1.7 Real Forms of Complex Algebraic Groups
1.7.1 Real Forms and Complex Conjugations
We begin with a definition that refers to subgroups ofGL(n,C). We will obtain a more general notion of areal formlater in this section.
Definition 1.7.1.LetG⊂GL(n,C)be an algebraic subgroup. ThenGis defined overRif the idealIGis generated byIR,G={f∈IG : f(GL(n,R))⊂R}. IfGis defined overR, then we setGR=G∩GL(n,R)and callGRthegroup ofR-rational pointsofG.
Examples
1.The groupG=GL(n,C)is defined overR(sinceIG=0), andGR=GL(n,R).
2.The groupG=Bnofn×ninvertible upper-triangular matrices is defined over R, sinceIGis generated by the matrix-entry functions{xi j :n≥i> j≥1}, which are real valued on GL(n,R). In this caseGR is the group ofn×n real invertible upper-triangular matrices.
Forg∈GL(n,C)we setσ(g) =g(complex conjugation of matrix entries). Then σis an involutive automorphism ofGL(n,C) as a real Lie group (σ2is the identity) and dσ(A) =A¯forA∈Mn(C). If f ∈O[GL(n,C)]then we set
f¯(g) =f(σ(g)).
Here the overline on the right denotes complex conjugation. Sincef is the product of det−k(for some nonnegative integerk) and a polynomialϕin the matrix-entry func- tions, we obtain the function ¯f by conjugating the coefficients ofϕ. We can write
f =f1+if2, where f1= (f+f¯)/2, f2= (f−f¯)/(2i), and i=√
−1. The functions f1and f2are real-valued onGL(n,R), and ¯f =f1−if2. Thus f(GL(n,R))⊂Rif and only if ¯f =f.
Lemma 1.7.2.Let G⊂GL(n,C)be an algebraic subgroup. Then G is defined over R if and only if IGis invariant under f7→f .¯
Proof. Assume thatGis defined overR. Iff1∈IR,Gthenf1=f¯1. Hencef1(σ(g)) = f1(g) =0 forg∈G. SinceIR,Gis assumed to generateIG, it follows thatσ(g)∈G for allg∈G. Thus for anyf ∈IGwe have ¯f(g) =0, and hence ¯f ∈IG.
1.7 Real Forms of Complex Algebraic Groups 63
Conversely, if IG is invariant under f 7→ f¯, then every f ∈IG is of the form f1+if2as above, where fj∈IR,G. ThusIR,GgeneratesIG, and soGis defined over
R. ut
Assume thatG⊂GL(n,C)is an algebraic group defined overR. Letg⊂Mn(C) be the Lie algebra ofG. SinceIR,G generatesIGandσ2 is the identity map, this implies that σ(G) =G. Hence σ defines a Lie group automorphism of G and dσ(A) =A¯∈gfor allA∈g. By definition,GR={g∈G:σ(g) =g}. HenceGRis a Lie subgroup ofGand
Lie(GR) ={A∈g: ¯A=A}.
If A∈gthen A=A1+iA2, where A1= (A+A)/2 and¯ A2= (A−A)/2i are in¯ Lie(GR). Thus
g=Lie(GR)⊕i Lie(GR) (1.61)
as a real vector space, so dimRLie(GR) =dimCg. Therefore the dimension of the Lie groupGRis the same as the dimension ofGas a linear algebraic group overC (see Appendix A.1.6).
Remark 1.7.3.If a linear algebraic groupG is defined overR, then there is a set Aof polynomials with real coefficients such thatGconsists of the common zeros of these polynomials inGL(n,C). The converse assertion is more subtle, however, since it is possible thatAdoes not generate the idealIG, as required by Definition 1.7.1. For example, the groupBnof upper-triangularn×nmatrices is the zero set of the polynomials{x2i j :n≥i> j≥1}; these polynomials are real onGL(n,R)but do not generateIBn (of course, in this case we already know thatBnis defined over R).
By generalizing the notion of complex conjugation we now obtain a useful cri- terion (not involving a specific matrix form ofG) forGto be isomorphic to a linear algebraic group defined overR. This will also furnish the general notion of areal formofG.
Definition 1.7.4.LetGbe a linear algebraic group and letτbe an automorphism of Gas a real Lie group such thatτ2is the identity. For f ∈O[G]define fτ by
fτ(g) =f(τ(g))
(with the overline denoting complex conjugation). Thenτis acomplex conjugation onGif fτ∈O[G]for all f ∈O[G].
WhenG⊂GL(n,C)is defined overR, then the mapσ(g) =gintroduced previ- ously is a complex conjugation. In Section 1.7.2 we shall give examples of complex conjugations whenGis a classical group.
Theorem 1.7.5.Let G be a linear algebraic group and letτbe a complex conjuga- tion on G. Then there exists a linear algebraic group H ⊂GL(n,C)defined over Rand an isomorphismρ:G //H such thatρ(τ(g)) =σ(ρ(g)), whereσis the conjugation ofGL(n,C)given by complex conjugation of matrix entries.
Proof. Fix a finite set{1,f1, . . . ,fm}of regular functions onGthat generateO[G]
as an algebra overC(for example, the restrictions toGof the matrix entry functions and det−1given by the defining representation ofG). SetC(f) = fτ for f ∈O[G]
and let
V=SpanC{R(g)fk,R(g)C fk :g∈Gand k=1, . . . ,m}.
ThenV is invariant underGandC, sinceCR(g) =R(τ(g))C. Letρ(g) =R(g)|V. It follows from Proposition 1.5.1 thatV is finite-dimensional and(ρ,V)is a regular representation ofG.
We note that ifg,g0∈Gandfk(g) = fk(g0)for allk, then f(g) = f(g0)for allf ∈ O[G], since the set{1,f1, . . . ,fm}generatesO[G]. Letting f run over the restrictions toGof the matrix entry functions (relative to some matrix form ofG), we conclude thatg=g0. Thus ifρ(g)f =f for all f∈V, theng=I, proving that Ker(ρ) ={I}. SinceC2is the identity map, we can decomposeV =V+⊕V−as a vector space overR, where
V+={f ∈V :C(f) =f}, V−={f ∈V :C(f) =−f}.
BecauseC(if) =−iC(f)we haveV−=iV+. Choose a basis (over R) ofV+, say {v1, . . . ,vn}. Then{v1, . . . ,vn}is also a basis ofV overC. If we use this basis to identifyV withCnthenCbecomes complex conjugation. To simplify the notation we will also writeρ(g)for the matrix ofρ(g)relative to this basis.
We now have an injective regular homomorphismρ:G //GL(n,C)such that ρ(τ(g)) =σ(ρ(g)), where σ denotes complex conjugation of matrix entries. In Chapter 11 (Theorem 11.1.5) we will prove that the image of a linear algebraic group under a regular homomorphism is always a linear algebraic group (i.e., a closed subgroup in theZariski topology). Assuming this result (whose proof does not depend on the current argument), we conclude thatH=ρ(G)is an algebraic subgroup of GL(n,C). Furthermore, ifδ ∈V∗is the linear functional f 7→ f(I), then
f(g) =R(g)f(I) =hδ,R(g)fi. (1.62) Henceρ∗(O[H]) =O[G], since by (1.62) the functions f1, . . . ,fmare matrix entries of(ρ,V). This proves thatρ−1is a regular map.
Finally, let f ∈IH. Then forh=ρ(g)∈Hwe have f¯(h) =f(σ(ρ(g))) = f(ρ(τ(g))) =0.
Hence ¯f ∈IH, so from Lemma 1.7.2 we conclude thatHis defined overR. ut Definition 1.7.6.LetGbe a linear algebraic group. A subgroupKofGis called a real formofGif there exists a complex conjugationτonGsuch that
K={g∈G:τ(g) =g}.
LetKbe a real form ofG. ThenKis a closed subgroup ofG, and from Theorem 1.7.5 and (1.61) we see that the dimension ofKas a real Lie group is equal to the dimension ofGas a complex linear algebraic group, and
1.7 Real Forms of Complex Algebraic Groups 65
g=Lie(K)⊕i Lie(K) (1.63)
as a real vector space.
One of the motivations for introducing real forms is that we can study the repre- sentations ofGusing the real form and its Lie algebra. LetGbe a linear algebraic group, and letG◦be the connected component of the identity of G(as a real Lie group). LetKbe a real form ofGand setk=Lie(K).
Proposition 1.7.7.Suppose(ρ,V)is a regular representation of G. Then a subspace W ⊂V is invariant underdρ(k)if and only if it is invariant under G◦. In particular, V is irreducible underkif and only if it is irreducible under G◦.
Proof. AssumeW is invariant underk. Since the mapX7→dρ(X)fromgto End(V) is complex linear, it follows from (1.63) thatW is invariant underg. LetW⊥⊂V∗ be the annihilator ofW. Thenhw∗,(dρ(X))kwi=0 forw∈W,w∗∈W⊥,X∈g, and all integersk. Hence
hw∗,ρ(expX)wi=hw∗,exp(dρ(X))wi=
∑
∞k=0
1
k!hw∗,dρ(X)kwi=0, soρ(expX)W⊂W. SinceG◦is generated by exp(g), this proves thatW is invariant underG◦. To prove the converse we reverse this argument, replacingX bytX and
differentiating att=0. ut