Differential of a Rational Representation

1.5 Rational Representations

1.5.2 Differential of a Rational Representation

LetG⊂GL(n,C)be a linear algebraic group with Lie algebrag⊂gl(n,C). Let (π,V)be a rational representation ofG. ViewingGandGL(V)as Lie groups, we can apply Proposition 1.3.14 to obtain a homomorphism (of real Lie algebras)

dπ:g //gl(V).

We call dπ thedifferentialof the representationπ. Sincegis a Lie algebra overC in this case, we haveπ(exp(tA)) =exp(dπ(tA))forA∈gandt∈C. The entries in the matrixπ(g)(relative to any basis forV) are regular functions onG, so it follows thatt7→π(exp(tA))is an analytic (matrix-valued) function of the complex variable t. Thus

dπ(A) = d

dtπ(exp(tA)) _t=0

,

and the mapA7→dπ(A)is complex linear. Thus dπis a homomorphism of complex Lie algebras whenGis a linear algebraic group.

This definition of dπ has made use of the exponential map and the Lie group structure on G. We can also define dπ in a purely algebraic fashion, as follows:

View the elements ofgas left-invariant vector fields onGby the mapA7→X_Aand differentiate the entries in the matrix forπusingXA. To express this in a basis-free way, recall that every linear transformationB∈End(V)defines a linear function fC

on End(V)by

f_C(B) =tr_V(BC) forB∈End(V).

The representative function f_C^π=f_C◦π onGis then a regular function.

Theorem 1.5.2.The differential of a rational representation (π,V)is the unique linear mapdπ:g //End(V)such that

X_A(f_C◦π)(I) =f_dπ(A)C(I) for all A∈gand C∈End(V). (1.45) Furthermore, for A∈Lie(G), one has

X_A(f◦π) = (X_dπ(A)f)◦π for all f ∈O[GL(V)]. (1.46) Proof. For fixedA∈g, the mapC7→X_A(f_C◦π)(I)is a linear functional on End(V).

Hence there exists a uniqueD∈End(V)such that

X_A(f_C◦π)(I) =tr_V(DC) =f_DC(I).

But f_DC=X_Df_C by equation (1.36). Hence to show that dπ(A) =D, it suffices to prove that equation (1.46) holds. Let f ∈O[GL(V)]andg∈G. Then

XA(f◦π)(g) = d

dtf π(gexp(tA)) _t=0

= d

dtf π(g)exp(tdπ(A)) _t=0

= (X_dπ(A)f)(π(g))

by definition of the vector fieldsXAonGandX_dπ(A)onGL(V). ut Remark 1.5.3.An algebraic-group proof of Theorem 1.5.2 and the property that dπ is a Lie algebra homomorphism (taking (1.45) as the definition of dπ(A)) is outlined in Exercises 1.5.4.

LetGandH be linear algebraic groups with Lie algebrasgandh, respectively, and letπ:G //Hbe a regular homomorphism. IfH⊂GL(V), then we may view (π,V)as a regular representation ofGwith differential dπ:g //End(V).

Proposition 1.5.4.The range ofdπ is contained inh. Hence dπ is a Lie algebra homomorphism fromgtoh. Furthermore, if K⊂GL(W)is a linear algebraic group andρ:H //K is a regular homomorphism, thend(ρ◦π) =dρ◦dπ. In particular, if G=K andρ◦π is the identity map, then dρ◦dπ is the identity map. Hence isomorphic linear algebraic groups have isomorphic Lie algebras.

Proof. We first verify that dπ(A)∈hfor allA∈g. Let f ∈IHandh∈H. Then

1.5 Rational Representations 51 (X_dπ(A)f)(h) =L(h⁻¹)(X_dπ(A)f)(I) =X_dπ(A)(L(h⁻¹)f)(I)

=X_A((L(h⁻¹)f)◦π)(I)

by (1.46). ButL(h⁻¹)f∈IH, so(L(h⁻¹)f)◦π=0, sinceπ(G)⊂H. Hence we have X_dπ(A)f(h) =0 for allh∈H. This shows that dπ(A)∈h.

Given regular homomorphisms

G−→^π H−→^ρ K,

we setσ=ρ◦πand takeA∈gandf ∈O[K]. Then by (1.46) we have (Xdσ(A)f)◦σ=X_A((f◦ρ)◦π) = (Xdπ(A)(f◦ρ))◦π= (Xdρ(dπ(A))f)◦σ. Taking f=f_CforC∈End(W)and evaluating the functions in this equation atI, we

conclude from (1.45) that dσ(A) =dρ(dπ(A)). ut

Corollary 1.5.5.Suppose G⊂H are algebraic subgroups ofGL(n,C). If(π,V)is a regular representation of H, then the differential of π|^G is dπ|^g.

Examples

1.LetG⊂GL(V)be a linear algebraic group. By definition ofO[G], the represen- tationπ(g) =gonV is regular. We call(π,V)thedefiningrepresentation ofG. It follows directly from the definition that dπ(A) =AforA∈g.

2.Let(π,V)be a regular representation. Define thecontragredient (ordual) repre- sentation(π^∗,V^∗)byπ^∗(g)v^∗=v^∗◦π(g⁻¹). Thenπ^∗is obviously regular, since

hv^∗,π(g)vi=hπ^∗(g⁻¹)v^∗,vi forv∈V andv^∗∈V^∗.

If dimV =d (thedegreeofπ) andV is identified withd×1 column vectors by a choice of basis, thenV^∗is identified with 1×drow vectors. Viewingπ(g)as ad×d matrix using the basis, we have

hv^∗,π(g)vi=v^∗π(g)v (matrix multiplication).

Thusπ^∗(g)acts by right multiplication on row vectors by the matrixπ(g⁻¹).

The space of representative functions forπ^∗consists of the functionsg7→f(g⁻¹), where f is a representative function forπ. IfW ⊂V is aG-invariant subspace, then

W^⊥={v^∗∈V^∗:hv^∗,wi=0 for allw∈W}

is a G-invariant subspace of V^∗. In particular, if (π,V) is irreducible then so is (π^∗,V^∗). The natural vector-space isomorphism (V^∗)^∗∼=V gives an equivalence (π^∗)^∗∼=π.

To calculate the differential ofπ^∗, letA∈g,v∈V, andv^∗∈V^∗. Then

hdπ^∗(A)v^∗,vi= d

dthπ^∗(exptA)v^∗,vi _t=0= d

dthv^∗,π(exp(−tA))vi _t=0

=−hv^∗,dπ(A)vi.

Since this holds for allvandv^∗, we conclude that

dπ^∗(A) =−dπ(A)^t forA∈g. (1.47) Caution: The notationπ^∗(g)for the contragredient representation should not be confused with the notationB^∗for the conjugate transpose of a matrixB. The pairing hv^∗,vibetweenV^∗andV iscomplex linearin each argument.

3.Let(π1,V1)and(π2,V2)be regular representations ofG. Define thedirect sum representationπ1⊕π2onV₁⊕V₂by

(π₁⊕π₂)(g)(v₁⊕v₂) =π₁(g)v₁⊕π₂(g)v₂ forg∈Gandv_i∈V_i. Thenπ₁⊕π₂is obviously a representation ofG. It is regular, since

hv^∗₁⊕v^∗₂,(π1⊕π2)(g)(v1⊕v2)i=hv^∗₁,π1(g)v1i+hv^∗₂,π2(g)v2i

for vi∈Vi andv^∗_i ∈V_i^∗. This shows that the space of representative functions for π1⊕π2 is E^π1⊕π2=E^π1+E^π2. If π=π1⊕π2, then in matrix form we have

π(g) =

π1(g) 0 0 π2(g)

.

Differentiating the matrix entries, we find that dπ(A) =

dπ₁(A) 0 0 dπ₂(A)

forA∈g. Thus dπ(A) =dπ₁(A)⊕dπ₂(A).

4.Let(π1,V1)and(π2,V2)be regular representations ofG. Define thetensor product representationπ1⊗π2onV₁⊗V₂by

(π1⊗π2)(g)(v1⊗v₂) =π1(g)v1⊗π2(g)v2

forg∈Gandv_i∈V. It is clear thatπ1⊗π2is a representation. It is regular, since hv^∗₁⊗v^∗₂,(π1⊗π2)(g)(v1⊗v₂)i=hv^∗₁,π1(g)vihv^∗₂,π2(g)v2i

forv_i∈Vandv^∗_i ∈V_i^∗. In terms of representative functions, we have E^π1⊗π2=Span(E^π1·E^π2)

(the sums of products of representative functions ofπ1 andπ2). Setπ =π1⊗π2. Then

1.5 Rational Representations 53

dπ(A) = d dt

n

exp tdπ₁(A)

⊗exp tdπ₂(A)o _t=0

=dπ₁(A)⊗I+I⊗dπ₂(A). (1.48)

5.Let(π,V)be a regular representation ofGand setρ=π⊗π^∗onV⊗V^∗. Then by Examples 2 and 4 we see that

dρ(A) =dπ(A)⊗I−I⊗dπ(A)^t. (1.49) However, there is the canonical isomorphismT:V⊗V^∗∼=End(V), with

T(v⊗v^∗)(u) =hv^∗,uiv.

Setσ(g) =Tρ(g)T⁻¹. IfY ∈End(V)thenT(Y⊗I) =Y T andT(I⊗Y^t) =TY. Henceσ(g)(Y) =π(g)Yπ(g)⁻¹and

dσ(A)(Y) =dπ(A)Y−Ydπ(A) forA∈g. (1.50) 6.Let(π,V)be a regular representation ofGand setρ=π^∗⊗π^∗onV^∗⊗V^∗. Then by Examples 2 and 4 we see that

dρ(A) =−dπ(A)^t⊗I−I⊗dπ(A)^t.

However, there is a canonical isomorphism betweenV^∗⊗V^∗and the space of bilin- ear forms onV, whereg∈Gacts on a bilinear formBby

g·B(x,y) =B(π(g⁻¹)x,π(g⁻¹)y).

IfVis identified with column vectors by a choice of a basis andB(x,y) =y^tΓx, then g·Γ=π(g⁻¹)^tΓ π(g⁻¹) (matrix multiplication) . The action ofA∈gonBis

A·B(x,y) =−B(dπ(A)x,y)−B(x,dπ(A)y).

We say that a bilinear formBisinvariantunderGifg·B=Bfor allg∈G. Likewise, we say thatBisinvariantundergifA·B=0 for allA∈g. This invariance property can be expressed as

B(dπ(A)x,y) +B(x,dπ(A)y) =0 for allx,y∈V andA∈g.