HOMOTOPY TYPE AND v1PERIODIC HOMOTOPY GROUPS OF
pCOMPACT GROUPS
DONALD M. DAVIS
Abstract.We determine the v1periodic homotopy groups of all
irreducible pcompact groups. In the most difficult, modular, cases,
we follow a direct path from their associated invariant polynomi
als to these homotopy groups. We show that, with several ex
ceptions, every irreducible pcompact group is a product of ex
plicit sphericallyresolved spaces which occur also as factors of p
completed Lie groups.
1.Introduction
In [4] and [3], the classification of irreducible pcompact groups was comple*
*ted.
This family of spaces extends the family of (pcompletions of) compact simple L*
*ie
groups. The v1periodic homotopy groups of any space X, denoted v11ss*(X)(p), *
*are
a localization of the portion of the homotopy groups detected by Ktheory; they*
* were
defined in [20]. In [17] and [16], the author completed the determination of th*
*e v1
periodic homotopy groups of all compact simple Lie groups. Here we do the same *
*for
all the remaining irreducible pcompact groups.1
Recall that a pcompact group ([22]) is a pair (BX, X) such that BX is pcomp*
*lete
and X = BX with H*(X; Fp) finite. Thus BX determines X and contains more
structure than does X. The homotopy type and homotopy groups of X do not take
into account this extra structure nor the group structure on X.
__________
Date: September 21, 2007.
2000 Mathematics Subject Classification. 55Q52, 57T20, 55N15.
Key words and phrases. v1periodic homotopy groups, pcompact groups, Adams
operations, Ktheory, invariant theory.
The author is grateful to Kasper Andersen and Clarence Wilkerson for valuable
suggestions related to this project.
1If the groups v11ssi(X) are finite, then pcompletion induces an isomorphism
v11ss*(X) ! v11ss*(Xp). ([9, p.1252])
1
2 DONALD M. DAVIS
According to [4, 1.1,11.1] and [3], the irreducible pcompact groups correspo*
*nd to
compact simple Lie groups2 and the padic reflection groups listed in [2, Table*
* 1]
for which the character field is strictly larger than Q. See [13, pp.430431] a*
*nd [25,
p.165] for other listings. We use the usual notation ((BXn)p, (Xn)p), where n i*
*s the
ShephardTodd numbering ([33] or any of the previouslymentioned tables) and p *
*is
the prime associated to the completion.
We will divide our discussion into four families of cases:
(1)The compact simple Lie groups_infinite family 1, part of in
finite family 2, and cases 28, 35, 36, 37 in the ShephardTodd
list.
(2)The rest of the infinite families numbered 2a, 2b, and 3.
(3)The nonmodular special cases, in which p does not divide the
order of the reflection group. This is cases 427 and 2934.
(4)The modular cases, in which p divides the order of the reflec
tion group. These are cases (X12)3, (X24)2, (X29)5, (X31)5, and
(X34)7. (Actually, we include (X12)3 in Case (3) along with the
nonmodular cases, and the DwyerWilkerson space (X24)2 was
handled in [6].)
Here is a brief summary of what we accomplish in each case. The author feels *
*that
his contributions here are nil in case (1)3, minuscule in case (2), modest in c*
*ase (3),
and significant in case (4).
(1)Spaces X1, X28, X35, X36, and X37 are, respectively, SU(n),
F4, E6, E7, and E8. These are pcompact groups for all primes
p, although for small primes they were excluded by Clark and
Ewing ([13]) because H*(BX; Fp) is not a polynomial algebra.
The exceptional Lie group G2is the case m = 6 in infinite family
2b. The spaces SO(n), Spin(n), and Sp(n) appear in the infinite
family 2a with m = 2. Simplification of the homotopy types
of many of these, when p is odd, to products of spheres and
__________
2Cases in which distinct compact Lie groups give rise to equivalent pcompact
groups are discussed in [4, 11.4].
3But he accomplished much in these cases in earlier papers such as [16], [17],
and [18].
pCOMPACT GROUPS 3
sphericallyresolved spaces was obtained in [29, (8.1),8.1]. The
v1periodic homotopy groups of these spaces were computed in
[18], [7], [17], [16], [8], and other papers. We will say no more
about these cases.
(2)In Section 2, we use work of Castellana and BrotoMoller to
show that the spaces in the infinite families can be decomposed,
up to homotopy, as products of factors of pcompletions of uni
tary groups, spheres, and sphere bundles over spheres. See 2.3,
2.5, and 2.7 for the specific results.
(3)In Table 3.2, we list the homotopy types of all cases (Xn)p which
are not products of spheres. There are 31 such cases. In each
case, we give the homotopy type as a product of spheres and
spaces which are spherically resolved with ff1 attaching maps.
In Remark 3.3, we discuss the easilycomputed v1periodic ho
motopy groups of these spaces.
(4)The most novel part of the paper is the determination of the
v1periodic homotopy groups of (X29)5, (X31)5, and (X34)7. We
introduce a direct, but nontrivial, path from the invariant poly
nomials to the v1periodic homotopy groups. En route, we de
termine the Adams operations in K*(BX; ^Zp) and K*(X; ^Zp).
In the case of (X34)7, we give new explicit formulas for the in
variant polynomials. We conjecture in 4.1 (resp. 5.17) that the
homotopy type of (X29)5 (resp. (X34)7) is directly related to
SU(20) (resp. SU(42)). We explain why it appears that an
analogous result is not true for (X31)5.
2. Infinite families 2 and 3
Family 3 consists of pcompleted4 spheres S2m1 with p 1 mod m, which is a
loop space due to work of Sullivan ([34]). The groups v11ss*(S2m1)(p), origin*
*ally due
to Mahowald (p = 2) and Thompson (p odd), are given in [19, 4.2].
__________
4All of our spaces are completed at an appropriate prime p. This will not alw*
*ays
be present in our notation. For example, we will often write SU(n) when we real*
*ly
mean its pcompletion.
4 DONALD M. DAVIS
Family 2 consists of spaces X(m, r, n) where m > 1, rm, and n > 1. The "de
grees" of X(m, r, n) are m, 2m, . .,.(n  1)m, m_rn. These are the degrees of i*
*nvariant
polynomials under a group action used in defining the space. The ClarkEwing ta*
*ble
doubles the degrees to form the "type," as these doubled degrees are the degree*
*s of
generators of H*(BX; Fp) in the cases which they consider. For most5 of the ir*
*re
ducible pcompact groups X, H*(X; Fp) is an exterior algebra on classes of grad*
*ing
2d  1, where d ranges over the degrees. Family 2b consists of spaces X(m, r, n*
*) in
which n = 2 and r = m, while family 2a is all other cases. The reason that thes*
*e are
separated is that 2b has more applicable primes. Indeed, for family 2a, there a*
*re p
compact groups when p 1 mod m, while for family 2b, these exist when p 1 m*
*od
m, and also p = 2 if m = 4 or 6, and p = 3 if m = 3 or 6. The case m = 6 in fam*
*ily
2b is the exceptional Lie group G2. Note that all primes work when m = 6. The c*
*ase
(p = 2, m = 4) has X = Sp(2), while (p = 3, m = 3) has X = SU(3) or P SU(3), the
projective unitary group. In this case, there are two inequivalent pcompact gr*
*oups
corresponding to the same Qpreflection group; however, since SU(3) ! P SU(3) i*
*s a
3fold covering space, they have isomorphic v1periodic homotopy groups.
The following results of Broto and Moller ([11]) and Castellana ([12]) will b*
*e useful.
They deal with the homotopy fixedpoint space XhG when G acts on a space of the
same homotopy type as a space X. Here and throughout, Cm denotes a cyclic group
of order m, and U(N) is the pcompletion of a unitary group.
Theorem 2.1. ([11, 5.2, 5.12]) If m(p  1), 0 s < m, and n > 0, then
U(mn + s)hCm ' X(m, 1, n)
and is a factor in a product decomposition of U(mn + s).
Theorem 2.2. ([11, 5.2, 5.14]) If m(p  1), m 2, r > 1, and n 2, then
X(m, r, n)hCm ' X(m, 1, n  1)
and is a factor in a product decomposition of X(m, r, n).
Corollary 2.3. If m(p  1) and r > 1, then
m_1
X(m, r, n) ' X(m, 1, n  1) x S2nr
__________
5According to [31], the only exclusions are certain compact Lie groups when p
is very small.
pCOMPACT GROUPS 5
and X(m, 1, n  1) is a factor in a product decomposition of U(m(n  1)).
Here X(m, 1, 1) is interpreted as S2m1.
Proof.We use Theorem 2.2 to get the first factor. By the Kunneth Theorem, the
m_1
other factor must have the same Fpcohomology as S2nr , and hence must have the
same homotopy type as this sphere. Now we apply Theorem 2.1 to complete the
proof. 
Remark 2.4. Our Corollary 2.3 appears as [12, 1.4], except that she has an appa*
*rent
typo regarding the dimension of the sphere. Also, neither she nor [11] have the
restriction r > 1, but it seems that the result is false for r = 1, since by in*
*duction it
would imply that X(m, 1, n) is a product of spheres, which is not usually true.
Remark 2.5. Let p be odd. By [29], for any N, pcompleted SU(N) splits as a
product of (p  1) spaces, each of which has H*(; Fp) an exterior algebra on o*
*dd
dimensional classes of dimensions b, b + q, . .,.b + tq, for some integers b an*
*d t. Here
and throughout q = 2(p1). Our space X(m, 1, n1) will be a product of (p1)=m
of these spaces for SU(m(n  1)). The v1periodic homotopy groups of these spac*
*es
can be read off from those of SU(m(n  1)), since the (p  1) factors have v1p*
*eriodic
homotopy groups in nonoverlapping dimensions. Thus, to the extent that [18] is
viewed as being a satisfactory description of v11ss*(SU(n))(p),6 Corollary 2.3*
* gives
v11ss*(X(m, r, n))(p)provided m(p  1).
Example 2.6. Let p = 7. Then X(2, 2, 6) ' X(2, 1, 5) x S11. There is a product
decomposition
(SU(10))7 ' B(3, 15) x B(5, 17) x B(7, 19) x S9 x S11x S13,
where B(2n + 1, 2n + 13) denotes a 7completed S2n+1bundle over S2n+13with at
taching map ff1. Then
X(2, 1, 5) ' B(3, 15) x B(7, 19) x S11.
__________
6[18, 1.4] states that v11ss2k(SU(n))(p)is a cyclic pgroup with exponent
min( p(j!S(k, j)) : j n), where S(, ) denotes the Stirling number of the se*
*cond
kind. In [21], more tractable formulas were obtained if n p2 p + 1. Here and
throughout, p() is the exponent of p.
6 DONALD M. DAVIS
What remains for Family 2 is the cases 2b when m(p + 1). These are the spaces
X(m, m, 2) with m(p + 1). Let B(3, 2p + 1) denote the pcompletion of an S3bu*
*ndle
over S2p+1with attaching map ff1.
Theorem 2.7. If m(p + 1), then
8
,
the matrix of _k is D = diag(k4, k8, k12, k20). On the basis (over ^Z5)
,
it is P 1DP , where
0 1
1 0 0 0
BB 1 C
_5 1 0 0CC
P = BBB16_ 8_ CC
@ 25 5 1 0A
_64_125_4_251_51
is the changeofbasis matrix, obtained using the linear terms in 4.8. The matr*
*ix in
the statement of the theorem is obtained by dividing P 1DP by k, since _k in K*
*1()
corresponds to _k=k in K1(). 
We can use Theorem 4.9 to obtain the v1periodic homotopy groups of (X29)5 as
follows.
22 DONALD M. DAVIS
Theorem 4.10. The groups v11ss*(X29)(5)are given by
8
>>>0 t 6 3 (4)
>>>
>>Z=5min(8,3+ 5(t74.54))t 7 (20)
>>> min(12,3+ (t114.58))
>>>Z=5 5 t 11 (20)
>: min(20,3+ (t1912.516))
Z=5 5 t 19 (20).
_ *
* !
*
* ( 5)T
Proof.We use the result of [9] that v11ss2t(X)(5)is presented by the matrix (*
* 2)T  2tI.
We form this matrix by letting k = 5 and 2 in the matrix of Theorem 4.9 and let*
*ting
x = 2t, obtaining
0 1
125 15600 31274880 9765631257408
BB 0 78125 78000000 3051773400000C
BB 0 0 48828125 3814687500000CC
BB CC
BB 0 0 0 19073486328125CC.
BB8  x 24 1344 268704 CC (4.11)
BB 0 128  x 3072 84480 CCC
@ 0 0 2048  x 104448 A
0 0 0 524288  x
Pivoting on the units (over Z(5)) in positions (5,2) and (7,4) and removing the*
*ir rows
and columns does not change the group presented. We now have a 6by2 matrix,
whose nonzero entries are polynomials in x of degree 1 ori2.jIf x 6 3 mod 5, w*
*hich
is equivalent to t 6 3 mod 4, the bottom two rows are u1B0u2with uiunits, and*
* so
the group presented is 0.
Henceforth, we assume x 3 mod 5. The polynomial in new position (5,2) is
nonzero mod 5 for such x, and so we pivot on it, and remove its row and column.
The five remaining entries are ratios of polynomials with denominator nonzero m*
*od
5. Let p1, . .,.p5 denote the polynomials in the numerators. The group v11ss2t*
*(X29)(5)
is Z=5e, where e = min( (p1(x)), . .,. (p5(x))), where x = 2t. We abbreviate 5*
*()
pCOMPACT GROUPS 23
to () throughout the remainder of this section. We have
p1 = 71122941747658752 + 9480741773824512x  74067383851199x2+ 33908441866x3
p2 = 66750692556800000 + 8897903174800000x  69512640100000x2+ 31789306250x3
p3 = 8327872 . 1010+ 11101145 . 109x  86731015625000x2+ 39736328125x3
p4 = 4 . 1019 533203125 . 1010x + 41656494140625000x2 19073486328125x3
p5 = 1099511627776  146567856128x + 1145324544x2 526472x3+ x4.
For values of m listed in the table, we compute and present in Table 4.13 the*
* tuples
(e0, e1, e2, e3) so that, up to units,
pi(2m + y) = 5e0+ 5e1y + 5e2y2 + 5e3y3 (4.12)
(plus y4 if i = 5). Considerable preliminary calculation underlies the choice o*
*f these
values of m.
Table 4.13. Exponents of polynomials
 i

_____m________1________2__________3___________4_________5____
3 3, 2, 1,10, 7, 6,15, 12, 12,110, 21, 20,119, 3, 2, 1
15 3, 2, 1,80, 7, 6,153, 12, 11,2102, 21, 20,419, 4, 3, 2
7 + 4 . 54 8, 2, 1,80, 7, 6,157, 12, 11,2106, 21, 21,819, 3, 2, 1
11 + 4 . 5812, 2, 1,102, 7,17,35, 12, 11,110, 21, 20,1192, 3, 2, 1
19 + 12 . 51623, 2,22,00, 7,26,15, 12, 11,2102, 21, 20,2190, 3, 2, 1
i
Recall that (24.5 1) = i + 1, as is easily proved by induction. Thus
p(2m+20j) = p(2m + 2m (220j 1)) = p(2m + 25j . u),
(4.14)
with u a unit. Hence
min { (pi(23+20j)) : 1 i 5} = 3
since
p1(23+20j) = p1(23+ 52ju) = 53+ 52. 52ju + 5(52ju)2+ (52ju)3,
omitting some unit coefficients. Here we have set y = 52ju in (4.12). Replacing*
* 3 by
15 yields an identical argument. This yields the second line of Theorem 4.10.
24 DONALD M. DAVIS
We use Table 4.13 to show
16+20j
min{ (pi(219+12.5 )) : 1 i 5} = min(20, 4 + (j)) = min(20, 3 + (20j)).
(4.15)
Indeed, for (j) 16, the minimum is achieved when i = 1, with the 4 coming as
2 + 2 with one 2 being from the 25 in (4.14) and the other 2 being the first 2 *
*in the
last row of Table 4.13. If (j) > 16, the minimum is achieved when i = 2, using*
* the
first 20 in the last row of 4.13. The last case of Theorem 4.10 follows easily*
* from
(4.15), and the other two parts of 4.10 are obtained similarly.
To see that v11ss2t1(X29) v11ss2t(X29), we argue in three steps. First, *
*the two
groups have the same order using [9, 8.5] and the fact that the kernel and coke*
*rnel
of an endomorphism of a finite group have equal_orders.! Second, by [16, 4.4], a
5 1
presentation of v11ss2t1(X29) is given by 2 2t , i.e. like that for v1 s*
*s2t(X29)
except that the two submatrices are not transposed. Third, we pivot on this mat*
*rix,
which is (4.11) with the top and bottom transposed, and find that we can pivot *
*on
units three times, so that the group presented is cyclic. 
Next we provide the evidence for Conjecture 4.1 in the following result.
Proposition 4.16. Let B0 = B(7, 15, 23, 39) denote the space constructed just b*
*efore
Conjecture 4.1. The Adams modules K*(X29; ^Z5) and K*(B0; ^Z5) are isomorphic.
Proof.We use the 5typical basis of QK1(SU(20)) using powers of y = ep(`0(x)).
Here ep() is the series inverse to `p(), and x is the usual generator of QK1(*
*SU(n)).
Then _k(y) = ep(k`p(y)) only involves powers y1+m(p1). See [32, pp.660661] fo*
*r a
discussion of this. We compute _2 on the basis to be gi*
*ven by
0 1
8 0 0 0 0
BB 72 128 0 0 0 CC
2 = BBB1368 2688 2048 0 0 CCC.
@32472 67200 67584 32768 0 A (4.17)
865152 1841280 2095104 1474560 524288
We find that (0, 0, 0, 1, 3)T is an eigenvector of 2 for ~ = 231. We take the *
*quotient
by this vector, using as the new basis. To obtain the matrix*
* of _2 on
this basis, we subtract 3 times the fourth row of (4.17) from the fifth, and re*
*move the
pCOMPACT GROUPS 25
fourth row and column, obtaining
0 1
8 0 0 0
B 72 128 0 0 C
2B= BB@1368 2688 2048 0 CCA.
962568 2042880 2297856 524288
Let 2X denote the transpose of the bottom half of (4.11) without the x part.
We desire an invertible matrix Q over Z(5)such that 2XQ = Q 2B. To find such
a Q we find matrices Q1 and Q2 whose columns are eigenvectors of 2Xand 2B,
respectively. These columns can be multiplied by any scalars. We can find scala*
*rs so
that Q := Q1Q12has entries in Z(5)with units along the diagonal. We find that
0 1
51 0 0 0
B21=2 1=2 0 0C
Q = BB@48 2 2 0CCA
30 16 18 4
works, and it also satisfies _kXQ = Q kBfor any k. 
Since v1periodic homotopy groups can be computed from Adams modules, we also
have v11ss*(X29)(5) v11ss*(B0)(5).
For the cases discussed in Table 3.2, we could always construct equivalences *
*from
a factor B(, , ) to (Xn)p because it was always the case that if the Bspace
had a dcell, then ssd1((Xn)p) = 0. This will not be the case when trying to r*
*elate
B0 = B(7, 15, 23, 39) and (X29)5, primarily because p = 5 is so small compared *
*to the
dimensions of some of the cells in B0. For example, localized at p = 5, ss37(S7*
*) Z=5,
the first unstable class. See, e.g., Diagram 3.4 or [5, 5.16]. This gives a n*
*onzero
element in ss37(X29) which provides a possible obstruction to extending a map f*
*rom
the 37skeleton of B0 over the 38cell representing the class x15x23. A similar*
* problem
occurs due to ss45(S7) ! ss45(X29) being nonzero, giving a possible obstruction*
* to
extending over x7x39. Also, ff1fi1 on S23gives an apparent element in ss68(X29)*
* which
is a possible obstruction to extending over x7x23x39. It is conceivable that m*
*ore
delicate arguments such as that on [29, p.661] might show that these obstructio*
*ns
can be removed. By [9, 3.4,8.1] and our Proposition 4.16, the associated v1per*
*iodic
spectra B0 and X29are equivalent. However, the fact that, as we shall see at *
*the
end of this section, there is not a space related to SU(N) which might be equiv*
*alent
26 DONALD M. DAVIS
to (X31)5 tempers a belief that there should be a general reason for B0 and (X2*
*9)5 to
be equivalent.
Note that the entry in position (4, 3) of the matrix of 4.9 implies that the *
*39cell
of (X29)5 is attached to the 23cell by ff2, which is not detected by primary S*
*teenrod
operations. Note also that the spherical resolvability of (X29)5 and also the s*
*paces
(X31)5 and (X34)7 follows from [15].
We can determine the Adams operations and v1periodic homotopy groups of (X31*
*)5
by an argument very similar to that used above for (X29)5. We shall merely sket*
*ch.
The analogue of Theorem 4.2 is
Theorem 4.18. There is an isomorphism H*(BX31; ^Z5) H*(BT ; ^Z5)G31, where`
10 00'
G31 has the four generators given for G29 in the proof of 4.2 and also 010000*
*10.
00 01
Then H*(BT ; ^Z5)G31is a polynomial ring on the generators f8, f12, and f20give*
*n in
4.2 together with
f24 = m(24) 66m(20,4)+ 1023m(16,8)+ 2180m(12,12)+ 1293156m(8,8,4,4)
+267096m(12,4,4,4)+ 2121984m(6,6,6,6)+ 620352m(10,6,6,2) 4032m(14,6,2,2)
190080m(10,10,2,2) 11892m(12,8,4) 4938m(16,4,4) 24534m(8,8,8)
2304m(18,2,2,2).
The analogue of Theorem 4.8 is
Theorem 4.19. Let f8, f12, f20, f24be as in 4.18, and let
Fj = fj(`0(x1), . .,.`0(x4)).
Then the following series are 5integral through grading 24.
F8 8_5F12 7_25F82 4_25F20 21_125F8F12 99_125F24 597_625F83 558_625F*
*122
F12 2_5F82 1_5F20 4_25F8F12 18_25F24 74_125F83 11_125F122
F20 3_5F24 2_5F83.
The analogue of 4.9 is
pCOMPACT GROUPS 27
Theorem 4.20. The Adams operation _k in K1(X31; ^Z5) on the generators z7, z11,
z19, and z23is given by the matrix
0 7 1
k 0 0 0
BB 8_7 8_11 11 C
BB 5k  5k k 0 0 CC.
B@ 12_k7  8_k11 4_k19 1_k11 1_k19 k19 0 CC
25 25 25 5 5 A
279_7 168_11 12_ 19 99_ 2321_113_ 19 18_23 3_19 3_23 23
125k  125k  125k  125k25k  25k  25k 5k  5k k
The analogue of Theorem 4.10 is
Theorem 4.21. The groups v11ss*(X31)(5)are given by
8
>>>0 t 6 3 (4)
>>>
>>Z=5min(8,3+ 5(t118.54))t 11 (20)
>>> min(12,3+ (t1916.58))
>>>Z=5 5 t 19 (20)
>: min(20,3+ (t2316.516))
Z=5 5 t 23 (20).
One might hope that, analogously to 4.1, there might be a space B(15, 23, 39,*
* 47)
related to SU(24) and equivalent to (X31)5. This cannot happen. Diagram 4.22, in
which straight lines denote ff1 attaching maps and curved lines ff2, shows the *
*ways
in which the generating cells of (X31)5 and one factor of SU(24)=SU(7) are atta*
*ched
to one another. Because the 31cell is attached to a lower cell and has a highe*
*r cell
attached to it, there cannot be a map in either direction between these spaces *
*sending
generators across. This is what puts a damper on any hope that all pcompact gr*
*oups
are related to unitary groups.
Diagram 4.22. Attaching maps of generating cells
oe___________ ,
oe , oe ,  oe ,  oe ,
15 _____23 39 _____47 (X31)5
aess aess aess aess
oe___________ ,
oe , oe , oe , oe ,  oe ,
15 _____23 _____ 31 39 _____47 SU(24)=SU(7)
aess aess aess aess aess
ae___________ ss
28 DONALD M. DAVIS
5. The 7primary modular case
In this section, we first give in Theorem 5.1 new explicit formulas for the s*
*ix poly
nomials which generate as a polynomial algebra the invariant ring of the complex
reflection group G34of [33], called the Mitchell group in [14]. Over ^Z7, the i*
*nvariant
ring of G34is also a polynomial algebra, but the generators must be altered sli*
*ghtly
from the complex case, as we show prior to 5.6. Next we use this information to*
* find
explicit generators for K*(BX34; ^Z7) in 5.6, and from this the Adams operation*
*s in
QK1(X34; ^Z7) in 5.15. These in turn enable us to compute the v1periodic homot*
*opy
groups v11ss*(X34)(7). Finally, we show in 5.18 that the Adams module QK1(X34;*
* ^Z7)
is isomorphic to that of a space formed from SU(42); similarly to 4.1, we conje*
*cture
in 5.17 that this isomorphism is induced by a homotopy equivalence.
Theorem 5.1. The complex invariants of the reflection group G34 (defined in the
proof) form a polynomial algebra
C[x1, . .,.x6]G34 C[f6, f12, f18, f24, f30, f42]
with generators given by
Xk i j X
f6k= (1+(1)k27k1.5)m(6k)+ 6k3s(1+(1)k+s27k1)m(6k3s,3s)+ (e)me,
s=1 e
where e ranges over all partitions e = (e1, . .,.er) of 6k with 3 r 6 satis*
*fying ei
ej mod 3 for all i, j, and ei 0 mod 3 if r < 6. Here also (e) denotes the mult*
*inomial
coefficient (e1 + . .+.er)!=(e1! . .e.r!), and me the monomial symmetric polyno*
*mial,
which is the shortest symmetric polynomial in x1, . .,.x6 containing xe11. .x.e*
*rr.
For example, we have
of6 = 4m(6)+ 40m3,3+i720m(1,1,1,1,1,1);jij
P
of12= 136m(12) 26 123m(9,3)+ 28 126m(6,6)+ (e)me, where e
ranges over
{(6, 3, 3), (3, 3, 3, 3), (2, 2, 2, 2, 2, 2), (7, 1, 1, 1, 1, 1), (4, 4, 1, 1,*
* 1, 1)}.
pCOMPACT GROUPS 29
of 2 i18j 2 i18j 2
i18=j(15.27 )m(18)+ 3 (1+27 )m(15,3)+ 6 (127 )m(12,6)+
18 2 P
9 (1 + 27 )m(9,9)+ (e)me, where e ranges over
{(12, 3, 3), (9, 6, 3), (9, 3, 3, 3), (6, 6, 6), (6, 6, 3, 3), (6, 3, 3, 3,*
* 3), (3, 3, 3, 3, 3, 3),
(13, 1, 1, 1, 1, 1), (10, 4, 1, 1, 1, 1), (7, 4, 4, 1, 1, 1), (4, 4, 4, 4, *
*1, 1), (7, 7, 1, 1, 1, 1),
(8, 2, 2, 2, 2, 2), (5, 5, 2, 2, 2, 2)}
Proof of Theorem 5.1.As described in [33], the reflection group G34is generated*
* by
reflections across the following hyperplanes in C6: xi xj = 0, x1  !x2 = 0, a*
*nd
x1+x2+x3+x4+x5+x6 = 0. Here ! = e2ssi=3. It follows easily that G34is generated
by all permutation matrices together with the following two:
0 1 0 1
0 !2 0 0 0 0 1 1 1 1 1 1
BB! 0 0 0 0 0C B 1 1 1 1 1 1C
BB CC BB CC
BB0 0 1 0 0 0CC, I  1_BB11 1 1 1 1CC
BB0 0 0 1 0 0CCC 3BBB11 1 1 1 1CCC(5.2)
@ 0 0 0 0 1 0A @ 1 1 1 1 1 1A
0 0 0 0 0 1 1 1 1 1 1 1
In [14], Conway and Sloane consider G34instead as the automorphisms of a cert*
*ain
Z[!]lattice in C6. The lattice has 756 vectors of norm 2. There are none of sm*
*aller
positive norm. 270 of these vectors are those with !a in one position, !b in a*
*nother,
and 0 in the rest. Here, of course, a and b can be 0, 1, or 2. The other 486 ar*
*e those
P
of the form __1_p_3(!a1, . .,.!a6) such that ai 0 mod 3.
As a partial verification that this lattice approach to G34 is consistent wit*
*h the
reflection approach, one can verify that the reflection matrices permute these *
*756
vectors. It is obvious that permutation matrices do, and easily verified for th*
*e first
matrix of (5.2). The second matrix of (5.2), which has order 2, sends
o(!, !2, 0, 0, 0, 0) to _1_p_3(!2, !, 1, 1, 1, 1);
o__1_p_3(1, 1, 1, 1, 1, 1) to __1_p_3(1, 1, 1, 1, 1, 1);
o _1_p_3(1, 1, 1, !, !, !) to __1_p_3(!, !, !, 1, 1, 1);
o__1_p_3(1, 1, !, !, !2, !2) to itself.
After permutation, negation, and multiplication by !, this takes care of virtua*
*lly all
cases.
30 DONALD M. DAVIS
Let
X
pm (x1, . .,.x6) = (v1x1+ . .+.v6x6)m ,(5.3)
(v1,...,v6)
where the sum is taken over the 756 vectors described above. Then pm is invaria*
*nt
under G34for every positive integer m. It is proved in [14, Thm.10] that the ri*
*ng of
complex invariant polynomials is given by
C[x1, . .,.x6]G34= C[p6, p12, p18, p24, p30,(p42].5.4)
In [14], several other lattices isomorphic to the above one are described, an*
*y of
which can be used to give a different set of vectors v and invariant polynomials
pm , still satisfying (5.4). The one that we have selected seems to give the si*
*mplest
polynomials; in particular, the only ones with integer coefficients.
P P 2 a b 6k
We have p6k= S1+ S2, where S1 = i6=ja,b=0(! xi ! xj) , with 1 i, j 6,
and
2 X2 a a a ...a 6k
S2 = ______ (! 1x1+ . .+.! 5x5+ ! 1 x56) .
(3)3kai=0
The coefficient of 2 on S2 is due to the 1. Note that the sum for S1 has 6 . 5*
* . 32
terms, while that for S2 has 35 terms. Next note that if a term T 6koccurs in e*
*ither
sum, then so does (!T )6k and (!2T )6k, and all are equal. Thus we obtain S1 =
P P 2 b 6k
3 i6=j b=0(xi ! xj) and
2 X2 a a a ...a 6k
S2 = 3______ (x1+ ! 2x2+ . .+.! 5x5+ ! 2 x56) .
(3)3ka2,...,a5=0
We simplify S1 further as
6kX i j X X2
S1 = 3 (1)` 6k` x`ix6k`j !b`
`=0 i6=j b=0
2kX i j X
= 9 (1)s 6k3s x3six6k3sj
s=0 i6=j
Xk i j
= 18(5m6k+ (1)s 6k3sm(6k3s,3s)).
s=1
P 2 b`
At the first step, we have used that b=0! equals 0 if ` 6 0 mod 3, and equa*
*ls 3 if
P 3s 6k3s
` 0 mod 3. At the second step, we have noted that i6=jxi xj equals m(6k*
*3s,3s)
if s 62 {0, k, 2k}, it equals 2m(3k,3k)if s = k, and equals 5m(6k)if s = 0 or 2*
*k.
pCOMPACT GROUPS 31
The sum S2 becomes
6 X X2 e e a X2 e e a e e
S2 = ______ (e) (! 2 6) 2. . . (! 5 6) 5x11. .x.66
(3)3k e a2=0 a5=0
6 X 4 e e
= ______ (e)3 x11. .x.66.
(27)ke1 ... e6 (3)
Then (27)k(S1+ S2)=486 equals the expression which we have listed for f6kin *
*the
statement of the theorem. We have chosen to work with this rather than p6kitsel*
*f for
numerical simplicity. It is important that the omitted coefficient is not a mul*
*tiple of
7.
For good measure, we show that (5.3) is 0 if m 6 0 (6). If m 6 0 mod 3, th*
*en
replacing terms T m by (!T )m leaves the sums like S1 and S2 for (5.4) unchanged
while, from a different perspective, it multiplies them by !m . Thus the sums a*
*re 0.
P 3s m3s
If m 3 mod 6, the term in S1 corresponding to xi xj occurs with opposite
P m3s s m
sign to that corresponding to xi xj, and so S1 = 0. For S2, the ( 1) will *
*cause
pairs of terms to cancel. 
Remark 5.5. The only other place known to the author where formulas other than
(5.3) for these polynomials exist is [26], where they occupy 190 pages of dense*
* text
when printed.
As pointed out by Kasper Andersen, f42(f6)7 is divisible by 7. This is easil*
*y seen
P 6 7
by expanding (f6)7 = ( (v1x1+ . .+.v6x6) ) by the multinomial theorem. The ne*
*ed
for this became apparent to Andersen, as the author had thought that the invari*
*ant
ring of G34 over ^Z7was ^Z7[f6, . .,.f42], and this would have led to an imposs*
*ible
conclusion for the Adams operations in QK1(X34; ^Z7).
Let h42 = 1_7(f42 (f6)7). Then we have the following result, for which we a*
*re
grateful to Andersen.
Theorem 5.6. The invariant ring of G34over ^Z7is given by
Z^7[x1, . .,.x6]G34= ^Z7[f6, f12, f18, f24, f30, h42].
Proof.A Magma program written and run by Andersen showed that each of these
asserted generators is indecomposable over Z=7. (This is what failed when f42was
used; it equals (f6)7 over Z=7.) Thus the result follows from (5.4). 
32 DONALD M. DAVIS
Since f36is invariant under G34, it follows from (5.4) that it can be decompo*
*sed
over C in terms of f6, f12, f18, f24, and f30. The nature of the coefficients *
*in this
decomposition was not so clear. It turned out that all coefficients were ratio*
*nal
numbers which are 7adic units. We make this precise in
Theorem 5.7. f36can be decomposed as
q1f6f30+ q2f12f24+ q3f218+ q4f26f24+ q5f6f12f18
+q6f312+ q7f36f18+ q8f26f212+ q9f46f12+ q10f66
with
q1 = 944610925401=15161583716
q2 = 733671261=19519520
q3 = 243068633=9781739
q4 = 133840666859131062549=73986709144034080
q5 = 1758887990521258018071215403=629320589839873719708800
q6 = 1602221942044323=4879880000000
q7 = 4011206338081535787030788541=114421925425431585401600
q8 = 701461342458322269763709951654931=15733014745996842992720000000
q9 = 11844219519446025955021712628669=22348032309654606523750000
q10 = 26589469730264682368719198549833=22348032309654606523750000
Each of these coefficients qi is a 7adic unit; i.e. no numerator or denominat*
*or is
divisible by 7.
Proof.The ten products, f6f30, . .,.f66, listed above are the only ones possibl*
*e. We
express each of these products as a combination of monomial symmetric polynomia*
*ls
me. We use Magma to do this. The length of m(e1,...,er)is defined to be r. We o*
*nly
kept track of components of the products of length 4. This meant that we only
had to include components of length 4 of the various f6kbeing multiplied.
There were 34 me's of length 4. These correspond to the partitions of 36 in*
*to
multiples of 3. (Note that monomials with subscripts 1 or 2 mod 3 only occur *
*for
us if the length is 6. Not having to deal with them simplifies our work conside*
*rably.)
Indeed, there was one of length 1, six of length 2, twelve of length 3, and fif*
*teen of
pCOMPACT GROUPS 33
length 4. Magma expressed each monomial such as f6f30or f66as an integer combi
nation of these, plus monomials of greater length. We just ignored in the outpu*
*t all
those of greater length. The coefficients in these expressions were typically 1*
*2 to 15
digits. We also wrote f36as a combination of monomial symmetric polynomials of
length 4, ignoring the longer ones. This did not require any fancy software, *
*just
the multinomial coefficients from Theorem 5.1.
Now we have a linear system of 34 linear equations with integer coefficients *
*in 10
unknowns. The unknowns are the coefficients qiin the equation at the beginning *
*of
5.7, and the equations are the component monomials of length 4. Miraculously,
there was a unique rational solution, as given in the statement of this theorem.
If it were not for the fact that the ConwaySloane theorem 5.4 guarantees tha*
*t there
must be a solution when all monomial components (of length 6) are considered,
then we would have to consider them all, but the fact that we got a unique solu*
*tion
looking at only the monomial components of length 4 implies that this solution
will continue to hold in the other unexamined components. 
Next we wish to modify the generators in 5.6 to obtain generators of QK1(X34;*
* ^Z7).
Similarly to 4.8, we let `0(x) = ln(1 + x), and
Fi= Fi(x1, . .,.x6) = fi(`0(x1), . .,.f6(`0(x6))).(5.8)
A major calculation is required to modify the classes Fi so that their coeffi*
*cients
are in ^Z7; i.e. they do not have 7's in the denominators. As observed after (4*
*.5), it
will be enough to accomplish this through grading 42 (with grading of xiconside*
*red
to be 1).
Theorem 5.9. The following expressions are 7integral through grading 42:
oF30+ 5_7F36+ 22_72F42;
oF24+ 4_7F30+ 45_72F36+ 104_73F42;
oF18+ 3_7F24+ 20_72F30+ 157_73F36+ 526_74F42;
oF12+ 2_7F18+ 45_72F24+ 109_73F30+ 1391_74F36+ 6201_75F42;
oF6+ 1_7F12+ 22_72F18+ 204_73F24+ 1107_74F30+ 9682_75F36+ 100682_76F42.
34 DONALD M. DAVIS
It was very surprising that just linear terms were needed here. Decomposable
terms were certainly expected. The analogue for G29in 4.8 involved many decom
posables. It would be interesting to know why Theorem 5.9 works with just linear
terms; presumably this pattern will continue into higher gradings.
Proof of Theorem 5.9.Similarly to the proof of 4.8, we define
Fei= eFi(x1, . .,.x6) = fi(`p(x1), . .,.`p(x6)),
and observe that a polynomial in the eFi's is 7integral if and only if the sam*
*e poly
nomial in the Fi's is.
Next note that in the range of concern for Theorem 5.9 `7(x) = x + x7=7. If we
define
hi= hi(x1, . .,.x6) = fi(x1+ x71, . .,.x6+ x76),
then 5.9 is clearly equivalent to
Statement 5.10. For t 1 and grading 42,
oh30+ 5h36+ 22h42 0 mod 7t in grading 30 + 6t;
oh24+ 4h30+ 45h36+ 104h42 0 mod 7t in grading 24 + 6t;
oh18+ 3h24+ 20h30+ 157h36+ 526h42 0 mod 7t in grading
18 + 6t;
oh12+ 2h18+ 45h24+ 109h30+ 1391h36+ 6201h42 0 mod 7t in
grading 12 + 6t;
oh6+ h12+ 22h18+ 204h24+ 1107h30+ 9682h36+ 100682h42 0
mod 7t in grading 6 + 6t.
We use Maple to verify 5.10. Our fi's are given in Theorem 5.1 in terms of me*
*'s. To
evaluate me(x1+x71, . .,.x6+x76), the following result keeps the calculation ma*
*nageable
(e.g. it does not involve a sum over all permutations). Partitions can be wri*
*tten
either in increasing order or decreasing order; we use increasing. If (a1, . .*
*,.ar) is
an rtuple of positive integers, let s(a1, . .,.ar) denote the sorted form of t*
*he tuple;
i.e. the rearranged version of the tuple so as to be in increasing order. For e*
*xample,
s(4, 2, 3, 2) = (2, 2, 3, 4).
pCOMPACT GROUPS 35
Proposition 5.11. The component of m(e1,...,er)(x1 + x71, . .,.x6 + x76) in gra*
*ding
P
ei+ 6t is
_ ! _ !
X P (e1+ 6j1, . .,.er+ 6jr)e e
_____________________ .1. .r ms(e1+6j1,...,er+6jr),
j P (e1, . .,.er) j1 jr
where j = (j1, . .,.jr) ranges over all rtuples of nonnegative integers summin*
*g to t,
and P (a1, . .,.ar) is the product of the factorials of repetend sizes.
For example, P (4, 2, 3, 3) = 2! because there are two 3's, P (3, 1, 3, 3, 1,*
* 2) = 3!2!,
and P (3, 4, 2, 1) = 1.
Example 5.12. We consider as a typical example, the component of
m(3,3,9,15)(x1+ x71, . .,.x6+ x76)
in grading 42. Table 5.13 lists the possible values of j and the contribution t*
*o the sum.
The final answer is the sum of everything in the right hand column.
Table 5.13. Terms for Example 5.12
_____j_______term_____
i j
(2, 0, 0, 0)3m3,9,15,15
i2j
(0, 2, 0, 0)3m3,9,15,15
i2j
(0, 0, 2, 0)9m3,3,15,21
i2j
(0, 0, 0, 2)15m3,3,9,27
2
(1, 1, 0, 0)3. 3 . 3m9,9,9,15
(1, 0, 1, 0)3. 9m3,9,15,15
(1, 0, 0, 1)3. 15m3,9,9,21
(0, 1, 1, 0)3. 9m3,9,15,15
(0, 1, 0, 1)3. 15m3,9,9,21
(0, 0, 1, 1)9. 15m3,3,15,21
Proof of Proposition 5.11.m(e1,...,er)(x1+ x71, . .,.x6+ x76) is related to
X e iej e +6 iej e +12 e iej e +6
(xo1e(1)+ 11xo1e(1)+ 21xo1e(1)+ . .).. .(.xore(r)+ 1rxore(r)+ . .).
oe (5.14)
summed over all permutations oe in r. If t values of ei are equal, then (5.14)*
* will
give t! times the correct answer. That is the reason that we divide by P (e). *
* If
(e1+6j1, . .,.er+6jr) contains s equal numbers, then the associated m will be o*
*btained
36 DONALD M. DAVIS
from each of s! permutations, which is the reason that P (e1+6j1, . .,.er+6jr) *
*appears
in the numerator. 
At first, mimicking 4.8, we were allowing for products of h's in addition to *
*the
linear terms which appear in 5.10, but it was turning out that what was needed *
*to
satisfy the congruences was just the linear term. If just a linear term was goi*
*ng to
work, the coefficients could be obtained by just looking at monomials of length*
* 1.
They were computed by Maple,iusingjthat, by 5.1 and 5.11, the coefficient of m(*
*6k+6t)
in h6kis (1 + (1)k27k1. 5) 6kt. Write the kth expression from the bottom of 5*
*.10
P P t
as j 0aj,kh6k+6j. We require that the coefficient of m(6k+6t)in j=0aj,kh6k+*
*6jis 0
mod 7t. But this coefficient equals
Xt i j
aj,k6k+6jtj(1 + (1)k+j27k+j1. 5).
j=0
We solve iteratively for aj,k, starting with a0,k= 1, and obtain the values in *
*5.10.
Note that it first gives a1,1 1 mod 7. If we had chosen a value such as 8 or *
*6
instead of 1, then the value of a2,1would be different than 22. So these numbe*
*rs
aj,kare not uniquely determined. These different choices just amount to choosin*
*g a
different basis for QK1(X34; ^Z7).
Verifying Statement 5.10 required running many Maple programs. For each line *
*of
5.10, a verification had to be made for each relevant tvalue, from two tvalue*
*s for
the first line down to six tvalues for the last line. Moreover, for each of th*
*ese pairs
(line number, tvalue), it was convenient to use a separate program for monomia*
*ls
of each length 2, 3, 4, and 5, and then, for monomials of length 6, it was done
separately for those with subscripts congruent to 0, 1, or 2 mod 3. Thus altoge*
*ther
(2 + 3 + 4 + 5 + 6)(4 + 3) = 140 Maple programs were run. The programs had enou*
*gh
similarity that one could be morphed into another quite easily, and a more skil*
*lful
programmer could incorporate them all into the same program.
Note that expanding from fj to hj does not change the number of components
in monomials, nor does it change the mod 3 value of the sum of the subscripts (*
*i.e.
exponents) in the monomials. This is simpler than the situation in the proof of*
* 5.7.
The algorithm is quite easy. For each combination of h's in 5.10, replace each *
*h6jby
the combination of me's in f6jin 5.1, but expanded using 5.11. 
pCOMPACT GROUPS 37
To obtain the Adams operations in QK1(X34; ^Z7), we argue similarly to the pa*
*ra
graph which precedes Theorem 4.9. First note that F36decomposes in terms of F6i*
*'s
exactly as does f36 in terms of f6i's in 5.7. We can7 modify by decomposables *
*in
dimensions greater than 42 to obtain 7integral classes G6, G12, G18, G24, and *
*G30
which agree with the classes of 5.9 (with F36 decomposed) through dimension 42.
There is also a 7integral class G42which agrees with 1_7(F42 (F6)7) in dimens*
*ion 42.
These generate K*(BX34; ^Z7) as a power series algebra. As in the preamble to 4*
*.9,
then zi:= B1e*(Gi+1) for i = 5, 11, 17, 23, 29, and 41 form a basis for QK1(X3*
*4; ^Z7),
and e* annihilates decomposables.
Similarly to the situation for (X29)5 in the proof of 4.9, if we let
0 1
1 0 0 0 0 0
BB 1_ C
BB 7 1 0 0 0 0CC
B 22_ 2_ 1 0 0 0CC
P = BBB49204745 3 CC,
BB___343__49_7 1 0 0CC
B@1107_109_ 20_ 4_ 1 0CC
2401 343 49 7 A
16647_1399_183_6_ 1_
16807 2401 343 49 7 1
then the matrix of _k on the basis {z5, z11, z17, z23, z29, z41} is
P 1diag(k5, k11, k17, k23, k29, k41)P.
The entries in the last row of P are 7 times the coefficients of F42in 5.9 redu*
*ced mod
1. Those coefficients were multiplied by 7 because z41is related to 1_7F42rathe*
*r than
to F42.
Using this, we compute the v1periodic homotopy groups, similarly to 4.10. No*
*te
the remarkable similarity with that result. Here, of course, () denotes the e*
*xponent
of 7 in an integer.
__________
7But we need not bother to do so explicitly.
38 DONALD M. DAVIS
Theorem 5.15. The groups v11ss*(X34)(7)are given by
8
>>>0 t 6 5 (6)
>>> 5
>>>Z=7 t 5, 35 (42)
>>> min(12,5+ (t1112.76))
>Z=7min(18,5+ (t1718.712))t 17 (42)
>>> min(24,5+ (t2318.718))
>>>Z=7 t 23 (42)
>>> min(30,5+ (t2912.724))
>>>Z=7 t 29 (42)
:Z=7min(42,5+ (t4124.736))t 41 (42).
_ !
(_7)T *
* x
Proof.The group v11ss2t(X)(7)is presented by (_3)T  3tI, since 3 generates Z*
*=49 .
We let x = 3tand form this matrix analogously to (4.11). Five times we can pivo*
*t on
units, removing their rows and columns, leaving a column matrix with 7 polynomi*
*als
in x. The 7exponent of v11ss2t(X)(7)is the smallest of that of these polynom*
*ials
(with x = 3t). This will be 0 unless x 5 mod 7, which is equivalent to t 5
mod 6. We find that two of these polynomials will always yield, between them, t*
*he
smallest exponent. Similarly to (4.12) and Table 4.13, we write these polynomia*
*ls as
pi(3m + y) for carefullychosen values of m. Much preliminary work is required*
* to
discover these values of m. Ignoring unit coefficients and ignoring higherpowe*
*r terms
whose coefficients will be sufficiently divisible that they will not affect the*
* divisibility,
these polynomials will be as in Table 5.16.
Table 5.16. Certain pi(3m + y), (linear part only)
_____m________p1________p2____
5, 35 75+ 74y7 12+ 711y
11 + 12 . 76712+ 74y712+ 711y
17 + 18 . 712718+ 74y718+ 711y
23 + 18 . 718725+ 74y724+ 711y
29 + 12 . 724731+ 74y730+ 711y
41 + 24 . 736743+ 74y742+ 711y
The claim of the theorem follows from Table 5.16 by the same argument as was
used in the proof of 4.10. For t in the specified congruence, if 3t = 3m + y, *
*then
(y) = (t  m) + 1 2, similarly to (4.14). For example, if t 11 mod 42, and
6 6 t *
* t
3t= 311+12.7+ y, then (y) = (t  11  12 . 7 ) + 1. Thus min( (p1(3 )), (p2(*
*3 )))
pCOMPACT GROUPS 39
will be determined by the 75 in p1 if t 5, 35 (42), while in the other cases,*
* it is
determined by the 74y in p1 or the constant term in p2.
The groups v11ss2t1(X34) are cyclic by an argument similar to the one descr*
*ibed at
the end of the proof of 4.10, and have the same order as v11ss2t(X34) for the *
*standard
reason described there. 
Similarly to the discussion preceding Conjecture 4.1, one of the factors in t*
*he prod
uct decomposition of SU(42)7 given in [29] is an Hspace B75(7) whose F7cohomo*
*logy
is an exterior algebra on classes of grading 11, 23, 35, 47, 59, 71, and 83, an*
*d which
is built from spheres of these dimensions by fibrations. Using [36], we can obt*
*ain a
degree1 map B75(7) ! S71. Let B7 := B(11, 23, 35, 47, 59, 83) denote its fiber.
Conjecture 5.17. There is an equivalence (X34)7 ' B7.
The evidence for this conjecture is the following analogue of Proposition 4.1*
*6.
Proposition 5.18. There is an isomorphism of Adams modules K*(X34; ^Z7) K*(B7*
*; ^Z7).
Proof.We argue exactly as in the proof of 4.16. We compute _3 on the basis
of QK1(SU(42)) where y = ep(`0(x)). Then (0,*
* 0, 0, 0, 0, 1, 5)T
is an eigenvector for ~ = 335. Quotienting out by this vector, we obtain as the*
* trans
pose of the matrix of _3 on this basis of QK1(B7; ^Z7) the following matrix (_3*
*)TB7.
(We write the transpose for typographical reasons.)
0 5 1
3 126360 118399320136947072600176770713576600354126788968985033040
BB0 311 202656168253117553832340318273704552711333125213838324912C
BB0 0 317 228319808184356407220575224855037442924642953872CC
B@0 0 0 323 225190483754184736156248630810154992CC
0 0 0 0 329 453306387710146810320A
0 0 0 0 0 341
Let _3Xdenote the matrix on X34. Using eigenvectors of the two matrices similar*
*ly
to the proof of 4.16, we find that the matrix Q below satisfies _kXQ = Q_kB7for*
* all
k, and has diagonal entries 6 0 mod 7. Finding such a matrix Q was by no means
automatic; it required simultaneous satisfying of many congruence equations.
0 1
5=2 0 0 0 0 0
BB95=2 66 0 0 0 0 C
BB CC
81005 138611 88219 0 0 0 C
Q = BB CC
BB232625=2 210177 173677 55946 0 0 CC
@253775 487751 507389 312395 85347 0 A
301425 592328 668516 528725 314505 24324
40 DONALD M. DAVIS

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Lehigh University, Bethlehem, PA 18015
Email address: dmd1@lehigh.edu