On irreducible subgroups of simple algebraic groups

Let $G$ be a simple algebraic group over an algebraically closed field $K$ of characteristic $p\geqslant 0$, let $H$ be a proper closed subgroup of $G$ and let $V$ be a nontrivial irreducible $KG$-module, which is $p$-restricted, tensor indecomposable and rational. Assume that the restriction of $V$ to $H$ is irreducible. In this paper, we study the triples $(G,H,V)$ of this form when $G$ is a classical group and $H$ is positive-dimensional. Combined with earlier work of Dynkin, Seitz, Testerman and others, our main theorem reduces the problem of classifying the triples $(G,H,V)$ to the case where $G$ is an orthogonal group, $V$ is a spin module and $H$ normalizes an orthogonal decomposition of the natural $KG$-module.


Introduction
Let G be a simple algebraic group defined over an algebraically closed field K , let H be a closed positive-dimensional subgroup of G, and let V be a nontrivial rational irreducible K G-module. We say that (G, H, V ) is an irreducible triple if V is irreducible as a K H-module. Triples of this form arise naturally in the investigation of maximal subgroups of classical algebraic groups, and their study can be traced back to work of Dynkin [7] in the 1950s, who considered the special case where H is connected and K = C. In the 1980s, Seitz [17] and Testerman [20] extended the analysis to arbitrary algebraically closed fields (still assuming H is connected), and more recent work of Ghandour [10] has completed the classification of irreducible triples for exceptional groups. Therefore, in this paper we focus on classical groups and their disconnected subgroups.
In [8,9], Ford determines the irreducible triples (G, H, V ) where G is a classical group, H is disconnected, H 0 is simple and the composition factors of the restriction V | H 0 are p-restricted. Our main aim is to extend Ford's analysis by removing the restrictive conditions on the structure of H 0 and the composition factors of V | H 0 . The cases for which V | H 0 is irreducible are easily deduced from the work of Seitz [17], so we focus on the situation where V | H is irreducible, but V | H 0 is reducible. By Clifford theory, the highest weights of the K H 0 -composition factors of V are H -conjugate and this severely restricts the possibilities for V . Since the triples with H maximal have recently been determined in [3,4], in this paper we will adopt the following hypothesis: Hypothesis 1. 1 The group G is a simply connected cover of a simple classical algebraic group defined over an algebraically closed field K of characteristic p 0, H is a closed positive-dimensional subgroup of G, and V is a nontrivial p-restricted irreducible tensor indecomposable rational K G-module such that the following conditions hold: H1. V = W τ for any automorphism τ of G, where W is the natural module; H2. H Z(G)/Z (G) is disconnected and non-maximal in G/Z (G).
Let G be a classical group as in Hypothesis 1.1, let n denote the rank of G and let {λ 1 , . . . , λ n } be a set of fundamental dominant weights for G (we adopt the standard labelling given in Bourbaki [2]). We will write V G (λ) for the irreducible K G-module with highest weight λ. Remark 1.2 Condition H1 in Hypothesis 1.1 is equivalent to assuming V = W, W * , and also V = V G (λ 3 4 . This hypothesis is unavoidable. For example, we cannot feasibly determine all the simple subgroups of G that act irreducibly on W or W * (indeed, even the dimensions of the irreducible modules for simple groups are not known, in general). In particular, H1 is a condition adopted in [3,17].
Suppose G is of type B n or D n , and let R(W ) = R be the radical of the corresponding bilinear form on W (recall that either R = 0, or p = 2, dim W is odd and dim R = 1). An orthogonal decomposition of W is a decomposition of the form where the W i are pairwise orthogonal subspaces of W . Note that if W = W 1 + · · · + W t is such a decomposition, then the W i are non-degenerate spaces such that W i ∩ ( j =i W j ) ⊆ R for each i (in particular, if p = 2 then W = W 1 + · · · + W t is a direct sum). We say that a subgroup H of G normalizes such a decomposition if it permutes the W i .  Table 1; or (b) G is of type B n or D n (or type C n if p = 2), V is a spin module and H is a decomposition subgroup.
Moreover, if (a) holds then V | H is irreducible. Remark 1.5 In case (i) of Table 1, T n denotes a maximal torus of G. In all cases, H 0 is the connected component of a maximal subgroup of G, with the exception of the cases labelled (ii) and (iii), where H is contained in a subgroup D 4 < G. Also note that in cases (iii)-(vi), W | H 0 is the tensor product of the natural modules of the simple components of H 0 . In case (ii), H is the image of a subgroup C 3 1 .X < C 4 as in (iii), under an isogeny ϕ : C 4 → B 4 . In cases (v) and (vi), we record H and V up to Aut(G)-conjugacy (so in case (vi) for example, ifH is the image of H under a graph automorphism of G, then V G (λ 8 ) is an irreducible KH -module). Finally, let us note that the situation in part (b) of Theorem 1.4 is very special and we refer the reader to Sect. 7 for further details. More precisely, we are either in the situation described in part (b) of Theorem 1.4, or in terms of Seitz's notation in [17, Table 1], one of the following holds (modulo scalars): Let us briefly describe our approach to the proof of Theorem 1.4. Suppose V | H 0 is reducible. If H is a maximal subgroup of G then (G, H, V ) can be read off from the main theorems of [3,4], so let us assume H < M < G with M maximal. Since V | H is irreducible, it follows that V | M is also irreducible and we can consider the possibilities for the irreducible triple (G, M, V ), which are determined in [17] (if M is connected) and [3,4] (if M is disconnected). We can then proceed by studying the possible embeddings of H in M.
Our next result is a combination of the main theorems in [3,4,17], together with Theorem 1.4. Note that we assume n 3 if G = B n , n 2 if G = C n , and n 4 if G = D n . Corollary 1.7 Let G be a simply connected cover of a simple classical algebraic group over an algebraically closed field K of characteristic p 0. Let H be a positivedimensional closed subgroup of G, and let V = V G (λ) be a nontrivial p-restricted irreducible tensor indecomposable rational K G-module such that V | H is irreducible. Then one of the following holds: (a) V = W τ for some automorphism τ of G, where W is the natural module; (b) G is of type B n or D n (or type C n if p = 2) and V is a spin module; (c) (G, H, V ) is recorded in Table 2. Remark 1.8 Let us make some comments on the cases in Table 2: (i) In Table 2, we write X for Sym 3 or Z 3 , Z for Z 2 or 1, and Y denotes any ktransitive subgroup of Sym n+1 . (ii) For G of type A n , the highest weight λ is recorded up to conjugacy by a graph automorphism of G (this is consistent with [17, Table 1]). For instance, in the first row of the table, we have G = A n , H = T n .Y and λ = λ k with 2 k (n+1)/2. By applying a suitable graph automorphism, we see that V G (λ k )| H is irreducible for all 2 k n − 1.
i=1 a i ω i + (a n−1 + 1)ω n 2 S e eR e m a r k1.8(v)(b) S e eR e m a r k1. 8(v)(c) C 28 E 7 λ 5 ω 2 + ω 3 1 p = 2, 3, 5 Table 2 continued (v) In the final column, we record various conditions on G, H and λ that are necessary and sufficient for the irreducibility of V | H . In a few cases, the conditions for irreducibility are rather complicated and so we record them here (the example in case (b) was discovered by Ford in [8]): Our final result concerns chains of irreducibly acting subgroups. Let G and V be given as in Hypothesis 1.1 and write = (G, V ) for the length of the longest chain of closed positive-dimensional subgroups We call such a sequence of subgroups an irreducible chain. If G is an orthogonal group (or a symplectic group with p = 2) and V is a spin module, then (G, V ) can be arbitrarily large, and it is easy to see that the same is true if V = W or W * . Theorem 1.9 Suppose G and V satisfy the conditions in Hypothesis 1.1 and assume V is not a spin module. Then either (G, V ) 5, or G is of type A n and λ ∈ {λ 2 , λ 3 , λ n−2 , λ n−1 }.
The upper bound in this theorem is best possible. In fact, if we exclude the exceptional cases then either (G, V ) 4, or G ∈ {B 4 , C 4 }, λ = λ 3 , p = 2 and is an irreducible chain of length 5 (see Theorem 8.1 for a more precise statement). The exceptions with G = A n are genuine in the strong sense that (G, V ) can be arbitrarily large. We refer the reader to Sect. 8 for further details.

Notation and terminology
Most of our notation is fairly standard. As in Hypothesis 1.1, let G be a simply connected cover of a simple classical algebraic group, which is defined over an algebraically closed field K of characteristic p 0. Fix a Borel subgroup B = U T of G, where T is a maximal torus of G and U is the unipotent radical of B. Let (G) = {α 1 , . . . , α n } be the corresponding base of the root system (G) of G, where n denotes the rank of G. Let {λ 1 , . . . , λ n } be the fundamental dominant weights for T corresponding to (G).
There is a bijection between the set of dominant weights of G and the set of isomorphism classes of irreducible K G-modules; if λ is a dominant weight then we use V G (λ) to denote the unique irreducible K G-module with highest weight λ. We also recall that if p > 0 then a dominant weight λ = i a i λ i is p-restricted if a i < p for all i. By Steinberg's tensor product theorem, every irreducible K G-module decomposes in a unique way as a tensor product ) is the K G-module obtained by preceding the action of G on V i by the endomorphism σ p i . It is convenient to say that every dominant weight is p-restricted if p = 0.
In addition, Lie(G) denotes the Lie algebra of G, and U α = {x α (t)|t ∈ K } is the root subgroup of G corresponding to a root α ∈ (G). If x ∈ G then t x : G → G is the inner automorphism of G induced by conjugation by x, so t x (g) = xgx −1 for all g ∈ G. We write T i for an i-dimensional torus. If H is a closed positive-dimensional subgroup of G and T H 0 is a maximal torus of [H 0 , H 0 ] contained in T , then we abuse notation by writing μ| H 0 to denote the restriction of a T -weight μ to the subtorus T H 0 . We define a partial order on the set of weights for T , where μ ν if and only if μ = ν − n i=1 c i α i for some non-negative integers c i (in this situation, we say that μ is under ν). Finally, we set N 0 = N ∪ {0}, we write Sym n and Alt n for the symmetric and alternating groups of degree n, and we denote a cyclic group of order m by Z m (or just m).

123
Recall that a map ϕ : G 1 → G 2 of algebraic groups is a morphism if it is a group homomorphism that is also a morphism of the underlying varieties. In particular, it is important to note that an injective morphism does not necessarily induce an isomorphism G 1 ∼ = ϕ(G 1 ) of algebraic groups. If G 1 is a semisimple algebraic group with root system , then we will say that G 1 is of type (and we will sometimes denote this by writing G 1 = ). For example, SL 2 (K ) and PGL 2 (K ) are both simple algebraic groups of type A 1 = B 1 = C 1 , and PSp 4 (K ) and SO 5 (K ) are both of type B 2 = C 2 . Finally, note that if H is a closed positive-dimensional subgroup of an algebraic group G, and ϕ : H → G is the inclusion map, then the differential dϕ : Lie(H ) → Lie(G) is an injective Lie algebra homomorphism (since ϕ : H → ϕ(H ) is an isomorphism of algebraic groups).

Diagonal embeddings
In the context of algebraic groups, the related notion of a diagonally embedded subgroup is defined as follows: Definition 2.1 Let H be a closed subgroup of G = G 1 × · · · × G t where the G i are isomorphic simply connected simple algebraic groups. We say that H is diagonally embedded in G if each projection π i : H → G i is a bijective morphism. Note that we do not require each projection map π i to induce an isomorphism H ∼ = G i of algebraic groups.
The next lemma is a well known result of Steinberg (see [18,Theorem 30] and [19, 10.13]), which describes the bijective endomorphisms of a simple algebraic group.
Here t x and σ q are defined as in Sect. 2.1, and we adopt Steinberg's definition of a graph automorphism of a simple algebraic group G (see [18,Sect. 10]). In particular, a graph automorphism is an isomorphism of algebraic groups unless (G, p) = (C 2 , 2), (G 2 , 3) or (F 4 , 2).

Lemma 2.2
Let G be a simple algebraic group over an algebraically closed field of characteristic p 0. Let ϕ : G → G be a bijective morphism. Then ϕ = t x σ q γ k for some x ∈ G, p-power q and integer k ∈ {0, 1}, where γ is a graph automorphism of G. Moreover, if G is classical and (G, p) = (C 2 , 2), then ϕ is an isomorphism of algebraic groups if and only if σ q = 1.

Lemma 2.3
Let ϕ : H → G be a surjective morphism of algebraic groups and let dϕ : Lie(H ) → Lie(G) be the corresponding differential map. Then dϕ(Lie(H )) is a K G-submodule of Lie(G), and hence also an ideal of Lie(G).
Proof Let Ad G : G → GL(Lie(G)) be the adjoint representation of G. We must consider Ad G (g)(dϕ(X )), for g ∈ G and X ∈ Lie(H ). As above, let t g : G → G denote conjugation by g.
Since ϕ is surjective, g = ϕ(h) for some h ∈ H , so we have Therefore, dϕ(Lie(H )) is Ad G -invariant and hence a K G-submodule of Lie(G).
Finally, let V be a K G-module with corresponding representation ρ : G → GL(V ), and let S be a G-invariant subspace of V . Then S is invariant under the action of dρ(Lie(G)). We conclude that dϕ(Lie(H )) is an ideal of Lie(G).
Recall that a morphism ϕ : H → G of algebraic groups is an isogeny if it is surjective with finite kernel. If such a map exists, we say that H is isogenous to G (this is not a symmetric relation).

Lemma 2.4
Let G be a simply connected simple classical algebraic group of rank m over an algebraically closed field K of characteristic p 0, let H be a connected algebraic group and let ϕ : H → G be an isogeny. Then ϕ is a bijection. Moreover, if dϕ = 0 then either ϕ is an isomorphism of algebraic groups, or one of the following holds: (i) G and H are both of type A m , with p dividing m + 1; (ii) G and H are both of type B m , C m or D m , with p = 2; Proof First we claim that H is also a simple group of rank m. Clearly, if N is a proper nontrivial connected normal subgroup of H , then ϕ(N ) is a proper nontrivial connected normal subgroup of G, which is not possible since G is simple. Therefore, H is simple. If T H is a maximal torus of H , then ϕ(T H ) is a maximal torus of G (see [1,Proposition 11.14], for example), and dim T H = dim ϕ(T H ). Therefore, H has rank m. Now, by comparing dimensions, we deduce that G and H have the same root system, unless (G, H ) = (B m , C m ) or (C m , B m ). Note that if p = 2 and G = B m then H is also of type B m because an isogeny from B m to C m only exists when p = 2. Similarly, if p = 2 and G = C m then H is of type C m .
To see that ϕ is a bijection (of abstract groups), first observe that ker(ϕ) Z (H ) since H is simple, so the claim is trivial if p = 2 and H = B m or C m . Now assume p = 2 if H = B m or C m . As above, G and H have the same root system. In particular, if H sc denotes the simply connected group with the same root system as H , then H sc and G are isomorphic algebraic groups (this follows from the classification of simple algebraic groups over K , using the fact that G is simply connected). Set ψ = ϕ • π , where π : H sc → H is the natural isogeny. Then ψ : H sc → G is an isogeny with kernel L Z (H sc ), so H sc /L ∼ = G as abstract groups. In particular, Z (H sc /L) ∼ = Z (G) ∼ = Z (H sc ), so L = 1 is the only possibility. Therefore ψ is injective, and thus ϕ is also injective. We conclude that ϕ is a bijection.
To complete the proof, we may assume that dϕ = 0 and (G, H, p) is not one of the cases labelled (i)-(iii) in the statement of the lemma. As above, G and H are both simple groups of the same type and rank. Since Lie(H ) is simple (see [11, Table 1]), it follows that dϕ is an isomorphism of Lie algebras and thus ϕ is an isomorphism of algebraic groups. Lemma 2.5 Let J be a closed connected subdirect product of G 1 × G 2 , where G 1 and G 2 are isomorphic simply connected simple classical algebraic groups. Then J is diagonally embedded in G 1 × G 2 , and either J = G 1 × G 2 or J ∼ = G 1 as algebraic groups.
Proof Let π i : J → G i be the i-th projection map and set L = ker(dπ 1 ) ∩ ker(dπ 2 ). Note that L = 0 since J is a closed positive-dimensional subgroup of G 1 × G 2 . Without loss of generality, we will assume that dπ 1 = 0.
First assume ker(π 1 ) is infinite. Since π 2 is injective on ker(π 1 ), we have dim ker(π 1 ) = dim π 2 (ker(π 1 )). Moreover, the surjectivity of π 2 implies that π 2 (ker(π 1 )) is an infinite normal subgroup of G 2 , so the simplicity of G 2 implies that π 2 (ker(π 1 )) = G 2 and thus dim ker(π 1 ) = dim G 2 . Therefore, dim J = dim G 1 + dim G 2 and we conclude that For the remainder, we may assume that ker(π 1 ) is finite, so π 1 is an isogeny and Lemma 2.4 implies that π 1 is a bijection and either J ∼ = G 1 , or (G 1 , J ) is one of the cases labelled (i)-(iii). In particular, J is simple and ker(π 2 ) is finite. By a further application of Lemma 2.4, we see that π 2 is also a bijection and thus J is diagonally embedded. To complete the proof it remains to show that J ∼ = G 1 as algebraic groups. Seeking a contradiction, let us assume that ker For m 2, the ideal structure of Lie(J ) is described in [6,Sect. 5]. Excluding the case where J = C m is simply connected and m 3 is odd, we observe that Lie(J ) has an irreducible socle S (as a K J-module), which immediately implies that L contains S. This is a contradiction, since L = 0. Now assume J = C m is simply connected and m 3 is odd. If G 1 = C m then J ∼ = G 1 since G 1 is simply connected, so let us assume G 1 = B m . The socle of Lie(J ) is of the form Z ⊕ M, where Z = Z (Lie(J )) is 1-dimensional and M is a nontrivial irreducible module. Without loss of generality, we may assume that M is not contained in ker(dπ 1 ), which implies that ker(dπ 1 ) = Z . Therefore im(dπ 1 ) is an ideal of Lie(G 1 ) of codimension 1, but this is not compatible with the ideal structure described in [6]. Therefore, once again we have reached a contradiction. Finally, if J = B 1 = C 1 is adjoint then Lie(J ) has an irreducible socle and we can repeat the argument given above.
Next suppose that G 1 and J are both of type A m , where p divides m + 1. We may assume that m 2. Seeking a contradiction, suppose that J is not simply connected. By inspecting [11, Table 1] we deduce that im(dπ i ) = Z (Lie(G i )) and ker(dπ i ) is the commutator subalgebra of Lie(J ) for i = 1, 2. But this implies that L = 0, which is a contradiction. To complete the proof, we may assume that G 1 and J are both of type D m , with p = 2. If m is odd then we can repeat the previous argument, using [11, Table 1], so let us assume m is even. If J is not simply connected then Lie(J ) has an irreducible socle S (see [6,Sect. 5]), which must be contained in L. Once again, this is a contradiction.
The next result is a natural generalization of Lemma 2.5.

Proposition 2.6
Let J be a closed connected subdirect product of G 1 × · · · × G t , where the G i are isomorphic simply connected simple classical algebraic groups. Then the following hold: There exists a positive integer r t such that J = J 1 · · · J r , where each J i is isomorphic to G 1 . (iii) There exist integers 0 = t 0 < t 1 < t 2 < · · · < t r = t such that J i is diagonally embedded in G t i−1 +1 × · · · × G t i .
Proof We use induction on t, noting that the case t = 1 is trivial, and Lemma 2.5 handles the case t = 2. Let us assume t 3, and let π i : J → G i be the i-th projection map. As in the proof of the previous lemma, we may assume that dπ 1 = 0. If ker(π i ) is finite for any i, then dim J = dim G 1 and thus ker(π 1 ) is also finite. Then by arguing as in the proof of Lemma 2.5, we deduce that J is diagonally embedded in G 1 × · · · × G t and J ∼ = G 1 . For the remainder, we may assume that ker(π i ) is infinite for all i. Since π i (R u (J )) is a proper normal subgroup of π i (J ) = G i for each i, it follows that J is reductive. Similarly, by considering π i (Z (J )), we deduce that Z (J ) is finite and thus J is semisimple. In particular, we may write We now consider two cases.
First assume ker(σ ) is infinite, so ker(σ ) 0 is a connected positive-dimensional normal subgroup of J . By relabelling the J i , if necessary, we may assume that ker(σ ) 0 = J 1 · · · J a for some a ∈ {1, . . . , r } (see [13,Theorem 27.5(c)], for example). Now π 1 (J i ) = G 1 for all 1 i a, so the injectivity of π 1 on ker(σ ) 0 , together with the simplicity of G 1 , implies that a = 1 and J 1 is of type G 1 . In particular, r > 1 since we are assuming that ker(π 1 ) is infinite. Also note that π i (J 2 · · · J r ) = G i for all i 2.
We claim that π 1 (J i ) = 1 for all 2 i r , so J 2 · · · J r G 2 ×· · ·× G t is a subdirect product and the result follows by induction. To justify the claim, let i ∈ {2, . . . , r } and consider π 1 | J 1 J i : It is easy to see that ker(π 1 | J 1 J i ) 0 = J i is the only possibility, so π 1 (J i ) = 1 as claimed.
To complete the proof, we may assume that ker(σ ) is finite. Now J/ ker(σ ) is connected and reductive, and it is isomorphic to a subdirect product of G 2 ×· · ·×G t . By induction, there exists s ∈ {1, . . . , t −1} such that J/ ker(σ ) = L 1 · · · L s and L i ∼ = G 1 for each i. But J = J 1 · · · J r and the J i are simple, so r = s and dim J i = dim G 1 for 123 all i (indeed, J i is isogenous to G 1 ), hence dim J = r dim G 1 . Note that r > 1 since ker(π 1 ) is infinite.

Irreducible triples
Define G, H and V as in Hypothesis 1.1. The next result records a basic observation (see [4,Remark 2]).

Lemma 2.7
If V | H is irreducible, then H does not normalize a nontrivial connected unipotent subgroup of G. In particular, H 0 is reductive.
where m 2 and the V i are irreducible K H 0 -modules that are transitively permuted under the induced action of H/H 0 .

Remark 2.8
Since the irreducibility of V | H implies that H 0 is reductive, we have acts as scalars on the V i so that they are also irreducible upon restriction to J . In particular, the irreducibility of V | H implies that the K J-composition factors of V | J are transitively permuted under the induced action of H/H 0 .
If the V i in (1) are isomorphic as K H 0 -modules, then V | H 0 is said to be homoge- We will also need the following lemma. Lemma 2.10 Let V 1 and V 2 be p-restricted irreducible K G-modules and set V = V 1 ⊗ V 2 . Then one of the following holds: (i) V is irreducible, G = B n or C n , p = 2 and V 1 and V 2 can be arranged so that V i = V G (μ i ) and μ 1 (respectively μ 2 ) has support on the short (respectively, long) roots; (ii) V has non-isomorphic composition factors.
Proof By [17, (1.6)], V is irreducible if and only if G and V satisfy the conditions in (i), so let us assume V is reducible. Write V i = V G (μ i ) and note that V has a composition factor of highest weight μ = μ 1 + μ 2 occurring with multiplicity 1. Then any other composition factor has highest weight ν = μ and the result follows.

A reduction theorem
Let G be a simple classical algebraic group with natural module W . Following [14, Sect. 1], we introduce six natural, or geometric, collections of closed subgroups, labelled C i for 1 i 6, and we set C = i C i . These subgroups are defined in terms of the underlying geometry of W , and a rough description of the subgroups in each C i collection is given in Table 3 (note that the subgroups in the collection C 5 are finite). There are two types of subgroups in the C 4 collection (indicated by the two rows in Table  3); following [4], we write C 4 = C 4 (i) ∪ C 4 (ii) accordingly. The following result is [14, Theorem 1] (we use the term non-geometric for the subgroups arising in part (ii)).
Normalizers of symplectic-type r -groups, r = p prime C 6 Classical subgroups 123

Geometric subgroups of GO(W )
In our inductive proof of Theorem 1.4, we will need to consider the subgroup structure of G = GO(W ), which is the full isometry group of a non-degenerate quadratic form on W . Here dim W = 2n 6 and thus G 0 = SO(W ) is a simple group of type D n . The notion of a geometric subgroup extends naturally to G and we can define the subgroup collections C 1 , . . . , C 6 as above. It is straightforward to check that the proof of the main theorem of [14] extends to this slightly more general situation (see [14, Theorem 1 ]), and thus Theorem 3.1 holds. In particular, any subgroup of G that is not contained in a geometric subgroup is said to be non-geometric, and these subgroups satisfy the conditions described in part (ii) of Theorem 3.1.
In the proofs of Propositions 5.10 and 5.19, we need information on the maximal non-parabolic geometric subgroups of G. This is given in the following proposition.  Table 4.
Proof Here M is a disconnected C i -subgroup of G, where i ∈ {1, 2, 3, 4} (recall that the subgroups in C 5 are finite, and there are no C 6 -subgroups in orthogonal groups). The structure of M is easily determined from its geometric description (see [4, Sect. 2.5], for example), and it is straightforward to determine whether or not M is contained in G 0 .
For example, if M ∈ C 1 then M = G U is the stabilizer of a subspace U of W (the natural K G-module), and one of the following holds (recall that M is non-parabolic, so U is not totally singular): (a) U is non-degenerate and dim U is even; (b) U is non-degenerate, dim U is odd and p = 2; (c) U is non-singular, dim U = 1 and p = 2.
These are the cases labelled (i), (ii) and (iii) in Table 4. Table 4 The non-parabolic geometric subgroups is a central product and the x i are certain involutions. More precisely, if a 1 and a 2 are both even, then we may assume that x 1 acts as a reflection on W 1 and centralizes W 2 (and vice versa for x 2 ). Therefore, x 1 , x 2 ∈ SO(W ) and thus M < G 0 . On the other hand, if a 1 is odd (so a 2 is even) then we can choose x 1 so that it acts as −1 on W 1 and centralizes W 2 , and x 2 is defined as above. Here Table 4.
The other cases are similar. For instance, suppose M is a C 3 -subgroup. Geometrically, M is the stabilizer of a decomposition W = U 1 ⊕ U 2 , where U 1 and U 2 are maximal totally singular subspaces of W , so dim U 1 = n and M = GL(U 1 ).2. Now G 0 contains an element interchanging U 1 and U 2 if and only if n is even, so M is contained in G 0 if and only if n is even, and this explains the parity condition on n recorded in Table 4 (see case (vi)). Table 4, and assume n 3 and p = 2. Set V = V G 0 (λ), where one of the following holds:

Then V extends to a representation of G, and V | M is reducible.
Proof First observe that λ is fixed under the induced action of an involutory graph automorphism of G 0 on the set of T -weights of G (where T is a maximal torus of G 0 ), so the representation V = V G 0 (λ) does indeed extend to a representation of G = G 0 .2. We will deal in turn with each of the relevant cases in Table 4 (note that case (iii) is not applicable, since we are assuming that p = 2). Seeking a contradiction, let us assume that V | M is irreducible. By Clifford theory (see Sect. Similar reasoning applies if there are four composition factors. By Clifford theory, if ν is the highest weight of a K M 0 -composition factor, then ν| M 2 = λ| M 2 or (τ 2 ·λ)| M 2 (here ν| M 2 denotes the restriction of ν to a suitable maximal torus of M 2 contained in T , and similarly for λ| M 2 and (τ 2 · λ)| M 2 ). However, μ = λ − α 1 − α 2 − · · · − α n−1 = −λ 1 + 2λ n affords the highest weight of a K M 0 -composition factor in case (a), but clearly μ| M 2 is not conjugate to λ| M 2 . Case (b) is entirely similar, using μ = λ − α 1 − · · · − α k . Therefore, in both cases we have reached a contradiction. Now assume l 2. As noted in the proof of [4, Lemma 3.2.3], up to conjugacy we have and M = M 0 τ 1 , τ 2 , where τ 1 and τ 2 act as involutory graph automorphisms on M 1 and M 2 , respectively. Let {ω 1,1 , . . . , ω 1,l } and {ω 2,1 , . . . , ω 2,n−l } be the fundamental dominant weights corresponding to the above bases of the root systems of M 1 and M 2 , respectively (here τ 1 acts as a transposition on {ω 1,1 , . . . , ω 1,l }, interchanging ω 1,l−1 and ω 1,l , and similarly τ 2 acts on {ω 2,1 , . . . , ω 2,n−l } by interchanging the weights ω 2,n−l−1 and ω 2,n−l ).
First consider (a). In terms of the above notation, we calculate that λ| M 0 = 2η l + 2ν n−l−1 . By considering the restrictions α i | M 0 , we see that λ − α n−1 and λ − α n both restrict to the weight λ| M 0 − γ n−l−1 . This weight has multiplicity 1 in the K M 0composition factor of V afforded by λ. Moreover, one checks that λ is the only Tweight μ in V such that λ| M 0 − γ n−l−1 μ| M 0 and λ| M 0 − γ n−l−1 = μ| M 0 , so there must be a K M 0 -composition factor with highest weight λ| M 0 − γ n−l−1 . However, this contradicts the homogeneity of V | M 0 . Now consider (b). Suppose k l, so l > 0 and λ| M 2 is trivial. Now the weight μ = λ − α k − · · · − α l affords the highest weight of a K M 0 -composition factor of V , but μ| M 2 is nontrivial and this contradicts the homogeneity of V | M 0 . Now assume k > l. Here λ| M 0 = 2η l + ν k−l+1 . However, the weight μ = λ − α l − α l+1 − · · · − α k affords the highest weight of a K M 0 -composition factor of V and Once again, this contradicts the homogeneity of V | M 0 .
Here n 3 is odd. Set L = (M 0 ) = A n−1 and note that V | L has exactly two composition factors. We may assume that L = U ±α 1 , . . . , U ±α n−1 . As in the proof of [4, Lemma 3.2.2], let V j be the sum of the T -weight spaces (T a maximal torus of every T -weight of V is of this form, with j = 0 or 1 (by [16,Theorem 1] and saturation; see [12,Sect. 13.4]). In particular, w 0 λ = −(τ · λ) is of this form (where w 0 is the longest word in the Weyl group of G 0 , and τ is an involutory graph automorphism of G 0 that interchanges the weights λ n−1 and λ n ). Therefore, if we and c n ∈ {0, 1}. By expressing the λ i in terms of the α i , we deduce that either λ = λ 1 , or n = 3 and λ = λ 2 or λ 3 . This immediately eliminates cases (a) and (b).
The projection of M 1 into each of the factors of this group is the natural embedding of a group of type B a in A 2a . We may assume that where α| M 0 denotes the restriction of α to a maximal torus Let P be the parabolic subgroup of M 0 , which contains the opposite Borel subgroup, with Levi factor M 1 T M 0 . We may assume that P is contained in the parabolic subgroup of G, which contains the opposite Borel subgroup of G, whose Levi factor has derived subgroup as given above. By comparing the flags of commutator subspaces of W with respect to the two unipotent radicals, we are able to determine the restrictions of sufficiently many T -weights to T M 0 in order to deduce the restrictions of the remaining simple roots. We get [17], V | M 0 is reducible, so Clifford theory implies that V has precisely two K M 0 -composition factors, with highest weights λ| M 0 and (σ · λ)| M 0 . In particular, if we set T M 1 = T M 0 ∩ M 1 , then every K M 1 -composition factor of V has highest weight λ| M 1 , whence every T M 1 -weight of V is of the form However, in case (a) we find that the weight λ − α n restricts to λ| M 1 + β 0 − β 1 , which contradicts (4).
where either r = 0 or r is a positive root of M 1 . Once again, this contradicts (4).
This completes the proof of Proposition 3.3. Table 4, and

Proposition 3.4 Let M be one of the subgroups of G = D n .2 listed in
is p-restricted. Assume that V is nontrivial and V = W, W * . Then V extends to a representation of G, and V | M is reducible.
Proof As before, λ is fixed by an involutory graph automorphism of G 0 , so V extends to a representation of G = G 0 .2. Seeking a contradiction, let us assume that V | M is irreducible. There are four cases to consider.
As in the proof of the previous proposition, we may assume that M 2 = U ±α 2 , . . . , U ±α n , and by arguing as in the proof of [4, Lemma 3.2.3] we quickly reduce to the case where V | M 0 has exactly four composition factors. Let k be minimal such that a k = 0. Then λ−α 1 −· · ·−α k affords the highest weight of a K M 2composition factor of V , which is not conjugate (via a graph automorphism of M 2 ) to the composition factor afforded by λ. This contradiction eliminates the case l = 1. Now assume l 2. As before, M = M 0 γ 1 , γ 2 , where γ 1 and γ 2 act as involutory graph automorphisms on M 1 and M 2 , respectively, and we may assume that M 1 and M 2 are as given in (2). Let {ω 1,1 , . . . , ω 1,l } and {ω 2,1 , . . . , ω 2,n−l } be the fundamental dominant weights corresponding to the bases (M 1 ) and (M 2 ) in (2), respectively. In view of (3), it is easy to see that Now, by arguing as in the proof of [4, Lemma 3.2.3] we quickly reduce to the case λ = a n−1 (λ n−1 +λ n ). By inspecting (2), we see that μ = λ−α l −α l+1 −· · ·−α n−2 −α n−1 affords the highest weight of a K M 0 -composition factor, but this is not conjugate to where V 1 and V 2 are irreducible K M 0 -modules. Since z is central, it follows that V | M 0 is homogeneous, but this is ruled out by Proposition 2.9. Finally, if l = 2 then we quickly get down to the cases λ ∈ {λ 1 +λ n−1 +λ n , λ n−1 +λ n }. The case λ = λ n−1 + λ n is ruled out as in loc. cit., and the other case is eliminated by arguing as in the final paragraph in the proof of [4,Lemma 4.3.8].
Here n 3 is odd and the argument given in the analysis of Case 5 in the proof of Proposition 3.3 can be applied.
This completes the proof of Proposition 3.4.

Geometric subgroups: the connected case
Suppose G, H and V = V G (λ) satisfy the conditions in Hypothesis 1.1. In this section and the next, we will establish Theorem 1.4 when H is contained in a geometric maximal subgroup M of G, excluding the special situation described in part (b) of Theorem 1.4. Assume V | H is irreducible, so H 0 is reductive by Lemma 2.7. Our first task is to determine the possibilities for the irreducible triple (G, M, V ). If M is connected then we can read off the relevant cases by applying [17,Theorem 1]; the cases that arise are recorded in Table 5. Similarly, if M is disconnected, we can appeal to the main theorem of [4], which yields the list of cases given in   [17] implies that the short (respectively, long) root A 2 in G 2 acts irreducibly on any p-restricted G 2 -module whose highest weight has support on the short (respectively, long) roots, but only the former is listed in the table. In addition, we also note that the highest weights in [17, Table 1] are only given up to conjugacy by a graph automorphism, and we adopt the same convention in Table  5. Note that if the graph automorphism introduces a Frobenius twist on the module, then we will list the irreducible action on the corresponding p-restricted module.

Remark 4.2
As noted in Remark 1.2, if W denotes the natural K G-module then any irreducible triple (G, M, V ) with V = W τ (for some τ ∈ Aut(G)) is also excluded in [17, Table 1]. In particular, Seitz does not list the cases (G, , with V a spin module for G. For example, the spin module for B 2 is 4-dimensional, and it corresponds to the natural symplectic representation of C 2 .

Remark 4.3
The triples of the form (G, M, V ), where (G, p) = (B n , 2) and M is a disconnected geometric maximal subgroup, are not stated explicitly in [4], but they are easily determined from the relevant list of cases in [4, Table 1] for the corresponding dual group of type C n . The only possibilities are M = B t l .Sym t (a C 2 -subgroup) with λ = λ n , or M = D n .2 and λ = λ n or i<n a i λ i . Note that G acts reducibly on W , so there are no triples involving non-geometric subgroups. Tables 5 and 6: (a) In case (ii) in Table 5  123 Table 5 M is a connected geometric subgroup  Table 5, so we may assume that k 2 (if k = 1 then V is a spin module). (c) In case (vi) in Table 5 we have G = D n , M = B m (with n = m + 1 4, p = 2) and λ = bλ k + aλ n−1 , where 1 k < n − 1, ab = 0 and a + b + n − 1 − k ≡ 0(mod p). (d) In Table 5, following [17, Table 1], for G of type A n we record the highest weight λ up to conjugacy by a graph automorphism. (e) Consider case (iv) in Table 6, where G = B n and M = D n .2 is a C 1 -subgroup.

Remark 4.4 Let us make a few comments on the cases in
Here p = 2 and the conditions on the highest weight λ = n i=1 a i λ i are given in part (b) of Remark 1.8(v). Since H is a decomposition subgroup, we may assume that a i = 0 for some i < n. (f) In case (vi) in Table 6 we have G = C n and M = C 2 m .2 is a C 2 -subgroup, where n = 2m. Moreover, λ = λ n−1 + aλ n , where 0 a < p and 2a + 3 ≡ 0(mod p). In particular, p = 2. The main result of this section is the following. First consider case (i) in Table 5 (this is the case labelled I 1 in [17, Table 1]). Here G = A n and M = C m is a C 6 -subgroup of G, where n = 2m − 1, m 2 and p = 2 (this is the natural embedding Sp(W ) < SL(W )). Note that λ = kλ 1 , k 2 and V | M = V M (kη 1 ) (see [17, We consider the irreducible triple (M, J, V M (kη 1 )). If V | J 0 is irreducible then the triple (M, J 0 , V M (kη 1 )) has to be in [17, Table 1], but it is easy to check that there are no compatible examples. Therefore J is disconnected and V | J 0 is reducible. In this situation, (M, J, V M (kη 1 )) must be one of the triples arising in the main theorems of [3,4], but once again we find that there are no such triples. We conclude that V | H is reducible in case (i). The other cases in Table 5 are very similar, although some extra care is required in case (vi). Here G = D n , M = B n−1 is a C 1 -subgroup (we can view M as the stabilizer in G of a 1-dimensional non-degenerate subspace of W ), and λ = bλ k +aλ n−1 satisfies the following conditions: (see case IV 1 in [17, Table 1]). We note that V | M = V M (bη k + aη n−1 ). As before, let J be a maximal subgroup of M containing H , and consider the irreducible triple (M, J, V M (bη k + aη n−1 )).
If V | J 0 is irreducible then by inspecting [17, Table 1] we see that the only possibility is the case labelled III 1 , where M = B 3 (so n = 4), J = G 2 , k = 2 and a + 2b + 2 ≡ 0(mod p). By (5), we also have a + b + 1 ≡ 0(mod p), so p divides b + 1, and thus p divides a, which is a contradiction since the highest weight λ = bλ k + aλ n−1 is p-restricted. Therefore J is disconnected and V | J 0 is reducible. We are now in a position to apply the main theorems in [3,4]. We deduce that the only possibility is the configuration found by Ford, with J = D n−1 .2 (see the case labelled U 2 in [8, Table II]; also see Remark 4.4(e)). Since the highest weight of V | M is bη k + aη n−1 , we must have a = 1 and 2a ≡ −2(n − 1 − k) − 1(mod p). But it is easy to see that this congruence condition is incompatible with the congruence condition in (5).   Table 7. Table 6, so G = C n , M = D n .2 and p = 2. In Proposition 5.19 we deduce that V | H is irreducible if and only if (n, p) = (4, 2), λ = λ 3 and H = C 3 1 .X with X = Z 3 or Sym 3 , and so this establishes Theorem 5.1 in this situation. The result for case (vii) will be obtained from the result for (v) via an isogeny. That is, V | H is irreducible if and only if (n, p) = (4, 2), λ = λ 3 and H = B 3 1 .Z 3 or B 3 1 .Sym 3 . Therefore, for the remainder of this section we will exclude case (v) in Table 6 from our analysis.
We begin with a couple of preliminary lemmas. Our first result will play an important role in the analysis of cases (vii), (x), (xi) and (xii) in Table 6. We will also need the following lemma when dealing with cases (ii) and (viii) in Table 6.

Lemma 5.4
Let G = A n or D n , and set μ 2 = γ · μ 1 , where μ 1 = i a i λ i is a Tweight of G and γ is an involutory graph automorphism of G. If we write μ 2 − μ 1 = Proof First assume G = A n . We may assume that γ interchanges the fundamental dominant weights λ i and λ n+1−i (1 i n/2), so μ 2 = i a n+1−i λ i . Set = n/2 . Since Table 1], for example) it follows that so for 1 j we have and if + 1 j n then We deduce that c j + c n+1− j = 0 for all 1 j . In addition, if n is odd then c +1 = 0. The result follows.
The case G = D n is very similar. Here we may assume that γ interchanges λ n−1 and λ n , so μ 2 − μ 1 = (a n − a n−1 )(λ n−1 − λ n ) and the desired result follows.
For the remainder of this section, we will assume that (G, H, V ) is given as in the statement of Theorem 1.4, so the conditions in Hypothesis 1.1 are satisfied. We will deal with each of the cases in Table 6 in turn, excluding case (v) as explained in Remark 5.2.

Proof of Theorem 5.1, part I
In this section, we will establish Theorem 5.1 in the case where H < M < G and (G, M, V ) is one of the cases labelled (i), (ii), (iii), (iv), (vi), (viii), (ix) or (xiii) in Table 6. The remaining cases, labelled (vii), (x), (xi) and (xii), will be handled in Sect. 5.2. H < M < G and (G, M, V ) is the case labelled (i) in Table 6, so G = A n , M = N G (T ) = T.Sym n+1 is the normalizer of a maximal torus T of G, and V = V G (λ k ) with 1 < k < n. Then V | H is irreducible if and only if H = T.X and X < Sym n+1 is -transitive, where = min{k, n + 1 − k}.

Proposition 5.5 Suppose
is the k-th exterior power of the natural K Gmodule W , and by duality we may assume that 1 < k (n + 1)/2. Let W(G) = N G (T )/T = Sym n+1 be the Weyl group of G.
Set S = H 0 and note that S T is a subtorus. Let S (W ) and S (V ) be the set of S-weights of W and V , respectively, so where W μ is the μ-weight space of W , and similarly V μ is the μ-weight space of V . There is a natural action of N G (S) on S (W ) and S (V ) given by (x · μ)(s) = μ(xsx −1 ). In particular, N G (S) permutes the S-weight spaces on W and V .
First assume S = T . Here the S-weight spaces on W and V are 1-dimensional, and V | H is irreducible if and only if H/T W(G) acts transitively on S (V ). This is equivalent to the condition that H/T is a k-transitive subgroup of W(G) = Sym n+1 . Indeed, we note that the S-weight vectors on V are of the form w i 1 ∧ · · · ∧ w i k , where the i j are distinct and {w 1 , . . . , w n+1 } is a basis of W consisting of S-weight vectors. This gives the desired result when S = T , so for the remainder let us assume that S is a proper subtorus of T .
Seeking a contradiction, suppose V | H is irreducible. Now H N G (S) (since S = H 0 ) and thus V | N G (S) is irreducible. In particular, N G (S) acts irreducibly on W (otherwise N G (S) lies in a parabolic subgroup of G, which would imply V | H is reducible), so N G (S) must transitively permute the set of S-weight spaces on W . Therefore, these S-weight spaces are equidimensional, whence N G (S) J < G, where J is a C 2 -subgroup of G. More precisely, J is the normalizer in G of the direct sum decomposition μ∈ S (W ) W μ . If we now consider the irreducible triple (G, J, V ) then the main theorem of [4] implies that the S-weight spaces on W are 1-dimensional, so S is a regular torus. In particular, As noted above, W(S) permutes the S-weight spaces V μ , and the irreducibility of V | N G (S) implies that this action is transitive. In particular, the S-weight spaces on V are equidimensional. In fact, we claim that they are 1-dimensional. To see this, let d denote the dimension of S and fix a basis {w 1 , . . . , w n+1 } of W comprising S-weight vectors. Then there exist integers c i, j , 1 i d, 1 j n + 1 such that for all (s 1 , . . . , s d ) ∈ (K * ) d ∼ = S and all 1 j n + 1. Without loss of generality, we may assume that the w j are ordered so that the d-tuples (c 1,1 , . . . , c 1,d ), . . . , (c n+1,1 , . . . , c n+1,d ) are in lexicographic order. (Note that these d-tuples are distinct since the S-weight spaces on W are 1-dimensional.) Then w 1 ∧ · · · ∧ w k ∈ V is an S-weight vector of weight In view of the lexicographic ordering of the tuples in (6), it follows that this S-weight has multiplicity 1, and this justifies the claim.
As previously observed, the irreducibility of V | N G (S) now implies that W(S) Sym n+1 is k-transitive, so to complete the proof of the proposition, it suffices to show that W(S) is not 2-transitive.
To see this, first let c be the codimension of S in T and let X (T ) ∼ = Z n and X (S) ∼ = Z n−c be the corresponding character groups. The sublattice S ⊥ is defined by

Now W(G) acts faithfully on X (T ) R , and W(S) = N G (S)/T stabilizes the c-
Let P be the pointwise stabilizer of S ⊥ R in W(G). By [15, Corollary A.29], P is a parabolic subgroup of W(G) = Sym n+1 , so it is a direct product of smaller degree symmetric groups. In particular, P is intransitive. Finally, we observe that W(S) normalizes P (since it stabilizes S ⊥ R ), so the intransitivity of P implies that W(S) is either intransitive, or transitive and imprimitive. In particular, W(S) is not 2-transitive. Table 6. Then V | H is reducible. . This is the case labelled U 2 in [8, Table II]. In particular, we note that a n = 1 and V | M 0 has exactly two composition factors, say
(with respect to fundamental dominant weights {η 1 , . . . , η n } for M 0 = D n ). As noted in Remark 4.4(e), we may assume that a i = 0 for some i < n.
Seeking a contradiction, let us assume that V | H is irreducible, so H M 0 since V | M 0 is reducible. Set H 1 = H ∩ M 0 and let J be a maximal subgroup of M 0 that contains H 1 . Then H = H 1 .2 and the irreducibility of V | H implies that V 1 | H 1 and We can now consider the irreducible triple (M 0 , J, V 1 ), which must be one of the cases recorded in [3,4,17]. Given the conditions on λ (in particular, the fact that a n = 1 and a i = 0 for some i < n), it is easy to see that there are no compatible examples in [3,4]. The only possible example in [17,Table1] is the case labelled IV 1 , with J = B n−1 , a = 1 and b = 0. However, we claim that the conditions in this configuration are incompatible with those that are given in Remark 4.4(e). Indeed, we have a n−1 = 0 and there is a unique k < n − 1 with a k = 0. In case IV 1 we have a k + n − k ≡ 0(mod p) and the conditions in Remark 4.4(e) yield 2a k ≡ −2(n − k) − 1(mod p). If both conditions hold, then p divides a k + n − k and 1 + 2a k + 2(n − k), so p divides 1 + a k + n − k. Clearly, this is impossible. Table 6. Then V | H is reducible.

Proposition 5.7 Suppose H < M < G and (G, M, V ) is the case labelled (vi) in
2 is a C 2 -subgroup and λ = λ n−1 + aλ n , where n = 2m, 0 a < p and 2a + 3 ≡ 0(mod p). In particular, note that p = 2 and a < p − 1. Seeking a contradiction, let us assume that V | H is irreducible. Set M 0 = C 2 m = M 1 M 2 (a direct product of two simply connected groups of type C m ) and let {ω i,1 , . . . , ω i,m } be fundamental dominant weights for M i .
Note that H 0 is a proper subgroup of M 0 since we are assuming that H is disconnected and non-maximal. Let π i : H 1 → M i be the i-th projection map and note that ker(dπ 1 ) ∩ ker(dπ 2 ) = 0 since H 1 is a closed positive-dimensional subgroup of M 0 .
Next we claim that π 2 is surjective. Suppose otherwise. Then there exists a positivedimensional maximal subgroup J 2 of M 2 such that π 2 (H 1 ) J 2 < M 2 . The irreducibility of V 1 | H 1 and V 2 | H 1 implies that π 2 (H 1 ) acts irreducibly on the K M 2modules with highest weights ω 2,m−1 + aω 2,m and (a + 1)ω 2,m , so we can consider the irreducible triples By inspecting [17, Table 1] we see that there are no compatible examples with J 2 connected. Similarly, by applying the main theorems in [3,4], there are no examples with J 2 disconnected. This is a contradiction, hence π 2 is surjective.
It follows that π 2 (H 0 1 ) = M 2 , so π 2 | H 0 : H 0 → M 2 is a bijective morphism. Moreover, ker(d(π 2 | H 0 )) = 0 since ker(dπ 2 ) = 0, so d(π 2 | H 0 ) is an isomorphism of Lie algebras and thus π 2 | H 0 is an isomorphism of algebraic groups. In particular, H 0 is a simply connected group of type C m . By Lemma 2.2 we may write π 2 | H 0 = t x for some x ∈ H 0 , where t x is an inner automorphism (conjugation by x). In addition, note that H N G (H 0 ) = H 0 C G (H 0 ) and thus V | H 0 is homogeneous. Let {η 1 , . . . , η m } be a set of fundamental dominant weights for H 0 . Then V | H 0 has composition factors isomorphic to V H 0 (η m−1 + aη m ) and V H 0 ((a + 1)η m ), which contradicts the fact that V | H 0 is homogeneous. We conclude that π 1 (H 1 ) is infinite, and similarly π 2 (H 1 ) is also infinite.
As before, we find that there are no compatible examples, which is a contradiction and thus π 1 is surjective. An entirely similar argument shows that π 2 is also surjective.

123
By the previous claim, it follows that π i (H 0 ) = M i for i = 1, 2, so H 0 is a subdirect product of the direct product M 0 = M 1 M 2 . By applying Lemma 2.5, noting that H 0 < M 0 , we deduce that H 0 ∼ = M 1 is diagonally embedded in M 1 M 2 , so we may write where τ i : Sp 2m (K ) → M i is a bijective morphism. By appealing to Lemma 2.2, we may write τ i = t x i σ q i for some x i ∈ H 0 and p-power q i (where σ q i is a standard field automorphism), and once again we note that V | H 0 is homogeneous. Note that at least one q i is equal to 1 (since H 0 is a closed subgroup M 0 ); without loss of generality we will assume q 2 = 1.
Let {η 1 , . . . , η m } be a set of fundamental dominant weights for H 0 . Then so V | H 0 has composition factors with highest weights Since V | H 0 is homogeneous, these highest weights must be equal and thus q 1 = 1. Now p < a − 1 so the modules V H 0 ((a + 1)η m ) and V H 0 (η m−1 + aη m ) are prestricted and thus Lemma 2.10 implies that V | H 0 is not homogeneous. This is a contradiction. Table 6. Then V | H is reducible.
Set H 1 = H ∩ M 0 and note that H 0 four composition factors and thus the irreducibility of V | H implies that |H : H 1 | = 4. Therefore, V | H 1 has exactly four composition factors, namely In order to proceed as in the proof of the previous proposition, we need to slightly modify our set-up. Indeed, G is the simply connected group of type D n , so M 0 = M 1 M 2 is a central product of spin groups of type D m . Since the K M 0 -module W lifts to a representation ρ : L → GL(W ), where L = L 1 L 2 is the direct product of two simply connected groups of type D m , we have M 0 = L/Y where Y = ker(ρ). In particular, there exist subgroups R R 1 L such that Let π i : R 1 → L i be the i-th projection map and note that ker(dπ 1 ) ∩ ker(dπ 2 ) = 0 since R 1 is a closed positive-dimensional subgroup of L.
The K M 0 -module V i lifts to a representation ρ i : L → GL(V i ), so we can consider and similarly π 2 (R 1 ) is irreducible on the K L 2 -modules Claim. π 1 (R 1 ) and π 2 (R 1 ) are infinite. We proceed as in the proof of Proposition 5.7. Suppose π 1 (R 1 ) is finite. Then π 2 (R 1 ) is infinite, ker(dπ 2 ) = 0 and R 0 1 ker(π 1 ), so π 2 | R 0 : R 0 → L 2 is injective. Suppose π 2 is not surjective. Then there exists a positive-dimensional maximal subgroup J 2 of L 2 such that π 2 (R 1 ) J 2 < L 2 , and we can consider the irreducible triples (L 2 , J 2 , U ) for the four K L 2 -modules U in (8). By applying the main theorems of [3,4,17] we find that there are no compatible examples (note that in the case labelled IV 1 in [17, Table 1], we require the parameters to be a = b = k = 1, hence the given congruence condition implies that m is even, which is false). This is a contradiction, hence π 2 is surjective.
Suppose π 1 is not surjective. Then there exists a positive-dimensional maximal subgroup J 1 of L 1 such that π 1 (R 1 ) J 1 < L 1 and we can consider the irreducible triples (L 1 , J 1 , U ) for the four K L 1 -modules U in (7). We have already noted that there are no compatible examples and thus π 1 is surjective. Similarly, π 2 is surjective.
We have π i (R 0 ) = L i for i = 1, 2, so R 0 is a subdirect product of L = L 1 L 2 and thus Lemma 2.5 implies that either R 0 = L, or R 0 ∼ = L 1 is diagonally embedded in L 1 L 2 . If R 0 = L then H 0 = M 0 and the irreducibility of V | H implies that H = M, which is false. Therefore R 0 is diagonally embedded, so and τ i : Spin 2m (K ) → L i is a bijective morphism. In particular, we may write τ i = t x i σ q i γ k i for some x i ∈ R 0 , p-power q i and k i ∈ {0, 1} (see Lemma 2.2). Again, we observe that (9) holds. Since R 0 is a closed subgroup of L, it follows that at least one q i is equal to 1. We may assume q 2 = 1.
Let {η 1 , . . . , η m } be a set of fundamental dominant weights for R 0 . By considering the restriction of V to R 0 , we deduce that V i | R 0 has a composition factor with highest weight μ i as follows: In view of (9), we deduce that q 1 = 1 in all four cases. By applying Lemma 2.10 it follows that V | R 0 is not homogeneous. More precisely, μ 1 affords the highest weight of a composition factor of V 1 | R 0 and if ν is the highest weight of any other composition factor of V 1 | R 0 , then ν = μ 1 and ν μ 1 . However, in view of Lemma 5.4, this is incompatible with (9). Table 6. Then V | H is reducible.

Proposition 5.9 Suppose H < M < G and (G, M, V ) is the case labelled (ii) in
Proof Here G = A n and M = A 2 m .2 is a C 4 (ii)-subgroup, where n = m(m + 2), p = 2 and m 2. Moreover, V = V G (λ) and λ = λ 2 or λ n−1 . By duality, we may assume that V = V G (λ 2 ) = 2 (W ). Seeking a contradiction, let us assume that V | H is irreducible.
Set H 1 = H ∩ M 0 and note that H 0 1 = H 0 and H = H 1 .2. Since V | H is irreducible it follows that V | H 1 has exactly two composition factors, namely Note that H 0 < M 0 since H is disconnected and non-maximal.
As in the proof of Proposition 5.8, we need to modify this initial set-up in order to proceed as we did in the proof of Proposition 5.7. Since W is a K M 0 -module, it lifts to a representation ρ : L → GL(W ), where L = L 1 L 2 is the direct product of two simply connected groups A m = SL m+1 (K ).
Suppose π 1 is not surjective. Then there exists a positive-dimensional maximal subgroup J 1 of L 1 such that π 1 (R 1 ) J 1 < L 1 and we can consider the irreducible triples (L 1 , J 1 , V L 1 (2ω 1,1 )) and (L 1 , J 1 , V L 1 (ω 1,2 )). As above, there are no compatible examples and thus π 1 is surjective. An entirely similar argument shows that π 2 is also surjective. Now π i (R 0 ) = L i for i = 1, 2, so R 0 is a subdirect product of L = L 1 L 2 and thus Lemma 2.5 implies that R 0 ∼ = L 1 is diagonally embedded in L 1 L 2 (note that H 0 < M 0 , so R 0 < L). Therefore where τ i : SL m+1 (K ) → L i is a bijective morphism. As before, we may write τ i = t x i σ q i γ k i for some x i ∈ R 0 , p-power q i and k i ∈ {0, 1}. Note that (9) holds. As before, we may assume that q 2 = 1.
Let {η 1 , . . . , η m } be a set of fundamental dominant weights for R 0 . Now V | R 0 = V 1 | R 0 ⊕ V 2 | R 0 and we calculate that V | R 0 has composition factors with the following highest weights μ 1 and μ 2 : In all four cases, (9) implies that q 1 = 1. Now V | R 0 is non-homogeneous by Lemma 2.10. More precisely, V 1 | R 0 has a composition factor of highest weight μ 1 as in the table (with q 1 = 1), occurring with multiplicity 1. If ν denotes the highest weight of any other composition factor of V 1 | R 0 , then ν = μ 1 and ν μ 1 (so μ 1 − ν = i c i α i for some c i ∈ N 0 ). Therefore, Lemma 5.4 implies that V | R 0 does not satisfy the homogeneity condition in (9) and this final contradiction completes the proof of the proposition. Table 6. Then V | H is reducible.

Proposition 5.10 Suppose H < M < G and (G, M, V ) is the case labelled (iii) in
(recall that H 0 is reductive; see Lemma 2.7). Also note that V = 2 (W ), where W is the natural K G-module. We claim that H 0 = A 1 . To see this, suppose S H 0 is a central torus. Then H N G (S) and thus the set of fixed points of S on V is H -invariant. But S lies in a maximal torus of M 0 , which has nontrivial fixed points on V , so this contradicts the irreducibility of V | H . This justifies the claim. Therefore and thus V | H 0 is homogeneous. By considering the possible embeddings of H 0 in M 0 , it follows that W | H 0 is the two-fold tensor product U ⊗U , where U is the natural We consider the irreducible triples (M 0 , J, V 1 ) and Table 1], and using the main theorems of [3,4], we deduce that J = B m−1 is the only possibility and {ξ 1 , . . . , ξ m−1 } are fundamental dominant weights for B m−1 ); see case IV 1 in [17, Table 1]. Note that if H 1 = J = B m−1 then the two K H 0 -composition factors of V | H 0 (namely V 1 | H 0 and V 2 | H 0 ) are isomorphic, but this is ruled out by Proposition 2.9. Therefore H 1 is a proper subgroup of J , so let L be a maximal subgroup of J that contains H 1 , in which case

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We now consider the irreducible triple (J, L , V B m−1 (2ξ m−1 )). In the usual way, by inspecting [3,4,17], we deduce that there are no compatible configurations and this completes the analysis of case (a).
Next consider case (b). First assume H M 0 . Since we are assuming H is disconnected and non-maximal, it follows that H J < M 0 for some maximal subgroup J of M 0 , and we may consider the irreducible triple (M 0 , J, V M 0 (η m−1 + η m )). By inspecting [3,4,17], it is easy to check that there are no compatible examples. In the same way, we deduce that H M 0 in case (c). Finally, let us consider cases (b) and (c), with H M 0 . Let J be a maximal subgroup of M = D m .2 such that Note that J is disconnected, and J is either geometric or non-geometric (as described in Sect. 3.2). Given the highest weight of V | M 0 , we can rule out the latter possibility by applying [3, Theorem 3], so we may assume J is geometric. (Note that we can appeal to [3, Theorem 3] since V | M 0 is irreducible.) The possibilities for J are determined in Proposition 3.2 and they are listed in Table 4. We now apply Proposition 3.3, which implies that V | J is reducible. This final contradiction completes the proof of the proposition. Table 6. Then V | H is reducible.

Proposition 5.11 Suppose H < M < G and (G, M, V ) is one of the cases labelled (ix) or (xiii) in
Proof First consider the case labelled (xiii). Here G = D 8 , M = C 2 2 .2 is a C 4 (ii)subgroup, p = 5 and V = V G (λ 7 ). Seeking a contradiction, let us assume that V | H is irreducible. We proceed as in the proof of Proposition 5.8.
We have π i (R 0 ) = L i for i = 1, 2, so R 0 is a subdirect product of L = L 1 L 2 and thus Lemma 2.5 implies that either R 0 = L, or R 0 ∼ = L 1 is simply connected and diagonally embedded in L. If R 0 = L then H 0 = M 0 and thus H = M (since H is disconnected), which is false. Therefore, R 0 ∼ = L 1 is diagonally embedded and thus where each τ i : Sp 4 (K ) → L i is a bijective morphism. By Lemma 2.2 we may write τ i = t x i σ q i γ k i for some x i ∈ R 0 , p-power q i and k i ∈ {0, 1}, where γ is a graph automorphism of C 2 if p = 2, otherwise γ = 1. We may assume that q 2 = 1. Since N G (H 0 ) = H 0 C G (H 0 ), it follows that V | R 0 is homogeneous.
The connected component M 0 = M 1 M 2 is a central product of two simply connected groups of type B 1 , and we note that Table 6.2]), where ω 1 and ω 2 are fundamental dominant weights for M 1 and M 2 , respectively. We leave the remaining details to the reader.

Proof of Theorem 5.1, part II
In order to complete the proof of Theorem 5.1, it remains to deal with the cases labelled (vii), (x), (xi) and (xii) in Table 6.
is one of the cases (x), (xi) or (xii). Here M is the normalizer of a tensor product decomposition W = W 1 ⊗ · · · ⊗ W t of the natural K G-module, with t = 3 or 4. Therefore, by combining Lemma 5.3 with our earlier work in Sects. 4 and 5.1, we deduce that W | H 0 is irreducible. Indeed, if W | H 0 is reducible then Lemma 5.3 implies that we may replace M by some other geometric maximal subgroup of G that does not normalize such a decomposition, in which case our earlier work implies that V | H is reducible.
In order to deal with cases (x) and (xi), we first establish some preliminary reductions. Proof Here M 0 = C 3 1 = M 1 M 2 M 3 is a central product of three simply connected groups of type C 1 . Let ω i be the fundamental dominant weight for M i , and note that where if λ = λ 2 , and if λ = λ 3 (see [4, Table 6.2]). As noted in Remark 5.12, the irreducibility of V | H implies that W | H 0 is irreducible, where W denotes the natural K G-module.
As a K M 0 -module, W lifts to a representation ρ : L → GL(W ), where L = L 1 L 2 L 3 is the direct product of three simply connected groups of type C 1 .
We need to show that R 0 is a subdirect product of L. Note that the irreducibility of W | H 0 implies that W | R 0 is also irreducible. Set J = [R 0 , R 0 ] and recall that H 0 is reductive (see Lemma 2.7), so R 0 is reductive and thus If J = C 3 1 then we are done, so assume otherwise. If J = 1 then R 0 is a torus, contradicting the irreducibility of W | R 0 . Finally, suppose J = C 2 1 or C 1 . Since R 0 is the product of J and a central torus, the irreducibility of W | R 0 implies that W | J is irreducible. This immediately implies that the projection maps π i : J → L i are surjective, so R 0 is a subdirect product of L as required. Table  6, where V = V G (λ 2 ) and p = 2. If V | H is irreducible, then H 0 = M 0 .

Lemma 5.14 Suppose H < M < G and (G, M, V ) is the case labelled (x) in
Proof As in the previous lemma, G = C 4 and M 0 = C 3 1 = M 1 M 2 M 3 is a central product of simply connected groups of type C 1 .
where the V i are given in (11). Also recall that the irreducibility of V | H implies that W | H 0 is also irreducible (see Remark 5.12).
By Lemma 5.13, R is a subdirect product of L 1 L 2 L 3 , so Proposition 2.6 implies that R 0 is isomorphic to a commuting product of simple groups of type C 1 . If R 0 is of type C 3 1 then H 0 = M 0 and we are done, so let us assume that R 0 is of type C 1 or C 2 1 . Suppose R 0 is of type C 1 . Let η 1 be the fundamental dominant weight for R 0 . By Proposition 2.6, R 0 ∼ = L 1 is simply connected and diagonally embedded in L, so we may write where τ i : Sp 2 (K ) → L i is a bijective morphism. By Lemma 2.2, τ i = t x i σ q i for some x i ∈ R 0 and p-power q i , and we may assume that q 3 = 1 (since R 0 is a closed subgroup of L). Then V | R 0 has composition factors with highest weights (2q 1 + 2q 2 )η 1 , (2q 1 + 2)η 1 and (2q 2 + 2)η 1 . Since V | R 0 is homogeneous (note that N G (H 0 ) = H 0 C G (H 0 )), it follows that q 1 = q 2 = 1. But now Lemma 2.10 contradicts the homogeneity of V | R 0 .
Finally, let us assume that R 0 = R 1 R 2 is of type C 2 1 . Let {η 1 , η 2 } be fundamental dominant weights for R 0 . Once again, Proposition 2.6 implies that R 1 and R 2 are simply connected groups of type C 1 , and without loss of generality we may assume that where τ i : Sp 2 (K ) → L i is a bijective morphism. As before, we may write τ i = t x i σ q i , so V | R 0 has composition factors with highest weights Since R 0 is a closed subgroup of L, at least one q i is equal to 1. By considering N G (H 0 ), it follows that V | R 0 is either homogeneous, or the homogeneous components of V | R 0 are conjugate under an involutory automorphism of R 0 interchanging R 1 and R 2 . However, this observation is incompatible with the weights recorded in (13). This is a contradiction.
The next lemma gives an analogous reduction for V = V G (λ 3 ) in case (x) in Table  6. Note that we include the additional case p = 2, which will be needed in Propositions 5.17 and 5.19. Proof This is entirely similar to the proof of Lemma 5.14, and we omit the details. In particular, we note that there are no additional difficulties when p = 2.
We are now in a position to settle cases (x) and (xi). Table 6 (12) holds (see [4, Table 6.2]). Let {ξ 1 , ξ 2 , ξ 3 , ξ 4 } be fundamental dominant weights for N . Since V is the restriction of the K N-module V N (ξ 3 ) to G (see the case labelled MR 4 in [17, Table 1]), Lemma 5.15 implies that H 0 = M 0 and thus H = C 3 1 .Z 3 is the only possibility. An entirely similar argument applies if λ = λ 1 + λ 4 , and once again we deduce that H = C 3 1 .Z 3 . To complete the proof of Theorem 5.1 it remains to consider cases (vii) and (xii) in Table 6. First we establish an important reduction for case (vii). Table  6 Table 4. By applying Proposition 3.4, we conclude that V | J is reducible, which is a contradiction. Now suppose H M 0 . Let J be a maximal subgroup of M 0 containing H (note that H < M 0 since we are assuming H is disconnected). Then V | J is irreducible and we can consider the possibilities for the irreducible triple (M 0 , J, V | M 0 ). By inspecting [3,4,17], given the highest weight of V | M 0 , we quickly deduce that n = 4 is the only possibility (note that if n = 3 then the highest weight of V | M 0 has at least two non-zero coefficients and it is easy to check that there are no compatible examples), Table 6

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Proof Suppose that V | H is irreducible. By the proof of the previous lemma, n = 4, λ = λ 3 and In addition, if {η 1 , . . . , η 4 } are fundamental dominant weights for M 0 = D 4 , then . Therefore, we have now reduced the problem to the case numbered (xi) in Table 6, which was handled in Proposition 5.17. In particular, we conclude that H = C 3 1 .Z 3 or C 3 1 .Sym 3 , as required.  Table 6. Table 6

Proposition 5.21 Suppose H < M < G and (G, M, V ) is the case labelled (xii) in
which is a central product of simply connected groups of type C 1 , and let ω i be the fundamental dominant weight for M i . Then [4, Table 6.
Assume that V | H is irreducible, so W | H 0 is also irreducible (see Remark 5.12), where W is the natural K G-module.
Since W is a K M 0 -module, it lifts to a representation ρ : L → GL(W ), where L = L 1 L 2 L 3 L 4 is the direct product of simply connected groups of type C 1 . Then M 0 = L/Y , where Y = ker(ρ), and there exists a subgroup R of L such that H 0 = R/Y .
Claim. H 0 is a subdirect product of M 0 .
We need to show that R 0 is a subdirect product of L. To do this, we proceed as in the proof of Lemma 5.13; the argument is very similar (using the irreducibility of W | H 0 ) and we omit the details.
Since R 0 is a subdirect product of L, Proposition 2.6 implies that R 0 is isomorphic to a commuting product of simply connected groups of type C 1 . If R 0 is of type C 4 1 then H 0 = M 0 , so we may assume that R 0 is of type C 1 , C 2 1 or C 3 1 . Suppose R 0 is of type C 1 . Let η be the fundamental dominant weight for R 0 . By Proposition 2.6, we may write where τ i : Sp 2 (K ) → L i is a bijective morphism. As before, Lemma 2.2 implies that τ i = t x i σ q i for some x i ∈ R 0 and p-power q i , and we may assume that q 4 = 1. Then V | R 0 has composition factors with highest weights (q 1 + q 2 + q 3 + 3)η, (q 1 + q 2 + 3q 3 + 1)η, (q 1 + 3q 2 + q 3 + 1)η, (3q 1 + q 2 + q 3 + 1)η.
Since N G (H 0 ) = H 0 C G (H 0 ), it follows that V | R 0 is homogeneous, so q i = 1 for all i. But Lemma 2.10 implies that V | R 0 is non-homogeneous, so we have reached a contradiction. Note that if p = 2 then one of the tensor factors in V i | R 0 is non-restricted, but we can still argue as in the proof of Lemma 2.10 by considering the tensor product of the restricted factors.
We have now reduced to the case H 0 = M 0 = C 4 1 . From the above description of V | M 0 it is clear that V | H is irreducible if and only if H = C 4 1 .X , where X < Sym 4 is transitive. The result follows.
This completes the proof of Theorem 5.1.

Non-geometric subgroups
In order to complete the proof of Theorem 1.4, it remains to determine the irreducible triples (G, H, V ) satisfying Hypothesis 1.1, where V | H 0 is reducible and H is not contained in a geometric subgroup of G. Here the latter condition implies that H is one of the non-geometric subgroups that arise in part (ii) of Theorem 3.1 in Sect. 3. In particular, W | H 0 is irreducible and tensor indecomposable, so we can apply the main theorem of [3]. (Note that if (G, p) = (C n , 2) and H fixes a non-degenerate quadratic form on W , then H is contained in a geometric C 6 -subgroup D n .2 < C n , which is a situation we dealt with in Proposition 5.19.) In view of Theorems 4.5, 5.1 and 6.1, the proof of Theorem 1.4 is complete.

Spin modules
In this section, we briefly consider the special case arising in part (b) of Theorem 1.4, where G is a simply connected group of type B n or D n (or type C n if p = 2), V is a spin module and H is a decomposition subgroup of G, as defined in the introduction. Recall that H normalizes an orthogonal decomposition W = W 1 + · · · + W t of the natural K G-module W , where the W i are pairwise orthogonal subspaces.
Our goal here is simply to highlight the difference between this very specific situation and the general case we have studied in Sects. 4, 5 and 6. We will do this by establishing a preliminary result (see Proposition 7.4); a detailed analysis of spin modules and decomposition subgroups will be given in a forthcoming paper.
Let G be a simply connected simple algebraic group of type B n or D n over an algebraically closed field K of characteristic p. For convenience, we will assume that p = 2. Let W be the natural K G-module. As before, fix a set of simple roots {α 1 , . . . , α n } and fundamental dominant weights {λ 1 , . . . , λ n } for G. We will assume that n 3 if G = B n and n 5 if G = D n (note that the spin modules for D 4 are excluded in Hypothesis 1.1; see Remark 1.2). We may write Similarly, for a subgroup J of G we set J = J Z/Z G. Let V be a spin module for G. In terms of highest weights, either V = V G (λ n ), or G = D n and V = V G (λ n−1 ) (in the latter case, note that V G (λ n−1 ) = V G (λ n ) τ , where τ is a graph automorphism of G). The next result is well known (see [4,Lemma 2.3.2] for a proof).
Let W = W 1 ⊥ W 2 be an orthogonal decomposition, where W 1 and W 2 are non-degenerate subspaces with dim W i 3. Let H be the stabilizer in G of this decomposition, so H = H 0 .2 and H 0 is a central product of two simply connected orthogonal groups. More precisely, where t 2 and each W i is a non-degenerate subspace with dim W i 3. Let H be the stabilizer in G of this decomposition, so If t = 2 then V | H is irreducible by Proposition 7.2, so let us assume t 3. We claim that V | H is still irreducible. To see this, we first handle the special case where the W i are equidimensional.
Here the proof of [4, Lemma 4.3.2] goes through unchanged (the symmetric group Sym t in the C 2 -subgroup plays no role in the argument) and we deduce that V | H is irreducible. Now assume d = 2l is even, so H = D t l .2 t−1 and we may write H 0 = X 1 · · · X t , where each X i = D l is simply connected. Here the elementary abelian 2-group 2 t−1 is generated by involutions z 1 , . . . , z t−1 , where z i acts as a graph automorphism on X i and X i+1 , and centralizes the remaining factors of H 0 . Now V 1 ⊗ · · · ⊗ V t is a composition factor of V | H 0 , where each V i is a spin module for X i . By repeatedly applying the z i ∈ H to conjugate this composition factor, we deduce that V | H 0 has at least 2 t−1 distinct, H -conjugate K H 0 -composition factors. Since 2 t−1 · dim(V 1 ⊗ · · · ⊗ V t ) = 2 t−1 · 2 t (l−1) = 2 tl−1 = dim V (see Lemma 7.1) we conclude that V | H is irreducible.
We can now establish our main result for spin modules and decomposition subgroups. (17), and assume dim W i 3 for each i. Then V | H is irreducible.

Proposition 7.4 Let H be the stabilizer in G of the decomposition in
Proof Let d 1 , . . . , d s be the distinct dimensions of the summands in (17), and let a i be the number of summands of dimension d i . If s > 1 then we may assume that d i < d i+1 for all 1 i < s. We may re-order and re-label the subspaces in (17) so that W = (W 1,1 ⊥ · · · ⊥ W 1,a 1 ) ⊥ · · · ⊥ (W s,1 ⊥ · · · ⊥ W s,a s ), (18) where dim W i, j = d i for all i, j. Then H = (GO(W 1,1 ) × · · · × GO(W 1,a 1 ) × · · · × GO(W s,1 ) × · · · × GO(W s,a s )) ∩ G.
We proceed by induction on s.
First assume U 1 and U 2 are odd-dimensional. Here [17, Table 1] indicates that V | M 0 = V 1 ⊗ V 2 is irreducible, where V i is the spin module for M i . The K M 0module V lifts to a representation ϕ : L → GL(V ), so we can consider V | R 1 . By induction, V i | π i (R 1 ) is irreducible for i = 1, 2, so V | R 1 is irreducible and thus V | H 1 is also irreducible. The result follows.
Next suppose dim U 1 is even and dim U 2 is odd, so [4, Proposition 3.1.1] implies that V | M 0 has exactly two composition factors, namely where V 1 and V 1 are the two spin modules for M 1 , and V 2 is the spin module for M 2 . Here the K M 0 -modules V 1 ⊗ V 2 and V 1 ⊗ V 2 lift to representations ϕ : L → GL(V 1 ⊗ V 2 ) and ϕ : L → GL(V 1 ⊗ V 2 ), so we can consider (V 1 ⊗ V 2 )| R 1 and (V 1 ⊗ V 2 )| R 1 . By induction, V 2 | π 2 (R 1 ) is irreducible, and π 1 (R 1 ) acts irreducibly on V 1 and V 1 . Therefore, V | R 1 has precisely two composition factors, which are interchanged by an element in R 1 .2 that acts as a graph automorphism on L 1 and centralizes L 2 . Therefore, V | H is irreducible. An entirely similar argument applies if dim U 1 is odd and dim U 2 is even.
Finally, suppose U 1 and U 2 are both even-dimensional. Here where V i and V i are the two spin modules for M i , and the inductive hypothesis implies that π i (R 1 ) acts irreducibly on V i and V i , for i = 1, 2. As before, it follows that V | R 1 has exactly two composition factors, which are interchanged by an element in R 1 .2 that acts simultaneously as a graph automorphism on both L 1 and L 2 . Therefore, the two K H 1 -composition factors of V are H -conjugate, and thus V | H is irreducible.
In view of the previous proposition, we can easily construct chains of positivedimensional closed subgroups H k < H k−1 < · · · < H 1 < G such that V | H i is irreducible for all i. For instance, take any sequence of successive refinements of a fixed orthogonal decomposition of W such that each refinement is also an orthogonal decomposition that only contains subspaces of dimension at least three. Then the stabilizers in G of these decompositions form a chain of subgroups with the desired irreducibility property. In particular, such a chain can be arbitrarily long. This is in stark contrast to the general situation, where the length of an irreducible chain is at most five (see Theorem 1.9, which will be proved in the next section).

Irreducible chains
In this final section we prove Theorem 1.9. Recall that if G is a simple algebraic group and V = V G (λ) is a nontrivial p-restricted irreducible K G-module, then we write = (G, V ) for the length of the longest chain of closed positive-dimensional subgroups H < H −1 < · · · < H 2 < H 1 = G such that V | H is irreducible.
As noted in the previous section, if G is an orthogonal group (or a symplectic group with p = 2) and V is a spin module, then (G, V ) can be arbitrarily large (one can simply take an appropriate chain of decomposition subgroups, for example). Similarly, if V = W or W * (where W is the natural K G-module) then (G, V ) is unbounded. For instance, if G = Sp 2n (K ) then set H i = Sp 2 (K ) X i , where X i Sym n is transitive. The transitivity of X i implies that W | H i is irreducible, and it is easy to see that if n is sufficiently large then we can find arbitrarily long chains of transitive subgroups X i < X i−1 < · · · < X 1 = Sym n .
In fact, if we choose n appropriately, then we may assume that each X i is 3-transitive (see Remark 8.2 below). Now let us assume that V = W, W * , and also assume that V is not a spin module. In this situation, it is natural to ask whether or not (G, V ) is bounded above by an absolute constant. Our main theorem is the following, which immediately yields Theorem 1.9. (Note that in Table 8, T is a maximal torus of G and M n is the simple Mathieu group of degree n). More precisely, excluding the cases in (ii), we have (G, V ) 3, unless (G, V ) is one of the cases listed in Table 8.

Remark 8.2
The cases in part (ii) of Theorem 8.1 are genuine exceptions; for suitable values of n, (G, V ) can be arbitrarily large. By duality, we only need to consider the cases λ = λ 2 and λ 3 . Recall that if H = T n .X < G, then V G (λ 3 )| H is irreducible if and only if X Sym n+1 is 3-transitive (and similarly, X has to be 2-transitive if λ = λ 2 ); see Proposition 5.5. Suppose n = q = 2 e for some positive integer e 2. The finite simple group PSL 2 (q) has a faithful 3-transitive action on the projective line F q ∪ {∞}, which extends to a faithful action of its automorphism group P L 2 (q) = PSL 2 (q).Z e . Therefore, PSL 2 (q).d P L 2 (q) is a 3-transitive subgroup of Sym n+1 for every divisor d of e. In particular, by choosing e appropriately we can construct arbitrarily long chains of 3-transitive subgroups of Sym n+1 , and each of the corresponding subgroups T n .X < G acts irreducibly on V G (λ 2 ) and V G (λ 3 ).
Using the classification of finite simple groups, it can be shown that Sym n+1 and the alternating group Alt n+1 (for n 5) are the only 4-transitive groups of degree n + 1, unless n ∈ {10, 11, 22, 23} in which case the Mathieu group M n+1 is also 4-transitive (see [5,Theorem 4.11], for example). Similarly, if t 5 then Sym n+1 and Alt n+1 (for n t + 1) are the only t-transitive groups of degree n + 1, with the single exception of M 24 when t = 5 and n = 23. Since s 4, it follows that either s = 4 and n ∈ {10, 11, 22, 23}, or s = 5 and n = 23. In each case, X 4 = M n+1 , X 3 = Alt n+1 , X 2 = Sym n+1 and no proper positive-dimensional subgroup of H 4 acts irreducibly on V . These special cases are recorded in Table 8.
To complete the proof of Theorem 8.1 for G = A n , we may assume that (G, H 4 , V ) is an irreducible triple with H 4 connected. The possibilities are recorded in [17, Table 1], and in view of (19) we see that the relevant cases therein are labelled I 1 , I 1 , I 2 , I 3 , I 4 , I 5 , I 12 .
Consider the irreducible triple (G, H 3 , V ). If H 3 is connected, then (G, H 3 , V ) also corresponds to one of the cases in (20) This completes the proof of Theorem 8.1 for G = A n . The other cases are similar, and we leave it to the reader to check the details.