DOCSLIB.ORG

11. Roots of Lie Algebras 1

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
11. Roots of Lie Algebras 1 11. Roots of Lie algebras 1 11. Roots of Lie algebras With the scalar product defined in terms of the inverse of the Killing form on the Cartan subalgebra, it can be proved that α α> 0, for any root α. It · has been shown that, for a semi-simple Lie algebra, if α is a root, then α is also a root. It can also be shown that 2α is not a root. − [The obvious result [Eα, Eα] = 0 is consistent with this statement, but is not sufficient to prove it, since it is conceivable that Nαα =0 but [Eα β, Eα β] = 0.] + − 6 The only multiples of a root α which are also roots are 0 and α. It is also possible to prove that all non-zero roots are non-degenerate,± i.e. there is only one root vector for each root. The degeneracy of the zero root, which corresponds to the vectors of the Cartan subalgebra, is equal to the rank of the algebra. Suppose α, β are roots, such that α + β is not a root, i.e. [Eα, Eβ] = 0. ′ ′ Then [E−α, Eβ] Eβ−α is proportional to Eβ−α and so [E−α, Eβ−jα] ′ ≡ ≡ Eβ−(j+1)α is proportional to Eβ−(j+1)α. (It is assumed that some normali- sation has been chosen for the root vectors Eα. The normalisation of the ′ vectors Eα is then fixed by their definition.) Since there is only a finite number of roots, there must be a maximum value of j, denoted M, such ′ ′ ′ that [E−α, Eβ−Mα]=0= Eβ−(M+1)α = 0. Now consider [Eα, Eβ−jα] = ′ ⇒ µjEβ−(j−1)α, which defines the factor of proportionality µj, since the normal- ′ isations of Eα and Eα are fixed. It follows that ′ ′ µj+1Eβ−jα = [Eα, Eβ−(j+1)α] ′ = [Eα, [E−α, Eβ−jα]] ′ ′ = [E−α, [Eβ−jα, Eα]] [Eβ−jα, [Eα, E−α]] − ′ − ′ = µj[E−α, Eβ−(j−1)α]+ gα,−α[Hα, Eβ−jα] ′ ′ = µjE + gα, αα (β jα)E , β−jα − · − β−jα which implies µj = µj +(α β jα α)gα, α. This is a recursion relation +1 · − · − for the coefficient µj, with the initial condition µ0 = 0. ′ [This follows from [Eα, Eβ−α]= µ1Eβ = [Eα, [E−α, Eβ]] = [E α, [Eβ, Eα]] [Eβ, [Eα, E α]] = gα, α[Hα, Eβ] − − − − − = gα, αα βEβ = µ = α βgα, α = µ + α βgα, α.] − · ⇒ 1 · − 0 · − The recursion relation has the solution µj = gα,−α (jα β α αj(j 1)/2), ′ ′ · − · − satisfying µ0 = 0. But µM+1Eβ−Mα = [Eα, Eβ−(M+1)α]=0= µM+1 = 0= α β = Mα α/2= M =2α β/α α. ⇒ ⇒ · · ⇒ · · Introductory Algebra for Physicists Michael W. Kirson 11. Roots of Lie algebras 2 To recapitulate, starting from a root β such that β + α is not a root, a sequence of roots β jα was generated, terminating with the root β Mα, where M =2α β/α− α. This sequence is called an α-string of roots.− · · In general, any root γ belongs to an α-string of roots, ranging from γ mα to γ + pα. Here, m and p are non-negative integers, which may be zero.− The “top” of this string is at β = γ + pα, and it ends at β Mα = γ mα, so M = m + p. Hence m + p = M = 2α (γ + pα)/α α−= 2α γ/α−α +2p. Hence · · · · m p =2α γ/α α. − · · This is a major result, with many implications and applications. It will be referred to from now on as the “magic formula”. According to the “magic formula”, for any two roots α, β, the quantity 2α β/α α is an integer. If β is at the top of an α-string (i.e. β + α is not a root,· so p· = 0), then the quantity is non-negative. If α α = 0, then α β =0 ij · · for all β, which implies Pi αig = 0. But this implies det g = 0, which is impossible for semi-simple Lie algebras. If β belongs to an α-string from β mα to β + pα, then the midpoint of − the string is at β (m p)α/2= β (β α/α α)α and the “mirror image” of β on the string− is β −2(β α/α α−)α,· and is· also a root. − · · β (m p)α/2 − 6− sssαα- - s α- α s sss αα β mα β (m p)α× β + pα − − − β (β + pα) (β mα) - − − (p + m)α/2 - This procedure for producing a root from a given root is called a Weyl re- flection, and the set of all such operations constitutes the Weyl group. There can be at most four roots in a string. [Suppose there are five roots, chosen without loss of generality as β 2α, β α,β,β + α, β +2α. Since β +2α β = 2α, and 2α − − − is not a root, and since β +2α + β = 2(β + α), and 2(β + α) is not a root, the root β +2α belongs to a β-string of length 1, so m =0= p and (β +2α) β = 0. Similarly, (β 2α) β = 0, hence β β = 0, which is impossible.]· − · · Introductory Algebra for Physicists Michael W. Kirson 11. Roots of Lie algebras 3 The only possible values of m p in a 4-string are 3 and 1; in a 3- string they are 2 or 0; in a 2-string− they are 1; and in± a 1-string± the only possible value is± 0. So, from the “magic formula”,± 2α β/α α can only take · · on the values 0, 1, 2, 3. ± ± ± It is possible to adopt a simple geometrical picture in which α α is the square of the length of the root α and α β is proportional to the cosine· of the · 2 2 angle between the roots α and β. In fact, cos φαβ =(α β) /(α α)(β β)= rs/4, where r, s are integers, the appropriate values of m· p for· the relevant· α- and β-strings. − 2 [Note that cos φαβ 1, because of the Schwarz inequality (α β)2 (α α)(β ≤β). This inequality also correlates the values of m· p≤for α· - and· β-strings.] − 2 It may be concluded that cos φαβ takes only the values 0, 1/4, 1/2, 3/4, 1, so the only possible angles between two roots are 90◦, 60◦ or 120◦, 45◦ or 135◦, ◦ ◦ 2 30 or 150 , with cos φαβ =1= β = α. Since r/s = β β/α α, it is easy to deduce that the ratio⇒ of lengths± of two roots is 1· if the· angle between them is 60◦ or 120◦, √2 if the angle is 45◦ or 135◦, √3 if the angle is 30◦ or 150◦ and indeterminate if the angle is 90◦. Each Lie algebra can then be represented by a vector diagram of roots in an ℓ-dimensional space, where ℓ is the rank of the algebra. If a basis set of roots α(i) is chosen in the root space, then any root can {(i) } be written as β = Pi qiα , where the expansion coefficients qi are rational numbers. (j) (j) (j) (i) (j) (j) (j) [2β α /α α = P qi2α α /α α is a set of linear · · i · · equations for the qi , with integer coefficients, so the solutions are rational.] { } This is consistent with the earlier comment that real linear combinations of roots constitute a sufficient root space. The scalar products α α and α β · · are also rational numbers. ij ik lj ik lj [α α = Pij αig αj = Pijkl αig gklg αj = Pijkl Pβ αig βkβlg αj = · 2 2 2 Pβ(α β)(β α) = Pβ(α β) = Pβ(m p)β(α α) /4 = · · 2 · − · ⇒ α α = 4/ Pβ(m p)β, which is rational and manifestly pos- itive,· justifying the− use of the term “scalar product”. Then α β =(m p)βα α/2, also rational.] · − · Introductory Algebra for Physicists Michael W. Kirson 11. Roots of Lie algebras 4 An ordering can be introduced into the real root space by choosing an arbitrary basis and defining a root as positive if the first non-zero coefficient in its expansion in terms of the basis is positive. Then α > β if α β > 0. − Since α> 0 α< 0, half of the non-zero roots will be positive. ⇐⇒ − A positive root is said to be simple if it cannot be written as the sum of two positive roots. If two roots α, β are simple, then their difference α β − is not a root. [If α β is a root, then α β > 0 or β α> 0. But α =(α β)+β and −β =(β α)+ α, so− either α or −β is not simple.] − − For the α-string containing β, m p =2α β/α α. But β α is not a root, so m = 0. Therefore α β 0 for− simple roots.· · − · ≤ The set of simple roots is linearly independent. (i) (i) [If the simple roots α were linearly dependent, then Pi ciα = { } (i) 0 for some coefficients ci . Since all α > 0, some of the ci { } (i) {(j)} are negative. Rewrite the dependence as Pi aiα = Pj bjα , (i) (i) (i) where all ai, bj > 0. Then (Pi aiα ) (Pi aiα )=(Pi aiα ) (j) (i) (j) · · (Pj bjα ) = Pi=6 j aibjα α . The left hand side is positive, while the right hand side is· a sum of negative terms.] It can be shown that the number of simple roots is equal to ℓ, the rank of the algebra, so the simple roots form a basis for the real root space ∗.
Load more

Recommended publications
  • Modules and Lie Semialgebras Over Semirings with a Negation Map 3
    MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP GUY BLACHAR Abstract. In this article, we present the basic definitions of modules and Lie semialgebras over semirings with a negation map. Our main example of a semiring with a negation map is ELT algebras, and some of the results in this article are formulated and proved only in the ELT theory. When dealing with modules, we focus on linearly independent sets and spanning sets. We define a notion of lifting a module with a negation map, similarly to the tropicalization process, and use it to prove several theorems about semirings with a negation map which possess a lift. In the context of Lie semialgebras over semirings with a negation map, we first give basic definitions, and provide parallel constructions to the classical Lie algebras. We prove an ELT version of Cartan’s criterion for semisimplicity, and provide a counterexample for the naive version of the PBW Theorem. Contents Page 0. Introduction 2 0.1. Semirings with a Negation Map 2 0.2. Modules Over Semirings with a Negation Map 3 0.3. Supertropical Algebras 4 0.4. Exploded Layered Tropical Algebras 4 1. Modules over Semirings with a Negation Map 5 1.1. The Surpassing Relation for Modules 6 1.2. Basic Definitions for Modules 7 1.3. -morphisms 9 1.4. Lifting a Module Over a Semiring with a Negation Map 10 1.5. Linearly Independent Sets 13 1.6. d-bases and s-bases 14 1.7. Free Modules over Semirings with a Negation Map 18 2.
    [Show full text]
  • LECTURE 12: LIE GROUPS and THEIR LIE ALGEBRAS 1. Lie
    LECTURE 12: LIE GROUPS AND THEIR LIE ALGEBRAS 1. Lie groups Definition 1.1. A Lie group G is a smooth manifold equipped with a group structure so that the group multiplication µ : G × G ! G; (g1; g2) 7! g1 · g2 is a smooth map. Example. Here are some basic examples: • Rn, considered as a group under addition. • R∗ = R − f0g, considered as a group under multiplication. • S1, Considered as a group under multiplication. • Linear Lie groups GL(n; R), SL(n; R), O(n) etc. • If M and N are Lie groups, so is their product M × N. Remarks. (1) (Hilbert's 5th problem, [Gleason and Montgomery-Zippin, 1950's]) Any topological group whose underlying space is a topological manifold is a Lie group. (2) Not every smooth manifold admits a Lie group structure. For example, the only spheres that admit a Lie group structure are S0, S1 and S3; among all the compact 2 dimensional surfaces the only one that admits a Lie group structure is T 2 = S1 × S1. (3) Here are two simple topological constraints for a manifold to be a Lie group: • If G is a Lie group, then TG is a trivial bundle. n { Proof: We identify TeG = R . The vector bundle isomorphism is given by φ : G × TeG ! T G; φ(x; ξ) = (x; dLx(ξ)) • If G is a Lie group, then π1(G) is an abelian group. { Proof: Suppose α1, α2 2 π1(G). Define α : [0; 1] × [0; 1] ! G by α(t1; t2) = α1(t1) · α2(t2). Then along the bottom edge followed by the right edge we have the composition α1 ◦ α2, where ◦ is the product of loops in the fundamental group, while along the left edge followed by the top edge we get α2 ◦ α1.
    [Show full text]
  • CLIFFORD ALGEBRAS Property, Then There Is a Unique Isomorphism (V ) (V ) Intertwining the Two Inclusions of V
    CHAPTER 2 Clifford algebras 1. Exterior algebras 1.1. Definition. For any vector space V over a field K, let T (V ) = k k k Z T (V ) be the tensor algebra, with T (V ) = V V the k-fold tensor∈ product. The quotient of T (V ) by the two-sided⊗···⊗ ideal (V ) generated byL all v w + w v is the exterior algebra, denoted (V ).I The product in (V ) is usually⊗ denoted⊗ α α , although we will frequently∧ omit the wedge ∧ 1 ∧ 2 sign and just write α1α2. Since (V ) is a graded ideal, the exterior algebra inherits a grading I (V )= k(V ) ∧ ∧ k Z M∈ where k(V ) is the image of T k(V ) under the quotient map. Clearly, 0(V )∧ = K and 1(V ) = V so that we can think of V as a subspace of ∧(V ). We may thus∧ think of (V ) as the associative algebra linearly gener- ated∧ by V , subject to the relations∧ vw + wv = 0. We will write φ = k if φ k(V ). The exterior algebra is commutative | | ∈∧ (in the graded sense). That is, for φ k1 (V ) and φ k2 (V ), 1 ∈∧ 2 ∈∧ [φ , φ ] := φ φ + ( 1)k1k2 φ φ = 0. 1 2 1 2 − 2 1 k If V has finite dimension, with basis e1,...,en, the space (V ) has basis ∧ e = e e I i1 · · · ik for all ordered subsets I = i1,...,ik of 1,...,n . (If k = 0, we put { } k { n } e = 1.) In particular, we see that dim (V )= k , and ∅ ∧ n n dim (V )= = 2n.
    [Show full text]
  • Introuduction to Representation Theory of Lie Algebras
    Introduction to Representation Theory of Lie Algebras Tom Gannon May 2020 These are the notes for the summer 2020 mini course on the representation theory of Lie algebras. We'll first define Lie groups, and then discuss why the study of representations of simply connected Lie groups reduces to studying representations of their Lie algebras (obtained as the tangent spaces of the groups at the identity). We'll then discuss a very important class of Lie algebras, called semisimple Lie algebras, and we'll examine the repre- sentation theory of two of the most basic Lie algebras: sl2 and sl3. Using these examples, we will develop the vocabulary needed to classify representations of all semisimple Lie algebras! Please email me any typos you find in these notes! Thanks to Arun Debray, Joakim Færgeman, and Aaron Mazel-Gee for doing this. Also{thanks to Saad Slaoui and Max Riestenberg for agreeing to be teaching assistants for this course and for many, many helpful edits. Contents 1 From Lie Groups to Lie Algebras 2 1.1 Lie Groups and Their Representations . .2 1.2 Exercises . .4 1.3 Bonus Exercises . .4 2 Examples and Semisimple Lie Algebras 5 2.1 The Bracket Structure on Lie Algebras . .5 2.2 Ideals and Simplicity of Lie Algebras . .6 2.3 Exercises . .7 2.4 Bonus Exercises . .7 3 Representation Theory of sl2 8 3.1 Diagonalizability . .8 3.2 Classification of the Irreducible Representations of sl2 .............8 3.3 Bonus Exercises . 10 4 Representation Theory of sl3 11 4.1 The Generalization of Eigenvalues .
    [Show full text]
  • Matrix Lie Groups and Their Lie Algebras
    Matrix Lie groups and their Lie algebras Alen Alexanderian∗ Abstract We discuss matrix Lie groups and their corresponding Lie algebras. Some common examples are provided for purpose of illustration. 1 Introduction The goal of these brief note is to provide a quick introduction to matrix Lie groups which are a special class of abstract Lie groups. Study of matrix Lie groups is a fruitful endeavor which allows one an entry to theory of Lie groups without requiring knowl- edge of differential topology. After all, most interesting Lie groups turn out to be matrix groups anyway. An abstract Lie group is defined to be a group which is also a smooth manifold, where the group operations of multiplication and inversion are also smooth. We provide a much simple definition for a matrix Lie group in Section 4. Showing that a matrix Lie group is in fact a Lie group is discussed in standard texts such as [2]. We also discuss Lie algebras [1], and the computation of the Lie algebra of a Lie group in Section 5. We will compute the Lie algebras of several well known Lie groups in that section for the purpose of illustration. 2 Notation Let V be a vector space. We denote by gl(V) the space of all linear transformations on V. If V is a finite-dimensional vector space we may put an arbitrary basis on V and identify elements of gl(V) with their matrix representation. The following define various classes of matrices on Rn: ∗The University of Texas at Austin, USA. E-mail: [email protected] Last revised: July 12, 2013 Matrix Lie groups gl(n) : the space of n
    [Show full text]
  • What Does a Lie Algebra Know About a Lie Group?
    WHAT DOES A LIE ALGEBRA KNOW ABOUT A LIE GROUP? HOLLY MANDEL Abstract. We define Lie groups and Lie algebras and show how invariant vector fields on a Lie group form a Lie algebra. We prove that this corre- spondence respects natural maps and discuss conditions under which it is a bijection. Finally, we introduce the exponential map and use it to write the Lie group operation as a function on its Lie algebra. Contents 1. Introduction 1 2. Lie Groups and Lie Algebras 2 3. Invariant Vector Fields 3 4. Induced Homomorphisms 5 5. A General Exponential Map 8 6. The Campbell-Baker-Hausdorff Formula 9 Acknowledgments 10 References 10 1. Introduction Lie groups provide a mathematical description of many naturally-occuring sym- metries. Though they take a variety of shapes, Lie groups are closely linked to linear objects called Lie algebras. In fact, there is a direct correspondence between these two concepts: simply-connected Lie groups are isomorphic exactly when their Lie algebras are isomorphic, and every finite-dimensional real or complex Lie algebra occurs as the Lie algebra of a simply-connected Lie group. To put it another way, a simply-connected Lie group is completely characterized by the small collection n2(n−1) of scalars that determine its Lie bracket, no more than 2 numbers for an n-dimensional Lie group. In this paper, we introduce the basic Lie group-Lie algebra correspondence. We first define the concepts of a Lie group and a Lie algebra and demonstrate how a certain set of functions on a Lie group has a natural Lie algebra structure.
    [Show full text]
  • Clifford Algebras and Lie Groups
    Clifford algebras and Lie groups Eckhard Meinrenken Lecture Notes, University of Toronto, Fall 2009. (Revised version of lecture notes from 2005) CHAPTER 1 Symmetric bilinear forms Throughout, K will denote a field of characteristic = 2. We are mainly interested in the cases K = R or C, and sometimes specialize6 to those two cases. 1. Quadratic vector spaces Suppose V is a finite-dimensional vector space over K. For any bilinear form B : V V K, define a linear map × → ♭ B : V V ∗, v B(v, ). → 7→ · The bilinear form B is called symmetric if it satisfies B(v1, v2)= B(v2, v1) ♭ ♭ for all v1, v2 V . Since dim V < this is equivalent to (B )∗ = B . The symmetric∈ bilinear form B is uniquely∞ determined by the associated quadratic form, QB(v)= B(v, v) by the polarization identity, (1) B(v, w)= 1 Q (v + w) Q (v) Q (w) . 2 B − B − B The kernel (also called radical ) of B is the subspace ker(B)= v V B(v, v ) = 0 for all v V , { ∈ | 1 1 ∈ } i.e. the kernel of the linear map B♭. The bilinear form B is called non- degenerate if ker(B) = 0, i.e. if and only if B♭ is an isomorphism. A vector space V together with a non-degenerate symmetric bilinear form B will be referred to as a quadratic vector space. Assume for the rest of this chapter that (V,B) is a quadratic vector space. Definition 1.1. A vector v V is called isotropic if B(v, v) = 0, and non-isotropic if B(v, v) = 0.
    [Show full text]
  • Matrices Lie: an Introduction to Matrix Lie Groups and Matrix Lie Algebras
    Matrices Lie: An introduction to matrix Lie groups and matrix Lie algebras By Max Lloyd A Journal submitted in partial fulfillment of the requirements for graduation in Mathematics. Abstract: This paper is an introduction to Lie theory and matrix Lie groups. In working with familiar transformations on real, complex and quaternion vector spaces this paper will define many well studied matrix Lie groups and their associated Lie algebras. In doing so it will introduce the types of vectors being transformed, types of transformations, what groups of these transformations look like, tangent spaces of specific groups and the structure of their Lie algebras. Whitman College 2015 1 Contents 1 Acknowledgments 3 2 Introduction 3 3 Types of Numbers and Their Representations 3 3.1 Real (R)................................4 3.2 Complex (C).............................4 3.3 Quaternion (H)............................5 4 Transformations and General Geometric Groups 8 4.1 Linear Transformations . .8 4.2 Geometric Matrix Groups . .9 4.3 Defining SO(2)............................9 5 Conditions for Matrix Elements of General Geometric Groups 11 5.1 SO(n) and O(n)........................... 11 5.2 U(n) and SU(n)........................... 14 5.3 Sp(n)................................. 16 6 Tangent Spaces and Lie Algebras 18 6.1 Introductions . 18 6.1.1 Tangent Space of SO(2) . 18 6.1.2 Formal Definition of the Tangent Space . 18 6.1.3 Tangent space of Sp(1) and introduction to Lie Algebras . 19 6.2 Tangent Vectors of O(n), U(n) and Sp(n)............. 21 6.3 Tangent Space and Lie algebra of SO(n).............. 22 6.4 Tangent Space and Lie algebras of U(n), SU(n) and Sp(n)..
    [Show full text]
  • Lie Algebras by Shlomo Sternberg
    Lie algebras Shlomo Sternberg April 23, 2004 2 Contents 1 The Campbell Baker Hausdorff Formula 7 1.1 The problem. 7 1.2 The geometric version of the CBH formula. 8 1.3 The Maurer-Cartan equations. 11 1.4 Proof of CBH from Maurer-Cartan. 14 1.5 The differential of the exponential and its inverse. 15 1.6 The averaging method. 16 1.7 The Euler MacLaurin Formula. 18 1.8 The universal enveloping algebra. 19 1.8.1 Tensor product of vector spaces. 20 1.8.2 The tensor product of two algebras. 21 1.8.3 The tensor algebra of a vector space. 21 1.8.4 Construction of the universal enveloping algebra. 22 1.8.5 Extension of a Lie algebra homomorphism to its universal enveloping algebra. 22 1.8.6 Universal enveloping algebra of a direct sum. 22 1.8.7 Bialgebra structure. 23 1.9 The Poincar´e-Birkhoff-Witt Theorem. 24 1.10 Primitives. 28 1.11 Free Lie algebras . 29 1.11.1 Magmas and free magmas on a set . 29 1.11.2 The Free Lie Algebra LX ................... 30 1.11.3 The free associative algebra Ass(X). 31 1.12 Algebraic proof of CBH and explicit formulas. 32 1.12.1 Abstract version of CBH and its algebraic proof. 32 1.12.2 Explicit formula for CBH. 32 2 sl(2) and its Representations. 35 2.1 Low dimensional Lie algebras. 35 2.2 sl(2) and its irreducible representations. 36 2.3 The Casimir element. 39 2.4 sl(2) is simple.
    [Show full text]
  • Determination of the Differentiably Simple Rings with a Minimal Ideal Author(S): Richard E
    Annals of Mathematics Determination of the Differentiably Simple Rings with a Minimal Ideal Author(s): Richard E. Block Source: The Annals of Mathematics, Second Series, Vol. 90, No. 3 (Nov., 1969), pp. 433-459 Published by: Annals of Mathematics Stable URL: http://www.jstor.org/stable/1970745 Accessed: 07/11/2010 19:03 Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use, available at http://www.jstor.org/page/info/about/policies/terms.jsp. JSTOR's Terms and Conditions of Use provides, in part, that unless you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, and you may use content in the JSTOR archive only for your personal, non-commercial use. Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at http://www.jstor.org/action/showPublisher?publisherCode=annals. Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission. JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Annals of Mathematics is collaborating with JSTOR to digitize, preserve and extend access to The Annals of Mathematics. http://www.jstor.org Determinationof the differentiablysimple ringswith a minimalideal* By RICHARD E.
    [Show full text]
  • The Jacobson Radical of a Semiring" in These PROCEEDINGS, 37, 163-170 (1951) Is in Part Incorrect
    462 ERRATA PROC. N. A. S. ERRA TA Dr. R. E. Johnson has called to my attention that Theorem 1 of my paper "The Jacobson Radical of a Semiring" in these PROCEEDINGS, 37, 163-170 (1951) is in part incorrect. This theorem should read as follows: If I is an ideal of a semiring S, then S is homomorphic to the difference semiring S = S - I. SAMUEL BOURNE Downloaded by guest on September 30, 2021 Vol.. 37, 1951 31MA THEMA TICS: S. BOURNE 163 a (W1(%qw(2) ..W(n)) However, the coefficient of u? is 1. Hence all the coefficients are rational numbers. Letting u2 = 1, ua = U4 = ... = = Owe seev(2) for (i = 1, 2, ...,n) are all algebraic numbers. Similarly, we have vj/) for i, j = 1, 2, ..., n are all algebraic numbers. Let K = R(vP2), 3(l)M, ..., P.(l)), where R is the rational numbers (i.e., K is the least algebraic extension of R containing v2M'), . ..,vI (1)). As ul + P2(1)u2 + . + Pn(l)un $ Ounless u, = u2 = ... = un =O, we see that (K:R) = n' n. Hence, consider the n' different isomorphisms of K leaving R fixed. As they leave R fixed they leave the coefficients of the u's in the expansion of (9) fixed. By Hilbert's theorem on the uniqueness of the irreducible factors of a polynomial in n variables, we see that the n' iso- morphisms of K must merely permute the linear factors on the right side of (9). As all isomorphisms of K act differently on ul + u2 2(1) + .
    [Show full text]
  • LIE ALGEBRAS 1 Definition of a Lie Algebra K Is a Fixed Field. Let L Be a K-Vector Space (Or Vector Space). We Say That L Is
    LIE ALGEBRAS 1 Definition of a Lie algebra k is a fixed field. Let L be a k-vector space (or vector space). We say that L is a k-algebra (or algebra) if we are given a bilinear map L×L −! L denoted (x; y) 7! xy. If L; L0 are algebras, an algebra homomorphism f : L −! L0 is a linear map such that f(xy) = f(x)f(y) for all x; y. We say that f is an algebra isomorphism if it is an algebra homomorphism and a vector space isomorphism. In this case f −1 : L0 −! L is an algebra isomorphism. If L is an algebra, a subset L0 ⊂ L is a subalgebra if it is a vector subspace and x 2 L0; y 2 L0 =) xy 2 L0. Then L0 is itself an algebra in an obvious way. If L is an algebra, a subset I ⊂ L is an ideal if it is a vector subspace; x 2 L; y 2 I =) xy 2 I; x 2 I; y 2 L =) xy 2 I. Then I is a subalgebra. Moreover, the quotient vector space V=I is an algebra with multiplication (x + I)(y + I = xy + I. (Check that this is well defined.) The canonical map L −! L=I is an algebra homomorphism with kernel I. Conversely, if f : L −! L0 is any algebra homomorphism then ker(f) = f −1(0) is an ideal of L. We say that the algebra L is a Lie algebra if (a) xx = 0 for all x; (b) x(yz) + y(zx) + z(xy) = 0 for all x; y; z.
    [Show full text]