Orthogonal $*$-basis of relative symmetry classes of polynomials

Transcription

1 Electronic Journal of Linear Algebra Volume 30 Volume 30 (2015) Article Orthogonal $*$-basis of relative symmetry classes of polynomials Kijti Rotes Department of Mathematics, Faculty of Science, Naresuan University, Follow this an aitional works at: Part of the Algebra Commons Recommene Citation Rotes, Kijti. (2015), "Orthogonal $*$-basis of relative symmetry classes of polynomials", Electronic Journal of Linear Algebra, Volume 30. DOI: This Article is brought to you for free an open access by Wyoming Scholars Repository. It has been accepte for inclusion in Electronic Journal of Linear Algebra by an authorize eitor of Wyoming Scholars Repository. For more information, please contact

2 ORTHOGONAL -BASIS OF SYMMETRY CLASSES OF POLYNOMIALS KIJTI RODTES Abstract. In this note, the existence of orthogonal -basis of the symmetry classes of polynomials is iscusse. Analogously to the orthogonal -basis of symmetry classes of tensor, some criteria for the existence of the basis for finite groups are provie. A conition for the existence of such basis of symmetry classes of polynomials associate to symmetric groups an some irreucible characters is also investigate. Key wors. Symmetry classes of polynomials, Orthogonal - basis. AMS subject classifications. 05E05, 15A Introuction. One of the classical areas of algebra, the theory of symmetric polynomials is well-known because of its role in branches of algebra, such as Galois Theory, representation theory an algebraic combinatorics. For a review of the theory of symmetric polynomials, one can see the book of Maconal, [6]. The relative symmetric polynomials as a generalization of symmetric polynomials are introuce by M. Shahryari in [11]. In fact, he use the iea of symmetry classes of tensors to introuce such notions. One of the most interesting topics about symmetry classes of tensors is the issues of fining a necessary conition for the existence of an orthogonal -basis for the symmetry classes of tensors associate with a finite group an an irreucible character. Many researchers pay a lot of attention to investigate conition state above. For example, M.R. Pournaki, [8], gave such a necessary conition for the irreucible constituents of the permutation character of the finite groups in which he extene a result of R.R. Holmes, [2]. Also, M. Shahryari provie an excellent conition for the existence of such basis in [10]. Furthermore, the existence of the special basis for particular groups have been iscusse by many authors, see, for example, [3, 4, 13]. Similar questions concerning about the existence of an orthogonal -basis arise in the context of relative symmetric polynomials as well, see, for example [9, 14, 15]. The general criterion is still an open problem, [11]. Receive by the eitors on December 17, Accepte for publication on March 29, Hanling Eitor: Tin-Yau Tam. Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok, 65000, Thailan (kijtir@nu.ac.th). Supporte by Naresuan University on the project R2558C

3 Orthogonal -Basis of Symmetry Classes of Polynomials 161 In this article, we provie some criteria for the existence of the special basis of symmetry classes of polynomials for finite groups an some corresponing permutation characters which are parallel to those of M.R. Pournaki in [8], R.R. Holmes in [2] an M. Shahryari in [10]. We also investigate some conition for the existence of such basis of symmetry classes of polynomials associate to symmetric groups an some irreucible characters, which are similar to the results of Y. Zamani in [12]. 2. Notations an backgroun. Let G be a subgroup of the full symmetric group S m an χ be an irreucible character of G. Let H [x 1,...,x m ] be the complex space of homogenous polynomials of egree with the inepenent commuting variables x 1,...,x m. Let Γ + m, be the set of all m-tuples of non-negative integers α = (α 1,...,α m ), such that m i=1 α i =. For any α Γ + m,, let Xα be the monomial x α1 1 xα2 2 xαm m. Then the set {Xα α Γ + m, } is a basis of H [x 1,...,x m ]. An inner prouct on H [x 1,...,x m ] is efine by X α,x β = δ α,β. (2.1) The group G, as a subgroup of the full symmetric group S m, acts on H [x 1,...,x m ] by (for σ G), It also acts on Γ + m, by q σ (x 1,...,x m ) = q(x σ 1 (1),...,x σ 1 (m)). σα = (α σ(1),...,α σ(m) ). Let be a set of representativesoforbits ofγ + m, uner the action ofg. Now consier the symmetrizer associate with G an χ T(G,χ) = χ(1) χ(σ)σ. (2.2) It is well known that T(G,χ) 2 = T(G,χ) an T(G,χ) = T(G,χ). The image of H [x 1,...,x m ] uner the map T(G,χ) is calle the symmetry class of polynomials of egree with respect to G an χ an it is enote by H (G;χ). For any q H [x 1,...,x m ], σ G q χ = T(G,χ)(q) is calle a symmetrize polynomial with respect to G an χ. Note that H (G;χ) = X α, χ ;α Γ + m,.

4 162 Kijti Rotes We write X α, instea of X α, χ unless it is necessary to avoi confusion. Definition 2.1. An orthogonal -basis (o-basis, for short) of a subspace U of H (G;χ) is an orthogonal basis of U of the form {X α1,,x α2,,...,x αt, } for some α i Γ + m,. Since the set {T(G,χ) : χ Irr(G)} is a complete set of orthogonal iempotents, where Irr(G) is the set of irreucible complex characters of G, we have the following orthogonal irect sum ecomposition (cf. Remark 2.3 in [11]) H [x 1,...,x m ] = H (G;χ). (2.3) χ Irr(G) Note that X α, is a generator of H (G;χ) if X α, 0, which can be checke from (χ,1) Gα, where (χ,φ) K is the inner prouct of characters χ an φ of an arbitrary group K, i.e. (χ,φ) K = 1 K σ K χ(σ)ψ(σ 1 ). Namely, (see, [9, 11]), X α, 0 if an only if (χ,1) Gα 0. (2.4) Also, for the inuce inner prouct on H (G;χ), we have (see, [9, 11]). X σ1α,,x σ2β, = { 0, if α / Orb(β) ; χ(1) σ σ 2G ασ 1 χ(σ), if α = β, 1 (2.5) where Orb(β) is the orbit of β in Γ + m, uner the action of G. Then the norm of Xα,, with respect to the inuce inner prouct, is given by X α, 2 = χ(1) (χ,1) G α [G : G α ]. (2.6) Accoring to (2.4), let Ω = {α Γ + m, : (χ,1) Gα 0}. Since H [x 1,...,x m ] = α X σα : σ G, we have the orthogonal irect sum H (G;χ) = α H α, (χ), (2.7) where = Ω an H α, (χ) = X σα, σ G. The imension of H α, (χ) can be calculate by using Freese s Theorem (see, e.g. [1], [9]) imh α, (χ) = χ(1)(χ,1) Gα = χ(1) G α σ G α χ(σ). (2.8)

5 Orthogonal -Basis of Symmetry Classes of Polynomials 163 As an immeiate consequence of (2.7) an (2.8), imh (G;χ) = χ(1) α (χ,1) Gα. (2.9) In particular, if χ is linear, then the set {X α, : α } is an orthogonal basis of H (G;χ) an imh (G;χ) =. Thus, the orthogonal -basis for H (G;χ) exists for any abelian group G. 3. Main criteria. Accoring to the notations in the previous section, H (G;χ) enotes the relative symmetry classes of polynomials of egree with respect to G an χ. This class is equippe with the inuce inner prouct as in (2.5). Let Λ be a set of m elements. Suppose G acts faithfully on Λ. So, we consier {f σ σ G} as the group G, where f σ : Λ Λ efine by f σ (λ) = σ λ, for all λ Λ. Namely, G can be viewe as a subgroup of S m in this way. We also enote the permutation character of G by θ. It is well known that θ(σ) = {λ Λ σ λ = λ}, for each σ G. The similar criterion as in the main theorem of [8] is shown below. Theorem 3.1. Let G be a finite group an let Λ be a set of m elements, m > 1. Assume that G acts transitively an faithfully on Λ. Let χ be an irreucible constituent of permutation character θ of G. If χ(1)(χ,θ) G > m 2, then H (G;χ) oes not have an orthogonal -basis. Proof. Suppose H (G;χ) has an orthogonal -basis. Then, by (2.7), H α, (χ) has an o-basis for each α. We now consier α = (,0,0,...,0) Γ + m, an choose to be the set of representatives of orbits of Γ + m, uner the action of G in which α. We can assume without loss of generality that Λ = {1,2,...,m} an thus G α = G 1, where G α refers to the stabilizer subgroup of α (when G acts on Γ + m, ) an G 1 refers to the stabilizer subgroup of 1 (when G acts on Λ). Since G acts transitively on Λ, (1 Gα ) G = (1 G1 ) G = θ, by Lemma 5.14 of [5]. Hence, by (2.8) an Frobenius reciprocity, we have that σ G α χ(σ) = G α (χ,1 Gα ) Gα = G α (χ,(1 Gα ) G ) G = G α (χ,θ) G. Since χ is an irreucible constituent of permutation character θ of G, (χ,θ) G 0 an σ G α χ(σ) 0. Thus α. So, H α, (χ) has an o-basis. By orbit-stabilizer theorem an transitive action of G on Λ, we have that m = Λ = Orb(1) = [G : G 1 ] = [G : G α ]. So, G = m i=1 σ ig α, where {σ 1,σ 2,...,σ m } is a

6 164 Kijti Rotes system of istinct representatives of left cosets of G α in G. Let imh α, (χ) = χ(1) χ(σ) = χ(1)(χ,θ) G := t. G α σ G α We can assume that {X σ1α,,x σ2α,,...,x σtα, } is an o-basis for H α, (χ). Define the m m complex matrix D = [D ij ] by D ij := X σiα,,x σjα,. Note that D is iempotent. In fact, m (D) 2 ij = D ik D kj = = k=1 m X σiα,,x σkα, X σkα,,x σjα, k=1 m χ(1) k=1 = χ(1)2 2 = χ(1)2 2 m k=1σ G α m σ σ k G ασ 1 i k=1λ σ k G α χ(σ) χ(1) τ σ jg ασ 1 k τ G α χ(σ k σσ 1 i )χ(σ j τσ 1 k ) µ G ασ 1 k Now, let µλ = δ G α. Then µ = δλ 1 an we have m (D) 2 ij = χ(1)2 2 = χ(1) = χ(1) k=1λ σ k G α δ G α δ G α ( χ(1) χ(λσ 1 i )χ(σ j µ). δ G α χ(λσ 1 i )χ(σ j δλ 1 ) λ G χ(λσ 1 i )χ(σ j δλ 1 ) ( χ(1) χ(σ)χ(σ j δσi 1 σ 1 ) σ G By orthogonal relations of irreucible character, we have (D) 2 ij = χ(1) χ(σ j δσi 1 ), δ G α which shows that D 2 = D. χ(τ) We note that m = θ(1) = χ Θχ(1)(χ,θ), where Θ is the set of all irreucible constituents of the permutation character θ. Since m > 1, Θ > 1 an hence m > t. We can now write D in the form [ ] D1 D 2, D 3 D 4 ) ).

7 Orthogonal -Basis of Symmetry Classes of Polynomials 165 where D 1,D 2,D 3 an D 4 are matrices of sizes t t,t (m t),(m t) t an (m t) (m t) respectively. On the matrix D 1, we have, by (2.5), that, for 1 i,j t, { (D 1 ) ij = X σiα,,x σjα, 0, if i j; = χ(1) σ G α χ(σ), if i = j { ( ) 0, if i j; t = t m, if i = j = m I t, ij [ ( t ) where I t is the t t ientity matrix. So, D = m It D 2 D 3 D 4 we get ( ) t D 2 D 3 = m t2 m 2 I t. Since t < m, ( ) t m t2 m 2 H α, (χ) has an o-basis, then ]. Now, using D 2 = D, 0 an hence D 2 D 3 is invertible. This means that if t = rankd 2 D 3 min{rankd 2,rankD 3 } min{t,m t} m t. Therefore, if χ(1)(χ,θ) G = t > m 2, then H (G;χ) oes not havean orthogonal -basis, by (2.7). We also obtain a similar results of Holmes in [2]. Corollary 3.2. (cf. [2, 8]) Let G be a 2-transitive subgroup of S m, m > 2. Let χ = θ 1 G, where θ is the permutation character of G. Then H (G;χ) oes not have an orthogonal -basis. Proof. Note that G has a canonical transitive an faithful action on the set Λ = {1,2,...,m}, given by σ i := σ(i) for each σ G S m an i Λ. Since G acts 2-transitively on Λ with permutation character θ, by Corollary 5.17 in [5], χ = θ 1 G is an irreucible constituent of θ. We compute that χ(1)(χ,θ) G = χ(1)(θ 1 G,θ) G = χ(1)[(θ,θ) G (θ,1 G ) G ] = χ(1)[2 1] = m 1. Since m > 2, m 1 > m 2 an hence χ(1)(χ,θ) G > m 2. Thus, by Theorem 3.1, the result follows. Example 3.1. (cf. [8]) Let G = Λ = A 4 be the alternatinggroup ofegree 4. We know that G acts transitively an faithfully on Λ by left multiplication. Then we can view G as a subgroup of S 12. Note that G has an irreucible character, χ, of egree 3 an the permutation character θ of G is regular. Thus χ is an irreucible constituent of θ of multiplicity 3. Hence, χ(1)(χ,θ) G = 9 > 12 2 = Λ 2 an then H (A 4 ;χ) oes

8 166 Kijti Rotes not have an orthogonal -basis, by Theorem 3.1. In this example, however, the action G on Λ is not 2-transitive. By using the same technique as in the proof of Theorem 3.1, we also obtain an analogous criterion of Shahryari in [10]. Theorem 3.3. Let G be a permutation group of egree m an χ be a non-linear irreucible character of G. If there is α Γ + m, such that 2 2 < Xα, χ < 1, then H (G;χ) oes not have an orthogonal -basis. Proof. Let α Γ + m,. Suppose the orbit of α uner the action of G is Orb(α) = {σ 1 α,σ 2 α,...,σ r α}. Then, by orbit-stabilizer theorem, r = [G : G α ] an G = r i=1 σ ig α is a partition. Now, we construct r r matrix D = [D ij ] by D ij := X σiα,,x σjα, which is iempotent as before. Next, suppose χ is a non-linear irreucible character of G an α an assume also that {X σ1α,,x σ2α,,...,x σtα, } is an o-basis for H α, (χ) in which t < r, where t = imh α, (χ). So, the matrix D has the block partition form [ ( t ) D = r It D 2 D 3 D 4 where D 2,D 3 an D 4 are matrices of sizes t (r t),(r t) t an (r t) (r t) respectively. By the same arguments as in the proof of Theorem 3.1, we reach to the conclusion that t r t or t r 2. Thus if t < r an t > r 2, then Hα, (χ) oes not have o-basis. Substituting r = [G : G α ], t = χ(1)(χ,1) Gα in the inequality r 2 < t < r an using (2.6) an (2.7), the result follows. 4. Symmetric groups. It is well known that there is a stanar one-to-one corresponence between the complex irreucible characters of the symmetric group S m an the partitions of m. Here, a partition π of m of length t, enote by π m, means an unorere collection of t positive integers that sum to m. In this article, we use the same symbol to enote an irreucible character of S m an the partition of m corresponing to it. Typically, we represent the partition by a sequence π = [π 1,π 2,...,π t ] in which π 1 π 2 π t > 0. A partition π = [π 1,π 2,...,π t ] is usuallyrepresentebyacollectionofmboxesarrangein trowssuchthat thenumber of boxes of row i is equal to π i, for i = 1,2,...,t. This collection is calle the Young iagram associate with π an enote by [π]. The Young subgroup corresponing to π m is the internal irect prouct ], S π = S π1 S π2 S πt.

9 Orthogonal -Basis of Symmetry Classes of Polynomials 167 We write 1 Sπ = 1 π for the principle character of S π. Note that 1 Sm π is a character of S m, so there must exist integers K µ,π such that 1 Sm π = µ mk µ,π µ. The numbers K µ,π = ( 1 Sm π,µ ) S m are calle Kostka coefficients. By Corollary 4.54 in [7], the Kostka coefficient K π,π = 1 for all π m. For each orere pair (i,j), 1 i t, 1 j π i, there is corresponing a box, B ij, in Young iagram [π]. Each B ij etermines a unique hook in [π] consisting of B ij itself, all the boxes in row i of [π] to the right of B ij an all boxes in column j of [π] below B ij. The hook length, h ij := (π i i)+(π j j)+1, where π j := {k {1,2,...,t} π k j} (a j part of conjugate partition of π), is the number of boxes in the hook etermine by B ij. By the Frame-Robinson-Thrall Hook Length Formula (see, e.g., Theorem 4.60 in [7]), if π is a partition of m, then the egree of the irreucible character of S m corresponing to π = [π 1,π 2,...,π t ] is π(1) = m! t πi i=1 j=1 h. (4.1) ij As a consequence of Theorem 3.3, we have an analogous result of Y. Zamani in [12]. Theorem 4.1. Let π be an irreucible character of S m of the cycle type; I. π = [m l,l], 0 mo l, 0 such m 3l, or II. π = [m l,l 1,1], r mo l,0 < r < l,l > 2, r such m > 3l+ 4 l 2. Then H (S m ;π) oes not have an orthogonal -basis. Proof. For the form I, we set α = (0,0,...,0,k,k,...,k), where k = }{{}}{{} l. Then m l l α Γ + m,. Uner the action of S m on Γ + m,, we choose a system of representatives such that α. Since 0, k 0 an (S m ) α = Sm l S l = S π, where (S m ) α is the stabilizer subgroup of α an S π is the Young subgroup corre-

10 168 Kijti Rotes sponing to π m. Hence, by Frobenius Reciprocity Theorem, 1 ( ) π(σ) = π,1(sm) (S m ) α α (S m) α σ (S m) α = (π,1 π ) Sπ = ( ) π,1 Sm π S m = π, µ mk µ,π µ S m This yiels α an, moreover, by (2.8), that imh α, (π) = π(1) (S m ) α = µ mk µ,π (π,µ) Sm = K µ,µ = 1 0. σ (S m) α χ(σ) = π(1). Now, we compute the prouct of the hook lengths of [π] which we get 2 π i h ij = (m l+1)(m l) (m 2l+2)(m 2l) (2)(1)l(l 1) (2)(1) i=1j=1 Hence, by (4.1), = (m l+1)!l! (m 2l+1). Now, using (2.6), we have imh α, (π) = π(1) = (m 2l+1)m! (m l+1)!l!. X α, 2 = imhα, (π) [S m : (S m ) α ] = m 2l+1 m l+1. Hence, 1 2 < Xα, 2 < 1 if an only if m 3l. Thus, the result for the first form follows from Theorem 3.3. For the form II, π = [m l,l 1,1], we set α = (0,0,...,0,k,k,...,k,k + r), }{{}}{{} m l l 1, we choose a system of representatives such that α. Since r 0, k 0 an k + r k an hence where k = r l. Then α Γ + m,. Uner the action of S m on Γ + m, (S m ) α = Sm l S l 1 S 1 = S π.

11 Orthogonal -Basis of Symmetry Classes of Polynomials 169 By the same arguments as the first form, we conclue that imh α, (π) = π(1). For the proucts of the hook lengths, we compute that 3 π i h ij = (m l+2)(m l) (m 2l+3)(m 2l+1) (1)l(l 2)(l 3) (1) i=1j=1 Then = (m l+2)!l! (m l+1)(m 2l+2)(l 1). imh α, (π) = π(1) = (m l+1)(m 2l+2)(l 1)m!. (m l+2)!l! Now, using (2.6) again, we have X α, 2 = imhα, (π) [S m : (S m ) α ] = (m 2l+2)(l 1). (m l+2)(l) It is now easy to show that 1 2 < Xα, 2 < 1 if an only if m > 3l + 4 l 2, because l > 2. The result for the secon form follows from Theorem 3.3. Acknowlegment. The author is grateful to the anonymous referees an Professor Tin-Yau Tam for their recommenations, an woul like to thank Naresuan University for financial support on the project R2558C030. REFERENCES [1] R. Freese. Inequalities for generalize matrix functions base on arbitrary characters. Linear Algebra an its Applications, 7: , [2] R.R. Holmes Orthogonal bases of symmetrize tensor spaces. Linear an Multilinear Algebra, 39: , [3] R.R. Holmes an T.Y. Tam. Symmetry classes of tensors associate with certain groups. Linear an Multilinear Algebra, 32:21 31, [4] M. Hormozi an K.Rotes. Symmetry classes of tensors associate with the Semi-Diheral groups SD 8n. Colloquium Mathematicum, 131:59 67, [5] I.M. Isaacs. Character Theory of Finite Groups. Acaemic Press, New York, [6] I.G. Maconal. Symmetric Functions an Orthogonal Polynomials. American Mathematical Society, [7] R. Merris. Multilinear Algebra. Goron an Breach Science Publishers, Amsteram, [8] M.R. Pournaki. On the orthogonal basis of the symmetry classes of tensors associate with certain characers. Linear Algebra an its Applications, 336: , [9] K. Rotes. Symmetry classes of polynomials associate to the Semiiheral group an o-bases. Journal of Algebra an its Applications, 13(8): , [10] M. Shahryari. On the orthogonal basis of symmetry classes. Journal of Algebra, 220: , [11] M. Shahryari. Relative symmetric polynomials. Linear Algebra an its Applications, 433: , 2010.

12 170 Kijti Rotes [12] Y. Zamani. On the special basis of certain full symmetry class of tensors. Pure Mathematics an Applications, 18(3/4): , [13] Y. Zamani an E. Babaei. The imensions of cyclic symmetry classes of polynomials. Journal of Algebra an its Applications, 132: , [14] Y. Zamani an E. Babaei. Symmetry classes of polynomials associate with the icyclic group. Asian-European Journal of Mathematics, 63: , [15] Y. Zamani an E. Babaei. Symmetry classes of polynomials associate with the iheral group. Bulletin of the Iranian Mathematical Society, to appear.