CHAPTER IX

Automorphisms

In this chapter we shall study the groups of automorphisms of semi-simple Lie algebras over an algebraically closed field of characteristic 0. If z is an element of a Lie algebra of characteristic 0 such that ad z is nilpotent, then we know that exp (ad z) is an automorphism. Products of automorphisms of this type will be called invariant automorphisms. These constitute a subgroup G₀() of the group of automorphisms G() of . If τ is any automorphism, then τ^–1(exp ad z)τ = exp ad z^τ and ad z^τ is nilpotent. It is clear from this that G₀ is an invariant subgroup of G.

The main problem we shall consider in this chapter is the determination of the index of G₀ in G for finite dimensional simple over an algebraically closed field Φ of characteristic zero. We show first that if is finite-dimensional over Φ (not necessarily simple), then G₀ acts transitively on the set of Cartan subalgebras, that is, if ₁ and ₂ are Cartan subalgebras then there exists a σ ε G₀ such that A conjugacy theorem of this type was first noted by Cartan in the case of semi-simple over the field of complex numbers and it was applied by him to the study of the automorphisms of these algebras. The extension and rigorous proof of the conjugacy theorem is due to Chevalley. This result reduces the study of the position of G₀ in G to the study of automorphisms which map a Cartan subalgebra into itself.

If is semi-simple, then the Weyl group plays an important role in our considerations. We shall require also some explicit calculations of invariant automorphisms due to Seligman. The final results we derive give the group of automorphisms for the Lie algebras A_l, B_l, C_l, D_l, l > 4, G₂ and F₄. It is noteworthy that similar results can be obtained in the characteristic p case (see Jacobson [8] and Seligman [4]). As usual, for the sake of simplicity we stick to the characteristic 0 case. In the next chapter we shall extend our final results for non-exceptional simple Lie algebras over algebraically closed fields to algebras of this type over arbitrary base fields (of characteristic 0).

1. Lemmas from algebraic geometry

Let be a finite-dimensional vector space with basis (u₁, u₂, …, u_m) over an infinite field Φ. Any x has a unique representation as x = Σξ_iu_i and the ξ_i are the coordinates of x relative to the basis (u_i)· Let f(λ₁, ···, λ_m) be an element of the polynomial ring Φ[λ₁, ···, λ_m], in the m indeterminates λ_i with coefficients in Φ. Then f(λ₁, ···, λ_m) and the basis (u_i) define a mapping f of into Φ by the rule that the image f(x) = f(ξ) = f(ξ₁, ···, ξ_m). (In this chapter we shall often use the notation f(x) rather than xf or x^f.) We call f a polynomial function on . If (u′₁, ···, u′₂) is a second basis, then it is readily seen that the function f can be defined by another polynomial in Φ[λ₁, ···, λ_m] with respect to (u′_i). In this sense the notion of a polynomial function is independent of the choice of the basis for . The set of polynomial functions is an algebra Φ[] relative to the usual compositions of functions. We recall that if f and g are functions on with values in Φ, then (f + g)(x) = f(x) + g(x), (αf)(x) = αf(x) for α in Φ, (fg)(x) = f(x)g(x). The mapping is a homomorphism of Φ[λ₁, ···, λ_m] into Φ []. Since Φ is infinite, f(ξ₁···, ξ_m) = 0 for all ξ_i in Φ implies f(λ₁, …, λ_m) = 0. Hence the homomorphism is an isomorphism. The isomorphism maps λ_i into the projection function π_i such that π_i(x) = ξ_i. Hence it is clear that the π_i generate the algebra Φ[].

Let be a second finite-dimensional space over Φ with the basis (v₁, v₂, ···, v_n). A polynomial mapping P of into is a mapping of the form where η_j = p_j (ξ₁, ···, _m), p_j(λ₁, ···, λ_m) ε Φ[λ₁, ···, λ_m]. This notion is independent of the basis. The set of polynomial mappings of into is a vector space under the usual addition and scalar multiplication of mappings. The resultant of a polynomial mapping P of to and a polynomial mapping Q of to is a polynomial mapping PQ of into . The notion of a polynomial function is the special case of a polynomial mapping in which the image space is the one-dimensional space Φ. Hence if P is a polynomial mapping of into and f is a polynomial function on , then Pf is a polynomial function on . The mapping f → Pf is a mapping of the algebra Φ [] of polynomial functions on into Φ [] We now look at the form of this mapping. Thus we have P(x) = y where x = Σ ξ_iu_i, y = Ση_jv_j and

P_j(λ₁, ···, λ_m) ε Φ[λ₁, ···, λ_m]. Also we have f(y) = f(η₁···, η_n), f(μ₁, ···, μ_n), εΦ[μ₁, ··· μ_n], μ_j indeterminates. Hence Pf maps x into

where p_j(ξ) = p_j (ξ₁, ξ₂, ···, ξ_m). If f, g ε Φ [], then

This shows that f → Pf is a homomorphism σ_p of Φ [] into Φ [].

Conversely, let σ be an algebra homomorphism of Φ [] into Φ[]. Consider the projection p_j: Σ η_kv_k → η_j. Suppose is the mapping Σ ξ_i u_i → f_j(ξ₁, ξ₂, ···, ξ_m) and let P be the polynomial mapping of into such that Σ ξ_iu_i → Σ η_iv_i where η_i = f_j(ξ). Then P_ρj maps Σ ξ_iu_i into f_j(ξ) so that Pρ_j = ρ^σ_j Since the ρ_j generate Φ[] it follows that σ coincides with the homomorphism σ_p determined by P. Thus every homomorphism of Φ [] into Φ [] (sending 1 into 1) is realized by a polynomial mapping of into . It is easy to see that if P and Q are two such polynomial mappings, then the homomorphism σ_p = σ_Q if and only if P = Q. Hence we have a 1 : 1 correspondence between the polynomial mappings of into and the homomorphisms of Φ[] into Φ[]

Of particular importance for us is the set of algebra homomorphisms of Φ[] into the base field Φ. This can be obtained in a somewhat devious manner by identifying Φ with the algebra Φ[] of polynomial functions on = 0 and applying the foregoing result. However, it is more straightforward to look at this directly. We note first that if y ε , then the mapping σ_y: f → f(y), the specialization of f at y, is a homomorphism of Φ[] into Φ. Conversely, if σ is any homomorphism of Φ[] into Φ, we let as before. Then it is immediate that σ = σ_y where y = Σ η_jv_j. It is clear also that if y₁ ≠ y₂ in , then σ_y1 ≠ σ_y2

Let f ε Φ[] and let a = Σ α_iu_i ε . Then we can define a linear function d_αf on by

It is easy to see that d_αf is independent of the basis used to define this mapping. The linear mapping d_αf is called the differential of f at a. We have the following properties

If t is an indeterminate and f is extended to _Φ(t) in the obvious way, then we have the following relation in the algebra Φ[t], which is a consequence of Taylor’s formula

More generally, let P be a polynomial mapping of into . Then we define a linear mapping d_aP, the differential of P at a by

Again, one can verify that this is independent of the bases chosen in and . Also one has the useful generalization of (5):

A set of generators for the image space (d_aP) is the set of vectors

Hence d_aP is surjective if and only if the Jacobian matrix

has rank n.

LEMMA 1. Assume Φ perfect. Then if d_aP is surjective for some a, the homomorphism σ_P is an isomorphism of Φ[] into Φ[].

Proof. Our hypothesis is that one of the n-rowed minors of the matrix (∂p_j/∂λ_i) is not zero. If σ_P is not an isomorphism, then there exists a polynomial f(μ₁,···, μ_n) ≠ 0 such that f(p₁(ξ), p₂(ξ), ···, p_nξ) = 0 for all ξ_i in Φ. This implies that f(p₁(λ), ···, p_n(λ))= 0, that is, the polynomials p₁(λ₁, ···, λ_m), ···, p_n(λ_i, ···, λ_m) are algebraically dependent. If this is the case, then we may assume that the polynomial f ≠ 0 giving the dependence is of least degree. The relation f(p₁(λ), ···, P_n(λ)) = 0 gives

This contradicts the hypothesis on the Jacobian matrix unless (∂f/∂μ_j)(p₁(λ), ···, p_n(λ)) = 0. Since the degree of f is minimal for algebraic relations in the p₁(λ), ···, p_m(λ) we must have (∂f/∂μ_j) = 0, j = 1, ···, n. This implies that f is a non-zero element of Φ—which is absurd—if the characteristic is 0. If the characteristic is p ≠ 0, we obtain that f is a polynomial in μ^p₁, μ^p₂, ···, μ^p_n Since Φ is perfect this implies that f = g^p, g a polynomial in the μ’s. Then g(p₁(λ), ···, p_m(λ) = 0 which again contradicts the minimality of the degree of f.

The main result we shall require for the conjugacy theorem for Cartan subalgebras is the following

THEOREM 1.Let Φ be algebraically closed and let P be a polynomial mapping of into such that d_aP is surjective for some a ε . If f is a non-zero polynomial function on , then there exists a non-zero polynomial function g on such that if y is any element of satisfying g(y) ≠ 0, then there exists an x in such that f(x) ≠ 0 and P(x) = y.

In geometric form this result has the following meaning: Given an “open” set in defined to be the set of elements x such that f(x) ≠ 0 for the non-zero polynomial f, then there exists an open set in defined by g(y) ≠ 0, g a non-zero polynomial which is completely contained in the image under P of the given open set in . (Suggestion: Draw a figure for this.) We shall see that Theorem 1 is an easy consequence of the following theorem on extensions of homomorphisms.

THEOREM 2. Let Φ be an algebraically closed field and let Γ be an extension field of Φ, a subalgebra of Γ and ′ an extension algebra of of the form ′ = [u₁, u₂, ···, u_r], u_i ε Γ. Let f be a non-zero element of ′. Then there exists a non-zero element g in such that if σ is any homomorphism of into Φ such that g^σ ≠ 0, then σ has an extension homomorphism τ of ^′ into Φ such that f^τ≠ 0.

Proof·. Suppose first that r = 1, so that ′ = [u], Case I: u is transcendental over the subfield Σ of Γ generated by . We write f = f_o + f₁u + ···, + f_mu^r, f_i ε , f_r ≠ 0. Let g = f_r and let σ be a homomorphism of into Φ such that g^σ ≠ 0. Consider the polynomial in Φ[λ], λ an indeterminate. Since , this polynomial has at most m roots in Φ so we can choose c in Φ so that Let τ be the homomorphism of ′ = [u] into Φ such that . This is an extension of σ and f^τ≠ 0 as required. Case II: u is algebraic over Σ. The canonical homomorphism of [λ], λ an indeterminate, onto ′ = [u](identity on , λ → u) has a non-zero kernel . Since f ε [u], f is algebraic over Σ also. Let p(λ), q(λ) be non-zero polynomials in [λ] of least degree such that p(u) = 0 and q(f) = 0. Then these polynomials are also of least degree in Σ[λ] such that p(u) = 0, q(f) = 0. Hence they are irreducible in Σ[λ]. Let g₁ be the leading coefficient of p(λ) and g₂ = q(0) and choose g = g₁ g₂. We shall show that g has the required property for f. Thus suppose σ is a homomorphism of into Φ satisfying g^σ = g^σ₁g^σ₂ ≠ 0 and suppose τ is any extension of σ to a homomorphism of ′ = [u] into Φ. Since g(f) = 0 we shall have g^σ(f^τ) = 0, which implies that f^τ≠ 0 since g^σ(0) ≠ 0. Thus we need to show only that the homomorphism σ can be extended to a homomorphism of ′. For this purpose let c be a root of p^σ(λ) = 0 and consider the homomorphism τ′ of [λ] which coincides with σ on and maps λ into c. Let h(λ) be any element in the ideal . Then the minimality of the degree of p(λ) implies that there exists a non-negative integer k such that is divisible in [λ] by p(λ). Since and since Hence h(λ)^τ′ = 0 and so is mapped into 0 by τ′. It follows that τ′ induces a homomorphism τ of ′ = [u] ≅ [λ]/ which is an extension of σ.

Now assume the result holds for r – 1. Let = [u_r] so that ′ = [u₁, ···, u_r-1]· Then there exists an element h ε such that any homomorphism p of such that h^ρ ≠ 0 has an extension τ to ′ such that f^τ ≠ 0. By the case r = 1, there exists g ε such that any homomorphism σ of into Φ such that g^σ ≠ 0 has an extension ρ to = [u_r] such that h^ρ ≠ 0. Hence τ is an extension of σ such that f^τ ≠ 0 as required.

We now give the

Proof of Theorem 1 : The hypothesis on d_aP implies that σ_P is an isomorphism of Φ[] into Φ[] Let = Φ[]^σP ′ = Φ[]. If π₁ ···,π_m are the projection mappings of into Φ we have ′ = Φ[π₁, ···, π_m] = [π₁, ···, π_m], Since ′ is an integral domain we may suppose it imbedded in its field of quotients Γ. Hence we can apply Theorem 2 to and ′. Let f be a non-zero element of ′ = Φ[]. Then there exists a non-zero element g ∈ Φ [] such that if y is an element of such that g(y) ≠ 0 then the homomorphism σ: h^σp → h(y) of = Φ[]^σp into Φ, which satisfies g^σpσ = g(y) ≠ 0, can be extended to a homomorphism τ of Φ[] into Φ satisfying f^τ≠0. We have seen that τ has the form k → k(x) where x is an element of . Then f(x) ≠ 0 and for every h ∈ , h^σp = h^σpσ. This means that h(P(x)) = h(y). Hence P(x) = y and the theorem is proved.

2. Conjugacy of Cartan subalgebras

Let be a finite-dimensional Lie algebra over an algebraically closed field of characteristic 0 and let be a Cartan subalgebra of . Let

be the decomposition of into root spaces corresponding to the roots 0, α, β, ··· of acting in . If h ∈ and e_α ∈ _α, then there exists an integer r such that e_α(ad h — α(h)1)^r = 0. This is equivalent to the condition that α(h) is the only characteristic root of the restriction of ad h to _α. The α are linear functions on . If x ∈ _ρ (ρ = 0 or ρ ≠0) then [xe_a] = 0 or ρ + α is a root and [xe_a] ∈_{ρ + α}. In the latter case either [[xe_a]e_a] = 0 or ρ + 2a is a root and [[xe_α]e_α] ∈ _ρ+2α. If we continue in this way and we take into account the fact that there are only a finite number of distinct roots we see that x(ad e_a)^k = 0 for sufficiently high k. This implies that ad e_α is nilpotent for every e_α € _α, α ≠ 0. It follows that if ∈ e_α1 ∈ _α1 ···, eα _k ∈ _αk, α₁, α₂, ···, α_k non-zero roots, then

is an invariant automorphism of .

Now let (h₁, h₂, ···, h_l, e_{l + 1} ···, e_n) be a basis for such that (h₁ h₂, · ·, h_n) is a basis for and the elements e_l+1, ···, e_n are in root spaces _α, α ≠ 0. Let λ₁,··· λ_n be indeterminates, P = Φ (λ₁, ···, λ_n) and form the element

where the p_i and p_j are polynomials in the λ’s. These determine a polynomial mapping

in .

The product αβ ··· ρ of the non-zero roots is a non-zero polynomial function. It follows that there exist such that if , the α(h⁰)β(h⁰), ···, ρ(h°) ≠ 0. Then the characteristic roots of the restriction of ad h⁰ to _α + _β + ··· + _ρ are all different from 0 and so this restriction of ad h⁰ is non-singular.

We shall now calculate the differential d_h⁰P of P at h⁰. For this purpose we let , let t be an indeterminate and we consider

If we compare this with (7) we we see that d_h⁰ P is the mapping

Since h → h and e → [h⁰e] are non-singular it follows that d_h⁰P is surjective. We are therefore in a position to apply Theorem 1. Accordingly, we have the following result: If f is a polynomial function ≠ 0 on , then there exists a polynomial function g ≠ 0 on such that if y ∈ and g(y) ≠ 0, then there is an x in such that P(x) = y and f(x) ≠ 0.

We recall the definition of a regular element a of as an element such that ad α has the minimum number l′ of 0 characteristic roots. We recall also that if a is regular, then the set of vectors h belonging to the characteristic root 0 of ad a is a Cartan subalgebra. It follows from this that if is a Cartan subalgebra and contains a regular element a, then is just the collection of elements h ∈ such that h(ad a)^r = 0 for some integer r. We shall need also the characterization given in § 3.1 of regular elements. For this we take the element in _p, P =Φ(λ, ···, λ_n) and we consider the characteristic polynomial

Then τ_n–l′(λ₁, ···, λ_n) is a non-zero homogeneous polynomial of degree n — l′ in the λ’s and if , then x→τ_n-l′(x) = τ_n-l (ξ₁, ···, ξ_n) is a polynomial function on . The element x is regular in if and only if τ_n-1′(x) ≠ 0. (It will be a consequence of the theorem we are going to prove that l′ = l.)

We now consider again the Cartan subalgebra and the basis (h₁, ···, h_l, e_l+1 ···,e_n) for . We apply Theorem 1 to the polynomial function f = τ^σp_n-l′ which is ≠ 0 since τ_n-l′ ≠ 0 and σ_p is an isomorphism. Accordingly we see that there is a non-zero polynomial function g on such that every y ∈ satisfying g(y) ≠ 0 has the form P(x) where . Hence every y such that g(y) ≠ 0 is regular and if , then y = P(x) = (Σξ_ih_i) (exp ad ξ_{l + 1} e_{l + 1}) ··· (exp ad ξ_ne_n) = h^η where h = Σξ_ih_i and η is an invariant automorphism. Thus y is the image of an element h ∈ under an invariant automorphism. It follows that h = y^η-1 is regular. It is now easy to prove the conjugacy theorem for Cartan subalgebras.

THEOREM 3. If ₁, and ₂ are Cartan subalgebras of a finitedimensional Lie algebra over an algebraically closed field of characteristic 0, then there exists an invariant automorphism η such that

Proof : There exists a non-zero polynomial function gi such that if y is an element satisfying gi(y) ≠ 0, then a regular element in i and ηi an invariant automorphism. Since g1 g2 ≠ 0 we can choose y so that g1(y) ≠ 0 and g2(y) ≠ 0. Then y = hη11 = h′22, hi a regular element of i, ηi an invariant automorphism. Then . Since hi is regular and is contained in i it follows that

Remarks. It is a consequence of the theorem that every Cartan subalgebra contains regular elements and that all Cartan subalgebras have the same dimensionality I which is the same as the number l′ indicated above. It is easy to see also that if the notations are as before, then the regular elements of belonging to are just the elements h° such that α(h⁰)β(h⁰) ··· ρ(h°) ≠ 0.

3. Non-isomorphism of the split simple Lie algebras

We shall apply the conjugacy theorem for Cartan subalgebras first to settle a point which has been left open hitherto, namely, that the split simple Lie algebras which were listed in §§4.5-4.6 are distinct in the sense of isomorphism. We recall that these were: A_l, l 1, B_l, l 2, C_l, l 3, D_l, l 4, G₂, F₄, E₆, E₇, and E₈. The dimensionalities of these algebras are given in the following table:

For the classical types and for G₂, F₄ and E₆ this was derived in §4.6. We have proved the existence of E₇ and E₈ in §7.5. The dimensionalities of these Lie algebras can be derived by determining the positive roots directly from the Cartan matrices. We shall not carry this out but we shall assume the result for these two Lie algebras.

To prove that no two of the Lie algebras we have listed are isomorphic it suffices to assume the base field algebraically closed. This is clear since ₁ ≅ ₂ implies _1P ≅ _2p for any extension P of the base field. We therefore assume Φ algebraically closed. The subscript l in the designation X_l (e.g., A_l, E₆) for our Lie algebras is the dimensionality of a Cartan subalgebra. The conjugacy theorem shows that this is an invariant. Hence necessary conditions for isomorphism of X_l and Y_l′ are l=l′ and dim X_l = dim Y_l′. A glance at the list of dimensionalities shows that the only possible isomorphisms which we may have are between B_l and C_l, l 3 and between B₆ and E₆ and between C₆ and E₆. The latter two have been ruled out in an exercise (Exercise 7.6). It therefore remains to show that B_l C_l, l 3.

Since C_l has an irreducible module in a 2l-dimensional space it will suffice to show that if is an irreducible module for B_l such that B_l ≠ 0, then dim 2l + 1. We could establish this, as in a similar discussion for G₂ (§ 7.6), by using the fact that the set of weights is invariant under the Weyl group. However, we can now obtain the result more quickly by using Weyl’s dimensionality formula. We observe first that (8.41) shows that if is an irreducible module of least dimension satisfying B_l ≠ 0, then is a basic module (cf. also Exercise 7.13). The dimensionalities of these are , k = 1, 2, ···, l – 1 and 2^l (cf. (8.45) and Exercise 8.12). Since l 3, these numbers exceed 2l.

This proves our assertion and completes the proof that B_l and C_l are not isomorphic if l 3.

4. Automorphisms of semi-simple Lie algebras over an algebraically closed ñeld

Let be a finite-dimensional semi-simple Lie algebra over an algebraically closed field of characteristic 0, a Cartan subalgebra, π = {α₁, ···, α_l} a simple system of roots relative to , e_i, f_i, h_i, i = 1, 2, ···, l, a set of canonical generators for determined by π. Thus the h_i form a basis for , e_i ∈ _αi, f_i ∈ _-αi and we have the following relations:

where (A_ij) is the Cartan matrix determined by π.

Let τ be an automorphism. Then ^τ is a second Cartan subalgebra. Hence there exists an invariant automorphism σ such that ^σ = ^τ. Then the automorphism τ′ = τσ ^-1 maps into itself. We now consider an automorphism τ(=τ′) which maps the Cartan subalgebra into itself. If e_α ∈ _α we have [e_αh] = α(h)e_α. Hence [e^τ_αh^τ] = α(h)e^τ_α. It follows that where β is a root. In this way we obtain a mapping α → β of the set of roots. Since e_α^τ ∈ _β we have [e^τ_αh h^τ] = β(h^τ)β^τ_α. Hence we see that

Let τ^* denote the transpose in ^* of the restriction of τ to . By definition, if ξ ∈ ^*, then ξ^τ* = ξ(h^τ). Then (19) implies that β^τ* = α or β = α^(τ*)-1. Hence we have the following

PROPOSITION 1. Let τ be an automorphism of such that ^τ = for a Cartan subalgebra of . Then if α is any root of in ,

where τ^* is the transpose in ^* of the restriction of τ to .

We note next that if π = {α₁, ···, α_l} and β_i = α_i^(τ*)-1, then π′ = {β ₁, ···, β_l} is a simple system of roots. Thus every root has the form ± Σ k_iα_i where the k_i are non-negative integers. If we apply (τ^*)^–1 we see that every root also has the form ± Σk_iβ_i. This guarantees that π′ is a simple system. We prove next

PROPOSITION 2. Let π and π′be simple systems of roots. Then there exists an invariant automorphism σ such that ^σ = and π ^(σ*)-1= π′.

Proof: Let π = {α₁, α₂, ···, α_l}, π’ = {β ₁, ···, β_l}. Then we know (Theorems 8.1, 8.2) that π′ = π S_αi1S_αi2, ··· S_αir for suitable Weyl reflections S_αi. It is therefore clear that it suffices to prove the result for π′ = πS_αi. For this purpose we introduce the invariant automorphisms exp ad ξf_i and exp ad ξe_i, ξ ∈ Φ. In our calculation we shall use the formulas for the irreducible representations of the three-dimensional split simple Lie algebra given in (36) of § 3.8. We note that the matrices of the restrictions of ad f_i, ad e_i to _i = Φe_i + Φh_i + Φf_i, using the basis (e_i, h_i, [h_i f_i]) are, respectively,

Hence for exp ad ξ f_i, exp ad ξe_i in _i we have the matrices

It follows that the matrix of the restriction to _i of

is

In particular, . Next let h ∈ satisfy α_i(h) = 0. Then [he_i]= 0 = [hf_i] and consequently h^σ_i(ξ) = h. Since is the direct sum of Φh_i and the subspace of vectors h such that α_i(h) = 0 it is clear that ^σ_i(ξ) Moreover, the restriction of σ_i(ξ) to is the reflection determined by h_i : h → h — [2(h, h_i)/(h_i, h_i)]h_i = h — [2(h, h_αi)/(h_αi, h_αi)] h_αi = h-[2αi(h)/ α_i(h_αi)]h_αi. If ρ ∈ ^*, then

This shows that the transpose inverse of the restriction to of σ_i() is the Weyl reflection S_αi in *. Hence the invariant automorphism σ_i() satisfies , as required.

Proposition 2 and the considerations preceding it show that if τ is an automorphism such that ^τ = , then there exists an invariant automorphism σ such that ^σ = and if τ′ = τ σ^-1, then π((τ′)*)^-1 = π for the simple system π, We simplify our notation again by writing τ for τ¹. Then we have a permutation i → i′ of i = 1, 2, ···, l such that α^(τ*)-1 = α_i′. Also we have . Since , we have . Then . Since and So that we have and Since Hence we have

The subgroup of the symmetric group on 1, 2, ···, l of the permutations i → i′ satisfying (25) will be called the group of automorphisms of the Cartan matrix (A_ij). If we recall the definition of the Dynkin diagram of the Cartan matrix, it is clear that if α₁, α₂, ···, α_l are the vertices of the diagram, then any element i → i′ of the group of automorphisms of (A_ij) defines an automorphism of the Dynkin diagram, that is, a 1:1 mapping α_i → α_i such that (α_iα_i) and for any i,j, the number of lines connecting α_i to α_j is the same as that connecting a_i′ and a_j′ (cf. §4.5). The argument used to show that the Dynkin diagram determines the Cartan matrix shows also that the converse holds, that is, if α_i → α_i′ is an automorphism of the Dynkin diagram, then i → i′ is an automorphism of the Cartan matrix.

Now suppose that τ is an automorphism such that , i = 1, 2, ···, l. Then we have the identity mapping i→i′ = i in the foregoing argument. This shows that Hence τ acts as the identity in . Conversely, Prop. 1 shows that if the restriction of τ to is the identity, then for every a so For these automorphisms we have the following

PROPOSITION 3. If τ is an automorphism such that h^τ = h for every h of a Cartan subalgebra of then τ is an invariant automorphism.

Proof: We have seen that Let σ_i(ξ) be the invariant automorphism defined by (23) and let

We have seen that if α_i(h) = 0. It follows that for all h. The matrix relative to (e_i, h_i, [h_if_i]) of the restriction of ω_i(ξ) to _i is the product of the matrix in (24) by the matrix obtained from this by taking ξ = 1. The result is

Hence we have

We wish to calculate next for j ≠ i. We have [f_je_i] = 0 and [f_jh_i] = — A_ijf_j. Hence f_i generates an irreducible module for _i. There are four possibilities —A_ij = 0, 1, 2 or 3. If A_ij = —2 the module is equivalent to _i and the argument just used shows that and this implies that . If A_ij = 0, [f_{i i}] = 0 and this implies that Next let A_ij = —1. Then the matrices of f_i, e_i acting in the module generated by f_i are, respectively

It follows that the matrix of σ_i(ξ) is

and that of ω_i(ξ) is diag{–ξ, -ξ^-1}. Hence we have and . Finally, let A_ij = — 3. Then the representing matrices for f_i and e_i are, according to to (36) of § 3.8:

It follows that the matrix of σ_i(ξ) as defined in (23) is

Consequently, the matrix of ω_i(ξ) in the module generated by f_i is diag {—ξ³, —ξ, —ξ —ξ^-1, —ξ^-3}. Hence and Thus in all cases we have

(j = i or j ≠ i). Now set

Then (33) implies that

We recall that the matrix (A_ij) is non-singular and its determinant is clearly an integer d. Hence we have an integral matrix (B_ij) such that (A_ij)(B_ij) = d1. Let j be fixed and set ξ_k k = 1, …, l. Then

It is clear that a suitable product of the invariant automorphisms defined here for j = 1,2,···, l coincides with the given automorphism τ in its action on the e_i, f_i, h_i. Since these are generators it is clear that τ is an invariant automorphism.

It is now easy to see that the index of the invariant subgroup G₀ in G does not exceed the order of the group of automorphisms of the Cartan matrix. In fact, we can display a subgroup K of G which is isomorphic to the group of automorphisms of the Cartan matrix such that every τ ∈ G is congruent modulo G₀ to a τ_P ∈ K. Let P: i → i′ be an automorphism of the Cartan matrix. Then we have the relations

Hence the isomorphism theorem for split semi-simple Lie algebras (Theorem 4.3) implies that there exists a (unique) automorphism τ_p of such that The set of these automorphisms is evidently a subgroup K of G isomorphic to the group of automorphisms of the Cartan matrix.

Now let τ be any automorphism of . We have seen that τ is congruent modulo G₀ to an automorphism τ₁ such that ^τ₁ = Also we know that τ₁ is congruent modulo G₀ to an automorphism τ₂ such that ^τ2 = and for the simple system of roots π = {α, α₂, ···, α_l} relative to . We have and i→i′ is an automorphism P of the Cartan matrix. Let be the corresponding element of K. Then satisfies . Hence σ is an invariant automorphism, by Prop. 3. Thus τ is congruent modulo G₀ to τ_P.

It can be shown that no element of K is in G₀, which means that G is the semi-direct product of K and G₀. This is equivalent to showing that the index of G₀ in G is the order of the group of automorphisms of the Cartan matrix. A proof of this will be indicated in the exercises. We shall now restrict our attention to simple. The result stated will follow for the Lie algebras A_l, B_l, C_l, D_l, l ≠ 4, G₂ and F₄ from the explicit determination of the groups of automorphisms for these Lie algebras which we shall give in the next section. The result will also be clear for E₇ and E₈.

We now examine the groups of automorphisms of the connected Dynkin diagrams. We recall that these are the ones which correspond to the simple Lie algebras . The types are A_l, l 1, B_l l 2, C_l, l 3, D_l, l 4, G₂, F₄, E₆, E₇, E₈. If we look at these diagrams as given in §4.5 we see that for A₁, B_l, C_l, G₂, F₄, E₇ and E₈ the group of automorphisms of the diagram is the identity.

For we have in addition to the identity mapping the automorphism α_i → α_l+1-i. For

we have the identity automorphism and the mapping α_i → α_i which is an automorphism. These are the only automorphisms if l 5. For l = 4 the diagram

has the group of automorphisms which permute α₁, α₃, α₄ and leave α₂ fixed. This is isomorphic to the symmetric group on three elements. For

the group of automorphisms consists of the identity mapping and the mapping α₆ → α₆, α_i → α_6-i, i 5.

It is now clear that we have the following

THEOREM 4. Let be finite-dimensional simple over an algebraically closed field of characteristic 0, G the group of automorphisms of , G₀ the invariant subgroup of invariant automorphisms. Then G = G₀ unless is of one of the following types: A_l, l > 1, D_l or E₆. In all of these cases except D₄, the index [G : G₀] 2 and for D₄, [G : G₀] 6.

Remark. The group G is an algebraic linear group (cf. Chevalley [2]). It is easy to see that G₀ is the algebraic component of the identity element of G. If Φ is the field of complex numbers, then G is a topological group and G₀ is the connected component of 1 in G.

5. Explicit determination of the automorphisms for the simple Lie algebras

Let be a semi-simple subalgebra of _L, the algebra of linear transformations of a finite-dimensional vector space over Φ. Let Z be an element of such that ad Z is nilpotent. Since the algebra of linear transformations ad is semi-simple it follows from Theorem 3.17 that there exists an element H ∈ such that [ZH] = Z. This implies (Lemma 4, § 2.5) that Z is nilpotent. We have ad Z = Z_R – Z_L (Z_R : X → XZ, Z_L : X → ZX). Hence

exp ad Z = exp (Z_R — Z_L) = exp Z_R exp (—Z_L),

since [Z_RZ_L] = 0. Then

If we set A = exp Z, then the automorphism exp ad Z of is the mapping

We now consider the simple Lie algebras of types A_l, B_l, C_l, D_l, G₂, and F₄.

A_l. Here is the Lie algebra of linear transformations of trace zero in an (l + l)-dimensional vector space. We can identify with the Lie algebra of matrices of trace 0 in Φ_l+1. If α is a nonsingular matrix, then X → A^-1XA is an automorphism of . Since the only matrices which commute with all matrices of trace 0 are the scalar matrices it is clear that the automorphisms X → A^-1XA, X → B^-lXB are identical if and only if B = ρA, ρ ∈ Φ. Besides the automorphisms X → A^-1XA we have the automorphism X→ — X′ of and, more generally, we have the set of automorphisms X → — A^-1X′A. Suppose the automorphism X → —X′ coincides with one of the automorphisms X → A^-1XA. Then we have

for all X of trace 0. This implies that

so that A(A′)^-1 commutes with all X. It follows that A′ = ρA; hence A′ = ±A. The condition (39) can be rewritten as A^-1X′A = —X. The set of X′s satisfying this condition is either the symplectic Lie algebra or the orthogonal Lie algebra. Accordingly the dimensionality is either (l + 1)(l + 2)/2 (odd l only) or l(l + l)/2 (l even or odd). Since the dimensionality of the space of (l + 1) x (l + 1) matrices of trace 0 is l² + 2l, we must have either l² + 2l = (l + 1)(l + 2)/2 or l² + 2l = l(l + l)/2. The only solution is l = 1 in which case l² + 2l = (l + 1) (l + 2)/2. Thus we see that X→ —X′ coincides with an automorphism of the form X → A^-lXA only if l = 1. If l = 1, then we have and

all X of trace 0.

The result obtained at the beginning of this section shows that every invariant automorphism of has the form X → A^-1XA where α is a product of exponentials of nilpotent matrices. For the Lie algebra A₁ this is the complete automorphism group. For A_l, l 1, we have the automorphism X → –X′ which is not invariant. Hence [G : G₀] = 2 and the automorphisms are the mappings X → A^-lXA and X → –A^-lX′A.

THEOREM 5. The group of automorphisms of the Lie algebra of 2 x 2 matrices of trace 0 is the set of mappings X → A^-1XA. The group of automorphisms of the Lie algebra of n x n matrices of trace 0, n > 2, is the set of mappings X → A^-1XA and X → — A^-1 X′ A. (The base field Φ is algebraically closed of characteristic 0.]

B_l, C_l, D_l. These algebras are the sets of matrices X satisfying S^-1X′S = —X where S = 1 for B_l and D_l and S′ = —S for C_l. The size of the matrices is 2l for C_l and 2l + 1 for B_l. We take l 2 for B_l, l 3 for C_l and l 4 for D_l. Let O be a matrix such that

where ρ ≠ 0 is in Φ. Then O is non-singular and if X ∈ , then Y = O¹XO satisfies

Hence Y ∈ and X → Y = O ^lXO is an automorphism of . Since the base field is algebraically closed we can replace O by ρ^-20 = 0, and we obtain O_l ^-1XO′₁ = Y, O′₁SO₁, = S. If we write O for O₁, then we see that for B_l and D_l, O is an orthogonal matrix (S = 1) and for C_l, O is a symplectic matrix. It is easy to verify that the enveloping associative algebra of is the complete matrix algebra. We leave it to the reader to prove this. It follows that the only matrices which commute with all the elements of are the scalars. Hence if for all X ∈ , then O₂ = ρO₁, ρ ∈ Φ.

Let Z be an element of such that ad Z is nilpotent. Then we have seen that Z is nilpotent and the automorphism exp ad Z has the form X → A^-1XA where A = exp Z. The nilpotence of Z implies that exp Z is unimodular (Exercise 5.4). Also, we have

Hence A = exp Z is orthogonal for B_l and D_l and symplectic for C_l. For B_l and C_l every automorphism is an invariant automorphism. Hence in these cases the automorphisms of have the form X → O^-1XO where O is unimodular and satisfies O′SO = S. For B_l this states that O is a proper orthogonal matrix (corresponding to a rotation). For C_l it is known that if O is symplectic (O′SO = S) then O is necessarily unimodular (see Artin [1], p. 139, or Exercise 12 below). Hence this condition can be dropped. Now consider D_l In this case there exist orthogonal matrices O of determinant —1 and we cannot have O = ρO₁ where O₁ is a proper orthogonal matrix. Thus if O = ρO₁, then ρ = ±1 and in either case det O = det ρO₁ = det O₁ = 1. This contradiction implies that the automorphism X → O^-1X0, where O is improper orthogonal, is not invariant and we see that G ⊃ G₀. If l > 4, then the index of G₀ in G is 2; hence this index is 2 and every automorphism of D_l has the form X → O^-lXO where O is orthogonal. We therefore have the following

THEOREM 6. Let Φ be algebraically closed of characteristic 0 and let be the Lie algebra of skew matrices or the “symplectic” Lie algebra of matrices X such that S^-lX′S = —X where S′ = —S. Assume the number of rows n 5 in the odd-dimensional skew case, n 6 in the symplectic case and n 10 in the even-dimensional skew case. Then the groups of automorphisms of in the skew cases consist of the mappings X→0^-lX0 where O is orthogonal. In the odd-dimensional case one can add the condition that O is proper. In the symplectic case the group of automorphisms is the set of mappings X → O^-1XO where O′SO = S.

The case of D₄ is not covered by this theorem. It can be shown that in this case the group G/G₀ is isomorphic to the symmetric group on three letters and the group G can be determined. This will be indicated in some exercises below.

We consider next the Lie algebra G₂. Here we use the representation of as the algebra of derivations of a Cayley algebra (§ 4.6). Now, in general, if is a non-associative algebra, D a derivation of , A an automorphism of , then A^-1 DA is a derivation. It follows that the mapping X → A^-lXA determined by an automorphism of is an automorphism of the derivation algebra = (). In the case of , the Cayley algebra over an algebraically closed field of characteristic 0, we know that every automorphism of is invariant and so it is a product of mappings of the form X → A^-lXA where A = exp Z, Z in and Z a nilpotent linear transformation in . Since Z ∈ , Z is a derivation of . It follows that A = exp Z is an automorphism of . We therefore see that every automorphism of has the form X→ A^–1XA where A is an automorphism in .

The same reasoning applies to the Lie algebra F₄. Here we represent as the derivation algebra of the exceptional Jordan algebra M₃⁸ and we obtain the result that the group of automorphisms of is the set of mappings X→ A^–1XA where A is an automorphism of M₃⁸.

THEOREM 7. Let be the Lie algebra of derivations of the Cayley algebra or of the Jordan algebra M₃⁸ over an algebraically closed field of characteristic 0. Then the automorphisms of have the form X → A^-lXA where A is an automorphism of or of M₃⁸.

Exercises

The base field in all of these exercises will be of characteristic 0 and all spaces are finite-dimensional.

1. Show that any non-singular linear transformation A can be written in the form A_uA_S where A_u is unipotent, that is, A_u = 1 + N, N nilpotent, and A_s is semi-simple and A_u and A_s are polynomials in A. Prove that if A = B_sB_u and B_u and B_s commute where B_u is unipotent and B_s is semisimple then A_u = B_u and A_s = B_s. (Hint: Use Theorem 3.16.)

2. Let τ be an automorphism of a non-associative algebra and let τ = τ_uτ_s as in 1. Prove that τ_s and τ_u are automorphisms. (Hint: Assume the base field is algebraically closed and use Exercise 2.5. Extend the field to obtain the result in the general case.)

3. Let be a finite-dimensional simple Lie algebra over an algebraically closed field and let G₀ be the group of invariant automorphisms of . Prove that is irreducible relative to G₀.

4. Let be the split Cayley algebra over an algebraically closed field, ₀ the subspace of elements of trace 0 in . We have seen that the derivation algebra (= G₂) acts irreducibly in ₀ (§ 7.6). Use this result to prove that ₀ is irreducible relative to the group of automorphisms G of .

5. Same as 4. with replaced by (cf. § 4.6).

Exercises 6 through 9 are designed to prove that if is semi-simple over an algebraically closed field, then G/G₀ is isomorphic to the group of automorphisms of the Cartan matrix or Dynkin diagram determined by a Cartan subalgebra of . In all of these exercises , , G, G₀, etc., are as indicated in the text.

6. If a ∈ and , then dim , the dimensionality of a Cartan subalgebra of . Sketch of proof: Note that dim _a = dim — rank (ad a). Show that if a is a regular element, that _a = _a Cartan subalgebra so dim _a = l. If (u₁, …, u_n) is a basis for over Φ and (ξ₁, … ξ_n) are indeterminates, then regular element of so rank (ad x) = n — l. If the specialization ξ_i = α_i shows that rank . Hence dim .

7. If τ is an automorphism in let I_τ be the set of fixed points of τ: b^τ = b. Prove that if τ is invariant, then dim . Sketch of proof. We have where Z_i = ad z_i is nilpotent and we have to show that rank (τ — I) n — l. Let ξ₁, …, ξ_r be indeterminates and let P be the field of formal power series in the ξ_i, that is, the quotient field of the algebra of formal power series in the ξ_i with coefficients in Φ. Then is an invariant automorphism of p. The matrix of τ(ξ) relative to the basis (u₁, …, u_n) of has entries which are polynomials in the ξ_i and the specialization ξ_i = 1 gives the matrix of τ relative to this basis. Hence if dim a specialization argurrent will show that (semi-simplicity and the rank are unchanged in passing from to p). Now the exponential formula can be used to show that where (Exercise 5.12). Then so dim by 6.

8. Let τ be an invariant automorphism such that and for a simple system of roots ·. Then h^τ = h for every h ∈ . Sketch of proof: (τ*)^–1 induces a permutation of the set of roots which we can write as a product of cycles . Since leaves the set of positive roots invariant; hence the β_i in a cycle (β₁, …, β_r) are either all positive or all negative. is invariant under τ. If eβ_i is a base element for β_i, then we have so that the characteristic polynomial of the restriction of τ to . If v₁ … v_r ≠ 1 then λ — 1 is not a factor of this and consequently . If this holds for every cycle, then I_τ ⊂ since Since dim I_τ l by 7. it will follow that I^τ = and h^τ = h for all h ∈ . Now let σ be the automorphism in such that and (e_i, f_i, h_i canonical generators). We have shown that σ is invariant. If a is a positive root then α = Σk_iα_i and and The automorphism τ′ = στ satisfies the same conditions as where if the β_i are all positive and non-negative integers, and if the β_i are all negative and Since s_i ≠ 0 for some i we can choose the μ′s so that v′₁ …, v′_l ≠ 1 for every cycle. Then the argument used before shows that h^z′ = h, h ∈ . Hence h^z = h, h ∈ .

9. Prove that G/G₀ is isomorphic to the group of automorphisms of the Cartan matrix.

10. Let = Φ′_l+1, l + 1 = 2r and let τ be an automorphism of the form X → — A^-1X′A in . Show that dim I_z r and that there exist τ such that dim I_z = r.

11. Let be semi-simple over an algebraically closed field and let H be the subgroup of G₀ of η such that ^η , K the subgroup of G₀ of ξ such that h^ξ = h for all h ∈ . Show that K is an invariant subgroup of H and that H/K W. (This gives a conceptual description of the Weyl group.)

12. Use the proof of Theorem 6 to prove that if O is a symplectic matrix with entries in a field of characteristic 0, then det O = 1.

13. Let be a split Cayley algebra and let = α1 - a₀ if a = α1 + a₀, a₀ ∈ _ρ. Set N(a) = a = a and Verify that (a, b) is a non-degenerate symmetric bilinear form of maximal Witt index. Prove that Hence show that if c ∈ ₀, then are in the orthogonal Lie algebra (D₄) of the space relative to the form (a, b). Show that every element of this Lie algebra has the form R_c + Σ [R_ciR_di] where c, c_i, d_i ∈ ₀.

14. Use the alternative law (eq. (4.79) to prove the following identity in

or 2(ab)R_c = (ac_L)b + a(bc_R). Use this to prove the principle of local triality : For every linear transformation A in which is skew relative to (a, b) there exist a unique pair (B, C) of skew linear transformations such that

15. Show that the mappings A → B and A → C determined in 14 are automorphisms of the orthogonal Lie algebra. Prove that every automorphism of this Lie algebra is of the form X → O^-1XO where O is orthogonal or is the product of one of the automorphisms defined in 14. by an automorphism X → O^-1X0.

16. Let ₁ and ₂ be two subalgebras of isomorphic to D₄. Prove that there exists an automorphism of D₄ mapping ₁ onto ₂.

17. Show that the automorphism in A_l, l > 1, such that e_i → f_i, f_i → e_i (cf. p.127) is not invariant. Show that for D_l, l 4, this automorphism is invariant if and only if l is even.

18. (Steinberg). If is a Cartan subalgebra let G_o() be group of automorphisms generated by all exp ad e where e belongs to a root space of 2 relative to (cf. Chevalley [7]). Let ₁ be a second Cartan subalgebra and let G_o(₁) be the corresponding group of automorphisms. Show that there exists such that Then This implies that so that

19. Prove that G_o() = G_o, the group of invariant automorphisms.