Smooth representations of unit groups of split basic algebras over non-Archimedean local fields

Let 𝒜 be a finite-dimensional split basic 𝕜-algebra. Then every subalgebra of 𝒜 is also a split basic 𝕜-algebra.

Proof

Let ℬ be a subalgebra of 𝒜, and let J ⁢ ( B ) denote the Jacobson radical of ℬ. As J ⁢ ( B ) = B ∩ J ⁢ ( A ) , it is clear that ℬ is a basic 𝕜-algebra; moreover, the 𝕜-algebra B / J ⁢ ( B ) is naturally isomorphic to the subalgebra ( B + J ⁢ ( A ) ) / J ⁢ ( A ) of the semisimple 𝕜-algebra A / J ⁢ ( A ) . Therefore, without loss of generality, we may assume that 𝒜 is a split basic semisimple 𝕜-algebra. Since ℬ is a basic semisimple 𝕜-algebra, there are non-zero orthogonal idempotents e 1 ′ , … , e m ′ ∈ B such that B = 𝕜 1 ⁢ e 1 ′ ⊕ ⋯ ⊕ 𝕜 n ⁢ e m ′ , where 𝕜 1 , … , 𝕜 n are finite field extensions of 𝕜. On the other hand, let e 1 , … , e n ∈ A be non-zero orthogonal idempotents such that A = 𝕜 ⁢ e 1 ⊕ ⋯ ⊕ 𝕜 ⁢ e n . It is straightforward to check that there exists a subset partition I 1 , … , I m of { 1 , … , n } such that e j ′ = ∑ i ∈ I j e i , and thus

It follows that 𝕜 j ⁢ e j ′ ⁢ e i = 𝕜 j ⁢ e i ⊆ A ⁢ e i = 𝕜 ⁢ e i for all i ∈ I j and all 1 ≤ j ≤ m , which clearly implies that 𝕜 j = 𝕜 for all 1 ≤ j ≤ m . ∎

Let 𝒜 be a finite-dimensional split basic 𝕜-algebra, let 𝒟 be a diagonal subalgebra of 𝒜, and let e 1 , … , e n ∈ A be non-zero orthogonal idempotents such that D = 𝕜 ⁢ e 1 ⊕ ⋯ ⊕ 𝕜 ⁢ e n . The following properties hold.

1. If 𝒱 is an (arbitrary) 𝒟-bimodule, then 𝒱 decomposes as a direct sum of the (non-zero) homogeneous sub-bimodules e i ⁢ V ⁢ e j for 1 ≤ i , j ≤ n .

2. For every sub-bimodule V 1 of a 𝒟-bimodule 𝒱, there exists a sub-bimodule V 2 of 𝒱 such that V = V 1 ⊕ V 2 .

3. Every 𝒟-bimodule decomposes as a direct sum of one-dimensional sub-bimodules.

4. If 𝒱 is a one-dimensional 𝒟-bimodule and v ∈ V is such that V = 𝕜 ⁢ v , then there exist uniquely determined 1 ≤ i , j ≤ n such that v = e i ⁢ v ⁢ e j .

Let M n ⁢ ( 𝕜 ) denote the 𝕜-algebra consisting of all n × n matrices with entries in 𝕜, and let A = B n ⁢ ( 𝕜 ) denote the Borel subalgebra of M n ⁢ ( 𝕜 ) consisting of all upper-triangular matrices; hence G = A × is the standard Borel subgroup B n ⁢ ( 𝕜 ) of the general linear group GL n ⁡ ( 𝕜 ) (consisting of all invertible matrices in M n ⁢ ( 𝕜 ) ). In this case, T ≤ G is the standard torus consisting of all diagonal matrices, and P = 1 + J ⁢ ( A ) is the standard unitriangular group; we note that J ⁢ ( A ) is the nilpotent ideal of 𝒜 consisting of all upper-triangular matrices with zeroes on the main diagonal. In this case, for each 1 ≤ i ≤ n , the idempotent e i ∈ A can be chosen to be the elementary matrices e i = e i , i having a unique non-zero entry (equal to 1) in the position ( i , i ) , and thus D = 𝕜 ⁢ e 1 ⊕ ⋯ ⊕ 𝕜 ⁢ e n is the subalgebra of 𝒜 consisting of all diagonal matrices.

Let 𝒜 be a finite-dimensional split basic algebra over a perfect non-Archimedean local field 𝕜, let G = A × be the unit group of 𝒜, and let 𝑉 be an irreducible smooth 𝐺-module. Then there exists a subalgebra ℬ of 𝒜 and a smooth character of the unit group H = B × such that V ≅ c - Ind H G ⁡ ( C ϑ ) .

Let 𝕜 be an arbitrary non-Archimedean local field, and let A = B 2 ⁢ ( 𝕜 ) be the standard Borel algebra of M 2 ⁢ ( 𝕜 ) ; hence G = A × is the standard Borel subgroup B 2 ⁢ ( 𝕜 ) of GL 2 ⁡ ( 𝕜 ) consisting of all upper-triangular matrices. It is obvious that 𝐺 is the semidirect product G = T ⁢ P , where T ≅ 𝕜 × × 𝕜 × is the subgroup of 𝐺 consisting of all diagonal matrices and P ≅ 𝕜 + is the abelian normal subgroup of 𝐺 consisting of all unipotent matrices.

We consider the conjugation action of 𝐺 on P ∘ : for every g ∈ G and every ϑ ∈ P ∘ , we define ϑ g ∈ P ∘ by ϑ g ⁢ ( x ) = ϑ ⁢ ( g ⁢ x ⁢ g - 1 ) for all x ∈ P . It is clear that there are exactly two 𝐺-orbits on P ∘ , namely, the singleton { 1 P } consisting of the trivial character of 𝑃, and its complement P ∘ ∖ { 1 P } . Both these 𝐺-orbits are locally closed, and thus by [14, Corollaire 2 au Théorème 3], there are two distinct families of irreducible smooth 𝐺-modules each one corresponding to one of these two 𝐺-orbits. On the one hand, one has the family consisting of one-dimensional 𝐺-modules corresponding to the smooth characters ϑ ∈ G ∘ which satisfy P ⊆ ker ⁡ ( ϑ ) ; indeed, 𝑃 lies in the kernel of every smooth character of 𝐺.

On the other hand, there is a family corresponding to a fixed non-trivial smooth character ϑ ∈ P ∘ . In this case, the centraliser^[1] G ϑ is the (internal) direct product G ϑ = Z ⁢ P , where Z = Z ⁢ ( G ) is the centre of 𝐺; note that Z ≅ 𝕜 × , whereas P ≅ 𝕜 + (hence G ϑ ≅ 𝕜 × × 𝕜 + ). Since G ϑ is abelian, every irreducible smooth G ϑ -module is one-dimensional. By Rodier’s result, for every smooth character τ ∈ ( G ϑ ) ∘ satisfying τ P = ϑ , the c-induced smooth G ϑ -module c - Ind G ϑ G ⁡ ( C τ ) is irreducible (and clearly of dimension at least 2), and moreover, the mapping τ ↦ c - Ind G ϑ G ⁡ ( C τ ) defines a one-to-one correspondence between smooth characters of G ϑ satisfying τ P = ϑ and irreducible smooth 𝐺-modules with dimension at least 2.

Now, let 𝒪 denote the ring of algebraic integers of 𝕜, let

where O × denotes the unit group of 𝒪, and consider the subgroup G 0 = T 0 ⁢ G ϑ of 𝐺; note that G 0 is an open (hence also a closed) subgroup of 𝐺. Let τ ∈ ( G ϑ ) ∘ be an arbitrary smooth character of G ϑ satisfying τ P = ϑ , and consider the c-induced 𝐺-module V = c - Ind G ϑ G ⁡ ( C τ ) . By the transitivity of c-induction (see [1, Proposition 2.25 (b)]), V = c - Ind G 0 G ⁡ ( V 0 ) , where V 0 = c - Ind G ϑ G 0 ⁡ ( C τ ) . Since 𝑉 is irreducible, it is obvious that V 0 is an irreducible smooth G 0 -module. Since G 0 is an open subgroup of 𝐺, it follows from [4, Lemma 2.5] that V 0 is naturally embedded as a vector subspace of 𝑉 and that we have a direct sum decomposition

where 𝒢 is a complete set of representatives of the coset space G / G 0 .

Let 𝐾 be a sufficiently small compact open subgroup of G 0 such that ( V 0 ) K ≠ 0 (hence V K ≠ 0 ); note that we can choose 𝐾 to be the compact open subgroup consisting of the matrices x ∈ G 0 such that x 1 , 1 , x 2 , 2 ∈ O × and x 1 , 2 ∈ ϖ r ⁢ O for a suitable r ∈ Z (here ϖ ∈ O is the uniformiser of 𝕜 so that ϖ ⁢ O is the unique maximal ideal of 𝒪). Let

then, for every m ∈ N , we have g m ∉ G 0 and g m ⁢ K ⁢ g m - 1 ⊆ K . It follows that ( g m ⁢ V 0 ) K = g m ⁢ V 0 K ≠ { 0 } for every m ∈ N , and thus

has infinite dimension. Therefore, the smooth 𝐺-module 𝑉 is not admissible, and hence it follows from Rodier’s theorem that an irreducible smooth 𝐺-module is admissible if and only if it is one-dimensional.

Henceforth, we fix the following notation which we will use repeatedly, without always recalling their meaning.

• 𝕜 is a perfect non-Archimedean local field.

• 𝒜 is a finite-dimensional split basic 𝕜-algebra.

• G = A × is the unit group of 𝒜.

• J = J ⁢ ( A ) is the Jacobson radical of 𝒜, and P = 1 + J .

• 𝒟 is a diagonal subalgebra of 𝒜, and T = D × is a diagonal subgroup of 𝐺.

Let 𝑉 be an arbitrary smooth 𝐺-module, and let 𝑄 be an ideal subgroup of 𝐺. Then Spec Q ⁡ ( V ) is a 𝐺-invariant subset of Q ∘ and

is a 𝐺-submodule of 𝑉 (that is, a 𝐺-invariant vector subspace of 𝑉). In particular, if 𝑉 is irreducible and Spec Q ⁡ ( V ) is non-empty, then [ Q , Q ] ¯ acts trivially on 𝑉, and hence 𝑉 becomes an irreducible smooth ( G / [ Q , Q ] ¯ ) -module in a natural way.

Proof

For the first assertion, it is enough to observe that V ⁢ ( ϑ g ) = g - 1 ⁢ V ⁢ ( ϑ ) for all ϑ ∈ Q ∘ and all g ∈ G . For the second assertion, we start by observing that V 0 ≠ V (otherwise, V ⁢ ( ϑ ) = V for all ϑ ∈ Spec Q ⁡ ( V ) ), and thus V 0 = { 0 } (because 𝑉 is irreducible). It follows that the natural linear map

is injective, and thus [ Q , Q ] ¯ acts trivially on 𝑉 (because [ Q , Q ] ¯ ⊆ ker ⁡ ( ϑ ) for all ϑ ∈ Q ∘ ). ∎

Let 𝑄 be an ideal subgroup of 𝐺, and let ϑ ∈ Q ∘ . Then the 𝐺-orbit ϑ G = { ϑ g : g ∈ G } is a locally closed subset of Q ∘ .

Proof

We claim the Q ∘ has the structure of an affine algebraic 𝕜-variety. To show this, we consider the affine algebraic group Q / [ Q , Q ] ¯ ; since 𝕜 is perfect, both 𝑄 and [ Q , Q ] ¯ are 𝕜-split, and hence Q / [ Q , Q ] ¯ is also 𝕜-split (see [2, Corollary 15.5]). On the one hand, if 𝕜 has characteristic zero, then the usual exponential map shows that Q ∘ is naturally isomorphic to the vector space over 𝕜 which is dual to the Lie algebra of Q / [ Q , Q ] ¯ , and hence it has a natural structure of affine algebraic 𝕜-variety. On the other hand, if 𝕜 has prime characteristic, then Q / [ Q , Q ] ¯ is isomorphic as a topological group to its dual Q ∘ (see [13, Corollary 5.16]), and thus Q ∘ also adquires the structure of an affine algebraic 𝕜-variety, as required. Since 𝐺 acts 𝕜-morphically on Q ∘ , it follows by [2, Proposition 6.7] that the 𝐺-orbit ϑ G is a locally closed subset of Q ∘ (with respect to the Zariski topology), and the result follows because the Zariski topology is coarser than the topology of Q ∘ as an ℓ c -group. ∎

Let 𝑉 be an irreducible smooth 𝐺-module, let 𝑄 be an ideal subgroup of 𝐺, and let ϑ ∈ Q ∘ be such that V ϑ ≠ { 0 } . Then Spec Q ⁡ ( V ) = ϑ G ; moreover, V ϑ is an irreducible smooth G ϑ -module and V ≅ c - Ind G ϑ G ⁡ ( V ϑ ) .

Proof

We consider the closure [ Q , Q ] ¯ of the commutator subgroup of 𝑄 and the quotient group G ¯ = G / [ Q , Q ] ¯ ; then G ¯ is an ℓ-group, and Q ¯ = Q / [ Q , Q ] ¯ is an abelian normal subgroup of G ¯ . By Lemma 3, 𝑉 is an irreducible smooth G ¯ -module; moreover, Lemma 4 clearly implies that every G ¯ -orbit is a locally closed subset of Q ¯ ∘ ≅ Q ∘ . Therefore, it follows from [14, Corollaire 1 au Théorème 3] that Spec Q ¯ ⁡ ( V ) = ϑ G ¯ and that V ϑ is an irreducible smooth G ¯ ϑ -module; moreover, [14, Corollaire 2 au Théorème 3] implies that

The result follows because ϑ G = ϑ G ¯ , [ Q , Q ] ¯ ⊆ G ϑ and G ¯ ϑ = G ϑ / [ Q , Q ] ¯ . ∎

Let 𝑉 be an irreducible smooth 𝐺-module, and let 𝑄 be an arbitrary algebra subgroup of 𝐺. Then the smooth 𝑄-module Res Q G ⁡ ( V ) has an irreducible quotient.

Proof

By [3, Theorem 1.3], every irreducible smooth 𝑄-module is admissible. Since 𝑄 is also an ℓ c -group, Res Q G ⁡ ( V ) has an irreducible quotient by [3, Corollary 4.8]. ∎

Let 𝑉 be an irreducible smooth 𝐺-module, and let 𝑊 be an irreducible quotient of Res P G ⁡ ( V ) . Suppose that 𝑊 is one-dimensional, and let ϑ ∈ P ∘ be the character afforded by 𝑊. Then 𝜗 is 𝐺-invariant if and only if 𝑉 is one-dimensional.

Proof

It is clear that 𝜗 is 𝐺-invariant whenever 𝑉 is one-dimensional. Conversely, suppose that 𝜗 is 𝐺-invariant. Then V ⁢ ( ϑ ) is a 𝐺-invariant vector subspace of 𝑉, and hence either V ⁢ ( ϑ ) = { 0 } or V ⁢ ( ϑ ) = V (because 𝑉 is irreducible). Since 𝑃 is an ℓ c -group, [1, Proposition 2.35] implies that W ϑ = W / W ⁢ ( ϑ ) is an epimorphic image of V ϑ = V / V ⁢ ( ϑ ) , and thus V ϑ ≠ 0 (because W ⁢ ( ϑ ) = { 0 } , and hence W ≅ W ϑ ). Therefore, we must have V ⁢ ( ϑ ) = { 0 } , and thus

It follows that the restriction Res T G ⁡ ( V ) of 𝑉 to the diagonal subgroup 𝑇 of 𝐺 is irreducible; indeed, if V ′ is 𝑇-submodule of Res T G ⁡ ( V ) , then V ′ is also a 𝐺-submodule of 𝑉 (because every vector subspace of 𝑉 is 𝑃-invariant), and so either V ′ = { 0 } or V ′ = V . Since 𝑇 is an abelian ℓ-group, Schur’s lemma implies that 𝑉 is one-dimensional, and this completes the proof. ∎

For every ideal subgroup 𝑄 of 𝐺 and every ϑ ∈ Q ∘ , the centraliser T ϑ is the unit group of a subalgebra of 𝒟.

Proof

If J ⊆ J ⁢ ( A ) is an ideal of 𝒜 and ϑ ∈ ( 1 + J ) ∘ , then it is straightforward to check that

is a subalgebra of 𝒟 with ( D ϑ ) × = T ϑ . By way of example, let d , d ′ ∈ D ϑ , and let a ∈ J be arbitrary. Then, since a ⁢ ( 1 + d ⁢ a ) - 1 = ( 1 + a ⁢ d ) - 1 ⁢ a , we deduce that

Let 𝑉 be an irreducible smooth 𝐺-module, and let 𝑊 be an irreducible quotient of Res P G ⁡ ( V ) . Suppose that 𝑊 is one-dimensional, and let ϑ ∈ P ∘ be the character afforded by 𝑊. Then G ϑ is the unit group of some subalgebra of 𝒜, and V ϑ is a one-dimensional smooth G ϑ -module such that

Proof

We have already proved that the smooth G ϑ -module V ϑ is one-dimensional and that V = c - Ind G ϑ G ⁡ ( V ϑ ) . If 𝒟 is a diagonal subalgebra of 𝒜 and T = D × , then the previous proposition ensures that T ϑ is the unit group of some subalgebra D ϑ of 𝒟. Since G ϑ = T ϑ ⁢ P and since J ⁢ ( A ) is an ideal of 𝒜, it follows that A ϑ = D ϑ ⊕ J ⁢ ( A ) is a subalgebra of 𝒜 and that G ϑ = ( A ϑ ) × is the unit group of A ϑ . ∎

Let ς ∈ N ∘ be 𝑃-invariant, and define

Then J ς is a subalgebra of 𝒥 satisfying J 2 ⊆ J ς and dim ⁡ J ς ≥ dim ⁡ J - 1 . Furthermore, if we define the map φ ς : P → Q ∘ by the rule

then φ ς is a group homomorphism with ker ⁡ ( φ ς ) = 1 + J ς and φ ς ⁢ ( P ) ⊆ N ⟂ , where N ⟂ = { τ ∈ Q ∘ : N ⊆ ker ⁡ ( τ ) } is the orthogonal subgroup of 𝑁 in Q ∘ ; hence φ ς naturally defines a group homomorphism φ ¯ ς : P → ( Q / N ) ∘ .

Proof

We first observe that the map φ ς is a well-defined group homomorphism. On the one hand, we have [ P , Q ] ⊆ [ 1 + J , 1 + J n - 1 ] ⊆ 1 + J n = N . On the other hand, since [ g , h ⁢ k ] = [ g , k ] ⁢ [ g , h ] k , we deduce that

for all g ∈ P and all h , k ∈ Q ; we recall that 𝜍 is 𝑃-invariant. It follows that, for every g ∈ P , the map φ ς ⁢ ( g ) : Q → C × is indeed a (smooth) character of 𝑄. Similarly, since [ g ⁢ h , k ] = [ g , k ] h ⁢ [ h , k ] , we have

for all g , h ∈ P and all k ∈ Q , and so φ ς is a group homomorphism.

Now, since [ P , N ] ⊆ ker ⁡ ( ς ) (because 𝜍 is 𝑃-invariant), the image φ ς ⁢ ( P ) clearly lies in N ⟂ ; moreover, it is obvious (by the definition) that ker ⁡ ( φ ς ) = 1 + J ς . Let a ∈ J and α ∈ 𝕜 be arbitrary. As in the proof of [3, Proposition 3.1 (b), p. 547], we deduce that

for all u ∈ J n - 1 . Since 𝜍 is 𝑃-invariant (and [ 1 + J , 1 + J n ] = [ P , N ] ), we conclude that

for all u ∈ J n - 1 , and this clearly implies that α ⁢ a ∈ J ς for all α ∈ 𝕜 and all a ∈ J ς . On the other hand, [9, Theorem 1.4] implies that

and thus J 2 ⊆ J ς . Since

we see that u + v ∈ J ς for all u , v ∈ J ς . It follows that J ς is an ideal of 𝒥 with J 2 ⊆ J ς .

Finally, note that N ⟂ ≅ ( Q / N ) ∘ and that

Therefore, we get an isomorphism of abelian groups N ⟂ ≅ ( 𝕜 + ) ∘ , and hence N ⟂ acquires the structure of a vector space over 𝕜, where the scalar multiplication is defined by

On the other hand, since 𝕜 is a self-dual field, there is a 𝕜-linear isomorphism 𝕜 ≅ ( 𝕜 + ) ∘ , and thus N ⟂ is one-dimensional. Furthermore, the argument used above can be repeated to show that the mapping a ↦ φ ς ⁢ ( 1 + a ) defines a 𝕜-linear homomorphism φ ^ ς : J → N ⟂ with kernel J ς ; we note that † implies that

It follows that dim ⁡ J - dim ⁡ J ς ≤ 1 , and this completes the proof. ∎

Let ς ∈ N ∘ be 𝐺-invariant, let ℐ be an ideal of 𝒜 with J 2 ⊆ I ⊆ J , and let J ς ⊆ J be defined as in Lemma 8. Then I ς = I ∩ J ς is an ideal of 𝒜.

Proof

The result is obvious in the case where J ς = J ; hence we assume that J ς ≠ J so that dim ⁡ J ς = dim ⁡ J - 1 (by the previous lemma). The result is also clearly true in the case where I ⊆ J ς ; thus we may assume that J ς + I = J , which implies that dim ⁡ I ς = dim ⁡ I - 1 . We now proceed by induction on dim ⁡ I , the result being obvious if dim ⁡ I = dim ⁡ J 2 + 1 ; indeed, since J 2 ⊆ I ∩ J ς ⊆ I , either

Therefore, we may assume that dim ⁡ I ≥ dim ⁡ J 2 + 2 and that the result is true whenever I ′ is an ideal of 𝒜 with J 2 ⊆ I ′ ⊆ J and dim ⁡ I ′ < dim ⁡ I .

Let I ς ′ be the unique ideal of 𝒜 which is maximal with respect to the condition I ς ′ ⊆ I ς ; hence we must prove that I ς ′ = I ς . Since I ς ′ is clearly a 𝒟-bimodule, Lemma 2 ensures that I = I ς ′ ⊕ V for some sub-bimodule 𝒱 of ℐ. Let V ς = V ∩ J ς , and note that I ς = I ς ′ ⊕ V ς ; hence I ς ′ = I ς if and only if V ς = { 0 } . By way of contradiction, we assume that V ς ≠ { 0 } ; note that V ς ≠ V (otherwise, I ς = I ). Since 𝑄 is a 𝑇-invariant subgroup of 𝐺 (because it is an ideal subgroup of 𝐺), we deduce that

for all a ∈ V ς , all t ∈ T and all h ∈ Q . Since V ς ⊆ J ς (and since 𝒱 is clearly 𝑇-invariant), it follows that V ς is a 𝑇-invariant vector subspace of 𝒱.

On the other hand, suppose that V ′ ≠ { 0 } is a proper sub-bimodule of 𝒱, and let I ′ = I ς ′ + V ′ . Then I ′ is an ideal of 𝒜 with I ′ ⊊ I , and thus I ′ ∩ J ς is an ideal of 𝒜 (by the inductive hypothesis). Since I ς ′ ⊆ I ′ ∩ J ς ⊆ I ∩ J ς = I ς , we conclude that I ′ ∩ J ς = I ς ′ (by the maximality of I ς ′ ), and thus

Therefore, the vector subspaces 𝒱 and V ς of ℐ satisfy the assumptions of [10, Lemma 2.2] (we note that the proof of this result holds for an arbitrary field). In particular, if e 1 , … , e n ∈ D are non-zero orthogonal idempotents such that D = 𝕜 ⁢ e 1 ⊕ ⋯ ⊕ 𝕜 ⁢ e n , then dim ⁡ e r ⁢ V ≤ 1 and dim ⁡ V ⁢ e r ≤ 1 for all 1 ≤ r ≤ n .

Next, we consider the ideal subgroup Q = 1 + L of 𝐺, and the group homomorphism φ ς : P → Q ∘ (as defined in the previous lemma); we recall that we have ker ⁡ ( φ ς ) = 1 + J ς . Since ℒ is an ideal of 𝒜 with dim ⁡ L = dim ⁡ J n + 1 , we have L = J n ⊕ 𝕜 ⁢ u , where u = e i ⁢ u ⁢ e j for some 1 ≤ i , j ≤ n ; hence Q = ( 1 + 𝕜 ⁢ u ) ⁢ N .