matrix coefficients of depth-zero supercuspidal representations of...

inv lvea journal of mathematics

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Matrix coefficients of depth-zero supercuspidalrepresentations of GL(2)

Andrew Knightly and Carl Ragsdale

2014 vol 7 no 5

mspINVOLVE 75 (2014)

dxdoiorg102140involve20147669

Matrix coefficients of depth-zero supercuspidalrepresentations of GL(2)Andrew Knightly and Carl Ragsdale

(Communicated by Michael E Zieve)

We give explicit formulas for matrix coefficients of the depth-zero supercuspidalrepresentations of GL(2) over a nonarchimedean local field highlighting the casewhere the test vector is a unit new vector We also describe the partition of theset of such representations according to central character and compute sums ofmatrix coefficients over all representations in a given class

Introduction

Let F be a nonarchimedean local field with integer ring o maximal ideal p=$oand residue field k = op of cardinality q The supercuspidal representations ofGL2(F) are precisely those irreducible admissible representations which do notarise as constituents of parabolic induction They are characterized by having matrixcoefficients which are compactly supported modulo the center

In this paper we explicitly compute the matrix coefficients of depth-zero super-cuspidal representations These are the supercuspidals with the smallest possibleconductor exponent namely 2 First discovered by Mautner [1964 Section 9] theyarise by compact induction from the (qminus1)-dimensional representations of GL2(o)

inflated from the cuspidal series of the finite group GL2(k)In the first section we show that the matrix coefficients of any supercuspidal

representation are expressible in terms of those of the finite-dimensional inducingrepresentation Thus the task at hand essentially reduces to a computation of thematrix coefficients of the cuspidal representations of GL2(k) The latter is achievedin Theorem 27 using the explicit model from [Piatetski-Shapiro 1983]

With global applications in mind in Section 3 we single out the case wherethe test vector in the supercuspidal matrix coefficient is a unit new vector Theresulting function given in (3-6) and Theorem 32 may be used to define an integraloperator on the global automorphic spectrum of GL2 which isolates those cuspidal

MSC2010 22E50Keywords supercuspidal matrix coefficients regular characters

669

670 ANDREW KNIGHTLY AND CARL RAGSDALE

newforms with depth-zero supercuspidal local type (see Section 34) Possibleapplications include various trace formulas involving newforms of level p2 whichare supercuspidal (as opposed to principal series or special) at p For a recentexample involving the supercuspidal representations of conductor p3 see [Knightlyand Li 2012]

It is often desirable to organize representations according to central characterFor example there are exactly 2(q minus 1) supercuspidal representations of GL2(F)of conductor p3 with a given central character (see [Bushnell and Henniart 2014Remark 22]) By contrast there are (q2)(q minus 1) distinct cuspidal representationsof GL2(k) and (q minus 1) possible central characters but obviously there cannot beq2 of each kind if q is odd We sort this out in Proposition 23 and use it to give aformula (4-1) for the number of supercuspidals of conductor p2 with a given centralcharacter It depends on the parity of q and the order of the central character Wethen give formulas for various sums of matrix coefficients over the set of depth-zerosupercuspidal representations with a given central character These computationsrely on sum formulas for primitive characters derived in Proposition 24 We closein the final section with some simple examples

1 Matrix coefficients of supercuspidal representations

In this section let G =GLn(F) or more generally any unimodular locally profinitegroup [Bushnell and Henniart 2006] with center Z Let HsubG be an open and closedsubgroup containing Z with HZ compact and let (ρ V ) be an irreducible smoothrepresentation of H Consider the compact induction π = c-IndG

H (ρ) It consistsof the functions φ G minusrarr V with compact support (mod Z ) for which φ(hg)=ρ(h)φ(g) for all h isin H g isin G with G acting on the space by right translationHere we show that as observed by Mautner [1964] the matrix coefficients of π areessentially those of ρ These matrix coefficients are compactly supported (moduloZ ) so if π is irreducible and admissible it is supercuspidal Conversely it isconjectured that all supercuspidal representations arise in this way This was provenby Bushnell and Kutzko [1993] for G = GLn(F) and more recently in great (butnot complete) generality in [Stevens 2008 Kim 2007]

We assume for simplicity that ρ has unitary central character so that by the factthat HZ is compact ρ is unitarizable Let 〈vw〉V denote an H -equivariant innerproduct on V Then the inner product on c-IndG

H (ρ) given by

〈φψ〉 =sum

xisinHG

langφ(x) ψ(x)

rangV

is convergent (in fact a finite sum) and well-defined Further for any g isin Glangπ(g)φ π(g)ψ

rang=

sumxisinHG

langφ(xg) ψ(xg)

rangV =

sumxisinHG

langφ(x) ψ(x)

rangV = 〈φψ〉

MATRIX COEFFICIENTS OF DEPTH ZERO SUPERCUSPIDALS 671

Thus π is unitary relative to this inner productFor v isin V and y isin G define a function fyv isin c-IndG

H (ρ) by

fyv(g)=ρ(h)v if g = hy isin H y0 if g isin H y

Then the set

fyv | y isin HG v isin V

spans the space c-IndGH (ρ) (Note that

fhyv = fyρ(hminus1)v)

Proposition 11 For y z isin G and vw isin V langπ(g) fyv fzw

rang=

langρ(h)vw

rangV if g = zminus1hy isin zminus1 H y

0 if g isin zminus1 H y

Proof By definition of the inner product

〈π(g) fyv fzw〉 =sum

xisinHG

langπ(g) fyv(x) fzw(x)

rangV

=

sumxisinHG

langfyv(xg) fzw(x)

rangV

=langfyv(zg) w

rangV

since fzw(x) vanishes unless x isin H z If g = zminus1hy isin zminus1 H y then the above isequal to lang

fyv(hy) wrangV =

langρ(h)vw

rangV

as needed If g isin zminus1 H y then zg isin H y so fyv(zg) = 0 and the inner productvanishes

If we let G=GZ then the formal degree dπ of π is a positive constant satisfyingintG

∣∣langπ(g) f frang∣∣2dg =

f 4

dπ(1-1)

for all f isin c-IndGH (ρ) It depends on the choice of Haar measure on G (The

existence of dπ is due to Godement see for example [Knightly and Li 2006Proposition 104])

Proposition 12 For any choice of Haar measure on G = GZ the associatedformal degree of π is given by

dπ =dim ρ

meas(H)

where H is the (open) image of H in G


Proof Let v isin V be a unit vector and consider the function f1v By Proposition 11langπ(g) f1v f1v

rang=

langρ(g)v v

rangV if g isin H

0 if g isin H

Therefore intG

∣∣langπ(g) f1v f1vrang∣∣2 dg =

intH

∣∣langρ(g)v vrangV ∣∣2 dg

=v4

dim(ρ)meas(H)=

meas(H)dim(ρ)

by the Schur orthogonality relations for irreducible representations of compactgroups By (1-1)

meas(H)dim(ρ)

= f1v

4

dπ

Therefore it suffices to show that f1v = 1 This can be done via a directcomputation

f1v2= 〈 f1v f1v〉 =

sumxisinHG

langf1v(x) f1v(x)

rangV

=langf1v(1) f1v(1)

rangV = 〈v v〉 = 1

as needed

2 Cuspidal representations of GL2(k)

Let q be a prime power let k be the finite field with q elements let L be the uniquequadratic extension of k and let G = GL2(k) Define the subgroups

U =(

1 b0 1

)isin G

Z =

( aa)isin G

B =

(a b0 d

)isin G

Note that Z is the center of G Recall that the cuspidal representations of G arethose that do not contain the trivial character of the unipotent subgroup U Theseare precisely the irreducible representations that do not arise via parabolic inductionThey have dimension qminus1 and are parametrized by the Galois orbits of the primitivecharacters of Llowast defined below

21 Primitive characters of Llowast For any finite abelian group H we write H forthe dual group consisting of the characters χ H minusrarr Clowast We recall that

H sim= H

Thus Llowast is a cyclic group of order q2minus1 A character ν isin Llowast is primitive if νq

6= νOtherwise ν is imprimitive Letting α = αq denote the Frobenius map the norm


map N L minusrarr k is given by

N (α)= αα = αq+1

Proposition 21 Let ν Llowastrarr Clowast be a character of Llowast Then the following areequivalent

(i) ν is imprimitive that is νq= ν

(ii) ν = χ N for some χ isin klowast

(iii) L1sub ker(ν) where L1 is the subgroup of norm 1 elements of Llowast

Proof Let θ be a generator of the cyclic group Llowast Then θq+1 is a generator of klowastIf νq

= ν then ν(θ) is a (q minus 1)-st root of unity so we may define χ isin klowast byχ(θq+1)= ν(θ) Then ν = χ N so (i) implies (ii) It is clear that (ii) implies (iii)On the other hand for any x isin Llowast N (xqminus1) = N (x)qminus1

= 1 so that xqminus1isin L1

Therefore if (iii) holds νq(x)= ν(xq)= ν(xqminus1x)= ν(xqminus1)ν(x)= ν(x) Hence(iii) implies (i)

The imprimitive characters thus correspond bijectively with the characters of klowastso there are q minus 1 of them It follows that there are (q2

minus 1)minus (q minus 1) = q2minus q

primitive characters of Llowast

Lemma 22 Let ν be a primitive character of Llowast Then for all α isin klowastsumxisinLlowast

N (x)=α

ν(x)= 0

Proof By Proposition 21 there exists λ isin L1 such that ν(λ) 6= 1 ThussumN (x)=α

ν(x)=sum

N (x)=α

ν(λx)= ν(λ)sum

N (x)=α

ν(x)

It follows thatsum

N (x)=αν(x)= 0

Next we examine the partition of primitive characters into classes accordingto their restrictions to klowast This will allow us to count the number of depth-zerosupercuspidal representations with a given central character (see (4-1))

Proposition 23 Suppose ω is a given character of klowast Let Pω denote the numberof primitive characters ν of Llowast for which ν|klowast = ω Then

Pω =

q minus 1 if q is odd and ω(qminus1)2 is trivialq + 1 if q is odd and ω(qminus1)2 is nontrivialq if q is even


Proof Let ξ be a generator of the cyclic group Llowast Let ν0 = ξqminus1 Note that for

α isin klowastν0(α)= ξ(α

qminus1)= ξ(1)= 1

In fact ν0 is a generator of the order q + 1 subgroup Llowastklowast of Llowast Let us considertwo characters of Llowast to be equivalent if they have the same restriction to klowast Thenthe equivalence class of a given character ν is the set

ν νν0 νν20 νν

q0

(2-1)

If ω= ν|klowast then Pω = q+1minus Aω where Aω is the number of imprimitive elementsof the above set Write ν = ξ b Without loss of generality (replacing ν by someννm

0 ) we may assume that 0le blt qminus1 A character ξa is imprimitive if and onlyif ξqa

= ξa or equivalently (q + 1) | a Suppose ννs0 and ννk

0 are both imprimitivefor 0le s le k le q Then b+ (q minus 1)s and b+ (q minus 1)k are both divisible by q + 1and strictly less than q2

minus 1 Their difference (k minus s)(q minus 1) ge 0 also has theseproperties and furthermore it is divisible by

lcm(q minus 1 q + 1)=(q2minus 1)2 if q is odd

q2minus 1 if q is even

It follows that kminus s = 0 if q is even and kminus s isin 0 (q + 1)2 if q is odd Thismeans that Aω le 1 if q is even and Aω le 2 if q is odd

Suppose q is odd and b is even Then there are two imprimitive elements namely

b+ 12 b(q minus 1)= 1

2 b(q + 1) (2-2)

giving ννb20 = ξ

b2 N and

b+b+ q + 1

2(q minus 1)=

b+ q minus 12

(q + 1) (2-3)

giving νν(b+q+1)20 = ξ (b+qminus1)2

N Hence Aω = 2 in this case Noting that bis even if and only if ω(qminus1)2

= 1 we obtain the first claim of the propositionPω = q + 1minus Aω = q minus 1

Suppose q is odd and b is odd Then for all k b+ k(q minus 1) is odd and hence itcannot be divisible by the even number q+1 So Aω = 0 and Pω = q+1 provingthe second claim of the proposition

If q is even then as shown above Aω le 1 If b is even then (2-2) is a solutionand if b is odd (2-3) is a solution Either way this shows that Aω ge 1 and henceAω = 1 proving the final claim that Pω = q + 1minus Aω = q when q is even

The character sums in the next proposition will be used in Section 4 when wesum matrix coefficients over all representations with a given central character


Proposition 24 Let ω be a character of klowast and let [ω] denote the set of primitivecharacters of Llowast extending ω

Suppose q is odd and ω(qminus1)2 is nontrivial Then for α isin Llowastsumνisin[ω]

ν(α)=

(q + 1)ω(α) if α isin klowast0 if α isin klowast

(2-4)

Suppose q is odd and ω(qminus1)2 is trivial Then for α isin Llowast

sumνisin[ω]

ν(α)=

(q minus 1)ω(α) if α isin klowastminus2ω(α(q+1)2) if α isin klowast α(q

2minus1)2= 1

0 if α(q2minus1)2=minus1

(2-5)

(Note that necessarily α isin klowast if α(q2minus1)26= 1)

Suppose q is even Then for α isin Llowastsumνisin[ω]

ν(α)=

qω(α) if α isin klowastminusω(N (α)12) if α isin klowast

(2-6)

Here we note that the square root is unique in klowast since the square function is abijection when q is even

Proof If α isin klowast then the sum is equal to Pωω(α) and the assertions follow fromthe previous proposition So we may assume that α isin klowast We use the notation fromthe previous proof Suppose q is odd and ω(qminus1)2 is nontrivial By the proof of theprevious proposition sum

νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)

where on the right-hand side ν is any fixed element of [ω] Noting thatqsum

m=0

νm0 (α)=

sumχisinLlowastklowast

χ(α)=

q + 1 if α isin klowast0 if α isin klowast

(2-7)

(2-4) followsNow suppose q is odd and ω(qminus1)2 is trivial By the proof of the previous

propositionsumνisin[ω]

ν(α)= ξ b(α)

( qsumm=0

νm0 (α)minus ν0(α)

b2minus ν0(α)

(b+q+1)2)

Since α isin klowast by (2-7) this is equal to

minusξ b(α)[ν0(α)

b2+ ν0(α)

b2ν0(α)(q+1)2]


Recalling that ν0 = ξqminus1 and writing ξ b

= ν this is

=minusν(α1+((qminus1)2))[1+ ξ(α(q2

minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)

Observe that α(q2minus1)2

= plusmn1 since its square is 1 If it is equal to +1 thenα(q+1)2

isin klowast since its (qminus 1)-st power is 1 and we immediately obtain the middleline of (2-5) Otherwise ξ

(α(q

2minus1)2

)= ξ(minus1)=minus1 and (2-8) vanishes

When q is even there exists a choice of ξ for which ω = ξ b with b even Thenusing (2-2) we find that when α isin klowastsum

νisin[ω]

ν(α)=minusξ b2(N (α))=minusω(N (α)12

)

22 Model for cuspidal representations There are various ways to construct thecuspidal representation ρν attached to a primitive character ν The action of Llowast onthe k-vector space L sim= k2 by multiplication gives an identification

Llowast sim= T (2-9)

of Llowast with a nonsplit torus T sub G with klowast sub Llowast mapping onto Z sub T Thecharacteristic polynomial of an element g isin G is irreducible over k if and only if gis conjugate to an element of T minus Z

Fix a nontrivial character of the additive group

ψ k minusrarr Clowast

viewed in the obvious way as a character of U Then one may define ρν implicitlyby

IndGZU (νotimesψ)= ρν oplus IndG

T ν (2-10)

see [Bushnell and Henniart 2006 Theorem 64] Although (2-10) allows for com-putation of the trace of ρν (see (2-20) below) it is not convenient for computing thematrix coefficients For this purpose we shall use the explicit model for ρν definedin [Piatetski-Shapiro 1983 Section 13] as follows1

Given a primitive character ν of Llowast and ψ as above let

V = C[klowast]

1There is a minus sign missing from the definition of j (x) in Equation (4) of Section 13 of[Piatetski-Shapiro 1983] (otherwise his identity (6) will not hold) Likewise a minus sign is missingfrom (16) on page 40 The expression four lines above (16) is correct (except K should be K lowast)


be the vector space of functions f klowastrarr C We define a representation ρν of Gon V as follows For any

(a b0 d

)isin B f isin V let[

ρν

(a b0 d

)f

](x)= ν(d)ψ(bdminus1x) f (adminus1x) (x isin klowast) (2-11)

and for g isin Gminus B define

(ρν(g) f )(x)=sumyisinklowast

φ(x y g) f (y) (x isin klowast) (2-12)

where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast

N (t)=xyminus1 det g

ψ

(minus

yc(t + t )

)ν(t) (2-13)

Theorem 25 If ν is a primitive character of Llowast (2-11) and (2-12) give a well-defined representation (ρν V ) which is cuspidal Furthermore every cuspidalrepresentation is isomorphic to some ρν and ρν sim= ρνprime if and only if ν prime isin ν νq

Inparticular there are (q2

minus q)2 distinct cuspidal representations

Proof See [Piatetski-Shapiro 1983 Section 13ndash14] where it is assumed that q gt 2throughout When q = 2 G is isomorphic to the symmetric group S3 The uniquecuspidal representation is the character sending each permutation to its sign It isreadily checked that the above construction defines this character as well so thetheorem remains valid when q = 2

Define an inner product on V by

〈 f1 f2〉 =sumxisinklowast

f1(x) f2(x) (2-14)

We will work with the orthonormal basis

B= fr risinklowast for fr (x)=

1 if x = r0 if x 6= r

Proposition 26 Let ν be a primitive character of Llowast and let ρν be the associatedcuspidal representation of G Then ρν is unitary with respect to the inner product(2-14)

Proof By linearity it suffices to prove that for all fr fs isinB and g isin G

〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)

By the Bruhat decomposition G = B cup BwprimeU for wprime =(

1minus1

)and the fact that ρν

is a homomorphism we only need to consider g isin B and g = wprime


Suppose first that g =(

a b0 d

)isin B Then by (2-11)lang

ρν(g) fr ρν(g) fsrang=

sumxisinklowast

ν(d)ψ(bdminus1x) fr (adminus1x)ν(d)ψ(bdminus1x) fs(adminus1x)

Using the fact that ν and ψ are unitary and replacing x by aminus1 dx we see that thisexpression equals sum

xisinklowastfr (x) fs(x)= 〈 fr fs〉

as neededIt remains to prove (2-15) for g = wprime =

( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast

φ(x ywprime) f (y)=minus1q

sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy

ψ(u+ u)ν(u)ν(yminus1) f (y)=sumyisinklowast

ν(yminus1) j (xy) f (y)

wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence

ρν(wprime) fr (x)=

sumyisinklowast

ν(yminus1) j (xy) fr (y)= ν(rminus1) j (r x)

We now see that

〈ρν(wprime) fr ρν(w

prime) fs〉 =sumxisinklowast

ν(rminus1) j (r x)ν(sminus1) j (sx)= ν(srminus1)sumxisinklowast

j (r x) j (sx)

= ν(srminus1)sumxisinklowast

j (rsminus1x) j (x)

Taking r prime = rsminus1 it suffices to prove thatsumxisinklowast

j (r primex) j (x)=

0 if r prime 6= 11 if r prime = 1

(2-17)

From the definition (2-16) we havesumxisinklowast

j (r primex) j (x)= 1q2

sumxisinklowast

sumN (α)=r primex

sumN (β)=x

ψ(α+ α)ν(α)ψ(minusβ minus β)ν(βminus1)

=1q2

sumβisinLlowast

sumN (α)=r primeN (β)

ψ(α+ αminusβ minus β)ν(αβminus1)


Since the norm map is surjective there exists z isin Llowast such that N (z) = r prime Thenα = zβu for some u isin L1 This allows us to rewrite the above sum assum

xisinklowastj (r primex) j (x)=

1q2

sumβisinLlowast

sumuisinL1

ψ(zβu+ zβuminusβ minus β

)ν(zu)

=ν(z)q2

sumuisinL1

ν(u)sumβisinLlowast

ψ(tr[(zuminus 1)β]

) (2-18)

Generally for c isin L the map R(β) = ψ(tr[cβ]) is a homomorphism from L toClowast If c 6= 0 then R is nontrivial since the trace map from L to k is surjective Itfollows that sum

βisinLlowastψ(tr[cβ]

)=

minus1 if c 6= 0q2minus 1 if c = 0

(2-19)

Suppose r prime 6= 1 Then N (zu)= N (z)= r prime 6= 1 so in particular zu 6= 1 Therefore(2-18) becomes sum

xisinklowastj (r primex) j (x)=minus

ν(z)q2

sumuisinL1

ν(u)= 0

where we have used the fact (Proposition 21) that ν is a nontrivial character of L1

since ν is primitiveNow suppose r prime = 1 Then we can take z = 1 so by (2-18) and (2-19)sum

xisinklowastj (x) j (x)=

1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1

ν(u)=minus1 again because ν is nontrivial on L1 This proves (2-17)

23 Matrix coefficients of cuspidal representations Let ν be a primitive charac-ter of Llowast Using (2-10) one finds that

tr ρν(x)=

(q minus 1)ν(x) if x isin Z minusν(z) if x = zu z isin Z u isinU u 6= 1minusν(x)minus νq(x) if x isin T x isin Z 0 if no conjugate of x is in T cup ZU

(2-20)

(see [Bushnell and Henniart 2006 (641)]) This is a sum of matrix coefficientsFor the coefficients themselves we use the model given in the previous section toprove the following (which can also be used to derive (2-20))


Theorem 27 Let g=(

a bc d

)isinG and fr fs isinB Let ρν be a cuspidal representation

of G If g isin B thenlangρν(g) fr fs

rang=

ν(d)ψ(bdminus1s) if r = adminus1s0 if r 6= adminus1s

If g isin B then langρν(g) fr fs

rang= φ(s r g)

where φ is defined in (2-13)

Proof First suppose g isin B (ie c = 0) Thenlangρν(g) fr fs

rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)

= ν(d)ψ(bdminus1s) fr (adminus1s)=ν(d)ψ(bdminus1s) if r = adminus1s0 if r 6= adminus1s

Now suppose g isin B Then[ρν(g) fr

](x)=

sumyisinklowast

φ(x y g) fr (y)= φ(x r g)

Therefore

〈ρν fr fs〉 =sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed

3 Depth-zero supercuspidal representations of GL2(F)

We move now to the p-adic setting When no field is specified G Z BU etcwill henceforth denote the corresponding subgroups of GL2(F) rather than GL2(k)Fix a primitive character ν and let ρν be the associated cuspidal representation ofGL2(k) We view ρν as a representation of K = GL2(o) via reduction modulo p

K minusrarr GL2(k)minusrarr GL(V )

The central character of this representation is given by z 7rarr ν(z(1+ p)) for z isin olowastExtend this character of olowast to Z sim= Flowast=

⋃nisinZ$

nolowast by choosing a complex numberν($) of absolute value 1 We denote this character of Flowast by ν This allows us toview ρν as a unitary representation of the group ZK Let

πν = c-IndGL2(F)ZK (ρν)

be the representation of G compactly induced from ρν This representation isirreducible and supercuspidal (see for example [Bump 1997 Theorem 481])


31 Matrix coefficients Define an inner product on c-IndGZK (ρν) by

〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)

where 〈 middot middot 〉V denotes the inner product on V defined in (2-14) As in Section 1this inner product is G-equivariant

The matrix coefficients of πν can now be computed explicitly by using Proposition11 in conjunction with Theorem 27 Likewise by Proposition 12 if we normalizeso that meas(K )= 1 then

dπν = dim ρν = q minus 1 (3-2)

Define a function φν G minusrarr C by

φν(x)=

tr ρν(x) if x isin ZK 0 otherwise

(3-3)

Then this is a pseudocoefficient of πν in the sense that for any irreducible temperedrepresentation π of G with central character ω

trπ(φν)=

1 if π sim= πν0 otherwise

(see [Palm 2012 Section 941]) The function φν may be computed explicitly using(2-20)

32 New vectors For an integer n ge 0 define the congruence subgroup

K1(pn)=

(a bc d

)isin K | c (d minus 1) isin pn

If π is a representation of G we let πK1(pn) denote the space of vectors fixed by

K1(pn) By a result of Casselman [1973] for any irreducible admissible repre-

sentation π of G there exists a unique ideal pn (the conductor of π) for whichdimπK1(p

n)= 1 and dimπK1(p

nminus1)= 0 A nonzero vector fixed by K1(p

n) is calleda new vector The supercuspidal representations constructed above have conductorp2 We shall give an elementary proof below and exhibit a new vector Moregenerally the new vectors for depth-zero supercuspidal representations of GLn(F)were identified by Reeder [1991 Example (23)]

Proposition 31 The supercuspidal representation πν defined above has conductorp2 If we let w isin C[klowast] denote the constant function 1 that is

w =sumrisinklowast

fr (3-4)


then the function f = f($1

)w

supported on the coset

ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w

is a new vector of πν

Proof To see that f is K1(p2)-invariant it suffices to show that

f((

$1)k)= w for all k isin K1(p

2)

Writing k =(

a bc d

)with c isin p2 and d isin 1+ p2 we have

f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w

as needed This shows that πK1(p2)

ν 6= 0 so the conductor divides p2 There arevarious ways to see that the conductor is exactly p2 When ngt1 it is straightforwardto show that a continuous irreducible n-dimensional complex representation ofthe Weil group of F has Artin conductor of exponent at least n (see for example[Gross and Reeder 2010 Equation (1)]) So by the local Langlands correspondencethe conductor of any supercuspidal representation of GLn(F) is divisible by pn giving the desired conclusion here when n = 2 For an elementary proof in thepresent situation one can observe that a function f isin c-IndG

ZK (ρν) supported on acoset ZK x is K1(p)-invariant if and only if ρν is trivial on K cap x K1(p)xminus1 Usingthe double coset decomposition

G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)

(we may use the representatives δ isin 1 cup( y 1

1 0

) ∣∣ y isin op see for example

[Knightly and Li 2006 Lemma 131]) it suffices to consider x =($ n

1

)δ and one

checks that in each case ρν is not trivial on K cap x K1(p)xminus1 so πK1(p)ν = 0


33 Matrix coefficient of the new vector Generally if π is a supercuspidal repre-sentation with unit new vector v and formal degree dπ the function

g 7rarr dπ 〈π(g)v v〉

can be used to define a projection operator onto CvIn the present context if f is the new vector defined in Proposition 31 one finds

easily that f 2 = (q minus 1) so with the standard normalization meas(K ) = 1 by(3-2) we have

8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)

for w as in (3-4) This is computed as follows

Theorem 32 Let h =(

a bc d

)isin G(k) = GL2(k) and let w isin V be the function

defined in (3-4) Then

〈ρν(h)ww〉V =

(q minus 1)ν(d) if b = c = 0minusν(d) if c = 0 b 6= 0minus

sumαisinLlowast

α+α=aN (α)det h +d

ν(α) if c 6= 0

Remark The sum may be evaluated using Proposition 33 below

Proof To ease notation we drop the subscript V from the inner product Supposeh =

(a0

bd

)isin B(k) Then applying Theorem 27lang

ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast

langρν(h) fadminus1s fs

rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives

〈ρν(h)ww〉 =(q minus 1)ν(d) if b = 0minusν(d) if b 6= 0


Now suppose h =(a

cbd

)isin G(k)minus B(k) Then by Theorem 27

〈ρν(h)ww〉 =sum

rsisinklowastφ(s r h)

=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast

N (α)=srminus1 det h

ψ(minusrcminus1(α+ α)

)ν(α)

Let l = srminus1 so s = rl From the previous display we have

〈ρν(h)ww〉 = minus1q

sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h

ν(α)sumrisinklowast

ψ(rcminus1(al + d minus (α+ α ))

)=minus

1q

sumlisinklowast

sumN (α)=l det h


ψ(r(al + d minus (α+ α))

)

There are two cases for the inner sumsumrisinklowast


)=

minus1 if al + d minus (α+ α) 6= 0q minus 1 if al + d minus (α+ α)= 0

Thereforelangρν(h)ww

rang=minus

1q

sumlisinklowast

( sumN (α)=l det hα+α 6=al+d

minusν(α)+sum

N (α)=l det hα+α=al+d

(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)

By Lemma 22 the first sum in the big parentheses vanishes Solangρν(h)ww

rang=minus

1q

sumlisinklowast

qsum


ν(α)=minussumαisinLlowast


ν(α) (3-7)

as claimed

This sum can be refined as follows

Proposition 33 For l isin klowast and h =(

a bc d

)isin G(k) with c 6= 0 define

phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast

phl (X) irred over k

sumrootsα of

phl (X) in Llowast

ν(α) minussumlisinklowast

phl (X)=(Xminusα)2in k[X ]

ν(α) (3-9)

Proof If α isin Llowast contributes to the sum (3-7) then it is a root of

X2minus

(aN (α)det h

+ d)

X + N (α)

Conversely given l isin klowast a root αl of phl(X) contributes to (3-7) if and only ifphl(X)= (Xminusαl)(Xminus αl) in L[X ] which is the case if and only if either phl(X)is irreducible or phl(X)= (Xminusαl)

2 with αl isin klowast The proposition now follows

34 Motivation Although specific global applications of the above formulas arebeyond the scope of this article perhaps a few words of motivation will be helpfulThe two functions φν and 8ν of (3-3) and (3-5) serve slightly different purposesThe former is simpler and hence easier to work with Taken as a local component ofa global test function it is well suited for use in the ArthurndashSelberg trace formulafor example if one is interested in detecting those automorphic representationsπ =

otimesw πw of the adelic group GL2(AQ) with the local condition πw sim= πν at a

given finite place w (eg to obtain a dimension formula for the associated space ofclassical newforms) This method was treated in detail recently by Palm [2012] Onthe other hand as mentioned in the Introduction the matrix coefficient 8ν givesrise to an operator projecting onto the span of the newforms attached to the globalrepresentations π as above (see [Knightly and Li 2012 Section 25]) It can be usedin variants of the trace formula to extract finer information like Fourier coefficientsor L-values of these newforms Of course the utility of 8ν in explicit computationis limited by the complexity of the sum in Theorem 32

4 Consideration of central character

The supercuspidal representations which have a given central character ω occurnaturally together as irreducible subrepresentations of the right regular representationof G(F) on the space of L2 functions f G(F)rarrC that transform under the centerby ω It is often the case in number theory that one can achieve a certain amountof simplification by simultaneously treating all objects in a family via averagingHere we sum the trace functions φν from (3-3) over all isomorphism classes ofdepth-zero supercuspidal πν with a given central character and similarly for thenew vector matrix coefficients 8ν of (3-5)

Let ω be a unitary character of Flowast and let Sω denote the set of isomorphismclasses of depth-zero supercuspidal representations of GL2(F)with central character


ω In order for Sω to be nonempty ω must be trivial on 1+ p In fact we have

|Sω| =

Pω2 if ω |(1+p) = 10 otherwise

(4-1)

for Pω as in Proposition 23 (Note that ν and νq have the same restriction to klowast

and ρν sim= ρνq )Assuming ω |(1+p) is trivial consider the sum of the trace functions φν defined

in (3-3) In view of the fact that φν = φνq we define

φω =12

sumνisin[ω]

φν (4-2)

with notation as in Proposition 24 We can make it explicit with the following

Theorem 41 Suppose q is odd and ω(qminus1)2 is nontrivial Then for x isin G(k)

sumνisin[ω]

tr ρν(x)=

(q2minus 1)ω(x) if x isin Z

minus(q + 1)ω(z) if x = zu z isin Z u isinU u 6= 10 if no conjugate of x is in ZU

Suppose q is odd and ω(qminus1)2 is trivial Then with T as in (2-9)

sumνisin[ω]

tr ρν(x)=

(q minus 1)2ω(x) if x isin Z minus(q minus 1)ω(z) if x = zu z isin Z u isinU u 6= 14ω(x (q+1)2) if x isin T minus Z x (q

2minus1)2= 1

0 if x isin T minus Z x (q2minus1)2=minus1 or if

no conjugate of x is in T cup ZU

Suppose q is even Then

sumνisin[ω]

tr ρν(x)=

q(q minus 1)ω(x) if x isin Z minusqω(z) if x = zu z isin Z u isinU u 6= 12ω(N (x)12) if x isin T minus Z 0 if no conjugate of x is in T cup ZU

Proof This follows immediately by examining the various cases using (2-20) andProposition 24

Likewise for a depth-zero supercuspidal representation π = πν let 8π =8ν bethe matrix coefficient defined in (3-5) Define a function 8ω on G by

8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)

with h isin ZK In principle this function can be used todefine an operator that projects the automorphic spectrum of GL2(AQ) onto the span


of those newforms of a given weight and level p2 that correspond to automorphicrepresentations which are unramified away from p and are supercuspidal (as opposedto special or principal series) at p

One can evaluate 8ω via the following

Theorem 42 Suppose h =( a

d)isin G(k) is diagonal Then

sumνisin[ω]

langρν(h)ww

rangV =

(q2minus 1)ω(d) if q is odd and ω(qminus1)2 is nontrivial

(q minus 1)2ω(d) if q is odd and ω(qminus1)2 is trivialq(q minus 1)ω(d) if q is even

(4-4)

If h =(

a b0 d

)isin B(k) with b 6= 0 then

sumνisin[ω]

langρν(h)ww

rangV =

minus(q + 1)ω(d) if q is odd and ω(qminus1)2 is nontrivialminus(q minus 1)ω(d) if q is odd and ω(qminus1)2 is trivialminusqω(d) if q is even

(4-5)

If g isin G(k)minus B(k) then the sum is given by (4-6) below

Proof Suppose h =(

a 00 d

)is diagonal Then by Theorem 32 we can writesum

νisin[ω]

〈ρν(h)ww〉 = (q minus 1)sumνisin[ω]

ν(d)

Applying Proposition 24 now gives (4-4) using the fact that d isin klowastSimilarly if h =

(a b0 d

)with b 6= 0 then applying Theorem 32sum

νisin[ω]

〈ρν(h)ww〉 = minussumνisin[ω]

ν(d)

Using Proposition 24 this gives (4-5)Now suppose h =

(a bc d

)isin G(k)minus B(k) By Proposition 33sum

νisin[ω]

〈ρν(h)ww〉 = minussumlisinklowast

phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)

where phl(X) = X2minus((al det h)+ d

)X + l isin k[X ] If we fix one root αl isin Llowast

of phl(X) for each l then the above is

=minus2sumlisinklowast

phl (X) irred

sumνisin[ω]

ν(αl) minus Pωsumlisinklowast

phl (X)=(Xminusαl )2

ω(αl) (4-6)

where Pω =∣∣[ω]∣∣ as in Proposition 23 We have used the fact thatsum

νisin[ω]

ν(αl)=sumνisin[ω]

ν(αl)


since ν and νq both belong to [ω] Once again (4-6) can be evaluated on a case-by-case basis using Proposition 24

For instance suppose q is odd If ω(qminus1)2 is nontrivial the first term of (4-6)vanishes If (al + d det h)= 0 then l makes no contribution to the second term Inparticular the term vanishes if a = d = 0 Generally at most two l contribute tothe second term when q is odd since 2αl = ((al det h)+ d) implies that l satisfiesthe quadratic equation l = α2

l =14

((al det h)+ d

)2On the other hand if q is even a given l isin klowast contributes to the second term of

(4-6) if and only if (al + d det h)= 0 since (X minusαl)2= X2

minusα2l and as remarked

earlier every l is a square when q is even

5 Examples

Take q = 2 By (4-1) there is a unique supercuspidal representation π of GL2(Q2)

of conductor 22 As mentioned before the cuspidal representation of GL2(F2)sim= S3

is the character ρ sending a matrix g to (minus1)|g|+1 where |g| is the order of gin the finite group Explicitly ρ sends

(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to

1minus1minus1minus1 1 1 respectively This one-dimensional representation of coursecoincides with its matrix coefficientlang

ρ(h)wwrang= ρ(h)〈ww〉 = ρ(h)

Indeed one may verify that Theorem 32 recovers ρ when q = 2 For exampleconsider h=

(0 11 1

) The polynomial (3-8) becomes ph1(X)= X2

+X+1 If θ isin Flowast4is a root then ν(θ)= exp(2π i3) defines a primitive character By (3-9)

ρ(h)=minus(ν(θ)minus ν(θ2)

)= 1

By (3-6) the matrix coefficient 8 attached to the new vector of π has the simpleexpression

8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)

Now consider k = F5 and L = F25 Then L = k[θ ] where θ2= 2 One finds that

1+ 2θ generates the cyclic group Llowast so the characters of Llowast are the maps

νn(1+ 2θ)= ζ n (n isin Z24Z)

where ζ = exp(2π i24) Note that νn is primitive if and only if 6 - n There areexactly four characters of klowast given by

ωn(3)= in (n isin Z4Z)


Noting that νn(3) = νn((1+ 2θ)6) = ζ 6n= in we see that νn|klowast = ωn So the

equivalence classes of primitive characters of Llowast are as follows

[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23

where each primitive character is listed alongside its conjugate The above illustratesProposition 23 As a simple illustration of Theorem 42 we now show thatsum

νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0

)the polynomial (3-8) becomes phl(X)= X2

+l The second term in(4-6) vanishes Thus an element l isin klowast contributes to (4-6) only if minusl is a quadraticnonresidue in k The roots of X2

+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15

The roots of X2+ 3 are

α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21

For any ν isin [ω0] ν(α j ) is a power of ζ 12=minus1 when 4|n Hence ν5(α j )= ν(α j )

Thus in this case (4-6) equals

minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)

]=minus4[minus1+ 1minus 1+ 1] = 0

Acknowledgements

This work is based on Ragsdalersquos Masterrsquos thesis at the University of Maine Wethank the referee for carefully reading the manuscript and making several sug-gestions and corrections which have greatly improved the exposition Ragsdalethanks the Department of Mathematics and Statistics of the University of Maine forsummer support during the final stages of writing this paper Both authors weresupported by NSF grant DMS 0902145

References

[Bump 1997] D Bump Automorphic forms and representations Cambridge Studies in AdvancedMathematics 55 Cambridge University Press 1997 MR 97k11080 Zbl 086811022

[Bushnell and Henniart 2006] C J Bushnell and G Henniart The local Langlands conjecture forGL(2) Grundlehren der Math Wiss 335 Springer Berlin 2006 MR 2007m22013 Zbl 110011041

[Bushnell and Henniart 2014] C J Bushnell and G Henniart ldquoLanglands parameters for epipelagicrepresentations of GLnrdquo Math Ann 3581ndash2 (2014) 433ndash463 arXiv 13024304

[Bushnell and Kutzko 1993] C J Bushnell and P C Kutzko The admissible dual of GL(N ) viacompact open subgroups Annals of Mathematics Studies 129 Princeton University Press 1993MR 94h22007 Zbl 078722016


[Casselman 1973] W Casselman ldquoOn some results of Atkin and Lehnerrdquo Math Ann 201 (1973)301ndash314 MR 49 2558 Zbl 023910015

[Gross and Reeder 2010] B H Gross and M Reeder ldquoArithmetic invariants of discrete Langlandsparametersrdquo Duke Math J 1543 (2010) 431ndash508 MR 2012c11252 Zbl 120711111

[Kim 2007] J-L Kim ldquoSupercuspidal representations an exhaustion theoremrdquo J Amer Math Soc202 (2007) 273ndash320 MR 2008c22014 Zbl 111122015

[Knightly and Li 2006] A Knightly and C Li Traces of Hecke operators Mathematical Surveysand Monographs 133 American Mathematical Society Providence RI 2006 MR 2008g11090Zbl 112011024

[Knightly and Li 2012] A Knightly and C Li ldquoModular L-values of cubic levelrdquo Pacific J Math2602 (2012) 527ndash563 MR 3001804 Zbl 06136442

[Mautner 1964] F I Mautner ldquoSpherical functions over b-adic fields IIrdquo Amer J Math 86 (1964)171ndash200 MR 29 3582 Zbl 013517204

[Palm 2012] M Palm Explicit GL(2) trace formulas and uniform mixed Weyl laws PhD thesisGoumlttingen 2012 arXiv 12124282

[Piatetski-Shapiro 1983] I Piatetski-Shapiro Complex representations of GL(2 K ) for finite fieldsK Contemporary Mathematics 16 American Mathematical Society Providence RI 1983 MR84m20046 Zbl 051320026

[Reeder 1991] M Reeder ldquoOld forms on GLnrdquo Amer J Math 1135 (1991) 911ndash930 MR 92i22018Zbl 075811027

[Stevens 2008] S Stevens ldquoThe supercuspidal representations of p-adic classical groupsrdquo InventMath 1722 (2008) 289ndash352 MR 2010e22008 Zbl 114022016

Received 2013-08-14 Revised 2013-11-12 Accepted 2013-11-16

knightlymathumaineedu Department of Mathematics and StatisticsUniversity of Maine 5752 Neville Hall Room 333Orono ME 04469-5752 United States

cwragsdasyredu Department of Mathematics and StatisticsUniversity of Maine 5752 Neville Hall Room 333Orono ME 04469-5752 United States

mathematical sciences publishers msp

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Kenneth S Berenhaut Wake Forest University USA berenhkswfuedu

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Joseph OrsquoRourke Smith College USAorourkecssmithedu

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Ram U Verma University of Toledo USAverma99msncom

John C Wierman Johns Hopkins University USAwiermanjhuedu

Michael E Zieve University of Michigan USAzieveumichedu

PRODUCTIONSilvio Levy Scientific Editor

See inside back cover or msporginvolve for submission instructions The subscription price for 2014 is US $120year for the electronic version and$165year (+$35 if shipping outside the US) for print and electronic Subscriptions requests for back issues from the last three years and changesof subscribers address should be sent to MSP

Involve (ISSN 1944-4184 electronic 1944-4176 printed) at Mathematical Sciences Publishers 798 Evans Hall 3840 co University of CaliforniaBerkeley CA 94720-3840 is published continuously online Periodical rate postage paid at Berkeley CA 94704 and additional mailing offices

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PUBLISHED BY

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httpmsporgcopy 2014 Mathematical Sciences Publishers


involve2014 vol 7 no 5

585Infinite cardinalities in the Hausdorff metric geometryALEXANDER ZUPAN

595Computing positive semidefinite minimum rank for small graphsSTEVEN OSBORNE AND NATHAN WARNBERG

611The complement of Fermat curves in the planeSETH DUTTER MELISSA HAIRE AND ARIEL SETNIKER

619Quadratic forms representing all primesJUSTIN DEBENEDETTO

627Counting matrices over a finite field with all eigenvalues in the fieldLISA KAYLOR AND DAVID OFFNER

647A not-so-simple Lie bracket expansionJULIE BEIER AND MCCABE OLSEN

657On the omega values of generators of embedding dimension-threenumerical monoids generated by an interval

SCOTT T CHAPMAN WALTER PUCKETT AND KATY SHOUR

669Matrix coefficients of depth-zero supercuspidal representations ofGL(2)

ANDREW KNIGHTLY AND CARL RAGSDALE

691The sock matching problemSARAH GILLIAND CHARLES JOHNSON SAM RUSH ANDDEBORAH WOOD

699Superlinear convergence via mixed generalized quasilinearizationmethod and generalized monotone method

VINCHENCIA ANDERSON COURTNEY BETTIS SHALABROWN JACQKIS DAVIS NAEEM TULL-WALKER VINODHCHELLAMUTHU AND AGHALAYA S VATSALA

involve2014

vol7no5

Introduction



21 Primitive characters of L

22 Model for cuspidal representations

23 Matrix coefficients of cuspidal representations


31 Matrix coefficients

32 New vectors

33 Matrix coefficient of the new vector

34 Motivation


5 Examples

Acknowledgements

References

mspINVOLVE 75 (2014)

dxdoiorg102140involve20147669

Matrix coefficients of depth-zero supercuspidalrepresentations of GL(2)Andrew Knightly and Carl Ragsdale

(Communicated by Michael E Zieve)

We give explicit formulas for matrix coefficients of the depth-zero supercuspidalrepresentations of GL(2) over a nonarchimedean local field highlighting the casewhere the test vector is a unit new vector We also describe the partition of theset of such representations according to central character and compute sums ofmatrix coefficients over all representations in a given class

Introduction

Let F be a nonarchimedean local field with integer ring o maximal ideal p=$oand residue field k = op of cardinality q The supercuspidal representations ofGL2(F) are precisely those irreducible admissible representations which do notarise as constituents of parabolic induction They are characterized by having matrixcoefficients which are compactly supported modulo the center

In this paper we explicitly compute the matrix coefficients of depth-zero super-cuspidal representations These are the supercuspidals with the smallest possibleconductor exponent namely 2 First discovered by Mautner [1964 Section 9] theyarise by compact induction from the (qminus1)-dimensional representations of GL2(o)

inflated from the cuspidal series of the finite group GL2(k)In the first section we show that the matrix coefficients of any supercuspidal

representation are expressible in terms of those of the finite-dimensional inducingrepresentation Thus the task at hand essentially reduces to a computation of thematrix coefficients of the cuspidal representations of GL2(k) The latter is achievedin Theorem 27 using the explicit model from [Piatetski-Shapiro 1983]

With global applications in mind in Section 3 we single out the case wherethe test vector in the supercuspidal matrix coefficient is a unit new vector Theresulting function given in (3-6) and Theorem 32 may be used to define an integraloperator on the global automorphic spectrum of GL2 which isolates those cuspidal

MSC2010 22E50Keywords supercuspidal matrix coefficients regular characters

669








H (ρ) given by

〈φψ〉 =sum

xisinHG

langφ(x) ψ(x)

rangV


rang=

sumxisinHG

langφ(xg) ψ(xg)

rangV =

sumxisinHG

langφ(x) ψ(x)

rangV = 〈φψ〉



H (ρ) by


Then the set





rang=

langρ(h)vw





xisinHG


rangV

=

sumxisinHG

langfyv(xg) fzw(x)

rangV

=langfyv(zg) w

rangV


fyv(hy) wrangV =

langρ(h)vw

rangV




f 4

dπ(1-1)




dπ =dim ρ

meas(H)




rang=

langρ(g)v v

rangV if g isin H

0 if g isin H

Therefore intG


intH


=v4

dim(ρ)meas(H)=

meas(H)dim(ρ)


meas(H)dim(ρ)

= f1v

4

dπ


f1v2= 〈 f1v f1v〉 =

sumxisinHG

langf1v(x) f1v(x)

rangV

=langf1v(1) f1v(1)


as needed



U =(

1 b0 1

)isin G

Z =

( aa)isin G

B =

(a b0 d

)isin G



H sim= H





N (α)= αα = αq+1













N (x)=α

ν(x)= 0


ν(x)=sum

N (x)=α

ν(λx)= ν(λ)sum

N (x)=α

ν(x)

It follows thatsum

N (x)=αν(x)= 0



Pω =





qminus1)= ξ(1)= 1



q0

(2-1)










b+ 12 b(q minus 1)= 1

2 b(q + 1) (2-2)

giving ννb20 = ξ

b2 N and

b+b+ q + 1

2(q minus 1)=

b+ q minus 12

(q + 1) (2-3)










ν(α)=


(2-4)


sumνisin[ω]

ν(α)=


2minus1)2= 1


(2-5)



ν(α)=


(2-6)



νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)


m=0

νm0 (α)=


χ(α)=


(2-7)



ν(α)= ξ b(α)

( qsumm=0


b2minus ν0(α)

(b+q+1)2)



b2+ ν0(α)

b2ν0(α)(q+1)2]



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References








H (ρ) given by

〈φψ〉 =sum

xisinHG

langφ(x) ψ(x)

rangV


rang=

sumxisinHG

langφ(xg) ψ(xg)

rangV =

sumxisinHG

langφ(x) ψ(x)

rangV = 〈φψ〉



H (ρ) by


Then the set





rang=

langρ(h)vw





xisinHG


rangV

=

sumxisinHG

langfyv(xg) fzw(x)

rangV

=langfyv(zg) w

rangV


fyv(hy) wrangV =

langρ(h)vw

rangV




f 4

dπ(1-1)




dπ =dim ρ

meas(H)




rang=

langρ(g)v v

rangV if g isin H

0 if g isin H

Therefore intG


intH


=v4

dim(ρ)meas(H)=

meas(H)dim(ρ)


meas(H)dim(ρ)

= f1v

4

dπ


f1v2= 〈 f1v f1v〉 =

sumxisinHG

langf1v(x) f1v(x)

rangV

=langf1v(1) f1v(1)


as needed



U =(

1 b0 1

)isin G

Z =

( aa)isin G

B =

(a b0 d

)isin G



H sim= H





N (α)= αα = αq+1













N (x)=α

ν(x)= 0


ν(x)=sum

N (x)=α

ν(λx)= ν(λ)sum

N (x)=α

ν(x)

It follows thatsum

N (x)=αν(x)= 0



Pω =





qminus1)= ξ(1)= 1



q0

(2-1)










b+ 12 b(q minus 1)= 1

2 b(q + 1) (2-2)

giving ννb20 = ξ

b2 N and

b+b+ q + 1

2(q minus 1)=

b+ q minus 12

(q + 1) (2-3)










ν(α)=


(2-4)


sumνisin[ω]

ν(α)=


2minus1)2= 1


(2-5)



ν(α)=


(2-6)



νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)


m=0

νm0 (α)=


χ(α)=


(2-7)



ν(α)= ξ b(α)

( qsumm=0


b2minus ν0(α)

(b+q+1)2)



b2+ ν0(α)

b2ν0(α)(q+1)2]



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



H (ρ) by


Then the set





rang=

langρ(h)vw





xisinHG


rangV

=

sumxisinHG

langfyv(xg) fzw(x)

rangV

=langfyv(zg) w

rangV


fyv(hy) wrangV =

langρ(h)vw

rangV




f 4

dπ(1-1)




dπ =dim ρ

meas(H)




rang=

langρ(g)v v

rangV if g isin H

0 if g isin H

Therefore intG


intH


=v4

dim(ρ)meas(H)=

meas(H)dim(ρ)


meas(H)dim(ρ)

= f1v

4

dπ


f1v2= 〈 f1v f1v〉 =

sumxisinHG

langf1v(x) f1v(x)

rangV

=langf1v(1) f1v(1)


as needed



U =(

1 b0 1

)isin G

Z =

( aa)isin G

B =

(a b0 d

)isin G



H sim= H





N (α)= αα = αq+1













N (x)=α

ν(x)= 0


ν(x)=sum

N (x)=α

ν(λx)= ν(λ)sum

N (x)=α

ν(x)

It follows thatsum

N (x)=αν(x)= 0



Pω =





qminus1)= ξ(1)= 1



q0

(2-1)










b+ 12 b(q minus 1)= 1

2 b(q + 1) (2-2)

giving ννb20 = ξ

b2 N and

b+b+ q + 1

2(q minus 1)=

b+ q minus 12

(q + 1) (2-3)










ν(α)=


(2-4)


sumνisin[ω]

ν(α)=


2minus1)2= 1


(2-5)



ν(α)=


(2-6)



νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)


m=0

νm0 (α)=


χ(α)=


(2-7)



ν(α)= ξ b(α)

( qsumm=0


b2minus ν0(α)

(b+q+1)2)



b2+ ν0(α)

b2ν0(α)(q+1)2]



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



rang=

langρ(g)v v

rangV if g isin H

0 if g isin H

Therefore intG


intH


=v4

dim(ρ)meas(H)=

meas(H)dim(ρ)


meas(H)dim(ρ)

= f1v

4

dπ


f1v2= 〈 f1v f1v〉 =

sumxisinHG

langf1v(x) f1v(x)

rangV

=langf1v(1) f1v(1)


as needed



U =(

1 b0 1

)isin G

Z =

( aa)isin G

B =

(a b0 d

)isin G



H sim= H





N (α)= αα = αq+1













N (x)=α

ν(x)= 0


ν(x)=sum

N (x)=α

ν(λx)= ν(λ)sum

N (x)=α

ν(x)

It follows thatsum

N (x)=αν(x)= 0



Pω =





qminus1)= ξ(1)= 1



q0

(2-1)










b+ 12 b(q minus 1)= 1

2 b(q + 1) (2-2)

giving ννb20 = ξ

b2 N and

b+b+ q + 1

2(q minus 1)=

b+ q minus 12

(q + 1) (2-3)










ν(α)=


(2-4)


sumνisin[ω]

ν(α)=


2minus1)2= 1


(2-5)



ν(α)=


(2-6)



νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)


m=0

νm0 (α)=


χ(α)=


(2-7)



ν(α)= ξ b(α)

( qsumm=0


b2minus ν0(α)

(b+q+1)2)



b2+ ν0(α)

b2ν0(α)(q+1)2]



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



N (α)= αα = αq+1













N (x)=α

ν(x)= 0


ν(x)=sum

N (x)=α

ν(λx)= ν(λ)sum

N (x)=α

ν(x)

It follows thatsum

N (x)=αν(x)= 0



Pω =





qminus1)= ξ(1)= 1



q0

(2-1)










b+ 12 b(q minus 1)= 1

2 b(q + 1) (2-2)

giving ννb20 = ξ

b2 N and

b+b+ q + 1

2(q minus 1)=

b+ q minus 12

(q + 1) (2-3)










ν(α)=


(2-4)


sumνisin[ω]

ν(α)=


2minus1)2= 1


(2-5)



ν(α)=


(2-6)



νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)


m=0

νm0 (α)=


χ(α)=


(2-7)



ν(α)= ξ b(α)

( qsumm=0


b2minus ν0(α)

(b+q+1)2)



b2+ ν0(α)

b2ν0(α)(q+1)2]



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References




qminus1)= ξ(1)= 1



q0

(2-1)










b+ 12 b(q minus 1)= 1

2 b(q + 1) (2-2)

giving ννb20 = ξ

b2 N and

b+b+ q + 1

2(q minus 1)=

b+ q minus 12

(q + 1) (2-3)










ν(α)=


(2-4)


sumνisin[ω]

ν(α)=


2minus1)2= 1


(2-5)



ν(α)=


(2-6)



νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)


m=0

νm0 (α)=


χ(α)=


(2-7)



ν(α)= ξ b(α)

( qsumm=0


b2minus ν0(α)

(b+q+1)2)



b2+ ν0(α)

b2ν0(α)(q+1)2]



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References




ν(α)=


(2-4)


sumνisin[ω]

ν(α)=


2minus1)2= 1


(2-5)



ν(α)=


(2-6)



νisin[ω]

ν(α)= ν(α)

qsumm=0

νm0 (α)


m=0

νm0 (α)=


χ(α)=


(2-7)



ν(α)= ξ b(α)

( qsumm=0


b2minus ν0(α)

(b+q+1)2)



b2+ ν0(α)

b2ν0(α)(q+1)2]



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



= ν this is


minus1)2)]=minusν

(α(q+1)2)[1+ ξ(α(q2

minus1)2)] (2-8)




(α(q

2minus1)2



νisin[ω]


)








T ν (2-10)



V = C[klowast]




(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



(a b0 d


ρν

(a b0 d

)f





where for g =(

a bc d

)isin Gminus B

φ(x y g)=minus1qψ

[ax + dy

c

] sumtisinLlowast


ψ

(minus

yc(t + t )

)ν(t) (2-13)







f1(x) f2(x) (2-14)



1 if x = r0 if x 6= r



〈ρν(g) fr ρν(g) fs〉 = 〈 fr fs〉 (2-15)


1minus1





a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



a b0 d



sumxisinklowast





( 0 1minus1 0

) By (2-12)

ρν(wprime) f (x)

=

sumyisinklowast


sumyisinklowast

sumtisinLlowast

N (t)=xyminus1

ψ(y(t + t))ν(t) f (y)

=minus1q

sumyisinklowast

sumuisinLlowast

N (u)=xy



wherej (t)=minus1

q

sumuisinLlowast

N (u)=t

ψ(u+ u)ν(u) (2-16)

Hence


sumyisinklowast


We now see that




j (r x) j (sx)


j (rsminus1x) j (x)


j (r primex) j (x)=


(2-17)



sumxisinklowast

sumN (α)=r primex

sumN (β)=x


=1q2

sumβisinLlowast






1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References




1q2

sumβisinLlowast

sumuisinL1


)ν(zu)

=ν(z)q2

sumuisinL1



) (2-18)



)=


(2-19)



ν(z)q2

sumuisinL1

ν(u)= 0




1q2

sumuisinL1


ψ(tr[(uminus 1)β]

)=

q2minus 1

q2 minus1q2

sumuisinL1

u 6=1

ν(u)=(

1minus1q2

)+

1q2 = 1

sincesum

uisinL1

u 6=1



tr ρν(x)=


(2-20)



Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References


Theorem 27 Let g=(

a bc d



rang=



rang= φ(s r g)



rang=

sumxisinklowast

[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)



](x)=

sumyisinklowast


Therefore


[ρν(g) fr

](x) fs(x)=

[ρν(g) fr

](s)= φ(s r g)

as needed





⋃nisinZ$






〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



〈 f1 f2〉 =sum

xisinZKG

langf1(x) f2(x)

rangV (3-1)





φν(x)=


(3-3)


trπ(φν)=




K1(pn)=

(a bc d





n)= 1 and dimπK1(p




w =sumrisinklowast

fr (3-4)



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



)w


ZK($ 00 1

)= ZK

($ 00 1

)K1(p

2)

and defined by

f(

zk($ 00 1

))= ρν(zk)w



f((


2)

Writing k =(

a bc d


f

(($

1

)k

)= f

(($

1

)(a bc d

)($minus1

1

)($

1

))

= ρν

((a $b

$minus1c d

))w = ρν

((a 00 1

))w

=

sumrisinklowast

ρν

((a

1

))fr =

sumrisinklowast

faminus1r = w




G =⋃nge0

ZK($ n

1

)K =

⋃nge0

⋃δisinKK1(p)

ZK($ n

1

)δK1(p)


1 0



1

)δ and one







8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References






8ν(g)def= dπν

langπν(g)

f f

f f

rang= 〈πν(g) f f 〉 (3-5)

By Proposition 11

Supp(8ν)=($minus1

1

)ZK

($

1

)

and for g =($minus1

1

)h($

1)isin Supp(8ν)

8ν(g)= 〈ρν(h)ww〉V (3-6)


Theorem 32 Let h =(

a bc d



〈ρν(h)ww〉V =


sumαisinLlowast


ν(α) if c 6= 0



(a0

bd


ρν(h)wwrang=

sumrsisinklowast

langρν(h) fr fs

rang=

sumsisinklowast


rang= ν(d)

sumsisinklowast

ψ(bdminus1s)

This gives



Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References


Now suppose h =(a

cbd




=minus1q

sumrsisinklowast

ψ(cminus1(sa+ rd)

) sumαisinLlowast



)ν(α)



sumrisinklowast

sumlisinklowast

ψ(cminus1(rla+ rd)

) sumN (α)=l det h


)ν(α)

=minus1q

sumlisinklowast

sumN (α)=l det h



)=minus

1q

sumlisinklowast

sumN (α)=l det h



)



)=



rang=minus

1q

sumlisinklowast


minusν(α)+sum


(q minus 1)ν(α))

=minus1q

sumlisinklowast

(minus

sumN (α)=l det h

ν(α)+ qsum


ν(α)

)


rang=minus

1q

sumlisinklowast

qsum




ν(α) (3-7)

as claimed



a bc d


phl(X)= X2minus

(al

det h+ d

)X + l isin k[X ] (3-8)


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References


Thenlangρν(h)ww

rang= minus

sumlisinklowast


sumrootsα of

phl (X) in Llowast



ν(α) (3-9)


X2minus

(aN (α)det h

+ d)

X + N (α)











|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References



|Sω| =


(4-1)




φω =12

sumνisin[ω]

φν (4-2)



sumνisin[ω]

tr ρν(x)=




sumνisin[ω]

tr ρν(x)=


2minus1)2= 1




sumνisin[ω]

tr ρν(x)=




8ω(g)=sumπisinSω

8π (g)=12

sumνisin[ω]

〈ρν(h)ww〉V (4-3)

for g =($minus1

1

)h($

1)







sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References






sumνisin[ω]

langρν(h)ww

rangV =



(4-4)

If h =(

a b0 d


sumνisin[ω]

langρν(h)ww

rangV =


(4-5)


Proof Suppose h =(

a 00 d


νisin[ω]


ν(d)


(a b0 d


νisin[ω]


ν(d)


(a bc d


νisin[ω]


phl (X) irred

sumroots α of

phl (X) in Llowast

sumνisin[ω]


phl (X)=(Xminusα)2

sumνisin[ω]

ν(α)





phl (X) irred

sumνisin[ω]



ω(αl) (4-6)


νisin[ω]


ν(αl)




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References




l =14

((al det h)+ d





5 Examples




(1 00 1

)(

0 11 0

)(

1 10 1

)(

1 01 1

)(

1 11 0

)(

0 11 1

)to




(0 11 1




)= 1


8(g)=ρ(h) if g = z

(2minus1

1

)h(

21

)isin Z

(2minus1

1

)K(

21

)

0 otherwise(5-1)









[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References




[ω0] = ν4 ν20 ν8 ν16 [ω1] = ν1 ν5 ν9 ν21 ν13 ν17

[ω2] = ν2 ν10 ν14 ν22 [ω3] = ν3 ν15 ν7 ν11 ν19 ν23


νisin[ω0]

langρν

(( 0 1minus1 0

))ww

rang= 0

For h=( 0 1minus1 0



+ 2 are

α2 = (1+ 2θ)3 and α2 = (1+ 2θ)15


α3 = θ = (1+ 2θ)9 and θ = (1+ 2θ)21



minus2[2ν4(α2)+ 2ν8(α2)+ 2ν4(α3)+ 2ν8(α3)


Acknowledgements


References






















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References


















































gladasmathuriedu





































PUBLISHED BY


















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References
















involve2014

vol7no5

Introduction








32 New vectors


34 Motivation


5 Examples

Acknowledgements

References

matrix coefficients of depth-zero supercuspidal representations of...

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