Branching Rules for $n$-fold Covering Groups of $\mathrm{SL}_2$ over a Non-Archimedean Local Field

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by  Camelia Karimianpour

BRANCHING RULES FOR n-FOLD COVERING GROUPS OF SL OVER A
2
NON-ARCHIMEDEAN LOCAL FIELD
CAMELIA KARIMIANPOUR
Abstract. Let G(cid:102)1 be the n-fold covering group of the special linear group of degree two, over a non-
Archimedeanlocalfield. Wedeterminethedecompositionintoirreduciblesoftherestrictionoftheprincipal
6 series representations of G(cid:102)1 to a maximal compact subgroup of G(cid:102)1.
1
0
2

n
1. Introduction
a
J
Inthispaper,coveringgroups,alsoknownintheliteratureasmetaplecticgroups,arecentralextensions
3
ofasimplyconnectedsimpleandsplitalgebraicgroup, overanon-ArchimedeanlocalfieldF, bythegroup

] ofthen-throotsofunity,µn. TheproblemofdeterminingthisclassofgroupswasstudiedbySteinberg[22]
T
and Moore [15] in 1968, and further completed by Matsumoto [12] in 1969 for simply connected Chevalley
R
groups. Around the same time, Kubota independently constructed n-fold covering groups of SL [10] and
2
.
h GL [11], by means of presenting an explicit 2-cocycle. Kubota’s cocycle is expressed in terms of the n-th
2
t
a Hilbert symbol.
m Since then, there have been a number of studies of representations of this class of groups from different
[ perspectives, among them being the work of H. Aritu¨rk [1], D. A. Kazhdan and S. J. Patterson [9], C.

Moen [14], D. Joyner [6, 7], G. Savin [20], M. Weissman and T. Howard [5], and P. J. McNamara [13].
1
v In this paper, we consider the principal series representations of the n-fold covering group S(cid:102)L2(F) of
33 SL2(F). The principal series representations of S(cid:102)L2(F) are those representations that are induced from
3 the inverse image B(cid:102)1 of a Borel subgroup B1 of SL (F). The construction of those representations of B(cid:102)1
2
0
that are trivial on the unipotent radical of B(cid:102)1 brings us to the study of the irreducible representations of
0
1. the metaplectic torus T(cid:102)1, i.e., the inverse image of the split torus, T1, of SL2(F) in S(cid:102)L2(F).
0 An important feature of T(cid:102)1, which differentiates the nature of its representations from those of a linear
6
torus, is that it is not abelian. However, it is a Heisenberg group and its irreducible representations are
1
: governedbytheStone-vonNeumanntheorem. TheStone-vonNeumanntheoremcharacterizesirreducible
v
representations of Heisenberg groups, according to their central characters. Indeed, given a character of
i
X the centre of a Heisenberg group that satisfies some mild conditions, the Stone-von Neumann theorem
r provides a recipe to construct the corresponding, unique up to isomorphism, irreducible representation
a
of the Heisenberg group. The construction involves induction from a maximal abelian subgroup of the
Heisenberg group. We only consider those characters of the centre of T(cid:102)1 where µ acts by a fixed faithful
n
character.
Once an irreducible representation ρ , with central character χ, of T(cid:102)1 is obtained, the principal series
χ
representationπχ ofS(cid:102)L2(F)isIndS(cid:102)L2(F)ρχ,whereρχ istriviallyextendedontheunipotentradicalsubgroup
B(cid:102)1
of B(cid:102)1. These representations admit several open questions. The question we consider, and answer, in this
paper is to decompose π upon the restriction to the inverse image K(cid:102)1 of a maximal compact subgroup
χ
Keywords: local field, covering group, representation, Hilbert symbol, K-type
2010 Mathematics Subject Classification: 20G05
1
2 C.KARIMIANPOUR
K1 of SL (F). We refer to this decomposition as the K-type decomposition. We assume n|q−1, where q
2
is the size of the residue field of F, so that the central extension S(cid:103)L2(F) splits over K1.
The study the decomposition of the restriction of representations to a particular subgroup is a common
technique in representation theory. In the theory of real Lie groups, restriction to maximal compact
subgroups retains a lot of information from the representation; in fact, such a restriction is a key step
towards classifying irreducible unitary representations. In the case of reductive groups over p-adic fields,
investigating the decomposition upon restriction to maximal compact subgroups reveals a finer structure
oftherepresentation, intheinterestsofrecoveringessentialinformationabouttheoriginalrepresentation.
The K-type problem for reductive p-adic groups is visited and solved in certain cases, including the
principal series representations of GL(3) [2, 3, 19], and SL(2) [16, 17], representations of GL(2) [4], and
supercuspidal representations of SL(2) [18].
Themainideaistoreducetheproblemtocalculatingthedimensionsofcertainfinite-dimensionalHecke
algebras. The key calculation for determining the decomposition is the determination of certain double
cosets that support intertwining operators for the restricted principal series representation (Proposition 3
and Proposition 4).
Our method is aligned with the one in [16] for the linear group SL (F); however, the technicalities in
2
the covering case are much more involved than the linear case, and the results are fairly different. For
instance, the K-type decomposition is no longer multiplicity-free (Corollary 2).
Thispaperisorganizedasfollows. InSection2, wepresentKubota’sconstructionofthecoveringgroup
of SL (F), in Section 3 we overview the structure of this covering group and compute some subgroups
2
of our interest. We compute the K-type decomposition for the principal series representations of S(cid:103)L2(F)
in Section 4. This decomposition is completed by considering a similar problem for the n-fold covering
group of GL (F) in Section 5. Our main result, Theorem 2, is stated in Section 6.
2
2. Notation and Background
Let F be a non-Archimedean local field with the ring of integers O and the maximal ideal p of O. Let
κ := O/p be the residue field and q = |κ| be its cardinality. Let O× denote the group of units in O.
We fix a unoformizing element (cid:36) of p. For every x ∈ F×, the valuation of x is denoted by val(x), and
|x| = q−val(x). Let n ≥ 2 be an integer such that n|q−1. Set n = n if n is odd, and n = n if n is even.
2
We assume that F contains the group µ of n-th roots of unity.
n
Set G = GL (F), and G1 = SL (F). Let B1 (B) be the standard Borel subgroup of G1 (G) and N1 (N)
2 2
beitsunipotentradical, andletT1 (T)bethestandardtorusinG1 (G). SetK1 = SL (O)(K = GL (O))
2 2
to be a maximal compact subgroup of G1 (G). By the Iwasawa decomposition, we have G1 = T1N1K1
(G = TNK). Our object of study is the central extension G(cid:102)1 of G1 by µ ,
n
(1) 0 → µ →i G(cid:102)1 →p G1 → 0,
n
where i and p are natural injection and projection maps respectively. The group G(cid:102)1, which we call the
n-fold covering group of G1, is constructed explicitly by Kubota [10]. In order to describe Kubota’s
construction, we need knowledge of the n-th Hilbert symbol (, ) : F××F× → µ . Under our assumption
n n
on n, the n-th Hilbert symbol is given via (a,b)n = cq−n1, where c = (−1)val(a)val(b)abvvaall((ab)), and c is the
image of c in κ×. We benefit from the properties of the n-th Hilbert symbol, which can be found in [21,
Ch XIV]. In particular, we benefit extensively from the following fact: (a,b) = 1 for all a ∈ F×, if and
n
only if b ∈ F×n.
BRANCHING RULES FOR n-FOLD COVERING GROUPS OF SL2 OVER A NON-ARCHIMEDEAN LOCAL FIELD 3
Define the map β : G1×G1 → µ by
n
(cid:40)
(cid:18) (cid:19) (cid:18)(cid:18) (cid:19)(cid:19)
X(g g ) X(g g ) a b c if c (cid:54)= 0
1 2 1 2
(2) β(g ,g ) = , , where X =
1 2 X(g ) X(g ) c d d otherwise.
1 2 n
In [10] Kubota proved that β is a non-trivial 2-cocycle in the continuous second cohomology group of G1
with coefficients in µ ; whence, G(cid:102)1 = G1×µ as a set, with the multiplication given via (g ,ζ )(g ,ζ ) =
n n 1 1 2 2
(g g ,β(g ,g )ζ ζ ), for all g ,g ∈ G1 and ζ ,ζ ∈ µ .
1 2 1 2 1 2 1 2 1 2 n
In 1969, Kubota extends the map β to a 2-cocycle β(cid:48) for G(cid:101) in [11], which defines the n-fold covering
group G(cid:101) ∼= F×(cid:110)G(cid:102)1 of G. The covering group G(cid:101) fits into the exact sequence 0 → µn →i G(cid:101) →p G → 0.
For all t,s ∈ F×, set dg(t) = (cid:0)t 0 (cid:1) ∈ T1, dg(t,s) = (t 0) ∈ T, and ι(t) = (dg(t),1) ∈ T(cid:102)1. Set
0 t−1 0 s
w = (cid:0) 0 1(cid:1), and w = (w,1) ∈ G(cid:102)1. Moreover, for matrices X and Y, with Y invertible, let XY := Y−1XY
−1 0 (cid:101)
and YX := YXY−1 denote the conjugations of X by Y.
3. Structure Theory
For any subgroup H of G1, the inverse image H(cid:101) := p−1(H) is a subgroup of G(cid:102)1. In particular, we are
interested in the subgroups T(cid:102)1, B(cid:102)1, and K(cid:102)1 of G(cid:102)1. We say the central extension splits over the subgroup
H of G1, if there exists an isomorphism that yields p(H)−1 ∼= H ×µ .
n
It is not difficult to see that T(cid:102)1 is not commutative, and hence, the central extension does not split
over T1 (and therefore neither over B1). Additionally, it is easy to see that the commutator subgroup
[T(cid:102)1,T(cid:102)1] ∼= µ is central in (1); which implies that T(cid:102)1 is a two-step nilpotent group, also known as a
n
Heisenberg group. Clearly, µ ∈ Z(T(cid:102)1), indeed, using the properties of the Hilbert symbol and some
n
elementary calculation, one can show that Z(T(cid:102)1) = {(dg(t),ζ) | t ∈ F×n,ζ ∈ µ }.
n
Lemma 1. The index of Z(T(cid:102)1) in T(cid:102)1 is n2.
Proof. Note that [T(cid:102)1 : Z(T(cid:102)1)] = [F× : F×n], which because F× ∼= O× × Z, is equal to n[O× : O×n].
Consider the homomorphism φ : O× → O×n. Then ker(φ) = {x ∈ O×| xn = 1}. Note that f(x) =
xn −1 = 0 has (n,q−1), which equals n under our assumption of n|q−1, solutions in the cyclic group
κ×. By Hensel’s lemma, any such root in κ× lifts uniquely to a root in O×. It follows that, |ker(φ)| = n.
Therefore, [O× : O×n] = |ker(φ)| = n, and the result follows. (cid:3)
InordertoconstructprincipalseriesrepresentationsofG(cid:102)1 inSection4,weneedtoconstructirreducible
representations of the Heisenberg group T(cid:102)1. To do so, we need to identify a maximal abelian subgroup
of T(cid:102)1. Set A1 = C (T(cid:102)1 ∩K(cid:102)1), to be the centralizer of T(cid:102)1 ∩K(cid:102)1 in T(cid:102)1. It is not difficult to calculate
T(cid:102)1
that A1 = {(dg(a),ζ) | a ∈ F×, n|val(a), ζ ∈ µ }, and see that it is abelian. Observe that T(cid:102)1∩K(cid:102)1 ⊂ A1
n
implies that A1 is a maximal abelian subgroup. Note that [T(cid:102)1 : A1] = [Z : nZ] = n.
Let N1 be the unipotent radical of B1. It follows directly from the Kubota’s formula for β that β|
N1
is trivial, so N1×{1} is a subgroup of G(cid:102)1. We identify N1 with N1×{1}. Under this identification, we
have the covering analogue of the Levi decomposition: B(cid:102)1 = T(cid:102)1(cid:110)N1.
Next, we describe a family of compact open subgroups of G(cid:102)1. It is proven in [11] that
(cid:40)
(cid:18)(cid:18) (cid:19)(cid:19)
a b (c,d) , 0 < val(c) < ∞
(3) K(cid:102)1 → K1×µ , (k,ζ) (cid:55)→ (k,s(k)ζ), where s = n
n c d 1, otherwise.
4 C.KARIMIANPOUR
isanisomorphism. TheimageofK1 inK(cid:102)1 undertheisomorphism(3)isthesubgroupK(cid:101)0 := {(k,s(k)−1) |
k ∈ K1} of K(cid:102)1. Consider the compact open congruent subgroups K1 := {g ∈ K1 | g ≡ I mod pj}, for
j 2
j ≥ 1, of K1.
Lemma 2. The central extension (1) splits trivially over each of the subgroups K1, j ≥ 1, T1∩K1, and
j
B1∩K1.
Proof. Using the Hensel’s lemma, it is easy to see that 1 + p ⊂ O×n. Then, it follows from (3) and
properties of the n-th Hilbert symbol that, for all i ≥ 1, s| is trivial. On the other hand, it follows
K1
j
directly from (3) that s| and s| are trivial. (cid:3)
T1∩K1 B1∩K1
We identify K1 ∼= K1 ×{1}, j ≥ 1, B1 ∩K1 ∼= (B1∩K1)×{1} and T1 ∩K1 ∼= (T1∩K1)×{1} as
j j
subgroups of K(cid:102)1.
In a similar way, we define the subgroups T(cid:101), B(cid:101) and K(cid:101) of G(cid:101) to be the inverse images of the standard
torus, Borel, and the maximal compact K = GL(O) subgroups of G respectively. The central extension
G(cid:101) does not split over T. Moreover, T(cid:101) is a Heisenberg group. It is not difficult to see that Z(T(cid:101)) =
{(dg(s,t),ζ) | s,t ∈ F×n,ζ ∈ µn}, and [T(cid:101) : Z(T(cid:101))] = n4. Moreover, set A = CT(cid:101)(T(cid:101)∩K(cid:101)) = {(dg(s,t),ζ) |
s,t ∈ F×,n|val(s),n|val(t),ζ ∈ µn}. Then, A is a maximal abelian subgroup of T(cid:101) and [T(cid:101) : A] = n2. In
addition, β(cid:48)| is trivial, where N is the unipotent radical of B. Hence, we can identify N with N ×{1}.
N
Under this identification, we have the Levi decomposition: B(cid:101) = T(cid:101)(cid:110)N. It is shown in [11] that the central
extension G(cid:101) splits over K. For j ≥ 1, let Kj denote the family of compact open congruent subgroups
{g ∈ K | g ≡ I2 mod pj} of K. Similar to Lemma 2, one can show that G(cid:101) splits over Kj, T ∩K and
B∩K.
4. Branching Rules for G(cid:102)1
First, we present the construction of the principal series representations of G(cid:102)1 following [13]. Fix a
faithful character (cid:15) : µ → C×. A representation of G(cid:102)1 is genuine if the central subgroup µ acts by (cid:15).
n n
Such representations do not factor through representations of G1. The construction of principal series
representations of G(cid:102)1 is based on the essential fact that T(cid:102)1 is a Heisenberg subgroup, and hence its
representations are governed by the Stone-von Neumann theorem, which we state here. See [13] for the
proof.
Theorem 1 (Stone-von Neumann). Let H be a Heisenberg group with center Z(H) such that H/Z(H)
is finite, and let χ be a character of Z(H). Suppose that ker(χ)∩[H,H] = {1}. Then there is a unique
(up to isomorphism) irreducible representation π of H with central character χ. Let A be any maximal
abelian subgroup of H and let χ be any extension of χ to A. Then π ∼= IndHχ .
0 A 0
Note that [T(cid:102)1 : Z(T(cid:102)1)] = n2 < ∞. Let χ be a genuine character of Z(T(cid:102)1), so that χ| = (cid:15). Thus,
µn
ker(χ)∩[T(cid:102)1,T(cid:102)1] is trivial. Hence Theorem 1 applies: genuine irreducible smooth representations ρ of T(cid:102)1
are classified by genuine smooth characters of Z(T(cid:102)1). Moreover, dim(ρ) = [T(cid:102)1 : A(cid:102)1] = n.
Let χ be a fixed extension of χ to A1; so that (ρ,IndT(cid:102)1χ ) is the unique smooth genuine irreducible
0 A1 0
representation of T(cid:102)1 with central character χ. Let us again write ρ for the genuine smooth irreducible
representation of T(cid:102)1, with central character χ, extended trivially over N1 to a representation of B(cid:102)1 =
T(cid:102)1 (cid:110)N1. Then the genuine principal series representation of G(cid:102)1 associated to ρ is IndG(cid:102)1ρ, where Ind
B(cid:102)1
BRANCHING RULES FOR n-FOLD COVERING GROUPS OF SL2 OVER A NON-ARCHIMEDEAN LOCAL FIELD 5
denotes the smooth (non-normalized) induction. In the rest of this section, we decompose Res IndG(cid:102)1ρ
K(cid:102)1 B(cid:102)1
into irreducible constituents. We drop the adjective “genuine” for simplicity.
Define the character
(4) ϑ : F× → µ , a (cid:55)→ ((cid:36),a) .
n n
Observe that ϑ is ramified of degree one. Set ϑ := ϑ| . Observe that a typical element of A1 can
O×2 O×2
be written as (dg(a(cid:36)rn),ζ), and a typical element of T(cid:102)1∩K(cid:102)1 can be written as (dg(a),ζ), where a ∈ O×,
r ∈ Z, and ζ ∈ µ .
n
Lemma 3. Let ρ be the unique irreducible representation of T(cid:102)1 with central character χ. Then Res ρ ∼=
A1
(cid:76)n−1χ , where the χ are n distinct characters of A1 defined by
i=0 i i
χ (dg(a(cid:36)nr),ζ) = χ (cid:0)dg(a(cid:36)nr),ϑ2i(a)ζ(cid:1),
i 0
for all a ∈ O×, r ∈ Z, ζ ∈ µ , and 0 ≤ i < n.
n
Proof. By Theorem 1, ρ ∼= IndT(cid:102)1χ . By Mackey’s theory, Res IndT(cid:102)1χ = (cid:76) IndA1 χ s, where
A1 0 A1 A1 0 s∈Sn A1∩sA1 0
S is a complete set of coset representatives for A1\T(cid:102)1/A1. It is not difficult to see that we can choose
n
S = {(cid:0)dg((cid:36)i),1(cid:1) |0 ≤ i < n}. Since A1 is stable under conjugation by S , IndA1 χ s = χ s. Let
n n A1∩sA1 0 0
(dg(a(cid:36)rn),ζ) ∈ A1, and s = (cid:0)dg((cid:36)i),1(cid:1) ∈ S . Then
n
s−1(dg(a(cid:36)rn),ζ)s = (cid:0)dg((cid:36)−i),((cid:36)i,(cid:36)i) (cid:1)(dg(a(cid:36)rn),ζ)(cid:0)dg((cid:36)i),1(cid:1)
n
= (cid:0)dg(a(cid:36)rn−i),(a(cid:36)rn,(cid:36)−i) ((cid:36)i,(cid:36)i) ζ(cid:1)(cid:0)dg((cid:36)i),1(cid:1) = (cid:0)dg(a(cid:36)rn),((cid:36)i,a(cid:36)rn−i) (a(cid:36)rn,(cid:36)−i) ((cid:36)i,(cid:36)i) ζ(cid:1)
n n n n n
= (cid:0)dg(a(cid:36)rn),((cid:36),a)2iζ(cid:1) = (cid:0)dg(a(cid:36)rn),ϑ2i(a)ζ(cid:1).
n
Hence, χ s((dg(a(cid:36)rn),ζ)) = χ (cid:0)(cid:0)dg(a(cid:36)rn),ϑ2i(a)ζ(cid:1)(cid:1). Denote this character χ . To show that the χ ,
0 0 i i
0 ≤ i < n, are distinct, it is enough to show that ϑ2i| = 1 if and only if i = 0. Observe that
O×
(q−1)2i
ϑ2i(a) = a−1 n , which is equal to 1 for all a ∈ O× if and only if n|2i. The result follows. (cid:3)
Thecharactersχ definedinLemma3areclearlydistinctwhenrestrictedtoT(cid:102)1∩K(cid:102)1 and, againwriting
i
χ for these restrictions,
i
n−1
(cid:77)
(5) Res ρ = χ .
T(cid:102)1∩K(cid:102)1 i
i=0
Proposition 1. Let χ , 0 ≤ i < n, denote also the trivial extension of the characters in (5) to B(cid:102)1∩K(cid:102)1.
i
Then
n−1
Res IndG(cid:102)1ρ ∼= (cid:77)IndK(cid:102)1 χ .
K(cid:102)1 B(cid:102)1 B(cid:102)1∩K(cid:102)1 i
i=0
Proof. By Mackey’s theorem, we have Res IndG(cid:102)1ρ ∼= (cid:76) IndK(cid:102)1 Res ρx, where X is a
K(cid:102)1 B(cid:102)1 x∈X B(cid:102)1x−∩1K(cid:102)1 B(cid:102)1x−∩1K(cid:102)1
completesetofdoublecosetrepresentativesofK(cid:102)1 andB(cid:102)1 inG(cid:102)1. TheIwasawadecompositionK(cid:102)1B(cid:102)1 = G(cid:102)1
impliesthatX = {(I ,1)}andhenceRes IndG(cid:102)1ρ = IndK(cid:102)1 Res ρ.Theresultfollowsfrom(5). (cid:3)
2 K(cid:102)1 B(cid:102)1 B(cid:102)1∩K(cid:102)1 B(cid:102)1∩K(cid:102)1
Hence, in order to calculate the K-types, it is enough to decompose each IndK(cid:102)1 χ , 0 ≤ i < n, into
i
B(cid:102)1∩K(cid:102)1
irreducible representations. Note that the induction space IndK(cid:102)1 χ is smooth and admissible. Fix
i
B(cid:102)1∩K(cid:102)1
6 C.KARIMIANPOUR
i ∈ {0,··· ,n − 1}. The smoothness of IndK(cid:102)1 χ implies that IndK(cid:102)1 χ = (cid:83) (cid:0)IndK(cid:102)1 χ (cid:1)Kl1.
B(cid:102)1∩K(cid:102)1 i B(cid:102)1∩K(cid:102)1 i l≥1 B(cid:102)1∩K(cid:102)1 i
Note that, by admissibility, (cid:0)IndK(cid:102)1 χ (cid:1)Kl1 is finite-dimensional for every l ≥ 1 and since K1 is normal
B(cid:102)1∩K(cid:102)1 i l
in K(cid:102)1, it is K(cid:102)1-invariant. Hence, to decompose IndK(cid:102)1 χ into irreducible constituents, it is enough to
i
B(cid:102)1∩K(cid:102)1
decompose each (cid:0)IndK(cid:102)1 χ (cid:1)Kl1 into irreducible constituents.
i
B(cid:102)1∩K(cid:102)1
For any character γ of any subgroup D of T(cid:102)1, we say γ is primitive mod m if m is the smallest strictly
positive integer for which Res γ = 1. From now on, let m ≥ 1 be a positive integer such that χ
D∩K1
m
is primitive mod m. Because 1 + p ⊂ F×n, Z(T(cid:102)1) ∩ K1 = T(cid:102)1 ∩ K1, for all m ≥ 1. Note that since
m m
χ | = χ, χ | = χ| . Hence, χ is primitive mod m if and only if the χ for 0 ≤ i < n
i Z(T(cid:102)1) i T(cid:102)1∩K1 Z(T(cid:102)1)∩K1 i
m m
are primitive mod m. Set B(cid:102)1 := (B(cid:102)1∩K(cid:102)1)K1.
l l
Lemma 4. For every 0 ≤ i < n,
(cid:40)
{0}, 0 < l < m
(cid:0)IndK(cid:102)1 χ (cid:1)Kl1 =
B(cid:102)1∩K(cid:102)1 i IndK(cid:102)1χ , otherwise.
i
B(cid:102)1l
Proof. Suppose 0 < l < m, and that f is a vector in (cid:0)IndBK(cid:102)(cid:102)11∩K(cid:102)1χi(cid:1)Kl1. Because χi(cid:12)(cid:12)B(cid:102)1∩Kl1 (cid:54)= 1 for l < m,
we can choose b ∈ B(cid:102)1∩K1 such that χ (b) (cid:54)= 1. Let g ∈ K(cid:102)1. Note that K1 is normal in K(cid:102)1 and
l i l
hence g−1bg ∈ K1. On the one hand, f(bg) = χ (b)f(g); on the other hand, f(bg) = f(gg−1bg) =
l i
(cid:0)g−1bg(cid:1)·f(g) = f(g), since f is fixed by K1. It follows that χ (b)f(g) = f(g). Our choice of b implies
l i
that f(g) = 0 and because g is arbitrary, f = 0. However, if l ≥ m then χ | = 0 and because K1
i K1 l
l
is normal in K(cid:102)1, it is not difficult to see that every K1-fixed vector f translates on the left by B(cid:102)1 and
l l
vice-versa. Hence the result follows. (cid:3)
Lemma4tellsusthat, inordertodecompose(IndK(cid:102)1 χi)Kl1 intoirreducibleconstituents, itisenough
B(cid:102)1∩K(cid:102)1
todecomposeIndK(cid:102)1χ . Hence,weareinterestedincountingthedimensionofHom (IndK(cid:102)1χ ,IndK(cid:102)1χ ).
B(cid:102)1l i K(cid:102)1 B(cid:102)1l i B(cid:102)1l i
By Frobenius reciprocity, this latter space is isomorphic to Hom (Res IndK(cid:102)1χ ,χ ). It follows from
B(cid:102)1l B(cid:102)1l B(cid:102)1l i i
Mackey’s theory that
Res IndK(cid:102)1χ ∼= (cid:77)IndB(cid:102)1l χx,
B(cid:102)1l B(cid:102)1l i x∈S B(cid:102)1lx−∩1B(cid:102)1l i
where S is a set of double coset representatives of B(cid:102)1 \K(cid:102)1/B(cid:102)1 . The set S is a lift to the covering group
l l
K(cid:102)1 of a similar set of double coset representatives calculated in [16]. Using the latter set, and because
µ ⊂ B(cid:102)1 , it is easy to see that
n l
(6) S = {(I2,1),w(cid:101),l(cid:101)t(x(cid:36)r) | x ∈ {1,ε},1 ≤ r < l},
where ε is a fixed non-square. For 0 ≤ i,j < n, let H be the Hecke algebra
i,j
H := H(B(cid:102)1 \K(cid:102)1/B(cid:102)1 , χ ,χ ) = {f : K(cid:102)1 → C|f(lgh) = χ (l)f(g)χ (h), l,h ∈ B(cid:102)1 ,g ∈ K(cid:102)1}.
i,j l l i j i j l
Proposition 2. Let 0 ≤ i,j < n. Then dimHom (IndK(cid:102)1χ ,IndK(cid:102)1χ ) = dimH .
K(cid:102)1 B(cid:102)1l i B(cid:102)1l j i,j
BRANCHING RULES FOR n-FOLD COVERING GROUPS OF SL2 OVER A NON-ARCHIMEDEAN LOCAL FIELD 7
Proof. On the one hand, observe that Hom (IndK(cid:102)1χ ,IndK(cid:102)1χ ) = (cid:76) Hom (IndB(cid:102)1l χx,χ ),
which by Frobenius reciprocity is equal to (cid:76)K(cid:102)1 HomB(cid:102)1l i (B(cid:102)χ1xl,χj ). LetxS∈S be thB(cid:102)e1lset oBf(cid:102)1axll−l∩1xB(cid:102)1∈l Si sujch
x∈S x−1 i j i,j
B(cid:102)1l ∩B(cid:102)1l
that χ (g) = χ (h), whenever h,g ∈ B(cid:102)1 and xgx−1 = h. Then dimHom (IndK(cid:102)1χ ,IndK(cid:102)1χ ) = |S |.
i j l K(cid:102)1 B(cid:102)1l i B(cid:102)1l j i,j
On the other hand, observe that for every x ∈ S, there exists a function f ∈ H with support on the
i,j
double coset represented by x if and only if h = xgx−1 implies χ (g) = χ (h) for all h,g ∈ B(cid:102)1 . Moreover,
i j l
the basis of H is parametrized by such double coset representatives. Hence, dimH = |S |. (cid:3)
i,j i,j i,j
Hence, in order to decompose (IndK(cid:102)1 χi)Kl1, we are interested in counting the dimension of Hi,i. Set
B(cid:102)1∩K(cid:102)1
(T1 ∩K1)2 := {dg(t2) | t ∈ O×}, T1 := {ι(t) | t ∈ O×(1+pl)}, and (T1)2 := {ι(t2) | t ∈ O×(1+pl)}. It
l l
is not difficult to see that T1 and (T1)2 are subgroups of (T(cid:102)1∩K(cid:102)1)K1.
l l l
(cid:40)
1+2(l−m), if χ | (cid:54)= 1;
Proposition 3. Let l ≥ m and 0 ≤ i < n. Then dimH = i (T1∩K1)2
i,i
2l, otherwise.
Proof. Assume l ≥ m. Note that f(bkb(cid:48)) = χ (b)f(k)χ (b(cid:48)) for all f ∈ H , b,b(cid:48) ∈ B(cid:102)1 and k ∈ K(cid:102)1.
i i i,i l
Hence, for every double coset representative x in (6), there exists a function f ∈ H , with support on
i,i
the double coset represented by x if and only if bxb(cid:48) = x implies that χ (bb(cid:48)) = 1 for all b,b(cid:48) ∈ B(cid:102)1 .
i l
The set of such double cosets parameterizes a basis for H . We now determine these double cosets.
i,i
Let b = (b,ζ) = (cid:0)(cid:0)t s (cid:1),ζ(cid:1), and b(cid:48) = (b(cid:48),ζ(cid:48)) = (cid:0)(cid:0)t(cid:48) s(cid:48) (cid:1),ζ(cid:48)(cid:1), where t,t(cid:48) ∈ O×(1+pl), s,s(cid:48) ∈ pl and
0 t−1 0 t(cid:48)−1
ζ,ζ(cid:48) ∈ µ denote arbitrary elements of B(cid:102)1 .
n l
The identity coset B(cid:102)1 : Afunctionf ∈ H hassupportonB(cid:102)1 ifandonlyiff(b) = χ (b),∀b ∈ B(cid:102)1 .
l i,i l i l
So there is always a function with support on the identity coset, namely f = χ .
i
The coset of w: For b and b(cid:48) in B(cid:102)1 , bwb(cid:48) = w implies, via a quick calculation, that b = b(cid:48) =
(cid:101) l (cid:101) (cid:101)
dg(t), for some t ∈ O×(1+pl) and ζ(cid:48) = ζ−1. Therefore, χ (bb(cid:48)) = χ (cid:0)(dg(t),ζ)(dg(t),ζ−1)(cid:1) =
i i
χ (cid:0)dg(t2),(t,t) (cid:1) = χ (cid:0)dg(t2),1(cid:1). So, H contains a function with support on this coset if and
i n i i,i
(cid:12)
only if χi(ι(t2)) = 1 for all t ∈ O×(1+pl); that is if and only if χi(cid:12)(T1)2 = 1. Observe that for
(cid:12) l
0 ≤ i < n, χi(cid:12)(T1)2 = 1, where l ≥ m, if and only if χi|(T1∩K1)2 = 1. Suppose χi|(T1∩K1)2 = 1, for
l
some 0 ≤ i < n. We show that in this case, m = 1. Suppose α ∈ 1+p, consider f(X) = X2−α.
Observe that f(1) = 0 mod p, and f(cid:48)(1) = 2(1) (cid:54)= 0 mod p. By Hensel’s lemma, f(X) has a
root in O; that is α ∈ O×2. Therefore 1+p ⊂ O×2, which implies χ | = 1, so m = 1.
i T(cid:102)1∩K1
1
The coset of l(cid:101)t(x(cid:36)r): For b and b(cid:48) in B(cid:102)1l, bl(cid:101)t(x(cid:36)r)b(cid:48) = l(cid:101)t(x(cid:36)r) implies that tt(cid:48) ∈ 1+pr and ζ =
(cid:16) (cid:17) (cid:16) (cid:17)
ζ(cid:48)−1. Therefore, χ (bb(cid:48)) = χ (bb(cid:48),1) = χ ( tt(cid:48) ts(cid:48)+st(cid:48)−1 ,1). Note that tt(cid:48) ts(cid:48)+st(cid:48)−1 ∈ B(cid:102)1 ∩K1.
i i i 0 t−1t(cid:48)−1 0 t−1t(cid:48)−1 r
Hence, χ (bb(cid:48)) = 1 if and only if B(cid:102)1 ∩K1 ⊆ ker(χ ). The latter holds if and only if r ≥ m, since
i r i
χ is primitive mod m.
i
Now, let us summarize our result. There is always one function with support on the identity coset, and
(cid:12)
2(l−m) functions on cosets represented by l(cid:101)t(x(cid:36)r), x ∈ {1,ε}, m ≤ r < l. If χi(cid:12)(T1∩K1)2 (cid:54)= 1, no function
in H has support on the double coset represented by w, otherwise, there exists an additional function
i,i (cid:101)
in H with support on the double coset represented by w. (cid:3)
i,i (cid:101)
Next two lemmas elaborate on the condition χ | = 1 that appears in Proposition 3.
i (T1∩K1)2
Lemma 5. For each 0 ≤ i < n, χ | = 1 if and only if χ | = (cid:15)◦ϑ−2i .
i (T1∩K1)2 0 (T1∩K1)2 O×2
8 C.KARIMIANPOUR
Proof. Let ι(s) ∈ (T1∩K1)2, so s ∈ O×2. By Lemma 3, χ (ι(s)) = χ (cid:0)dg(s),ϑ(s)2i(cid:1) = χ (ι(s))(cid:15)(ϑ(s)2i),
i 0 0
which is equal to 1 if and only if χ | = (cid:15)◦ϑ−2i . (cid:3)
0 (T1∩K1)2 O×2
Lemma 6. If 4 (cid:45) n then the characters ϑ−2i , 0 ≤ i < n are distinct. Otherwise, the ϑ−2i , 0 ≤ i < n,
O×2 O×2 4
are distinct; for n ≤ i < n, ϑ−2i = ϑ−2(i−n4).
4 2 O×2 O×2
(q−1)2i
Proof. By definition of ϑ in (4), ϑ−2i(s) = 1 for all s ∈ O×2 if and only if t2 n = 1 for all t ∈ O×, or
equivalently when n|4i. Therefore, the equality holds only for i = 0 unless 4|n, in which case the equality
holds for both i = 0 and i = n. (cid:3)
4
For l > m, let W(cid:102)i,l denote the l-level representations W(cid:102)i,l := (IndK(cid:102)1 χi)Kl1/(IndK(cid:102)1 χi)Kl1−1. More-
B(cid:102)1∩K(cid:102)1 B(cid:102)1∩K(cid:102)1
over, for 0 ≤ i < n, set V(cid:101)i := IndK(cid:102)1 χi.
B(cid:102)1∩K(cid:102)1
Corollary 1. Assume l ≥ m. We can decompose Res IndG(cid:102)1ρ as follows:
K(cid:102)1 B(cid:102)1
n−1(cid:32) (cid:33)
Res IndG(cid:102)1ρ ∼= (cid:77) V(cid:101)Km1 ⊕(cid:77)(cid:16)W(cid:102)+⊕W(cid:102)−(cid:17) ,
K(cid:102)1 B(cid:102)1 i i,l i,l
i=0 l>m
where W(cid:102)i+,l ⊕W(cid:102)i−,l ∼= W(cid:102)i,l. All the pieces are irreducible, except when m = 1 and χ0|(T1∩K1)2 = (cid:15)◦ϑ−O2×i2
for some 0 ≤ i < n, in which case, we are in one of the following situations:
(1) If 4 (cid:45) n then there is exactly one 0 ≤ i < n for which V(cid:101)K11 decomposes into two irreducible
i
constituents. All other constituents are irreducible.
(2) If 4|n then there are exactly two 0 ≤ i,k < n, |i−k| = n for which V(cid:101)K11 decomposes into two
4 i
irreducible constituents. All other constituents are irreducible.
Proof. It follows from Lemma 4 and Proposition 3 that for l > m, dimHom(W(cid:102)i,l,W(cid:102)i,l) = 2. Hence, W(cid:102)i,l
decomposes into two inequivalent irreducible subrepresentations. Moreover,
(cid:40) (cid:12)
(7) dimHom(V(cid:101)Km1 ,V(cid:101)Km1 ) = 1, if χi(cid:12)(T1∩K1)2 (cid:54)= 1
i i
2, otherwise.
By Lemma 5, χi(cid:12)(cid:12)(T1∩K1)2 = 1 is equivalent to χ0|(T1∩K1)2 = (cid:15)◦ϑO−2×i2, which also implies that m = 1.
Hence, V(cid:101)iKm1 is irreducible except when m = 1 and χ0|(T1∩K1)2 = (cid:15)◦ϑO−2×i2, where it decomposes into two
irreducible constituents. If the latter is the case, by Lemma 6, there is exactly one 0 ≤ i < n satisfying
χ | = (cid:15)◦ϑ−2i if 4 (cid:45) n, and there are exactly two 0 ≤ i < n satisfying χ | = (cid:15)◦ϑ−2i if
0 (T1∩K1)2 O×2 0 (T1∩K1)2 O×2
4|n. (cid:3)
Next we determine the multiplicity of each constituent in the decomposition in Corollary 1. To do so,
(cid:16) (cid:17)
we count the dimension of Hom IndK(cid:102)1χ ,IndK(cid:102)1χ , which is equal to the dimension of the Hecke
K(cid:102)1 B(cid:102)1l k B(cid:102)1l i
algebra H = H(B(cid:102)1 \K(cid:102)1/B(cid:102)1 ,χ ,χ ).
k,i l l k i
Proposition 4. Let l ≥ m, 0 ≤ k,i < n, and i (cid:54)= k. Then
(cid:40)
−(k+i)
2l−1, if χ | = (cid:15)◦ϑ
dimH = 0 (T1∩K1)2 O×2
k,i
2(l−m), otherwise.
BRANCHING RULES FOR n-FOLD COVERING GROUPS OF SL2 OVER A NON-ARCHIMEDEAN LOCAL FIELD 9
Proof. Similar to the proof of Proposition 3, we determine which double cosets in B(cid:102)1 \K(cid:102)1/B(cid:102)1 support a
l l
function in H . For every double coset representative x in Lemma (6), there exists a function f ∈ H
k,i k,i
with support on the double coset represented by x if and only if bxb(cid:48) = x, b,b(cid:48) ∈ B(cid:102)1 , implies that
l
χ (b)χ (b(cid:48)) = 1. Let t,t(cid:48) ∈ O×(1+pl), s,s(cid:48) ∈ pl and ζ,ζ(cid:48) ∈ µ , so that b = (b,ζ) = (cid:0)(cid:0)t s (cid:1),ζ(cid:1) and
k i n 0 t−1
b(cid:48) = (b(cid:48),ζ(cid:48)) = (cid:0)(cid:0)t(cid:48) s(cid:48) (cid:1),ζ(cid:48)(cid:1) are arbitrary elements of B(cid:102)1 .
0 t(cid:48)−1 l
Because χ (cid:54)= χ , there is no function in H with support on the identity double coset.
k i k,i
For the double coset of w, bwb(cid:48) = w implies that b = b(cid:48) = dg(t), for some t ∈ O×(1+pl) and ζ(cid:48) = ζ−1.
(cid:101) (cid:101) (cid:101)
Therefore, χ (b)χ (b(cid:48)) = χ (dg(t),ζ)χ (cid:0)dg(t),ζ−1(cid:1) equals
k i k i
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
χ dg(t),ϑ(t)2kζ χ (cid:0)dg(t),ϑ(t)2iζ−1(cid:1) = χ dg(t2),ϑ(t)2(k+i) = χ ι(t2)(I ,ϑ(t2)k+i)
0 0 0 0 2
(cid:16) (cid:17)
= χ (cid:0)ι(t2)(cid:1)(cid:15) ϑ(t2)k+i .
0
Therefore, because l ≥ m, χ (b)χ (b(cid:48)) = 1 if and only if χ | = (cid:15)◦ϑ−(k+i). In this case, m = 1
k i 0 (T1∩K1)2 O×2
and w supports a function in H .
(cid:101) k,i
Finally, for the double cosets represented by l(cid:101)t(x(cid:36)r), x ∈ {1,ε}, 1 ≤ r < l, b l(cid:101)t(x(cid:36)r)b(cid:48) = l(cid:101)t(x(cid:36)r)
implies that ζ(cid:48) = ζ−1, and t+s(cid:36)r = t(cid:48)−1 mod pl, or equivalently, t = t(cid:48)−1 mod pr, and t−1(cid:36)r = (cid:36)rt(cid:48)−1
mod pl, or equivalently t−1 = t(cid:48)−1 mod pl−r. Observe that, in general, χ (b)χ (b(cid:48)) is equal to
k i
(cid:16) (cid:17) (cid:16) (cid:17)
(8) χ (dg(t),ζ)χ (cid:0)dg(t(cid:48)),ζ(cid:48)(cid:1) = χ dg(t),ϑ(t)2kζ χ (cid:0)dg(t(cid:48)),ϑ(t(cid:48))2iζ(cid:48)(cid:1) = χ dg(tt(cid:48)),ϑ(t)2kϑ(t(cid:48))2iζζ(cid:48)
k i 0 0 0
(cid:16) (cid:17)
= χ (cid:0)ι(tt(cid:48))(cid:1)(cid:15) ϑ(t)2kϑ(t(cid:48))2iζζ(cid:48) .
0
Note that ϑ is primitive mod one. Observe that r ≥ 1 and l −r ≥ 1. Therefore, t = t(cid:48)−1 mod p and
t = t(cid:48) mod p, which implies that t = t(cid:48) = α mod p where α ∈ {±1}. Hence, ϑ(t)2 = ϑ(t(cid:48))2 = 1, and (8)
simplifies to χ (ι(tt(cid:48)))(cid:15)(ζζ(cid:48)). We are in one of the following situations:
0
Case 1: Suppose r ≥ m. Then we have ζ(cid:48) = ζ−1, and t = t(cid:48)−1 mod pm; that is tt(cid:48) ∈ 1+pm. Hence,
χ (ι(tt(cid:48)))(cid:15)(ζζ(cid:48)) = χ (tt(cid:48)) = 1, because χ is primitive mod m. Therefore, in this case, there is
0 0 0
always a function in H with support on these double cosets.
k,i
Case 2: Suppose r < m. Then ζ(cid:48) = ζ−1, so χ (ι(tt(cid:48)))(cid:15)(ζζ(cid:48)) = χ (tt(cid:48)), which equals one if and only
0 0
if tt(cid:48) ∈ 1+pm, which is not the case in general. Hence, in this case, there is no function in H
k,i
with support on these double cosets.
Tosummarizetheresult,thecosetrepresentedbywsupportsafunctioninH ifandonlyifχ | =
(cid:101) k,i 0 (T1∩K1)2
(cid:15)◦ϑ−(k+i). If r ≥ m then the cosets represented by lt(x(cid:36)r) support a function in H ; otherwise, there
O×2 k,i
is no function in H with support on these double cosets. (cid:3)
k,i
Corollary 2. In the decomposition of Res IndG(cid:102)1ρ given in Corollary 1,
K(cid:102)1 B(cid:102)1
(1) For each 0 ≤ i < n and l > m, there exists a way of decomposing W(cid:102)i,l as W(cid:102)i+,l⊕W(cid:102)i−,l such that for
l > m, W(cid:102)+ ∼= W(cid:102)+ and W(cid:102)− ∼= W(cid:102)− for all 0 ≤ i,j < n.
i,l j,l i,l j,l
(2) For l = m, {(IndK(cid:102)1 χi)Km1 | 0 ≤ i < n} consists of mutually inequivalent representations,
B(cid:102)1∩K(cid:102)1
except when m = 1 and χ0|(T1∩K1)2 = (cid:15)◦ϑ−Oj×2, for some 0 ≤ j < n, where V(cid:101)iK11 ∼= V(cid:101)kK11, exactly
when i+k ≡ j mod n.
10 C.KARIMIANPOUR
(cid:16) (cid:17)
Proof. It follows from Proposition 4 that for l > m, dimHomK(cid:102)1 W(cid:102)i,l,W(cid:102)k,l = 2, and when i+k ≡ j
mod n
(cid:40)
(cid:16) (cid:17) 1, χ | = (cid:15)◦ϑ−j
dimHom V(cid:101)Km1 ,V(cid:101)Km1 = 0 (T1∩K1)2 O×2
K(cid:102)1 i k 0, otherwise,
and hence the result. (cid:3)
In order to further investigate the irreducible spaces W(cid:102)i+,l and W(cid:102)i−,l, we will show that W(cid:102)i,l, 0 ≤ i < n, is
the restriction to K(cid:102)1 of an irreducible representation of the maximal compact subgroup K(cid:101) of the covering
group G(cid:101) of GL2(F).
5. Branching Rules for G(cid:101)
We define the genuine principal series representations of G(cid:101) similarly by starting with a genuine smooth
irreducible representation ρ(cid:48) of T(cid:101) with the central character χ(cid:48), which is constructed via the Stone-von
Neumann theorem. Observe that dimρ(cid:48) = [T(cid:101) : A] = n2. Then, after extending ρ(cid:48) trivially over N, the
genuine principal series representation π(cid:48) of G(cid:101) is IndG(cid:101)ρ(cid:48). Applying a similar machinery as in Section 4,
B(cid:101)
we obtain the K-type decomposition for Res π(cid:48). Since the argument in Section 4 goes through almost
K(cid:101)
exactly, here we only overview the main steps and point out the differences. For detailed calculations,
see [8].
Similar to Lemma 3, it follows that Res ρ(cid:48) ∼= (cid:76)n−1 χ(cid:48) , where the χ(cid:48) denote n2 distinct characters
A i,j=0 i,j i,j
of A, defined by χ(cid:48) (dg(a(cid:36)un,b(cid:36)vn),ζ) = χ(cid:48) (cid:0)dg(a(cid:36)un,b(cid:36)vn),ϑ(a)−jϑ(b)−iζ(cid:1) where a,b ∈ O×, u,v ∈ Z
i,j 0
and ζ ∈ µ and ϑ(a) = ((cid:36),a) was defined in (4), and χ(cid:48) is a fixed extension of χ(cid:48) to A. The χ(cid:48) remain
n n 0 i,j
distinct when restricted to T(cid:101)∩K(cid:101), and again writing χ(cid:48) for there restrictions, Res ρ(cid:48) ∼= (cid:76)n−1 χ(cid:48) .
i,j T(cid:101)∩K(cid:101) i,j=0 i,j
Then similar to Proposition 1, we have Res (IndG(cid:101)ρ(cid:48)) ∼= (cid:76)n−1 IndK(cid:101) χ(cid:48) . This latter isomorphism
K(cid:101) B(cid:101) i,j=0 B(cid:101)∩K(cid:101) i,j
reduces the problem of decomposing the K-type to the one of decomposing each IndK(cid:101) χ(cid:48) , which, by
B(cid:101)∩K(cid:101) i,j
smoothness, can be written as the union of its K , l ≥ 1, fixed points.
l
Suppose χ(cid:48) is primitive mod m. It follows that the χ(cid:48)i,j are also primitive mod m. Set B(cid:101)l = (B(cid:101)∩K(cid:101))Kl.
(cid:16) (cid:17)K
It can be seen that each level l representation IndK(cid:101) χ(cid:48) l = IndK(cid:101) χ(cid:48) if l ≥ m, and is zero if l < m.
B(cid:101)∩K(cid:101) i,j B(cid:101)l i,j
SimilartoProposition2,onecanseethatdimHomK(cid:101)(IndKB(cid:101)(cid:101)lχ(cid:48)i,j,IndKB(cid:101)(cid:101)lχ(cid:48)i,j) = dimH(cid:48)i,j(B(cid:101)l\K(cid:101)/B(cid:101)l,χ(cid:48)i,j,χ(cid:48)i,j).
We count the dimension of H(cid:48) using a method similar to the one we used in Proposition 3. To do so,
i,j
we need to calculate a set of double coset representatives of B(cid:101)l in K(cid:101).
Lemma 7. A complete set of double coset representatives of B(cid:101)l in K(cid:101) is given by {(I2,1),w(cid:101),l(cid:101)t((cid:36)r) | 1 ≤
r < l}.
Proof. Note that this set is a subset of the set S in (6). Observe that under the isomorphism
(9) F×(cid:110)G(cid:102)1 ∼= G(cid:101), (y,(g,ζ)) (cid:55)→ (dg(1,y)g,ζ),
O× ×K(cid:102)1 maps to K(cid:101) and O× ×B(cid:102)1l maps to B(cid:101)l. For every k(cid:48) ∈ K(cid:101), let (y,k) be the inverse image of k(cid:48)
under the isomorphism (9), and let b ,b ∈ B(cid:102)1 be such that b xb = k, for some x ∈ S. Let b(cid:48) and b(cid:48)
1 2 l 1 2 1 2
be the image of (y,b ) and (y,b ) under (9) respectively. It follows from the multiplication of F× (cid:110)G(cid:102)1
1 2
and the isomorphism map (9), that b(cid:48)1xb(cid:48)2 = k(cid:48). Thus, K(cid:101) = (cid:83)x∈SB(cid:101)lxB(cid:101)l. A short calculation shows that
(cid:0)dg(ε−1,1),1(cid:1)l(cid:101)t((cid:36)r)(dg(ε,1),1) = (lt(ε(cid:36)r),((cid:36)r,ε)n(ε,(cid:36)r)n) = l(cid:101)t(ε(cid:36)r),