A Cohomological Proof that Real Representations of Semisimple Lie Algebras Have $\mathbb{Q}$-Forms

A Lie algebra $\mathfrak{g}_\mathbb{Q}$ over $\mathbb{Q}$ is said to be $\mathbb{R}$-universal if every homomorphism from $\mathfrak{g}_\mathbb{Q}$ to $\mathfrak{gl}(n,\mathbb{R})$ is conjugate to a homomorphism into $\mathfrak{gl}(n,\mathbb{Q})$ (for every $n$). By using Galois cohomology, we provide a short proof of the known fact that every real semisimple Lie algebra has an $\mathbb{R}$-universal $\mathbb{Q}$-form. We also provide a classification of the $\mathbb{R}$-universal Lie algebras that are semisimple.

1 Introduction Definition 1.1. Let g be a Lie algebra over Q. (All Lie algebras and all representations are assumed to be finite-dimensional.) 1. g is universal for real representations (or R-universal, for short) if every real representation of g has a Q-form [6,Defn. 7.1]. This means that if ρ : g → gl(n, R) is any (Q-linear) Lie algebra homomorphism, then there exists M ∈ GL(n, R), such that M ρ(x) M −1 ∈ gl(n, Q), for every x ∈ g.
2. g is a Q-form of a real Lie algebra g R if g ⊗ Q R ∼ = g R .
This note uses Galois cohomology to present a short proof of the following known result, which was first obtained by M. S. Raghunathan [9, §3] in the important special case where g R is compact. Corollary 1.3. Let G be a connected, semisimple Lie group with finite center. Then there is a discrete subgroup Γ of G, such that 1. G/Γ has finite volume (so Γ is a "lattice" in G), and 2. if ρ : G → GL(n, R) is any finite-dimensional representation of G, then ρ(Γ) is conjugate to a subgroup of GL(n, Z).

Proof of the main result
We begin by recalling a result of J. Tits that uses Galois cohomology to characterize the irreducible representations of semisimple algebraic groups over fields that are not algebraically closed.
Definition 2.1 ([11, §4.2]). Suppose g is a semisimple Lie algebra over a subfield F of C, and let G be the corresponding simply connected, semisimple algebraic group over F . It is well known that there is a (unique) quasi-split algebraic group G q over F , and a 1-cocycle Letting Z(G q ) be the center of G q , the short exact sequence e → Z(G q ) → G q → G q → e yields a corresponding long exact sequence of Galois cohomology sets, including a connecting map δ * : H 1 (F ; G q ) → H 2 F ; Z(G q ) . Hence, we have a cohomology class δ * [ξ] ∈ H 2 F ; Z(G q ) . Now, fix a maximal F -torus T of G q that contains a maximal F -split torus, and suppose λ is a weight of T that is invariant under the * -action of the Galois group Gal(F /F ). Then the restriction of λ to Z(G q ) is a Gal(F /F )-equivariant homomorphism from Z(G q ) to the group µ of roots of unity in C, so it induces a homomorphism λ * : H 2 F ; Z(G q ) → H 2 F ; µ . Therefore, we may define [11,Thm. 7.2 and Lem. 7.4]). Suppose g is a semisimple Lie algebra over a subfield F of C, and λ is a dominant weight. Then:

Proposition 2.2 (Tits
1. There is an irreducible representation F ρ λ : g → gl(n, F ), for some n, such that F ρ λ ⊗ F C has an irreducible summand with highest weight λ. Furthermore, F ρ λ is unique up to isomorphism.   if λ is invariant under the * -action of Gal(C/R), and β g,R (λ) = 0, then λ is also invariant under the * -action of Gal(Q/Q), and β g,Q (λ) = 0. (2.4) Proof. We prove only (⇐), but the argument is reversible. We wish to show that Q ρ λ is a Q-form of R ρ λ , for every dominant weight λ. By combining (2.4) with Proposition 2.2(3), we see that Q ρ λ ⊗ Q C is irreducible if and only if R ρ λ ⊗ R C is irreducible. Furthermore, since g Q splits over a quadratic extension, we know that Q ρ λ ⊗ Q C is either irreducible or the direct sum of two irreducibles [6, Cor. 3.2(2)]. Therefore, Q ρ λ ⊗ Q C and R ρ λ ⊗ R C have the same number of irreducible constituents. (Namely, either they are both irreducible, or they are both the direct sum of 2 irreducibles.) Since R ρ λ is a summand of Q ρ λ ⊗ Q R, We will also use the following (weak form of an) important fact in the theory of Galois cohomology: [7,Prop. 6.17, p. 337]). If G is a connected, semisimple algebraic group over an algebraic number field F ⊂ R, and G splits over a finite, Galois extension L of F , with L ⊂ C, but L ⊂ R, then the restriction map Proof of Proposition 1.2. Suppose g R is a real semisimple Lie algebra, and let G be the simply connected, semisimple R-algebraic group whose Lie algebra is g R . Write G = ξ G q , where ξ : Gal(C/R) → G q is a 1-cocycle and G q is quasi-split. Let L = Q[i]. By choosing an appropriate Q-form, we may assume that G q is a quasi-split Q-group that splits over L, and that the * -action of Gal(L/Q) is the same as the * -action of Gal(C/R).
Let σ be the nontrivial element of Gal(C/R), fix a representative a ∈ G q of ξ(σ) ∈ G q , and let z = a σ a. Since σ 2 is trivial and ξ is a 1-cocycle, we know that z is trivial in G q , which means z ∈ Z(G q ). This implies that a commutes with σ a, so σ fixes z, which means z ∈ Z(G q )(R).
Let H be the product of the almost simple factors of G q that are absolutely almost simple and of type 2 A n (more concretely, H is the product of the factors that are isomorphic to SU(k, ℓ), for some k and ℓ), We claim that we may assume z ∈ Z(G q )(Q). While proving this, we may consider each simple factor individually, so there is no harm in assuming G q is almost simple. This allows us to furthermore assume that G q is absolutely almost simple. (Otherwise, since every C-group is split, we could assume ξ is trivial.) Also, since |Gal(C/R)| = 2, we may assume, by replacing a with aw for an appropriately chosen w ∈ z , that |z| is a power of 2. Assuming, as we may, that z is nontrivial, this implies that G q is not of type 1,2 E 6 .
Then Z(G q )(R) = Z(G q )(Q): if G q is not of type 2 A n , then this can be seen from the table on page 332 of [7], but if not, then the definition of G q implies |Z(G q )| ≤ 2, so every element of Z(G q ) is defined over Q. This completes the proof of the claim.
The claim of the preceding paragraph implies that the cyclic subgroup z generated by z is defined over Q. Hence, the quotient G q = G q / z, Z(H) 2 is a semisimple Q-group. Now, since a σ a = z is trivial in G q , we know that ξ lifts to a 1-cocycle ξ : Gal(C/R) → G q . Then Proposition 2.5 implies that, after replacing ξ with a cohomologous cocycle, we may assume ξ is the restriction of a 1-cocycle ζ : To complete the proof, we show that g is R-universal, by verifying (2.4). To this end, let λ be a Gal(C/R)-invariant dominant weight, such that β g,R (λ) = 0. Since G q is L-split and the * -action of Gal(L/Q) is the same as the * -action of Gal(C/R) (by the choice of the Q-form of G q ), we know that λ is invariant under Gal(Q/Q).
Since ζ is a 1-cocycle into G q = G q / z, Z(H) 2 , we know that, in the notation of Definition 2.1 with F = Q, we have δ * [ζ] ∈ H 2 Q; z, Z(H) 2 . Therefore, in order to show that β g,Q (λ) = λ * δ * [ζ] = 0, it suffices to show that λ is trivial on both z and Z(H) 2 . Note that, in the notation of Definition 2.  . Suppose g R is a compact, simple Lie algebra over R. There is a Q-form g Q of g R , such that g Q splits over some quadratic extension of Q, but is not R-universal, if and only if either a. g R ∼ = su(n), for some n that is divisible by 4, or b. g R ∼ = so(n), for some n ≡ 3, 5 (mod 8) (with n ≥ 6). , where ℓ is required to only be odd, whereas it actually needs to be ≡ 3 (mod 4). This means that g, the compact real form of type A ℓ , is isomorphic to su(n), for some n that is divisible by 4. In [6, §7], it is incorrectly stated that n only needs to be even, not divisible by 4. Corollary 3.3. Suppose g R is a compact, simple Lie algebra over R. If g R is of type C n , Proof. Every Lie algebra of any of these types (over an algebraic number field) splits over an appropriate quadratic extension [7, Prop. 6.16, p. 335], and g R does not appear in Proposition 3.1. (Lie algebras of type B n also split over a quadratic extension, but they are the Lie algebras in 3.1(b) with n odd.) In the remainder of this section, we determine exactly which Q-forms are R-universal for each of the other types of compact simple Lie algebras: 2 A n (Proposition 3.7), B n (Proposition 3.9), 1,2 D n (Lemma 3.12 and Proposition 3.13), and 2 E 6 (Proposition 3.14). The results for classical groups can be obtained quite easily from the calculations of β g,F (λ) in [4, §27.B, pp. 378-379], and the answer for 2 E 6 is immediate from observations of Tits [11, §6.4].
We will use the following concrete interpretation of β g,F (λ): Since every root of G q is trivial on the center, the following observation is immediate from Definition 2.1. Proof. We may assume that λ is not in the root lattice, for otherwise Lemma 3.5 tells us that β g,F (λ) = 0.
Let L = F √ a and b = det A. Proof. (⇐) From (1), we see that g = su n (A; L, τ ), so g is L-split. Hence, we may assume n is divisible by 4, for otherwise Proposition 3.1 implies the desired conclusion that g is R-universal. Then, since Gal(C/R) and Gal(L/Q) have the same * -action (and det A is a norm in L), Lemma 3.6 implies that β g,Q (λ) = 0 for every Gal(C/R)-invariant weight λ. This establishes (2.4), so g is R-universal.
(⇒) The natural representation ρ : g ֒→ Mat k×k (D) is irreducible over Q. Since g is R-universal, this representation must remain irreducible over R. By Schur's Lemma, this implies that D ⊗ Q R has no zero divisors. On the other hand, D ⊗ Q R is split (since its center is L ⊗ Q R = C). Therefore D must be a field, so D = L. This establishes (1).
Let λ be a Gal(C/R)-invariant weight that is not in the root lattice. If n is divisible by 4, then (−1) n(n−1)/2 det I = 1 is a square in R, so Lemma 3.6 tells us that β g,R (λ) = 0. Then, since g is R-universal, we must have β g,Q (λ) = 0. So Lemma 3.6 tells us that det A must be a norm in L. This establishes (2).
3B Q-forms of so(n) Notation 3.8. Any symmetric matrix A ∈ GL k (Q) determines a nondegenerate quadratic form on Q k . We use Cliff 0 Q (A) to denote the corresponding even Clifford algebra [5, p. 104]. It is well known that Cliff 0 Q (A) is either a simple algebra or the direct sum of two isomorphic simple algebras over Q [5,Thms. 2.4 and 2.5,p. 110]. If A has been diagonalized, then it is straightforward to determine whether this simple algebra is split (in which case, we also say that Cliff 0 Q (A) is split). Namely, the simple algebra is Brauer equivalent to a quaternion algebra that can be calculated from the eigenvalues of A (cf. [5,Cor. 3.14, p. 117]). Proposition 3.9. Suppose g is a Q-form of so(n), for some odd n ≥ 5, so g = so n (A; Q), where A is a symmetric matrix in GL n (Q). Then g is R-universal if and only if either n ≡ ±3 (mod 8) or Cliff 0 Q (A) is split. Proof. Since g is of type B k (where 2k + 1 = n), we know that it splits over a quadratic extension L of Q [7, Prop. 6.16(2), p. 335]. Hence, we may assume n ≡ ±1 (mod 8), for otherwise Proposition 3.1 implies that g is R-universal.
If λ is any dominant weight of g that is not in the root lattice, then the proof of Proposition 3.1 tells us that β g,R (λ) = 0. (This can also be deduced from [ 1. k is odd and Cliff 0 Q (A) is split, or 2. k ≡ 2 (mod 4), and det A is a square in Q, or 3. k is divisible by 4, det A is a square in Q, and Cliff 0 Q (A) is split. Proof. As in [4, §27B, type D n , p. 379], let λ, λ + , λ − be dominant weights that represent the three nonzero classes modulo the root lattice. The weight λ corresponds to the natural representation of g on Q 2k , so β g,Q (λ) = [Q] = 0.
Suppose, first, that k is even. Then so(2k) is an inner form, which means that Gal(C/R) acts trivially on Λ + . Hence, Lemma 3.11 tells us that if g is R-universal, then g must be inner, which (since k is even) means that det A is a square in Q. Therefore, Cliff 0 Q (A) is a direct sum of two algebras C + and C − that are Brauer equivalent to the full Clifford algebra [5, Thm. 2.5(3), p. 110]. Furthermore, from [4, §27B, type D n , p. 379], we know that β g,Q (λ ± ) = [C ± ]. Since k is even, the Clifford algebra of so(2k) is split if and only if k is divisible by 4 [5, p. 123]. Therefore, Lemma 3.10 shows that an inner form g is: • automatically R-universal, when k ≡ 2 (mod 4), but • R-universal if and only if Cliff 0 Q (A) is split, when k is divisible by 4. Assume, now, that k is odd. This means that so(2k) is an outer form (so g is obviously also outer). Then Gal(C/Q) interchanges λ + and λ − , so Proposition 3.4(1) implies D g,R (λ ± ) = C. Let L be the (unique, imaginary) quadratic extension of Q over which g becomes inner. From [4, §27B, type D n , p. 379], we see that D g,Q (λ ± ) is Brauer equivalent to Cliff 0 Q (A), which is central simple over L. Hence, Lemma 3.10 implies that g is R-universal if and only if Cliff 0 Q (A) is split.
3C Q-forms of 2 E 6 Proposition 3.14. Suppose g is a Q-form of the compact real Lie algebra of type 2 E 6 , and let L be the unique quadratic extension of Q over which g is inner. Then g is R-universal if and only if it splits over L.
Proof. Let λ be a weight that is not in the root lattice. Then λ is not fixed by the * -action of Gal(C/R), so ρ λ R ⊗ R C is the direct sum of two irreducible representations [11,Lem. 7.4]. Hence, g is R-universal if and only if ρ λ Q ⊗ C is also the direct sum of only two irreducible representations; in other words, β g,L (λ) = 0. This obviously holds if g splits over L. Now, assume g does not split over L. Since g is outer over R, but inner over L, we know that L is an imaginary extension, so g obviously splits at the infinite place of L. Then, by inspection of the possible Tits indices of type 1 E 6 over a nonarchimedean local field [10, p. 58], we see that the central vertex of the Tits index is circled at every place, so it must be circled over L [2,Satz 4.3.3]. Therefore, g must be of type 1 E 16 6,2 over L (since it is not split). From [11, 6.4.5], we see that this implies β g,L (λ) = 0. So g is not R-universal.