18.745 Introduction to Lie Algebras Fall 2004

Lecture 10 — October 12, 2004

Prof. Victor Kač Scribe: Steven Sivek

Recall that a trace form on a Lie algebra 𝔤, given a representation π of 𝔤 in a finite dimensional vector space V, is defined as (a, b)_{V} = tr_{V} (π(a)π(b)).

Definition 1. The Killing form on a finite-dimensional Lie algebra 𝔤 is the trace form K(a, b) = tr_{𝔤}((ad a)(ad b)) in the adjoint representation of 𝔤.

Exercise 10.1. Show that the trace form for the defining representation of gl_{n}(𝔽), sl_{n}(𝔽), so_{n}(𝔽), sp_{n}(𝔽) is nondegenerate.

Solution. Let e_{xy} denote the nxn matrix with the element in row x, column y equal to 1 and all other elements 0. We note that tr(e_{ij}e_{kl}) = tr(δ_{jk}e_{il}) = δ_{jk}δ_{il}. To show that this form is nondegenerate for a Lie algebra 𝔤 ⊂ gl_{n}(𝔽), it suffices to show for any a ∈ 𝔤 that tr(ab) ≠ 0 for some b ∈ 𝔤, so this is how we will proceed. (We will assume that char(𝔽) = 0 for simplicity, since e.g. the trace form for sl_{n}(𝔽) is degenerate when char(F) | n.)

𝔤 = gl_{n}(𝔽): Take nonzero a = (a_{ij}) ∈ 𝔤, and pick x, y such that a_{xy} ≠ 0. Then tr(ae_{yx}) = a_{xy} ≠ 0.

𝔤 = sl_{n}(𝔽): Choose nonzero a = (a_{ij}) ∈ 𝔤. If a is not a diagonal matrix, then we may pick x ≠ y such that a_{xy} ≠ 0, and as before we have tr(ae_{yx}) ≠ 0. Otherwise, let a = diag(b_{1}, …, b_{n}). Then a ≠ αI_{n} for any α, since otherwise we would have either a = 0 or tr(a) = nα with n, α ≠ 0. It follows that for some k < n we must have b_{k} ≠ b_{k+1}; we calculate that tr(a(e_{kk} − e_{k+1}k + 1)) = b_{k} − b_{k+1} ≠ 0.

𝔤 = so_{n}(𝔽): Pick a basis for 𝔽^{n} such that so_{n}(𝔽) is the algebra of skew-symmetric matrices, i.e. so_{n}(𝔽) = {a ∈ gl_{n}(𝔽) | a^{T} = −a}. Then 𝔤 has as a basis B = {e_{ij} − e_{ji} | i < j}. We may easily calculate that for any b_{1}, b_{2} ∈ B, we have tr(b_{1}b_{2}) = −2 if b_{1} = b_{2} and tr(b_{1}b_{2}) = 0 otherwise. Then take nonzero a = (a_{ij}) ∈ 𝔤, and pick x < y such that a_{xy} ≠ 0; we calculate that tr(a(e_{xy} − e_{yx})) = −2a_{xy} ≠ 0.

𝔤 = sp_{n}(𝔽): Note that n mus be even, and write n = 2m. If we pick as a representative skewsymmetric form the matrix ( \begin{matrix}0 & I_{m} \\ −I_{m} & 0 \end{matrix} ), then 𝔤 consists of matrices expressible in block form as

a = ( \begin{matrix}A & B \\ C & −A^{T} \end{matrix} ), where each block is mxm, and B and C are symmetric. Take a ∈ sp_{n}(𝔽) of

this form; if B = C = 0, then tr(a · ( \begin{matrix}D & 0 \\ 0 & −D^{T} \end{matrix} )) = 2tr(AD), and since A ≠ 0 and the trace form on gl_{m}(𝔽) is nondegenerate, we can find D ∈ gl_{m}(𝔽) such that 2tr(AD) ≠ 0. Otherwise, assume without loss of generality that B ≠ 0. Then tr(a · ( \begin{matrix}0 & 0 \\ E & 0 \end{matrix} )) = tr(BE), so we wish to find a symmetric matrix E such that tr(BE) ≠ 0. Take the basis B = {e_{ii}}_{i=1}^{m} ∪ {e_{ij} + e_{ji} | i < j} of the mxm symmetric matrices; then we can compute that for b_{1}, b_{2} ∈ B, tr(b_{1}b_{2}) ≠ 0 if and only if b_{1} = b_{2}. Write B = ∑_{b_{i}∈B} c_{i}b_{i} for some constants c_{i} ∈ 𝔽, and fix k such that c_{k} ≠ 0; then tr(B · b_{k}) = c_{k}tr(b_{k}^{2}) ≠ 0, as desired.

Exercise 10.2. Show that the Killing form on sl_{n}(𝔽), char(𝔽) ≠ 2, is nondegenerate if and only if char(𝔽) |̸ n.

Solution. Assume that K is nondegenerate but that char(𝔽) | n. Then tr(I_{n}) = n = 0 in 𝔽, so I_{n} ∈ sl_{n}(𝔽). But ad I_{n} = 0, so for all g ∈ sl_{n}(𝔽) we have K(I_{n}, g) = tr_{𝔤}(ad I_{n})(ad g) = tr_{𝔤}0 = 0, contradicting the nondegeneracy of K. Thus if K is nondegenerate, we must have char(𝔽) |̸ n.

Now suppose that char(𝔽) |̸ n. Let 𝔤 = sl_{n}(𝔽) have basis B = {e_{xy} | x ≠ y} ∪ {e_{ii} − e_{nn} | i < n}; then for any a, b ∈ g, K(a, b) = tr_{𝔤}(ad a)(ad b) is the sum of the b_{i}-components of (ad a)(ad b)b_{i} over all b_{i} ∈ (B). Let a = (a_{xy}) ∈ 𝔤. We first calculate K(a, e_{xy}) for x ≠ y by computing that [a, [e_{xy}, e_{pq}]] has e_{pq}-component δ_{py}a_{px} + δ_{xq}a_{yq} = (δ_{py} + δxq)a_{yx}, and that [a, [e_{xy}, e_{ii} − e_{nn}]] has (e_{ii} − e_{nn})-component (δ_{iy}(1 + δ_{nx}) + δ_{ix}(1 + δ_{ny}))a_{yx}, so that

K(a, e_{xy}) = ∑_{p≠q} ∑_{q=1}^{n}(δ_{py} + δ_{xq})a_{yx} + ∑_{i=1}^{n−1}(δ_{iy}(1 + δ_{nx}) + δ_{ix}(1 + δ_{ny}))a_{yx}
= 2(n − 1)a_{yx} + (1 + δ_{nx} − δ_{ny})a_{yx} + (1 + δ_{ny} − δ_{nx})a_{yx} = 2na_{yx}.

Hence if a is not diagonal, we may pick some x ≠ y such that a_{xy} ≠ 0, and then K(a, e_{yx}) = 2na_{xy} ≠ 0 as long as char(𝔽) |̸ n. A similar calculation yields K(a, e_{ii} −e_{nn}) = 2n(a_{ii} −a_{nn}). If a ≠ αI_{n} for any α, then there is some k < n such that a_{kk} ≠ a_{nn}, and so K(a, e_{kk} − e_{nn}) = 2n(a_{kk} − a_{nn}) ≠ 0. Otherwise, take α such that a = αI_{n}; then tr(a) = nα = 0, so (since char(𝔽) |̸ n) we must have α = 0 and hence a = 0. Since we can thus find a basis element b for any nonzero a ∈ 𝔤 such that K(a, b) ≠ 0, it follows that K is nondegenerate. Therefore the Killing form on 𝔤 = sl_{n}(𝔽) is nondegenerate if and only if char(𝔽) |̸ n, as desired.

Having computed these examples, we now proceed to our first practical application of trace forms. We will need the following lemma:

Lemma 1 (Cartan’s lemma). Let 𝔤 be a finite-dimensional Lie algebra over an algebraically closed field 𝔽 of characteristic 0, and let π be a representation of 𝔤 in a finite-dimensional vector space V over 𝔽. Let 𝔥 be a Cartan subalgebra of 𝔤, yielding generalized root and weight space decompositions 𝔤 = ⊕_{α∈𝔥^{∗}}𝔤_{α} and V = ⊕_{λ∈𝔥^{∗}}V_{λ}. Take e ∈ 𝔤_{α}, f ∈ 𝔤_{−α}, so that if h = [e, f] then h ∈ 𝔤_{0} = 𝔥. Suppose that V_{λ} ≠ 0 for some λ ∈ 𝔥^{∗}. Then λ(h) = r · α(h), where r is a rational number which depends only on α and λ.

Proof. Let U = ⊕_{n∈ℤ}V_{λ+nα}. Then U is invariant with respect to the operators π(e), π(f), since π(𝔤_{±α})V_{μ} ⊂ V_{±α+μ}. Hence we can restrict π to U and compute that tr_{U}(π(h)) = tr_{U}(π([e, f])) = 0. On the other hand, each V_{λ+nα} is invariant with respect to h ∈ 𝔤_{0}, so we can compute tr_{U}π(h) = ∑_{n} tr_{V_{λ+nα}}π(h). But π(h)|_{V_{μ}} has an upper triangular matrix with all diagonal entries μ, so we get

0 = tr_{U}(π(h)) = ∑_{n} dim(V_{λ+nα})(λ + nα)(h)
= λ(h) ∑_{n} dim(V_{λ+nα}) + α(h) ∑_{n}(n dim(V_{λ+nα})).

Let the two sums be P and Q, respectively; they are clearly both integers, since each of the summands is, and P = dim(U) > 0, so λ(h) = \frac{−Q}{P}α(h), where the ratio \frac{−Q}{P} depends only on λ and α, as desired.

We may use this to describe solvable subalgebras of gl_{V} in several ways in terms of trace forms.

Theorem 1 (Cartan’s criterion). Let 𝔤 be a subalgebra of gl_{V} for V a finite-dimensional vector space over a an algebraically closed field 𝔽 of characteristic 0. Then the following are equivalent:

1. (𝔤, [𝔤, 𝔤])_{V} = 0.

2. (a, a)_{V} = 0 for any a ∈ [𝔤, 𝔤].

3. 𝔤 is a solvable Lie algebra.

Proof. (1) ⇒ (2): Take a ∈ [𝔤, 𝔤], and write a = [b, c]. Then since (a, [b, c])_{V} = 0, we have (a, a)_{V} = 0.

(2) ⇒ (3): Suppose that 𝔤 is not solvable. Then the derived series of 𝔤 stabilizes to some nonzero subalgebra 𝔭 = 𝔤^{(N)} = 𝔤^{(N+1)} = ⋯, so [𝔭, 𝔭] = 𝔭. In particular, since 𝔭 ⊂ 𝔤, we have (a, a)_{V} = 0 for any a ∈ [𝔭, 𝔭] = 𝔭 by our assumption. Let 𝔥 be a Cartan subalgebra of 𝔭, and consider the root and weight space decompositions

𝔭 = ⨁_{α∈𝔥^{∗}} 𝔭_{α}, V = ⨁_{λ∈𝔥^{∗}} V_{λ}.

Since [𝔭_{α}, 𝔭_{β}] ⊂ 𝔭_{α+β} and [𝔭, 𝔭] = 𝔭, we conclude that 𝔥 = 𝔭_{0} = ∑_{α}[𝔭_{α}, 𝔭_{−α}]; that is, 𝔥 is a span of elements of the form h_{α,i} = [e_{α,i}, f_{α,i}], where e_{α,i} ∈ 𝔭_{α} and f_{α,i} ∈ 𝔭_{−α}.

Suppose that V_{λ} ≠ 0 for some fixed λ ∈ 𝔥^{∗}. By Cartan’s lemma, we can write λ(h_{α,i}) = r_{α,λ}α(h_{α,i}) for all α and i, where r_{α,λ} ∈ ℚ. Since we are assuming (2), (h_{α,i}, h_{α,i})_{V} = 0. But we compute that (h_{α,i}, h_{α,i})_{V} = ∑_{λ} λ(h_{α,i})^{2} dim(V_{λ}) by considering the restriction of h_{α,i} to each subspace V_{λ} as in the proof of Cartan’s lemma; this is then equal to ∑_{λ} r_{α,λ}^{2}α(h_{α,i})^{2} dim(V_{λ}) by the lemma, so

α(h_{α,i})^{2} (∑_{λ} r_{α,λ}^{2} dim(V_{λ})) = 0. Hence, since all the r_{α,λ} are rational numbers, either α(h_{α,i}) = 0, or r_{α,λ} = 0 whenever V_{λ} ≠ 0, and in either case we have λ(h_{α,i}) = r_{α,λ}α(h_{α,i}) = 0 whenever V_{λ} ≠ 0.

For any λ ∈ 𝔥^{∗} such that V_{λ} ≠ 0, since the elements {h_{α,i}} span 𝔥, and λ = 0 on all of these elements, we conclude that λ ≡ 0 on 𝔥. Therefore the weight space decomposition of V is V = V_{0}. But for any α ≠ 0 we have 𝔭_{α}V_{0} ⊂ V_{α} = 0, so 𝔭_{α} = 0; hence 𝔭 = 𝔭_{0} = 𝔥. But if 𝔭 = 𝔥 then 𝔭 is nilpotent, and so [𝔭, 𝔭] is a proper subset of 𝔭. This is a contradiction, so 𝔤 must in fact be solvable.

(3) ⇒ (1): By a corollary to Lie’s theorem, we may pick a basis of V such that all of the matrices in 𝔤 are upper triangular, hence all matrices in [𝔤, 𝔤] are strictly upper triangular. Then tr_{V} (ab) = 0 for any a ∈ 𝔤 and b ∈ [𝔤, 𝔤], since ab is strictly upper triangular, and so (𝔤, [𝔤, 𝔤])_{V} = 0.

This theorem leads almost immediately to a characterization of solvable Lie algebras, as follows:

Corollary 1. A finite dimensional Lie algebra 𝔤 over an algebraically closed field of characteristic 0 is solvable if and only if K(𝔤, [𝔤, 𝔤]) = 0.

Proof. We know that 𝔤 is solvable if and only if 𝔤/center(𝔤) and center(𝔤) are both solvable, hence if and only if 𝔤/center(𝔤) is solvable. Consider the adjoint representation ad : 𝔤 → gl_{𝔤}. Since ker(ad) = center(𝔤), this proof reduces to consideration of the subalgebra ad 𝔤 ⊂ gl_{𝔤}; we now apply the fact that conditions (1) and (3) of Cartan’s criterion are equivalent, and we are done.

Up until this point, we have restricted many of our results to Lie algebras over algebraically closed fields 𝔽 of characteristic zero, since root space and weight space decompositions exist for such algebras. Requiring 𝔽 to be algebraically closed is not always necessary, as we shall see:

Remark 1. Let 𝔽 be a field of characteristic zero which is not necessarily algebraically closed, and let 𝔤 be a Lie algebra over 𝔽. Then the following are true:

1. Cartan’s criterion and the corollary which followed it are both true over 𝔽.

2. 𝔥_{0}^{a} is a Cartan subalgebra of 𝔤 if a ∈ 𝔤 is regular.

3. [𝔤, 𝔤] is nilpotent if 𝔤 is finite-dimensional and solvable.

In order to show this, we will introduce some notation: let \overline{𝔽} be the algebraic closure of 𝔽, and let \overline{𝔤} = \overline{𝔽} ⊗_{𝔽} 𝔤.

Exercise 10.3. (a) Show that 𝔤 is solvable (resp. nilpotent, abelian) if and only if \overline{𝔤} is, and that \overline{[𝔤, 𝔤]} = [\overline{𝔤}, \overline{𝔤}]. (b) Prove the above remark for 𝔽 not algebraically closed.

Solution. Given two Lie algebras 𝔤 and 𝔥 over 𝔽, we claim that [\overline{𝔤}, \overline{𝔥}] = \overline{[𝔤, 𝔥]}. Clearly since 𝔤 ⊂ \overline{𝔤} and 𝔥 ⊂ \overline{𝔥}, we have \overline{[𝔤, 𝔥]} ⊂ \overline{[\overline{𝔤}, \overline{𝔥}]} = [\overline{𝔤}, \overline{𝔥}]. Given bases \{g_{α}\}_{α∈I} of 𝔤 and \{h_{β}\}_{β∈J} of 𝔥, these same bases generate \overline{𝔤} and \overline{𝔥} when combined over \overline{𝔽} rather than 𝔽. Hence the set \{[g_{α}, h_{β}]\}_{α∈I,β∈J} generates [\overline{𝔤}, \overline{𝔥}] over \overline{𝔽}. But this set also generates [𝔤, 𝔥] with coefficients in 𝔽, so when we extend this to \overline{F} we get [\overline{𝔤}, \overline{𝔥}] ⊂ \overline{[𝔤, 𝔥]}. Thus [\overline{𝔤}, \overline{𝔥}] = \overline{[𝔤, 𝔥]}.

Part (a) now follows easily, since we note that \overline{𝔤}^{i} = \overline{𝔤^{i}} and \overline{𝔤}^{(i)} = \overline{𝔤^{(i)}} for all i. Then 𝔤 is solvable if and only if 𝔤^{(n)} = 0 for some n; but this is equivalent to \overline{𝔤^{(n)}} = 0, or \overline{𝔤}^{(n)} = 0, and so 𝔤 is solvable iff \overline{𝔤} is. The same is true of nilpotency, as we see by replacing g^{(n)} with g^{n} in this argument, and abelianness, which comes from the solvability argument with n = 1. In particular, letting 𝔥 = 𝔤 above yields \overline{[𝔤, 𝔤]} = [\overline{𝔤}, \overline{𝔤}].

For part (b), we first look at Cartan’s criterion. Since 𝔤 is solvable if and only if \overline{𝔤} is, we have the equivalent conditions (1) (\overline{𝔤}, [\overline{𝔤}, \overline{𝔤}]_{V} = 0; (2) (a, a)_{V} = 0 for all a ∈ [\overline{𝔤}, \overline{𝔤}]; and (3) 𝔤 is solvable. But (\overline{𝔤}, [\overline{𝔤}, \overline{𝔤}])_{V} = (\overline{𝔤}, \overline{[𝔤, 𝔤]})_{V} = \overline{(𝔤, [𝔤, 𝔤])_{V}}, so (𝔤, [𝔤, 𝔤])_{V} = 0 if and only if (\overline{𝔤}, [\overline{𝔤}, \overline{𝔤}])_{V} = 0. Furthermore, (a, a)_{V} = 0 for all a ∈ [\overline{𝔤}, \overline{𝔤}] = \overline{[𝔤, 𝔤]} iff it does for all a ∈ [𝔤, 𝔤], since [𝔤, 𝔤] and \overline{[𝔤, 𝔤]} share the same basis over different fields and thus (a, a)_{V} = 0 for all a in one iff it does for all a in the other. Therefore Cartan’s criterion holds for 𝔤 given that it does for \overline{𝔤}; the corollary’s proof over 𝔤 is identical to its original proof, since it only needed algebraic closure to satisfy Cartan’s criterion.

Next, we consider the proof that 𝔥_{0}^{a} is a Cartan subalgebra if a is regular. If a ∈ 𝔤 is regular, it is regular in \overline{𝔤}, so we know that \overline{𝔥_{0}^{a}} is a Cartan subalgebra of \overline{𝔤}, or \overline{𝔥_{0}^{a}} = N\overline{𝔤}(\overline{𝔥_{0}^{a}}). Since [b, \overline{𝔥_{0}^{a}}] ⊂ \overline{𝔥_{0}^{a}} if and only if [b, 𝔥_{0}^{a}] ⊂ 𝔥_{0}^{a} for any b ∈ 𝔤, we conclude that N_{𝔤}(𝔥_{0}^{a}) = 𝔥_{0}^{a}, and so 𝔥_{0}^{a} is a Cartan subalgebra of 𝔤.

Finally, assume 𝔤 is finite dimensional and solvable. Then \overline{𝔤} is finite dimensional and solvable, so we know that [\overline{𝔤}, \overline{𝔤}] is nilpotent. But [\overline{𝔤}, \overline{𝔤}] = \overline{[𝔤, 𝔤]}, so \overline{[𝔤, 𝔤]} is nilpotent, hence [𝔤, 𝔤] is nilpotent.

Remark 2. The basic properties of a trace form are that it is symmetric (i.e. (a, b)_{V} = (b, a)_{V}) and invariant (i.e. ([a, b], c)_{V} = (a, [b, c])_{V}). The basic results on trace forms (like Cartan’s criterion) fail, however, if we assume only that the bilinear forms involved are symmetric and invariant.

We can construct an example of this as follows:

Exercise 10.4. Consider the 4-dimensional Lie algebra 𝔇 = 𝔽p + 𝔽q + 𝔽c + 𝔽d, where [p, q] = c, c is central, [d, p] = p, and [d, q] = d. Construct a nondegenerate symmetric invariant bilinear form on 𝔇.

Solution. Define a bilinear form B(x, y) on 𝔇 by the values B(p, q) = B(q, p) = 1; B(c, d) = B(d, c) = 1; B = 0 on all other pairs of basis elements; and all other values of B follow by symmetry

and bilinearity. Then B is nondegenerate, since it has matrix ( \begin{matrix}0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{matrix} ) with respect to the

basis {p, q, c, d} and this matrix is invertible. Invariance is also easy to check, so we are done.

The algebra 𝔇 is solvable, since [𝔇, 𝔇] = H_{1} is solvable, so by Cartan’s criterion, any trace form on 𝔇 must satisfy (𝔇, H_{1})_{V} = 0. But B(d, c) = 1, so B cannot be a trace form.

We conclude the lecture by defining a new class of Lie algebras:

Definition 2. A Lie algebra 𝔤 is called semisimple if it contains no nonzero solvable ideals. Equivalently, 𝔤 is semisimple if it contains no nonzero abelian ideals.

Equivalence can be proved as follows: If 𝔤 has a nonzero abelian ideal, then it has a nonzero solvable ideal, since abelian ideals are solvable. Conversely, if 𝔤 has a nonzero solvable ideal 𝔥, take n such that 𝔥^{(n)} ≠ 0 but 𝔥^{(n+1)} = 0; then 𝔥^{(n)} is a nonzero abelian ideal.

In the next lecture, we’ll prove that a finite-dimensional Lie algebra 𝔤 over a field of characteristic zero is semisimple if and only if the Killing form on 𝔤 is nondegenerate.