CHAPTER IV

Split Semi-Simple Lie Algebras

In this chapter we shall obtain the classification of simple Lie algebras over an algebraically closed field of characteristc 0. This was given first by Killing, modulo some errors which were corrected by Cartan. Later simplifications are due to Weyl, van der Waerden, Coxeter, Witt, and Dynkin. In our discussion we shall follow Dynkin’s method which is fairly close to Cartan’s original method. However, we shall formulate everything in terms of “split” semi-simple Lie algebras over an arbitrary base field of characteristic 0. It is easy to see that the assumption of algebraic closure in the classical treatments is used only to ensure the existence of a decomposition of the algebra as = ⊕ _α ⊕ _β … _δ where is a Cartan subalgebra and the _α are the root spaces relative to . This can be achieved by assuming the existence of a “splitting Cartan subalgebra” (cf. § 1). It appears to be clearer and more natural to employ this hypothesis in place of the stronger one of algebraic closure of the base field. For the benefit of a reader who has some familiarity with the associative theory it might be remarked that the split simple Lie algebras which are singled out in the classical theory are the analogues of the simple matrix algebras Φ_n of the associative theory.

A part of the results of this chapter (the isomorphism and existence theorems) will be derived again in Chapter VII in a more sophisticated way. In Chapter X we shall take up the problem of extending the classification from algebraically closed base fields—or from split Lie algebras—to simple Lie algebras over any field of characteristic 0. It should be noted that the classification we shall give is valid also for characteristic p ≠ 0 under fairly simple hypotheses which are stronger than simplicity. This has been shown by Seligman and, in an improved form, by Mills and Seligman.

1. Properties of roots and root spaces

We shall call a Cartan subalgebra of a finite-dimensional Lie algebra a splitting Cartan subalgebra (abbreviated s.c.s.) if the characteristic roots of every adh, h ε , are in the base field. We shall say that is split if it has a splitting Cartan subalgebra. If the base field Φ is algebraically closed, any finite-dimensional is split and any Cartan subalgebra is a s.c.s.

Example. Let = Φ_nL. and let denote the subalgebra of diagonal matrices. We have seen (§3.3) that is a s.c.s. in Φ_nL, so Φ_nL is split. Next let Φ be the field of real numbers and let A be a matrix whose characteristic roots ξ_i (in the complex field) are distinct but not every ξ_i ε Φ. Then the polynomial algebra Φ[A] is a Cartan subalgebra of . The characteristic polynomial of ad A is and some of its roots are not in Φ. Hence Φ[A] is not a s.c.s.

In the remainder of this chapter will denote a split finitedimensional semi-simple Lie algebra over a field Φ of characteristic 0, will be a splitting Cartan subalgebra of and we shall write (a, b) for f(a, b) = tr ad a ad b, the Killing form on . We know that (a, b) is non-degenerate. Our assumption on is that ad is a split algebra of linear transformations. Hence we know that we can decompose as

where α, β, …, δ are the non-zero roots. These are linear functions on and _α is the set of elements x_α ε such that x_α(ad h — a(h))^r = 0 for some r = r(h), h ε (cf. §2.4 and §3.2). In the same way = ₀, the Fitting null component of relative to ad . We have [_{α β}] _{α + β} if α + β is a root while [_αβ] = 0 if α + β is not a root. Our first task will be to obtain additional information on , on the roots α and on the corresponding root spaces _α. We shall number these results by Roman numerals.

I. If α and β are any two roots (including 0) and β ≠ — α, then α and _β are orthogonal relative to the Killing form.

Proof: We show first that = ₀ ⊥ _a if a ≠ 0. Let h ε e_α ε _α, and choose h′ so that α(h′) ≠ 0. Then the restriction of ad h′ to _α is a non-zero scalar plus a nilpotent and so this is nonsingular. It follows that for any k = 1, 2, … we can find an such that Since the Killing form Is invariant we have

Since is nilpotent, k can be chosen so that [h′ … [h′h]…] = 0. Then the above relation implies that (e_α, h) = 0. Thus ⊥ _α for a ≠ 0. Now let β ≠ — α and let e_β ε _β. As before, write e_α = Then (e_α, e_β) = (, e_β) = – . If a + β is a root it is non-zero and ε _β+α. Hence = 0. If a + β is not a root, = 0 and again ,) = 0. Hence (e_α, e_β) = 0 and _α ⊥ _β.

II. is a non-isotropic subspace of (relative to (a, b)). If α is a root, then —α is a root and _α, and _-α are dual spaces relative to (a, b).

Proof. If z ε and z ⊥ , then z ⊥ since, by I, z ⊥ for all α ≠ 0. Then z = 0 by the non-degeneracy of (a, b). If α is a root and —α. is not a root, then _α ⊥ _β for every root β. Then _α ⊥ contrary to the non-degeneracy of the Killing form. Also the argument used for α = 0 shows that if z ≠ 0 is in _α, then there exists a w ε _-α such that (z, w) ≠ 0. Similarly, if w ≠ 0 is in _-α then there exists a z ε _α, such that (z, w) ≠ 0. This shows that _α and _-α are dual spaces relative to (a, b).

We recall that the matrices of the restrictions of ad h, h ε , to _α can be taken simultaneously in the form

(§2.6). If dim _α = n_α and h, k ε , then this gives the formula

We recall also that α(h′) = 0 for every h′ ε ′. These results imply our next two results.

III. There are l linearly independent roots where l = n₀ = dim .

Proof. The roots are linear functions and so belong to the conjugate space ^* of . We have dim ^*= l. Hence if the assertion is false, then the subspace of ^* spanned by the roots has dimensionality l′ < l. This implies that there exists a non-zero vector h ε such that α(h) = 0 for every root α. Then (3) implies that (h, k) = 0 for every k ε , contrary to II.

IV. is abelian.

Proof: If h′ ε ′ = [], then α(h′) = 0 for all α. Then (h′, k) = 0 for all k and h′ = 0. Hence ′ = 0 and is abelian.

The fact that the restriction of the Killing form to is non-degenerate implies that if ρ(h) is any element of ^*. that is, any linear function, then there exists a unique vector h_ρε such that

The mapping ρ → h_ρ of into ^* is surjective and 1:1. If ρ,σ ε ^* we define

Then

and it is immediate that (ρ, σ) is a non-degenerate symmetric bilinear form on ^*.

V. Let e_α be an element of _α such that [e_αh] = α(h)e_a, h ε , and let e_-α, be any element of _-α. Then

proof: We have

Hence (7) follows from the non-degeneracy of (h, k) in .

VI. Every non-zero root α is non-isotropic relative to the bilinear form (ρ, σ) in ^*.

Proof: We note first that ad e_α is nilpotent if α is a non-zero root and e_α ε _α. For this it suffices to show that if x_βε _β, then there exists a positive integer k such that . Consider the sequence

The vectors in this sequence are either 0 or belong respectively to the roots β + α, β + 2α, β + 3α, … Since there are only a finite number of distinct roots for it follows that for a suitable k. We can choose e_α ≠ 0 in _α so that [e_αh] = α(h′)e_α (cf. (2)) and e_-α ε __α so that (e_α, e_-a) = 1. The latter choice can be made since _α, _-α are dual spaces relative to the Killing form. If ρ is any element of ^* we define _ρ = 0 if ρ is not a root and _ρ is the root space corresponding to ρ if ρ is a root. Set

It is clear from the rule on the product of root spaces that _–ka is a nilpotent subalgebra of . Also is a subalgebra containing _–ka as ideal. Hence is solvable. If x_-α ε _-α, then [e_αx_-α] is a multiple of h_α, by (7). Also if k > l, then [e_α-_kα] _-(k-1)α and [e_αh_α] = α(h_α) e_α = (α, α)e_α. This shows that is a subalgebra of and that if (α, α) = 0, then is an ideal in . Since this ideal is solvable and has one-dimensional difference algebra, (α, α) = 0 implies that is solvable. Then ad is a solvable algebra of linear transformations acting in . Since ad e_α is nilpotent, this element is in the radical of the enveloping associative algebra of ad (Corollary 2 to Theorem 2.8). Hence the same is true of (ad e_α) (ad e_-α), which implies that (e_α,e-_α) = tr ad e_α ad_-α = 0, contrary to (e_α, e-_α) = 1. Hence (α, α) ≠ 0.

VII. If α is a non-zero root, then n_α = dim _α = 1. Moreover, the only integral multiples kα of α which are roots are α, 0 and — α.

Proof: Let e_α, e_-α, be defined as in the proof of VI (cf. (8)). Then is an invariant subspace of relative to ad h, h ε , since [e_αh] = α(h)e_α, [h_αh] = 0 and [_-kαh] _-kα. The restriction of ad h to _-kα has the single characteristic root —ka(h). Hence we have

and, in particular

Since is a subalgebra containing e_α and e_-α it is invariant under ad e_α and ad e_-α and since [e_-αe_α] = h_α, [ade_-α, ade_œ] = adh_α. Hence tr (adh_α) = 0. Since (α, α) ≠ 0 this and (9) imply that 1 — n_-α — 2n_-2α —…= 0. This occurs only if n_-α = 1, n_-2α = n_-3α = … = 0. Thus —2α., —3α, … are not roots and n_-α = 1. Since we can replace α by —α in the argument, we have also that n_a = 1 and 2α, 3α, … are not roots.

If α is a non-zero root, _α = Φe_α and [e_ah] = α(h)e_α. Moreover, (7) shows that if we choose e_α and e_-α so that (e_α, e_-α) = — 1 then [e_αe_-α] = h_α. Then [e_αh_α] = (α, α)e_α, [e_αh_α] = — (α, α)e_α. If we set

Thus (e′_α, e′ _-α, h′_α) is the type of normalized basis we have considered before for a split three-dimensional simple algebra. This proves

VIII. If α is a non-zero root then Φh_α + _α + _-αis a split threedimensional simple subalgebra of .

At this point we have all the information we need on the multiplication in , the products of elements of · by elements of the _α and the products [e_αe_-α], e_α ε _α, e_-α ε _-α. It remains to investigate products of the form [e_αe_-α] where α and β are non-zero roots and β ≠ — α. We shall obtain the results required after we have established in the next section a basic result on representations of semi-simple Lie algebras.

2. A basic theorem on representations and its consequences for the structure theory

We shall need only a part of the following theorem at this point. Later the full result will play an important role in the representation theory.

THEOREM 1. Let be a finite-dimensional split semi-simple Lie algebra over a field of characteristic 0, a splitting Cartan subalgebra, = + Σ Φe_α the decomposition of into root spaces relative to . Let h_α and (ρ, σ) for ρ, σ in the conjugate space ^* be defined as before (cf. (4), (5)) and let e_α and e _-α be chosen so that [e_αe_-α] = h_α. Let be a finite-dimensional module for , R the representation. Then is a split module for and if _Λ is the weight module of corresponding to a weight Λ, then the linear transformation induced by h^R in , is the scalar Λ(h) 1. Let ^(α) = + Φe_α + Φe_-α, Then ^(α) is a subalgebra and is a completely reducible ^(α)module. Any irreducible ^(α)-submodule of has a basic (y₀, y₁, … y_m) such that

where M is a weight and 2(M, α)/(α, α) = m. Moreover, if y is any non-zero vector such that yh = M(h)y and ye_α = 0, then y generates an irreducible ^(a)-submodule of . Let Λ be a weight of in and let Σ be the collection of weights of the form Λ + iα, i an integer, α a fixed non-zero root. Then Σ is an arithmetic progression with first term Λ — ra, difference α, and last term Λ + qα and we have

If x is a non-zero vector such that xh = (Λ + qα)(h)x, then x generates an irreducible ^(α) -submodule and every weight belonging to Σ occurs as weight of in this submodule. If Λ is a weight, then

is also a weight and

Proof: By III we can find a basis for ^* consisting of l roots α₁, α₂ … α_l. The corresponding elements h_α1 h_α2 … h_αl form a basis for . Also _i = Φh_αi + Φe_αi+ Φe_-αi is a split three-dimenional simple Lie algebra and, by (10), e′_αi, e′_-αi = 2e_-αi/(α_i, α_i), h′_αi = 2h_αi/(α_i,α_i) is a canonical basis for _i. If we recall the form of the irreducible modules for such an algebra given in § 3.8 and use complete reducibility of finite-dimensional modules, we see that there exists a basis for such that h′^R_αi hence also = (1/2)(α_i, α_i)h′^R_αi has a diagonal matrix relative to this basis. This is equivalent to two statements about : the characteristic roots of are in Φ and a semi-simple linear transformation. Since the commute, a standard argument shows that there exists a basis (u₁, u₂, … u_N) for such that every has a diagonal matrix relative to this basis (cf. the proof of Theorem 3.10). Since the h_αi form a basis for it follows that h^R has a diagonal matrix relative to (u_j), that is, we have: u_jh = Λ_j((h)u_j, h ε , j = 1, 2, ···, N. Thus it is clear that ^R is a split abelian Lie algebra of linear transformations and that the Λ_j = Λ_j((h) are the weights of . Moreover, h^R is the scalar Λ(h)1 in the weight space _Λ (whose basis is the set of u_j such that Λ_j = Λ). This proves the first statement. Now let α be a non-zero root and consider the subspace ^(α) = + Φe_α + Φe_-α Since is an abelian subalgebra and [e_αhh] = α(h)e_α, [e-_αh] = — α(h)e-_α, ^(α) is a subalgebra. Let _o be the subspace of defined by α(h) = 0. Then = ₀ + Φh_α and ^(α) = ₀ ⊕ , where is the split three-dimensional simple Lie algebra with basis (e_α, e_-α, h_α). We have [₀] = 0, so ₀ is the center of ^(α). Now we have seen that h^R is semi-simple for every h ε . Hence the main criterion for complete reducibliity (Theorem 3.10) shows that is completely reducible as an ^(α)-module. Let be an irreducible ^(α) -submodule of . Then contains a vector y ≠ 0 such that yh = M(h)y where M is a weight. If we replace y by one of the vectors in the sequence y, ye^R_α, y(e^R_α)², … we may suppose also that where (e′_α, e′_-α, h′_α) is the basis for given in (10). Then

and yE = 0. The argument of § 3.8 shows that 2(M, α)/(α, α) is a nonnegative integer m (strictly speaking m. 1) and that (y, yF, …, yF^m) is a basis for an irreducible -module. Also we have yF^m+1 = 0 and induction on i shows that (yFⁱ)h = (M — iα)(h)yFⁱ Hence ΣΦyFⁱ is an ^(a) -submodule of and since is irreducible as -module, = ΣΦyFⁱ. We now modify the basis for by replacing yFⁱ by

Then we have y_ih = (M — iα)(h)y_i, y_ie_-α = y_i+1 for i = 0,1, ···, m — 1 and y_me_-α = 0. Also the last formula of (3.36):

becomes

which is the same as the last equation in (12). Thus we have established (12). The argument shows also that any non-zero y in satisfying yh = M(h)y, ye_α = 0 generates an irreducible ^(α)-submodule of . This proves our assertions about ^(α). Now let Λ be any weight and let Σ be the set of weights of the form Λ + iα, α a fixed non-zero root, l an integer. The weight Λ is a weight for in one of the irreducible ^(α)-modules into which can be decomposed and we may suppose that is generated by y such that yh = M(h)y, ye_α = 0. Then Λ = M — kα where 2(M, α)/(α, α) = m and 0 k m. Let q be the largest integer such that Λ + qα is a weight. Then if x ≠ 0 is chosen so that xh = (Λ + qα)(h)x, we have xe_α = 0, since Λ + (q + 1) α is not a weight. Hence x generates an irreducible ^(α) -submodule of dimensionality s + 1 where

On the other hand, M = Λ + kα is a weight, so k q, and we have m = 2(M, α)/(α, α) = 2(Λ + kα, α)/(α, α) = 2(Λ, α)/(α,α) + 2k. Then

so k — m q — s. Now the weights in the module generated by y and those in the module generated by x are, respectively,

Since k q and q — s k — m,. all those in the first progression are contained in the second and since Λ was arbitrary in Σ, it follows that Σ coincides with the second sequence. Since Λ is contained in this sequence we have q — s 0. If we set r = — (q —s), then the last term becomes Λ — rα where r 0. Also 2(Λ, α)/(α,α) = s — 2q = r — q, which proves (13). We have Λ′ = Λ — [2(Λ, α)/(α, α)]α = Λ + (2q — s)α and —r 2q — s q by our inequalities. Hence Λ′ ε Σ and so Λ′ is a weight. It remains to prove (15). We observe that the argument just used shows that Λ′ also occurs in the first sequence in (17), that is, Λ′ is a weight in every irreducible ^(α)-submodule which has Λ as weight. We have dim _Λ = 1 = dim _Λ′. Hence if = Σ ⊕ _j, is a direct decomposition of into irreducible ^(α)-submodules, then dim _Λ is the number of _j which have Λ as weight for . Hence dim_Λ = dim_Λ′.

This result applies in particular to the adjoint representation of . Our result states that if β and α are any two roots, α ≠ 0, then the roots of the form β + iα, i an integer, form an arithmetic progression with difference α. We call this the α-string of roots containing β. If the string is β — rα, β — (r — 1)α, · · ·, β + qα, then

A number of consequences can now be drawn from Theorem 1. We continue the numbering of § 1.

IX. If α, β and, α + β are non-zero roots, then [e_αe_β] ≠ 0 for any e_α, ≠ 0 in α and any e_β ≠ 0 in _β.

Proof: The «-string containing β is β — rα, …, β + qα and q 1. None of these roots is 0 since no integral multiple of α is a root except 0, ±α. Since the root spaces corresponding to nonzero roots are one-dimensional this holds for our string. Let x be a non-zero element of _β+qα. Then [xh] = (β + qa)(h)x. If we choose e_-α so that [e_αe_-α] = h_α, then the theorem shows that x, x ad e_-a, x(ade_-α)², ···, x(ad e_-α)^r+q are non-zero and these span the spaces _β+qa, _β+(q-1)α. In particular, e_β is a non-zero multiple of x(ade_-α)^q. Formula (12) implies that

which is not zero r 0 since and q > 0. Hence [e_βe_α] ≠ 0.

We see also that [[e_βe_α]e_-α] = — (q(r + l)/2)(α, α)e_β. This formula is valid also if β + α is not a root since in this case [e_βe_β] = 0 and q = 0. Also, if β = — α, then the α-string containing β is α, 0, § α, so r = 0, q = 2. The right-hand side of the formula is —(α, α)e_-α and the left-hand side is [[e-_αe_β]e-§] = — [h_χβ_-α] = [e_-α h h_α] = —(α, α)e_-α, so again the formula holds. We have therefore proved

X. Let a and β be non-zero roots and e_β ε _β, e_β ε _β, e_-α ε _-β satisfy [e_βe_-β] = h_α. Suppose the α-string containing β is β — rα, … β, β + qα. Then

We remark also that it a and —a are interchanged, r and q are interchanged and we have [e_-αe_α] = — h_α = h_α, (—α, —α) = (α, α). Hence another form of (20) is

XI. No multiple of a non-zero root a is a root except 0, a, — a.

Proof : Let β = kα be a root. Then 2(α, β)(α, α) = 2k is an integer and we may assume this is odd and positive. Then it is immediate that the α-string containing β contains γ = (1/2) α. This contradicts the fact that α = 2γ cannot be a root.

XII. The a-string containing β (α, β ≠ 0) contains at most four roots. Hence 2(α, β)(α, α) = 0, ±1, ±2, ±3.

Proof. We may assume that β ≠ α, —α since the α-string containing α consists of the three roots α, 0, —α. Assume we have at least five roots. By re-labelling these we may suppose that β — 2α, β — α, β, β + α, β + 2α are roots. Then 2α = (β + 2α) — β and 2(β + α) = (β + 2α) + β are not roots. Hence the β-string containing β + 2α has just the one term β + 2α. Hence (β + 2α, β) = 0. Similarly β — 2α — β and β — 2α + β are not roots, so that (β — 2α, β) = 0. Adding we obtain (β, β) = 0, contradicting the fact that the non-zero roots are not isotropic. We now have 2(α, β)/(α, α) = (r — q) while r + q + 1 4. Hence r 3, q 3 and 2(α, β)/(α, a) = 0, ±1, ±2, ±3.

From now on we shall identify the prime field of Φ with the field Q of rational numbers. As before, let * be the conjugate space of and now let * denote the Q-space spanned by the roots. We have seen that the roots span the space *. We now prove

XIII. dim ₀^* = l = dim .

Proof. It suffices to show that if (α₁, α₂, ···,α_l) is a basis for * consisting of roots, then every root β is a linear combination of the α_i with rational coefficients. We have β = Σ λ_iα_i, λ_i ε Φ. Hence (β, α_j) = Σ λ_i(α_i, α_j), j = 1, 2, …, l and

This system of equations has integer coefficients [2(α_iα_j)/(α_j, α,_j)], [2(β, α_j)/(α_j, α_j,)] and the determinant

since (ρ, σ) is non-degenerate and the α_i are a basis for *. Hence (21) has a unique solution which is rational. Thus the λ_i are rational numbers.

If Λ is a weight of for a re_βresentation of , then we have seen that 2(Λ, α)/(α, α) is an integer for every non-zero root α. The argument just given for roots shows that Λ is a rational linear combination of the roots α_j, j = 1, 2, ···, l. Thus the weights Λ ε *₀.

XIV. (ρ, σ) is a rational number for ρ, σ ε *₀ and (ρ, σ) is a positive definite symmetric bilinear form on *₀.

Proof: We recall the formula (3) for (h, k) h, k ε . If we take into account the fact that the _α are one-dimensional we can rewrite this as

Now let β be a non-zero root. Then (β, β) = Σ_α(β, α)². Let the β-string containing α be

Then, by (18), 2(α, β)/(β, β) = r_αβ - q_αβ and (α, β) = [(r_{α β} — q_αβ)/2](β, β). Hence

Since (β, β) ≠ 0, Σ_α(r_αβ — q_αβ)² ≠ 0 and

is a rational number and (α, β) = [(r_{α β} — q_αβ)/2](β, β). is rational for any roots roots. Then and (ρ, ρ) = 0 implies that every (ρ, α) = 0. Then ρ = 0 since the roots α span *.

We recall that in a space with a non-degenerate bilinear form (ρ, σ), if α is a non-isotropic vector, then the mapping S_α

is a linear transformation which leaves fixed every vector in the hyperplane orthogonal to α and sends α into —α. We call this the reflection determined by α. This belongs to the orthogonal group of the form (ρ, σ).

We have seen (Theorem 1) that if Λ is a weight of in a representation of , then Λ′ = ΛS_α = Λ — [2(Λ, α)/(α, α)](α) is also a weight. The reflections Sα, α a root, generate a group of linear transformations in ₀* called the Weyl group W of (relative to ). This group plays an important role in the representation theory for . The result we have just noted is that the weights of a particular re_βresentation are a set of vectors which is invariant under the Weyl group. In particular, this holds for the roots. If two elements of W produce the same permutation of the roots they are identical since the roots span *₀. Since there are only a finite number of roots it follows that W is a finite group.

3. Simple systems of roots

We now introduce an ordering in the rational vector space ₀*. For this purpose we choose a basis of roots α₁ α₂…. α_l and we call positive if the first non-zero λ_i is positive. The set of positive vectors is closed under addition and under multiplication by positive rationals. If σ, ρ ε *₀ we write σ > ρ if σ — ρ > 0. Then * is totally ordered in this way and if σ > ρ then a + τ > ρ + τ and λσ > λρ or λσ < λρ according as λ > 0 or λ < 0. We shall refer to the ordering of ₀* just defined as the lexicographic ordering determined by the ordered set of roots (α₁, α₂ ··α_l) (which form a basis for ₀*).

LEMMA 1. Let Suppose the ρ_i > 0 and (ρ_i, ρ_j) 0 if i ≠ j. Then the ρ’s are linearly independent over Q.

Proof: Suppose Since We have . Hence

On the other hand, which is a contradiction. Hence the ρ’s are Q-independent.

DEFINITION 1. Relative to the ordering defined in *₀ we call a root α simple if α > 0 and α cannot be written in the form β + γ where β and γ are positive roots.

XV. Let π be the collection of simple roots relative to a fixed lexicographic ordering of *₀. Then:

(i) If α, β ε π, α ≠ β, then α — β is not a root.

(ii) If α, β ε π and α ≠ β, then (α, β) 0.

(iii) The set π is a basis for ₀* over Q. If β is any positive root, then β = Σαεπk_αα where the k_αare non-negative integers.

(iv) If β is a positive root and β π, then there exists an α ε π such that β — α is a positive root.

Proof: (i) If α, β ε π and α — β is a positive root, then α = β + (α — β) contrary to the definition of π. If α — β is a negative root we again obtain a contradiction on writing β = (β — α) + α. (ii) Let β — rα, β — (r — 1 )α, …, β + qα be the α-tsring containing β. Then 2(α, β)/(α, α) = r — q. Since r = 0 by (i) and (α, α) > 0, (α, β) 0. (iii) The linear independence of the roots contained in π is clear from (ii) and the lemmα. Let β be a positive root and suppose we already know that every root γ such that β > γ > 0 is of the form Σ_αεπk_αα, k_α a non-negative integer. We may suppose also that β π, so that β = β_ι + β ₂, β_i > 0. Then β > β_i and non-negative integers; hence β = Σ (k′_α + k′′_α) which is the required form for β. If β is a negative root then — β is a positive root. Hence β = Σk_αα where the k_α are integers 0. The first statement of (iii) follows from this and the linear independence of the elements of π. (iv) Let β be a positive root not in π. The lemma and (iii) imply that there is an α ε π such that (β, α) > 0. Then 2(β, α)/(α, α) = r — q > 0 (r, q as before). Hence r > 0 and β — α is a root. If β — α < 0, then α — β > 0 and α = β + (α — β) contrary to: α ε π. Hence β — α > 0 and β = (β — α) + α where α ε π.

We now write π = (α_l α₂, …, α_l) and we call this the simple system of roots for relative to and the given ordering in _o^*. We have seen that every root β = Σk_iα_i where the k_i are integers and either all k_i 0 or all k_i 0. This property is characteristic of simple systems. Thus let π = (α₁, α₂, …, α_l) where l = dim and the are roots such that every root β = Σ k_iα_i where the k_i are integers of like sign (rationale would do too). Evidently our hypothesis implies that π is a basis for ₀*. We introduce the lexicographic ordering of ₀* based on the

Then the positive roots β = Σk_iα_i are those such that the k_i 0 and some k_i > 0. It is clear that no α_i is a sum of positive roots. Hence the α_i are simple. Since any simple system consists of l roots the set π = (α₁ α₂, …, α_l) is the simple system defined by the ordering.

If π = (α₁ α₂, …, α_l) is a simple system of roots, the matrix (A_ij), A_ij = 2(α_i, α_j)/(α_i, α_i) is called a Cartan matrix for (relative to ). The diagonal entries of this matrix are A_ii = 2 and the off diagonal entries are A_ij = 0, —1, —2 or —3. (XII and XV (ii)). If i ≠ j the α_i and α_j are linearly independent so that if θ_ij is the angle between α_i and α_j then This gives This implies that either both A_ij and A_ji are 0 or one is —1 while the other is § 1, —2, or —3. The determinant of the Cartan matrix (A_ij) is a non-zero multiple of that of ((α_i, a_j)). Hence det (A_ij) ≠ 0.

We now choose and we now write

These elements have the canonical multiplication table for a split three-dimensional simple Lie algebra: [e_ih_i] = 2e_i, [f_ih_i] = —2f_i,[e_if_i] = h_i. Also we have [e_if_j] = 0 if i ≠ j since α_i — α_j is not a root, [e_ih_j] = 2(α_i α_j)/(α_j, α_j)e_i = A_jie_i and [f_ih_j] = — A_jif_i. The last relations include [e_ih_i] = 2e_i, [f_ih_i] = — 2f_i. Thus we have the following relations for the e_i, f_i, h_i:

i, j = 1,2, ··· l

We wish to show that the 3l elements e_i, f_i, h_i (or the 2l elements e_i, f_i, since h_i = [e_if_i]) generate . Moreover, we shall show that we can obtain a basis for whose multiplication table is completely determined by the A_ij. This will prove that is determined by the Cartan matrix.

If β = Σ k_iα_i is a root then we define the level |β| = Σ |k_i|. The level is a positive integer and the positive roots of level one are just the α_i ε π.

XVI. The set of roots is determined by the simple system π and the Cartan matrix. In other words, the sequences (k₁ k₂, …, k_l) such that Σ k_iα_i are roots can be determined from the matrix (A_ij).

Proof: It suffices to determine the positive roots. The positive roots of level one are just the α_i, ε π. Now suppose we already know the positive roots of level n, n a positive integer. We give a method for determining the positive roots of the next level. By XV these are of the form β = α + α_j, α > 0 of level n, α_j ε π. Hence the problem is to determine for a given α > 0 of level n the α_j ε π such that α + α_j is a root. If α = α_j, α + α_j is not a root. Hence we may assume that α = Σ k_iα_i and some k_i > 0 for i ≠ j. Then the linear forms α — α_j, α — 2α_j, … which are roots are positive of level less than n. Hence one knows which of these are roots. Thus the number r such that the α-string containing α is α —rα_j, ···,α, ···, α, + qα_j is known. We have

hence q can be determined by the Cartan matrix. Since α + α_j is a root if and only if q > 0 this gives a method of ascertaining whether or not α + α_j is a root.

Example. Suppose the Cartan matrix is

that is,

Since α₁ — α₂ is not a root these relations imply that the α₁-string containing α₂ and the α₂-string containing α are, respectively,

The only positive root of level two is α₁ + α₂. Since α₂ + 2α_i is not a root the only positive root of level three is α₁ + 2α₂,. Since 2α₁ + 2α₂ is not a root the only positive root of level four is α₁ + 3α₂. We have 2(α₁+ 3α₂, α₁)/(α₁, α₁ = 2 – 3 = – 1, which implies that (α₁ + 3α₂) + α₁ = 2α₁ + 3α₂ is a root. Since α₁ + 4α₂ is not a root 2α₁ + 3α₂ is the only positive root of level five. Since (2α₁ + 3α₂) + α₁ = 3(α₁ + α₂) and (2α₁ + 3α₂) + α₂ = 2(α₁ + 2α₂) are not roots there are no roots of level six or higher. Hence the roots are

A simple induction on levels shows that any positive root β can be written as

α_ij ε π in such a way that every partial sum

is a root. The number k in (β2) is the level of β. We shall now abbreviate Then the result IX implies that

hence this element spans _β. It follows also that

and this element spans _–β. since the α_i ε π torm a basis for, the *, the h_αi and hence the h_i form a basis for . Since = + Σ (_β + _–β) summed over the positive roots, this proves

XVII. Let π — (α₁, α₂, …, α_l) be a simple system of roots for relative to and let e_i, f_i, h_i be as in (27). Then the 3l elements e_i, f_i, h_i generate . For each positive root β we can select a representation of β = α_i1 + α_i2, + … + α_ik so that α_i1 + · … + α_im is a root for every m k. Then the elements

determined by the positive roots β form a basis for .

We shall be interested in the multiplication table of the basis (36). For this we require

XVIII. Let β be a positive root and let the sequence i₁, i₂, …, i_k be determined by β as in XVII. Let 1′, 2′, …, k′ be a permutation of 1, 2, … k. Then [e_i1.e_i2, … e_ik,] is a rational multiple of [e_i1.e_i2, … e_ik,], the multiplier being determined by the A_ij. A simliar statement holds for the f’s.

Proof: This is clear for k = 1, so we use induction and assume the result for positive roots of level k — 1. If i_k = i_k, = j, then we may assume that the sequence chosen for the root β — α_j is i₁, i₂ … i_k–1 Then the result for k — l implies that [e_i1, … e_i(k–1),] = t[e_i1 … e_ik–1] where t is a rational number determined by the A_ij. Then Next suppose i_k = j ≠ i_k and write

where the displayed e_j is the last one occurring in the expression. If any of the partial sums α_i1 + … + a _im, is not a root, then [e_i1 … e_ik,] = 0. Since this fact can be ascertained from the knowledge of the Cartan matrix the result holds in this case. Now suppose every α_i1, + … + α_im, is a root. Then [e_i1, … e_ik,] ≠ 0. Since and

By (19), , is a root ≠ 0) where the q and r are integers, q > 0, r 0 and q and r are determined by the α_j string containing β = α_i1+ … + a_ir′. This string is known from (A_ij). A similar argument shows that

where s is a non-zero integer which can be determined from the A_ij. It follows that

Thus where t is a rational number which can be determined from the A_ij. This reduces the discussion to the first case. The f’s can be treated similarly.

We can now prove the following basic theorem.

THEOREM 2. Let π = (α₁ α₂, … α_l) be a simple system of roots for a split semi-simple Lie algebra relative to a splitting Cartan subalgebra . Let the e_i, f_i, h_i i = 1, 2, …, l, generators of defined in (27) and let the basis h_i, [e_i1 … e_ik], [f_i1 … f_ik] for be as specified in XVII. Then the multiplication table for this basis has rational coefficients which are determined by the Cartan matrix (A_ij).

Proof: We have [h_ih_j] = 0 and by (28). similarly, . It remains to consider products of e terms and f terms. Since [e_i1, … e_ik = and . This can be obtained by operating on x with a certain (non-commutative) polynomial in ad e_i1, …, ad e_ik. It follows from this and a similar argument for the f’s that it suffices to show that [[e_i1 … e_ik]e_j], [[f_i1 … f_ik]e_j], [[e_i1 … e_ik]f_j], [[f_i1 … f_ik]f_j] are rational combinations of our base elements where the coefficients can be determined by the A_ij. The argument is the same for the last two as for the first two so we consider the first two only. To evaluate [[e_i1 … e_ik]e_j] we ascertα_i n first whether or not β + α_j, β = α_i1 + … + α_ik is a root. If not, then the product is 0. On the other hand, if γ = β + α_j is a root then, by XVIII, [[e_i1 … e_ik]e_j] is a rational multiple of the e-base element associated with γ, the multiplier determined by the Cartan matrix. Next consider [[f_i1 … f_ik]e_j]. If k = 1 the product is 0 unless i₁ = j, in which case [f_i1e_j] = — h_j. If k 2 we shall show by induction that the product is a rational linear combination of f-base elements. Thus, if k = 2 the product is 0 unless j = i₁ or j = i₂. For [[f_i1f_j]e_j] we have [[f_i1f_j]e_j] = [f_i1[f_je_j]] + [[f_i1e_j]f_j] – [f_i1h_j] = A_ji1f_i1, since i₁ + j and so [f_i1e_j] = 0. The relation just derived implies also that [[f_i,f_i1]e_j] = — A_ji1f_i1. Now assume k > 2. If no i_r = j, then the product is 0. Otherwise, let i_r+l be the last index in [f_i1, ··· f_ik] which equals j. Then

The first term is a rational multiple determined by the A_ij of an f-base element. The induction hypothesis establishes the same claim for the second element. This completes the proof.

SUMMARY. Before continuing our analysis, it will be well to summarize the results which we have obtained. For any split semi-simple Lie algebra with splitting Cartan subalgebra we have obtained a canonical set of generators e_i, f_i, h_i, i = 1, 2, …, l satisfying the defining relations (28). These were obtained by choosing a simple system π of roots. The characteristic property of π is that every root , where the k_i are all either non-negative or non-positive integers. Such systems are obtained by introducing lexicographic orderings in the space ^*₀ of rational linear combinations of the roots and selecting the positive roots which are not of the form a + β, α, β > 0, in such an ordering. A canonical set of generators associated with π is h_i = 2h_α/(α_i, α_i), e = e_αi, f_i = 2e_-αi/(α_i, α_i) where e_αi is any non-zero element of _βi and e_-αi is chosen in _-αi so that [e_αi e_-αi] = h_αi. We observe that h_i is uniquely determined by α_i while e_i can be replaced by any μ_ie_i, μ_i ≠ 0, in Φ. Then f_i will be replaced by . The elements e_i, f_i, h_i constitute a canonical basis for a split three-dimensional simple Lie algebra _i. Moreover, it is easy to check that if is any canonical basis for _i such that

If e_i, f_i, h_i are canonical generators we obtain a canonical basis for in the following manner: the basis h_i for and for each non-zero root β a base element e_β = [e_i1 … e_ik] or [f_i1 … f_ik] according as β > 0 or β < 0 where (i₁ …, i_k) is a sequence such that α_i1 + … + α_ik = ± β and every partial sum α_i1 … α_im + a _im is a root.

The multiplication table for the canonical basis is rational and is determined by the Cartan matrix (A_ij), A_ij = 2(α_i, α_j)/(α_i, α_i). The A_ij are integers, A_ij = 2 and if i ≠ j, then either A_ij = 0 = A_ji or one of the numbers A_ij, A_ji is —1 and the other is —1, —2 or —3. We observe also that the group of orthogonal linear transformations generated by the reflections

which is a subgroup of the Weyl group W is finite. (Later we shall see that this subgroup coincides with W.) The reflection S_i can be described by the Cartan matrix. Thus if we take the basis (α₁, α₂, · α_l) for ^*₀, then S_i is completely described by

The conditions we have noted on the A_ij are redundant. However, sometimes one subset of these conditions will be used while at other times another will be used.

4. The isomorphism theorem. Simplicity

Theorem 2 of the last section makes the following isomorphism theorem almost obvious.

THEOREM 3.Let , ′ be split semi-simple Lie algebras over a field Φ of characteristic 0 with splitting Cartan subalgebras , ′ of the same dimensionality l. Let (α₁, α₂, … α_l), (α′₁, α′₂, …, α′_l) be simple systems of roots for and ′ respectively. Suppose the Cartan matrices (2(α_i, α_i))/(2(α′_i, α′_j)/(α′_i, α′_i)) are identical. Let e_i, f_i, h_i, e′_i, f′_i, h′_i, i = 1,2, ···,l, be canonical generators for and ′ as in (27). Then there exists a unique isomorphism of onto ′ mapping e¿ on e_i, on e′_i, f_i on f′_i, h_i on h′_i.

Proof: By XVI we know that Σk_iα_i is a root for if and only if Σk_iα′_i is a root for ′. If β is a positive root for we write β = α_i1 + … α_ik so that α_i1 + … α_im is a root, m k. Then β′ = α′_i1 + … α′_ik and every α′_i1 + … + α′_im is a root. We can choose as base elements in the canonical basis (36) for and ′ the elements Then Theorem 2 shows that the coefficients in the multiplication table for the basis for and ′ are identical. Hence the linear mapping which matches these base elements is the required isomorphism. The uniqueness is clear since the e_i, f_i, h_i are generators.

The result we have just proved is basic for the problem of determining the simple Lie algebras. It is useful also for the study of automorphisms (which we shall consider later) of a single Lie algebrα. We note here that the result implies that there exists an automorphism of mapping e_i →, f_i, f_i → e_i, h_i → —h_i. This follows by observing that α′_i = —α_i, i = 1, ···,l, is a simple system and e′_i = f_i, f′_i = e_i, h′_i = —h_i is a corresponding set of generators. We shall need this in § 7.

DEFINITION 2. A simple system of roots π = (α₁, α₂, …, α_l) is called indecomposable if it is impossible to partition π into non- vacuous non-overlapping sets π′, π′′ such that A_ij = 0 for every α_i ε π′, α_j ε π′′.

THEOREM 4. is simple if and only if the associated simple system, π of roots is indecomposable.

Proof. Suppose first that π = (α₁, … α_k) ∪ (α_k+1, ···, α_l) 1 k < l, so that A_ij = 0, i k, j > k. Choose canonical generators e_i, f_i, h_i and let ₁ denote the subalgebra generated by the e_j, f_j, h_j, j k. It is readily seen that where ₁ is the subspace of spanned by the h_j and the summation is taken over the roots γ which are linearly dependent on the α_j,. Hence 0 ⊂ ₁ ⊂ . If r > k, j k, then A_jr = 0 and since α_j — α_r is not a root, this implies that α_j + α_j is not a root. Hence [e_je_r] = 0 as well as [f_j, e_r] = 0. Also [h_je_r] = 0 and consequently e_r is in the normalizer of _i. similarly, f_r is in the normalizer of ₁. Since ₁ is contained in its own normalizer it follows that is the normalizer of ₁. Hence ₁ is an ideal and is not simple. Conversely, suppose is not simple. Then = ₁ ⊕ ₂ where the _i are non-zero ideals. Let α be a non-zero root and let e_α ε _α. Then and Since _α is one dimensional this implies that either _{α 1} or _{α 2}. Since [₁₂] = 0 and [_α-α] ≠ 0 we have either . In particular, we may order the canonical generators and e_i, f_i so that e₁ f₁, …, e_k, f_k ε ₁, e_k+1, f_k+1, …, e_l, f_l ε ₂. Since the _i are non-zero ideals, l k < l and 0 = [e_j[e_rf_r]] = [e_jh_r] = A_rje_j if j k and r > k. Hence A_jr = A_rj = 0 and the simple system of roots is decomposable.

5. The determination of the Cartan matrices

The results of the last section reduce the problem of determining the simple Lie algebras to the following two problems: (1) determination of the Cartan matrices (A_ij) corresponding to indecomposable simple systems of roots and (2) determination of simple algebras associated with the Cartan matrices. We consider (1) here and (2) in the next section. We observe first that the indecomposability condition amounts to saying that it is impossible to order the indices (or the α_i) so that the matrix has the block form

where B, C are not vacuous.

We now associate a diagram—the Dynkin diagram — with the Cartan matrix A_ij. We choose l points α₁, α₂, ··· α_l and we connect α_i to α₁, i ≠ j, by A_ijA_ji lines. Also we attach to each point α_i the weight (α_i, α_i). If A_ij = 0 = A_ji, α_i and α_j, are not connected and if A_ij ≠ 0, A_ji ≠ 0, then A_ij/ A_ji = (α_i, α_i)/(α_j, α_j). Hence A_ji/A_ij and A_ijA_ji can be determined from the diagram. Since A_ij is nonpositive this information determines A_ij and A_ji. Thus the matrix can be reconstructed from the diagram of points, lines and the weights. We consider two examples:

For G₂ we have A₂₁/A₁₂ = (α₁ α₁)/(α₂α₂) = 3, A₁₂A₂₁ = 3 which implies thatA₁₂ = –1, A₂₁ = –3. Hence the Cartan matrix is the matrix in (29). For A_i we have and all the other A_ij = 0. Hence the matrix is

In determining the Dynkin diagrams we at first drop the weights (α_i, α_i) on the points and consider only the collection of points and the lines joining these. We have l points α₁, α ₂, …, α_i, and α_i and α_j, i ≠ j, are not connected if A_ijA_ji = 0 and are connected by A_ijA_ji = 1, 2, or 3 lines if A_ijA_ji ≠ 0. The α_i are linearly independent vectors in a Euclidian space E₀ over the rationals. This can be imbedded in the Euclidian space E = E_0R over the field R of real numbers. If θ_ij denotes the angle between α_i and α_j then A_ijA_ji = 4 cos² θ_ij and cos θ_ij 0.

Any finite set α₁ α₂, …, α_l of linearly independent vectors in a Euclidian space (over the reals) will be called an allowable configuration (a.c.) if 4 cos² θ_ij = 4(α_i, α_j)² / (α_i, α_i) (α_j, α_j) = 0, 1, 2 or 3 and cos θ_ij 0 for every i, j, i ≠ j. Thus cos or and accordingly θ_ij = 90°, 120°, 135°, or 150°. We may replace α_i by the unit vector u_i which is a positive multiple of α_i. Then the conditions are

The Dynkin diagram (without weights) of an α.c. π is a collection of points u_i, i = 1, ···, l, and lines connecting these according to the rule given before: u_i and u_j are not connected if (u_i, u_j) = 0 and u_i and u_j are connected by 4(u_i, u_j)² = 1, 2 or 3 lines otherwise. An α.c. is indecomposable if it is impossible to partition π into non-overlapping non-vacuous subsets π′, π′′ such that (u_i, u_j) = 0 if u_i ε π′, u_j ε π′′. The corresponding condition on the Dynkin diagram is connectedness′. If u, v ε π, then there exists a sequence u_i1 = u, u_i2, …, u_ik = v such that u_ij and u_ij+1 are connected in the diagram. If the Dynkin diagram is known, then all (u_i, u_j) will be known. We shall determine the Dynkin diagrams for all the connected α.c. after a few simple observations as follows.

1. If S is a Dynkin diagram, the diagram obtained by suppressing a number of points and the lines incident with these is the Dynkin diagram of the α.c. obtained by dropping the vectors corresponding to the points.

2. If l is the number of vertices (points) of a Dynkin diagram, then the number of pα_i rs of connected points (u, v, (u, v) ≠ 0) is less than l.

Proof. Let u = Σ^l₁u_i.Then

If (u_i, u_j) ≠ 0, then 2(u_iu_j) — 1. Hence the inequality shows that the number of pairs u_i, u_j with (u_i, u_j) ≠ 0 is less than l.

3. A Dynkin diagram of an α.c. contains no cycles. (A cycle is a sequence of points u₁ ···, u_k such that u_i is connected to u_i+1, i k — 1 and u_k is connected to u_i.)

Proof. The subset forming a cycle is a diagram of an α.c. violating 2.

4. The number of lines (counting multiplicities) issuing from a vertex does not exceed three.

Proof. Let u be a vertex, v₁, v₂, …, v_k the vertices connected to u. No two v_i are connected since there are no cycles. Hence (v_i, v_j) = 0, i ≠ j. In the space spanned by u and the v_i we can choose a vector v₀ such that (v₀, v₀) = 1 and v₀, v₁ …, v_k are mutually orthogonal. Since u and the v_i, i 1, are linearly independent, u is not orthogonal to v₀ and so (u, v₀) ≠ 0. Since u = ∑^k₀(u, v_j)v_j,

Hence . Since 4(u, v_i)² is the number of lines connecting u and v_i we have our result.

5. The only connected α.c. containing a triple line is

This is clear from 4.

6. Let π be an a.c. and let v₁, v₂, …, v_K be vectors of π such that the corresponding points of the diagram form a simple chain in the sense that each one is connected to the next by a single line. Let π′ be the collection of vectors of π which are not in the simple chain, v₁, …, v_k together with the vector . Then π′ is an a.c.

Proof: We have 2(v_i, v_i+1) = −1, i = 1, …, k− 1. Hence (v, v) = k + 2 ∑_i<j(v_i, v_j). Since there are no cycles (v_i, v_j) = 0 if i < j unless j = i + 1. Hence (v, v) = k − (k − 1) = 1 and v is a unit vector. Now let u ∈ π, u ≠ v_i. Then u is connected with at most one of the v_i, say v_j since there are no cycles. Then and 4(u, v)² = 4(u, v_j)² = 0,1, or 3 as required.

The diagram of π′ is obtained from that of π by shrinking the simple chain to a point: Thus we replace all the vertices v_i by the single vertex v and we join this to any u ∈ π, u ≠ v_i by the total number of lines connecting u to any one of the v_j in the original diagram. Application of this to the following graphs

reduces these respectively to

Since the center vertex in each of these has four lines from this vertex, these cannot be diagrams of a.c., by 4. Hence we have

7. No Dynkin diagram contains a subgraph of the form (40).

8. The only possible connected Dynkin diagrams have one of the following forms

Proof: If a connected Dynkin diagram S contains a triple line then it must by G₂ by 5. If S contains a double line it contains only one such line and it contains no node, that is, graph of the form This is clear from 7. Also it is clear that S cannot contain two nodes. This reduces the possibilities to those of (42).

We now investigate the possibilities for p, q, r in the second and third types in (42). For the second type set Since 2(u_i, u_i+1) = − 1 and 2(v_i, v_j+1) = −1 we have

Since pq > 0 this gives (p + 1) (q + 1) >2pq which is equivalent to (p − 1) (q − 1) < 2. Hence the only possibilities for the positive integers p, q are

The first two cases differ only in notation. Hence we have

9. The only connected Dynkin diagrams of the second type in (42) are

Finally we consider the third case in (42). Set The vectors u, v, w are mutually orthogonal and z is not in the space spanned by these vectors. Hence if θ₁, θ₂, θ₃ respectively, are the angles between z and u, v and w then cos² θ₁ + cos² θ₂ + cos² θ₃ < 1 (cf. the proof paragraph 4 above). Now

Similarly so that we have

We may suppose p q r ( 2). Then p⁻¹ q⁻¹ r⁻¹ and (49) implies that 3r⁻¹ > 1. Since r 2. this gives r = 2. Then (49) gives p⁻¹ + q⁻¹ . Hence 2q^−l > and q < 4. Hence 2 q < 4. If q = 2, then the condition is that p⁻¹ which holds for all p. If q = 3, then the condition is p−1 > 1/6 and p < 6. Hence in this case p = 3, 4, 5. Thus the solutions for p, q, r are

This proves

10. The only connected Dynkin diagrams of the third type in (42) are

We have now completed the proof of the following

THEOREM 5. The only connected Dynkin diagrams are A_l l 1, B_l = C_l, l 2, D, l 4 and the five “exceptional” diagrams G₂, F₄, E₆, E₇, E₈, given in (42), (48), (51), and (52).

We now re-introduce the weights on the diagrams. This will give all the possible Cartan matrices: We recall that in the Dynkin diagram obtained from the simple system π = (α₁, α₂, …, α₁), A_ij A_ji, i ≠ j, is the number of lines connecting α_i and α_j. If A_ij ≠ 0, A_ji ≠ 0, then A_ji/A_ij = (α_i, α_i)/(α_j, α_j) and A_ij or A_ji = −1 while the other of these is −1, −2 or −3. Since nothing is changed in multiplying all the α_i by a fixed non-zero real number, we may take one of the α_i to be a unit vector. If the diagram has only single lines, then all the (α_i, α_i) = 1 since the diagram is connected. Hence the weighted diagrams for A_l, D_l, E₆, E₇, E₈ are

For G₂ we have chosen the notation so that

For F₄ we may take the weights as follows:

For B_l and C_l we take the following diagrams:

These diagrams give the possible Cartan matrices.

6. Construction of the algebras

The time has now come to reveal the identity of the principal characters of our story—the split simple Lie algebras. With every connected Dynkin diagram of an α.c. which we determined in § 5 we obtain a corresponding Cartan matrix (A_ij) and there exists for this matrix a split simple Lie algebra with canonical generators e_i, f_i, h_i, i = 1,2, …,l such that [e_ih_j] = A_jie_i We shall give the simplest (linear) representation of the algebras corresponding to diagrams A_l, B_l, C_l, D_l, G₂, F₄ and E₆. Later (Chapter VII) another approach will be used to prove the existence of split simple Lie algebras corresponding to the diagrams E₇ and E₈.

We recall that if is an irreducible Lie algebra of linear transformations in a finite-dimensional vector space over a field of characteristic 0, then where ₁ is semi-simple and is the center. Hence such an algebra is semi-simple if and only if = 0. We note also that if is semi-simple and contains an abelian subalgebra such that … where α, β, … are non-zero mappings of into Φ such that [e_αh] = α(h)e_αh ∊ is a splitting Cartan subalgebra for and is split. We shall use these facts in our constructions.

Let be the algebra of linear transformations in an (l + 1)- dimensional vector space over , l 1. It is well known (and easy to prove) that acts irreducibly on . We have where is the derived algebra and is the set of linear transformations of trace 0. Evidently, any -invariant subspace is -invariant. Hence is irreducible. Also the decomposition shows that the center of is 0. Hence is semisimple.

We now identify with the algebra _l+1 of (l + 1) × (l + 1) matrices with entries in , with the set of matrices of trace 0. We introduce the usual matrix basis (e_ij), i,j = 1, …, l + 1, in _l+1 so that

A basis for is

Set . Then the set of h′s is an abelian subalgebra of l dimensions and

The l² + l linear functions are distinct and are non-zero weights of ad. We have where α runs over these weights. It follows that is a splitting Cartan subalgebra and the α are the non-zero roots. Set

Then

This shows that every root has the form ∑k_iα_i where the k_i are integers and k_i 0 for all k_i 0 for all l. Hence the α_i form a simple system of roots for relative to . Equations (57) and (58) show that α_iΛ+_j; is not a root, 1 i l − 1, α_i + 2_{αi + 1} is not a root, and α_i + α_j is not root if j > l + 1. This means that the Cartan integers A_ij have the following values:

Hence the Dynkin diagram is the connected simple chain

It follows that is simple of type A_l (with Dynkin diagram A_l). Hence we have

THEOREM 6. Let be the Lie algebra of linear transformations of trace zero in an (l + 1)-dimensional vector space over . Then is a split simple Lie algebra of type A_l.

Assume next that is n-dimensional over and is equipped with a non-degenerate bilinear form (x, y) which is either symmetric or skew. Let be the Lie algebra of linear transformations A which are skew relative to (x,y), that is, (xA, y) = −(x, yA) (cf. § 1.2). We require the following

LEMMA 2. is irreducible if n 3.

Proof: Let be a non-zero invariant subspace relative to and let z be a non-zero vector in . Let u be any vector in the orthogonal complement . Choose and consider the linear transformation A: x →(x, y) − (v, x)u One checks that A ε . Moreover, zA = –(v, z)u ≠ 0 is in . Hence contains for every z ε . It follows that dim n − 1 and dim = n unless for every z ε . Then is totally isotropic and dim [n/2], (cf. the author’s Lectures in Abstract Algebra II, p. 170). This implies that n = 2, contrary to our assumption.

We shall now distinguish the following three cases: B. (x, y) is symmetric and n = 2l + 1, l 1. C. (x, y) is skew so necessar2y n is even dimensional, say, n = 2l, l > 1. D. (x,y) is symmetric, n = 2l, > 1. Moreover, we shall assume that in the symmetric cases (B and D) the bilinear form has maximal Witt index. This means that contains a totally isotropic subspace of l dimensions.

B. Let (u₁, u₂, …, u_n) be a basis for and let (u_i, u_j) = σ_ij, s = (σ_ij). A linear transformation A ε if and only if (u_iA, u_j) = −(u_i,u_jA) for i,j = l, …,n. If u_iA = ∑_kα_ik,u_jk then these conditions are that ∑_kα_ik,σ_kj = –∑_kα_ik,σ_jkor in matrix form,

Since the form is of maximal Witt index the basis can be chosen so that

where 1_l denotes the identity matrix of l rows and columns and σ is a non-zero element in (cf. the author [2], vol. II, p. 168). Since nothing is changed by re_βlacing the form by a non-zero multiple of the form we may assume that

If we partition a in the same way as s:

where a_ij ε Ф_l, u₁ u₂ are l × 1 matrices and v₁,v₂ are 1 × l matrices, then a simple computation shows that (60) holds if and only if

These conditions imply that the following set of elements is a basis for identified with the algebra of matrices satisfying (60)

where i, j = 1, …, l. The set is an abelian subalgebra of . The linear forms which are subscripts can be identified with the linear mappings h → α(h), where h = ∑ ω_ih_i and we have [e_αh] = αe_α, α = ω_i − ω_j, ω_i + ω_j etc. It follows that is a splitting Cartan subalgebra and the α are the non-zero roots. acts irreducibly on and if z = h₀ + Σρ_αe_α is a center element, then [zh] = 0 gives Σ ρ_αα(h)e_α = 0. Since the e_α are linearly independent this implies that ρ_αα(h) = 0 for all h. Since α ≠ we have ρ_α = 0. Hence z = h₀. But then [ze_α] = 0 gives α(h₀) = 0 for all α. Since there are l linearly inde_βendent α (e.g. the ω_i, l = 1, …) we have h₀ = 0. Hence the center is 0 and is semi-simple and split.

We now assume l 12 and we set

One checks that this is a simple system of roots with Dynkin diagram B_i:

Hence is simple of type B_i. We therefore have the following

THEOREM 7. Let be the Lie algebra of linear transformations in a (2l + 1)-dimensional space, l 2, which are skew relative to a nondegenerate symmetric linear form, of maximal Witt index. Then is a split simple Lie algebra of type B_i.

C. Let be of dimension 2l, l 1, (x, y) non-degenerate skew linear form in . Let be the Lie algebra of linear transformations which are skew relative to (x, y). We can choose a basis (u₁, u₂, …, u_2l) so that the matrix q = ((u_i, u_j)) is

As in B, one sees that can be identified with the Lie algebra of matrices α ε _2l such that aq = −qa^′. This implies that

where

Hence has the basis

where i, j = 1, 2, …,l. As in B, one proves that = {∑ ω_i h_i} is a Cartan subalgebra and the subscripts in (68) define the roots relative to . Also has center 0 and so, by Lemma 2, is semi-simple. The roots

form a simple system with Dynkin diagram C_l if l 3. This proves

THEOREM 8. Let be the Lie algebra of linear transformations in a 2l-dimensional space, l 3, which are skew relative to a nondegenerate skew linear form (the symplectic Lie algebra.) Then is a split simple Lie algebra of type C_l.

D. , 2l-dimensional, l 2, (x, y) symmetric of maximal Witt index. Here we can choose the basis (u_i) so that t = ((u_i, u_j)) has the form

and can be identified with the Lie algebra of matrices a satisfying at = −ta_′. These are the matrices

such that

Then has the basis

where i,j = 1, 2, …,l. = {Σ ω_ih_i} is a splitting Cartan subalgebra and the subscripts in (73) define the roots. The center of is 0; hence is semi-simple. Set

Then these α_i form a simple system of roots which has the Dynkin diagram D_l if l 4. We therefore have

THEOREM 9. Let be the Lie algebra of linear transformations in a 2l-dimensional space, l 2, which are skew symmetric relative to a non-degenerate symmetric linear form of maximal Witt index. Then if l 4, is a split simple Lie algebra of type D_l.

The four classes of Lie algebras α, B, C and D are called the “great” classes of simple Lie algebras. These correspond to the linear groups which Weyl has called the classical groups in his book with this title. It is easy to see directly, or from the bases, that we have the following table of dimensionalities

The determination of the simple systems and the general isomorphism theorem (Theorem 3) and the criterion for simplicity (Theorem 4) yield a number of isomorphisms for the low dimensional orthogonal and symplectic Lie algebras. The verifications are left to the reader. These are

1. The orthogonal Lie algebra in 3-space (three-dimensional space ) defined by a form of maximal Witt index and the symplectic Lie algebra in 2-space are split three-dimensional simple and so are isomorphic to the algebra of matrices of trace 0 in 2-space.

2. The orthogonal Lie algebra of a form of maximal Witt index in 4-space is a direct sum of two ideals isomorphic to split threedimensional simple Lie algebras.

3. The symplectic Lie algebra in 4-space is isomorphic to the orthogonal Lie algebra in 5-space defined by a symmetric form of maximal Witt index.

4. The orthogonal Lie algebra in 6-space of a form of maximal Witt index is isomorphic to the Lie algebra of linear transformations of trace 0 in 4-space.

The remaining split simple Lie algebras: types G₂, F₄, E₆, E₇, and E₈, are called exceptional. We shall give irreducible re_βresentations for G₂, F₄, E₈ but we shall be content to state the results without proofs, even though some of these are not trivial. A complete discussion can be found in a forthcoming article by the author ([11]).

Our realizations of G₂ and G₄ whill be as the derivation algebras of certain non-associative algebras, namely, an algebra of Cayley numbers and an exce_βtional simple Jordan algebra .

Following Zorn, the definition of the split Cayley algebra or vector-matrix algebra is as follows. Let V be the three-dimensional vector algebra over . Thus V has basis i, j, k over and has bilinear scalar multiplication and skew symmetric vector multiplication × satisfying: i,j,k are orthogonal unit vectors and

Let be the set of 2 × 2 matrices of the form

Addition and multiplication by elements of are as usual, so that is eight-dimensional. We define an algebra product in by

The split Cayley algebra is defined to be together with the vector space operations defined before and the multiplication of (78). is not associative but satisfies a weakening of the associative law called the alternative law:

Let be the Lie algebra of derivations in . The unit matrix 1 is the identity in and since 1² = 1, 1D = 0 for every derivation D. The space ₀ of elements of trace 0 (α + β. = 0) coincides with the space spanned by the commutators [xy] = xy − yx, x, y ε . Hence ₀D ₀ for a derivation. Thus ₀ is a seven-dimensional subspace of which is invariant under The re_βresentation in ₀ is faithful and irreducible.

If T is a linear transformation of trace 0 in V, and T* is its adjoint relative to the scalar multiplication, then it can be verified that

is a derivation in . The set of these derivations is a subalgebra ₀ isomorphic to In any alternative algebra any mapping of the form D_{α, b} = [a_Lb_L] + [a_Lb_R] + [a_Rb_R], where α, b are in the algebra and a_L, a_R denote the left and the right multiplications (x → ax, x → xa) determined by α, is a derivation. In any derivation has the form D_e1.a12 + D_e2.b21 + D₀ where

and D₀ ε ₀. If is a Cartan subalgebra of ₀, is a Cartan subalgebra of . If we identify ₀ with , we can take to be the set of matrices of the form ω₁h₁ + ω₂h₂, h₁ = e₁₁ − e₃₃, h₂ = e₂₂ − e₃₃. Then is a splitting Cartan subalgebra of and the roots of in are: The center of is 0 and since acts irreducibly in ₀, is semisimple. The roots α₁ = ω₁ − ω₂, α₂ = ω₂ form a simple system with Dynkin diagram G₂. Hence is split simple of type G₂.

The Cayley algebra has an anti-automorphism of period two such that . Let denote the space of 3 × 3 hermitian Cayley matrices defined by this anti-automorphism. Thus is the set of matrices of the form

If X, Y ε , then

where the product XY is the usual matrix product. Then is a non-associative algebra relative to the usual vector space operations and the multiplication (83). Moreover, the multiplication in satisfies

These identities are the defining properties of a class of algebras called Jordan algebras. A consequence of these identities is that if R_α denotes the mapping X → X·A = A·X then

is a derivation in the algebrα.

Let denote the Lie algebra of derivations in and let ₀ denote the subalgebra of those derivations which map the elements

into 0. Then ₀ is isomorphic to an orthogonal Lie algebra of a symmetric form of maximal Witt index in 8-space. Let be a Cartan subalgebra of ₀ corresponding to the Cartan subalgebra selected above for the orthogonal Lie algebra (type D₄). Then is a splitting Cartan subalgebra of .

The derivation algebra maps the 26-dimensional space J₀ of matrices of of trace 0 into itself. The representation in J₀ is faithful and irreducible. The center of is 0, so is semi-simple. The roots of in are

The roots

form a simple system with the diagram of type F₄. Hence is split simple of type F₄.

We now use the notation for the derivation algebra of and we let denote the set of linear transformations in which have the form

This is clearly a subspace of the space of linear transformations in Moreover, if tr B = 0 and E ε , then

Since these relations imply that is a Lie algebra of linear transformations. Now acts irreducibly in and has 0 center. Hence is semi-simple. Let denote the Cartan subalgebra of defined above and let . Then is a Cartan subalgebra of with roots

where A_i and M_i are as in (87). The roots

form a simple system of type E₆. Hence is simple of type E₆.

An enumeration of the roots gives the following table of dimensionalities:

We state also (without proof) that E₇ and E₈ exist and have dimensionalities 133 and 248 respectively.

Conclusion. Every split simple Lie algebra over an arbitrary field of characteristic 0 is isomorphic to one of the Lie algebras A_l, l 1, B_l, l 2, C_l, l 3, D_l, l 4 constructed above or to an exce_βtional Lie algebra G₂, F₄, E₆, E₇ or E₈. We shall see later (Chapter IX) that the algebras listed here are not isomorphic. Hence the results give a complete classification of split simple Lie algebras over any field of characteristic 0. In particular, we have a complete classification of simple Lie algebras over any algebraically closed field of characteristic 0.

7. Compact forms

One of the most fruitful and profound ideas in the theory of Lie groups is the method introduced by Weyl for transferring problems on representations of Lie groups and algebras to the case of compact Lie groups. Weyl gave this method the striking title of the “unitary trick” and he used it to give the first proof of complete reducibility of the representations of semi-simple Lie algebras over algebraically closed fields of characteristic 0. He used it also for studying the irreducible representations, determining the characters and dimensionalities of these representations. We shall consider this later on. Weyl’s method permits the application of analysis−via the theory of compact groups−to Lie algebras. The essence of the method has been formalized by Chevalley and Eilenberg in the following form.

A property P of Lie algebras is called a linear property if (1) P holds for implies that P holds for _Ω, Ω any extension field of the base field of L and (2) P holds for _Ω for some extension field Ω implies that P holds for . A trivial example is the statement that dim = n. Cartan’s criterion implies that the property of semi-simplicity is a linear property for Lie algebras of characteristic 0. Also it can be proved that the property that the finite-dimensional representations of such algebras are completely reducible is a linear property.

We now introduce a certain class of semi-simple Lie algebras over the field R of real numbers and we shall see that the validity of a linear property P for all of these algebras implies the validity of P for all semi-simple Lie algebras of characteristic zero. The class we require is given in the following

DEFINITION 3. A Lie algebra over the field R of real numbers is called compact if its K2ling form is negative definite. (This implies semi-simplicity.)

The importance of compact Lie algebras is that the groups associated with these algebras in the Lie theory are compact groups. The following result is an immediate consequence of the classification-due to Cartan−of simple Lie algebras over the field of real numbers. The proof we shall give, which does not make use of this structure theory, is due to Weyl.

THEOREM 10. Let be a semi-simple Lie algebra over the field C of complex numbers. Then there exists a compact Lie algebra _u over the field of real numbers such that the “complexification” .

Proof: Since C is algebraically closed our structure theory applies. Let e_i, f_i, h_i be canonical generators for and let the canonical basis be chosen as in §4. We have seen that there exists an automorphism σ of such that . Then and since the e_i f_i generate , σ² = 1. If we recall the form of the canonical basis determined by the e_i, f_i, h_i it is clear that for every non-zero root α. Hence if we choose any e_α ≠ 0 in _α then . If we replace we obtain and, since we are working in an algebraically closed field, we may choose λ_α so that . Hence we may suppose that for the positive roots α. If we choose for the positive roots α (relative to some ordering in . We now choose as basis for : (h₁, h₂, …, h_l, e_α, e_−α, …) where and these elements satisfy

for every non-zero root α. If α and β are non-zero roots and β ≠ ± α then

if α + β is a root and [e_αe_β] = 0 otherwise. If we apply σ to this relation we obtain

We know also that (e_α, e_−α) = − 1 implies that [e_αe_−α] = h_α and this element is a rational linear combination of the h_i. Now if α + β is also a root, then

On the other hand,

by (20). This can be continued to

Since α and β are linearly inde_βendent, h_α and β_β are linearly inde_βendent. Hence the last relation can be compared with (95) to see that is a positive rational number. It follows that all the N_αβ, α, β roots, are real numbers. Since [e_αh_l = α(h)e_α and α(h_i) is rational it is now clear that the multiplication table for the basis we have chosen has real coefficients. Hence the set of real linear combinations of this basis is an algebra ₁ over the real field R such that _1c = . Since induces an automorphism of period two in ₁. We now modify this automorphism by intertwining it with the standard automorphism of the complex field. Thus if we denote the chosen basis by (u_i) we let τ be the mapping: This is a semi-automorphism of in the sense that τ is a semi- linear transformation and [xy] ^τ = [x^τy^r], In fact, we have (x +y^τ) = x^τ + y^τ, . The fact that τ satisfies [xy]^τ = [x^τy^τ] is clear from the reality of the multiplication table of the u_i. It is clear also that τ is an automorphism of considered as an algebra over R; hence the set _u of fixed points under τ is an R- subalgebra of . We shall show that _u is the required compact form of . It is clear that τ² = 1. Hence . follows that every element of _u is a real linear combination of the elements . If we use the form of the we see that every element of _u is a real linear combination of the elements

where α ranges over the positive roots. These elements form a basis for over C and this implies that . Also we see that any element of _u has the form

where the ξ_j are real, ρ_α complex, the conjugate. Since = , the Killing form for _u is the restriction to _u of the Killing form in . Hence we may use the orthogonality properties and the relations (e_α, e_−α) = − 1 to calculate

where h = ∑ ξ_ih_i. Since (h, h) > 0 unless h = 0, it is clear that (x, x) < 0 unless x = 0. Hence the Killing form is negative definite.

Now let P be a linear property of Lie algebras which is valid for every compact Lie algebra over the field R of real numbers. Let be a semi-simple Lie algebra over a field of characteristic 0 and let Ω be the algebraic closure of . Then _Ω is split and so this has the form _0Ω where ₀ is a Lie algebra over the field Q of rational numbers. Then _0c is a semi-simple Lie algebra over the complex field C; hence, by Theorem 10, there exists a compact Lie algebra _u over the reals such that . Since ρ is satisfied by _u, it holds for hence for ₀ and for . It follows that P holds for .

Exercises

In these exercises all base fields are of characteristic 0 and all algebras and modules are finite-dimensional. Unexplained notations are as in the text.

1. Show that is a Cartan subalgebra of a semi-simple Lie algebra if and only if: (1) is maximal commutative and (2) is reductive in , that is, ad is completely reducible.

2. Let be a Cartan subalgebra of the semi-simple Lie algebra . Show that if R is any finite-dimensional re_βresentation of , then h^R is semisimple for every h ε .

3. Let ₁ be a semi-simple subalgebra of the semi-simple Lie algebra and let ₁ be a Cartan subalgebra of ₁. Show that ₁ can be imbedded in a Cartan subalgebra of .

4. Use the method of § 3 to obtain a canonical basis and multiplication table for A₂ and G₂.

5. Use the method of § 3 to obtain the roots for the split Lie algebra of type E₇ (assuming it exists) and show in this way that the dimensionality of this Lie algebra is 133.

6. Without using the determination of the connected Dynkin diagrams show that cannot be the Dynkin diagram of a Cartan matrix. (Hint: Show that the set of roots determined by this contains a positive root β such that 2β is a root.)

7. Show that the constants of multiplication of a canonical basis (§ 3) of a split semi-simple Lie algebra are rational numbers with denominators of the form 2^k,3^l, k, l integers.

8. Prove that any split semi-simple Lie algebra can be generated by two elements.

9. Let , the algebra of linear transformations in an (l + 1)-dimensional vector space and let (A, B) be the Killing form in . Show that

(Hint: Use Exercise 3.9)

10. Express the Killing forms of the Lie algebras of types B_i, C_i and D_i in terms of the trace form tr AB where the re_βresentation is the one given in §6.

11. Show that if is split, a splitting Cartan subalgebrα, then is a maximal solvable subalgebra and , is a maximal n2potent subalgebra of .

12. Determine the Weyl group for the split Lie algebra of type G₂ and the split Lie algebra of type A_l.

13. Let α_l be a simple root and let α be a positive root ≠ α_i. Show that every root of the form α + k_ai, k an integer, is positive. Show that where the summation is taken over the k such that α + k_ai is a root and α is any positive root. Hence prove that .