Chapter 3 Simple Lie algebras. Classification and representations. Roots and weights 3.1 Cartan subalgebra. Roots. Canonical form of the algebra We consider a semi-simple (i.e. with no abelian ideal) Lie algebra. We want to construct a canonical form of commutation relations modeled on the case of SU(2) [J z,j ± ]=±J ± [J +,J ]=2J z. (3-1) It will be important to consider the algebra over C, at the price of complexifying it if it was originally real. The adjoint representation will be used. As it is a faithful representation for a semi-simple algebra, (i.e. ad X =0 X = 0, see exercise B of Chap. 2), no information is lost. It may also be useful to remember that the complex algebra has a real compact version, in which the real structure constants lead to a negative definite Killing form, and, as the representations can be taken unitary, the elements of the Lie algebra (the infinitesimal generators) may be taken as Hermitian (or antihermitian, depending on our conventions). 3.1.1 Cartan subalgebra We define first the notion of Cartan subalgebra. This is a maximal abelian subalgebra of g such that all its elements are diagonalisable (hence simultaneously diagonalisable) in the adjoint representation. That such an algebra exists is non trivial and must be established, but we shall admit it. If we choose to work with the unitary form of the adjoint representation, the elements of g are Hermitian matrices, and assuming that the elements of h are commuting among themselves ensures that they are simultaneously diagonalizable. This Cartan subalgebra is non unique, but one may prove that two distinct choices are related by an automorphism of the algebra g. 71
72 CHAPTER 3. CLASSIFICATION OF SIMPLE ALGEBRAS. ROOTS AND WEIGHTS. For instance if g is the Lie algebra of a Lie group G and if h is a Cartan subalgebra of g, any conjugate ghg 1 of h by an arbitrary element of G is another Cartan subalgebra. Let h be a Cartan subalgebra, call l its dimension, it is independent of the choice of h and it is called the rank of g. For su(2), this rank is 1, (the choice of J z for example); for su(n), the rank is n 1. Indeed for su(n), a Cartan algebra is generated 1 by diagonal traceless matrices, a basis of which is given by the n 1 matrices H 1 = diag (1, 1, 0,, 0), H 2 = diag (0, 1, 1, 0,, 0),,H n 1 = diag (0,, 0, 1, 1). (3-2) An arbitrary matrix of the Lie algebra, (in that representation), (anti-)hermitian and traceless, is diagonalisable by a unitary transformation; its diagonal form is traceless and is thus expressed as a linear combination of the h j ; the original matrix is thus conjugate by a unitary transformation of a linear combination of the h j. This is a general property, and one proves (Cartan, see [Bu], chap. 16) that If g is the Lie algebra of a group G, any element of g is conjugate by G of an element of h. Application. Canonical form of antisymmetric matrices. Using the previous statement, prove the Proposition If A = A = A T is a real skew-symmetric matrix ( of dimension ) N, one may find a real orthogonal matrix O such that A = ODO T 0 µ j where D = diag ( ) if N = 2n and D = µ j 0 ( ) 0 µ j diag (0, ) if N =2n +1, with real µ j. µ j 0 If one allows ( the complexification ) of orthogonal matrices, ( one ) may fully diagonalise the matrix A in the form iµ j 0 iµ j 0 D = diag ( ) or D = diag (0, ). For a proof making only use of matrix 0 iµ j 0 iµ j theory, see for example [M.L. Mehta, Elements of Matrix Theory, p 41]. 3.1.2 Canonical basis of the Lie algebra Let H i, i =1,,l be a basis of h. It is convenient to choose the H i Hermitian. By definition [H i,h j ] = 0, (abelian subalgebra) or more precisely, since we are in the adjoint representation, [ad H i, ad H j ] = 0. (3-3) We may thus diagonalise simultaneously these ad H i. We already know (some?) eigenvectors of vanishing eigenvalue since i, j, ad H i H j = 0, and we may complete them to make a basis by finding a set of eigenvectors E α linearly independent of the H j ad H i E α = α (i) E α (3-4) i.e. a set of elements of g such that [H i,e α ]=α (i) E α, (3-5) 1 We use momentarily the representation of definition (made of n n matrices) rather than the adjoint representation.
3.1. CARTAN SUBALGEBRA. ROOTS. CANONICAL FORM OF THE ALGEBRA 73 with the α (i) not all vanishing (otherwise the subalgebra h would not be maximal). The space h. In these expressions, the α (i) are eigenvalues of the operators ad H i. Since we chose Hermitian ad H i, their eigenvalues α (i) are real. By linearity, for an arbitrary element of h written as H = i hi H i, ad (H)E α = α(h)e α, (3-6) and the eigenvalue of ad (H) on E α is α(h) := i hi α (i), which is a linear form on h. In general linear forms on a vector space E form a vector space E, called the dual space of E. One may thus consider the root α, of components α (i), as a vector of the dual space of h, hence α h, the root space. Note that α(h i )=α (i). Roots enjoy the following properties 1. if α is a root, α in another root; 2. the eigenspace of the eigenvalue α is of dimension 1 (no multiplicity); 3. if α is a root, the only roots of the form λα are ±α; 4. roots α generate all the dual space h. For proofs of 1., 2., 3., see below, for 4. see exercise A. Number of roots. Since the H j are diagonalisable, the total number of their eigenvectors E α and H i must be equal to the dimension of the space, here the dimension d of the adjoint representation, i.e. of the Lie algebra g. As any (non vanishing by definition) root comes along with its opposite, the number of roots α is even and equal to d l (with l = rank(g)). We denote the set of roots. In the basis {H i,e α } of g, the Killing form takes a simple form (H i,e α ) = 0 (E α,e β ) = 0 unless α + β =0. (3-7) To show that, we write (H, [H,E α ]) = α(h )(H, E α ), and also, using the definition of the Killing form and the cyclicity of the trace (H, [H,E α ]) = tr (ad H[ad H, ad E α ]) = tr ([ad H, ad H ] ad E α ) = 0 (3-8) since [ad H, ad H ] = 0. It follows that H, H h, α(h )(H, E α ) = 0, hence that (H, E α ) = 0. Likewise ([H, E α ],E β )=α(h)(e α,e β )= (E α, [H, E β ]) = β(h)(e α,e β ) (3-9) again by the cyclicity of the trace, and thus (E α,e β ) = 0 if H : (α + β)(h) 0, i.e. if α + β 0. Note that the point 1. above follows simply from (3-7): if α were not a root, E α would be orthogonal to all elements of the basis hence to any element of g, and the form would be degenerate, contrary to the hypothesis of semi-simplicity (and Cartan s criterion). For an elegant proof of points 2. et 3., see [OR, p. 29]. The restriction of this form to the Cartan subalgebra is non-degenerate, since otherwise one would have H h, H h : (H, H ) = 0, but (H, E α ) = 0, thus X g, (H, X) = 0 and the form would be degenerate, contrary to the hypothesis of semi-simplicity (and Cartan s criterion, Chap. 1, 4.4). The Killing form being non-degenerate on h, it induces an isomorphism between h and h : to α h one associates the unique H α h such that H h α(h) =(H α,h). (3-10)
74 CHAPTER 3. CLASSIFICATION OF SIMPLE ALGEBRAS. ROOTS AND WEIGHTS. (Or said differently, one solves the linear system g ij h j α = α (i) which is of Cramer type since g ij =(H i,h j ) is invertible.) One has also a bilinear form on h inherited from the Killing form α, β =(H α,h β ), (3-11) which we are going to use in 2 to study the geometry of the root system. It remains to find the commutation relations of the E α among themselves. Using the Jacobi identity, one finds that ad H i [E α,e β ]=[H i, [E α,e β ]] = [E α, [H i,e β ] [E β, [H i,e α ]] = (α + β) (i) [E α,e β ]. (3-12) Invoking the trivial multiplicity (=1) of roots, one sees that three cases may occur. If α + β is a root, [E α,e β ] is proportional to E α+β, with a proportionality coefficient N αβ which will be shown below to be non zero (see 2.1 and exercise B). If α + β 0 is not a root, [E α,e β ] must vanish. Finally if α + β = 0, [E α,e α ] is an eigenvector of all ad H i with a vanishing eigenvalue, thus [E α,e α ]=H h. To determine that H, let us proceed like in (3-9) (H i, [E α,e α ]) = tr (ad H i [ad E α, ad E α ]) = tr ([ad H i, ad E α ] ad E α ) = α (i) (E α,e α ) = (H i,h α )(E α,e α ) (3-13) hence [E α,e α ] = (E α,e α )H α. (3-14) To recapitulate, we have constructed a canonical basis of the algebra g [H i,h j ] = 0 [H i,e α ]= α (i) E α N αβ E α+β if α + β is a root [E α,e β ]= (E α,e α ) H α if α + β =0 0 otherwise (3-15) Up to that point, the normalisation of the vectors H i and E α has not been fixed. It is common to choose (H i,h j )=δ ij (E α,e β )=δ α+β,0. (3-15) (Indeed, the restriction of the Killing form to h, after multiplication by i to make the ad H i Hermitian, is positive definite.) With that normalisation, H α defined above by (3-10) satisfies also Note that E α,e α and H α form an su(2) subalgebra H α = α.h := α (i) H i. (3-16) [H α,e ±α ]=± α, α E ±α [E α,e α ]=H α. (3-17)
3.5. YOUNG TABLEAUX AND REPRESENTATIONS OF GL(N) AND SU(N) 99 Exercises for chapter 3 A. Cartan algebra and roots 1. Show that any element X of g may be written as X = x i H i + α xα E α with the notations of 3.1.2. For an arbitrary H in the Cartan algebra, determine the action of ad (H) on such a vector X; conclude that ad (H)ad (H )X = α xα α(h)α(h )E α and taking into account that the eigenspace of each root α has dimension 1, cf point 2. du 3.1.2, that the Killing form reads (H, H ) = tr (ad (H)ad (H )) = α(h)α(h ). (3-81) α 2. One wants to show that roots α defined by (3-5) or (3-6) generate all the dual space h of the Cartan subalgebra h. Prove that if it were not so, there would exist an element H of h such that α α(h) = 0. (3-82) Using (3-81) show that this would imply H h, (H, H ) = 0. Why is that impossible in a semi-simple algebra? (see the discussion before equation (3-10)). 3. Variant of the previous argument: under the assumption of 2. and thus of (3-82), show that H would commute with all H i and all the E α, thus would belong to the center of g. Prove that the center of an algebra is an abelian ideal. Conclude in the case of a semi-simple algebra. B. Computation of the N αβ 1. Show that the real constants N αβ satisfy N αβ = N βα and, by complex conjugation of [E α,e β ]=N αβ E α+β that N αβ = N α, β. (3-83) 2. Consider three roots satisfying α + β + γ = 0. Writing the Jacobi identity for the triplet E α,e β,e γ, show that α (i) N βγ + cycl. = 0. Derive from it the relation N αβ = N β, α β = N α β,α. (3-84) 3. Considering the α-chain through β and the two integers p et q defined in 3.2.1, write the Jacobi identity for E α, E α and E β+kα, with p k q, and show that it implies α, β + kα = N α,β+kα N α,β+(k 1)α + N β+kα,α N α,β+(k+1)α. Let f(k) := N α,β+kα N α, β kα. Using the relations (3-84), show that the previous equation may be recast as α, β + kα = f(k) f(k 1). (3-85) 4. What are f(q) et f(q 1)? Show that the recursion relation (3-85) is solved by f(k) = (N α,β+kα ) 2 =(k q) α, β + 1 (k + q + 1). (3-86) 2 What is f(p 1)? Show that the expression (3-21) is recovered. Show that (3-86) is in accord with (3-23). The sign of the square root is still to be determined..., see [Gi].