the isotropic triples in dimension 2, which are all symplecticallyindecomposable. Since many of them are linearly decomposable, we will be using thedescription in Example 3.1 of the symplectifications of the three nested pairs in k .The following are the possibilities for a triple, in the symplectic plane k ∗ ⊕ k , ofpairs each consisting of a coisotropic subspace containing its isotropic orthogonal. Theseare necessarily symplectically indecomposable, but only the final example is linearlyindecomposable. The rest are symplectifications of 1-dimensional linear representations. We note that, for indecomposable sextuples of the type ∆(1 , λ ) with λ ∈ C \{ , } , the separationinto two families (according to the absolute value of λ and the sign of its imaginary part) is somethingwe have introduced “artificially” for this table in order to emphasize a separation into dual pairs. Infact, the values of λ are not intrinsic; the values used here are simply one of many possible ways toparametrize the “moduli space” of continuous-type indecomposable sextuples.
32. Three copies of k ∗ ⊕ k ⊇ ⊕
0. (The isotropics are all zero.) This is the symplec-tification of the sextuple in k consisting of three copies of k ⊇
0, whose dimensionvector is (1; 1 ,
0; 1 ,
0; 1 , A (1 , k ∗ ⊕ k ⊇ ⊕ ⊕ k ⊇ ⊕ k . (Two isotropicsare zero, and one is a line.) This is the symplectification of two copies of k ⊇ k ⊇ k (or its dual 0 ⊇ B r (1 ,
1) (or its dual B r (1 , − r = 1, 2, or 3. Thus there are three possibilities here, dependingupon the value of r (i.e., on which of the three isotropics is a line). The B r (1 , ± B r (3 k + 1 , ± B (1 , ,
0; 1 ,
0; 1 , B (3 k + 1 , −
1) has dimensionvector (1; 1 ,
0; 1 ,
0; 0 , k ∗ ⊕ k ⊇ ⊕ k ∗ ⊕ ⊇ k ∗ ⊕
0, and 0 ⊕ k ⊇ ⊕ k . (One isotropicis zero, and the other two are different lines.) This is the symplectification of onecopy each of k ⊇
0, 0 ⊇
0, and k ⊇ k . In terms of the DF classification, the linearrepresentation being symplectified is D (1 , D (1 , D (1 ,
0) (or the dual D (1 , D (1 , D (1 ,
0) respectively). Again, there are three possibilities,depending upon which of the three isotropics is zero. The D ij (1 ,
0) are the firstmembers of the families D ij (3 k + 1 , k ∗ ⊕ k ⊇ ⊕ ⊕ k ⊇ ⊕ k . (One isotropic is zero,and the other two are identical lines.) This is the symplectification of one copyof k ⊇ k ⊇ k . In the DF classification, this is C r (1 ,
2) (or itsdual C r (1 , − r = 1, 2, or 3, leading again to three possibilities, dependingupon which isotropic is zero. The C r (1 , ±
2) are the first members of the families C r (3 k + 1 , ± ⊕ k ⊇ ⊕ k . (All three isotropics are the same line.) Thisis the symplectification of three copies of k ⊇ k (or its dual 0 ⊇ A (1 ,
3) (or A (1 , − ,
1; 1 ,
1; 1 ,
1) (or (1; 0 ,
0; 0 ,
0; 0 , A (1 , ±
3) are the first members of thefamilies A (3 k + 1 , ± ⊕ k ⊇ ⊕ k and one copy of k ∗ ⊕ ⊇ k ∗ ⊕
0. (Two of theisotropics are the same line, and the third one is a different line.) This is thesymplectification of two copies of k ⊇ k and one copy of 0 ⊇ C r (1 ,
1) (or its dual C r (1 , − r = 1, 2, or 3.Again we have three possibilities, depending upon which line is distinct from theother two. The C r (1 , ±
1) are the first members of the families C r (3 k + 1 , ± A (2 , ,
1; 1 ,
1; 1 , k . We may find a symplectic basis ( e , f ) whose elementsspan I and I , respectively. I is then spanned by e + af for some nonzero a ∈ k . If we change the basis to be , b − f , then I is spanned by be + baf = be + b a ( b − f ). This implies that the set of isomorphism classes of triples oflines may be parametrized (taking the case a = 1 as “basepoint”) by the squareclass group k × / k × introduced in Remark 3.27. Thus, when k is algebraicallyclosed, there is just one isomorphism class of this type, while in the case k = R ,there are two. In this case, the isomorphism class is invariant only under cyclicpermutations of the three lines, and the
Maslov index of the triple distinguishesthe two possibilities . A (2 ,
0) is the first member of the family A (3 k + 2 ,
0) whosemembers for k even admit compatible symplectic structures. Those for odd k require symplectification.We conclude that the number of isomorphism classes of isotropic triples in dimension2 is 1 + 3 + 3 + 3 + 1 + 3 + k × / k × ), or 14 + k × / k × ), where the last term is theorder of the square class group. All of the isotropic triples in dimension 2 were listed in the previous subsection. Again,they are the symplectifications, for the case k = 0, of A (3 k + 1 , B r (3 k + 1 , ± D ij (3 k + 1 , C r (3 k + 1 , ± A (3 k + 1 , ± C r (3 k + 1 , ± A (2 , k , the members of the family A (3 k + 2 ,
0) are always self-dual (see Section5 below), and they admit symplectic forms if only if k is even. Similarly, the members ofthe family A (3 k +1 ,
0) are all self-dual and admit symplectic forms if and only if k is odd.Thus, for k = 1 or k = 2 one finds non-split indecomposable isotropic triples arisingfrom compatible forms for A (3 + 1 ,
0) and A (6 + 2 , A (3 + 2 ,
0) and A (6 + 1 , A (3 + 1 , k = 0, the symplectifications of A (3 k + 2 , ± C r (3 k + 2 , ± B r (3 k + 2 , ± C r (3 k + 2 , ± D ij (3 k + 2 , k = 0, the symplectifications of the discrete sextuples C r (3 k + 3 , ± B r (3 k + 1 , ± C r (3 k + 3 , ± k = 1 the symplectificationsof the non self-dual continuous-type indecomposable sextuples. In addition, for k = 2there are the non-split isotropic triples arising from compatible symplectic forms forthe self-dual continuous-type sextuples. As we will see later in this paper, non-split For information about the square class group of other fields, we refer to [3], [16], and [26]. Forinstance, the square class group of a finite field has order 1 or 2 according to whether the characteristicis even (i.e. 2) or odd. For the p-adic numbers, the order of the square class group is 8 for p = 2, and4 otherwise. Here we are referring to the Maslov index for Lagrangian triples, also known as the Kashiwaraindex, see [9], [19].
In this section we outline briefly how another problem in linear symplectic geometry canbe treated using symplectic poset representations of + + + , and how this problemcan in turn be encoded in isotropic triples. As a result, we will get our first example ofisotropic triple of continuous type.Set P = + + + and consider the involution ⊥ on P which exchanges the firsttwo elements and leaves the last two elements fixed. This involution is trivially order-reversing, since all elements of P are incomparable. Symplectic poset representationsof ( P, ⊥ ) are then subspace systems ( V ; U , U , U , U ) where V is symplectic, U and U are mutually orthogonal, and U and U are lagrangian.The problem we will discuss is that of classifying linear hamiltonian vector fieldsup to conjugation by linear symplectomorphisms; in other words, the classification ofthe orbits of the Lie algebra sp( V, ω ) under the adjoint action of the symplectic groupSp(
V, ω ). This is a problem whose solution is well-known and has a long history (goingback to Williamson [35] in the 1930s). It has since been treated by various authors,in particular with a view toward finding special normal forms adapted to applications;see, for example, [15], [18].Let (
V, ω ) be a symplectic vector space. A linear hamiltonian vector field X on V is an element of the Lie algebra sp( V, ω ); i.e. it is a linear map X : V → V suchthat ˜ ω ◦ X = − X ∗ ◦ ˜ ω . One wishes to understand equivalence classes, where onelinear hamiltonian vector field ( V , ω , X ) is equivalent to another, ( V , ω , X ), if thereexists a linear symplectomorphism φ : V → V such that X ◦ φ = φ ◦ X . There isa natural notion of direct sum, and any linear hamiltonian vector field is the directsome of indecomposable pieces. We wish, here, to point out how one may view linearhamiltonian vector fields as symplectic poset representations of ( P, ⊥ ). For this weproceed in two steps: first, in the following lemma, we reformulate hamiltonian vectorfields in terms of certain kinds of subspaces. Lemma 4.5.
There is a bijective correspondence between linear hamiltonian vector fields ( V, ω, X ) and linear maps f : V → V ∗ such that graph ( f ) ⊆ V × V ∗ is a symplecticsubspace with respect to the canonical symplectic form on V × V ∗ . Proof
We give only a sketch. Given (
V, ω, X ), it is readily checked that the graph of f X := (˜ ω ◦ X ) + ˜ ω is a symplectic subspace.Conversely, if f : V → V ∗ is a linear map whose graph is a symplectic subspace,then the asymmetric part f a will be invertible, and hence defines a symplectic structure35 f on V . Setting X f := f − a f s , one finds that X f is hamiltonian with respect to ω f .It is straightforward to check that the two operations are mutually inverse to oneanother. Corollary 4.6.
A linear hamiltonian vector field ( V, ω, X ) can be encoded in the sym-plectic representation of ( P, ⊥ )( V × V ∗ ; graph ( f X ) , graph ( f X ) ⊥ , V × , × V ∗ ) . Although we do not show it here, the passage from a linear hamiltonian vector fieldto the associated symplectic representation of ( P, ⊥ ) is functorial and compatible withthe respective notions of direct sum.Next we show how the objects above can be encoded in isotropic triples. Observethat the symplectic poset representation of ( P, ⊥ ) which we associated to a linear hamil-tonian vector field is such that the first two subspaces, which are mutually orthogonal,are symplectic subspaces; in other words, they are independent to each other. Thelast two subspaces, which are lagrangian, are also independent, and all four subspaceshave the same dimension. In the following we will consider only those symplectic posetrepresentations of ( P, ⊥ ) which are of this kind.Given such a symplectic representation ϕ = ( V ; S, S ⊥ , L , L ), let ω V denote thesymplectic form on V , ω S the restriction of ω V to the symplectic subspace S , and let S denote a copy of S equipped with the symplectic form − ω S . From ϕ we constructthe following isotropic triple in the ambient symplectic space V × S , with form ω := ω V × − ω S : I = L × C = L × SI = L × C = L × SI = { ( x, x ) | x ∈ S } C = I + ( S ⊥ ×
0) (6)The passage from ϕ to (6) is also functorial and compatible with direct sums. Thus,combining the above, we obtain a way of turning any linear hamiltonian vector fieldinto an isotropic triple. We will not analyze in full detail here exactly which typesof isotropic triples can be built from linear hamiltonian vector fields. The following,though, describes a large class of isotropic triples which can. Proposition 4.7.
Let ϕ = ( V ; C i , I i ) be an isotropic triple such that1. V = I ⊕ I ⊕ I , with dim I i = 1 / V for i = 1 , , ,2. I i + I j is a symplectic subspace for all i = j .Then we can construct from ϕ a symplectic form f a and a linear hamiltonian vector field X on I such that ϕ is isomorphic to the isotropic triple (6) obtained from ( I , f a , X ) . Remark 4.8.
Since the geometric description of the isotropic triples in this propositionis invariant under permutation of the indices, the choice of I for carrying the hamilto-nian vector field is arbitrary. We make this particular choice in order to have coherencewith certain types of normal forms which we will use later.36 roof We can write V as the sum S ⊕ S ′ of I ⊕ I and its symplectic orthogonal. I and I being a lagrangian decomposition of S, we can identify I with I ∗ and, hence, S with the “cotangent bundle” T ∗ I .Now I cannot intersect S ′ , or its sums with I and I would not be symplecticsubspaces. So I is the graph of a map g : S ← S ′ which is antipresymplectic, since I is isotropic. This means that this map g pulls back the symplectic form on S to thenegative of that on S ′ . This implies that g is injective and that its image g ( S ′ ) is asymplectic subspace of S which is independent of I and I . Thus, g ( S ′ ) is the graphof a map f : I → I ∗ , which can be considered as a bilinear form on I . Since g ( S ′ )is symplectic rather than isotropic, the form is not symmetric; in fact, it has a nonde-generate antisymmetric part f a . Writing f = f a + f s as the sum of its antisymmetricand symmetric parts, since f a is invertible, we can form the product f − a f s , which is alinear hamiltonian vector field on the symplectic space ( I , f a ).To see that the isotropic triple ϕ is isomorphic to the one of the form (6) associatedto ( I , f a , X ), observe that g defines a symplectomorphism S ← S ′ . It then easy tocheck that the direct sum of g with the ‘identity map’ on S = I ⊕ I defines a sym-plectomorphism from ϕ to (6).Let us look at an example, to see that isotropic triples of the kind in Proposition 4.7do exist. In fact the indecomposable ones come in families dependent on a parametertaking a continuum of values. Indeed, the isomorphism class of the isotropic triple (6)built from a linear hamiltonian vector field X depends on X up to conjugation of X bysymplectomorphisms, so the spectrum of X is an invariant of the isotropic triple. Example 4.1.
Let V = R with symplectic basis ( f , f , f , e , e , e ). Set I =
In ambient dimension 6, it is easy to see directly that isotropic triplesof the kind in Proposition 4.7 are symplectically indecomposable. If there were a de-composition, it would be an orthogonal splitting of the form k ⊕ k . Looking at thepossible ways in which each of the isotropics decomposes, it is not hard to check thatthe induced 2 form on one of the sums must have rank 2 rather than 4, a contradiction. Having given an overview of some background material in the representation theory ofposets and quivers, and its connection to isotropic triples, we begin now with the detailsof our classification. In this section, we study those sextuples which are self-dual andof discrete type, and how they may give rise to non-split isotropic triples. Discrete-type sextuples are somewhat simpler to study than the continuous-type sextuples; thelatter are studied in the subsequent, remaining sections of this paper (except for thelast section). 37ecall that the indecomposable discrete-type sextuples are uniquely characterized bytheir dimension vector. In particular, self-dual indecomposable discrete-type sextuplesare precisely those whose dimension vector is self-dual, which means here that c j + i j = v for j = 1 , , . Thus, such self-dual sextuples may be read off from the classificationin [8]. As stated already in Proposition 4.4, the discrete indecomposable sextuples withself-dual dimension vector are denoted A (3 k + i, k ∈ Z + and i equal to 1 or 2.The dimension vectors are(3 k + 1; 2 k + 1 , k ; 2 k + 1 , k ; 2 k + 1 , k ) for A (3 k + 1 , , and(3 k + 2; 2 k + 1 , k + 1; 2 k + 1 , k + 1; 2 k + 1 , k + 1) for A (3 k + 2 , . By Lemmas 3.15 and 3.17, each of these representations admits a compatible form.The degree of uniqueness of such forms is specified in Theorem 5.9 below. Of course,for the symplectic case, non-split istropic triples can only arise in cases of k odd for A (3 k + 1 ,
0) and k even for A (3 k + 2 , V even-dimensional. In fact, whenever V is even-dimensional, the compatible formsgranted by Lemma 3.17 are symplectic; this follows from Theorem 5.3 and Theorem5.9. The self-dual discrete-type sextuples having odd ambient dimension, on the otherhand, lead to isotropic triples via their symplectification. We give here brief geometric descriptions of the lowest-dimensional non-split discrete-type isotropic triples. These follow, for example, from the normal forms given in thesubsections below.In the previous section, we already saw the first example of a non-split isotropictriple: the underlying sextuple is of type A (2 ,
0) (it belongs to the A (3 k + 2 , A (4 , A (3 k + 1 , ,
1; 3 ,
1; 3 , I i in a 4-dimensional space V , and each of theirorthogonal subspaces C i is independent from the other two isotropics. Furthermore,( C i ∩ C j ) ∩ ( I i + I j ) = 0 for any i = j , so any two of the isotropics span a symplecticsubspace. Since the three isotropics are independent, their sum C has codimension 1and is hence coisotropic. Its orthogonal C ⊥ = C ∩ C ∩ C ⊆ I + I + I is a linewhich must be pairwise independent with each of the I i : if not, then C ⊥ = I i wouldbe the case for some i , and hence C = C i and so C i would contain all the istropics, acontradiction. Thus the I i and C ⊥ are four lines in general position in V . Symplecticreduction via the coisotropic C gives a non-split isotropic triple of the type A (2 , See Lemma 3.15. A (8 , A (3 k + 2 , ,
3; 5 ,
3; 5 , I i in an 8 dimensional symplectic space V . Though the isotropics I i are pairwise independent, they are not fully independent: for each distinct tripleof indices, Q i := I i ∩ ( I j + I k ) is a line. The three lines Q , Q , Q are themselvespairwise-independent, and span the 2-dimensional space I = ( I + I ) ∩ ( I + I ) ∩ ( I + I ) . This space I is contained in its orthogonal, C = ( C ∩ C ) + ( C ∩ C ) + ( C ∩ C )and is hence isotropic. Symplectic reduction by the coisotropic C gives a non-splitisotropic triple of the type A (4 , A (10 , A (3 k + 1 , ,
3; 7 ,
3; 7 , C which is therefore coisotropic.Its orthogonal, C ⊥ = ( I + I + I ) ⊥ = C ∩ C ∩ C is a line which is pairwise independent with each of the I i . Symplectic reduction via C gives a non-split isotropic triple of the type A (8 , A (3 k + 1 , Given k ≥ β = ( e , . . . e k +1 , f , . . . , f k , g , . . . , g k ) of a vector space V β , asextuple ( V β ; C βi , I βi ) of type A (3 k + 1 ,
0) is given by I β = h e − f , ..., e k − f k i ,I β = h g , ..., g k i ,I β = h f − g , ..., f k − g k i ,C β = h e , ..., e k +1 , f , ..., f k i C β = h e , ..., e k +1 , g , ..., g k i C β = h e , e − f , ..., e k +1 − f k , f − g , ..., f k − g k i . (7)Note that C β = I β + h e , ..., e k +1 i C β = I β + h e , ..., e k +1 i C β = I β + h e , e − f , ..., e k +1 − f k i = I β + h e , e − g , ..., e k +1 − g k i .
39n view of Remark 4.3, to show that this really does define an isotropic triple of type A (3 k + 1 ,
0) it suffices to observe that the above sextuple has the required dimensionvector dim V β = 3 k + 1 , dim I βi = k, dim C βi = 2 k + 1and that its endomorphism algebra is local. The latter follows from Remark 3.9 andthe following. Lemma 5.1.
Let ψ be an indecomposable sextuple of type A (3 k + 1 , . ThenEnd ( ψ ) ≃ End (( U, η )) where η is an indecomposable nilpotent endomorphism and dim U = k + 1 . In particular,End ( ψ ) is local and End ( ψ ) = k id ⊕ Rad.
Proof
Let ψ be given in the normal form (7). Consider the End( ψ )-invariant subspace U := C β ∩ C β = h e , ..., e k +1 i and the indecomposable nilpotent endomorphism η of U defined by η ( e ) = 0 and η ( e i +1 ) = e i , for i = 2 , ..., k . The endomorphism algebra of ( U, η ), i.e. the algebra ofendomorphisms of U which commute with η , is local because η is indecomposable . Wewill see now that this algebra is isomorphic to the endomorphism algebra of the sextuple(7), hence the latter is local as well. To do this, we use the following map. Given anendomorphism a of U which commutes with η , we can extend it to an endomorphism a of (7) by defining linear isomorphisms f : h e , ..., e k i → h f , ..., f k i , f ( e i ) := f i ∀ i = 1 , ..., k,g : h f , ..., f k i → h g , ..., g k i , g ( f i ) := g i ∀ i = 1 , ..., k, and setting ¯ a := f af − on h f , ..., f k i , ¯ a := g ¯ ag − on h g , ..., g k i . (8)(Note that domain of f is invariant under a .)Since a is defined via its action on the basis β , it is easily checked directly that a is an endomorphism of (7). To see this, note in particular that I β = { x − f ( x ) | x ∈h f , ..., f k i} and I β = { x − g ( x ) | x ∈ h g , ..., g k i} .The map a a has as its inverse the operation of taking an endomorphism b of (7)and restricting it to U = C β ∩ C β (which will necessarily be an invariant subspace of b , since by assumption C β and C β are b -invariant). To see this, notice that such a b necessarily decomposes as the direct sum of its restrictions to the subspaces U = h e , ..., e k +1 i , F := h f , ..., f k i , G := h g , ..., g k i See Remark 3.9. There it is also noted that this endomorphisms algebra E is such that E = k id ⊕ Rad E . V and must be invariant under b : U = C β ∩ C β , F = I β , G = C β ∩ ( I β + I β ) . The invariance of I β and I β under b then enforces that b | U is related to b | F and b | G via(8), and together with the invariance of C β ∩ C β it is ensured that b | U commutes with η . Thus if we restrict b to U and then extend to b , we recover b . Conversely, if we startwith an endomorphism a of U , the restriction of a to U is of course, by definition, again a . Finally, it is clear from (8) that the operation a a is a morphism of algebras. A (3 k + 2 , We will in fact work with A (3( k −
1) + 2 ,
0) = A (3 k − , A (3 k + 1 ,
0) for k >
0, so that the construction of compatible forms for sextuples ofboth types A (3 k + 1 ,
0) and A (3 k + 2 ,
0) can be treated uniformly.Given k ≥ γ = ( e , . . . e k , f , . . . , f k , g , . . . , g k ) of a vector space V γ , asextuple ( V γ ; C γi , I γi ) of type A (3 k − ,
0) is given by I γ = h f , f − e , ..., f k − e k i I γ = h g , ..., g k i ,I γ = h f − g , ..., f k − g k i ,C γ = h e , ..., e k , f , ..., f k i C γ = h e , ..., e k , g , ..., g k i C γ = h f − e , ..., f k − − e k , f − g , ..., f k − g k i . (9)Note that C γ = I γ + h e , ..., e k i = I γ + h f , ..., f k i C γ = I γ + h e , ..., e k i C γ = I γ + h f − e , ..., f k − − e k i . Again, in view of Remark 4.3, it suffices to observe that such a sextuple has therequired dimension vectordim V γ = 3 k − , dim I γi = k, dim C γi = 2 k − A (3 k + 1 , Lemma 5.2.
Let ψ be an indecomposable sextuple of type A (3 k − , . ThenEnd ( ψ ) ≃ End (( U, η )) where η is an indecomposable nilpotent endomorphism and dim U = k . In particular,End ( ψ ) is local and End ( ψ ) = k id ⊕ Rad. roof Let ψ be given in the normal form (9). Consider the End( ψ )-invariant subspace U := h f , ..., f k i = C γ ∩ ( I γ + I γ )and the indecomposable nilpotent endomorphism η of U defined by η ( f ) = 0 and η ( f i +1 ) = f i , for i = 2 , ..., k . Similarly as in the previous section, the algebra ofendomorphisms of U which commute with η is isomorphic to the endomorphism algebraof (9).To show this, we begin with an endomorphism a of U which commutes with η andextend it to an endomorphism a of V γ by defining maps f , g , and h by f : h e , ..., e k i → h f , ..., f k i , f ( e i ) = f i for i = 2 , .., k,g : h f , ..., f k i → h g , ..., g k i , g ( f i ) = g i for i = 1 , .., k,h : h f , ..., f k i → h e , ..., e k i , h ( f ) = 0 , h ( f i ) = e i for i = 2 , .., k. and setting a := haf on h e , ..., e k i .a := gag − on h g , ..., g k i . (10)Note that I γ = { x − h ( x ) | x ∈ h f , ..., f k i} and I γ = { x − g ( x ) | x ∈ h f , ..., f k i} ,and that E = h e , ..., e k i , U = h f , ..., f k i , G = h g , ..., g k i are subspaces invariant under the endomorphism algebra of (9) since E = C γ ∩ C γ , U = C γ ∩ ( I γ + I γ ) , G = I γ . For an endomorphism b of (9), the operation b b | F γ is inverse to the operation a a . Indeed, such a b decomposes as the direct sum b = b | E ⊕ b | U ⊕ b | G , and therelations (10) are enforced by the invariance of I γ and I γ under b .Finally, that the map a a is a morphism of algebras is evident from (10).The bases β and γ are referred to as standard bases for the respective types ofsextuple. For each fixed k > I β ⊆ C β ⊆ V β , I β = h e i , C β = h e , ..., e k , f , ..., f k , g , ..., g k i , and letting V γ = C β /I β , where the basis γ is induced (modulo re-indexing) by theelements of the basis β which span C (i.e. all elements except e k +1 ). Then we have theidentifications I γi = ( I βi ∩ C β ) /I β and C γi = ( C βi ∩ C β ) /I β for i = 1 , , . In fact, we can also view a sextuple of type A (3( k −
2) + 1 ,
0) = A (3 k − ,
0) as asubquotient of a sextuple of type A (3 k − , I γ = h f , g i , C γ = h e , ..., e k , f , ..., f k − , f , ..., g k − i , C γ /I γ gives a sextuple of type A (3( k −
2) + 1 , k = 0 k = 1 k = 2 k = 3 . . .A (3 k + 1 , A (1 , A (4 , A (7 , A (10 , . . .A (3 k + 2 , A (2 , A (5 , A (8 , A (11 , . . . Theorem 5.3.
Any sextuple of type A (3 k ± , admits ε -symmetric forms where ε = ( − k . With respect to standard bases there is a recursion providing compatible ε -symmetric forms with coefficients in the prime subfield. The existence will be shown by constructing matrices for these compatible formswith respect to standard bases. These matrices will be of the shape (11), i.e. having ablock structure and built using a (smaller) matrix which we call “ A ”.Fix k ∈ N , let ε = ( − k , and let A = A k ∈ k ( k +1) × ( k +1) . We consider the followingrelations on the entries of A :(1) a ij = 0 for i + j < k + 2, a ij = 0 for i + j = k + 2(2) A = εA t (3) a i,j +1 = a ij − a i +1 ,j for i, j = 1 , . . . , k (4) a kk = a k,k +1 + a k +1 ,k .Further we define ˚ A = ˚ A k as the minor of A given by restricting to row and columnindices in { , . . . , k } and we set c i = a i,k +1 , c ′ i = a k +1 ,i . If (1) holds, then the matrix A has the following form A = · · · · · · a ,k +1 · · · · · · a ,k a ,k +1 · · · · · · a ,k − a ,k a ,k +1 ... ... ... ... ... ...... ... 0 ... ... ...0 0 a k − , · · · · · · a k − ,k − a k − ,k a k − ,k +1 a k a k · · · · · · a k,k − a kk a k,k +1 a k +1 , a k +1 , a k +1 , · · · · · · a k +1 ,,k − a k +1 ,k a k +1 ,k +1 . H = H k := · · · c · · · · · · c A ... ... ˚ A c k c ′ c ′ . . . c ′ k c k +1 c ′ c ′ · · · c ′ k · · · · · · c · · · · · · c c c ... ˚ A ... ... ˚ A ˚ A ...0 c k c k A · · · c ′ c ′ · · · c ′ k · · · (11)and we interpret this matrix to be the coordinate matrix of a bilinear form B on V β (c.f. Section 5.2) with respect to the standard basis β = { e , ..., e k +1 , f , ..., f k , g , ..., g k } .The double lines in the matrix are visual aids for seeing the block structure H = H H H H H H H H H related to the subspaces h e , ..., e k +1 i , h f , ..., f k i , and h g , ..., g k i . Claim 5.4.
If (1) and (2) hold, then B is non-degenerate and ε -symmetric Proof
That B is ε -symmetric follows directly from (2). To see that B is non-degenerate, note that the blocks of the matrix H are such that H , H and H are zero and where H , H , and H are square matrices having non-zero entries onthe anti-diagonal, and zeros above the anti-diagonal; in particular, the latter blocks areinvertible. Because H is zero and because a ‘copy’ of H is contained in H
1, wecan use row operations to turn transform H to zero in such a way that only H isadditionally changed under these operations. In a similarly manner we can also turn H to zero using column operations. At this point our block matrix has the followingform (tilde indicates that there are changes)˜ H = H H H H H is · · · A − · · · A − · · · A = − · · · A . Now, clearly we can add columns from the third block column to the second blockcolumn to transform ˜ H into the zero matrix, and this leaves all other blocks unchanged,since H and H are zero. (Equivalently, we could have used row transformations usingrows from the bottom block row.) This puts our block matrix in the form H H H which shows non-degeneracy, since the non-zero blocks are non-degenerate. Claim 5.5.
If (2) holds, then B ( I β , C β ) = B ( I β , C β ) = B ( I β , h e , e − g , ..., e k +1 − g k i ) = 0 . Proof • B ( e i , e j − f j ) = a ij − a ij = 0 for i = 1 , . . . , k + 1 and j = 1 , . . . , k . • B ( f i , e j − f j ) = a ij − a ij = 0 for i, j = 1 , . . . , k . • B ( e i , g j ) = 0 for i = 1 , . . . , k + 1, j = 1 , . . . , k . B ( g i , g j ) = 0 for i, j = 1 , . . . , k . B ( e , f j − g j ) = B ( e , f j ) − B ( e i , g j ) = 0 − i, j = 1 , . . . , k . • B ( f i − g i , e j +1 − g j ) = B ( f i , e j +1 ) − B ( f i , g j ) − a i,j +1 − a i,j +1 = 0 for i, j = 1 , . . . , k . Claim 5.6.
If (1) through (4) hold, then B is ε -symmetric and a compatible form forthe sextuple ( V β ; C βi , I βi ). Proof
It remains to show B ( I β , C β ) = 0. Consider B ( f i − g i , f j − g j ) = B ( f i , f j ) − B ( f i , g j ) − B ( g i , f j ) + B ( g i , g j ) =: x ij . Direct checking gives: • x = 0 + 0 + 0 + 0 = 0. • x j = 0 + 0 + a j + 0 = 0 for j = 2 , . . . , k − • x ij = a ij − a i,j +1 − a i +1 ,j + 0 = 0 for i, j = 2 , . . . , k − x ik = a ik − c i − a i +1 ,k + 0 = 0 for i = 2 , . . . , k − • x k = 0 − c − a k + 0 = 0 by (3). • x kk = a kk − c k − c ′ k = 0 by (4).In view of Claim 5.5 and that I β = h f − g , ..., f k − g k i C β = h e , e − f , ..., e k +1 − f k , f − g , ..., f k − g k i we have compatibility of B . Claim 5.7.
For each k = 0 , , , ... there exist matrices A k satisfying (1)–(4). Proof
We proceed by induction on k . Let A = ( c ) , A = (cid:18) c − c (cid:19) , c = 0Assume that A k − is given satisfying (1)–(4). We define A = A k to have minor ˚ A = A k − with respect to row and column indices 2 , . . . , k . Thus, we have(1’) for 2 ≤ i, j ≤ k : a ij = 0 if i + j < k + 2, a ij = 0 if i + j = k + 2(2’) ˚ A = ε ˚ A t (3’) a i,j +1 = a ij − a i +1 ,j for i, j = 2 , . . . , k − a k − ,k − = a k − ,k + a k,k − .It remains to grant(1”) a j = a i = 0 for i, j = 1 , . . . , k and a ,k +1 = 0, a k +1 , = 0(2”) a k +1 ,i = εa i,k +1 for i = 1 , . . . , k + 1(3”) a i,j +1 = a ij − a i +1 ,j for j = k, i = 1 , . . . resp. i = k, j = 1 , . . . , k − a kk = a k,k +1 + a k +1 ,k .Define a ,k +1 = − a ,k a i,k +1 = a i,k − a i +1 ,k for i = 2 , . . . , k − a k,k +1 = a k,k if k is even a k,k +1 = arbitrary if k is odd a k +1 ,k +1 = arbitrary if k is even a k +1 ,k +1 = 0 if k is odd a k +1 ,i = εa i,k +1 for i = 1 , . . . , k. j = k , while for thosecases with i = k we have a k,j +1 = εa j +1 ,k = ε ( a j,k − a j,k +1 ) = a kj − a k +1 ,j . Finally, if k is odd then a kk = εa kk = 0 = a k,k +1 + εa k,k +1 = a k,k +1 + a k +1 ,k ; and if k is even then a kk = a kk + a kk = a k,k +1 + a k +1 ,k . Proof of Theorem 5.3 . For the A (3 k + 1 , A (3 k − , V γ as a subspace of V β and define on it a form given by a matrix H ′ = H ′ k obtained from H = H k by omitting all rows and columns indexed by e or e k +1 . Then H ′ defines an ε -symmetric form B ′ on V γ . To see this, let S β := h e , e k +1 i and note that H | Sβ is non-degenerate and ε -symmetric, and that we can make theidentification V γ = ( S β ) ⊥ ⊆ V β . Hence H ′ = H | ( Sβ ) ⊥ is also non-degenerate and ε -symmetric. It remains to show that H ′ is compatible with the A (3 k − ,
0) sextuple in V γ obtained from the A (3 k + 1 ,
0) sextuple in V β .Recall that if X ⊆ V β is an element of the sextuple in V β , then the correspondingsubspace of the sextuple in V γ is given by X := π ( X β ∩ C β ), where π denotes theprojection onto the second factor of the (orthogonal) decomposition V β = S β ⊕ V γ .Since C β = ( C β ) ⊥ ⊕ V γ is coisotropic, this is an instance of coisotropic reduction; inparticular B ′ ( X, Y ) = B ( X ∩ C β , Y ∩ C β )for any X, Y ⊆ V β . Hence B ( X, Y ) = { } implies that B ′ ( X, Y ) = { } . Since we knowthe dimensions of all the subspaces involved in our sextuples, this shows that for anyelement X of the sextuple in V β , the orthogonal of X in V γ is the same subspace as X ⊥ . Remark 5.8.
Suppose we are given sextuples of the types A (3 k + 1 ,
0) and A (3 k − , A (3 k + 1 ,
0) is givenby a matrix H k which is built from a matrix A as above, and such that A satisfiesthe relations (1)–(4). The analogous statement is true for compatible forms on A (3 k − ,
0) in terms of matrices of the form H ′ k , where H ′ k is obtained from H k by deleting the 1st and ( k + 1) th rows and columns.(ii) A parametrization of such matrices H k resp. H ′ k is given by the parameters a k +1 ,j , j = 1 , . . . , k + 1 resp. a k,j , j = 2 , . . . , k Moreover, the matrix entries are obtained via linear expressions from the param-eters.
Proof (i) According to the proofs of Claims 5.5 and 5.6, the structure of B and therelations (1)–(4) are forced by the requirement of admissibility.47ii) This follows from the recursive construction of the matrices A k ; in each step onefree parameter can be chosen. Example 5.1.
Let k = 1. A sextuple of type A (3 k + 1 ,
0) has ambient dimension 4.To construct a compatible symplectic form (with respect to a standard basis) we beginwith the matrix A = (cid:18) c − c (cid:19) where c is any non-zero scalar. From this we obtain the matrix (11) of a compatiblesymplectic form; in this example it is H = c − c − c c c − c (note that ˚ A is the empty matrix). To obtain a compatible symplectic form for thesextuple A (3 k − , A (3 k + 1 , st and ( k + 1) th rows and columns of the compatible form H given above. This gives H ′ = (cid:18) c − c (cid:19) . Example 5.2.
Let k = 3. A sextuple of type A (3 k +1 ,
0) has ambient dimension 10. Toconstruct a compatible symplectic form (with respect to a standard basis) we proceedsimilarly as in the previous example. Following the recursion recipe given in Claim 5.7above, we build from A the matrix A = c − c − c c c − c c − c where c is a scalar that we may freely choose. Now from A we obtain the matrix (11)of a compatible symplectic form; in this example it is H = c − c − c − c c c c − c c − c − c c − c c c − c − c c − c − c c c − c c c − c c − c c − c .
48o obtain a compatible symplectic form for the sextuple A (3 k − , st and ( k + 1) th rows and columns of H . This gives H ′ = − c − c c c c − c c − c − c c − c c c − c c − c c − c . Theorem 5.9.
The ( − k -symmetric forms on an A (3 k ± , are unique up to isomet-ric automorphism and multiplication by scalars; there are no ( − k +1 -symmetric formson an A (3 k ± , .If B is a compatible ε -symmetric form on A (3 k ± , and c ∈ k , then there is anautomorphism η of the sextuple which is an isometry, in the sense η ∗ B = cB , if andonly if c is a square in k . Proof
Dealing with A (3 k + 1 , η of A (3 k + 1 ,
0) such that f ∗ B = cB ,and let η U to be its restriction to U = C β ∩ C β , which is invariant under η . We mayassume that B is given by a matrix B k in terms of a standard basis as above. Thus, U = h e , ..., e k +1 i is non-degenerate and the restriction of B to U has matrix A which iszero above the anti-diagonal. By Observation 3.9, η U has upper triangular matrix ( η ij )with diagonal entries all the same. It follows that cB ( e k +1 , e ) = B ( ηe k +1 , ηe ) = B ( k +1 X i =1 η i,k +1 e i , η e )= B ( η k +1 ,k +1 e k +1 , η e ) = B ( η e k +1 , η e ) = η B ( e k +1 , e )whence c = η . (Conversely, given a square c = b ∈ k , one can of course always findan isometry between B and cB : simply take b · id.)For k >
1, a similar reasoning works for A (3 k − , U = C γ ∩ C γ = h e , ..., e k i . By definition, continuous-type indecomposable sextuples are those indecomposable sex-tuples which have dimension vectors of the form (3 k ; 2 k, k ; 2 k, k ; 2 k, k ), for integers k ≥
1. Note that this is a self-dual dimension vector.49s mentioned above, and proven in [8, Section 4.6], when k is algebraically closed the(isomorphism types of) continuous-type indecomposable sextuples in a given dimension(indicated by k ) consist of: • a collection ∆( k ; λ ), labeled by scalars λ ∈ k \{ , } , • so-called exceptional continuous-type sextuples, which are labeled∆ j ( k,
1) for j ∈ { , } , ∆ j ( k,
0) for j ∈ { , , } , ∆ j ( k, ∞ ) , for j ∈ { , , } . In the case of more general fields, the continuous-type indecomposable sextuples∆( k ; λ ) must be replaced by indecomposable sextuples ∆( k ; γ ), where γ ranges overall indecomposable endomorphisms of k k which do not have 0 or 1 as an eigenvalue.The exceptional continuous-type indecomposable sextuples remain the same. Thus theisomorphism classes of continuous-type indecomposable sextuples in a given dimension3 k are parametrized by the disjoint union of the sets { γ | γ indecomposable endomorphism of k k , with 0 , / ∈ spec( γ ) } and { , , , , , ∞ , ∞ , ∞ } , where the latter set consists of formal labels for the exceptional types.We review the classification and structure of continuous-type sextuples in Subsection6.1 below. To further analyze continuous-type sextuples, we recall the notion of a framein Subsection 6.2 and use these in Subsection 6.3 to build continuous-type sextuplesfrom underlying linear endomorphisms. In Subsection 6.4 we give a detailed descriptionof morphisms of such sextuples, and in Subsection 6.5 we identify which indecomposablecontinuous-type sextuples are dual to which. This sets the stage for the analysis of self-dual continuous indecomposable sextuples and their compatible forms in Sections 7 and8. Naturally, when k is not algebraically closed, the classification of indecomposablesextuples and associated isotropic triples becomes more complicated. First of all, theindecomposable endomorphisms γ underlying the sextuples have a richer structure.Furthermore, self-dual continuous-type indecomposable sextuples may admit more com-patible forms when k is not algebraically closed. The general question of uniqueness ofcompatible forms is treated in Subsection 8.5.As always, if we are looking for compatible symplectic structures, we can restrictour attention to the cases where the integer k is even. With the sole exception being the type ∆ ( k ; 1), all indecomposable continuous-typesextuples (up to isomorphism) may be obtained via a functor S from the “continuous-type” indecomposable representations of the extended Dynkin quiver ˜ A (with a certain50rientation, as reflected in the diagram (12)) . Indecomposable representations of this˜ A quiver have the following normal form X Y Y X X Y α β α β α β (12)where X i = Y i = k k for some k ∈ Z > , and i ∈ { , , } , and the linear maps α i and β i are all the identity map on k k , with one exception. The map which is anexception – we call it γ – is a linear endomorphism of k k , and the representation of ˜ A isindecomposable if and only γ is an indecomposable linear endomorphism. The followingare the continuous-type indecomposable representations of ˜ A , up to isomorphism (ineach case, k runs over Z > ): • Ξ( k ; γ ): γ runs over isomorphism classes of automorphisms of k k and we assume γ = β . Any other choice of “position” of γ , e.g. γ = α , leads to a representationwhich is isomorphic to Ξ( k ; γ ′ ) for some γ ′ = β . • Ξ i ( k ; 0) and Ξ i ( k ; ∞ ): γ is the unique indecomposable nilpotent endomorphismon k k in Jordan normal form (for nilpotent γ this normal form always exists), and i ∈ { , , } . The convention is that Ξ i ( k ; 0) denotes the case when γ = β i andΞ i ( k ; ∞ ) denotes the case when γ = α i .Although DF [8] work over algebraically closed fields, their classification of the indecom-posable representations of ˜ A does not depend on this, and admits the straightforwardgeneralization above, where single Jordan blocks are replaced with the condition ofindecomposability. Our notation is a slight modification of their notation.The image under S of a continuous-type ˜ A representation is the sextuple V = X ⊕ X ⊕ X , C i = X i ⊕ X i +1 , I i = Im ( α i × β i ) , (13)where indices are understood modulo 3. It will be convenient for us to cast these normalforms in slightly different notation. We set X = k k , X = X × × , X = 0 × X × , X = 0 × × X, and V = X × X × X . We call an endomorphism exceptional if it has 0 or 1 aseigenvalue. Otherwise, it is non-exceptional . Note that any direct summand of anon-exceptional endomorphism is again non-exceptional. See [8], pages 44 and 46 γ , the sextuples which are isomorphic to S Ξ( k ; − γ ) will becalled of type ∆( k ; γ ). Normal forms for these sextuples are I = { ( x, − γx, | x ∈ X } C = X × X × I = { (0 , x, x ) | x ∈ X } C = 0 × X × XI = { ( x, , x ) | x ∈ X } C = X × × X. (14)For indecomposable γ having eigenvalue 1, the sextuple S Ξ( k ; − γ ) is called of type∆ ( k ; 1); its normal form is the same as above. The only continuous-type indecompos-able sextuple not isomorphic to a sextuple in the image of the functor S is the type∆ ( k ; 1). It is obtained from ∆ ( k ; 1) via certain functor θ + (see [8], p. 38 and 46); anormal form for ∆ ( k ; 1) is: I = { (0 , , x ) | x ∈ X } C = { ( y, γy, x ) | x, y ∈ X } I = { ( x, , | x ∈ X } C = { ( x, y, − y ) | x, y ∈ X } I = { (0 , x, | x ∈ X } C = { ( − y, x, y ) | x, y ∈ X } , (15)where γ is indecomposable and with eigenvalue 1. Finally, for the cases when γ isnilpotent we set ∆ i ( k ; 0) := S Ξ i ( k ; 0) and ∆ i ( k ; ∞ ) := S Ξ i ( k ; ∞ ), for i = 1 , ,
3. Fornormal forms we take the same spaces C , C , C as in (14), and • for ∆ ( k ; 0) and ∆ ( k ; ∞ ), respectively: I = { ( x, γx, | x ∈ X } I = { ( γx, x, | x ∈ X } I = { (0 , x, x ) | x ∈ X } and I = { (0 , x, x ) | x ∈ X } I = { ( x, , x ) | x ∈ X } I = { ( x, , x ) | x ∈ X }• for ∆ ( k ; 0) and ∆ ( k ; ∞ ): I = { ( x, x, | x ∈ X } I = { ( x, x, | x ∈ X } I = { (0 , x, γx ) | x ∈ X } and I = { (0 , γx, x ) | x ∈ X } I = { ( x, , x ) | x ∈ X } I = { ( x, , x ) | x ∈ X }• for ∆ ( k ; 0) and ∆ ( k ; ∞ ): I = { ( x, x, | x ∈ X } I = { ( x, x, | x ∈ X } I = { (0 , x, x ) | x ∈ X } and I = { (0 , x, x ) | x ∈ X } I = { ( γx, , x ) | x ∈ X } I = { ( x, , γx ) | x ∈ X } . The following will be useful for identifying isomorphism types.
Lemma 6.1.
Let ( V ; I i , C i ) be an indecomposable continuous-type sextuple with dim V =3 k . Consider the following eight subspaces: I + I + I and C ∩ C ∩ C , I ∩ C and I ∩ C , I ∩ C and I ∩ C , I ∩ C and I ∩ C . Let ǫ = ( ǫ , ..., ǫ ) be the corresponding -vector of the dimensions of these spaces. The different possible types of indecomposablecontinuous-type sextuple have the following associated -vectors ǫ . The change of sign in front of γ here follows the conventions of DF [8]. . ∆( k ; γ ) , then ǫ = (3 k, , , , , , , ∆ ( k ; 1) , then ǫ = (3 k − , , , , , , , ∆ ( k ; 1) , then ǫ = (3 k, , , , , , , ∆ ( k ; 0) , then ǫ = (3 k, , , , , , , ∆ ( k ; ∞ ) , then ǫ = (3 k, , , , , , , ∆ ( k ; 0) , then ǫ = (3 k, , , , , , , ∆ ( k ; ∞ ) , then ǫ = (3 k, , , , , , , ∆ ( k ; 0) , then ǫ = (3 k, , , , , , , ∆ ( k ; ∞ ) , then ǫ = (3 k, , , , , , , Proof
Consider first the sextuples for which γ is an isomorphism. It is straightforwardto check, e.g. using the normal forms above, that for such sextuples dim I j ∩ C l = 0 forall j = l . Thus ǫ through ǫ are zero for the types ∆( k ; γ ), ∆ ( k ; 1) and ∆ ( k ; 1).Furthermore, if a sextuple is of type ∆( k ; γ ) or ∆ ( k ; 1), then from the normal form(14) we see that C ∩ C ∩ C = 0 and that( x, y, z ) ∈ I ∩ ( I + I ) ⇔ ( x, y, z ) = ( x, − γx,
0) with x − γx = 0 , so I ∩ ( I + I ) = 0 if and only if γ has 1 as eigenvalue. In the case ∆ ( k ; 1) when γ does have 1 as eigenvalue, the corresponding eigenspace has dimension 1 (because γ isindecomposable) and so dim I ∩ ( I + I ) = 1. Thus in this casedim( I + I + I ) = dim I + dim( I + I ) − k − V − . So, we have found that ( ǫ , ǫ ) = (3 k,
0) for ∆( k ; γ ) and ( ǫ , ǫ ) = (3 k − ,
0) for ∆ ( k ; 1).For sextuples of type ∆ ( k ; 1), it follows from the normal form (15) that I ∩ ( I + I ) = 0 and that C ∩ C ∩ C = { ( x, γx, − γx ) | x = γx } . Since for the type ∆ ( k ; 1) the map γ has a 1-dimensional eigenspace for the eigenvalue1, we find that dim C ∩ C ∩ C = 1. So, in this case ( ǫ , ǫ ) = (3 k, ( k ; 0). The same arguments as for the case ∆( k ; γ ) showhere that ( ǫ , ǫ ) = (3 k, I ∩ C = { ( x, γx, ∈ I | γx = 0 } = ker γ, which is 1-dimensional since γ is an indecomposable nilpotent map. From the normalform for ∆ ( k ; 0) is easily check that the other intersections I j ∩ C l , j = l , are zero.Thus ǫ = 1, and ǫ through ǫ are zero.The remaining cases are very similar to the case ∆ ( k ; 0) and may be treated anal-ogously. 53 orollary 6.2. Suppose we are given an indecomposable continuous-type sextuple withambient dimension k . The sextuple is of type1. ∆( k ; γ ) if and only if dim( I + I + I ) = dim V and dim( C ∩ C ∩ C ) = 0 .2. ∆ ( k ; 1) if and only if dim( I + I + I ) = dim V − ∆ ( k ; 1) if and only if dim( C ∩ C ∩ C ) = 1 ∆ i ( k ; 0) if and only if dim( I i ∩ C i − ) = 1 .5. ∆ i ( k ; ∞ ) if and only if dim( I i ∩ C i +1 ) = 1 . Following von Neumann [24], we introduce an abstract kind of coordinate system. Givena vector space V , a frame for V is a collection of five subspaces A , A , A , A , A satisfying the following relations: V = A ⊕ A ⊕ A (16) A + A = A ⊕ A = A ⊕ A (17) A + A = A ⊕ A = A ⊕ A (18)As a shorthand notation, we refer to a frame as ¯ A . The notions of morphism andisomorphism of frames are the obvious ones (i.e. view a frame as a special kind of posetrepresentation.)The following shows that the definition could also be phrased in a way that is moresymmetrical. Lemma 6.3.
Suppose we are given a frame A , A , A , A , A ⊆ V . We define A by A = ( A + A ) ∩ ( A + A ) . (19) Then A + A = A ⊕ A = A ⊕ A . (20) Proof
To see that A ∩ A = 0, note that dim V = 3 n for some n ∈ N , and A ∩ A = A ∩ [( A + A ) ∩ ( A + A )] = A ∩ ( A + A ) . Thusdim( A ∩ A ) = dim A + dim( A + A ) − dim( A + A + A ) = 3 n − n = 0since, via (16), (17), (18), and (19), A ∩ A = 0 and A + A + A = A + A + A = A + A + A = V. Similar dimension arguments can be used to show that A ∩ A = 0 and that dim A = n . 54 orollary 6.4. If A , A , A , A , A ⊆ V is a frame, and A defined as above, then A = ( A + A ) ∩ ( A + A ) , (21) A = ( A + A ) ∩ ( A + A ) . (22) Proof If { A , A , A , A , A } is a frame, then by Lemma 6.3 also { A , A , A , A , A } is a frame. Application of Lemma 6.3 to this latter frame gives (21); an analogous ar-gument gives (22).The relations (16), (17), (18), and (20) imply that dim A = dim A = dim A =1 / V and that each A ij can be interpreted as the negative graph of a linearisomorphism h ij : A i → A j , i.e. A ij = { x − h ij ( x ) | x ∈ A i } , where ij ∈ { , , } . We set h ii := id and h ji := h − ij . Lemma 6.5.
1. Given a frame in V , the associated maps satisfy h jk ◦ h ij = h ik for any indices i, j, k ∈ { , , } .2. Any frame is isomorphic to one built from a vector space U in the following way: V = U × U × UA = U × × A = { ( x, − x, | x ∈ U } A = 0 × U × A = { (0 , x, − x ) | x ∈ U } A = 0 × × U A = { ( x, , − x ) | x ∈ U } Proof
1. Since we are dealing only with invertible maps, equations of the form h jk ◦ h ij = h ik are equivalent to ones obtained by applying, to both sides of an equation, theoperations of inversion, or pre- or post-composition with one of the “ h ” maps.This allows one to reduce to the case of showing a single identity, say h ◦ h = h . (23)The negative graph of h is A (this subspace is the negative graph of h , andhence also of the inverse h ). So it is sufficient to show that the negative graphof h ◦ h is A . Butgraph( − h ◦ h ) = { x − z | x ∈ A , z ∈ A , z = ( h ◦ h )( x ) } = { x − h ( x ) + h ( x ) − h ( h ( x )) | x ∈ A , z ∈ A }⊆ ( A + A ) ∩ ( A + A ) = A , and the last inclusion is, for dimension reasons, actually an equality. Following von Neumann, we use negative graphs for symmetry reasons when dealing with frames.
55. Suppose we are given a frame ¯ A in some vector space W . Set U := A and let v , ..., v n be a basis of A . A basis of A is defined via u i := h ( v i ), i = 1 , ..., n ,and a basis of A is defined by w i := h ( v i ). Since (23) is equivalent to h ◦ h ◦ h = id A , (24)it follows that u i = h ( w i ) for each i . Now the linear isomorphism which sends thebasis u , ..., u n , v , .., v n , w , .., w n to the basis of V := U × U × U built canonicallyfrom v , ..., v n has, as its image, a frame of the desired form. Remark 6.6.
One might think of (24) as a kind of cocycle condition which says thatthat the endomorphism obtained from the “loop” A h −→ A h −→ A h −→ A is trivial.Given a frame ¯ A on V , with dim V = 3 k , we define a frame basis for ¯ A to be anordered basis { u , ..., u k , v , ..., v k , w , ..., w k } of V such that • { u , ..., u k } is a basis of A , { v , ..., v k } is a basis of A , { w , ..., w k } is a basis of A , • h ( u i ) = v i , h ( v i ) = w i , h ( w i ) = u i for all i = 1 , ..., k .A frame basis always exists (c.f. the proof of Lemma 6.5).We define an augmented frame in V to be a frame ¯ A in V together with a subspace C ⊆ V such that A + A = A ⊕ C . This latter condition says that C is the negativegraph of a linear map h : A → A (which is uniquely determined by C ). In particular,an augmented frame determines uniquely an endomorphism η := h ◦ h : A −→ A which we call the underlying endomorphism of the augmented frame. If η is the un-derlying endomorphism of an augmented frame we write this as ( ¯ A, η ). As was the casefor frames, augmented frames can be viewed as special kinds of poset representations,with the inherited notions of morphism and isomorphism.
Lemma 6.7.
Let ( ¯
A, η ) in V and ( ¯ A ′ , η ′ ) in V ′ be augmented frames. There is a bijectivecorrespondence between morphisms ( A , η ) → ( A ′ , η ′ ) and morphisms ( ¯ A, η ) → ( ¯ A ′ , η ′ ) .When ( A , η ) = ( A ′ , η ′ ) , this gives an isomorphism between the endomorphism al-gebras of ( A , η ) and ( ¯ A, η ) . roof Suppose f : V → V ′ is a morphism of augmented frames. In particular then f ( A ) ⊆ A ′ , f ( A ) ⊆ A ′ , f ( A ) ⊆ A ′ , and f ( C ) ⊆ C ′ , which implies that thediagrams A A ′ A A ′ A A ′ A A ′ f | A h h ′ f | A h h ′ f | A f | A commute. Stacking the second diagram under the first, we obtain that A A ′ A A ′ f | A η η ′ f | A commutes, as desired.Conversely, if we have a linear map f : A → A ′ such that f ◦ η = η ′ ◦ f , then itextends to a morphism ˆ f of augmented frames by setting ˆ f = h ′ ◦ f ◦ h on A andˆ f = h ′ ◦ f ◦ h on A .It is easy to see that the thus defined operations “restriction from V to A ” and“extension from A to V ” are mutually inverse.Now assume ( A , η ) = ( A ′ , η ′ ). To show that the mutually inverse “extension” and“restriction” maps define algebra isomorphisms between the respective endomorphismalgebras of ( A , η ) and ( ¯ A, η ), it is sufficient to check that one of these maps is a mor-phism of algebras. This is easiest to check for the restriction map: the operation ofrestriction is clearly compatible with composition, addition, scalar multiplication, andthe units in the respective endomorphism algebras.
Given an augmented frame ( ¯
A, C ) = ( ¯
A, η ) in V we define an associated sextuple S η in V by I = A C = A + A I = A C = A + A I = ( C + A ) ∩ ( A + A ) C = A + I (25)A sextuple which is isomorphic to one of this type will be called a framed sextuple . Lemma 6.8.
Let ( ¯
A, η ) and ( ¯ A ′ , η ′ ) be augmented frames in V and V ′ , respectively .1. Given a sextuple S η , the underlying augmented frame can be recovered via A = I A = C ∩ C A = C ∩ C A = ( I + I ) ∩ C A = I A = ( A + A ) ∩ ( A + A ) C = ( I + I ) ∩ C (26)57 . If S is a sextuple such that the expressions (26) define an augmented frame, then S is a framed sextuple, i.e. of the form (25).3. A linear map f : V → V ′ is a morphism ( ¯ A, η ) −→ ( ¯ A ′ , η ′ ) if and only if it is amorphism S η −→ S η ′ .4. Any sextuple S η built from an augmented frame ( ¯ A, η ) is isomorphic to one withthe following form, where U = A : V = U × U × UI = U × × C = U × U × I = 0 × × U C = 0 × U × UI = { ( − ηx, x, − x ) | x ∈ U } C = { ( x, − x, | x ∈ U } + I . (27) Proof
1. This can be checked using elementary linear algebra resp. modular lattice calcu-lations. For example, C ∩ C = ( A + A ) ∩ ( A + A ) = A , since by assumption V = A ⊕ A ⊕ A . To see that C = ( I + I ) ∩ C , we plug in the definitions of I , I and C and calculate( I + I ) ∩ C = ( A + [( C + A ) ∩ ( A + A )]) ∩ ( A + A ) ⊆ [( C + A ) ∩ ( A + A + A )] ∩ ( A + A )= [( C + A ) ∩ V ] ∩ ( A + A ) = C. The obtained inclusion is actually an equality, sincedim( I + I ) ∩ C = dim( I ⊕ I )+dim C − dim( I + I + C ) = 1 / V = dim C. The equations for A and A can be checked in a similar manner, and theequation for A holds by the definition of a frame.2. Assuming that (26) defines an augmented frame, we need to show that the rela-tions (25) hold.The expressions for I and I are trivially satisfied, so it remains to show theexpressions for I , C , C , and C .We first make some intermediate observations: A + C = A + C ∩ ( A + I ) = C ∩ ( A + A + I ) , (28)using the modular law for the last equality; from (28) and the assumption thatwe have an augmented frame, V = A + A + A = A + C + A = C ∩ ( A + A + I ) + A = ( C + A ) ∩ ( A + A + I ) , (29)58gain via the modular law for the last equality. But (29) implies that V = C + A = A + A + I , (30)and using (30) we find that A + A = C ∩ C + A = C ∩ ( C + A ) = C , (31)where the application of modular law is justified since A = I ⊆ C by assump-tion. This is the desire expression for C .The expression for C now follows from (30) and A + A = A + C = A + C ∩ ( I + I ) = C ∩ ( A + I + I ) = C , (32)using modularity via the fact that A = I ⊆ C .To obtain the expression for I , we first note that C + A = ( I + I ) ∩ C + A = ( I + I ) ∩ ( C + A ) = I + I (33)since A = I ⊆ I + I and, via (30), C + A = V . Now,( C + A ) ∩ ( A + A ) = ( C + A ) ∩ ( A + ( I + A ) ∩ C ) ( ) = ( C + A ) ∩ ( A + ( I + A ) ∩ ( A + A )) mod = ( C + A ) ∩ ( I + A ) ∩ ( A + A + A ) ( ) = ( C + A ) ∩ ( A + I ) ( ) = I + ( C + A ) ∩ A = I + ( C ∩ A ) = I , using for the second-to-last equality that C ⊆ A ⊕ A and V = A ⊕ A ⊕ A ,and for the last equality that C ∩ A = 0 by definition of an augmented frame.Finally, the desired expression for C follows from A + I = ( C ∩ C ) + I = C ∩ ( C + I )= C ∩ ( A + A + ( C + A ) ∩ ( A + A )) mod = C ∩ ( A + ( A + C + A ) ∩ ( A + A ))= C ∩ ( A + A + A ) = C , using in the second line the already-obtained expressions for C and I , and inthe last line the fact that A + C + A = A + A + A = V .3. Suppose first that f is a morphism of augmented frames. Since the subspaces I i , C i and I ′ i , C ′ i of the respective associated sextuples can be expressed entirelyvia lattice terms built only of subspaces in the respective augmented frames, itfollows that f ( I i ) ⊆ I ′ i and f ( C i ) ⊆ C ′ i for each i .Similarly, by part 1 above the underlying augmented frames can be expressed vialattice terms built from subspaces in the respective sextuples, so f is a morphismof augmented frames if it is a morphism of the associated sextuples.59. From Lemma 6.5, it is clear that any augmented frame ( ¯ A, η ) is isomorphic to oneof the following form: V = U × U × UA = U × × A = { ( x, − x, | x ∈ U } A = 0 × U × A = { (0 , x, − x ) | x ∈ U } A = 0 × × U A = { ( x, , − x ) | x ∈ U } C = { ( − ηx, x, | x ∈ U } (34)The sextuple associated to this augmented frame is precisely (27), and by part 3of this lemma, it is isomorphic to S η . Remark 6.9.
Together, parts 1 and 2 give a complete characterization of framed sex-tuples in lattice theoretic terms: a sextuple is a framed sextuple if and only if theexpressions (26) define an augmented frame, and augemented frames are themselvesdefined in lattice-theoretic terms.
Proposition 6.10.
Let ( U, η ) , ( U ′ , η ′ ) be spaces with endomorphisms.1. There is a bijective correspondence between morphisms ( U, η ) → ( U ′ , η ′ ) and mor-phisms S η → S ′ η . When ( U, η ) = ( U ′ , η ′ ) , this correspondence is an isomorphismof the respective endomorphism algebras. In particular, ( U, η ) is indecomposableif and only if S η is.2. Let u , ..., u n , v , ..., v n , w , .., w n and u ′ , ..., u ′ n , v ′ , ..., v ′ n , w ′ , .., w ′ n be frame basesof the underlying frames of sextuples S η and S η ′ . Let ˆ f : S η → S η ′ be a morphismand f its restriction f : A = h v , ..., v n i → A ′ = h v ′ , ..., v ′ n i . If the coordinatematrix of f with respect to the bases h v , ..., v n i and h v ′ , ..., v ′ n i is M , then thecoordinate matrix of ˆ f with respect to the respective frame bases is M M
00 0 M . (35) Proof
1. The said correspondence is the one defined in the proof of Lemma 6.7. By Lemma6.7 and Lemma 6.8, 2., it maps morphisms (
U, η ) → ( U ′ , η ′ ) to morphisms S η → S ′ η , and when ( U, η ) = ( U ′ , η ′ ), this correspondence is an isomorphism of algebras.The statement about indecomposability follows then from the fact that ( U, η ) and S η are each indecomposable if and only if their respective endomorphism algebrasare local, and the fact that “local-ness” is preserved under isomorphism.60. By definition, ˆ f = ( h ′ ◦ f ◦ h ) ⊕ f ⊕ ( h ′ ◦ f ◦ h ) : A ⊕ A ⊕ A → A ′ ⊕ A ′ ⊕ A ′ (c.f. Lemma 6.8). The form (35) follows now from the fact that, with respect toframe bases, all of the maps h , h , h ′ , h ′ have coordinate matrices which arethe identity matrix. Remark 6.11.
The correspondence in Proposition 6.10, is functorial.
In this section we identify which continuous-type indecomposable sextuples are isomor-phic to a framed sextuple.From Lemma 6.1 and Corollary 6.2 we see that, for η indecomposable, S η ≃ ∆ ( k ; 0)when η is nilpotent, and S η ≃ ∆ ( k ; ∞ ) when η has eigenvalue 1. Indeed, if η isnilpotent, then I ∩ C = { ( − ηx, x, − x ) | x ∈ U } ∩ (0 × U × U ) ⊇ { (0 , x, − x ) | x ∈ ker η } 6 = 0 . And if η has eigenvalue 1 with associated eigenspace U , then I ∩ C = (0 × × U ) ∩ { ( y − ηx, x − y, − x ) | x, y ∈ U } ⊇ { (0 , , − x ) | x ∈ U } 6 = 0 . We will call an indecomposable endomorphism exceptional if it is nilpotent or has1 as eigenvalue. Note that any non-exceptional indecomposable endomorphism is anautomorphism.It remains now to identify the sextuples S η , with η non-exceptional, in terms of theclassification discussed in the previous section. Proposition 6.12.
On the level of isomorphism classes (and for fixed ambient di-mension k ), there is a one-to-one correspondence between the sextuples S η , and thesextuples ∆( k ; γ ) , where γ and η are non-exceptional indecomposable endomorphisms.If one views both η and γ as endomorphisms of k k , the correspondence is given by η = γγ − or, in inversely, γ = ηη − . Proof
We start by considering an indecomposable continuous-type sextuple ∆( k ; γ ),with γ non-exceptional. From Section 6.1, we know that ∆( k ; γ ) is isomorphic to asextuple which has the form V = X ⊕ X ⊕ X , C i = X i + X i +1 , I i = Im ( α i ⊕ β i ) = Γ( β i α − i )where α i : Y i → X i and β : Y i → X i +1 are all invertible, and γ = β ◦ α − . In particular,we can identify the Y i with the I i (since the maps α i ⊕ β i are injective), and we have C i = X i ⊕ Y i = X i +1 ⊕ Y i i = 1 , , . (37)61o show that ∆( k ; γ ) ≃ S η for some endomorphism η , we “guess” an underlyingaugmented frame (using Lemma 6.8 to make our ansatz), and we show that this is anaugmented frame whose associated sextuple is isomorphic to ∆( k ; γ ). In this case weknow that η must be non-exceptional, since S η ≃ ∆ ( k ; 0) or S η ≃ ∆ ( k ; ∞ ) when η isexceptional.As our ansatz for the underlying augmented frame associated to ( V ; C i , I i ), we set A = Y A = ( X + X ) ∩ ( X + X ) = X A = ( X + X ) ∩ ( X + X ) = X A = ( Y + Y ) ∩ ( X + X ) A = Y A = ( A + A ) ∩ ( A + A ) C = ( Y + Y ) ∩ ( X + X )and check that this defines an augmented frame. In order to verify a ⊕ -relation, itsuffices to check the +- or the ∩ -relation, if the dimensions of the subspaces involvedare known and add up properly. Thus, to see that V = A ⊕ A ⊕ A , it suffices to notethat A + A + A = Y + X + Y = X + X + Y = X + X + X = V. Similarly, A ⊕ A = A ⊕ A = A + A follows from A + A = X + Y = X + X = A + A A + A = X + X = A + A . To see A ⊕ A = A ⊕ A = A + A , note that A + A = ( Y + Y ) ∩ ( X + X ) + X = ( Y + Y + X ) ∩ ( X + X )= X + X = X + Y = A + A A + A = ( Y + Y ) ∩ ( X + X ) + Y = ( Y + Y + Y ) ∩ ( X + X )= X + X = X + Y = A + A using modularity (and that X , Y ⊆ X + X ) to obtain the second equality in eachline, respectively. At this point it is not yet clear that the lefthand sums are direct, sincethe dimension of A is not yet determined. This can be checked directly, for example: Y + Y + X = X + Y + X = X + X + X = V whence A ∩ A = ( Y + Y ) ∩ X = 0and A ∩ A = ( Y + Y ) ∩ Y = 0 . Finally, A ⊕ C = A + A because A + C = Y + ( Y + Y ) ∩ ( X + X ) = ( Y + Y + Y ) ∩ ( X + X ) = X + X = A + A A ∩ C = Y ∩ ( Y + Y ) = 0using modularity in the first line (with the fact that Y ⊆ X + X ).This establishes that we have an augmented frame. It remains to verify that theoriginal sextuple ( V, I i , C i ) is the one associated to this augmented frame, i.e. we checkthat the equations (25) hold: 62 I = Y = A • I = Y = A • I = Y = ( Y + Y ) ∩ ( Y + Y ) = ( C + A ) ∩ ( A + A ),using for the last equality that Y + Y = ( Y + Y ) ∩ ( X + X + X ) = ( Y + Y ) ∩ ( X + X + Y )= ( Y + Y ) ∩ ( X + X ) + Y = C + A and Y + Y = ( Y + Y ) ∩ ( X + X + X ) = ( Y + Y ) ∩ ( Y + X + X )= Y + ( Y + Y ) ∩ ( X + X ) = ( A + A ) • C = A + A = Y + X = X + X = C • C = A + A = X + Y = X + X = C • C = A + ( C + A ) ∩ ( A + A ) = A + I = X + Y = X + X = C This establishes that every sextuple ∆( k ; γ ) is isomorphic to some S η , with η non-exceptional. It remains now to show that every S η with η non-exceptional is isomorphicto some ∆( k ; γ ). For this it is sufficient to prove the formula (36) and note that thisformula defines a bijection (in fact an involution) of the set of non-exceptional indecom-posable endomorphisms of k k .Consider again a sextuple of type ∆( k ; γ ). Set X = k k . We’ll use normal formswhich are isomorphic to the ones (13): let V = X × X × X , and X = X × × , X = 0 × X × , X = 0 × × X. Thus normal forms for these sextuple are I = Γ( β α − ) = { ( x, − γx, | x ∈ X } C = X × X × I = Γ( β α − ) = { (0 , x, x ) | x ∈ X } C = 0 × X × XI = Γ( β α − ) = { ( x, , x ) | x ∈ X } C = X × × X. (38)Set g i := β i α − i and g := g g g . Note that g = − γ when these are viewed as maps X → X .For such a sextuple, we compute the endomorphism η underlying the associatedaugmented frame. By definition, η = h ◦ h , so we need to compute h and h . Fromthe first part of this proof we know that for this frame A = I A = X A = I A = Γ( − h ) = X C = Γ( − h ) = ( I + I ) ∩ ( X + X ) .
63n particular one finds easily that h : A → A , ( x, g x, (0 , g x, ,C = { ( − g g x, x, | x ∈ X } , and h : A → A , (0 , x, ( g g ( g + 1) − x, g ( g + 1) − x, . It follows that η : A → A , (0 , x, (0 , g ( g + 1) − x, . Thus, viewed as endomorphisms of X , η = γ − γ . Remark 6.13.
Since η and γ commute, if γ has an eigenvalue, say λ γ , then η will alsohave an eigenvalue, say λ η , and any eigenvector for λ γ will also be an eigenvector for λ η . In this case, λ η = λ γ / ( λ γ −
1) and λ γ = λ η / ( λ η − In this section, we identify the duals of indecomposable continuous-type sextuples. Werecall: • The dual of a sextuple ( V ; C i , I i ) is ( V ∗ ; I ◦ i , C ◦ i ), where, for U ⊆ V , the subspace U ◦ = { f ∈ V ∗ | f ( U ) = 0 } is the annihilator of U . • A sextuple is self-dual if it admits an isomorphism (of poset representations) toits dual. A pair of sextuples is called mutually dual if each is isomorphic to thedual of the other. • The operation of taking the annihilator obeys the rules( U + U ) ◦ = U ◦ ∩ U ◦ and ( U ∩ U ) ◦ = U ◦ + U ◦ , for any subspaces U , U .From the structure of the dimension vector of indecomposable continuous-type sex-tuples it follows that the dual of a continuous-type sextuple is again of continuous type.Identifying the duals of the exceptional continuous-type sextuples is easiest. Lemma 6.14.
The indecomposable sextuples of type ∆ ( k ; 1) and ∆ ( k ; 1) are mutuallydual. Proof
This follows from Corollary 6.2: if ( V ; C i , I i ) is a sextuple of type ∆ ( k ; 1), thendim( I + I + I ) = dim V −
1. Thus for the dual ( V ′ ; C ′ i , I ′ i ) will hold C ′ ∩ C ′ ∩ C ′ = I ◦ ∩ I ◦ ∩ I ◦ = ( I + I + I ) ◦ = 1. This implies, by Corollary 6.2 that the dual is of type∆ ( k ; 1) (the dual must be indecomposable and of continuous type).64 emma 6.15. The indecomposable sextuples ∆ i ( k ; 0) and ∆ i − ( k ; ∞ ) are mutuallydual. In other words, in each ambient dimension k , we have the following three pairsof mutually dual indecomposable sextuples: ∆ ( k ; 0) and ∆ ( k ; ∞ ) , ∆ ( k ; 0) and ∆ ( k ; ∞ ) , ∆ ( k ; 0) and ∆ ( k ; ∞ ) . Proof
Suppose ( V ; C i , I i ) is a sextuple of type ∆ i ( k ; 0). By Corollary 6.2, this sextuplewill satisfy dim( I i ∩ C i − ) = 1. In particular thendim( I i + C i − ) = dim I i + dim C i − − dim( I i ∩ C i − ) = k + (2 k ) − V − . The dual sextuple ( V ′ ; C ′ i , I ′ i ) will therefore satisfydim( I ′ i − ∩ C ′ i ) = dim( C ◦ i − ∩ I ◦ i ) = dim( I i + C i − ) ◦ = 1 . This implies, via Corollary 6.2, that ( V ′ ; C ′ i , I ′ i ) is of type ∆ i − ( k ; ∞ ) Remark 6.16.
From Section 6.4 we know that ∆ ( k ; 0) and ∆ ( k ; ∞ ) are isomorphic,respectively, to the indecomposable framed sextuples S η where η is either nilpotent (inthe first case) or has eigenvalue 1 (in the second case). Thus the above shows that thesole two exceptional framed sextuples are dual to one another.Now we consider the non-exceptional indecomposable continuous-type sextuples∆( k ; γ ). From Corollary 6.2 it follows that the dual of such a sextuple is again anon-exceptional indecomposable continuous-type sextuple; let ∆( k ; γ ′ ) be the type ofthe dual. Our goal now is to determine the relationship between γ and γ ′ .From Proposition 6.12 we know that ∆( k ; γ ) and ∆( k ; γ ′ ) are isomorphic, respec-tively, to sextuples S η and S η ′ , with η and η ′ non-exceptional. Proposition 6.17.
Consider a non-exceptional indecomposable endomorphism η ∈ End ( U ) with associated sextuple S η . Let β := ( u , ..., u k , v , ..., v k , w , .., w k ) be a framebasis for this sextuple. Then:1. The dual of S η is isomorphic to S η ′ , with η ′ = ( Id − η ) ∗ ∈ End ( U ∗ ) . Moreover, ( − w ∗ , ..., − w ∗ k , v ∗ , ..., v ∗ k , − u ∗ , ..., − u ∗ k ) is a frame basis for S η ′ , where β ∗ := ( u ∗ , ..., w ∗ k ) is the dual basis of β .2. A bijective correspondence B : S η ∼ −→ S η ′ ←→ b : ( U, η ) ∼ −→ ( U ∗ , Id − η ∗ ) is given by restricting maps B to U .With respect to the bases β and β ∗ , any isomorphism B : S η → S η ′ has coordinatematrix of the form H M := O O − MO M O − M O O , here M ∈ k k × k is the coordinate matrix of the corresponding isomorphism b :( U, η ) → ( U ∗ , η ′ ) with respect to the respective bases ( v , ..., v k ) and ( v ∗ , ..., v ∗ k ) of U and U ∗ .Note, in particular, that B is (skew)-symmetric if and only if the correspondingmap b is. Proof
By definition, S η has ambient space V = U × U × U , and the associatedaugmented frame ( ¯ A, η ) is (34); in particular A = U × × A = 0 × U × A =0 × × U . We view the dual sextuple S ∗ η = S η ′ = ( V ′ ; C ′ i , I ′ i ) as having ambient space V ′ = U ∗ × U ∗ × U ∗ , paired with V = U × U × U via ( ξ , ξ , ξ ) : ( u , u , u ) ξ ( u ) + ξ ( u )+ ξ ( u ). The sextuple S η ′ , expressed in terms of the underlying augmented frameof S η , is I ′ = C ◦ = A ◦ ∩ A ◦ = 0 × × U ∗ C ′ = I ◦ = A ◦ = 0 × U ∗ × U ∗ I ′ = C ◦ = A ◦ ∩ A ◦ = U ∗ × × C ′ = I ◦ = A ◦ = U ∗ × U ∗ × I ′ = C ◦ = A ◦ ∩ [( C ◦ ∩ A ◦ ) + ( A ◦ ∩ A ◦ )] C ′ = I ◦ = ( C ◦ ∩ A ◦ ) + ( A ◦ ∩ A ◦ )(39)We compute the underlying augmented frame ( ¯ A ′ , η ′ ) of this sextuple, using (34) as anaid for the calculations: A ′ = I ′ = 0 × × U ∗ A ′ = C ′ ∩ C ′ = 0 × U ∗ × A ′ = I ′ = U ∗ × × A ′ = C ′ ∩ C ′ = ( I + I ) ◦ = { ( y, x, − x ) | x, y ∈ U } ◦ = { (0 , ξ, ξ ) | ξ ∈ U ∗ } A ′ = ( I ′ + I ′ ) ∩ C ′ = [( C ∩ C ) + I ] ◦ = { ( ξ, ξ, | ξ ∈ U ∗ } A ′ = ( A ′ + A ′ ) ∩ ( A ′ + A ′ ) C ′ = ( I ′ + I ′ ) ∩ C ′ = [( C ∩ C ) + I ] ◦ = { (0 , ξ, (id − η ) ∗ ξ ) | ξ ∈ U ∗ } (40)For the computation of C ′ , for example: C ∩ C = { ( y − ηx, x − y, − x ) | x, y ∈ U, y − ηx = 0 } , and I + C ∩ C = { ( y, x − ηx, − x ) | x, y ∈ U } , so C ′ is precisely those( ξ , ξ , ξ ) ∈ U ∗ × U ∗ × U ∗ such that ξ = 0 and ξ = ξ ◦ (id − η ).We can now read off the maps h ′ i,i +1 and h ′ associated to this augmented frame: • A ′ = Γ( − h ′ ) implies that h ′ : A ′ → A ′ , (0 , , ξ ) (0 , − ξ, • A ′ = Γ( − h ′ ) implies that h ′ : A ′ → A ′ , (0 , ξ, ( − ξ, , • h ′ = ( h ′ ◦ h ′ ) − : A ′ → A ′ , ( ξ, , (0 , , ξ ). • C ′ = Γ( − h ′ ) implies that h ′ : A ′ → A , (0 , ξ, (0 , , − (id − η ) ∗ ξ )In particular, η ′ = h ′ ◦ h ′ = (id − η ) ∗ . We also see that, taking ( − v ∗ , ..., − v ∗ k ) as a basisof A ′ = 0 × U ∗ ×
0, a frame basis is given by ( w ∗ , ..., w ∗ k , − v ∗ , ..., − v ∗ k , u ∗ , ..., u ∗ k ), since w ∗ i = h ′ ( − v ∗ i ) and u ∗ i = h ′ ( − v ∗ i ), i = 1 , .., k . The remaining statement of the currentproposition follows now from Proposition 6.10. In particular, any isomorphism S η → S η ′ has, with respect to the bases β and β ∗ , coordinate matrix of the form H M , where M is the coordinate matrix of an isomorphism U → U ∗ such that M AM − = id − A t .66 orollary 6.18. The sextuples ∆( k ; γ ) and ∆( k ; ( γ − ) ∗ ) are mutually dual. Proof
Viewing γ and η as endomorphisms of the same space, and similarly for γ ′ and η ′ , we have by (36) that η = γγ − and γ ′ = η ′ η ′ − . Substituting η ′ = 1 − η ∗ in the latterequation, using the former equation and simplifying gives γ ′ = ( γ − ) ∗ . Corollary 6.19.
A sextuple ∆( k ; γ ) is self-dual if and only if ( k k , γ ) and (( k k ) ∗ , ( γ − ) ∗ ) are isomorphic endomorphisms (in the sense of Section 3.2). In this case, if γ has aneigenvalue λ γ , then λ γ = − . Proof
The first part follows from Corollary 6.18 and the fact that, by Propositions6.12 and 6.10, ∆( k ; γ ) and ∆( k ; γ ) are isomorphic sextuples if and only if γ and γ areisomorphic endomorphisms. The statement about eigenvalues follows from the fact that γ ∗ has the same eigenvalue as γ (when such exists), and that γ cannot have eigenvalue1 (by the assumption that γ is non-exceptional). C and R We turn now to the classification of non-split isotropic triples of continuous type. Thespecial cases where the ground field k is either C or R are of particular interest. Wetreat these cases first, which allows for a simpler analysis. Furthermore, the case k = R provides basic intuition for understanding the more involved classification over perfectfields, which we undertake in the next section.Throughout this and the next section, N will always denote a nilpotent squarematrix which has “1” everywhere on the first upper off-diagonal and all other entries“0”, and whose size will be specified, or clear, depending on the context. We call N the standard indecomposable nilpotent matrix (of a given size). For example, if the sizehappens to be 3 ×
3, then N = . For block matrices, we will use the following convention, which generalizes multipli-cation of matrices by scalars: if M is a block matrix with square blocks of size l × l ,and if K is an l × l matrix, then KM will denote the block matrix whose blocks consistof the blocks of M each multiplied on the left by K . We define M K similarly.
Given a 3 k -dimensional indecomposable sextuple S η , we call a frame basis( u , ..., u k , v , ..., v k , w , ..., w k )a Jordan basis for S η if ( v , ..., v k ) is a Jordan basis for η for some (unique) λ ∈ k , i.e. η ( v ) = λv , η ( v j ) = v j − + λv j for j >
1. Note that any Jordan basis for η extends toone for S η . 67 heorem 7.1. Consider an indecomposable sextuple S η of dimension k with underly-ing endomorphism η having eigenvalue λ ∈ k ; fix a Jordan basis for S η . Let M ∈ k k × k be of the form . . . − j +1 . . . . . . . . . . . .. . . . . . − . . . . . . i.e. the entries of M satisfy m ij = ( − j +1 for i = k − j + 1 , and m ij = 0 else.1. The sextuple is isomorphic to its dual if and only if λ = .Now assume λ = .2. The matrix H M = O O − MO M O − M O O defines a compatible form B which is symplectic if k is even, symmetric if k isodd.3. Up to isomorphism and multiplication with a scalar, B is the only such form. Foreven k there is no symmetric compatible form, for odd k no symplectic one.4. For any = c ∈ k , cB and B are isometric via an automorphism of the sextupleif and only if c is a square in k .5. A complete list of compatible symplectic resp. symmetric forms is given by thematrices H MQ , where Q ∈ k k × k is of the form P k − i =0 a i N i , with a = 0 and a i = 0 for i even. Remark 7.2.
Observe that the particular value λ = arises from the way the endo-morphism is related to the sextuple. Other ways would give a different (but also unique)value.Also note that, setting ζ := η − , we have that η = + ζ underlies an indecomposableself-dual framed sextuple if and only if ζ is similar to − ζ ∗ . Furthermore, η has aneigenvalue if and only if ζ does, and in the case of self-duality, the unique eigenvalue of ζ is 0. The description “ η = + ζ ” is helpful to keep in mind in the proof below, andin the subsequent sections. Proof of Theorem 7.1
With respect to v , . . . , v k , the matrix of η is A = λI + N . Inparticular, A is similar to A t via the permutation matrix P corresponding to reversingthe order of the basis. Proof of 1 and 2 . Assume that S η is isomorphic to its dual. Then by Proposition6.17, I − A t is similar to A , and hence to A t . Thus, A is similar to I − A . In particular, I − A necessarily also has λ as its unique eigenvalue. On the other hand, given an68igenvector v for A , we have ( I − A ) v = v − λv = (1 − λ ) v , which means that 1 − λ isan eigenvalue of I − A . By indecomposablilty, I − A can only have one eigenvalue, thus λ = 1 − λ , i.e. λ = .Now, assume λ = and define B by H M . It remains to show that B is compatible,i.e. that H M defines, with respect to the given basis and its dual, an isomorphism from S η onto its dual. By Proposition 6.17 this means to show that M defines an isomorphism( A , η ) → ( A ∗ , (id − η ) ∗ ), i.e. M AM − = ( I − A ) t . Here, observe that M = DP , where P = P − has anti-diagonal entries 1, zero else, and where D = D − is diagonal withentries d jj = ( − j +1 . Now, P AP = A t = I + N t and DN t D = − N t , and hence M AM − = I + DN t D = I − N t = I − A t . Proof of 3 – 5
By Remark 3.9, the algebra E of endomorphisms of A commutingwith A = I + N is local and E = k id ⊕ Rad E ; Proposition 6.10 establishes anisomorphism of E onto the endomorphism algebra E ′ of S η which shows that E ′ islocal, too, with E ′ = k id ⊕ Rad E ′ . Thus, Lemma 3.26 applies, proving uniqueness. Theproof of 4. can be copied from that of Theorem 5.9. Moreover, units in the ring E ′ are given by block matrices as in (35). Thus, due to 2, any compatible form is givenby a matrix H MQ where Q ∈ k k × k is the coordinate matrix of an automorphism of( A , η ). In particular, this means that Q is of the form P kl =0 a l N l with a = 0. Thus M Q = P kl =0 a l M N l . Note that M N l is skew-symmetric if l = k mod 2 and symmetricif l = k mod 2:( M N l ) ij = X p m i,p ( N l ) p,j = m i,k − i +1 ( N l ) k − i +1 ,j = ( − k − i for i + j = l + k + 1( M N l ) ij = 0 else , and so, for i + j = l + k + 1,( M N l ) tij ( M N l ) ij = ( M N l ) ji ( M N l ) ij = ( − k − j ( − k − i = ( − k − l − . It follows that
M Q = P i even a i M N i + P i odd a i M N i is the (unique) decomposition of M Q into its symmetric and skew-symmetric parts: when k is odd, the first summandis the symmetric part and the second summand skew-symmetric; when k is even, thereverse is the case. The second summand is non-invertible, while the first summand isinvertible (since a = 0). Thus all compatible (non-degenerate) forms are of the form H MQ , where M Q = P i even a i M N i . For k odd they are symmetric; for k even, theyare skew-symmetric. Example 7.1.
Let k = 4, and consider the self-dual sextuple S η with underlying inde-composable endomorphism having eigenvalue λ = . Then S η admits only compatiblesymplectic forms. These are parametrised by the set { ( a , a ) ∈ k | a = 0 } and given,with respect to a fixed Jordan basis and its dual basis, by the matrices H A = O O − AO A O − A O O , A = − a a − a − a a a . R Theorem 7.3.
Let S η be an indecomposable sextuple over R of dimension k withunderlying η having no real eigenvalue.1. The sextuple admits an isomorphism onto its dual if and only if η has, in C ,eigenvalues λ = ± √− r , with real r > (both of geometric multiplicity ).2. Assume the case of self-duality. Then k = 2 l and both symmetric and symplecticcompatible forms exist. Furthermore:(a) There is a frame basis ( u i , v i , w i ) such that the matrix A of η is in real Jordannormal form with respect to the v i , and there are “canonical” compatibleforms B and B ′ , where “ B ′ = √− B ”. If l is odd, B is symmetric and B ′ is symplectic; if l is even, B is symplectic and B ′ is symmetric.With respect to the frame basis, B is given by the matrix H = H M where M ∈ k k × k has × -blocks M ij with M ij = ( − j +1 (cid:18) (cid:19) for i + j = l + 1 , M ij = (cid:18) (cid:19) elseand the matrix H ′ of B ′ is obtained from H by setting H ′ = ℑ H , where ℑ isthe matrix ℑ = (cid:18) −
11 0 (cid:19) . Multiplication with ℑ is understood in sense of “scalar multiplication by‘ √− ’ for × block matrices”.(b) All symmetric resp. symplectic compatible forms are given, up to isomorphyand multiplication with ± , respectively by H resp. ℑ H if l is odd, ℑ H resp. H if l is even.In particular, multiplying with ℑ changes the symmetry of a compatible form.There is no automorphism which is an isometry from H to − H or from ℑ H to −ℑ H .(c) A complete list of compatible symmetric, respectively symplectic, forms isgiven by the matrices H MQ resp. ℑ H MQ where M is as above and where Q ∈ R k × k is of the form P l − i =0 a i N i , a i ∈ R , a = 0 , and a i = 0 for i even. These are symmetric, respectively symplectic, if l is odd; symplectic,respectively symmetric, if l is even. roof With respect to a suitable basis of A , η has coordinate matrix in real Jordanform A = ZI + N , where Z = (cid:18) a − bb a (cid:19) and a ± √− b are the complex eigenvalues of η ( b = 0 by hypothesis). Here, ZI isthe block diagonal matrix with diagonal blocks Z and N the standard indecomposablenilpotent matrix. In view of Proposition 6.17, self-duality means that ( I − A ) t is similarto A . Now, I − A and ( I − A ) t have the same complex eigenvalues, and these are alsothose of A , by similarity. Thus, λ is an eigenvalue of A if and only if so is 1 − λ and soboth have the same real part which then must be . This proves that Z = (cid:18) − rr (cid:19) , r = 0On the other hand, any such Z satisfies ( I − Z ) t = Z , and thus corresponds to a self-dualsextuple.Now, to prove that H S is a compatible form, we can mimic the proof of 2 in Theorem7.1, replacing scalar matrix entries there by 2 × − by B , and m ij by M ij . Thus H S defines anisomorphism onto the dual. The further statements are then obvious. Proof of (2b). Observe that matrices Z = (cid:18) x − yy x (cid:19) belong to a subring F of R × which is isomorphic to C . F is the field in R × obtainedby adjoining, for example, (cid:0) −
11 0 (cid:1) to the field { ( x x ) | x ∈ R } ⊆ R × . For any Z ∈ F ,let ZI denote the block diagonal matrix in R k × k with diagonal blocks Z . Now, C is inthe (coordinatized) endomorphism algebra E of ( A , η ) if and only if it commutes withthe matrix A of η . By Proposition 8.11 below, E consists of the matrices C of the form C = l − X i =0 Z i N i , Z i ∈ F. By Proposition 6.10, S η has endomorphism algebra consisting of the block-diagonalmatrices CI , with C ∈ E (i.e. there are three diagonal blocks, each one a copy of agiven C ∈ E ). In particular, these matrices commute with H M . Now the proof ofLemma 3.26 generalizes to yield uniqueness of compatible forms up to isomorphismand multiplication with Z ∈ F ; that is, up to isomorphism, compatible forms are of theform ZH M = H ZM with Z = (cid:18) x − yy x (cid:19) ∈ F \{ } . For a more thorough discussion, see Subsection 8.5. y = 0, then ZM has blocks (cid:18) x x (cid:19) , (cid:18) − x − x (cid:19) , (cid:18) x x (cid:19) . . . along its anti-diagonal, and so ZH M = xH M . In this case, the scaling map 1 / p | x | · Idis an isometric isomorphism to from xH M to sign( x ) H M .If x = 0, we have blocks (cid:18) − yy (cid:19) , (cid:18) y − y (cid:19) , (cid:18) − yy (cid:19) . . . on the anti-diagonal of M , and so ZH M = y ℑ H M . The scaling map 1 / p | y | · Id is anisometric isomorphism to from y ℑ H M to sign( y ) ℑ H M .If x = 0 and y = 0 then we have (cid:18) x − yy x (cid:19) , (cid:18) − x y − y − x (cid:19) , (cid:18) x − yy x (cid:19) . . . on the diagonal of ZM , and neither symmetry nor skew-symmetry.It remains to show that there is no isometric isomorphism from H M to − H M (sucha map would also give an isometric isomorphism between ℑ H and H ). Assume that f were such an automorphism and C the matrix of f | A . From the above description of C ∈ E one reads off that C is upper block triangular with c = c k − , k − , c = c k, k − . On the other hand, by inspection of H M we have B ( v , v k − ) = B ( v , v k ) = 1 , B ( v , v i ) = 0 for i = 2 k − , B ( v , v i ) = 0 for i = 2 k. Thus, one would get the contradiction − − B ( v , v k − ) = B ( f v , f v k − ) = B ( c v + c v , k X i =1 c i, k − v i )= B ( c v , c , k − v k − ) + B ( c v , c k, k − v k ) = c + c . Proof of (2c). As in the proof of 4. in Theorem 7.1 one obtains matrices as statedabove, but initially with a i ∈ F . To have symmetry or skew-symmetry in M Q , though,all a i have to be in R I or R ℑ I . Conversely, this grants symmetry resp. skew symmetry,depending on the parity of l . Example 7.2.
Let k = 6, and let η be an indecomposable endomorphism over R , withcomplex eigenvalues ± √− r , r >
0. The corresponding sextuple S η is self-dual (and72t lives in ambient dimension 3 k = 18); compatible symmetric resp. symplectic formsare given by matrices H M , where M is of the form a
00 0 0 0 0 a − a − a a a a a resp. − a a
00 0 0 a − a − a − a a a for ( a , a ) ∈ R , a = 0. R from non-split framed sextuples In this section we return to the connection, discussed in Section 4.8, between lin-ear hamiltonian vector fields and isotropic triples. We show here that each non-splitisotropic triple (over R ) has an associated linear hamiltonian vector field.Let S η be an indecomposable self-dual continuous-type sextuple over R . We knowthat the spectrum in C of η must be { ± √− r } for some value of r ≥
0. Supposethat we make S η into an isotropic triple by choosing a frame basis and choosing acompatible symplectic form given by a matrix H MQ as in Theorem 7.1 or Theorem 7.3(depending on whether η has the eigenvalue or not). Set T := M Q and let A denotethe coordinate matrix of η with respect to the frame basis. Recall that when H T isskew-symmetric, as we have assumed, then so is T . From the normal form given inLemma 6.8, part 4, we can assume that S η has the form V = U × U × UI = U × × C = U × U × I = 0 × × U C = 0 × U × UI = { ( − ηx, x, − x ) | x ∈ U } C = { ( x, − x, | x ∈ U } + I , (41)and from Proposition 6.17 it follows that T is the coordinate matrix of an isomorphism( U, η ) → ( U ∗ , − η ∗ ). In other words, T A = (1 − A t ) T .We claim now that an isotropic triple whose underlying sextuple is (41) satisfiesthe hypotheses of Proposition 4.7, which gives sufficient conditions for constructing anassociated hamiltonian vector field. To verify the hypotheses, note first that clearly V = I ⊕ I ⊕ I . Second, we must check that I I + I j is a symplectic subspace for all i = j . For this, we check that( I i + I j ) ∩ ( I i + I j ) ⊥ = ( I i + I j ) ∩ C i ∩ C j = 0 . For i = 1 , j = 2, ( I + I ) ∩ C ∩ C = ( U × × U ) ∩ (0 × U ×
0) = 0 . For i = 3 , j = 1,( I + I ) ∩ C ∩ C = { ( − ηx + y, x, − x ) | x, y ∈ U } ∩ { ( z, − z, | y ∈ U } = 0 , − ηx + y, x, − x ) of this intersection would need to satisfy − x = 0, andhence − ηx = 0. But then ( y, ,
0) = ( z, − z,
0) holds for some z ∈ U only if y = 0.Finally, for i = 2 , j = 3,( I + I ) ∩ C ∩ C = { ( − ηx, x, y ) | x, y ∈ U } ∩ { (0 , − ηz + z, − z ) | z ∈ U } = { (0 , x, y ) | x, y ∈ U, ηx = 0 , x = ηy − y } = 0 . Indeed, ηx = 0 implies that x = 0, since η is invertible, and so ηy = y must hold. Byassumption η cannot have eigenvalue 1, so also y = 0.Now we will follow the proof of Proposition 4.7 in order to find the hamiltonianvector field associated to (41). We have the symplectic decomposition V = ( I ⊕ I ) ⊕ ( I ⊕ I ) ⊥ = ( U × × U ) ⊕ (0 × U × I is the graph of the map g : ( I ⊕ I ) ⊥ → I ⊕ I given by 0 × U × → U × × U, (0 , x, ( − ηx, , − x ) , so the image of g is the graph of the map I → I , (0 , , x ) ( ηx, , A , and the matrix of theidentification of I with I ∗ is T . Thus the image of g corresponds to a map f : I → I ∗ whose coordinate matrix is T A . Using
T A = (1 − A t ) T and that T is skew-symmetric,we find that the antisymmetric part f a is given by ( T A − A t T t ) = ( T A − T t (1 − A )) = T ( A + (1 − A )) = T, and for the symmetric part f s ( T A + A t T t ) = ( T A + T t (1 − A )) = T ( A − (1 − A )) = T (2 A − . Thus we obtain the hamiltonian vector field X = f − a f s given in coordinates by thematrix 2 A −
1. We have proved the following:
Proposition 7.4.
Non-split isotropic triples satisfy the hypotheses of Proposition 4.7:we can construct an associated hamiltonian vector field.If ϕ is such a triple, we may assume that its underlying sextuple is a framed sextuple S η , built from an endomorphism ( U, η ) . In this case, the symplectic form ω of ϕ inducesan isomorphism U → U ∗ which defines a symplectic form ω U on U . The hamiltonianvector field associated to ϕ is ( U, ω U , η − id ) . Remark 7.5.
Normal forms for indecomposable linear hamiltonian vector fields over R are given, for example, in [18]. These come in both “split” and “non-split” types,and are labeled by their (complex) eigenvalues. For a non-split isotropic triple as abovewith underlying indecomposable endomorphism η , the associated hamiltonian vectorfield 2 η − id is also indecomposable and non-split. The possible such η are parametrizedby ± √− · r with r ∈ [0 , ∞ ); the corresponding hamiltonian vector fields 2 η − idcorrespond in [18] to the non-split ones labeled by complex eigenvalues ±√− · ν (setting ν = 2 r in order to use their notation), where ν ∈ [0 , ∞ ).74 xample 7.3. Let k = 4, and let η be an indecomposable endomorphism over R withcomplex eigenvalue , i.e. with respect to a Jordan basis η has the coordinate matrix A = / / / / . We can make the corresponding self-dual sextuple S η into an isotropic triple by choosingthe compatible symplectic form given, with respect to a Jordan frame basis, by thematrix H T , with T = − − . The associated hamiltonian vector field on the symplectic space ( R , T ) is given, withrespect to a Jordan basis for η , by the matrix . Example 7.4.
Let k = 4, and let η be an indecomposable endomorphism over R withcomplex eigenvalue ± √− ν , where ν >
0, i.e. with respect to a (real) Jordan basis η has the coordinate matrix A = 12 − ν ν − ν ν . Again choosing the compatible symplectic form for S η given by H T with T = − − , we obtain the associated hamiltonian vector field on ( R , T ) given by − ν ν − ν ν . Continuous non-split isotropic triples over perfect fields
We continue working with a ground field k with char( k ) = 2, and additionally we assumethat k is perfect , in order that we may use the normal forms discussed in Proposition8.7 below. No other assumptions are made on k .According to Proposition 6.17, isomorphisms of a sextuple S η onto its dual arisefrom matrices M such that M − AM = ( I − A ) t where A is the matrix of η with respectto some basis; and the induced compatible form is (skew-)symmetric if and only if M is. As indicated by the results over the complex or real number field, the crucial case iswhen η has irreducible minimal polynomial q ( x ). We deal with this case first. Let q ( x ) = P li =0 a i x i ∈ k [ x ] be an irreducible polynomial, with a l = 1. Adjunctionto k of a zero λ of q yields an extension field k ( λ ) = k [ λ ] ∼ = k [ x ] /q ( x ) which as a k -vector space has basis 1 , λ, λ , . . . , λ l − . With respect to this basis, the k -linear map m λ defined by m λ ( r ) := λr has as its coordinate matrix the Frobenius matrix N t + C . . . − a − a − a l − . . . − a l − , where C = C q denotes the matrix whose last column is − a , − a , − a , . . . , − a l − andhas all other entries zero.In particular, by the Cayley-Hamilton theorem the above applies to any endomor-phism η of an vector space U having characteristic polynomial q ( x ), where η plays therole of λ above . Here we view k ( η ) as a subring of End( U ) with subfield { a id | a ∈ k } ∼ = k . In this situation, any v = 0 extends (uniquely) to a basis { v , ηv , ..., η l − v } such that the corresponding coordinate matrix of η is the Frobenius matrix of q ( x ). Inother words, if A ∈ k l × l has q ( x ) = det( xI − A ) irreducible, then k ( A ) = k [ A ] is asubring of k l × l which is an extension field of { aI | a ∈ k } ∼ = k with primitive element A , a zero of q ( x ). Lemma 8.1.
Let q ( x ) ∈ k [ x ] be an irreducible monic polynomial with deg q = l > .Let λ be a zero of q in some extension field of k , and set E = k ( λ ) .Suppose q (1 − λ ) = 0 . Then there exists µ ∈ E and an automorphism g of E over k such that k ( µ ) = E and g ( µ ) = − µ . This implies that: A field k is perfect if every algebraic extension of k is separable. Examples of perfect fields includeall finite fields, and all fields of characteristic zero. Recall that for indecomposable endomorphisms, the characteristic and minimal polynomials coin-cide. Also note that, when the characteristic polynomial is irreducible, the corresponding endomorphismis necessarily indecomposable. . Only even powers of x occur in the minimal polynomial r ( x ) of µ over k ; inparticular deg r = l must be even.2. With respect to the basis { , µ, ..., µ l − } of E over k , g has diagonal coordinatematrix D , with entries d ii = ( − i , for i = 0 , ..., l − .3. g is a k -isomorphism ( E , m − µ ) → ( E , m µ ) , i.e. m µ g = gm − µ . Proof
Since q ( λ ) = q (1 − λ ) = 0, there exists an automorphism g of E = k ( λ ) over k such that g ( λ ) = 1 − λ . Setting µ := λ − , we have λ = + µ and 1 − λ = − µ ; inparticular k ( λ ) = k ( µ ) and g ( µ ) = − µ . Note that − µ = µ , since − µ = µ would imply λ = 1 / ∈ k , and hence that [ k ( λ ) : k ] = deg( q ) = l = 1, contrary to the assumption l >
1. For the minimal polynomial r ( x ) = a l x l + a l − x l − + ... + a x + a (with a l = 1)of µ over k it follows that r ( − µ ) = r ( g ( µ )) = g ( r ( µ )) = 0, using that g fixes k . Thisimplies that only even powers of x occur in r ( x ), by comparing the coefficients of r ( − µ )and r ( µ ). Indeed, X k odd a k µ k = 12 ( r ( µ ) − r ( − µ )) = 0since r ( µ ) = r ( − µ ) = 0, so µ is a zero of δ ( x ) := P k odd a k x k , and hence r ( x ) | δ ( x ). If r ( x ) were to have odd degree, then δ ( x ) and r ( x ) would have the same degree and hencewould be equal (since they are both monic). But this would contradict the irreducibilityof r ( x ), because for a polynomial with only odd powers of x , one can always factor outthe polynomial p ( x ) = x . Therefore it must be that δ ( x ) ≡
0, i.e. a k = 0 for all odd k .The statement about the coordinate matrix of g is clear. For point 3., we check onthe basis elements µ i of E : m µ ( g ( µ i )) = m µ (( − i µ i ) = ( − i µ i +1 = − ( − i +1 µ i +1 = g ( m − µ ( µ i )) . Proposition 8.2.
Let = A ∈ k l × l such that q ( x ) = det( xI − A ) is irreducible. Thenthe following are equivalent:1. A and ( I − A ) t are similar to each other2. A and I − A are similar to each other3. q ( λ ) = q (1 − λ ) = 0 for some λ in some extension field of k .4. For any λ in any extension field of k : if q ( λ ) = 0 , then q (1 − λ ) = 0 .Suppose now that any of these equivalent conditions holds. Then deg q ( x ) is even.Moreover, setting ˜ A = A − I , the minimal polynomial r ( x ) of ˜ A satsifies r ( − ˜ A ) = 0 and, for any invertible T , T AT − = ( I − A ) t ⇔ T ˜ AT − = − ˜ A t . roof That 1. and 2. are equivalent follows from the fact that a matrix and itstranspose are always similar, and that similarity is a transitive relation. 2. implies that q ( I − A ) = 0, and thus that 3. holds in the field k ( A ) = k [ A ] ∈ k l × l . To see that 3.implies 4., it is enough to consider extensions k ( λ ) of k such that q ( λ ) = 0; these areall isomorphic via isomorphisms fixing k , and hence in each of them also the equation q (1 − λ ) = 0 is satisfied. Finally we show that 4. implies 2.. Since q ( A ) = 0, 4. implies q ( I − A ) = 0 and thus that q ( x ) is also the unique elementary divisor of I − A . Themultiset of elementary divisors is a complete invariant for similarity, so 2. follows. Thatdeg q is even and that r ( x ) satisfies r ( − ˜ A ) follows from Lemma 8.1. Remark 8.3.
In the subsequent, we’ll use the following fact. Let E ⊇ k be a finitefield extension, so E is a finite-dimensional vector space over k , and let τ : E → k be anon-zero k -linear map. Then the bilinear form on E defined by( u, v ) τ ( uv )is non-degenerate. Indeed, if for fixed v ∈ E \{ } we have τ ( uv ) = 0 for all u ∈ E , thenit would follow that τ ( u ′ ) = 0 for all u ′ ∈ E , since any u ′ can be written as u ′ v − v . Thiswould contradict the assumption that τ = 0. Proposition 8.4.
Let l > , and let = A ∈ k l × l be a Frobenius matrix with irreduciblecharacteristic polynomial r ( x ) such that r ( − A ) = 0 . Then there are skew-symmetric aswell as symmetric invertible T ∈ k l × l such that T AT − = − A t . Proof
We may restate the task as follows. We are given V = k ( µ ), dim V = l even,with irreducible monic r ( x ) = P li =0 a i x i such that r ( µ ) = r ( − µ ) = 0, in particular a i = 0 for odd i . Let m µ ( v ) = µv for v ∈ V . The task is to find an isomorphism β : V → V ∗ such that βm µ = − m ∗ µ β, (42)and such that the matrix T of β with respect to the basis 1 , µ, . . . , µ l − and its dualbasis 1 ∗ , µ ∗ , . . . is skew-symmetric resp. symmetric. Because of these bases, within thescope of this proof we use indices which always live between 0 and l − S such that SAS − = A t . Namely, one has the linear form τ ∈ V ∗ defined on the basis 1 , µ, . . . , µ l − by τ ( µ i ) = 1 if i = l − , τ ( µ i ) = 0 elseand we take S to be the matrix of the non-degenerate symmetric bilinear form ( u, v ) τ ( uv ), i.e. the entries of S are s ij = τ ( µ i + j ) 0 ≤ i, j ≤ l − . Let V be the k -subspace of V spanned by { µ i | ≤ i < l, i even } ; let V be the k -subspace spanned by { µ i | ≤ i < l, i odd } .78 laim 8.5. For any i ∈ N , µ i ∈ V if i is even, µ i ∈ V if i is odd. In particular τ ( µ i ) = 0 for all even i ∈ N . Proof of the claim . We use induction. Clearly, the claim holds for i < l . Also, notethat µ l = − P j
1. Thus T has entries t ij = s ij if j is even, t ij = − s ij if j is odd. In particular then, t ij = 0 if i + j is even (since s ij = 0if i + j is even) and t ij = − t ji if i + j is odd (since i and j must have different paritywhen i + j is odd). So T is skew symmetric.With respect to the basis 1 , µ, µ , . . . and its dual basis, S provides (according to[25] Prop. 4.4) an isomorphism ( V, m µ ) → ( V ∗ , m ∗ µ ), while by Lemma 8.1 D providesan isomorphism ( V, − m µ ) → ( V, m µ ). Thus T = SD gives an isomorphism ( V, − m µ ) → ( V ∗ , m ∗ µ ), as desired.With the skew-symmetric case done, we now turn to the symmetric case. Recall thatwe want to find β such that (42) holds; for this we work with the coordinate matrix T of β with respect to the basis 1 , µ, . . . , µ l − and its dual basis.For convenience, we write T = ( t ij ) i,j =0 ,...,l − . Summations will be for 0 , . . . , l − Claim 8.6.
Equation (42) holds for T if and only if all of the following are satisfied:1. t k,j +1 = − t k +1 ,j for j, k < l − P h − a h t kh = − t k +1 ,l − for j = l − , k < l − t l − ,j +1 = P h a h t hj for j < l − , k = l − − P h a h t l − ,h = P h a h t h,l − for j = k = l − Proof of Claim.
Let j, k ≤ l −
1. The matrix entries corrsponding to β ˆ µ and − ˆ µ ∗ β ,respectively, are x jk :=( β ˆ µ ( µ j ))( µ k ) = ( β ( µ j +1 ))( µ k ) y jk :=( − ˆ µβ ( µ j ))( µ k ) = − X i t ij ˆ µ ∗ ( µ i ∗ )( µ k ) = − X i t ij µ i ∗ (ˆ µ ( µ k )) = − X i t ij µ i ∗ ( µ k +1 ) , x jk = y jk for all j, k ≤ l −
1. Now, x jk = X i t i,j +1 µ i ∗ ( µ k ) = t k,j +1 if j < l − y jk = − X i t ij µ i ∗ ( µ k +1 ) = − t k +1 ,j if k < l − .x l − ,k = β ( µ l )( µ k ) = β ( X h − a h µ h )( µ k ) = X i,h − a h t ih µ i ∗ ( µ k ) = X h − a h t kh y j,l − = − X i t ij µ i ∗ ( µ l ) = − X i t ij µ i ∗ ( X h − a h µ h ) = X i,h t ij a h µ i ∗ ( µ h ) = X h t hj a h Thus, that x jk = y jk for all j , k is equivalent to the equations stated in the claim.Define ˜ τ ∈ V ∗ by ˜ τ ( µ i ) = 1 if i = l − , ˜ τ ( µ i ) = 0 else(note that we use that l > S whose entries are ˜ s ij =˜ τ ( µ i + j ) is invertible. Note that ˜ τ ( µ k ) = 0 for odd k ∈ N , since ˜ τ | V = 0. In particular,( − i +1 ˜ τ ( µ i + j ) = ( − j +1 ˜ τ ( µ i + j ) for all 0 ≤ i, j ≤ l −
1, since both sides are zero when i and j have different parity. Thus, if we define t ij := ( − i +1 ˜ τ ( µ i + j ) = ( − j +1 ˜ τ ( µ i + j )we obtain a symmetric matrix T . Note that T = − ˜ SD , so T is invertible. The entries t ij only depend on i + j , up to sign. For for i + j < l − i + j = l − t ij = ( − i = ( − j .It remains now only to check that this T fulfills the conditions of Claim 8.6:1. t k,j +1 = ( − k +1 ˜ τ ( µ k + j +1 ) = − t k +1 ,j for j, k < l − t k +1 ,l − = ( − k ˜ τ ( µ k + l ) = ( − k P h − a h ˜ τ ( µ k + h ) = P h a h ( − k +1 ˜ τ ( µ k + h ) = P h a h t kh for j = l − , k < l − t l − ,j +1 = ( − j ˜ τ ( µ j + l ) = ( − j P h − a h ˜ τ ( µ j + h ) = P h a h ( − j +1 ˜ τ ( µ j + h ) = P h a h t hj for j < l − , k = l − − P h a h t l − ,h = − P h even a h t l − ,h = 0 since l − h is odd for even h .Similarly, P h a h t h,l − = P h even a h t h,l − = 0 since l − h is odd for even h .80 .2 Generalized Jordan blocks In this section we recall some well-known facts about normal forms for endomorphisms.Let m, l ∈ N be natural numbers, and consider ml × ml -matrices A with l × l -blocks A ij . Let N be the standard nilpotent matrix of size ml . Then N l has blocks I l as firstupper off-diagonal of blocks, and all other entries zero, i.e. N l = O I l OO O I l . . . . . . . . . . Note that for any invertible matrix K ∈ k l × l , one has ( KI ) N l ( K − I ) = N l , where KI = K O OO K O . . . . . . . . . . Proposition 8.7.
Let η be an indecomposable endomorphism of the n -dimensionalvector space V over a perfect field k . Then1. There is unique m ∈ N and irreducible monic q ( x ) ∈ k [ x ] such that q ( x ) m is theminimal (= characteristic) polynomial of η ; that is, q ( x ) m is the unique elementarydivisor of η . In particular, ml = n where l = deg q ( x ) . Moreover, as an k [ x ] -module, V is isomorphic to k [ x ] /q ( x ) m ; in particular it is cyclic.2. Let q ( x ) m be the unique monic elementary divisor of η . There is a basis ¯ v of V with respect to which η has matrix A = ZI + N l = Z I l O · · · O Z I l . . . O O Z . . .... . . . . . . . where q ( x ) = det( xI i − Z ) . Conversely, for any basis with respect to which thecoordinate matrix of η is of the form A = ZI + N l , one has det( xI l − Z ) = q ( x ) and det( xI − A ) = q ( x ) m . Proof
1. This follows from the theory of a linear operator on a finite-dimensionalvector space. In general, the k [ x ]-module V is isomorphic to the direct sum of the k [ x ] /d i ( x ) where the d i ( x ) = q i ( x ) m i are the elementary divisors of η , the q i ( x ) beingcoprime irreducibles. The characteristic polynomial is Q i d i ( x ) = det( xI − A ). That η is indecomposable means that there is only a single elementary divisor q ( x ) m , and sothe characteristic polynomial and minimal polynomial of A coincide with q ( x ) m .2. For the existence of such a basis and corresponding normal form we use ourassumption that k is perfect; see [23] for a discussion of this normal form, and [27] and814] regarding necessary and sufficient conditions for its existence. Conversely, given sucha normal form for η , it follows from the behaviour of block matrices and determinantsthat q ( x ) m = det( xI − A ) = (det( xI l − Z )) m . By the uniqueness of factorization ofmonic polynomials into irreducible polynomials, and by comparing degrees, it followsthat det( xI l − Z ) = q ( x ). Remark 8.8.
We refer to the normal form given in part 2 above as “generalized Jordannormal form”. The matrix Z is, in coordinate form, a “generalized eigenvalue” of η . Theorem 8.9.
Given an indecomposable sextuple S η with no eigenvalue in k , the fol-lowing statements are equivalent:1. it is self-dual2. it admits compatible symplectic forms3. it admits compatible symmetric forms.In the case of self-duality, compatible ε -symmetric forms can be given, with respect tosuitable bases, by matrices of the form H M = O O − MO M O − M O O , with M = ... ... O O TO − T O · · ·
T O O · · · ∈ k ml × ml , where T ∈ k l × l and q ( x ) m is the characteristic polynomial of η , with q ( x ) irreducibleand deg q ( x ) = l . Clearly, ε ( H M ) = ε ( M ) = ( − m +1 ε ( T ) . Suitable bases are framebases { ¯ u, ¯ v, ¯ w } where the basis ¯ v = v , . . . , v ml renders η in normal form ( I l + ˜ Z ) I + N l with Frobenius matrix ˜ Z ∈ k l × l . Here, T can be chosen according to Proposition 8.4. Proof
Given compatible forms, self-duality is obvious. Conversely, consider self-dualindecomposable S η and consider a basis ¯ v such that the corresponding matrix of η is A = ZI + N l as in 2. of Proposition 8.7. Let q ( x ) m be the unique elementary divisorof η , where q is irreducible over k with deg q = l .By Proposition 6.17, A is similar to ( I − A ) t . Thus, q ( x ) m is also the uniqueelementary divisor of ( I − A ) t , implying in particular that det( xI l − ( I l − Z ) t ) = q ( x )(c.f. the proof of part 2. in Proposition 8.7). So Z and ( I l − Z ) t share q ( x ) as theirunique elementary divisor, and are hence similar. Now by Proposition 8.2, ˜ Z = Z − I l has irreducible characteristic polynomial r ( x ) with r ( − ˜ Z ) = 0.Since r is irreducible, there exists invertible K such that K ˜ ZK − is Frobenius.Note that KZK − = I l + K ˜ ZK − and KAK − = I + K ˜ ZK − I + N l . Thus we mayassume directly that η has matrix A = ZI + N l with Z = I l + ˜ Z where ˜ Z is a Frobenius82atrix with irreducible characteristic polynomial r ( x ) satisfying r ( − ˜ Z ) = 0. Moreover, T ZT − = ( I l − Z ) t if and only if T ˜ ZT − = − ˜ Z t for any invertible T (c.f. Proposition8.2). By Proposition 8.4, there exist T , skew-symmetric as well as symmetric, whichsatisfy T ˜ ZT − = − ˜ Z t . Choose such a T and let M be defined as above in the statementof this proposition. By Proposition 6.17, it suffices to show that M AM − = ( I − A ) t .For this, observe first that M = T QJ = QT J where J is block-diagonal withdiagonal blocks J jj = ( − j +1 I l and Q is block-anti-diagonal with blocks Q ij = I l for i + j = m + 1, and all other entries being 0. Also, observe that Q = I and that Q acts as permutation of blocks: acting on the right it exchanges block-columns j and m + 1 − j , acting on the left it exchanges block-rows i and m + 1 − i . Furthermore, thefollowing properties are easily seen: • QDQ = D if D is block-diagonal of the form XI for some X ∈ k l × l • QN l Q = ( N l ) t • J = I , and J DJ = D if D is block-diagonal • J N l J = − N l • T ( N l ) t T − = ( N l ) t since ( N l ) t has only unit and zero blocks.It follows that M ZIM − = QT J ZIJ T − Q = QT ZIT − Q = Q ( I l − Z ) t IQ = ( I l − Z ) t IM N l M − = T QJ N l J QT − = T Q ( − N l ) QT − = − T ( N l ) t T − = − ( N l ) t and hence M AM − = M ZIM − + M N l M − = ( I l − Z t ) I − ( N l ) t = I − A t , as desired. Corollary 8.10.
Up to isomorphism, the indecomposable sextuples underlying non-splitcontinuous-type isotropic triples are paremetrized by indecomposable linear endomor-phisms η of the form η = + ζ where ζ is an indecomposable endomorphism which lives in an even-dimensional spaceand is such that ζ is similar to − ζ ∗ . Such endomorphisms ζ are themselves parametrizedby the set { r m | m ∈ Z > , r monic irreducible and r = p ( x ) with p ∈ k [ x ] } ∪ { ( x ) m | m ∈ Z > } . Proof
From Proposition 6.17 we know that (isomorphism classes of) indecomposableself-dual continuous-type sextuples are in one-to-one correspondence with (isomorphism83lasses of) indecomposable endomorphisms η such that η is similar to 1 − η ∗ . Setting ζ := − η it easily seen that ζ is indecomposable if and only if η is, and that η similar to 1 − η ∗ ⇔ ζ similar to − ζ ∗ . Moreover, Theorem 8.9 and Theorem 7.1 imply that the underlying sextuples of non-split isotropic triples, up to isomorophism, are parametrized by all endomorphisms η as above, with the exception of those which have an eigenvalue in the ground field andlive in an odd-dimensional space.In the case when η has no eigenvalue in the ground field, it follows from the aboveproof of Theorem 8.9 and from Lemma 8.1 that the corresponding endomorphism ζ has a minimal polynomial of the form r ( x ) m , with r irreducible and with only evenpowers of the variable appearing, i.e. r ( x ) = p ( x ) for some p ∈ k [ x ]. (The irreduciblepolynomial r here is the minimal polynomial of the “eigenvalue” ˜ Z of ζ which appearsin the proof of Theorem 8.9; for Lemma 8.1 we use ˜ Z in the role of µ ).In the case when η does have an eigenvalue in the ground field, the correspondingendomorphism ζ is nilpotent, and hence, if it lives in an even-dimensional space, it hasa minimal polynomial of the form r ( x ) = ( x ) m for some m ∈ Z > In order to address the question of uniqueness of compatible forms, we first recall thestructure of endomorphism algebras of indecomposable sextuples S η . We’ve seen thatthese are isomorphic to endomorphism algebras of vector spaces with indecomposableendomorphism, i.e. endomorphism algebras of k [ x ]-modules V k [ η ] , where η is an inde-composable endomorphism of a vector space V over k . These are well-known; we recallsome basic facts. Proposition 8.11.
Given η, Z, A, ¯ v as in Proposition 8.7, the following hold:1. The endomorphism algebra E of V k [ η ] is isomorphic to the k -algebra k [ x ] /q ( x ) m ,which is of dimension n = ml = dim k V , where l is the degree of the irreducibleminimal polynomial q ( x ) of η . In particular, E is local with radical E · q ( x ) .2. With respect to ¯ v , E is given by the matrices C such that CA = AC . We have C ∈ E if and only if C = P k − i =0 Z i N li with Z i ∈ k ( Z ) , and this representationis unique. In particular, C is upper block triangular with diagonal blocks C = . . . = C kk = Z and C is invertible in E if and only if Z = 0 . Proof
1. Rephrasing the definitions, E is the collection of k -linear endomorphismsof the vector space V which commute with η . For η indecomposable, E is simply k [ η ] ⊆ End( V ), and hence is isomorphic to k [ x ] /q ( x ) m . One way to see this is to notethat since η is indecomposable, there exists a basis of V of the form { v, ηv, ..., η n − v } for some v ∈ V . For any f ∈ E , observe that f v = P n − i =0 c i η i v for some c i ∈ k ; itfollows then that in fact f = P n − i =0 c i η i since the latter expression coincides with f onthe above given basis of V . 84. The first statement is clear. To prove the first “if” it suffices to observe that, for W ∈ k ( Z ), one has W Z = ZW since k ( Z ) is commutative and W IN l = N l W I since N l has zero and unit blocks only. Thus, matrices C = P k − i =0 Z i N li form a subalgebra E ′ of E .To prove the “only if”, we show that E ′ and E have the same dimension over k .For this, we compute the dimension of E ′ as an k ( Z )-algebra. Since N lm = 0, induc-tion on m − h > I, N l , N l , . . . , N l ( m − are independent over k ( Z ): if P m − j = h Z j N lj = 0 then, applying N l , we get P m − j = h Z j N l ( j +1) = 0 and hence, using theinduction hypothesis, that Z h = ... = Z m − = 0. Thus, E ′ has dimension m over k ( Z )and so dimension ml = n = dim E over k , since k ( Z ) has dimension l over k . Thus, E ′ = E . In this section we generalize Lemma 3.26 in a way which applies to all poset represen-tations and also to endomorphisms (
V, η ). Recall (see Section 3.2) that for the latterthere is a natural notion of direct sum, morphism, etc., and that for an indecomposableendomorphism (
V, η ), its endomorphism algebraEnd(
V, η ) = { f : V → V | f η = ηf } is always local. For our application, we will use the following notion of the dual of anendomorphism: we define ( V, η ) ∗ := ( V ∗ , Id − η ∗ ) . Analogous to the definition for poset representation, a compatible form for (
U, η ) isan isomorphism B : ( V, η ) → ( V ∗ , Id − η ∗ ) which defines either to a symmetric orskew-symmetric bilinear form on V .Within the current subsection, the word “representation” and the notation “ ψ ” willbe used to denote a poset representation or an endomorphism (unless further specifica-tion is given) and similarly for compatible forms, etc.. Given a bilinear form B we write ε ( B ) = 1 if B is symmetric, and ε ( B ) = − B is skew-symmetric. Similar notationalso applies for matrices, we set A − t = ( A − ) t , and we use A ∗ = A t interchangeably.Consider now the following situation. Assume that we are given an indecomposablerepresentation ψ in V of dimension ml , a basis of V , and a subfield F of k l × l such thatthe endomorphism algebra E of ψ is given by block matrices (with l × l -blocks) of theform m − X i =0 Z i N li with unique Z i ∈ F and such that ZN l = N l Z for all Z ∈ F . In particular, E iscommutative and has radical rad E = EN l = P m − i =1 X i N li .Further, assume that we are given a compatible form for ψ , which we call B .Denote the corresponding coordinate matrix by H , which is itself necessarily also ablock matrix as above. 85 efinition 8.12. Let ψ , F , H be as above. For A ∈ k l × l , set A † := H − AH . Theanti-involution “ ( − ) † ” restricts to one on F via Z † := ( ZI ) † . We define F + H := { Z ∈ F | Z † = Z } F − H := { Z ∈ F | Z † = − Z } . Note that F + H = { Z ∈ F | Z t H = H Z } F − H = { Z ∈ F | Z t H = − H Z } . Remark 8.13. F = F + H ⊕ F − H by the usual argument: any Z ∈ F has a uniquedecomposition Z = ( Z + Z † ) + ( Z − Z † )into selfadjoint and anti-selfadjoint parts. Remark 8.14.
If 0 = Z ∈ F + H , then H := H Z determines a compatible form B suchthat ε ( B ) = ε ( B ), since in this case( H Z ) t = Z t H t = ε Z t H = ε H Z. Similarly, given 0 = Z ′ ∈ F − H , then H ′ := H Z defines an compatible form B ′ suchthat ε ( B ′ ) = − ε ( B ). Lemma 8.15.
Consider, in addition to B , another compatible form B for ψ , withassociated coordinate matrix H . Set ε := ε ( B ) , ε := ε ( B ) and ǫ = ε ε .If ǫ = 1 , there exists = Z ∈ F + H and an automorphism f of ψ which is an isometryfrom B to the compatible form given by H Z . Notation : In this case we say that B and B are equivalent up to automorphisms of ψ and multiplication with “scalars” in F . Proof
Let † denote the antiautomorphism given by the operation of adjoint withrespect to H , i.e. A † = H − A t H . Note that when A is in the endomorphism algebra E of ψ , then so is A † . Note also that ( H − ) † = H − t .Observe that H − H determines an automorphism of ψ , so H − H = C I − R for some invertible C ∈ F and R ∈ rad E. It follows that( C I ) † − R † = ( H − H ) † = H † ( H − ) † = H − H t H H − t = H − ε H H ε H − = ǫH − H = ǫC I − ǫR . Since E = F I ⊕ RadE and this decomposition is preserved under taking adjoints,( C I ) † = ǫC I and R † = ǫR ; in particular C ∈ F ǫH .Let ǫ = 1. Since A A † is an anti-automorphism of E , ( C − I ) † = ǫC − I = C − I .Set R := C − R = R C − , H := H C − , C := H − H = I − R R † = ( C − ) † R † = C − R = R and since R is nilpotent we can proceed as inLemma 3.26 and construct a unit h ∈ E such that h ∗ H h = H (where H here playsthe role of H in that Lemma). Setting f := h − and using that E is commutative, weobtain f ∗ H f = f ∗ H C f = f ∗ H f C = H C . Lemma 8.16.
Let B be as above, and assume that its matrix H is zero above the l × l -block anti-diagonal. Let B be another compatible form, with matrix H Z for somenon-zero Z ∈ F .There is an automorphism f of ψ which is an isometry from B to B if and onlyif there exist X ∈ F + H and Y ∈ F − H such that Z = X − Y . Proof If Z = X − Y with X ∈ F + H and Y ∈ F − H then define f by ( X + Y ) I . Inthis case, f ∗ H Zf = ( X + Y ) t H Z ( X + Y ) = ( X t + Y t ) H ( X + Y ) Z = ( X t + Y t )( X t − Y t ) H Z = Z t H Z = H , using in the last step that Z = X − Y implies that Z ∈ F + H .Conversely, let f ∈ End( ψ ) be an isometry from B to B . The matrix A of f is ofthe form A = P i Z i N i with Z i ∈ F and A t H A = H Z . Observe that N commuteswith the Z i and that applying N on the right of a matrix moves all columns of blocks onestep to the right (with overspill) and sets the first column to zero. Similarly, applicationof N t on the left shifts block rows downward. Thus, since H is zero above the blockanti-diagonal, ( N t ) j H N i has zero block anti-diagonal if i + j >
0. Now H Z = A t H A = X i,j Z tj ( N t ) j H N i Z i and H Z is zero above the block anti-diagonal, and on the block anti-diagonal the blocksare T ij Z = Z t T ij Z , with i + j = m + 1where the T ij denote the blocks of H . By hypothesis, Z = X + Y for some X ∈ F + H and Y ∈ F − H . It follows that Z = T − ij ( X t + Y t ) T ij ( X + Y ) = ( X − Y )( X + Y ) = X − Y . .6 Uniqueness for sextuples S η For indecomposable self-dual framed sextuples S η we obtain the following uniquenessresult for compatible forms by applying Lemma 8.15 and Lemma 8.16 to the underlyingendomorphism ( U, η ). minimal polynomial
Theorem 8.17.
Let S η be an indecomposable self-dual framed sextuple such that η hasno eigenvalue in k , and let q ( x ) m be the minimal polynomial of η , with q ( x ) irreducible, deg q ( x ) = l .Let B and B be compatible (skew)symmetric forms for S η . Set ε := ε ( B ) , ε := ε ( B ) , and ǫ := ε ε . Furthermore: • choose a basis of U which is as in Proposition 8.7 and extend this to a frame basisfor S η ; • define a subfield F ⊆ k l × l by F = k ( Z ) , with Z from Proposition 8.7.In terms of the respective coordinate matrices H and H of the compatible forms B and B , we have:1. If ǫ = 1 , then H and H are equivalent up to automorphisms of S η and multipli-cation with “scalars” in F + H .2. Given = K ∈ F , there is an automorphism f of S η which is an isometry from H to H K if and only if there are X ∈ F + H and Y ∈ F − H such that K = X − Y . Proof
In view of Proposition 8.7 and Propositions 8.11, we can apply Lemma 8.15to the underlying endomorphism (
U, η ). It follows then from Proposition 6.10 andProposition 6.17 that we can transfer the uniqueness statement from Lemma 8.15 tothe corresponding statement in part (1) above for compatible forms for S η .We turn to proving part 2. If K = X − Y for some X ∈ F + H and Y ∈ F − H , thenan isometry is given by ( X + Y )Id as in the proof of Lemma 8.16. So assume that H is equivalent (isometric) to H K .Without loss of generality we can assume that H is given by one of the canonicalcompatible forms defined in Theorem 8.9. Indeed, suppose that the statement to beproved were true for those canonical forms (and let H represent, for a moment, a formwhich is not necessarily such a canonical one). Suppose moreover that there exists anisometry f between H and H K . Choose the appropriate canonical compatible form H such that ε ( H ) = ε ( H ) = ε ( H K ). Then, by part 2. above, there exists some C ∈ F + H and an isometry g : H → H C . This will also be an isometry g : H K → H CK = H KC . From all this we obtain the isometry g − f g : H → H K. By assumption, this implies that K = X − Y for some X ∈ F + H and Y ∈ F − H .So we can assume H is a canonical compatible form. Now we wish to proceed in ananalogous manner as we did in proving parts 1 and 2, but this time using Lemma 8.16.It remains only to check that the hypotheses of Lemma 8.16 are satisfied by ( U, η ).88ote that from Theorem 8.9 and Proposition 6.10 it follows that the coordinatematrix H | U is of the form ... ... O O TO − T O · · ·
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