Unipotent Elements

2.2 Unipotent Elements

Unipotent elements give an algebraic relation between a linear algebraic group and its Lie algebra, since they are exponentials of nilpotent elements and the exponential map is a polynomial function on nilpotent matrices. In this section we exploit this property to prove the connectedness of the classical groups.

2.2.1 Low-Rank Examples

We shall show that the classical groupsSL(n,C),SO(n,C), andSp(n,C)are gen- erated by their unipotent elements. We begin with the basic caseG=SL(2,C). Let N⁺={u(z):z∈C}andN⁻={v(z):z∈C}, where

u(z) = 1 z 0 1

and v(z) = 1 0 z1

.

The groupsN⁺andN⁻are isomorphic to the additive group of the fieldC. Lemma 2.2.1.The groupSL(2,C)is generated by N⁺∪N⁻.

Proof. Letg=_{a b}

c d

withad−bc=1. Ifa6=0 we can use elementary row and column operations to factor

g=

1 0

a⁻¹c 1

a 0 0 a⁻¹

1 a⁻¹b

0 1

.

Ifa=0 thenc6=0 and we can likewise factor g=

0−1 1 0

c 0 0 c⁻¹

1c⁻¹d

0 1

.

Finally, we factor 0−1

1 0

= 1−1

0 1 1 0 1 1

1−1 0 1

, a 0

0 a⁻¹

= 1−a

0 1

1 0

(a⁻¹−1) 1 1 1 0 1

1 0

(a−1) 1

,

to complete the proof. ut

The orthogonal and symplectic groups of low rank are closely related toGL(1,C) andSL(2,C), as follows. Define a skew-symmetric bilinear formΩ onC²by

Ω(v,w) =det[v,w],

where[v,w]∈M₂(C)has columnsv,w. We have det[e₁,e₁] =det[e₂,e₂] =0 and det[e1,e2] =1, showing that the formΩ is nondegenerate. Since the determinant function is multiplicative, the formΩ satisfies

Ω(gv,gw) = (detg)Ω(v,w) forg∈GL(2,C).

Hence g preserves Ω if and only if detg =1. This proves that Sp(C²,Ω) = SL(2,C).

Next, consider the groupSO(C²,B)withBthe bilinear form with matrixs₂in (2.5). We calculate that

g^ts2g=

2ac ad+bc ad+bc 2bd

for g= a b c d

∈SL(2,C).

Sincead−bc=1, it follows thatad+bc=2ad−1. Henceg^ts2g=s2if and only ifad=1 andb=c=0. ThusSO(C²,B)consists of the matrices

a 0 0a⁻¹

for a∈C^×. This furnishes an isomorphism SO(C²,B)∼=GL(1,C).

Now consider the groupG=SO(C³,B), whereBis the bilinear form onC³with matrixs₃as in (2.5). From Section 2.1.2 we know that the subgroup

H={diag[x,1,x⁻¹]:x∈C^×}

of diagonal matrices inGis a maximal torus. SetGe=SL(2,C)and let He={diag[x,x⁻¹]:x∈C^×}

be the subgroup of diagonal matrices inG.e

We now define a homomorphismρ:Ge //Gthat mapsHeontoH. Set V={X∈M2(C): tr(X) =0}

and letGeact onV byρ(g)X=gX g⁻¹(this is the adjoint representation ofGe). The symmetric bilinear form

ω(X,Y) =1 2tr(XY)

is obviously invariant underρ(G), since tr(XYe ) =tr(Y X)for allX,Y∈M_n(C). The basis

v₀= 1 0 0−1

, v₁=

0√ 2 0 0

, v₋₁=

√0 0 2 0

for V is ω isotropic. We identify V with C³ via the map v₁7→e₁,v₀7→e₂, and v₋₁7→e₃. Thenω becomesB. From Corollary 1.6.3 we know that any element of

2.2 Unipotent Elements 79

the subgroupN⁺ orN⁻in Lemma 2.2.1 is carried by the homomorphism ρ to a unipotent matrix. Hence by Lemma 2.2.1 we conclude that det(ρ(g)) =1 for all g∈G. Hencee ρ(G)e ⊂Gby Lemma 2.2.1. Ifh=diag[x,x⁻¹]∈H, thene ρ(h)has the matrix diag[x²,1,x⁻²], relative to the ordered basis {v₁,v₀,v₋₁} forV. Thus ρ(H) =e H.

Finally, we considerG=SO(C⁴,B), whereBis the symmetric bilinear form on C⁴with matrixs₄as in (2.5). From Section 2.1.2 we know that the subgroup

H={diag[x1,x2,x₂⁻¹,x⁻₁¹]:x1,x2∈C^×}

of diagonal matrices inGis a maximal torus. SetGe=SL(2,C)×SL(2,C)and let He be the product of the diagonal subgroups of the factors ofG. We now define ae homomorphismπ:Ge //Gthat mapsHeontoH, as follows. SetV=M2(C)and letGeact onV byπ(a,b)X=aX b⁻¹. From the quadratic formQ(X) =2 detX onV we obtain the symmetric bilinear form β(X,Y) =det(X+Y)−detX−detY. Set

v₁=e₁₁, v₂=e₁₂, v₃=−e₂₁,and v₄=e₂₂.

Clearlyβ(π(g)X,π(g)Y) =β(X,Y)forg∈G. The vectorse v_j areβ-isotropic and β(v₁,v₄) =β(v₂,v₃) =1. If we identifyVwithC⁴via the basis{v₁,v₂,v₃,v4}, then β becomes the formB.

Letg∈Gebe of the form(I,b)or(b,I), wherebis either in the subgroupN⁺or in the subgroupN⁻of Lemma 2.2.1. From Corollary 1.6.3 we know thatπ(g)is a unipotent matrix, and so from Lemma 2.2.1 we conclude that det(π(g)) =1 for all g∈G. Hencee π(G)e ⊂SO(C⁴,B). Givenh= (diag[x1,x⁻₁¹],diag[x2,x⁻₂¹])∈H, wee have

π(h) =diag[x1x⁻₂¹,x₁x₂,x⁻₁¹x⁻₂¹,x⁻₁¹x₂].

Since the map(x1,x2)7→(x1x⁻₂¹,x1x2)from(C^×)²to(C^×)²is surjective, we have shown thatπ(H) =e H.

2.2.2 Unipotent Generation of Classical Groups

The differential of a regular representation of an algebraic groupGgives a repre- sentation of Lie(G). On the nilpotent elements in Lie(G)the exponential map is algebraic and maps them to unipotent elements inG. This gives an algebraic link from Lie algebra representations to group representations, provided the unipotent elements generateG. We now prove that this is the case for the following families of classical groups.

Theorem 2.2.2.Suppse that G isSL(l+1,C),SO(2l+1,C), orSp(l,C)with l≥1, or that G isSO(2l,C)with l≥2. Then G is generated by its unipotent elements.

Proof. We haveG⊂GL(n,C)(wheren=l+1,2l, or 2l+1). LetG⁰be the sub- group generated by the unipotent elements ofG. Since the conjugate of a unipotent element is unipotent, we see thatG⁰ is a normal subgroup ofG. In the orthogonal or symplectic case we take the matrix form of Gas in Theorem 2.1.5 so that the subgroupHof diagonal matrices is a maximal torus inG. To prove the theorem, it suffices by Theorems 1.6.5 and 2.1.7 to show thatH⊂G⁰.

Type A:WhenG=SL(2,C), we haveG⁰=Gby Lemma 2.2.1. ForG=SL(n,C) withn≥3 andh=diag[x1, . . . ,x_n]∈Hwe factorh=h⁰h⁰⁰,where

h⁰=diag[x₁,x⁻₁¹,1, . . . ,1], h⁰⁰=diag[1,x₁x₂,x₃, . . . ,x_n].

LetG₁∼=SL(2,C)be the subgroup of matrices in block form diag[a,I_n−2] with a∈SL(2,C), and letG2∼=SL(n−1,C)be the subgroup of matrices in block form diag[1,b] withb∈SL(n−1,C). Thenh⁰∈G1andh⁰⁰∈G2. By induction onn, we may assume thath⁰andh⁰⁰are products of unipotent elements. Hencehis also, so we conclude thatG=G⁰.

Type C:LetΩ be the symplectic form (2.6). From Section 2.2.1 we know that Sp(C²,Ω) =SL(2,C). Hence from Lemma 2.2.1 we conclude that Sp(C²,Ω) is generated by its unipotent elements. For G=Sp(C^2l,Ω)withl >1 andh= diag[x₁, . . . ,x_l,x⁻_l ¹, . . . ,x⁻₁¹]∈H, we factorh=h⁰h⁰⁰, where

h⁰=diag[x₁,1, . . . ,1,x⁻₁¹], h⁰⁰=diag[1,x₂, . . . ,x_l,x⁻_l¹, . . . ,x⁻₂¹,1]. We splitC^2l=V₁⊕V₂, whereV₁=Span{e₁,e_2l}andV₂=Span{e₂, . . . ,e_2l−1}. The restrictions of the symplectic formΩ toV₁and toV₂are nondegenerate. Define

G1={g∈G:gV1=V1 and g=I on V2}, G₂={g∈G:g=I on V₁ and gV₂=V₂}.

ThenG₁∼=Sp(1,C), whileG₂∼=Sp(l−1,C), and we haveh⁰∈G₁andh⁰⁰∈G₂. By induction onl, we may assume thath⁰andh⁰⁰ are products of unipotent elements.

Hencehis also, so we conclude thatG=G⁰.

Types B and D: LetB be the symmetric form (2.9) on Cⁿ. Suppose first that G=SO(C³,B). LetGe=SL(2,C).In Section 2.2.1 we constructed a regular ho- momorphismρ:Ge //SO(C³,B)that maps the diagonal subgroupHe⊂Geonto the diagonal subgroupH⊂G. Since every element ofHe is a product of unipotent elements, the same is true forH. HenceG=SO(3,C)is generated by its unipotent elements.

Now letG=SO(C⁴,B)and setGe=SL(2,C)×SL(2,C). LetHbe the diagonal subgroup ofGand letHebe the product of the diagonal subgroups of the factors of G. In Section 2.2.1 we constructed a regular homomorphisme π:Ge //SO(C⁴,B) that maps He ontoH. Hence the argument just given forSO(3,C) applies in this case, and we conclude thatSO(4,C)is generated by its unipotent elements.

2.2 Unipotent Elements 81

Finally, we consider the groupsG=SO(Cⁿ,B)withn≥5. EmbedSO(C^2l,B) intoSO(C^2l+1,B)by the regular homomorphism

a b c d

7→



 a0b 0 1 0 c0d



. (2.14)

The diagonal subgroup of SO(C^2l,B) is isomorphic to the diagonal subgroup of SO(C^2l+1,B)via this embedding, so it suffices to prove that every diagonal element inSO(Cⁿ,B)is a product of unipotent elements whennis even. We just proved this to be the case whenn=4, so we may assumen=2l≥6. For

h=diag[x₁, . . . ,x_l,x⁻_l ¹, . . . ,x⁻₁¹]∈H we factorh=h⁰h⁰⁰,where

h⁰ =diag[x1,x2,1, . . . ,1,x⁻₂¹,x⁻₁¹], h⁰⁰ =diag[1,1,x3, . . . ,x_l,x⁻_l ¹, . . . ,x⁻₃¹,1,1]. We splitCⁿ=V1⊕V2, where

V₁=Span{e₁,e₂,e_n−1,en}, V₂=Span{e₃, . . . ,e_n−2}. The restriction of the symmetric formBtoV_iis nondegenerate. If we set

G₁={g∈G:gV₁=V₁andg=IonV₂},

thenh∈G₁∼=SO(4,C). LetW₁=Span{e₁,en}andW₂=Span{e₂, . . . ,e_n−1}. Set G₂={g∈G:g=IonW₁andgW₂=W₂}.

We have G₂∼=SO(2l−2,C)andh⁰⁰∈G₂. Since 2l−2≥4, we may assume by induction thath⁰andh⁰⁰are products of unipotent elements. Hencehis also a product

of unipotent elements, proving thatG=G⁰. ut

2.2.3 Connected Groups

Definition 2.2.3.A linear algebraic groupGisconnected(in the sense of algebraic groups) if the ringO[G]has no zero divisors.

Examples

1.The ringsC[t]andC[t,t⁻¹]obviously have no zero divisors; hence the additive group Cand the multiplicative group C^× are connected. Likewise, the torus D_n of diagonal matrices and the groupN_n⁺ of upper-triangular unipotent matrices are connected (see Examples 1 and 2 of Section 1.4.2).

2.IfGandHare connected linear algebraic groups, then the groupG×His con- nected, sinceO[G×H]∼=O[G]⊗O[H].

3.IfGis a connected linear algebraic group and there is a surjective regular homo- morphismρ:G //H, thenHis connected, sinceρ^∗:O[H] //O[G]is injective.

Theorem 2.2.4.Let G be a linear algebraic group that is generated by unipotent elements. Then G is connected as an algebraic group and as a Lie group.

Proof. Suppose f₁,f₂∈O[G], f₁6=0, and f₁f₂=0. We must show that f₂=0.

Translating f₁and f₂by an element ofGif necessary, we may assume that f₁(I)6= 0. Let g∈G. Sinceg is a product of unipotent elements, Theorem 1.6.2 implies that there exist nilpotent elementsX₁, . . . ,X_ringsuch thatg=exp(X₁)···exp(X_r).

Define ϕ(t) =exp(tX1)···exp(tX_r) fort∈C. The entries in the matrixϕ(t)are polynomials int, andϕ(1) =g. SinceXjis nilpotent, we have det(ϕ(t)) =1 for all t. Hence the functionsp1(t) = f1(ϕ(t))andp2(t) = f2(ϕ(t))are polynomials int.

Sincep₁(0)6=0 whilep₁(t)p2(t) =0 for allt, it follows thatp₂(t) =0 for allt. In particular, f₂(g) =0. This holds for allg∈G, so f₂=0, proving thatGis connected as a linear algebraic group. This argument also shows thatGis arcwise connected,

and hence connected, as a Lie group. ut

Theorem 2.2.5.The groupsGL(n,C),SL(n,C),SO(n,C), andSp(n,C)are con- nected (as linear algebraic groups and Lie groups) for all n≥1.

Proof. The homomorphismλ,g7→λgfromC^××SL(n,C)toGL(n,C)is surjec- tive. Hence the connectedness ofGL(n,C)will follow from the connectedness of C^×andSL(n,C), as in Examples 2 and 3 above. The groupsSL(1,C)andSO(1,C) are trivial, and we showed in Section 2.2.1 thatSO(2,C)is isomorphic toC^×, hence connected. For the remaining cases use Theorems 2.2.2 and 2.2.4. ut Remark 2.2.6.The regular homomorphisms ρ :SL(2,C) //SO(3,C) and π : SL(2,C)×SL(2,C) //SO(4,C)constructed in Section 2.2.1 have kernels±I;

hence dρ and dπ are bijective by dimensional considerations. Since SO(n,C) is connected, it follows that these homomorphisms are surjective. After we intro- duce the spin groups in Chapter 6, we will see that SL(2,C)∼=Spin(3,C) and SL(2,C)×SL(2,C)∼=Spin(4,C).

We shall study regular representations of a linear algebraic group in terms of the associated representations of its Lie algebra. The following theorem will be a basic tool.

Theorem 2.2.7.Suppose G is a linear algebraic group with Lie algebra g. Let (π,V)be a regular representation of G and W⊂V a subspace.

1. Ifπ(g)W⊂W for all g∈G thendπ(A)W ⊂W for all A∈g.

2. Assume that G is generated by unipotent elements. Ifdπ(X)W ⊂W for all X∈g thenπ(g)W ⊂W for all g∈G. Hence V is irreducible under the action of G if and only if it is irreducible under the action ofg.