EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 4, 2021, 1108-1111 ISSN 1307-5543 – ejpam.com Published by New York Business Global On Eigenvectors of Nilpotent Lie Algebras of Linear Operators Morris W. Hirsch1,2,∗, Joel W. Robbin2 1 Department of Mathematics, University of California at Berkeley, Berkeley, CA 94720- 384, USA 2 Department of Mathematics, University of Wisconsin at Madison, WI 53706, USA Abstract. We give a condition ensuring that the operators in a nilpotent Lie algebra of linear operators on a finite dimensional vector space have a common eigenvector. 2020 Mathematics Subject Classifications: 22E25, 22E60, 47C05 Key Words and Phrases: Nilpotent, Lie algebras of Lie groups, Linear operators in algebras 1. Introduction Throughout this paper V is a vector space of positive dimension over a field f and ≫ is a nilpotent Lie algebra over f of linear operators on V . An element u ∈ V is an eigenvector for S ⊂≫ if u is an eigenvector for every operator in S. If V has a basis (e1, . . . , en) representing each element of ≫ by an upper triangular matrix, then e1 is an eigenvector for≫. Such a basis exists when f is algebraically closed and≫ is solvable (Lie’s Theorem), and also when every element of ≫ is a nilpotent operator (Engel’s Theorem). Our results are further conditions guaranteeing existence of eigenvectors. The minimal and characteristic polynomials of a linear operator A on V are denoted respectively by πA, µA ∈ f [t] = the ring of polynomials over f . The cardinality of a set S is written #S. Let k be a Galois extension field of f of degree d := [k : f ], and define M ⊂ to be the additive monoid generated by zero and the prime divisors d. Consider the conditions: (C1) µA splits in k for every A ∈≫ (C2) dimV /∈ M ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4086 Email addresses: mwhirsch@chorus.net (M. W. Hirsch), robbin@math.wisc.edu (J. W. Robbin) http://www.ejpam.com 1108 © 2021 EJPAM All rights reserved. M. W. Hirsch, J. W. Robbin / Eur. J. Pure Appl. Math, 14 (4) (2021), 1108-1111 1109 2. Results Our main result is: Theorem 1. If(C1) and(C2) hold then ≫ has an eigenvector. The proof is preceded by some applications. When(C1) holds, Theorem 1 shows that there is an eigenvector in every invariant subspace whose dimension is not in M. This is exploited to yield the following two results: Corollary 1. If a nilpotent Lie algebra of linear operators on n does not have an eigen- vector, every nontrivial invariant subspace has odd dimension. Proof. When f is the real field and k is the complex field , M consists of the positive even integers. Corollary 2. Let(C1) hold. Assume ≫ preserves a direct sum decomposition V = ⊕iWi, and let D ⊂ denote the set of dimensions of the subspaces Wi. (i) If ≫ does not have an eigenvector then D ⊂ M. (ii) If V ′ ⊂ V is a maximal subspace spanned by eigenvectors of ≫ then dim(V ′) ≥ #{D \M}. Proof. Assertion (i) follows from Theorem 1. To prove (ii) order the Wi so that W1, . . . ,Wm are the only summands whose dimensions are not in M. For each j ∈ {1, . . . ,m} we choose an eigenvector ej ∈ Wj by Theorem 1. The ej are linearly in- dependent and belong to V ′ by maximality of V ′, whence (ii). Example 1. Assume n /∈ M and let α ∈ f [t] be a monic polynomial that splits in k[t]. Denote by A(α) the set of n × n matrices T over f such that α(T ) = 0. Then every pairwise commuting family T ⊂ A(α) has an eigenvector in fn. This follows from Theorem 1 applied to the Lie algebra ≫ of linear operators on fn generated by T . Being abelian, ≫ can be triangularized over k, hence(C1) holds. Example 2. The assumption that n ∈ M is essential to Theorem 1. For instance, take f =, k =, V = 2. The abelian Lie algebra of 2× 2 of real skew symmetric matrices. does not have an eigenvector in 2. Example 3. The hypothesis of Theorem 1 cannot be weakened to ≫ being merely solvable. For a counterexample with f =, k =, take ≫ to be the solvable 3-dimensional real Lie algebra with basis (X,U, V ) such that [X,U ] = −V, [X,V ] = U, [U, V ] = 0. A Lie algebra β over f is supersolvable if the spectrum of the linear map ad A : β → β lies in f for all A ∈ β. If β is not supersolvable it need not have an eigenvector, as is shown by Example 3. We don’t know if Theorem 1 extends to supersolvable Lie algebras, except for the following special case: M. W. Hirsch, J. W. Robbin / Eur. J. Pure Appl. Math, 14 (4) (2021), 1108-1111 1110 Theorem 2. A supersolvable Lie algebra β of linear transformations of 3 has an eigen- vector. Proof. Lacking an algebraic proof, we use a dynamical argument. Let G ⊂ GL(3, ) be the connected Lie subgroup having Lie algebra β. The natural action of G on the projective plane ¶2 of lines in 3 through the origin fixes some L ∈ ¶2. This follows from supersolvability because dim(¶2) = 2, the action on ¶2 is effective and analytic, and the Euler characteristic of ¶2 is nonzero (Hirsch & Weinstein [1]). The nonzero points of L are eigenvectors for β. 2.1. Proof of Theorem 1 We rely on Jacobson’s Primary Decomposition Theorem [2, II.4, Theorem 5]. This states that V has a ≫-invariant direct sum decomposition ⊕Vi where each primary com- ponent Vi has the following property: For each A ∈≫ the minimal polynomial of A|Vi is a prime power in f [t]. Condition(C2) implies the dimension of some primary component is /∈ M. To prove Theorem 1 it therefore suffices to apply the following result to such a primary component: Theorem 3. Assume(C1) and(C2). If πA is a prime power in f [t] for each A ∈≫ then the following hold: (a) πA(t) = (t− rA) n, rA ∈ f (b) there is a basis putting ≫ in triangular form Assertion (a) is equivalent to πA having a root rA ∈ f . Therefore (a) follows from: Lemma 1. Let α ∈ f [t] be a polynomial of degree n that splits in k[t]. If n /∈ M then α has a root in f , and the sum of the multiplicities of such roots is /∈ M. Proof. Let R ⊂ k denote the set of roots of π, and Rj ⊂ R the set of roots of multiplicity j. The Galois group Γ has order [k : f ] and acts on R by permutations. The cardinality of each orbit divides [k : f ], and R ∩ f is the set of fixed points of this action. Each Rj is a union of orbits, as is Rj\f . It follows that #(Rj \ f) ∈ M. Let k ≤ n denote the sum of the multiplicities of the roots that are not in f . Then k = n∑ j=2 j ·#(Rj \ f) Therefore k ∈ M because M is closed under addition. By hypothesis n /∈ M, hence n − k /∈ M and n − k > 0. As n − k is the sum of the multiplicities of the roots in f , the conclusion follows. Now that (a) of Theorem 3 is proved, assertion (b) is a consequence of the following result: REFERENCES 1111 Lemma 2. Let be a nilpotent Lie algebra of linear operators on V . Assume that for all A ∈ there exists rA ∈ f such that πA(t) = (t− rA) n. Then V has a basis putting in triangular form. Proof. Every A ∈ can be written uniquely as rAI + NA with NA nilpotent and I the identity map of V . It is easy to see that the set comprising the NA is closed under commutator brackets. Therefore V has a basis triangularizing all the NA (Jacobson [2, II.2, Theorem 1′]), and such a basis triangularizes . This completes the proof of Theorem 1. References [1] M Hirsch and A Weinstein. Fixed points of analytic actions of supersoluble Lie groups on compact surfaces. Ergod. Th. Dyn. Sys., 21(6):1783–1787, 2001. [2] N Jacobson. Lie Algebras. Interscience Tracts in Pure Mathematics No. 10, John Wiley, New York, 1962.