Let | | be the valuation on F such that |π| = q−1. In this section we fix a split or irregular γ∈ [Γ ] with rational form rγ = diag(1, a, b), where ordπb ordπa 0.
7.1. Minimal lengths of homotopy cycles in [γ]
We begin by proving the first assertion of Theorem 5.4.1 for the split and irregular cases.
Theorem 7.1.1. Suppose γ∈ Γ is split or irregular with rγ = diag(1, a, b), where ordπb ordπa 0. Then
(1) lA([γ]) = ordπa + ordπb = minκγ(gKZ)∈[γ]lA(κγ(gKZ)), and (2) lG([γ]) = ordπb = minκγ(gKZ)∈[γ]lG(κγ(gKZ)).
Proof. For γ split, the centralizer CG(rγ) consists of the diagonal matrices in G so that, by Iwasawa decomposition, G = CG(rγ)U K, where y runs through the determinant of all i× i minors of g−1rγg. Consequently,
e1= min
and
e1+ e2+ e3= ordπa + ordπb. (7.3) In particular, e3 ordπb from the last two inequalities. Moreover, we have, for any g∈ G,
lA
κγ(PγgKZ)
= e3+ e2+ e1− 3e1= ordπa + ordπb− 3e1
ordπa + ordπb = lA
[γ]
(7.4) and
lG
κγ(PγgKZ)
= e3− e1 ordπb− e1 ordπb = lG
[γ]
. (7.5)
As noted before, the equalities in(7.4)and (7.5)hold for g ∈ CG(rγ). Therefore lA
[γ]
= min
κγ(gKZ)∈[γ]lA
κγ(gKZ)
and lG
[γ]
= min
κγ(gKZ)∈[γ]lG
κγ(gKZ) .
For γ irregular, we have either a = b or a = 1, and the centralizer CG(rγ) is isomorphic to GL2(F )× Z and G = CG(rγ)U0K, where U0consists of the elements in U with z = 0 (when a = b) or x = 0 (when a = 1). The above argument still holds. This proves the theorem. 2
The proof above shows that if κγ(PγgKZ) is algebraically minimal, then e1= 0; and it is geometrically minimal if the additional condition e1+ e2 = ordπa is satisfied. By (7.2), this obviously holds when ordπa = 0, i.e., γ has type 1. The proof above also shows that for γ irregular of type 1, a tailless κγ(PγgKZ) has g∈ CG(rγ)K. We record this in
Corollary 7.1.2. Suppose [γ] has type 1. Then algebraically minimal cycles in [γ] are geometrically minimal, hence they agree with the tailless cycles in [γ]. Moreover, if γ is irregular and has type 1, then the tailless cycles in [γ] are κγ(PγgKZ) with g∈ CG(rγ)K.
7.2. Counting homotopy cycles in [γ] in algebraic length
As discuss in Section 5.7, the number of algebraically minimal cycles in [γ] is the cardinality of CP−1
γ Γ Pγ(rγ)\ΔA([γ]) with ΔA([γ]) defined by (5.3). We showed in the previous section that for γ irregular, ΔA([γ]) = CG(rγ)K/KZ so that the number of algebraically tailless cycles in [γ] is equal to vol([γ]) given by(5.4).
The following theorem, stated in terms of a formal power series, counts the number of homotopy cycles in [γ] with given algebraic length.
Theorem 7.2.1. Suppose γ ∈ [Γ ] is split or irregular with rγ = diag(1, a, b). Then
# CP−1
γ Γ Pγ(rγ)\ΔA
[γ]
= vol [γ]
ω[γ]
and
κγ(gKZ)∈[γ]
ulA(κγ(gKZ))=
$vol([γ])· ω[γ]· ulA([γ]) 1−u1−q3u33 if γ splits, vol([γ])· ω[γ]· ulA([γ]) 1−u1−q2u33 if γis irregular.
Here vol([γ]) is given by (5.4), ω[γ] = (|1 − a||a − b||b − 1|)−1 for γ split, and ω[γ] = 1 for γ irregular.
Proof. If γ is split, the stabilizer CG(rγ) is the diagonal subgroup of G and G = CG(rγ)U KZ; while if γ is irregular, the stabilizer CG(rγ) is GL2(F )Z, where GL2(F ) is imbedded in G as diagonal block GL2(F )× {1} (when a = 1) or {1} × GL2(F ) (when a = b), and G = CG(rγ)U0KZ. Here U and U0are as in the proof ofTheorem 7.1.1. Put W = U or U0 according as γ split or irregular. Then (CG(rγ)∩ KZ)W KZ = W KZ.
Suppose S represents the double cosets in CP−1
γ Γ Pγ(rγ)\CG(rγ)/(CG(rγ)∩ KZ), then S has cardinality vol([γ]) by(5.4), and
G =
h∈S
CP−1
γ Γ Pγ(rγ)h
CG(rγ)∩ KZ
W KZ =
h∈S
CP−1
γ Γ Pγ(rγ)hW KZ.
Lemma 7.2.2. For γ∈ Γ split or irregular, the elements hu with h ∈ S and u ∈ W , where S and W are defined above, are double coset representatives of CP−1
γ Γ Pγ(rγ)\G/KZ.
Proof. Suppose CP−1
γ Γ Pγ(rγ)huKZ = CP−1
γ Γ Pγ(rγ)huKZ for h, h∈ S and u, u ∈ W . Then there is some c ∈ CPγ−1Γ Pγ(rγ) such that huKZ = chuKZ, i.e., u−1h−1chu ∈ KZ. This together with the definition of W implies that h−1ch∈ CG(rγ)∩ KZ. There-fore h and h in S represent the same double coset of CG(rγ), hence h = h. On the other hand, since Γ intersects gZKg−1 trivially for all g∈ G by assumption, the same holds for its conjugate h−1Pγ−1Γ Pγh. Now h−1ch ∈ (h−1Pγ−1Γ Pγh)∩ KZ, hence is equal to the identity in G. So c = id and consequently uKZ = uKZ. This implies u = u by definition of W , as desired. 2
Since κγ(PγhgKZ) and κγ(PγgKZ) have the same algebraic length for h∈ CG(rγ) and g ∈ G, we get
κγ(PγgKZ)∈[γ]
ulA(κγ(PγgKZ))= vol
[γ]
v∈W
ulA(κγ(PγvKZ)),
where W = U or U0according to γ split or irregular.
To proceed, we compute the sum on the right hand side. First assume γ split so that W = U . Given v∈ U, write v =1 x y
1 z 1
. As computed in the proof ofTheorem 7.1.1,
(Pγv)−1γPγvzγ = v−1rγv =
⎛
⎝1 x(1− a) y(1 − b) + xz(b − a)
a z(a− b)
b
⎞
⎠ = (vi,j).
For fixed m 0, we count the number of v’s such that lA(κγ(PγvKZ)) lA([γ]) + 3m.
By(7.4), the constraints are|vij| qmfor all 1 i, j 3. In other words,
x(1− a) qm, z(a− b) qm and y(1− b) + xz(b − a) qm. (7.6)
This implies
|x| qm|1 − a|−1 and |z| qm|a − b|−1
so that the numbers of x and z in F/OF are qm|1 − a|−1 and qm|a − b|−1, respectively.
Further, for chosen x and z, there are qm|1 − b|−1 choices of y satisfying the above constraint. We have shown
#
v∈ U lA
κγ(PγvKZ)
= lA [γ]
=
|1 − a||a − b||b − 1|−1
= ω[γ] (7.7) and, for m > 0,
#
v∈ U lA
κγ(PγvKZ)
= lA
[γ] + 3m
=
q3m− q3m−3
ω[γ]. (7.8) Put together, this gives
v∈U
ulA(κγ(PγvKZ))= ω[γ]ulA([γ])
1 +
m1
q3m− q3m−3 u3m
= ω[γ]ulA([γ])
1− u3 1− q3u3
.
Next consider the case γ irregular so that W = U0. Recall that U0consists of elements in U with z = 0 (when a = b) or x = 0 (when a = 1). Note that ordπb > 0, for otherwise γ would lie in the intersection of Γ with a conjugate of K, which is trivial. Consequently, 1− b is a unit in OF so that|1 − b| = 1. The argument above restricted to elements in U0 goes through as before, but the three inequalities in(7.6) are reduced to two with either x(1− a) = 0 or z(a − b) = 0. This then shows that the number of nonzero x or z is qm− 1 and the number of y is qm. Hence we obtain
#
v∈ U0lA
κγ(PγvKZ)
= lA
[γ] + 3m
= q2m− q2m−2, (7.9) which in turn gives
v∈U0
ulA(κγ(PγvKZ)) = ulA([γ])
1 +
m1
q2m− q2m−2 u3m
= ω[γ]ulA([γ])
1− u3 1− q2u3
. 2
7.3. Counting homotopy cycles of type 1 in [γ]
The theorem below gives the number of type 1 homotopy cycles in [γ] of given algebraic length. The result depends on the type of [γ].
Theorem 7.3.1. With the same notation as in Theorem 7.2.1, we have:
(A) If [γ] splits and is not of type 1, then
κγ(gKZ)∈[γ], type 1
ulA(κγ(gKZ))= vol [γ]
ω[γ]ulA([γ])
1− q−11 − q2u3 1− q3u3
.
Moreover, no type 1 cycles in [γ] are geometrically minimal.
(B) If [γ] splits and has type 1, then
κγ(gKZ)∈[γ], type 1
ulA(κγ(gKZ))= vol [γ]
ω[γ]ulA([γ])
q−1+
1− q−11 − q2u3 1− q3u3
.
(C) Suppose γ ∈ Γ is irregular. Then [γ] contains no cycles of type 1 if [γ] is not of type 1; while if [γ] has type 1, then
κγ(gKZ)∈[γ], type 1
ulA(κγ(gKZ))= vol [γ]
ω[γ]ulA([γ]).
Proof. For γ ∈ Γ split or irregular, we have rγ = diag(1, a, b), so [γ] has type (ordπb− ordπa, ordπa) and lA([γ]) = ordπb + ordπa. It has type 1 if and only of ordπa = 0. The argument is similar to the proof ofTheorem 7.2.1; the difference is that we only need to consider those v ∈ W such that κγ(PγvKZ) has type 1. Here W = U or U0 according as γ split or irregular. So we determine the cardinality of the set
v∈ W lG
κγ(PγvKZ)
= lA
κγ(PγvKZ)
= lA
[γ]
+ 3m = ordπb + ordπa + 3m for each m 0. As before, writing v as1 x y
1 z 1
and following the proofs ofTheorem 7.2.1 and Theorem 7.1.1, we arrive at the following constraints on x, y, z∈ F/OF:
(1) min{0, ordπx(1− a), ordπz(a− b), ordπ(y(1− b) + xz(b − a))} = −m, and (2) min{ordπa, ordπ[x(1− a)z(a − b) − a(y(1 − b) + xz(b − a))]} = −2m.
For m > 0, the two constraints are equivalent to
(3) ordπx(1− a) = −m = ordπz(a− b) and ordπ(y(1− b) + xz(b − a)) −m.
First assume γ splits. The number of x is (1− q−1)qm|1 − a|−1, the number of z is (1− q−1)qm|a − b|−1, and the number of y is qm|1 − b|−1 so that the total number of v is (1− q−1)2q3mω[γ]. For m = 0 and ordπa > 0, the same constraint (3) holds. In this case the number of x is|1 − a|−1 = 1, the number of y is|1 − b|−1= 1 and the number of z is (1− q−1)|a − b|−1 so that the total number of v is (1− q−1)ω[γ]. Finally, when m = ordπa = 0, the constraints (1) and (2) are equivalent to
(4) ordπx(1− a) 0, ordπz(a− b) 0 and ordπ(y(1− b) + xz(b − a)) 0.
Hence the numbers of x, y and z are|1 − a|−1,|1 − b|−1 and|a − b|−1, respectively, so that the number of v is ω[γ]. Note that y = z = 0 in this case.
Since vol([γ])ω[γ] is present in all cases, it suffices to compute 1
vol([γ])ω[γ]
κγ(gKZ)∈[γ], type 1
ulA(κγ(gKZ)).
In case ordπa > 0, namely [γ] does not have type 1, this sum is equal to
ulA([γ])
1− q−1+
m1
1− q−12
q3mu3m
= ulA([γ])
1− q−11 − q2u3 1− q3u3
,
and in case ordπa = 0, namely [γ] has type 1, it is equal to
ulA([γ])
1 +
m1
1− q−12
q3mu3m
= ulA([γ])
q−1+
1− q−11 − q2u3 1− q3u3
.
This proves (A) and (B).
When γ is irregular, either a = 1 or a = b, so (3) never holds and there are no cycles in [γ] of type 1 and algebraic length > lA([γ]). Further, there are vol([γ]) cycles in [γ]
with algebraic length equal to lA([γ]) and they have the same type as [γ]. This proves the assertion (C). 2
Contained in the proof above is the following statement.
Corollary 7.3.2. Suppose γ ∈ Γ is split or irregular with rγ = diag(1, a, b). Assume that γ has type 1, a∈ OF× and n = ordπb. Let δ = δ([γ]) = ordπ(1− a) for γ split, and δ = 0 for γ irregular. Then
ΔA
[γ]
=
⎧⎨
⎩hvxKZ h∈ CG(rγ)/
CG(rγ)∩ KZ , vx=
⎛
⎝1 x 1
1
⎞
⎠ with x ∈ π−δOF/OF
⎫⎬
⎭.