Homotopy cycles in [γ] for γ rank-one split

⎧⎨

⎩hvxKZ h∈ CG(r_γ)/

C_G(r_γ)∩ KZ , v_x=

⎛

⎝1 x 1

1

⎞

⎠ with x ∈ π^−δOF/OF

⎫⎬

⎭.

8. Homotopy cycles in [γ] for γ rank-one split

In this section we fix a rank-one split γ ∈ [Γ ] whose eigenvalues generate a quadratic extension L = F (λ) of F . Here λ is a unit or uniformizer in L according as L is unramified or ramified over F , i.e., γ is unramified or ramified rank-one split. Let rγ =

a e dc d e+db

be a rational form of γ as in Section 4.3. Fix a matrix Pγ so that P_γ⁻¹γPγzγ = rγ for some z_γ ∈ Z.

8.1. The centralizers of r_γ for γ rank-one split Embed L^× in GL2(F ) as the subgroup

%u vc

v u + vb

  u,v ∈ F, not both zero&

, (8.1)

which is further imbedded in GL3(F ) as '1 u vc v u+vb

(

. Note that CG(rγ) = L^×Z.

Further C_P−1

γ Γ Pγ(r_γ)\CG(r_γ)/(C_G(r_γ)∩ KZ) has cardinality vol([γ]) by(5.4).

The group of units UL of L^× is contained in K. If L is unramified over F , then L^×=πULso that CG(rγ)K/KZ is represented by the vertices diag(πⁿ, 1, 1)KZ, n∈ Z, on a line inB, and CPγ⁻¹Γ Pγ(r_γ)\CG(r_γ)/(C_G(r_γ)∩KZ) represented by diag(πⁿ, 1, 1)KZ, n mod vol([γ]). If L is ramified over F , then L^× =πLUL, where the uniformizer π_L does not lie in F and π²_L differs from π by a unit multiple. In this case CG(rγ)K/KZ is represented by the vertices diag(πⁿ, 1, 1)KZ and diag(πⁿ, 1, 1)πLKZ, n∈ Z, lying on two lines in B. There are two possibilities for CPγ⁻¹Γ Pγ(r_γ):

Case (i). The vertices in C_P−1

γ Γ Pγ(r_γ)KZ/KZ are contained in the line diag(πⁿ, 1, 1)KZ, n∈ Z. Then vol([γ]) is even so that C_Pγ⁻¹_{Γ Pγ}(rγ)\CG(rγ)/(CG(rγ)∩ KZ) is represented by the vertices diag(πⁿ, 1, 1)KZ and diag(πⁿ, 1, 1)πLKZ, n mod vol([γ])/2.

Case (ii). C_P−1

γ Γ Pγ(r_γ)KZ/KZ contains a vertex on the line diag(πⁿ, 1, 1)π_LKZ, n∈ Z. Let y ∈ CPγ⁻¹Γ Pγ(r_γ) be such that yKZ = diag(π^N, 1, 1)π_LKZ has the least non-negative N . Then y generates the group C_P−1

γ Γ Pγ(rγ), y²KZ = diag(π^2N−1, 1, 1)KZ, vol([γ]) = 2N−1 is odd, and CPγ⁻¹Γ Pγ(r_γ)\CG(r_γ)/(C_G(r_γ)∩KZ) is represented by the vertices diag(πⁿ, 1, 1)KZ, 0  n  N − 1 = (vol([γ]) − 1)/2, and diag(πⁿ, 1, 1)π_LKZ, 0 n  N − 2 = (vol([γ]) − 3)/2.

8.2. Double coset representatives of CG(rγ)\G/KZ

Proposition 8.2.1. The double cosets in CG(rγ)\G/KZ are represented by elements in

S =

Proof. Write an element g ∈ G as wk for some upper triangular w and some k ∈ K.

Since CG(r_γ) = L^×Z, modulo the center Z, we may assume that w =1 x y

First we check the disjoint union. Suppose otherwise. Then there exist m= n and g satisfying

Replacing g by its inverse if necessary, we may assume m > n. Write g =

x yπ⁻ⁿ a nonunit and hence z and y should both be units, but then yπ⁻ⁿ= czπ^mcannot hold by checking the order of both sides.

Next we prove equality. Let w =_{1 z}

represent the same double coset. Since ord_πc = 0 or 1, only such diagonal matrices with m  0 are needed as double coset representa-tives. Thus we assume ordπz < 0. It suffices to reduce w to a diagonal matrix via left multiplication by elements in L^× and right multiplication by elements in GL2(OF).

Case (I). 0 > ord_πz m+ordπc. Choose v∈ OF with ord_πv+m+ord_πc = ord_πz and

Case (II). m + ordπc > ordπz. Choose u ∈ OF with ordπu + ordπz = m + ordπc and v a unit such that uz =−vcπ^m. Thenu vc

v u+vb

w =_{u 0}

v vz+(u+vb)π^m

=_{u 0}

0 z

k for some k ∈ GL2(OF).

In both cases we have shown that w lies in the right hand side of(8.2), therefore(8.2) holds. This proves the proposition. 2

8.3. Minimal lengths of cycles in [γ]

The type of [γ], as defined in Section 5.4, is (n, m) such that rγ ∈ Tn,m = K diag(1, π^m, π^n+m)KZ. Observe that ord_πdet γ ≡ ordπdet r_γ ≡ ordπa(e + dλ)(e + d¯λ) ≡ 0 mod 3 by the assumption on Γ . Hence if e + dλ is a unit in L, then at least one of e, d is a unit and a is not a unit. Consequently, [γ] has type (ordπa, 0). Next assume e + dλ is not a unit. We distinguish two cases. If L is unramified over F (hence λ is a unit), then both e and d are non-units and a is a unit; in this case [γ] has type (0, min(ordπe, ordπd)). If L is ramified over F (hence λ is a uniformizer of L), then there are two possibilities:

(i) ord_π(e + dλ)(e + d¯λ) = 1. This happens if and only if e is a non-unit, d is a unit, and ordπa 2; in this case [γ] has type (ordπa− 1, 1).

(ii) ordπ(e + dλ)(e + d¯λ) > 1. This happens if and only if both e and d are non-units and a is a unit; in this case [γ] has type (0, ord_πe) if ord_πe ordπd, and type (1, ord_πd) if ord_πe > ord_πd.

This proves the first assertion of

Theorem 8.3.1. Let γ be a rank-one split element in [Γ ] with rational form rγ =

^a

e dc d e+db

. Suppose that rγ ∈ K diag(1, π^m, π^m+n)KZ. Then (1) The type (n, m) of [γ] is as follows.

(1.i) If ordπc = 0, then (n, m) = (ordπa, min{ordπe, ordπd}).

(1.ii) If ordπc = 1, then (n, m) = (ordπa, ordπe) provided that ordπe  ordπd, otherwise (n, m) = (max{ordπa− 1, 1}, max{ordπd, 1}).

(2) lA([γ]) = minκγ(gKZ)∈[γ]lA(κγ(gKZ)) = ordπa(e²+ edb− cd²) = n + 2m.

(3) lG([γ]) = min_{κγ(gKZ)∈[γ]}lG(κγ(gKZ)) = n + m.

This theorem combined withTheorem 7.1.1completes the proof of Theorem 5.4.1.

Remark. If γ is ramified rank-one split and [γ] has type (n, 1), then [γ²] has type (2n + 1, 0).

Proof. It remains to show that the algebraic and geometric lengths of the cycles in [γ]

are at least those of [γ] since, as observed before, the cycles κ_γ(P_γgKZ) with g∈ CG(r_γ) have the same algebraic and geometric lengths as [γ]. ByProposition 8.2.1, it suffices to

compute (Pγg)⁻¹γPγgzγ = g⁻¹rγg for g∈ S. Let g =1 x y 1 0 πⁱ

, where x, y∈ F/OF and i 0. Then

g⁻¹rγg =

⎛

⎝1 −x −yπ⁻ⁱ

1 0

π⁻ⁱ

⎞

⎠

⎛

⎝a

e dc

d e + db

⎞

⎠

⎛

⎝1 x y

1 0

πⁱ

⎞

⎠

=

⎛

⎝a (a− e)x − dyπ⁻ⁱ (a− e − db)y − cdxπⁱ

e dcπⁱ

dπ⁻ⁱ e + db

⎞

⎠

∈ K

⎛

⎝π^e1 π^e2

π^e3

⎞

⎠ K.

Here e1 e2 e3, and as in the proof of Theorem 7.1.1, we have

e₁ min{ordπa,−i + ordπd, ord_πe}  min{ordπa, ord_πd, ord_πe} = 0, (8.3) e₁+ e₂ min

ord_πae,−i + ordπad, ord_π

e²+ bed− cd²

 min

ord_πae, ord_πad, ord_π

e²+ bed− cd²

= m, (8.4)

and

e1+ e2+ e3= ordπa

e²+ bed− cd²

= n + 2m, (8.5)

in which the last upper bound for e1+e2can be verified using the statement (1). Therefore lA(κγ(PγgKZ)) = e1+ e2+ e3− 3e1 e1+ e2+ e3= n + 2m = lA([γ]) since e1 0. The inequalities(8.4) and (8.5) together give the lower bound e3  n + 2m − m = n + m, which in turn implies l_G(κ_γ(P_γgKZ)) = e₃− e1 n + m. This proves the theorem. 2 As shown in the proof above, if [γ] has type 1, i.e. m = 0, then an algebraically minimal cycle in [γ] satisfies e₁= 0, which implies e₁+ e₂ 0 and hence e1+ e₂= 0 by (8.4)and e₃= n by(8.5). This proves

Corollary 8.3.2. Suppose γ∈ [Γ ] is rank-one split. If [γ] has type 1, then the algebraically minimal cycles in [γ] coincide with the geometrically minimal (hence tailless) cycles in [γ].

8.4. Counting the number of cycles in [γ] in algebraic length

As observed before, given s∈ S, the cycles κγ(P_γgKZ) have the same algebraic length for all gKZ ∈ CG(r_γ)sK/KZ. Since S represents the double coset C_G(r_γ)\G/KZ, to count the number of cycles in [γ] of a given length, we need to determine the cardinality of C_P−1

γ Γ Pγ(rγ)\CG(rγ)sK/KZ for s∈ S. For this, we may take as representatives the

product of representatives of C_P−1

γ Γ Pγ(rγ)\CG(rγ)/(CG(rγ)∩KZ) (independent of s) by the representatives of (CG(rγ)∩ KZ)sK/KZ. The number of the former representatives is vol([γ]) by(5.4).

It remains to compute the cardinality of the latter. Recall that L^×∩ K consists of the units in L^×, which are identified with the matrices

UL= reduced to counting, for each m 0, the cardinality of UL

₁

#

We summarize the above discussion in Corollary 8.4.2. For each s = 1 x y

Now we are ready to state the main result of this section.

Theorem 8.4.3. Suppose γ ∈ [Γ ] is rank-one split with rational form rγ =

(A) If γ is unramified rank-one split, then the following hold.

(A1)

(B) If γ is ramified rank-one split, then the following hold.

(B1)

Moreover, in each case, if [γ] does not have type 1, none of the type 1 cycles in [γ] are geometrically minimal.

Remarks. 1. μ = 0 unless a, e, c are all nonunits, in which case it is 1 and δ = 0.

2. μ = 0 when [γ] has type 1.

3. δ > 0 in case (A2), while δ may be zero in case (A3).

Proof. Recall that the algebraic length of a cycle in [γ] is equal to lA([γ]) + 3m for some m  0. We shall follow the same notation and computation as in the proof of Theorem 8.3.1, letting g run through all elements in the double coset representatives S and computing, for each m 0, the number of cycles κγ(P_γgKZ) with l_A(κ_γ(P_γgKZ)) lA([γ]) + 3m using Corollary 8.4.2. As g =

1 x y 1 0 πⁱ

, this amounts to computing the number of x, y∈ F/OF and i 0 such that

e1= min ordπ

(a− e)x − dπ⁻ⁱy , ordπ

−cdπⁱx + (a− e − db)y

,−i + ordπd

 −m.

This is equivalent to 0 i  m + ordπd, (a− e)x − dπ⁻ⁱy ∈ π^−mOF and −cdπⁱx + (a− e − db)y ∈ π^−mOF. Denote ordπd by δ for short. So for each 0 i  m + δ, we solve the following system of linear equations

α β



=

 a− e −dπ⁻ⁱ

−cdπⁱ a− e − db

 x y



= M

x y



(8.6) for α, β ∈ π^−mOF and count the distinct pairs (x, y) ∈ F/OF × F/OF. Recall that a, e, d are integral, at least one of them is a unit, and a and e cannot be both units since ordπdet rγ > 0. Let

μ := ordπdet M = ordπ

(a− e)²− db(a − e) − cd² , which is 0 unless a, e and c are all nonunits, in which case it is 1. Put

ε := min

ordπ(a− e), −i + δ, ordπ(a− e − bd) ,

which is equal to−i + δ if δ  i  m + δ, and 0 if 0  i < δ. Then the coefficient matrix M = k1diag(π^ε, π^μ−ε)k2 for some k1, k2 ∈ GL2(OF). Thus system (8.6) has the same number of solutions as the system

α β



=

π^ε π^μ−ε

 x y



(8.7) for α, β∈ π^−mOF and (x, y)∈ F/OF× F/OF. We get the solutions x∈ π^−m−εOF/OF

and y∈ π^−m−μ+εOF/OF so that there are q^2m+μ different pairs (x, y) for each 0 i  m + δ. To proceed, we distinguish two cases.

Case (A) ordπc = 0, that is, γ is unramified rank-one split. Then μ = 0. By Corol-lary 8.4.2, the number of classes in [γ] with algebraic length at most lA([γ]) + 3m is

vol

Combined with Corollary 8.4.2, we see that the number of type 1 cycles κγ(PγgKZ) with l_A(κ_γ(P_γgKZ)) = l_A([γ]) + 3m is vol([γ])(q− 1)q^2m−1(q^δ+m+ q^δ+m−1).

Next consider the case m = 0. Under the assumption ord_πc = 0, we know from Theorem 8.3.1 that [γ] has type (ordπa, min{ordπe, ordπd}). Therefore it has type 1 if and only if ordπa > 0, in which case all cycles in [γ] with algebraic length equal to lA([γ]) have type 1, and the number of such cycles is vol([γ])^qδ+1_q−1^+qδ−2, as computed above. If [γ] does not have type 1, then δ = ordπd > 0; the condition e1= e2= 0 implies i = δ

and only one solution (x, y) = (0, 0). In this case the number of type 1 cycles in [γ] with while if [γ] does not have type 1, then

 computation as in Case (A) together with Corollary 8.4.2 shows that the number of classes in [γ] with algebraic length at most lA([γ]) + 3m is

Now we compute the number of type 1 cycles κ_γ(P_γgKZ) with algebraic length l_A(κ_γ(P_γgKZ)) = l_A([γ]) + 3m. First consider the case m  1. Following the same argument as in Case (A) and applyingCorollary 8.4.2, we see that the number of such cycles is vol([γ])(q− 1)q^2m+μ−1q^δ+m.

Next we discuss the remaining case m = 0. By Theorem 8.3.1, [γ] has type 1 if and not have type 1, the number of type 1 cycles in [γ] with algebraic length equal to lA([γ]) is vol([γ])q^δif μ = 0, and vol([γ])(q−1)q^δif μ = 1. In other words, it is vol([γ])q^δ(q^μ−μ).

Summing up, we have proved the following:

If [γ] has type 1, then

while if [γ] does not have type 1, then



This completes the proof of the theorem. 2

Contained in the proofs ofCorollary 8.4.2andTheorem 8.4.3is the proposition below, in which while if γ is unramified rank-one split, then

ΔA

[γ]

={hgi,j,uKZ| h and gi,j,u as above}

∪

hgi,zKZh as above, 1 i  δ, z ∈ πOF/πⁱOF

.

Consequently, the number of algebraically minimal cycles in [γ] is

# C_P−1

γ Γ Pγ(rγ)\ΔA

[γ]

= vol [γ]

ω_[γ], where

ω_[γ]=

$_qδ+1+q^δ−2

q−1 if [γ] is unramified rank-one split,

q^δ+1−1

q−1 if [γ] is ramified rank-one split.

We end this subsection by comparing Δ_G([γ]) and Δ_G([γ²]), where Δ_G([γ]) is defined by (5.5). Suppose [γ] is of type (m, n). If [γ²] is of type (2m, 2n), then a geometri-cally minimal 1-geodesic κ_γ(P_γgKZ) repeated twice is still geometrically minimal, hence Δ_G([γ])⊆ ΔG([γ²]).

If [γ] is not of type (2m, 2n), then γ is ramified rank-one split of type (n, 1) or (1, n).

Assume first that γ is of type (n, 1) so that μ = 1 and δ = 0 in Theorem 8.4.3. In this case, there are q · vol([γ]) algebraically minimal geodesics in [γ]. Among these, (q− 1) vol([γ]) of them are of type 1, and vol([γ]) of them are of type (n, 1). The latter ones are also geometrically minimal. On the other hand κγ(PγgKZ) is geometrically minimal for all g ∈ CG(rγ). We conclude that κγ(PγgKZ) is geometrically minimal if and only if g∈ CG(r_γ). As C_G(r_γ)⊆ CG(r²_γ), any g∈ CG(r_γ) gives rise to a geometrically minimal cycle κ_γ²(P_γgKZ). This shows Δ_G([γ])⊆ ΔG([γ²]) if γ is of type (n, 1).

Finally, note that κ_γ−1(P_γgKZ) and κ_γ(P_γgKZ) have the same geometric length but opposite types, so the same conclusion holds for γ of type (1, n). We have shown Proposition 8.4.5. For γ∈ Γ we have ΔG([γ])⊆ ΔG([γ²]).

8.5. Counting the number of tailless cycles in X_Γ of given algebraic length

Recall that Nn(XΓ) counts the number of tailless cycles of type 1 in XΓ with algebraic length n. These cycles fall in the disjoint union of [γ] as [γ] runs through type 1 conjugacy classes of Γ , andTheorem 7.2.1andProposition 8.4.4give the number of such cycles in each [γ]. Combined with Proposition 6.1.1 we obtain the following explicit expressions of the edge zeta functions.

Theorem 8.5.1. For i= 1, 2 we have u d

dulog Z1,i(XΓ, u) = 

γ∈[Γ ], [γ] of type 1

i vol [γ]

ω_[γ]u^ilA([γ]),

where vol([γ]) is defined by (5.4)and ω_[γ] is as inTheorem 7.2.1 andProposition 8.4.4.

在文檔中 Zeta functions of complexes arising from PGL(3) (頁 29-41)