On a new approach to classification of associative algebras

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1 arxiv:math/ v1 [math.ra] 2 Mar 2002 On a new approach to classification of associative algebras Contents Vladimir Dergachev November 7, Introduction 1 2 The method 2 3 Properties of the spaces V k (α) 8 4 Multiplicative functionals, ideals A 9 5 Jacobson s radical spaces V k (α) 11 6 Analysis of matrix algebra 11 7 Analysis of B(H) 12 8 Regular functionals 13 1 Introduction The study of associative algebras has usually centered on studying subalgebras, ideals, homomorphisms identities - i.e. properties describable within the category of algebras their modules. One of the most famous results include the study of commutative algebras by use of multiplicative functionals primitive ideals the classification theorem stating that 1

2 any finite dimensional associative algebra is a semi-product of its Jacobson radical direct sum of matrix algebras over a skew-field. Unfortunately, these results do not help in a situaton when the researcher has a multiplication table for the algebra wishes to learn as much as possible about it. In this paper we introduce a method by using which one can decompose a finite dimensional associative algebra into subspaces which provide valuable information about the structure of the algebra. These subspaces generally will not be ideals, or even algebras, their value is in simplifying the operation of multiplication when considered in basis subordinate to this decomposition. 2 The method We start with a finite dimensional associative algebra A, which we will assume contains unity 1. Pick any basis {e i } in A. The matrix A = {e i e j } is the multiplication table of A written in the basis {e i }. Pick a linear functional A. This choice will affect the properties of the resulting decomposition. We define A to be the matrix {(e i e j )}. Let ker L A = {x A : y A A(xy) = 0} ker R A = {x A : y A A(yx) = 0} Nil = ker L A ker R A Theorem 1 The spaces ker L A, ker R A Nil obey the following relations with respect to multiplication in A: 1. ker L A A ker L A 2. A ker R A ker R A 3. ker L A ker R A Nil 4. ker L A Nil Nil 5. Nil ker R A Nil 6. Nil A ker L A 7. A Nil ker R A 2

3 Proof 1. Let x ker L A, y A. Then for all z A we have ((xy)z) = (x(yz)) = 0 2. Let x A, y ker R A. Then for all z A we have (z(xy)) = ((zx)y) = 0 3. True because the image should be a subset of both ker L A ker R A. 4. A special case of A special case of A special case of A special case of 1. Define N = dima K = dima dimnil. Let us now change the basis in such a way that the last N K vectors span Nil. Upon rewriting A in the new basis we discover that elements in the last N K columns in the last N K rows are all zero. Let Ã be the submatrix formed by the first K columns the first K rows. Since Nil was defined as two-sided kernel of A the matrix λ Ã + µ ÃT is non-degenerate can be considered as a matrix of a bilinear form on A/Nil. We define the characteristic polynomial χ (λ,µ) of A as ( ) χ (λ,µ) = det λ Ã +µ ÃT The characteristic polynomial is a homogeneous polynomial. The points (λ,µ) of the projective line in which it vanishes form the spectrum of A. or convenience we assume the following parameterization of P 1 : WenowdefineStab (α) = ker (λ,µ) = (1, α) (Ã α ),withstab ÃT ( ) = ker (ÃT ). The sign in front of alpha was chosen so that Stab (1) would correspond to the stabilizer of in Lie algebra A L. 3

4 Since, in general, Ã ÃT do not commute we should expect the need to use Jordan spaces. And indeed, even though Stab (α) are transversal, their sum is commonly smaller than A/Nil. Since χ (λ,µ) is non-zero there exists a value α 0 in which it does not vanish. We have Ã α ÃT = (Ã α 0 Ã T ) (α α 0 ) ÃT = ( ) = (Ã α 0 Ã T 1 )(α α 0 ) α 0 Ã T 1 ) ÃT α α 0 (Ã We define Ṽ k (α) to be the k-level Jordan space of the operator (Ã ) 1 α 0 Ã T ÃT corresponding to the eigenvalue Λ = 1 α α 0. Recall, that by definition we have ( 1 (Ã ) Ṽ 0 (α) = ker α 0 Ã T 1 ) ÃT α α 0 Ṽ k+1 (α) = { ( 1 x : y Ṽ k (α) α α 0 Simple transformations show that (Ã Ṽ 0 (α) = Stab (α) ) α 0 Ã T 1 } ÃT )x = y { } Ṽ k+1 (α) = x : y Ṽ k (α) (Ã α )x ÃT = (Ã α 0 Ã T )y with the special case { } Ṽ k+1 ( ) = x : y Ṽ k ( ) ÃT x = (Ã α 0 Ã T )y Since A is finite-dimensional the spaces Ṽ k (α) stabilize, we will use Ṽ(α) to denote Ṽ dima (α). We have A/Nil = α Ṽ(α) 4

5 We now define V(α) (V k (α)) to be the preimage of (V)(α) (correspondingly Ṽ k (α)) under the projection A A/Nil. Restating the conditions defining Ṽ k (α) we obtain: V 0 (α) = Stab (α) V k+1 (α) = { x A : y V k (α) z A (xz) α(zx) = (yz) α 0 (zy) } with the special case V k+1 ( ) = { x A : y V k ( ) z A (zx) = (yz) α 0 (zy) } Also, from the Jordan decomposition definition of V(α) we have α : Nil V(α) A/Nil = α V(α)/Nil Theorem 2 The spaces V(α) do not depend on the choice of α 0. urthermore, when considering V(α) for a particular α we can use any α 0 α. Also for α we can define V k+1 (α) as { } V k+1 (α) = x : y Ṽ k (α) (Ã α )x ÃT = ÃT y Proof Indeed, all statements hold for V 0 (α) = Stab (α) by definition. We have for a k 0 { } V k+1 (α) = x : y Ṽ k (α) (Ã α )x ÃT = (Ã α 0 Ã T )y Starting with (Ã α 0 Ã T )y 5

6 we obtain (Ã α 0 Ã T )y = = = = (Ã α 1 Ã T )y (α 0 α 1 ) ÃT y = (Ã α 1 Ã T )y + α 0 α 1 α α 1 (Ã α )y ÃT α 0 α 1 α α 1 (Ã α 1 Ã T )y = (Ã α 1 Ã T )y + α 0 α 1 α α 1 (Ã α 1 Ã T )z α 0 α 1 α α 1 (Ã α 1 Ã T )y = (Ã ) α 1 Ã T )(y + α 0 α 1 α α 1 z α 0 α 1 α α 1 y where z is either zero (when k = 0) or such that (Ã α )y ÃT = (Ã α 1 Ã T )z (from definition of V k (α)). or the special case α = we have (Ã (Ã α 0 Ã T )y = α 1 Ã T )y (α 0 α 1 ) ÃT y = (Ã = α 1 Ã T )y (α 0 α 1 )(Ã α 1 Ã T )z = (Ã = α 1 Ã T )(y (α 0 α 1 )z) The last statement is proved similarly. Starting with { } V k+1 (α) = x : y Ṽ k (α) (Ã α )x ÃT = (Ã α 0 Ã T )y we derive (Ã α 0 Ã T )y = (Ã α ÃT )y +(α α 0 ) ÃT y = = ÃT z +(α α 0 ) ÃT y = = ÃT (z +(α α 0 )y) (Ã where z is either 0 (when k = 0) or such that α )y ÃT = ÃT z. In the opposite direction, given { } V k+1 (α) = x : y Ṽ k (α) (Ã α )x ÃT = ÃT y 6

7 we compute ) Ã T 1 y = α 0 (Ã α α ÃT y 1 ) 1 = α 0 (Ã α α 0 Ã T z 1 (Ã ) = α 0 Ã T z y α 0 α α 0 α (Ã α 0 α (Ã where z is either 0 (when k = 0) or is such that (Ã α )y ÃT = (Ã α 0 Ã T )z α 0 Ã T )y = α 0 Ã T )y = 7

8 3 Properties of the spaces V k (α) We will now investigate how the spaces V k (α) interact with respect to the operation of multiplication. Theorem 3 or all α,β we have V k (α) V m (β) V k+m (αβ) Proof irst, let us establish this property for V 0 (α) = Stab (α) V 0 (β) = Stab (β). We have Stab (α) = {x 1 A : z A (x 1 z) α(zx 1 ) = 0} Stab (β) = {x 2 A : z A (x 2 z) β(zx 2 ) = 0} or all z A we have ((x 1 x 2 )z) αβ(z(x 1 x 2 )) = = (x 1 (x 2 z)) α((x 2 z)x 1 )+α(x 2 (zx 1 )) αβ((zx 1 )x 2 ) = 0 thus (x 1 x 2 ) Stab (αβ). Now let us apply induction. Assume that for all k+m < T the statement is true consider the case k +m = T. By theorem 2 we have V k (α) = { x 1 A : y 1 V k 1 (α) : z A (x 1 z) α(zx 1 ) = (zy 1 ) } V m (β) = { x 2 A : y 2 V m 1 (β) : z A (x 2 z) β(zx 2 ) = (zy 2 ) } or any z A we have: (x 1 x 2 z) αβ(zx 1 x 2 ) = = (x 1 (x 2 z)) α((x 2 z)x 1 )+α(x 2 (zx 1 )) αβ((zx 1 )x 2 ) = = (x 2 zy 1 )+α(zx 1 y 2 ) = = β(zy 1 x 2 )+(zy 1 y 2 )+α(zx 1 y 2 ) = = (z(βy 1 x 2 +y 1 y 2 +αx 1 y 2 )) By induction hypothesis we have (βy 1 x 2 +y 1 y 2 +αx 1 y 2 ) V T 1 (αβ) thus x 1 x 2 V T (αβ). 8

9 Theorem 4 or all α 0,β 0 we have V k (α) V m (β) V k+m (αβ) Proof Let us notice that if we create a new algebra A with new multiplication operation a b := b a the space V A (α) corresponding to the algebra A is exactly the space V A (1/α) corresponding to the algebra A. Thus, the statement of this theorem is a consequence of theorem 3. The symmetry α 1/α is also reflected in the dimensions of spaces V(α) Stab (α). Because of the fact that ( ) det λ Ã +µ ÃT = det ( ) λ Ã T ) +µ ÃT = det (λ ÃT +µ Ã theory of Jordan decomposition (stating that multiplicities of roots of characteristic polynomial are equal to dimv(α)/nil ) we must have dimv(α)/nil = dimv(1/α)/nil dimv(α) = dimv(1/α) Similarly for Stab (α) we have ( ) dimstab (α) = dimker λ Ã +µ ÃT = ( ) dim ker λ Ã T +µ ÃT = ( ) dim ker λ ÃT +µ Ã = dimstab (1/α) due to the well known fact ranka = ranka T of linear algebra. 4 Multiplicative functionals, ideals A An important classical method of exploring associative algebras is the study of ideals homomorphisms. Multiplicative functionals (homomorphisms 9

10 into the base field) have long played an important role in commutative algebras - especially in commutative C -algebras. It is easy to see that when is a multiplicative functional A is symmetric has rank 1. It turns out that all functionals with rank A = 1 are multiplicative. Theorem 5 Assume that is such thatker L A = ker R A = Nil. Then Nil is an ideal. Proof Indeed this is so because of statements 6 7 of theorem 1. Theorem 6 Let be a functional on A such that (1) = 1 rank A = 1. Then is multiplicative. Proof Let us count the dimensions. Since rank A = 1 it must be that dima/ker L A = dima/ker R A = 1 dima/nil 2. Also, since 1 is always an element of Stab (1) (Nil ) = 0 it must be that dimstab (1)/Nil 1. Yet, since the factor spaces ker L A /Nil, ker R A /Nil Stab (1)/Nil are transversal we must have dimker L A /Nil +dimker R A /Nil +dimstab (1)/Nil 2 The only way this is possible is when dimker L A /Nil = dimker R A /Nil = 0 Thus Nil is an ideal ker = Nil. Consequently, as (1) = 1, must be multiplicative. Note that this theorem holds even for an infinite-dimensional algebra, the cause being that, due to the condition rank A = 1, the bulk of A is residing inside Nil hence all factor spaces have finite dimension. 10

11 5 Jacobson s radical spaces V k (α) The following famous result (see [?]) describes the structure of finite dimensional associative algebras: Theorem 7 Any finite dimensional associative algebra is a semi-direct sum of its Jacobson radical direct sum of matrix rings over skew-fields. Since full matrix rings are well-studied the question arises whether the spaces V k (α) provide anything beyond this result. or example, the Jacobson s radical would be contained within Nil for any produced by the pullback from the dual space to the factor of A by its Jacobson s radical. However, it is easy to see that must vanish completely on any space V(α), when α 1. As Jacobson s radical is an invariant feature of an associative algebra there are plenty of functionals which do not vanish on it. 6 Analysis of matrix algebra We will now consider the case when A = Mat n (C). The dual space A can be modeled as Mat n (C), with (X) = tr(x) Consider the case of generic, i.e. when is invertible has distinct eigenvalues {ν i }. Observe that conjugation is an algebra isomorphism - thus similar produce equivalent decompositions. (This is not specific to Mat n (C), but common to all algebras). Pick a basis that diagonalizes. Let {e ij } be the matrix elements in this basis. We have: e ij e km = δ jk e im 11

12 The matrix (e ij e km ) has the form: e ii e + ij e ij ν ν n e ii ν j 0 e + ij ν j ν i 0 e ij ν i where e + ij denote those e ij with j > i e ij denote e ij with j < i. It is easy to see that the Nil = 0. The characteristic polynomial is χ (λ,µ) = ( 1) n(n 1) 2 (λ+µ) n ν i (λν i +µν j ) = ( 1) n(n 1) 2 (λν i +µν j ) i i,j We find that V(1) is spanned by e ii, V(ν i /ν j ) = e ij C. 7 Analysis of B(H) i j We will now consider the algebra B(H) of bounded operators on countable Hilbertian space. The dual space of continuous functionals on this algebra is dependent upon the topology defined on B(H): for normed topology it consists of all operators of trace class, with the familiar formula (X) = tr(x). or weak strong operator topologies (which are both weaker than normed one) the dual space is restricted to operators of finite rank. Obviously, in the latter case, the Nil space will have finite codimension - the rest of analysis will follow closely the case of Mat n (C). The case of normed topology is also similar to Mat n (C). Operators of trace class are compact. Assume to have point spectrum with simple multiplicities 0 to be the only value outside of point spectrum. The characteristic polynomial is not well-defined. However the spaces V(α) are. V(1) is the space of all bounded operators that commute with. The space V(α = ν i /ν j ), where ν i,ν j belong to the point spectrum of, has dimension 1. 12

13 The remarkable property of this situation is that the only functionals available are degenerate ones - i.e. those for which λ A + µ A T is never invertible. 8 Regular functionals ollowing [3] we first prove the following lemma: Lemma 8 Let W be a vector ( subspace of ) A. ix λ µ. The set R of all A such that W ker λ A +µ A T 0 is closed in A. Proof Let e 1,...,e n be the basis of A such that the first p vectors form the basis of W. The system of equations is equivalent to ( ) p λ A +µ A T ǫ i e i = 0 i=1 p ǫ i (λ(e k e i )+µ(e i e k )) = 0 i=1 where k ranges from 1 to n. The matrix of the latter system has entries that are linear in. The existence of the non-zero solution is equivalent to requirement that all p-minors vanish. This proves that the set R can be defined as a solution to the system of polynomial equations. Hence it is closed. Theorem 9 Let S be a subspace ( of A. Let ) be such that for a fixed pair λ 0,µ 0 the dimension of ker λ 0 A +µ 0 A T is the smallest among functionals in small neighbourhood of inside( affine set +S. ) Then for any G S, any x ker λ 0 A +µ 0 A T any y ( ) ker µ 0 A +λ 0 A T we have G(λ 0 xy +µ 0 yx) = 0 13

14 Proof Pick any G S. Let ǫ = +ǫg. Choose any ( subspace W ) A of complementary dimension that is transversal to ker λ 0 A +µ 0 A T. ) The set of ǫ for which V ǫ = ker (λ 0 A ǫ +µ 0 A Tǫ remains transversal to W is open by lemma 8 thus contains a small neighbourhood of 0. Let {e i } be the basis( of A with first p) vectors forming the basis of W. ) Pickx ker λ 0 A +µ 0 A T. AsW ker (λ 0 A ǫ +µ 0 A Tǫ form a decomposition of A there exists w W such that ) x w ker (λ 0 A ǫ +µ 0 A Tǫ I.e. for all i Hence λ 0 ǫ ((x w)e i )+µ 0 ǫ (e i (x w)) = 0 λ 0 ǫ (we i )+µ 0 ǫ (e i w) = λ 0 ǫ (xe i )+µ 0 ǫ (e i x) The homogeneous part) of this system is the linear system that defines W ker (λ 0 A ǫ +µ 0 A Tǫ. As this intersection is trivial the rank of it must be p. Noticing that the right h part can be derived by substituting w = x (i.e. our system is of the type Qw = Qx) we conclude that there is a unique solution w ǫ. As ǫ is linear ( in ǫ w ǫ is a rational ) function in ǫ. Pick now any y ker λ 0 A +µ 0 A T. We have: Differentiating by ǫ produces: λ 0 ǫ ((x w ǫ )y)+µ 0 ǫ (y(x w ǫ )) = 0 λ 0 G ǫ ((x w ǫ )y)+µ 0 G ǫ (y(x w ǫ )) (λ 0 ǫ (w ǫ y)+µ 0 ǫ (yw ǫ )) = 0 After setting ǫ = 0 we have: λ 0 G(xy)+µ 0 G(yx) = 0 Corollary 1. Let be a functional such that Stab (α) has the smallest dimension in a neighbourhood of. (α is fixed). Then for all x Stab (α) y Stab (1/α) we have: xy αyx = 0 14

15 Corollary 2. Let be a functional with the smallest dimension of Stab (1). Then Stab (1) is a commutative subalgebra of A. Corollary 3. Let be a functional with the smallest dimension of Stab (0). Then Stab (0) Stab ( ) = {0} InthiscaseNil isasubalgebraofawithtrivial(identically 0)multiplication law. References 1. Vladimir Dergachev Alexre Kirillov, Index of Lie algebras of Seaweed type, to appear in Journal of Lie theory. 2. Vladimir Dergachev, Some properties of index of Lie algebras, math.rt/ J.Dixmier, Enveloping algebras, American Mathematical Society, Providence (1996) Elashvili, A. G, robenius Lie algebras, unktsional. Anal. i Prilozhen. 16 (1982), , On the index of a parabolic subalgebra of a semisimple Lie algebra, Preprint, I.M.Gelf A.A.Kirillov, Sur les corps liés aux algébres enveloppantes des algébres de Lie, Publications mathématiques 31 (1966) R.S. Pierce, Associative algebras, Springer-Verlag, New York,

Some properties of index of Lie algebras

arxiv:math/0001042v1 [math.rt] 7 Jan 2000 Some properties of index of Lie algebras Contents Vladimir Dergachev February 1, 2008 1 Introduction 1 2 Index of Lie algebras 2 3 Tensor products 3 4 Associative