On the closures of orbits of fourth order matrix pencils

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1 On the closures of orbits of fourth order matrix pencils Dmitri D. Pervouchine Abstract In this work we state a simple criterion for nilpotentness of a square n n matrix pencil with respect to the action of SL n (C) SL n (C) SL 2 (C). The orbits of matrix pencils are classified explicitly for n = 4 and the hierarchy of closures of nilpotent orbits is described. Also, we prove that the algebra of invariants of the action of SL n (C) SL n (C) SL 2 (C) on C n C n C 2 is isomorphic to the algebra of invariants of binary forms of degree n with respect to the action of SL 2 (C). 1. Introduction Throughout this paper we focus on the natural linear representation of the group G = SL n (C) SL n (C) SL 2 (C) in the complex vector space V = C n C n C 2. If bases in C n, C n, and C 2 are chosen, then the components T ijk of a tensor T V form two square matrices X and Y, whose entries are T ij1 and T ij2, respectively. An element of V can be regarded as a pair of complex n n matrices. Then it is called a matrix pencil of order n and is denoted by λx + µy, where λ and µ are varying coefficients. The polynomial morphism ϕ: V W = S n (C 2 ) that takes each matrix pencil λx + µy to the binary form det(λx +µy ), is dominant and G-equivariant if we assume that SL n (C) acts on W trivially. Hence, the corresponding morphism ϕ : C[W ] G C[V ] G of the algebras of invariants is an embedding. Two matrix pencils of order n are said to be equivalent (respectively, strictly equivalent) if they are equivalent with respect to the action of SL n (C) SL n (C) SL 2 (C) (respectively, the action of SL n (C) SL n (C)). Here we state some basic results on matrix pencils. The direct sum of a n 1 m 1 matrix pencil P 1 = λx 1 +µy 1 and a n 2 m 2 matrix pencil P 2 = λx 2 +µy 2 is the (n 1 +n 2 ) (m 1 +m 2 ) matrix pencil P 1 + P 2 = λ(x 1 X 2 ) + µ(y 1 Y 2 ), where Z 1 Z 2 denotes the diagonal block matrix composed of Z 1 and Z 2. A matrix pencil is said to be indecomposable if it cannot be represented as a direct sum of non-trivial matrix pencils. We also consider n 0 and 0 m matrix pencils; by this we mean that if such pencil is present in a direct sum, then the corresponding matrices are given rows or columns of zeroes [3]. A matrix pencil λx + µy is called regular if it is square and det(λx + µy ) is not equal to zero identically. Otherwise, the pencil is called singular. It is known that every matrix pencil is strictly equivalent to the pencil L ε L εq + L ε L ε q + R R s, (1) where ε 1,..., ε q and ε 1,..., ε q are the minimal indices of columns and rows, respectively (we recall their definitions in the next section); L r is an indecomposable singular r (r + 1) 1

2 matrix pencil......, (2) and R 1,..., R s are indecomposable regular pencils [3]. The set of indecomposable pencils in (1) is defined unambiguously up to a transposition. A matrix pencil is said to be completely singular if there are no regular terms in (1). Every matrix pencil P is strictly equivalent to the pencil P r + P s, where P r is a regular matrix pencil, and P s is a completely singular matrix pencil. We recall the definition of minimal indices of rows and columns. For any matrix pencil P = λx + µy we consider the C[λ, µ]-module Ker P that consists of all x(λ, µ) (C[λ, µ]) n such that (λx + µy )x(λ, µ) = 0 (3) for any λ, µ C. The module Ker P is free [3]. A fundamental set of solutions of (3) is a minimal system of homogeneous generators of Ker P. The degrees ε 1,..., ε k of (any) fundamental set of solutions of (3) sorted in ascending order are called the minimal indices of columns of λx + µy. The minimal indices of columns of P = λx + µy are called the minimal indices of rows of λx + µy. 2. Nilpotent matrix pencils From now on we discuss only square matrix pencils. Consider a linear representation of a reductive algebraic group H in a vector space M. An element x of M is called nilpotent if the closure O x of its orbit O x contains zero element. The set of all nilpotent elements of M is called the nullcone and is denoted by N M. The nullcone plays an important role in the theory of invariants: it has the greatest dimension among all fibers of the factorization map, and the modality of the action on any fiber does not exceed the modality of the action on the nullcone [1]. Lemma 1. A singular matrix pencil is nilpotent. Proof. First of all, if ε 1 = 0 or ε 1 = 0, then the canonical form of the pencil contains a row or a column of zeroes. Therefore, the pencil is nilpotent. Now we prove that if r = ε 1 > 0 and s = ε 1 > 0, then the closure of the pencil s orbit contains zero element. Multiply the columns of L r by t s, the columns of L s by t r 1, the rows of L r by t s, the rows of L s by t r+1 except for the last, and the last row of L s by t s. This transformation affects only the last element of the last row of L s : it is multiplied by t 1 r s. Taking the limit t we get a row of zeroes. This completes the proof. Theorem 1. The regular pencil λx +µy is nilpotent if and only if the binary form det(λx + µy ) is nilpotent. Proof. It is obvious that ϕ(n V ) N W. Now we prove that ϕ 1 (N W ) N V. Suppose that λx + µy is mapped to a nilpotent binary form. We need to show that λx + µy N V. 2

3 0 -q q Figure 1: The mechanical interpretation of (4). If λx + µy is singular, then λx + µy N V. Otherwise, λx + µy is equivalent to the pencil λe + µd, where E is the identity matrix and D is an upper triangular matrix with λ 1,..., λ n on the diagonal. Since O λe+µd O λe+µd, where D = diag(λ 1,..., λ n ), it is enough to prove that O λe+µd 0. It is well known that a binary from is nilpotent if and only if it contains a linear factor whose multiplicity is greater than the half of the degree of the form. Therefore, at least s = [n/2] + 1 of λ 1,..., λ n are equal to each other. We can assume that these are λ 1,..., λ s, that is, λ 1 =... = λ s. With a suitable transformation D D λ 1 E, they are equal to zero. Now we prove that λe + µd is nilpotent. Consider the one-parameter subgroup g t = (A t, B t, C t ), where A t = diag(t α 1,..., t α n ), B t = E, C t = diag(t q, t q ), and α i = 0. Then g t (λe + µd) = λ (diag(t q+α 1,..., t q+α n )) + µ (diag(0,..., λ s+1 t q+α s+1,..., λ n t q+α n )). In order to make g t (λe + µd) approach zero when t, we need to choose α 1,..., α n so that all powers of t in the previous formula are negative. Thus, the following conditions must hold: α 1,..., α n < q α s+1,..., α n < q. (4) α α n = 0 We can confirm that α 1,..., α n and q satisfying this condition exist using the ballance scale shown in figure 1. We shall distribute n material points of equal mass on it so that the scale stays in equilibrium. The first s points shall be placed to the left of the point q on the right arm, and the remaining n s points shall be placed to the left of the point q on the left arm. It is clear that the desired distribution exists if the right arm gets more points than the left one (s > n/2). 3. Classification of orbits In [2] we classified the nilpotent orbits of fourth order matrix pencils using embedding of the original representation in the adjoint representation of the simple Lie algebra of type E 7. In this section we develop a method that is applicable to arbitrary matrix pencils. Lemma 2. If the matrix pencil P = λx + µy is completely singular, then it is strictly equivalent to the pencil P = λ(t 1 X) + µ(t 2 Y ) for any t 1, t 2 C. Proof. Without loss of generality we can assume than P is L r + L s. It is enough to prove that λx + µy is strictly equivalent to λx + µ(ty ). We put γ = t r+s 2, A = diag(γ 1, γ 1 t 1,..., γ 1 t 1 r, γ, γt,..., γt s ) B = diag(γ, γt,..., γt r, γ 1, γ 1 t 1,..., γ 1 t 1 s ). (5) 3

4 It is clear that (A, B) SL n (C) SL n (C). Under the action of (A, B) the second matrix of λx + µy, that is, Y, is multiplied by t, and the first matrix doesn t change. Theorem 2. Singular matrix pencils P 1 and P 2 are equivalent if and only if P1 s is equivalent to P2 s and P 1 r is equivalent to kp 2 r for some k C. Proof. Let P 1 and P 2 be equivalent. It follows from [4, theorem 3] that P1 s is equivalent to P2 s with respect to the action of GL n(c) GL n (C), and P1 r is equivalent to P 2 r with respect to the action of GL n (C) GL n (C) GL 2 (C). Hence, P1 s is equivalent to P 2 s by lemma 2 and P1 r is equivalent to kp 2 r for some k C. Conversely, if P 1 r is equivalent to kp 2 r for some k C and P1 s is non-trivial, then P 1 is mapped to tp2 s + P 2 r with a suitable transformation that belongs to SL n (C). To conclude the proof, it remains to apply lemma 2. The classification of orbits of matrix pencils is now reduced to the classification of orbits of completely singular matrix pencils and the classification of orbits of regular matrix pencils. Since a completely singular matrix pencil is defined unambiguously by the minimal indices of its rows and columns, the former problem is reduced to combining the blocks L r and L s to fill a square matrix of a given size. The classification of orbits of regular matrix pencils can be obtained easily using the idea of elementary divisors [4]. However, we will need another method in order to describe the closures of the orbits. Every regular matrix pencil λx +µy is equivalent to the pencil λe +µd, where E is the identity matrix, and D is a Jordan matrix. Two matrix pencils λe + µd and λe + µd are equivalent with respect to the action of G if and only if there exist matrices A, B SL n (C), and C SL 2 (C) such that { A(c11 E + c 12 D)B 1 = E A(c 21 E + c 22 D)B 1 = D (6), where c ij are the entries of C 1. This condition is equivalent to the existence of the matrices A SL n (C) and C SL 2 (C) such that A(ψ C (D))A 1 = D and where ψ C is a fractional transformation defined by det(c 11 E + c 12 D) = 1, (7) ψ C (X) = (c 21 E + c 22 X)(c 11 E + c 12 X) 1. (8) Lemma 3. If matrix pencils λe + µd and λe + µd are equivalent and D is a Jordan matrix with eigenvalue α, then D also is a Jordan matrix, whose eigenvalue is ψ C (α). Proof. Let D be αe + N, where N is a nilpotent matrix. Then D = ψ C (D) = ( (c 21 + c 22 α)e + N )( (c 11 + c 12 α)e + N ) 1 = ψc (α)(e + γ 1 N + γ 2 N ). Since Ker N Ker N k, it follows that Ker (D αe) Ker (ψ C (D) ψ C (α)e). We have dim Ker (D αe) = dim Ker (ψ C (D) ψ C (α)e) = 1, since ψ C 1(ψ C (D)) = D. Therefore, D is a Jordan matrix, whose eigenvalue is ψ C (α). This lemma proves the following Theorem 3. The matrix pencils λe + µd and λe + µd are equivalent if and only if the eigenvalues of D can be mapped to the eigenvalues of D by a fractional transformation that satisfies (7) and the dimensions of the corresponding Jordan blocks are equal. 4

5 We conclude that the equivalence class of a matrix pencil is defined by 1) the set (α 1,..., α s ) of (distinct) eigenvalues defined up to a fractional transformation that satisfies (7) and the set (k 1,..., k s ) of their multiplicities 2) the partition (k i1,..., k ili ) of each of k i by the corresponding Jordan blocks. In the next section we will describe the hierarchy of closures of nilpotent orbits when n 4 (if n > 4, then the number of nilpotent orbits is infinite [5]). By theorem 1, a regular nilpotent matrix pencil of order n 4 has up to two distinct eigenvalues, and any pair of the eigenvalues can be mapped to another pair by a fractional transformation that satisfies (7). Therefore, the equivalence class of a regular nilpotent matrix pencil is defined by the partitions (k 1,..., k s ) and (k i1,..., k ili ) only. We denote it by R k1 (k 11,..., k 1l1 ) R ks (k s1,..., k sls ). The integers k ij that are equal to 0 or 1 are omitted in this expression. For instance, the matrix pencil that has Jordan blocks of order 1 and 2 with the eigenvalue α 1 and a Jordan block of order 1 with the eigenvalue α 2 α 1 is denoted by R 3 (2) + R 1. The classification of the orbits of the fourth order matrix pencils is shown in table 1. Here dim denotes the dimension of the orbit, R and r are the rank invariants (see section 4.), ε i and ε i are the minimal indices of rows and columns, respectively. In the canonical form of a matrix pencil the terms L 0 and L 0 are omitted, and P + P is abbreviated as 2P. For instance, 2L 1 stands for L 1 +L 1 +L 0 +L 0. The dimension of the orbit of a matrix pencil can be calculated if the dimension of its stationary subgroup is known. The later is equal to the dimension of the centralizer of the pencil in the tangent algebra of G. We calculated (using a computer) the rank of the linear system that defines the centralizer, thereby obtaining its dimension. 4. Closures of nilpotent orbits The hierarchy of the orbits closures is represented as a graph, whose vertices are orbits. The vertices O 1 and O 2 are connected with an arrow if and only if the closure of O 1 contains O 2 and there are no intermediate orbits, that is, orbits O such that O 1 O O 2. We obtained the graph for the fourth order matrix pencil using the following lemmas and the fact that the closure of an orbit can contain only lower-dimensional orbits. Lemma 4. If the closure of the orbit of P contains P, then dim Ker m P dim Ker m P, where Ker m Q denotes the vector space of the homogeneous elements of C[λ, µ]-module Ker Q of degree m. Proof. Let P t O P be a sequence of matrix pencils that converges to P. All dim Ker m P t are equal to each other, as P t belong to the same orbit. Since the Grassman manifold is compact, there exist a converging subsequence of Ker m P t, whose limit point belongs to Ker P. Therefore, dim Ker m P dim Ker m P. Corollary 1. If the orbit of P contains P and ε 1 (P ) = ε 1 ( P ),..., ε k 1 (P ) = ε k 1 ( P ), then ε k (P ) ε k ( P ). For a matrix pencil P = λx + µy we put r(p ) = min rk(λx + µy ) (λ,µ) (0,0) R(P ) = max rk(λx + µy ). (9) (λ,µ) 5

6 Canon. form dim R r ε i ε i Canon. form dim R r ε i ε i 1 R 1 + R 3 (3) L 1 + R R 4 (4) L 1 + R R 1 + R 3 (2) R 3 (3) R 4 (3) R 1 + R L 1 + L R 3 (2) L 1 + L L L L 1 + L L L L 1 + L 1 + R L L 2 + R L L 2 + R L 1 + R R 1 + L L 1 + R R 1 + L R R 4 (2, 2) R R R R 1 + R R 2 (2) L 1 + R 2 (2) R L 1 + R 2(2) L R 1 + R 2 (2) L R 4 (2) R Table 1: Nilpotent matrix pencils of order 4. It is clear that r(p ) and R(P ) are invariant under G. Lemma 5. If the orbit of P contains P, then r(p ) r( P ) and R(P ) R( P ). Proof. Let sequence P t = λx t + µy t O P converge to P = λ X + µȳ. It is obvious that r(p t ) = r(p ) and R(P t ) = R(P ). For every t we put M t = {(λ : µ) P 1 rk(λx t + µy t )) = r(p )}. The sequence M t of non-empty sets has at least one limit point (λ 0 : µ 0 ) in P 1. Then, we have r(p ) rk(λ 0 X + µ0 Ȳ ) r( P ). The inequality R(P ) R( P ) is trivial. Lemma 6. Let P = P r + Z m, where Z m is a zero matrix pencil of order m. If the closure of the orbit of P contains P = P r + Z m, then the number of distinct eigenvalues of P r is smaller than or equal to the number of distinct eigenvalues of P r. Proof. It follows from (6) that if the sequence g t = (A t, B t, C t ) of elements of G is such that lim g t P = P, then lim g tp r = P r, where A t and B t are main submatrices of A t and B t, and g t = (A t, B t, C t ). Therefore, the closure of SL 2 (C)-orbit of the binary form ϕ(p r ) contains ϕ( P r ). The number of distinct eigenvalues of P r is equal to the number of linear factors in ϕ(p r ), which does not increase when we take the limit. 6

7 R 4 (4) R 1 +R 3 (3) L 3 L 2 + L 1 T R 4 (3) R 1 +R 3 (2) L 2 + R 1 L 1 +L 1T + R 1 R 4 (2,2) L 1 + 2R 1 R 4 (2) R 1 +R 3 L 1 + R 2 (2) L 1 + R 2 3R 1 2L 1 R 1 + R 2 (2) L 2 L 1 +L 1 T R 3 (3) R 3 (2) R 1 + R 2 L 1 + R 1 R 4 2R 1 R 3 L 1 R 2 (2) R 1 R 2 0 Figure 2: The hierarchy of closures of nilpotent orbits. 7

8 We build the hierarchy of closures as follows. Running over all orbits sorted in ascending order by their dimensions, we check whether an orbit of smaller dimension belongs to the closure of the given orbit. In order to do this, we either construct the sequence that converges to the desired orbit or prove that it is impossible using lemmas 4-6. Here we give some examples: 1. L 2 / L 1 + L 1, since dim Ker 0 L 2 = 2 and dim Ker 0 (L 1 + L 1 ) = R 1 + R 3 / R 4 (4), since R 1 + R 3 has two distinct eigenvalues, while R 4 (4) has three. 3. 2L 1 / R 4 (2), since r(2l 1 ) > r(r 4 (2)). 4. R n (k) R n 1 (k 1) + R 1 and, in particular, R 4 (2) R 1 + R 3. t 2 λ λ λ + tµ λ λ when t 0. λ + tµ λ 5. L 2 2L 1. If ε 1 (P ) = ε 2 (P ) = 1, the fundamental set of solutions for P is {(µ, 0, λ, λ), (0, µ, µ, µ 1)}, ε 1 ( P ) = 2 and the fundamental set of solutions for P is {(µ 2 1, µ, λ, 0)}, then ( ) ( ) µ µ when t 0. tµ 0 The hierarchy of orbits closures is symmetric with respect to the operation of transposition of matrix pencil. It follows from theorem 3 that if the matrix pencil P is regular, then P is equivalent to P. Therefore, it is enough to apply transposition to the singular part of the pencil. In figure 2 we save some space by showing only a part of the graph. To get the complete graph one needs to transpose all pencils shown in figure 2, add a copy of the original graph to the graph obtained, and identify the vertices that have the same labels. New edges do not appear when two parts of the graph are merged. However, it doesn t imply that if P 1 and P 2 belong to the different parts of the graph, then O P1 is not contained in the closure of O P2 or vice versa. For instance, the closure of L 1 + 2R 1 contains L 1 + L 1, whose closure, in turn, contains both L 1 + R 1 and L 1 + R Invariants In this section we prove that the algebra of invariants of G acting in the space of square matrix pencils of order n is isomorphic to the algebra of invariants of SL 2 (C) acting in the space of the binary forms of degree n. Lemma 7. The extension ϕ : C[W ] G C[V ] G is integral. Proof. Denote by I the ideal of C[V ] G that is generated by the image under ϕ of the maximal homogeneous ideal of C[W ] G. It follows from theorem 1 and Hilbert s Nullstellensatz that the radical of I coincides with the ideal of C[V ] G that is generated by all its homogeneous elements of positive degrees. Then by Hilbert s theorem [6] the extension ϕ : C[W ] G C[V ] G is integer. 8

9 Theorem 4. The extension ϕ : C[W ] G C[V ] G is an isomorphism. Proof. It follows from theorem 3 that a regular generic matrix pencil can be transformed to the pencil λe + µd, where E is the identity matrix, and D is diagonal matrix whose eigenvalues are distinct. The matrix pencil λe + µd is equivalent to the matrix pencil λe+µd if and only if there exist a fractional transformation that satisfies (7) and transforms the diagonal elements of D to the diagonal elements of D. The binary forms (λ + µx i ) and (λ + µy i ) are equivalent if and only if {x 1,..., x n } and {y 1,..., y n } are obtained from each other by a fractional transformation that preserves the highest coefficient of the form. It is clear that the later condition is equivalent to (7). Therefore, ϕ establishes oneto-one correspondence between generic orbits of G in V and W. Hence, it induces the isomorphism of the fields of rational G-invariants on V and W. These fields coincide with the corresponding quotient fields, as C[W ] and C[V ] are factorial and G doesn t have nontrivial characters [1, theorem 3.3]. To conclude the proof, it now remains to apply lemma 7 and note that C[W ] G is integrally closed. The author acknowledges Steven Epstein for carefully reading the translated manuscript. References [1] E. B. Vinberg and V. L. Popov, Invariant theory, Itogi Nauki i Tekhniki. Sovrem. problemy Matematiki, Fundamental nye napravleniya, vol. 55, VINITI, Moscow 1989, pp [2] D. D. Pervushin, Invariants and orbits of the standard (SL 4 (C) SL 4 (C) SL 2 (C))- module, Izv. Ross. Akad. Nauk Ser. Mat. 64:5 (2000), ; English transl., Izvestiya Math. 64:5, [3] Gantmacher F. R., The Theory of Matrices, Chelsea, New York, [4] Ja ja J., An addendum to Kronecker s theory of pencils, SIAM J. Appl. Math. 37:3 (1979), [5] Kac V. G., Some remarks on nilpotent orbits, Journal of Algebra 64 (1980), [6] D. Hilbert, Über die vollen Invarianten Systeme, Math. Ann. 42 (1893)