Seminar Sophus Lie 1 (1991) 243–245

On complex Lie algebras with a simple real form

S. S. L. Darmstadt

The following observations, motivated by a question of Lawson’s [1], does not seem to be readily accessible in the literature.

Let 𝔞 denote a complex Lie algebra and 𝔤 a real subalgebra. Then 𝔤+i𝔤 and 𝔤 ∩ i𝔤 are complex subalgebras. If X ∈ 𝔤 then [X, 𝔤 ∩ i𝔤] ⊆ [X, 𝔤] ∩ i[X, 𝔤] ⊆ 𝔤 ∩ i𝔤. Then [iX, 𝔤 ∩ i𝔤] = i[X, 𝔤 ∩ i𝔤] ⊆ i(𝔤 ∩ i𝔤) = 𝔤 ∩ i𝔤. Thus 𝔤 ∩ i𝔤 is an ideal of 𝔤 + i𝔤.

Remark 1. If 𝔤 is simple, then either 𝔤 = 𝔤 ∩ i𝔤, i.e., i𝔤 = 𝔤 and 𝔤 is a complex subalgebra, or else 𝔤 ∩ i𝔤 = {0}, i.e. 𝔤 + i𝔤 = 𝔤 ⊕ i𝔤 ≅ ℂ ⊗ 𝔤 is the complexification of 𝔤 embedded into 𝔞.

Lemma 2. Assume 𝔤_{1} ⊕ i𝔤_{1} ≅ ℂ ⊗ 𝔤_{1} and 𝔤_{2} ⊕ i𝔤_{2} ≅ ℂ ⊗ 𝔤_{2} , and suppose that there is a morphism f: 𝔤_{1} → 𝔤_{2} of real Lie algebras. Then F: 𝔤_{1} ⊕i𝔤_{1} → 𝔤_{2} ⊕i𝔤_{2} , F (X + iY) = f(X) + if(Y) is a morphism of complex algebras.

Proof. Compute or observe that under the isomorphisms 𝔤_{j} + i𝔤_{j} ≅ ℂ ⊗ 𝔤_{j} the map F corresponds to the map id_{ℂ} ⊗f: ℂ ⊗ 𝔤_{1} → ℂ ⊗ 𝔤_{2} !

If 𝔞 is the complexification of a real simple Lie algebra 𝔤 which itself is the underlying real vector space of a complex Lie algebra then 𝔞 = 𝔤 ⊕ i𝔤 has a real involution τ given by τ(X + iY) = X − iY which is a complex conjugate automorphism of 𝔞, satisfying τ(iZ) = −iτ(Z) for all Z ∈ 𝔞. The complex Lie algebra 𝔞 is isomorphic to the direct sum 𝔤^{+} ⊕ 𝔤^{−} of two simple complex ideals which are conjugate complex (i.e., g^{−} = τ(g^{+})). These may nevertheless be isomorphic as complex algebras. Each real isomorphism f: 𝔤^{+} → 𝔤^{−} gives a subalgebra 𝔤_{f} = {X ⊕ f(X): X ∈ 𝔤^{+}} which is isomorphic to 𝔤^{+} ≅ 𝔤. We call these graph subalgebras of 𝔞. If we define σ: 𝔤^{+} → 𝔤^{−} by σ(X) = τ(X), i.e., by restricting and corestricting τ, then 𝔤 = 𝔤_{σ}. The graph subalgebra 𝔤_{f} is complex if and only if f: 𝔤^{+} → 𝔤^{−} is a complex isomorphism. A real automorphism of 𝔞 will be called special if it leaves the ideals 𝔤^{±} invariant.

Lemma 3. For two graph subalgebras 𝔤_{f} and 𝔤_{g} there is a special real automorphism F of 𝔞 with F (𝔤_{f}) = 𝔤_{g}. If f = σ and 𝔤_{g} is complex, then F is neither complex nor complex conjugate.

Proof. We note that gf^{−1}: 𝔤^{−} → 𝔤^{−} is an automorphism α of 𝔤^{−}, and g = α∘f. We define a special automorphism F: 𝔞 → 𝔞 by F (X ⊕Y) = X ⊕α(Y). Now we have 𝔤_{g} = {X ⊕ g(X): X ∈ 𝔤^{+}} = {X ⊕ α(f(X)): X ∈ 𝔤^{+}} = F {X ⊕ f(X): X ∈ 𝔤} = F (𝔤_{f}). Suppose that f = σ and that g is complex, then F |𝔤^{+} = id_{𝔤^{+}} is complex and F |𝔤^{−} = α is complex conjugate. Thus F is neither complex nor complex conjugate.

Proposition 4. Let 𝔞 be a complex Lie algebra with a simple real form 𝔤 and a real subalgebra 𝔤^{∗} which is isomorphic to 𝔤. Then one of the following mutually exclusive cases occurs:

(A) There is a complex automorphism F: 𝔞 → 𝔞 with F (𝔤) = 𝔤^{∗}.

(B) 𝔞 is not simple, and there is a special real autmorphism F: 𝔞 → 𝔞 which

is neither complex nor complex conjugate with F (𝔤) = 𝔤^{∗}. (C) 𝔞 is not simple, and 𝔤^{∗} is one of the two unique ideals 𝔤^{+} or 𝔤^{−}

with 𝔞 = 𝔤^{+} ⊕ 𝔤^{−}. There is no isomorphism of 𝔞 mapping 𝔤 to 𝔤^{∗}, but the projection onto 𝔤^{∗} is a complex endomorphism of 𝔞 mapping 𝔤 isomorphically onto 𝔤^{∗}.

Proof. Since 𝔤^{∗} ≅ 𝔤, the real subalgebra 𝔤^{∗} is simple. By Remark 1, we have two cases: Case (i): 𝔤^{∗} ⊕ i𝔤^{∗}. Case (ii): i𝔤^{∗} = 𝔤^{∗}.

In Case (i), Lemma 2 shows the existence of an endomorphism F of 𝔞 with F (𝔤) = 𝔤^{∗}. If 𝔞 is simple, then F must be an automorphism. If 𝔞 = 𝔤^{+}⊕𝔤^{−} then the complex endomorphism F must respect the two summands. If it were zero on one of the summands, then 𝔤 would have to be contained in the other, but 𝔤 is contained in neither. Hence F is an automorphism and we are in Case (A).

In Case (ii), because of 𝔤 ≅ 𝔤^{∗}, the algebra 𝔞 is the complexification of the underlying real algebra of a complex simple Lie algebra and is therefore of the form 𝔤 = 𝔤^{+} ⊕ 𝔤^{−} according to remarks preceding Lemma 3. The following mutually exclusive cases occur: (a) 𝔤^{∗} = 𝔤^{+} . (b) 𝔤^{∗} = 𝔤^{−}. (c) 𝔤^{∗} is a complex graph subalgebra. In Cases (a) and (b) we have Case (C) of the Proposition. Then 𝔤^{∗} is an ideal while 𝔤 is not. Hence no real or complex automorphism of 𝔞 maps 𝔤 onto 𝔤^{∗}. In Case (c) we apply Lemma 3 to the two graph subalgebras 𝔤 = 𝔤_{σ} and 𝔤^{∗} and find ourselves in Case (B) of the Proposition.

Corollary 5. If 𝔞 is a simple complex Lie algebra and 𝔤 a real form, and if 𝔤^{∗} is a real subalgebra of 𝔞 which is isomorphic to 𝔤 then there is a complex automorphism of 𝔞 mapping 𝔤 onto 𝔤^{∗}.

We remark, that the automorphism group Aut(𝔤) of a complex simple Lie algebra is a semidirect product of the normal connected subgroup Int(𝔤) = ⟨e^{ad X}: X ∈ 𝔤⟩ of inner automorphisms by a finite group E, namely, the automorphism group of a basis of a root system. The group E is of order 2 for A_{n}, n > 1, D_{n}, n ≠ 4, E_{6} and of order 3 for D_{4} , and it is trivial in all other cases. (See [3] or [2], notably p. 298.)

Corollary 6. Let A be a complex simple and simply connected Lie group with complex conjugation κ, and let G denote the real form of elements fixed under κ. Let H denote any real analytic subgroup of A whose Lie algebra is isomorphic to 𝔤 = L(G). Then H is conjugate to G under a complex automorphism of A. In particular, A is closed and isomorphic to H.

Proof. The Lie algebra 𝔞 of A is the complexification of 𝔤 and is simple. Hence we are in Case (A) of Proposition 4 and find a complex automorphism F: 𝔞 → 𝔞 with F (g) = 𝔥. We have Aut 𝔤 = (Inn 𝔤)× E = Ad(G)× E with a finite group E or order 1,2, or 3 according to the list above. Since A is simply connected, there is a unique automorphism f: A → A inducing F. The group H is connected (see e.g. [2], p. 214, Theorem 9). The analytic groups H and f(G) have the same Lie algebra and are both connected. Hence they agree.

If the group of automorphisms of 𝔞 fixing 𝔤 as a whole contains an automorphism which is not inner, or if there are no outer automorphisms of 𝔞 then the group of inner automorphisms acts transitively on the set of subalgebras isomorphic to 𝔤. The algebra 𝔰𝔩(nℂ) (realizing type A_{n−1} ) for instance admits a non-inner involution of the form X ↦ −X^{⊤} which leaves 𝔰𝔩(n, ℝ) invariant. A similar statement holds for D_{n}, n odd, and E_{6} (see [2], 297f.) In such cases, in Corollary 6, the groups G and H are conjugate under an inner automorphism.

References

[1] Lawson, J. D., A question on the simple real forms of the Lie algebra sp(2,ℂ), e-mail message, October 1991.

[2] Onishchik, A. L., and E. B. Vinberg, “Lie Groups and Algebraic Groups”, Springer-Verlag Berlin etc., 1990.

[3] Tits, J., “Tabellen zu den einfachen Lie Gruppen und ihren Darstellungen”, (Springer) Lecture Notes in Mathematics 40 (1967).

Karl H. Hofmann Fachbereich Mathematik Technische Hochschule Darmstadt Schloßgartenstraße 7 D-6100 Darmstadt, Germany

Received October 10, 1991