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arXiv:math/0510377v1 [math.RT] 18 Oct 2005

COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS

GEORGE J. MCNINCH AND DONNA M. TESTERMAN

Abstract. Let K be any field, and let G be a semisimple group over K. Suppose the

characteristic of K is positive and is very good for G.

homomorphisms φ : SL2 → G whose image is geometrically G-completely reducible – or

G-cr – in the sense of Serre; the description resembles that of irreducible modules given by

Steinberg’s tensor product theorem. In case K is algebraically closed and G is simple, the

result proved here was previously obtained by Liebeck and Seitz using different methods.

A recent result shows the Lie algebra of the image of φ to be geometrically G-cr; this plays

an important role in our proof.

We describe all group scheme

1. Introduction

Let K be an arbitrary field of characteristic p > 0.

rated K-scheme of finite type. An algebraic group will mean a smooth and affine K-group

scheme; a subgroup will mean a K-subgroup scheme, and a homomorphism will mean a K-

homomorphism. A smooth group scheme G is said to be reductive if G/Kalg is reductive in

the usual sense – i.e. it has trivial unipotent radical – where Kalgis an algebraic closure of

K. The Lie algebra g = Lie(G) may be regarded as a scheme over K; we permit ourselves to

write g for the set of K-points g(K).

For G a reductive group, a subgroup H ⊂ G is said to be geometrically G-completely

reducible – or G-cr – if whenever k is an algebraically closed field containing K and H/kis

contained in a parabolic k-subgroup P of G/k, then H/k⊂ L for some Levi k-subgroup L of

P; see §2.3 for more details. The notion of G-cr was introduced by J-P. Serre; see e.g. [Ser

05] for more on this notion. It is our goal here to describe all homomorphisms φ : SL2→ G

whose image is geometrically G-cr; this we achieve under some assumptions on G which are

described in §2.4. For the purposes of this introduction, let us suppose that G is semisimple.

Then our assumption is: the characteristic of K is very good for G (again see §2.4 for the

precise definition of a very good prime).

Let F : SL2 → SL2 be the Frobenius endomorphism obtained by base change from the

Frobenius endomorphism of SL2/Fp; cf. §2.8 below. We say that a collection of homomor-

phisms φ0,φ1,...,φr: SL2→ G is commuting if

By a scheme we mean a sepa-

imφi⊂ CG(imφj)for all 0 ≤ i ?= j ≤ r.

Let?φ = (φ0,...,φr) where the φi are commuting homomorphisms SL2 → G, and let ? n =

(n0 < ··· < nr) where the ni are non-negative integers. Then the data (?φ,? n) determines

a homomorphism Φ?φ,? n: SL2 → G given for every commutative K-algebra Λ and every

g ∈ SL2(Λ) by the rule

g ?→ φ0(Fn0(g)) · φ1(Fn1(g))···φr(Fnr(g)).

Date: October 18, 2005.

Research of McNinch supported in part by the US National Science Foundation through DMS-0437482.

Research of Testerman supported in part by the Swiss National Science Foundation grant PP002-68710.

1

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2GEORGE J. MCNINCH AND DONNA M. TESTERMAN

We say that Φ = Φ?φ,? nis the twisted-product homomorphism determined by (?φ,? n).

A notion of optimal homomorphisms SL2→ G was introduced in [Mc 05]; see §2.7 for the

precise definition. When G is a K-form of GL(V ) or SL(V ), a homomorphism f : SL2→ G is

optimal just in case the representation (f/Ksep,V ) is restricted and semisimple, where Ksepis a

separable closure of K; see Remark 18. We will say that the list of commuting homomorphisms

?φ = (φ0,φ1,...,φr) is optimal if each φiis an optimal homomorphism.

Theorem 1. Let G be a semisimple group for which the characteristic is very good, and let

Φ : SL2→ G be a homomorphism. If the image of Φ is geometrically G-cr, then there are

commuting optimal homomorphisms?φ = (φ0,...,φr) and non-negative integers ? n = (n0 <

n1 < ··· < nd) such that Φ is the twisted-product homomorphism determined by (?φ,? n).

Moreover,?φ and ? n are uniquely determined by Φ.

We actually prove the theorem for the strongly standard reductive groups described below

in 2.4; see Theorem 39.

In case K is algebraically closed and G is simple, this theorem was obtained by Liebeck

and Seitz [LS 03, Theorem 1]; cf. Remark 17 to see that the notion of restricted – or good –

A1-subgroup used in [LS 03] is“the same”as the notion of optimal homomorphism used here.

Note that Liebeck and Seitz prove a version of Theorem 1 where SL2is replaced by any

quasisimple group H. If G is a split classical group over K in good characteristic, the more

general form of Theorem 1 found in [LS 03] is a consequence of Steinberg’s tensor product

theorem [Jan 87, Cor. II.3.17]; cf. [LS 03, Lemma 4.1]. The proof given by Liebeck and

Seitz of Theorem 1 for a quasisimple group G of exceptional type relies instead on detailed

knowledge of the subgroup structure – in particular, of the maximal subgroups – of G; see

e.g. [LS 03, Theorem 2.1, Proposition 2.2, and §4.1] for the case H = SL2. In contrast, when

p > 2, our proof uses in an essential way the complete reducibility of the Lie algebra of a G-cr

subgroup of G [Mc 05a]; cf. the proofs of Lemma 24, Proposition 25, and Lemma 29 [when

p = 2, we have essentially just used the proof of Liebeck and Seitz].

We obtain also the converse to Theorem 1, though we do so only under a restriction on

p. Write h(G) for the maximum value of the Coxeter number of a simple k-quotient of G/k,

where k is an algebraically closed field containing K.

Theorem 2. Let G be semisimple in very good characteristic, and suppose that p > 2h(G)−2,

let?φ = (φ0,...,φd) be commuting optimal homomorphisms SL2→ G, and let ? n = (n0< n1<

··· < nd) be non-negative integers. Then the image of the twisted-product homomorphism

Φ : SL2→ G determined by (?φ,? n) is geometrically G-cr.

Again, this result is proved for a more general class of reductive groups; see Theorem 43.

The assumption on p made in the last theorem is unnecessary if G is a classical group – or

a group of type G2– in good characteristic; see Remark 44. However, it is not clear to the

authors how to eliminate the prime restriction in general.

The first named author would like to acknowledge the hospitality of the Centre Interfacul-

taire Bernoulli at the´Ecole Polytechnique F´ ed´ erale de Lausanne during a visit in June 2005;

this visit permitted much of the collaboration which led to the present manuscript.

2. Preliminaries

2.1. Reduced subgroups. Let k be a perfect field – in the application we take k to be

algebraically closed. Let B be a group scheme of finite type over k.

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COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS3

Lemma 3. There is a unique smooth subgroup Bred ⊂ B which has the same underlying

topological space as B. If A is any smooth group scheme over k and f : A → B is a k-

morphism, then f factors in a unique way into a k-morphism A → Bred followed by the

inclusion Bred→ B.

Proof. Use [Li 02, Prop. 2.4.2] to find the reduced k-scheme Bredwith the same underlying

topological space as B; the result just quoted then yields the uniqueness of Bred. It is clear

that Bredis a k-group scheme, and the assertion about A and f follows from loc. cit. Prop

2.4.2(d). Since k is perfect, apply [KMRT, Prop. 21.9] to see that a k-group is smooth if and

only if it is geometrically reduced if and only if it is reduced. Thus Bredis indeed smooth

.

?

We are going to consider later some group schemes which we do not a priori know to be

smooth, and we want to choose maximal tori in these group schemes. The following example

explains why in those cases we first extend scalars to an algebraically closed field (see e.g.

§3.2 below).

Example 4. If B is a group scheme over an imperfect field K, and if k is a perfect field

containing K, then a maximal torus of B/k,redneed not arise by base-change from a K-

subgroup of B. Let us give an example.

Let A = Gm⋉ Ga where Gm acts on Ga“with weight one”; i.e. K[A] = K[T,T−1,U]

where the comultiplication µ∗is given by

µ∗(T±1) = (T ⊗ T)±1

and

Suppose that K is not perfect, and let L = K(β) where βp= α ∈ K but β ?∈ K. Consider

the subgroup scheme B ⊂ A defined by the ideal I = (αTp− Up− α) ⊳ K[A].

If k is a perfect field containing K, notice that the image¯f ∈ k[B] of f = βT −U−β ∈ k[A]

satisfies¯fp= 0 but¯f ?= 0; thus B/kis not reduced. The subgroup B/k,red⊂ A/kis defined by

J = (βT −U−β), so that B/k,red≃ Gm/kis a torus. The group of k-points B/k,red(k) ⊂ A(k)

may be described as:

B/k,red(k) = {(t,βt − β) ∈ Gm(k) ⋉ Ga(k) | t ∈ k×}.

µ∗(U) = U ⊗ T + 1 ⊗ U.

Note that B/k,reddoes not arise by base change from a K-subgroup of A, e.g. since the

intersection B/k,red(k) ∩ A(K) consists only in the identity element [where the intersection

takes place in the group A(k)].

2.2. Cocharacters and parabolic subgroups. A cocharacter of an algebraic group A is a

homomorphism γ : Gm→ A. We write X∗(A) for the set of cocharacters of A.

A linear representation (ρ,V ) of A yields a linear representation (ρ◦γ,V ) of Gmwhich in

turn is determined by the morphism

(ρ ◦ γ)∗: V → K[Gm] ⊗KV = K[t,t−1] ⊗KV.

Then V is the direct sum of the weight spaces

(2.2.1)V (γ;i) = {v ∈ V | (ρ ◦ γ)∗v = ti⊗ v}

for i ∈ Z.

Consider now the reductive group G. If γ ∈ X∗(G), then

PG(γ) = P(γ) = {x ∈ G | lim

t→0γ(t)xγ(t−1) exists}

is a parabolic subgroup of G whose Lie algebra is p(γ) =?

for the notion of limit used here. Moreover, each parabolic subgroup of G has the form P(γ)

for some cocharacter γ; for all this cf. [Spr 98, 3.2.15 and 8.4.5].

i≥0g(γ;i); see e.g. [Spr 98, §3.2]

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4GEORGE J. MCNINCH AND DONNA M. TESTERMAN

We note that γ “exhibits” a Levi decomposition of P = P(γ). Indeed, P(γ) is the semi-

direct product CG(γ)·U(γ), where U(γ) = {x ∈ P | limt→0γ(t)xγ(t−1) = 1} is the unipotent

radical of P(γ), and the reductive subgroup CG(γ) = CG(γ(Gm)) is a Levi factor in P(γ); cf.

[Spr 98, 13.4.2].

2.3. Complete reducibility, Lie algebras. Let G be a reductive group, and write g for its

Lie algebra.

A smooth subgroup H ⊂ G is geometrically G-cr if whenever k is an algebraically closed

field containing K and H/k⊂ P for a parabolic k-subgroup P ⊂ G/k, then H/k⊂ L for some

Levi k-subgroup L ⊂ P.

Similarly, if h ⊂ g is a Lie subalgebra, we say that h is geometrically G-cr if whenever

k is an algebraically closed field containing K and P ⊂ G/kis a parabolic k-subgroup with

h/k= h ⊗Kk ⊂ Lie(P), then h/k⊂ Lie(L) for some Levi k-subgroup L ⊂ P.

Lemma 5. Let X and Y be schemes of finite type over K, and let f : X → Y be a (K-)

morphism. The following are equivalent:

i) f is surjective,

ii) f/k: X(k) → Y (k) is surjective for all algebraically closed fields k containing K, and

iii) f/k: X(k) → Y (k) is surjective for some algebraically closed field k containing K.

Proof. This follows from [DG70, I §3.6.10]

?

Lemma 6. Fix an algebraically closed field k containing K. Let G be a reductive group, let

J ⊂ G be a smooth subgroup, and let h ⊂ g be a Lie subalgebra. Then

(1) J is geometrically G-cr if and only if J/kis G/k-cr.

(2) h is geometrically G-cr if and only if h/kis G/k-cr.

Proof. We prove (1); the proof of (2) is essentially the same. We are going to apply the

previous Lemma.

First let P be the scheme of all parabolic subgroups of G, and let Y = PJbe the fixed

point scheme for the action of J; thus Y is the closed subscheme of those parabolic subgroups

containing J.1

Let also PL be the scheme such that for each commutative K-algebra Λ, the Λ-points

PL(Λ) are the pairs P ⊃ L where P is a parabolic of G/Λand L is a Levi subgroup of P;

cf. [SGA3, Exp. XXVI, §3.15]. Let X = (PL)Jbe the scheme of those pairs P ⊃ L where L

contains J.

There is an evident morphism PL → P given by (P ⊃ L) ?→ P; cf. [SGA3, Exp. XXVI,

§3.15]. By restriction one gets a morphism f : X → Y . Then f is surjective if and only if J

is G-cr, and (1) follows from the preceding Lemma.

?

Proposition 7. Let G be reductive, and let M ⊂ G be a Levi subgroup. Suppose that J ⊂ M

is a smooth subgroup, and that h ⊂ Lie(M) is a Lie subalgebra. Then J is geometrically

G-cr if and only if J is geometrically M-cr and h is geometrically G-cr if and only if h is

geometrically M-cr.

1For assertion (2), one should instead regard P as the scheme of parabolic subalgebras of g, which may be

regarded as a closed subscheme of a product of Grassman schemes Grd(g) for various d. Now the subscheme

X ⊂ P of parabolic subalgebras containing h is the intersection of P with the subscheme Z of the product of

Grassman schemes consisting of those subspaces containing h. Since Z is closed in the product, Y is closed in

P. Similar remarks apply to the definition of the subscheme Y ⊂ PL to be given in the next paragraph.

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COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS5

Proof. For the proof, it is enough to suppose that K is algebraically closed. The proof for J

is found in [BMR 05, Theorem 3.10]. The proof for h is deduced from [Ser 05, 2.1.8]; see [Mc

05a, Lemma 2] for the argument.

?

The following theorem was proved in [Mc 05a].

Theorem 8. Let H ⊂ G be a smooth subgroup which is geometrically G-cr. Then h = Lie(H)

is geometrically G-cr.

We recall a similar result of B. Martin [Ser 05, Th´ eor` eme 3.6].

Theorem 9 (Martin). Let H ⊂ G be a smooth subgroup which is geometrically G-cr, and let

H′⊳ H be a smooth normal subgroup. Then H′is geometrically G-cr as well.

Finally, we note:

Lemma 10. Let π : G → G1 be a central isogeny where G1 is a second reductive group, let

J ⊂ G be a smooth subgroup, and let h ⊂ g = Lie(G) be a Lie subalgebra. Then

(1) J is geometrically G-cr if and only if π(J) is geometrically G1-cr, and

(2) h is geometrically G-cr if and only if dπ(h) is geometrically G1-cr.

Proof. We may and will suppose that K is algebraically closed for the proof. It is clear that

J is contained in a parabolic subgroup P of G if and only if π(J) is contained in the parabolic

subgroup π(P) of G1, and similarly h is contained in Lie(P) if and only if dπ(h) is contained

in dπ(Lie(P)) = Lie(π(P)), the result follows since P ?→ π(P) determines a bijection between

the parabolic subgroups of G and those of G1.

?

2.4. Strongly standard reductive groups. If G is geometrically quasisimple with absolute

root system R2, the characteristic p of K is said to be a bad prime for R in the following

circumstances: p = 2 is bad whenever R ?= Ar, p = 3 is bad if R = G2,F4,Er, and p = 5 is

bad if R = E8. Otherwise, p is good. [Here is a more intrinsic definition of good prime: p is

good just in case it divides no coefficient of the highest root in R].

If p is good, then p is said to be very good provided that either R is not of type Ar, or

that R = Arand r ?≡ −1 (mod p).

There is a possibly inseparable central isogeny3

r?

i=1

for some torus T and some r ≥ 1, where for 1 ≤ i ≤ r there is an isomorphism Gi≃ RLi/KHi

for a finite separable field extension Li/K and a geometrically simple, simply connected Li-

group scheme Hi; here, RLi/KHidenotes the “Weil restriction”of Hito K.

Then p is good, respectively very good, for G if and only if that is so for Hi for every

1 ≤ i ≤ r. Since the Hi are uniquely determined by G up to central isogeny, the notions

of good and very good primes depend only on the central isogeny class of the derived group

(G,G). Moreover, these notions are geometric in the sense that they depend only on the

group G/kfor an algebraically closed field k containing K.

(2.4.1)

Gi× T → G

2The absolute root system of G is the root system of G/Ksep where Ksepis a separable closure of K.

3Indeed, the center of the reductive group G is a smooth subgroup scheme; this follows e.g. from [SGA3, II

Exp. XII Th´ eor` eme 4.1] since for reductive G, the center is the same as the “centre r´ eductif”. The radical

R(G) is the maximal torus of the center of G, so R(G) is a smooth torus, and we take T = R(G) in (2.4.1).

Now, multiplication gives a central isogeny G′× R(G) → G where G′is the derived group of G. So (2.4.1)

follows from the corresponding result for semisimple groups; see e.g. [KMRT, Theorems 26.7 and 26.8] or

[TW 02, Appendix (42.2.7)].

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6GEORGE J. MCNINCH AND DONNA M. TESTERMAN

One says that a smooth K-group D is of multiplicative type if D/K′ is diagonalizable for

some algebraic extension K ⊂ K′; i.e. that D/K′ ≃ Diag(Γ) for some commutative group Γ.

See [Jan 87, I.2.5] for the definition of Diag(Γ) – it is implicitly defined in [Spr 98, Corollary

3.2.4] as well. A torus is of multiplicative type, as is any finite, smooth, commutative subgroup

all of whose geometric points are semisimple.

If G is reductive and if D ⊂ G is a subgroup of multiplicative type, then CG(D) is a

reductive subgroup containing a maximal torus of G – use [SGA3, II Exp. XI, Cor 5.3] to see

that CG(D) is smooth, use [Spr 98, Theorem 3.2.3] to see that DK′ lies in a maximal torus

of GK′, and finally use [SS 70, II. §4.1] to see that CG(D) is reductive.

Consider reductive groups of the form

(∗)H = H1× S

where H1is a semisimple group for which the characteristic of K is very good, and where S

is a torus. We say that G is strongly standard if there is a group H as in (∗), a subgroup of

multiplicative type D ⊂ H, and a separable isogeny between G and the reductive subgroup

CH(D) of H.

Remark 11. This definition of strongly standard is more general than that given e.g. in [Mc

05]. It follows from Proposition 12 below that the main result of loc. cit. in fact applies to a

strongly standard group in this stronger sense.

Proposition 12. Let G be strongly standard.

(1) If D ⊂ G is a subgroup of multiplicative type, then the reductive group CG(D) is

strongly standard.

(2) The characteristic of K is good for the derived group of G, and there is a non-

degenerate, G-invariant bilinear form on Lie(G).

(3) Each conjugacy class and each adjoint orbit is separable. In particular, if g ∈ G(K)

and X ∈ g(K), then CG(g) and CG(X) are smooth.4

Remark 13. The centralizers considered in (3) – and elsewhere in this paper – are the scheme-

theoretic centralizers. Thus e.g. CG(X) is the group scheme with Λ-points CG(X)(Λ) = {g ∈

G(Λ) | Ad(g)X = X} for each commutative K-algebra Λ.

Proof of Proposition 12. For the proofs of (1) and (2), we may replace G by a separably

isogenous group and suppose G to be the centralizer of a subgroup of multiplicative type

D1⊂ H where H has the form (∗).

For (1), note that since D1centralizes D, the group D2= D · D1⊂ H is of multiplicative

type and (1) is immediate.

To prove (2), note first that the characteristic of K is good for the derived group of G by

[Mc 05, Lemma 1](2). Now, there is a non-degenerate H-invariant bilinear form β on a group

H of the form (∗) by [Mc 05, Lemma 1](1). Moreover, it suffices to see that the restriction

of β to Lie(G) is nondegenerate after making a field extension; thus, we may suppose that

D1≃ Diag(Γ). We have Lie(H) =?

e.g. [Jan 87, §I.2.11]. The subspaces Lie(H)γ and Lie(H)τ are evidently orthogonal unless

γ · τ = 1 in Γ5. Since Lie(G) = Lie(H)D1= Lie(H)1, the restriction of β to Lie(G) must

remain non-degenerate.

In view of (2), the proof of [Mc 05, Prop. 5] yields (3).

γ∈ΓLie(H)γ where D1acts on Lie(H)γ through γ; see

?

4In older language, these centralizers are defined over K.

5We are writing Γ multiplicatively

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COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS7

2.5. Nilpotent elements and associated cocharacters. Let G be a reductive group, and

let X ∈ g = Lie(G) be nilpotent. A cocharacter Ψ ∈ X∗(G) is said to be associated with X

if the following conditions hold:

(A1) X ∈ g(Ψ;2), and

(A2) there is a maximal torus S of CG(X) such that Ψ ∈ X∗(L1) where L = CG(S) and

L1= (L,L) is its derived group.

Assume now that G is strongly standard.

Proposition 14. Let X ∈ g be nilpotent.

(1) There is a cocharacter Ψ associated with X.

(2) If Ψ is associated to X and P = P(Ψ) is the parabolic determined by Ψ, then CG(X) ⊂

P. In particular, cg(X) ⊂ Lie(P).

(3) If Ψ,Φ ∈ X∗(G) are associated with X, then Ψ = Int(x)◦Φ for some x ∈ CG(X)(K).

(4) The parabolic subgroups P(Ψ) for cocharacters Ψ associated with X all coincide.

Proof. (1) is [Mc 04, Theorem 26], (2) is [Ja 04, Prop. 5.9]. (3) follows from [Mc 05, Prop/Defn

21(4)], and (4) is [Mc 05, Prop/Defn 21(5)].

?

Let Ψ be a cocharacter associated with X as in the previous Proposition. Then the para-

bolic subgroup P(X) = P(Ψ) of (4) is known as the the instability parabolic of X.

Let X ∈ g, and let [X] ∈ P(g)(K) be the K-point which “is” the line determined by X in

the corresponding projective space.

Proposition 15. Write NG(X) = StabG([X]).

(1) NG(X) is a smooth subgroup of G.

(2) For each maximal torus T of NG(X), there is a unique cocharacter λ ∈ X∗(T) asso-

ciated to X.

Proof. Recall that NG(X) is the scheme-theoretic stabilizer of the point [X] ∈ P(g)(K). (1)

follows from [Mc 04, Lemma 23], and in view of Proposition/Definition 14(1), assertion (2)

follows from [Mc 04, Lemma 25].

?

2.6. Notation for SL2. We fix here some convenient notation for SL2. We first choose the

“standard”basis for the Lie algebra sl2:

?0

0010

E =

1

?

,F =

?00

?

,and[E,F] =

?10

0−1

?

.

Now consider the homomorphisms e,f : Ga→ SL2given for each commutative K-algebra Λ

and each t ∈ Ga(Λ) = Λ by the rules

?1

01

e(t) =

t

?

andf(t) =

?10

1t

?

.

Finally, write T for the “diagonal”maximal torus of SL2; we fix the cocharacter

?t

0

(t ?→

0

t−1

?

) : Gm→ T

and use this cocharacter to identify T with Gm.

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8GEORGE J. MCNINCH AND DONNA M. TESTERMAN

2.7. Optimal homomorphisms. We will use without mention the notation of §2.6. Let G

be a reductive group. We say that a homomorphism φ : SL2→ G is optimal for X = dφ(E),

or simply that φ is an optimal SL2-homomorphism, if λ = φ|T is a cocharacter associated

with X.

Theorem 16 ([Mc 05]). Suppose that G is strongly standard. Let X ∈ g satisfy X[p]= 0,

and let λ ∈ X∗(G) be associated with X. There is a unique homomorphism

φ : SL2→ G

such that dφ(E) = X and φ|T= λ. Moreover, the image of φ is geometrically G-cr.

Proof. In view of Proposition 12(3), this follows from [Mc 05, Theorem 47 and Prop. 52].

?

Remark 17. Seitz has introduced a notion of “good A1-subgroup” of a quasisimple group in

[Sei 00]; in [LS 03], these subgroups are called “restricted”. Refer to [Mc 05, §8.5] to see that

a subgroup of type A1 of a quasisimple group G is restricted if and only if it is the image

of an optimal homomorphism SL2→ G. It is not hard to see that the image of an optimal

homomorphism is restricted; cf. [Mc 05, Prop. 30]. The proof that a restricted A1-subgroup

is the image of an optimal homomorphism is more involved.

Remark 18. If V is a finite dimensional vector space, a homomorphism φ : SL2→ SL(V ) is

optimal if and only if V is a restricted semisimple SL2-module. Indeed, if V is restricted and

semisimple, one sees at once that φ|T is associated with dφ(E) so that φ is indeed optimal.

On the other hand, if λ = φ|T is associated to X = dφ(E), then the the character of the

SL2-module V is determined by the cocharacter λ; it follows that the composition factors of

V as SL2-module are restricted. If 0 ≤ n < p, write L(n) for the restricted simple SL2-module

of highest weight n [Jan 87, §II.2]. The linkage principle [Jan 87, Corollary II.6.17] implies

that Ext1

SL2(L(n),L(m)) = 0 whenever 0 ≤ n,m < p. Thus, V is semisimple as well.

Proposition 19. Let S be the image of the optimal SL2-homomorphism φ, and let λ = φ|T∈

X∗(G). Write X = dφ(E) and Y = dφ(F). Then:

(1) CG(imdφ) = CG(S).

(2) The scheme-theoretic intersection CG(S) = CG(X) ∩ CG(λ) is a smooth subgroup of

G.

Proof. For (1), recall that if ε : Ga→ G is given by ε = φ ◦ e, then CG(X) = CG(imε) by

[Mc 05, Prop. 35]. Similarly, CG(Y ) = CG(imφ◦f). Since imdφ is spanned by X and Y and

since S is generated as a group scheme by the image of φ ◦ e and the image of φ ◦ f, we have

CG(imdφ) = CG(X) ∩ CG(Y ) = CG(imφ ◦ e) ∩ CG(imφ ◦ f) = CG(S).

For (2), the inclusion CG(S) ⊂ CG(X)∩CG(λ) is clear. To prove the other inclusion, let Λ

be a commutative K-algebra, and let g ∈ G(Λ) be such that Ad(g)X = X and Int(g)◦λ = λ.

By (1), it is enough to show that g centralizes Y . Since Y ∈ g(λ;−2)(K) and since Int(g)◦λ =

λ, we have Ad(g)Y ∈ g(λ;−2)(Λ). Notice that

[X,Ad(g)Y ] = [Ad(g−1)X,Y ] = [X,Y ].

Thus [X,Y −Ad(g)Y ] = 0 so that Y − Ad(g)Y ∈ cg(X)(λ;−2)(Λ). Since cg(X) ⊂ Lie(P(λ))

by Proposition 14, we have cg(X)(λ;−2) = 0 so that Y = Ad(g)Y as required. Now, CG(X)

is smooth by Proposition 12, hence CG(X) ∩ CG(λ) is smooth by [SGA3, II Exp. XI, Cor

5.3].

?

Page 9

COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS9

Remark 20. In the notation of the previous Proposition, we have imdφ = LieS whenever

p > 2, since the adjoint representation of SL2is irreducible for p > 2.

Proposition 21. Let G be a strongly standard reductive group.

(1) Let L ⊂ G be a Levi subgroup, and assume that φ : SL2→ L is a homomorphism.

Then φ is an optimal homomorphism in G if and only if it is an optimal homomor-

phism in L.

(2) Let π : G1 → G be a central isogeny, let f : SL2 → G be a homomorphism, and

suppose that?f : SL2→ G1satisfies π ◦?f = f. Then f is optimal if and only if?f is

Proof. We first prove (1). In view of Theorem 16, it suffices to prove the following: Let

X ∈ Lie(L) be nilpotent and let λ ∈ X∗(L) be a cocharacter with X ∈ Lie(L)(λ;2). Then λ

is associated to X in L if and only if λ is associated to X in G.

Note that λ ∈ X∗(NL(X)), so the image of λ normalizes CL(X). In particular, we may

choose a maximal torus S0 of CL(X) centralized by the image of λ, and we may choose a

maximal torus S of CG(X) with S0⊂ S. Notice that S0– and hence also S – contains the

center of L. Since L is the centralizer in G of the connected center of L, we have S ⊂ L so

that S = S0. Moreover, since CG(S) ⊂ L, it is clear that CG(S) = CL(S) = M. Moreover, it

is clear that λ ∈ X∗(M), and the Proposition follows since the condition that λ be associated

to X is just that λ ∈ X∗((M,M)); this condition is the same for L and for G.

We now prove (2). Note first that it suffices to prove (2) in case K is algebraically closed.

Let X = df(E) and? X = d?f(E). Then π induces a surjective morphism NG1(? X)red→ NG(X).

torus T of CG(X) centralized by imf|Tsuch that π(?T) = T.

Levi subgroup L = CG(T). Since the maximal tori in CG1(? X) are all conjugate, one sees that

tori in CG(X) are all conjugate, f|Tis associated with X if and only if imfT lies in the derived

group of L. Since π induces a central isogeny L1→ L, it follows that?f|Tis associated with

optimal.

6We may thus choose a maximal torus?T of CG1(? X)redcentralized by im?f|Tand a maximal

Now,? X is distinguished in the Levi subgroup L1= CG1(?T) and X is distinguished in the

?f|Tis associated with? X if and only if im?f|Tlies in the derived group of L1; since the maximal

? X if and only if f|Tis associated with X; (2) is an immediate consequence.

2.8. Frobenius endomorphisms. Let H be a connected, split, quasi-simple algebraicgroup;

recall that H arises by base change from a corresponding group scheme H/Fpover the prime

field Fp. There is a Frobenius endomorphism F : H → H which arises by base change from

the corresponding Frobenius endomorphism of H/Fp.

?

Proposition 22. Let G be an algebraic group, and let φ : H → G be a homomorphism. The

following are equivalent:

(1) dφ = 0

(2) there is a unique integer t ≥ 1 and a unique homomorphism ψ : H → G such that

dψ ?= 0 and φ = ψ ◦ Ft.

Proof. (1) =⇒ (2) is a consequence of [Mc 05, Cor. 20]. (2) =⇒ (1) is straightforward.

?

Of course, the above result hold in particular when H is the group SL2.

6Note that G1 need not be strongly standard.

Page 10

10GEORGE J. MCNINCH AND DONNA M. TESTERMAN

3. The tangent map of a G-completely reducible SL2-homomorphism

3.1. The set-up. Now fix a homomorphism φ : SL2→ G whose image is geometrically G-cr.

Assume that dφ ?= 0, and write

X = dφ(E),Y = dφ(F),H = dφ([E,F]) ∈ g,

Also put s = imdφ, and write λ = φ|T. Consider the smooth subgroups NG(X),NG(Y ) ⊂ G

which are the stabilizers of the points [X],[Y ] ∈ P(g)(K). Then λ is evidently a cocharacter

of NG(X) ∩ NG(Y ).

Consider the group schemes C(X,Y ) = (CG(X) ∩ CG(Y )) and N(X,Y ) = (NG(X) ∩

NG(Y )). We observe the following:

Lemma 23. C(X,Y ) is a normal subgroup scheme of N(X,Y ).

In particular, the image of λ normalizes C(X,Y ).

3.2. Working geometrically. Fix an algebraically closed field k containing K and consider

G/k, s/k= s ⊗Kk etc. In this section, we are forced to consider the reduced subgroups

corresponding to various subgroup schemes; recall the results of 2.1. Thus, for the remainder

of §3.2, we replace K by k and so suppose that K is algebraically closed.

According to §2.1, the image of λ normalizes C(X,Y ) and hence also C(X,Y )red. Thus,

we may choose a maximal torus T ⊂ C(X,Y )redcentralized by the image of λ.

Consider now the Levi subgroup M = CG(T) of G; M is a strongly standard reductive

group by Proposition 12(1). Since X and Y are centralized by T, and since s is generated as

a Lie algebra by X and Y , we have s ⊂ Lie(M). Of course, the image of the homomorphism

φ need not lie in M.

Lemma 24. s is not contained in Lie(P) for any proper parabolic subgroup P ⊂ M.

Proof. Any torus T1 ⊂ M centralizing s of course centralizes X and Y ; thus T lies in a

maximal torus of C(X,Y )red. Since T is central in M, T1centralizes T. Since T is a maximal

torus of C(X,Y )red, we find that T1⊂ T hence T1is central in M. Since s is the Lie algebra

of a G-cr subgroup of G, the Lie algebra s is itself G-cr by Theorem 8. Hence s is also M-cr

by Proposition 7 . If s is contained in Lie(P) for a parabolic subgroup P ⊂ M, then s is

contained in Lie(L) for some Levi subgroup L of P. But then any central torus of L is central

in M, so that P = M.

?

Proposition 25. Let T1be a maximal torus of NM(X) with λ ∈ X∗(T1), and let λ0∈ X∗(T1)

be the unique cocharacter of T1associated to X [see Proposition 15(2)]. Let φ0: SL2→ M be

the optimal homomorphism determined by X and λ0[Theorem 16]. Write µ = λ0−λ for the

cocharacter

t ?→ λ0(t) · λ(t−1)

of T1. Then the image of µ is central in M.

Proof. We have H ∈ m(λ0;0) ∩ m(λ;0) by the choice of T1; thus H ∈ m(µ;0). We have also

X ∈ m(λ0;2) ∩ m(λ;2) so that X ∈ m(µ;0) as well.

Write Y =?

cm(X), so that

Y = Y−2+

j∈ZYjwith Yj∈ m(λ0;j). Since [X,Y ] = H ∈ m(λ0;0), we have Y −Y−2∈

?

j≥0

Yj.

Since the images of λ and λ0commute, and since Y ∈ m(λ;−2), we have Yj∈ m(λ;−2) for all

j. Thus, Yj∈ m(λ0− λ;j + 2) = m(µ;j + 2) for all j, hence Y ∈?

ℓ≥0m(µ;ℓ) = LiePM(µ).

Page 11

COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS11

Since X,Y,H ∈ LiePM(µ), we have proved that s = imdφ lies in LiePM(µ). Thus by

Lemma 24, we have PM(µ) = M; we conclude that the image of µ is central in M.

?

Proposition 26. Let T1 be a maximal torus of NM(X), and write φ0 : SL2 → M for the

optimal homomorphism determined by the cocharacter λ0∈ X∗(T1) associated with X as in

Proposition 25. Then dφ = dφ0.

Proof. Recall that T is a fixed maximal torus of C(X,Y )red, and M = CG(T). Using (2.4.1),

one finds a (possibly inseparable) central isogeny

π : Gsc→ G,

where the derived group of Gscis simply connected. There is a torus?T ⊂ Gscwith π(?T) = T;

to a central isogeny π : Msc→ M.

Since SL2is simply connected, there are homomorphisms

?φ : SL2→ Gsc

such that φ = π ◦?φ and φ0= π ◦?φ0. We write?λ and?λ0 for the cocharacters obtained by

µ = λ0− λ1

and

Since λ,λ0∈ X∗(T1), we know that?λ,?λ0are cocharacters of Msc. Since π ◦ ? µ = µ, it follows

Since T centralizes imφ0and imdφ, it is clear that the images of d?φ0and d?φ lie in Lie(Msc).

H0. Indeed, by Proposition 25, the image of µ = λ0− λ centralizes Y ; thus, we have Y ∈

m(λ0;−2) and Y − Y0 ∈ m(λ0;−2). If H = H0, then [X,Y − Y0] = H − H0 = 0 so that

Y −Y0∈ cm(X). Since λ0is associated to X, we have cm(X) ⊂ LieP(λ0) =?

Proposition/Definition 14. We may thus conclude that Y = Y0so that dφ and dφ0are indeed

equal.

It remains now to show that H = H0. Let? H = d?φ([E,F]) and? H0= d?φ0([E,F]). It is

Now, since the derived group M′

scof Msc is simply connected, one knows that Msc is a

direct product

Msc= Zo(Msc) × M′

where Zo(Msc) is the connected component of the center of Msc(it is a torus). Thus also

Lie(Msc) = Lie(Zo(Msc)) ⊕ Lie(M′

The derived subalgebra [Lie(Msc),Lie(Msc)] is contained in Lie(M′

we have

imd?φ ⊂ Lie(M′

In particular,

(3.2.1)

? H0−? H ∈ Lie(M′

On the other hand, im ? µ lies in Zo(Msc), and so

(3.2.2)

Since Lie(Zo(Msc)) ∩ Lie(M′

This completes the proof.

then the Levi subgroup Msc= CGsc(?T) has simply connected derived group, and π restricts

and

?φ0: SL2→ Msc

restricting these homomorphisms to the maximal torus T of SL2. Moreover, write

? µ =?λ0−?λ1.

from Proposition 25 that the image of ? µ is central in Msc.

Write H0 = dφ0([E,F]). We claim that the Proposition will follow if we show that H =

i≥0m(λ0;i) by

clearly enough to show that? H =? H0.

sc

sc).

sc); since sl2= [sl2,sl2],

sc) and imd?φ0⊂ Lie(M′

sc).

sc).

? H0−? H ∈ imd? µ ⊂ Lie(Zo(Msc)).

sc) = 0, we deduce that? H0=? H by applying (3.2.1) and (3.2.2).

?

Page 12

12GEORGE J. MCNINCH AND DONNA M. TESTERMAN

3.3. The tangent map over any field. We now suppose that K is an arbitrary field of

characteristic p > 0. As in the previous section, we fix a homomorphism φ : SL2→ G whose

image is geometrically G-cr, and we assume that dφ ?= 0.

We also fix an algebraically closed field k containing K.

Corollary 27. The K-subgroup C(X,Y ) = CG(X) ∩ CG(Y ) is smooth.

Proof. Working over the algebraically closed field k ⊃ K, let φ0: SL2/k→ G/kbe any optimal

k-homomorphism as in Proposition 26. Write S0⊂ G/kfor the image of φ0, and recall that

s/k= imdφ/k= imdφ0. Then we know that

CG/k(S0) = CG/k(s/k) = CG/k(X) ∩ CG/k(Y )

by Proposition 19, hence CG/k(X) ∩ CG/k(Y ) = (CG(X) ∩ CG(Y ))/kis smooth. But then

CG(X) ∩ CG(Y ) is smooth, since that is so after extension of the ground field.

?

Corollary 28. There is a cocharacter λ0 of G associated to X such that if φ0 : SL2 → G

is the optimal homomorphism determined by X and λ0, then dφ = dφ0. Moreover, φ0 is

uniquely determined by φ.

Proof. Since CG(X)∩CG(Y ) is smooth by the previous corollary, we can find a maximal torus

T of CG(X)∩CG(Y ) centralized by the image of the torus λ = φ|T. Then the Levi subgroup

M = CG(T) is a strongly standard reductive K-subgroup. As in Proposition 26(1), we may

find maximal tori [now over K] of CM(X) centralized by the image of λ; Proposition 26 then

gives the first assertion of the Corollary. According to Theorem 16, an optimal homomorphism

is uniquely determined by its tangent mapping; the uniqueness assertion follows at once.

?

4. Proof of the main theorem

4.1. A general setting. Let H be a connected and simple algebraic group. For each strongly

standard reductive group G, suppose that one is given a set CGof homomorphisms H → G

with the properties to be enumerated below.

Let G be strongly standard and let f0∈ CGbe arbitrary; write S0for the image of f0. We

assume the following hold for each f0:

(C1) S0is geometrically G-cr.

(C2) CG(S0) is a smooth subgroup of G, and CG(S0) = CG(Lie(S0)).

(C3) Lie(S0) = imdf0.

We also suppose:

(C4) Given any homomorphism f : H → G for which df ?= 0 and for which imf is

geometrically G-cr, there is a unique f0∈ CGsuch that df = df0.

(C5) If f : H → G is a homomorphism and if L ⊂ G is a Levi subgroup with imf ⊂ L,

then f ∈ CGif and only if f ∈ CL.

The following Lemma gives a useful application of (C1) and (C2).

Lemma 29. Let G be a reductive group and let S ⊂ G be a subgroup with the property

CG(S) = CG(Lie(S)). Suppose that S is geometrically G-cr. If K ⊂ k is any field extension

and P ⊂ G/kis a k-parabolic subgroup, then S/k⊂ P if and only if Lie(S)/k⊂ Lie(P),

Proof. Since the Lemma follows once it is proved for algebraically closed extensions k, it

suffices to suppose that K itself is algebraically closed and to prove the conclusion of the

Lemma for a (K-)parabolic subgroup P ⊂ G.

Lie(S) ⊂ Lie(P).

First notice that if S ⊂ P, then clearly

Page 13

COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS 13

Now suppose that s = Lie(S) ⊂ Lie(P). Since S is G-cr, Theorem 8 shows that s is also

G-cr. Thus, we may find a Levi subgroup L ⊂ P with s ⊂ Lie(L). Then L = CG(T) where

T = Z(L), and we see that

T ⊂ CG(s) = CG(S);

thus T centralizes S, so that S ⊂ CG(T) = L ⊂ P, as required.

?

We now observe:

Proposition 30. Let p > 2, let H = SL2, and for each strongly standard reductive group G,

let CGbe the set of optimal homomorphisms SL2→ G. Then conditions (C1) – (C5) of §4.1

hold for the sets CG.

Proof. (C1) follows from Theorem 16, (C2) is Proposition 19, (C4) is Corollary 28, and (C5)

is Proposition 21(1).

Since p > 2, the adjoint representation of SL2is irreducible; since any optimal homomor-

phism f : SL2→ G has df(E) ?= 0, the map df must be injective and so (C3) is immediate.

?

4.2. Some results about twisted-product homomorphisms. Let H be a reductive group

and let CG be a collection of homomorphisms H → G for each strongly standard group G

which satisfies (C1)–(C5) of §4.1.

We are going to prove several technical results about twisted product homomorphisms; to

avoid repetition in the statements, we fix the following notation:

Let?h = (h0,h1,...,hr) with hi∈ CGbe commuting homomorphisms [as in the introduc-

tion], and let ? n = (n0< n1< ··· < nr) be non-negative integers; the data (?h,? n) determines

a twisted-product homomorphism Φ = Φ?h,? n: H → G given for each commutative K-algebra

Λ and each g ∈ H(Λ) by the rule

(4.2.1)g ?→ h0(Fn0g) · h1(Fn1g)···hr(Fnrg).

Several of the results proved in this section hold only assuming a subset of the conditions

(C1)–(C5); for simplicity of exposition, we assume all five conditions hold – we don’t bother

to identify the subset.

Lemma 31. Let G be strongly standard, let?h, ? n, and Φ = Φ?h,? nbe as in the beginning of §4.2.

Then dΦ = 0 if and only if n0> 0. If dΦ = 0 let Ψ be the twisted-product homomorphism

determined by (?h,(0,n1− n0,...,nr− n0)). Then Φ = Ψ ◦ Fn0and dΨ ?= 0. Moreover,

imΦ = imΨ.

Proof. Straightforward and left to the reader.

?

Proposition 32. Let G be strongly standard, let (f1,f2) be commuting homomorphisms H →

G with f1 ∈ CG. Let (n1 < n2) be non-negative integers, and let f be the twisted-product

homomorphism determined by (f1,f2) and (n1,n2). Write Si for the image of fi, i = 1,2,

and write S for the image of f. Then:

(1) for each field extension K ⊂ k and each parabolic subgroup P ⊂ G/k, we have S/k⊂ P

if and only if S1/k⊂ P and S2/k⊂ P.

(2) S1· S2is geometrically G-cr if and only if S is geometrically G-cr.

Proof. Note that (1) will follow once it is proved for algebraically closed extension fields k of

K. Thus, we suppose that k = K is algebraically closed, and prove the conclusion of (1) for

parabolic subgroups P ⊂ G.

Page 14

14 GEORGE J. MCNINCH AND DONNA M. TESTERMAN

If the parabolic subgroup P ⊂ G contains S1 and S2, it is clear by the definition of a

twisted-product homomorphism that P contains S. Suppose now that P contains S; we show

it contains also S1and S2.

Applying Lemma 31, one knows that if g : H → G is the twisted-product homomorphism

determined by (f1,f2) and (0,n2− n1), then img = S as well. We may thus suppose that

n1= 0, so that df ?= 0.

It is clear that df = df1. Since imdf1= Lie(S1), it follows that Lie(S) = Lie(S1). Since

S ⊂ P, we have Lie(S1) = Lie(S) ⊂ Lie(P); since (C1) and (C2) hold, we may apply Lemma

29 and conclude that S1⊂ P. Since f2is given by the rule

g ?→ f1(g)−1f(g)

it is then clear that S2⊂ P as well. This proves (1).

Since (2) is a geometric statement, we may again suppose that K is an algebraically closed

field. Write X for the building of G; cf. [Ser 05, §2 and §3.1]. Then X is a simplicial

complex whose simplices are in bijection with the parabolic subgroups of G. We have shown

the equality of fixed-point sets: XS= (XS2)S1= XS1·S2.

According to [Ser 05, Th´ eor` eme 2.1], the group S is G-cr if and only if XS= XS1S2is

contractible if and only if S1· S2is G-cr. This proves (2).

?

Corollary 33. Let G be strongly standard, and let?h, ? n, Φ = Φ?h,? nbe as in the beginning of

§4.2. Write S for the image of h and Si for the image of hi. If P ⊂ G/kis a k-parabolic

subgroup for an extension field K ⊂ k, then S/k⊂ P if and only if Si/k⊂ P for i = 1,...,r.

Proof. It is enough to give the proof assuming that K = k is algebraically closed. If Si⊂ P

for each i, it is clear by construction that S ⊂ P.

Now suppose that S ⊂ P. To prove that each Si⊂ P, we proceed by induction on r. If

r = 1, the result is immediate. Suppose that r > 1, and let Ψ : H → G be the twisted-product

homomorphism determined by (h2,...,hr) and (n2−1,...,nr−1). Then f may be regarded

as the twisted-product homomorphism determined by (h1,Ψ) and (0,1). Thus we may apply

Proposition 32(1) to see that S1⊂ P and imΨ ⊂ P. Now apply the induction hypothesis to

Ψ to learn that Si⊂ P for 2 ≤ i ≤ r. This completes the proof.

?

Proposition 34. Let G be strongly standard, and let?h, ? n, Φ = Φ?h,? nbe as in the beginning of

§4.2. If S denotes the image of Φ and Sithe image of hi, then CG(S) ⊂ CG(Si) for 1 ≤ i ≤ r.

Proof. If r = 1, the result is immediate. Suppose r > 1, write Ψ for the homomorphism

determined by (h2,...,hr) and (n2,...,nr) and write T = imΨ. It suffices by induction on

r to show that CG(S) ⊂ CG(S1) and CG(S) ⊂ CG(T), since then for 2 ≤ i ≤ r we have

CG(S) ⊂ CG(T) ⊂ CG(Si)

by the induction hypothesis.

Applying Lemma 31, we may assume that n1= 0 and dh ?= 0 without changing S. Thus

Lie(S) = Lie(S1). By (C2), we have

CG(S) ⊂ CG(Lie(S)) = CG(Lie(S1)) = CG(S1).

Finally, it remains to check that CG(S) ⊂ CG(T). Write Ψ∗,h∗

comorphisms of Ψ, h1, and Φ. Then by construction, Ψ∗is given by the composition

1,Φ∗: K[G] → K[H] for the

K[G]

µ

− → K[G] ⊗KK[G]

ι⊗id

− − − → K[G] ⊗KK[G]

h∗

− − − − → K[H] ⊗KK[H]

1⊗Φ∗

∆

− → K[H]

where the map µ defines the multiplication in G, the map ι defines the inversion in G, and ∆

is given by multiplication in K[H].

Page 15

COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS 15

Let g ∈ CG(S)(Λ) for some commutative K-algebra Λ. To show that g ∈ CG(T)(Λ),

it is enough to argue that the inner automorphism Int(g) of G induces the identity on the

subgroup scheme T/Λ. Since T is defined by the ideal kerΨ∗⊳ K[G], it is enough to require

that Ψ∗(Int(g)∗f) = Ψ∗(f) for each f ∈ Λ[G]. [Note: we write Ψ∗rather than Ψ∗

simplicity.]

Since g ∈ CG(S)(Λ) and g ∈ CG(S1)(Λ), we know for each f ∈ Λ[G] that

/Λfor

h∗

1(Int(g)∗f) = h∗

1(f)andΦ∗(Int(g)∗f) = Φ∗(f).

If f1⊗ f2∈ Λ[G] ⊗ΛΛ[G], then

(h∗

1⊗ Φ∗)((Int(g)∗⊗ Int(g)∗)(f1⊗ f2)) = h∗

1(Int(g)∗f1) ⊗ Φ∗(Int(g)∗f2)

= h∗

= (h∗

1(f1) ⊗ Φ∗(f2)

1⊗ Φ∗)(f1⊗ f2).

It follows for any f1∈ Λ[G] ⊗ΛΛ[G] that

(4.2.2)(h∗

1⊗ Φ∗)((Int(g)∗⊗ Int(g)∗)f1) = (h∗

1⊗ Φ∗)f1

Since Int(g) is an automorphism of G, we have for f ∈ Λ[G] that

(4.2.3) ((ι ⊗ id) ◦ µ)(Int(g)∗f) = (Int(g)∗⊗ Int(g)∗)((ι ⊗ id) ◦ µ)(f).

Combining (4.2.2) and (4.2.3), we see that Ψ∗(Int(g)∗f) = Ψ∗(f) for each f ∈ Λ[G], as

required. This completes the proof.

?

Remark 35. With notation as before, one can even show that

CG(S) =

r?

i=1

CG(Si).

The inclusion CG(S) ⊂?

inclusion may be proved by showing for each commutative K-algebra Λ that if g ∈ CG(Si)(Λ)

for each i, then Φ∗(Int(g)∗f) = Φ∗(f) for each f ∈ Λ[G]; the proof is like that used for the

Proposition.

iCG(Si) follows from the previous Proposition, and the reverse

4.3. Finding the twisted factors of a homomorphism with G-cr image. Let H be

a reductive group and let CG be a collection of homomorphisms H → G for each strongly

standard group G which satisfies (C1)–(C5) of §4.1. In this section, we are going to give the

proof of Theorem 1.

We first have the following:

Proposition 36. Fix a strongly standard reductive group G, and let the homomorphism

f : H → G have geometrically G-cr image S. Assume that df ?= 0 and let f0 ∈ CG be the

unique map – as in (C4) – such that df = df0. Then:

(1) the map f1: H → G given by the rule g ?→ f0(g−1) · f(g) is a group homomorphism.

(2) S1= imf1⊂ CG(imf0).

(3) df1= 0.

(4) S1is geometrically G-cr.

Proof. Write S0= imf0, and write f1: H → G for the morphism defined by the rule in (1).

Let Λ be an arbitrary commutative K-algebra, let g ∈ H(Λ), and let X ∈ Lie(H)(Λ). Since

df = df0, we know that

Ad(f(g))df0(X) = Ad(f(g))df(X) = df(Ad(g)X) = df0(Ad(g)X) = Ad(f0(g))df0(X).

Page 16

16 GEORGE J. MCNINCH AND DONNA M. TESTERMAN

It follows that Ad(f1(g)) = Ad(f0(g−1)f(g)) centralizes df0(X) for each X ∈ Lie(H)(Λ).

Since imdf0= Lie(S0) by (C3), it follows that the image of f1 lies in CG(Lie(S0)). If now

g,h ∈ H(Λ), then we see that

f1(gh) = f0(h−1)f0(g−1)f(g)f(h) = f0(h−1)f1(g)f(h) = f1(g)f0(h−1)f(h) = f1(g)f1(h).

Thus f1 is a homomorphism, so that (1) and (2) are proved. By construction, the tangent

map of f1is df − df0= 0; this proves (3).

Note that (2) implies that S0· S1 is a subgroup. Since S is geometrically G-cr, we may

apply Proposition 32 to see that S0· S1is geometrically G-cr. Since S1⊳ S0· S1is a normal

subgroup, it follows from the result of B. Martin (Theorem 9) that S1 is G-cr; this proves

(4).

?

Corollary 37. Let H be quasisimple and suppose that the homomorphism f : H → G has ge-

ometrically G-cr image. Then there are uniquely determined commuting CG-homomorphisms

h0,h1,...,hrand uniquely determined non-negative integers n0< n1··· < nr such that f is

the twisted-product homomorphism determined by (?h,? n).

Proof. We may use Proposition 22 to find a homomorphism h : H → G and an integer

t ≥ 0 such that f = h ◦ Ftwhere F is the Frobenius endomorphism of H.

dh ?= 0. If the conclusion of the Theorem holds for h, we claim that it holds for f as

well. Indeed, if h is the twisted-product homomorphism determined by the commuting CG-

homomorphisms?h = (h0,h1,...,hr) and the non-negative integers ? n = (0 = n0< ··· < nr),

then f is the commuting-product homomorphism determined by?h and the non-negative

integers ? m = (t < n1+ t < ··· < nr+ t). If f had a second representation as a commuting-

product homomorphism determined by?h′= (h′

using Lemma 31 one deduces that m′

0= t and one finds a representation of h as the twisted

product homomorphism determined by?h′and (0 < m′

? m = ? m′; this proves the claim. So we may and will suppose that df ?= 0.

Let us first prove the uniqueness assertion; namely, suppose that?h = (h0,h1,...,ht) and

?h′= (h′

? n = (n0< ··· < nt) and ? n′= (n′

and suppose that f = Φ?h,? n= Φ?h′,? n′. We must argue that s = t,?h =?h′and ? n = ? n′. We know

that df = dh0= dh′

h0= h′

Moreover,

0,...,h′

t) and ? m′= (m′

0< ··· < m′

t) then

1− t < ··· < m′

t). Thus?h =?h′and

0,h′

1,...,hs) are commuting homomorphisms with hi,h′

0< ··· < n′

j∈ CG, and suppose that

s) are non-negative integers with 0 = n0= n′

0,

0. Since h0∈ CGis the unique mapping with df = dh0by (C4), we have

0. It then follows that

Φ(h1,...,ht),(n1<···<nt)= Φ(h′

1,...,h′

t),(n′

1<···<n′

t)

so by induction on min(s,t), we find that s = t, hi = h′

completes the proof of uniqueness.

For the existence, we choose by (C4) the unique map f0∈ CGsuch that df = df0. We now

write f1: H → G for the homomorphism of Proposition 36(1). Thus f is given by the rule

iand ni = n′

ifor 1 ≤ i ≤ t; this

(4.3.1)g ?→ f0(g) · f1(g)

Write S for the image of f, and write S0and S1for the respective images of f0and f1.

We proceed by induction on the semisimple rank r of G. If r is smaller than the rank of

the simple group H, there are no homomorphisms H → G. If the semisimple rank of G is the

same as the rank of H, then apply Lemma 38 to S0⊂ G′, where G′is the derived group of G.

One deduces that CG′(S0) has no non-trivial torus, hence that any torus in CG(S0) is central

in G. Since SL2is its own derived group, imf1lies in G′; thus imf1is contained in CG′(S0)

Page 17

COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS17

by Proposition 36(2) and it follows that the map f1is trivial. We conclude in this case that

f = f0∈ CG.

We now suppose that the semisimple rank of G is strictly greater than the rank of H.

Since S1 ⊂ CG(S0), a maximal torus of S0 centralizes S1. Thus, the image of the G-cr

homomorphism f1 lies in some proper Levi subgroup L. Since the semisimple rank of L

is smaller than that of G, we may apply induction; we find commuting homomorphisms

h1,...,hr∈ CL, and non-negative integers n1< n2< ··· < nr such that f1is the twisted-

product map determined by (?h,? n). Since df1= 0, we have 0 < n1. It follows from (C5) that

h1,...,hr∈ CG.

Since imf0 = S0 ⊂ CG(S1), it follows from Proposition 34 applied to f1 that S0 ⊂

CG(imhi) for 1 ≤ i ≤ r. Thus the homomorphisms (f0,h1,...,hr) are commuting. In

view of (4.3.1), f is the twisted-product homomorphism determined by (f0,h1,...,hr) and

(0 < n1< ··· < nr).

?

Lemma 38. Let X and Y be semisimple groups of the same rank, and suppose that X ⊂ Y .

Then CY(X) contains no non-trivial torus.

Proof. Let S ⊂ Y be any torus centralizing X, and let T be a maximal torus of X. Since T

is centralized by S and is also maximal in Y , we have S ⊂ T so that S ⊂ X. Thus S is a

central torus in X. Since X is semisimple, S is trivial as required.

?

We can now prove the following; note that Theorem 1 is a special case.

Theorem 39. Let G be a strongly standard reductive group, and let Φ : SL2 → G be a

homomorphism. If the image of Φ is geometrically G-cr, then there are commuting optimal

homomorphisms?φ = (φ0,...,φr) and non-negative integers ? n = (n0< n1< ··· < nd) such

that Φ is the twisted-product homomorphism determined by (?φ,? n). Moreover,?φ and ? n are

uniquely determined by Φ.

Proof. For a strongly standard reductive group G, write CGfor the set of optimal homomor-

phisms SL2→ G. Suppose first that p > 2. Then Theorem 1 is a consequence of Proposition

30 together with Corollary 37.

Now suppose that p = 2. Use (2.4.1) to find a central isogeny π : Gsc→ G where the derived

group of Gscis simply connected. Since SL2is simply connected, there is a homomorphism

?Φ : SL2 → Gsc with Φ = π ◦?Φ. It follows from Lemma 10 that?Φ has geometrically G-cr

if π ◦ f : SL2→ G is optimal.

If?Φ is the twisted product homomorphism determined by the optimal homomorphisms

that Φ is the twisted product homomorphism determined by the optimal homomorphisms

?φ′= (π ◦ φ0,...,π ◦ φr) and ? n. Moreover, the uniqueness of?φ implies the uniqueness of?φ′;

thus it suffices to prove the theorem after replacing G by Gsc. So we now assume that the

derived group of G is simply connected.

Assume first that K is separably closed. Recall that since p = 2 is good for the derived

group of G, each of its simple factors has type Amfor some m. Since G is split and simply

connected, we find that G ≃ T ×?t

vector space Vi. Write πi: G → Gifor the i-th projection, and Φi= πi◦Φ. Steinberg’s tensor

product theorem [Jan 87, Cor. II.3.17] shows that Φi may be written as a twisted-product

homomorphism for a unique collection of commuting optimal homomorphisms and a unique

increasing list of non-negative integers; see [LS 03, Lemma 4.1]. The same then clearly holds

for Φ, so the Theorem is proved in this case.

image. Proposition 21(2) shows that a homomorphism f : SL2→ Gscis optimal if and only

?φ = (φ0,...,φr) and the non-negative integers ? n = (n0 < n1 < ··· < nr), it is then clear

i=1Giwhere T is a central torus, and Gi≃ SL(Vi) for a

Page 18

18GEORGE J. MCNINCH AND DONNA M. TESTERMAN

For general K, the above argument represents the base-changed morphism Φ/Ksep as the

twisted product homomorphism Φ?φ,? nfor unique commuting optimal Ksep-homomorphisms

?φ = (φ0,...,φt) and unique ? n = (n0 < n1 < ··· < nt). Since?φ and ? n are unique, we

may apply Galois descent to see that each φi arises by base change from an optimal K-

homomorphism, and the proof is complete.

?

5. Proof of a partial converse to the main theorem

In this section, we will prove Theorem 2, which is a geometric statement – it depends only

on G and H over an algebraically closed field. Thus we will suppose in this section that K is

algebraically closed, and we write “G-cr”rather than “geometrically G-cr”.

We begin with a result on G-cr subgroups.

Proposition 40. Let G be reductive, let h(G) be the maximum Coxeter number of a simple

quotient of G, and suppose that p > 2h(G)−2. Let A,B ⊂ G be smooth, connected, and G-cr,

and suppose that B ⊂ CG(A). Then A · B is G-cr.

Proof. Under our assumptions on p, it follows from [Ser 05, Corollaire 5.5] that a subgroup

Γ ⊂ G is G-cr if and only if the representation of Γ on Lie(G) is semisimple.

Since a smooth, connected G-cr subgroup is reductive [Ser 05, Prop. 4.1], the proposition

is now a consequence of the lemma which follows.

?

Lemma 41. Let G1,G2⊂ GL(V ) be connected and reductive, and suppose G2⊂ CGL(V )(G1).

Then V is semisimple for G1· G2.

Proof. Write H = G1·G2. Since H is a quotient of the reductive group G1×G2by a central

subgroup, H is reductive.

Since G1and G2 commute, G2 leaves stable the G1-isotypic components of V . Thus we

may write V as a direct sum of H-submodules which are isotypic for both G1and G2. Thus

we may as well assume that V itself is isotypic for G1and for G2.

Let Bi⊂ Gibe Borel subgroups and let Ti⊂ Bibe maximal tori for i = 1,2. Note that the

choice of a Borel subgroup determines a system of positive roots in each X∗(Ti); the weights

of Ti on Ui= Ru(Bi) are positive. Our hypothesis means that there are dominant weights

λi∈ X∗(Ti) such that each simple Gi-submodule of V is isomorphic to LGi(λi), the simple

Gi-module with highest weight λi.

Now, B = B1·B2is a Borel subgroup of H, and T = T1·T2is a maximal torus of B. Since

T1∩ T2lies in the center of H, one knows that there is a unique character λ ∈ X∗(T) such

that λ|Ti= λifor i = 1,2. Moreover, it is clear that λ is dominant. Put U = U1·U2= Ru(B).

It follows from [Jan 87, II.2.12(1)] that there are no non-trivial self-extensions of simple

H-modules; thus the Lemma will follow if we show that all simple H-submodules of V are

isomorphic to LH(λ).

Let L ⊂ V be a simple H-submodule; we claim that L ≃ LH(λ). Since L is simple, the

fixed point space of U on L satisfies dimKLU= 1 and our claim will follow once we show

that LU⊂ LT;λsince then LU= LT;λand L ≃ LH(λ); for all this, see [Jan 87, Prop. II.2.4]

[we are writing LT;λ for the λ weight space of the torus T on L]. Since L is semisimple

and G1-isotypic, LU1= LT1;λ1. Since G2⊂ CGL(V )(G1), LU1is a G2-submodule. Since LU1

is semisimple and isotypic as G2-module, we know that LU= (LU1)U2= (LU1)T2;λ2. Thus

LU⊂ LT1;λ1∩ LT2;λ2so indeed LU= LT;λas required.

?

Suppose that H is a simple group, and that for each strongly standard group G, one has

a set CGof homomorphisms H → G satisfying (C1)–(C5) of §4.1.

Page 19

COMPLETELY REDUCIBLE SL(2)-HOMOMORPHISMS 19

Theorem 42. Let G be strongly standard and assume that p > 2h(G) − 2. Let h0,...,hrbe

commuting CG-homomorphisms, and let n0< n1< ··· < nr be non-negative integers. Then

the image of the twisted-product homomorphism h determined by (?h,? n) is geometrically G-cr.

Proof. Write Sifor the image of hi, 0 ≤ i ≤ r. By (C1), Siis G-cr for 0 ≤ i ≤ r. In view of

our assumption on p, it follows from Proposition 40 that the subgroup A = S0· S1···Sr is

G-cr.

Write X for the building of G. If S = imh, Corollary 33 shows that XS= XA. Since A is

G-cr, XA= XSis not contractible, so that S is G-cr [Ser 05, Th´ eor` eme 2.1].

?

Theorem 43. Let G be a strongly standard reductive group, suppose that p > 2h(G) − 2, let

?φ = (φ0,...,φd) be commuting optimal homomorphisms SL2→ G, and let ? n = (n0< n1<

··· < nd) be non-negative integers. Then the image of the twisted-product homomorphism

Φ : SL2→ G determined by (?φ,? n) is geometrically G-cr.

Proof. As in the proof of Theorem 1, write CGfor the set of optimal homomorphisms SL2→ G

for a strongly standard group G. Note that the condition p > 2h(G) − 2 implies that p > 2.

Then Theorem 2 is a consequence of Proposition 30 and Theorem 42.

?

Of course, Theorem 2 is a special case of the previous result.

Remark 44. Let G be one of the following groups: (i) GL(V ), (ii) the symplectic group Sp(V ),

(iii) the orthogonal group SO(V ), or (iv) a group of type G2. In cases (ii), (iii) assume p > 2

while in case (iv) assume that p > 3; then p is very good for G. In case (iv), write V for

the 7 dimensional irreducible module for G; thus in each case V is the “natural” module for

G. Then a closed subgroup H ⊂ G is G-cr if and only if V is semisimple as an H-module;

see [Ser 05, 3.2.2]. Thus, the conclusion of Theorem 2 holds for G (with no further prime

restrictions). Indeed, in view of Lemma 41, one finds that the conclusion of Proposition 40

is valid with no further assumption on p by using V rather than the adjoint representation

of G. Now argue as in the proof of Theorem 42 when p > 2, or just use Steinberg’s tensor

product theorem when p = 2 (since we are supposing G = GL(V ) in that case).

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[Mc 05]

[Mc 05a]

[Sei 00]

[Ser 05]

Department of Mathematics, Tufts University, 503 Boston Avenue, Medford, MA 02155, USA

E-mail address: george.mcninch@tufts.edu

Institut de g´ eom´ etrie, alg` ebre et topologie, Bˆ atiment BCH,´Ecole Polytechnique F´ ed´ erale de

Lausanne, CH-1015 Lausanne, Switzerland

E-mail address: donna.testerman@epfl.ch