AN ELEMENTARY INVARIANT PROBLEM AND GENERAL
LINEAR GROUP COHOMOLOGY RESTRICTED TO THE
DIAGONAL SUBGROUP
MARIAN F. ANTON
Abstract.Conjecturally, for p an odd prime and R a certain ring of p-
integers, the stable general linear group GL(R) and the etale model for *
*its
classifying space have isomorphic mod p cohomology rings. In particular,*
* these
two cohomology rings should have the same image with respect to the rest*
*ric-
tion map to the diagonal subgroup. We show that a strong unstable version
of this last property holds for any rank if p is regular and certain hom*
*ology
classes for SL2(R) vanish. We check that this criterion is satisfied for*
* p = 3
as evidence for the conjecture.
1.Introduction
Let p be a prime and R = Z[i, 1=p] the ring of p-integers containing a primit*
*ive
p-th root of unity i. A great deal of interest has been generated by the mod p *
*co-
homology of the stable general general linear group GL(R). Dwyer and Friedlander
[4] constructed a topological space X(R) based on the etale topological type of*
* the
scheme Spec(R) and a continuous map
f : BGL(R) ! X(R)
where BGL(R) is the classifying space of the discrete group GL(R). They showed
that f is injective on mod p cohomology if p is regular [5, 6.4]. Voevodsky [9]*
* proved
a conjecture of Milnor having as consequence the fact that f is a p-equivalence*
* (i.e.
induces an isomorphism on mod p cohomology) for p = 2. Quillen and Lichtenbaum
[6] have studied a conjectural relationship between the algebraic K-theory of R*
* and
the values of an associated zeta function. Their work suggests that f should be*
* a
p-equivalence for any prime p.
In particular, let D(R) be the subgroup of diagonal matrices inside GL(R) and
consider the following diagram on mod p cohomology H* = H*(-, Fp):
* res
H*X(R) f-!H*BGL(R) --! H*BD(R)
where f* is induced by f and res by the inclusion D(R) GL(R). If f is a
p-equivalence, then res and res O f* have the same image (see also Remark 4.11).
In this article we show that a strong unstable version of the last property h*
*olds
under certain conditions. Namely, let n be a non-negative integer and GLn(R) the
general linear group of rank n over R. Then we can replace GL, D, X, f, res abo*
*ve
respectively by GLn, Dn, Xn, fn, resn and keep everything else unchanged.
____________
Date: November 12, 2002.
1991 Mathematics Subject Classification. Primary 57T10, 20J05; Secondary 19D*
*06, 55R40.
Key words and phrases. etale model, linear group, cohomology, invariants.
1
2 MARIAN F. ANTON
Theorem 1.1. Let p be an odd regular prime, Mn the image of resn O f*n, and In
the image of resn. If M2 = I2, then Mn = In for all n 0.
We can actually strengthen this theorem by weakening the condition M2 = I2
and requiring instead that certain homology classes for SL2(R) vanish (see 4.10*
*).
As an example when M2 = I2 holds we can cite our previous result (see also 4.4)
Corollary 1.2 ([1, 1.3]). If p = 3, then Mn = In for all n 0.
The Theorem 1.1 was inspired by [3] where a similar statement was proven
in case p = 2. Our generalization is based on two observations: (i) the mod p
cohomology ring of the etale model Xn(R) has a particularly simple structure for
p odd and regular [5] and (ii) the maps dual to resn assemble for n 0 into a
ring homomorphism on homology with respect to the ring structure induced by
matrix block-multiplication GLn(R) x GLm (R) ! GLn+m (R). Based on this last
ring structure, we formulate and solve a general invariant problem 3.4 having as
consequence 1.1. Surprisingly, the solution to the invariant problem is element*
*ary
and purely combinatorial, although quite tricky.
The article is organized as follows. In x2 we review the mod p cohomology ring
of the etale model Xn(R). In x3 we formulate and solve the invariant problem and
deduce the Theorem 1.1. In x4 we strengthen and generalize our theorem.
We acknowledge the hospitality of the Max-Planck-Institut für Mathematik in
Bonn during the preparation of this article and thank an anonymous referee for
accuracy and presentation suggestions.
2.Etale model cohomology
In what follows we assume that p is an odd regular prime and keep the notatio*
*ns
from the Introduction. In [5, 6.3] the graded ring associated to a filtration *
*of
H*Xn(R) was computed and we review this calculation here from a slightly differ*
*ent
perspective:
Theorem 2.1. Let p be an odd regular prime and n a non-negative integer. Then
Or
H*Xn(R) Fp[c1, c2, ..., cn] (ei,1, ei,2, ..., ei,n)
i=1
where r = (p + 1)=2, is the tensor product over Fp, and is the symbol for an
exterior Fp-algebra. For 1 j n and 1 i r, cj is the Chern class of degr*
*ee
|cj|= 2j associated with a complex embedding of R and ei,jis a class of degree
|ei,j|= 2j - 1.
The following corollary is essentially [5, 6.2] in a form more suitable for t*
*he
purpose of this article.
Corollary 2.2. With the same notations and hypotheses as in 2.1,
Or
H*BDn(R) Fp[x1, x2, ..., xn] (yi,1, yi,2, ..., yi,n)
i=1
where the degrees of the generators are given by |xj|= 2 and |yi,j|= 1 and the *
*map
resn O f*nis a ring monomorphism sending cj to the symmetrization of x1...xj and
ei,jto the symmetrization of x1...xj-1yi,jfor 1 j n and 1 i r.
Remark 2.3. For n 0, let n be the symmetric group of {1, 2, ..,.n}Then n
acts in an obvious manner on the diagonal subgroup Dn(R). This action induces an
action of n on H*Dn(R) via ring homomorphims x 7! xoesending xj to xoe(j)and
AN ELEMENTARY INVARIANT PROBLEM 3
yi,jto yi,oe(j)for 1 i r and 1 j n where oe 2 n. Then the symmetrizati*
*on
of an element x 2 H*Dn(R) isPthe sum without repetitions of all the elements xoe
with oe 2 n, i.e. the sum xoeover the orbit of x.
2.1. Etale models. Before sketching the proof of 1.1 and its corollary we review
some ingredients. Let n be a non-negative integer, p a prime, and R0 = Z[1=p].
Then the etale model Xn(S) at p is a topological space naturally assigned to ea*
*ch
noetherian R0-algebra S and the map
fn : BGLn(S) ! Xn(S)
is a natural transformation [4].
Proposition 2.4 ([1, 2.3], [5, 2.8, 2.11]). The following properties hold for a*
*ny
prime p and non-negative integer n:
(1) If S is a complete local ring with residue field k of characteristic dif*
*ferent
from p, then the induced map Xn(S) ! Xn(k) is a p-equivalence.
(2) If S is a finite field of order prime to p, then fn is a p-equivalence.
(3) If S is the field of complex numbers, then Xn(S) is p-equivalent to the
classifying space BGLtopn(S) of the Lie group GLtopn(S).
We recall that the cohomology ring of BGLtopn(C), the infinite complex Grass-
mannian, is a polynomial ring in the universal Chern classes c1, c2,..., cn. Al*
*so, if
qi is a prime ideal of R with the residue field ki of order 1 mod p and qi-a*
*dic
completion ^Rqi, then the cohomology ring of BGLn(ki) was calculated by Quillen
in [8]. As a corollary of these known facts and Proposition 2.4 we have:
(2.1) H*Xn(C) H*BGLtopn(C) Fp[c1, c2, ..., cn]
(2.2) H*Xn(R^qi) H*BGLn(ki) Fp[c1, c2, ..., cn] (ei,1, ei,2, ..., ei,n)
where |cj|= 2j and |ei,j|= 2j - 1 for 1 j n.
2.2. Proofs. Now we are ready to sketch the proofs of 2.1 and 2.2.
Proof of 2.1.According to [2] there are r = (p + 1)=2 prime ideals of R, say qi*
* for
1 i r, and a commutative diagram in the category of noetherian R0-algebras
(2.3) ^Rq1
"">> AA
"" AA
"" AA
" __A
R A____//_...__//_C>>
AA """
AAA """
A__ "
^Rqr
such that Xn((2.3)) is a homotopy cartesian diagram in the category of topologi*
*cal
spaces. Moreover, the residue field of qi, say ki, has order 1 mod p for 1 *
* i r.
In particular, the formula (2.2)holds for all 1 i r and combining these for*
*mulae
with (2.1)we deduce that the Eilenberg-Moore spectral sequence associated with
the homotopy cartesian diagram Xn((2.3)) collapses to a ring isomorphism
Or
H*Xn(R) H*Xn(R^qi)
i=1
4 MARIAN F. ANTON
Here the tensor product is over H*Xn(C) and hence the theorem follows from (2.2)
and (2.1).
Proof of 2.2.By Dirichlet's Unit Theorem, D1(R) is a product of r - 1 copies of
the integers Z where r = (p + 1)=2 and a finite cyclic group C with its p-prima*
*ry
part generated by the primitive p-th root of unity i. By the Künneth theorem, it
follows that H*BDn(R) is a tensor product of n(r-1) copies of H*(Z) and n copies
of H*(C) and therefore has the desired ring structure by standard calculations.
In order to prove the second part of the corollary, we can apply the same met*
*hod
as in the proof of 2.1. With the same notations, the restriction maps
H*BGLn(ki) ! H*BDn(ki), 1 i r
can be identified according to [8, p. 564] with the symmetrization maps
si: Fp[c1, ..., cn] (ei,1, ..., ei,n) ! Fp[x1, ..., xn] (yi,1, ...*
*, yi,n)
Let us take the tensor product of all si, 1 i r, with respect to Fp[c1, c2,*
* ..., cn]
in the source and Fp[x1, x2, ..., xn] in the target. Then we can identify the s*
*ource
of ri=1si with H*Xn(R) by 2.1, the target with H*BDn(R) by the first part of
the corollary, and hence, the map ri=1si itself with resn O f*nby naturality. *
*The
injectivity of ri=1si can be shown as in [8, Lemma 9].
3.An invariant problem
In this section, we deduce the Theorem 1.1 from a more general invariant prob*
*lem
which we formulate and solve after some preliminary steps.
3.1. Preliminary steps. In order to prove 1.1 we need to use the matrix block-
multiplication GLm (R) x GLn(R) ! GLm+n (R) for m 0 and n 0 (see Remark
3.5). This multiplication induces a mod p homology ring structure on each of the
direct sums appearing in the following diagram (compare to [3])
1M 1M M1
H*BDn(R) -resn*----!H*BGLn(R) -fn*--! H*Xn(R)
n=0 n=0 n=0
where resn* is dual to resn and fn* is dual to f*nfor n 0 (the homology and
cohomology are dual as vector spaces over Fp). With respect to this ring struct*
*ure
the maps 1n=0resn* and 1n=0fn* are ring homomorphisms. This diagram induces
a diagram of rings and ring homomorphisms
M1 M1 M1
(3.1) H*BDn(R) æ-! I0n-! M0n
n=0 n=0 n=0
where V 0denotes the degreewise dual of the graded vector space V over Fp and t*
*he
maps in the diagram are direct sums of maps dual to the inclusions Mn In and
In H*BDn(R) for n 0 (same notations as in 1.1). These properties suggest an
invariant problem which we formulate in the next subsection.
AN ELEMENTARY INVARIANT PROBLEM 5
3.2. Formulating the invariant problem. Let r be a non-negative integer and
F a field of characteristic different from 2. Both r and F are fixed throughout*
* the
discussion. For each n 0 we define the following graded algebra over F
Or
(3.2) An = F [x1, x2, ..., xn] (yi,1, yi,2, ..., yi,n)
i=1
with |xj|= 2 and |yi,j|= 1 for 1 i r and 1 j n. The symmetric group n
acts on An as described in Remark 2.3. With respect to this action we define the
following subring of invariants
Or
(3.3) En = F [c1, c2, ..., cn] (ei,1, ei,2, ..., ei,n)
i=1
where cj and ei,jare the symmetrizations of x1...xj and x1...xj-1yi,jfor 1 j *
* n
and 1 i r.
Let V 0denote the degreewise dual of the gradedLvector space V over F and
define an algebra structure on the direct sum 1n=0A0nas follows. For each n *
* 0,
let S(n) be the set of all sequences
(3.4) I = (k1, ..., kn; ffl1,1, ..., ffl1,n; ...; fflr,1, ..., fflr,*
*n)
of non-negative integers with ffli,j2 {0, 1}for 1 i r and 1 j n. The
monomials r
Y ffl ffl
xI = xk11...xknn yi,1i,1...yi,ni,n
i=1
with I 2 S(n) as in (3.4)form an additive basis for An and let (uI) be the dual
basis in A0n.
Definition 3.1. If I 2 S(n) is of the form (3.4)and J 2 S(m) of the form
J = (l1, ..., lm ; OE1,1, ..., OE1,m; ...; OEr,1, ..., OEr,m)
then we define the concatenation IJ 2 S(n + m) by:
IJ = (k1, ..., kn, l1, ..., lm ; ffl1,1, ..., ffl1,n, OE1,1, ..., OE1,m; ...;*
* fflr,1, ..., fflr,n, OEr,1, ..., OEr,m)
Definition 3.2. We define the product of any two basis elements uI and uJ by
the following equations
uIuJ = ffluIJ, xIJ = fflxIOxOJ
where I 2 S(n), J 2 S(m), O stands for sequences of the form (0, 0, ..., 0) such
that both IO and OJ belong to S(n + m), and ffl 2 {-1, 1}is chosen such that the
second equation is an identity in An+m . L
Lemma 3.3. With respect to the product defined in 3.2, the direct sum 1n=0A0n
is an algebra isomorphic to the tensor algebra generated by A01.
Proof.Let ` be the linear map from the tensor algebra generated by A01which sen*
*ds
each basis element of the form uI1 uI2 ...uIn to ffluI where n 0, Ik 2 S(1)*
* for
1 k n, I = I1I2...In, and ffl 2 {-1, 1}is chosen such that the following eq*
*uation
xI = fflxI1O...OxOI2O...OxO...OIn
is an identity in An. Here O 2 S(1) is a sequence of the form (0, 0, ..., 0) oc*
*curring
n - 1 times in each exponent. Then ` is the desired algebra isomorphism.
Now we can formulate the following invariant problem:
6 MARIAN F. ANTON
Theorem 3.4. For each n 0, let Vn be a graded linear subspace of An whichL
contains En and assume that there is a ring structure on the direct sum 1n=0V*
*n0
such that the linear map
1M 1M M1
g = gn : A0n-! Vn0
n=0 n=0 n=0
is a ring homomorphism with respect to 3.2 where gn is dual to the inclusion Vn
An for each n 0. Then E2 = V2 implies En = Vn for all n 0.
Remark 3.5. Let us drop the hypothesis that g is a ring homomorphism and assume
instead that Vn is a subring of the invariants Ann which contains En and whose
image with respect to the map which kills all generators xj and yi,jfor j > 2 (*
*and
keeps the others) becomes the subring E2 A2. It is a pleasant warm up exercise
to show that in case r = 0 and r = 1 the invariants (and therefore Vn) agree wi*
*th
the ring En. However, for r > 1 the invariants are bigger and it is not true th*
*at
En = Vn. For instance, let n 3 and yn the symmetrization of the monomial
y1,1y1,2y1,3y2,1y2,2y2,3in An. Then the subring Vn = En[yn] generated by En and
yn is a counterexample.
3.3. The proof of 1.1. We assume the notations and hypotheses of 1.1. By 2.1
and 2.2 the rings H*BDn(R) and Mn can be identified respectively with the rings
An and En as defined in (3.2)and (3.3)for F = Fp, r = (p + 1)=2, and n 0.
With these identifications, In is a graded linear subspace of An which contains
En for n 0. Therefore, the map æ from (3.1)can be identified with the map
g from 3.4 for Vn = In. According to our discussion inLx3.1, the map æ is a ring
homomorphism with respect to the algebra structure on 1n=0A0ninduced from the
matrix block-multiplication. By the Künneth theorem on homology, this algebra
structure agrees with the algebra structure defined in 3.2 (compare to the Lemma
3.3). Hence, 1.1 follows from 3.4.
3.4. The proof of 3.4. We assume now the notations of x3.2. Let us compose the
map g with the linear map
1M 1M M1
h = hn : Vn0-! E0n
n=0 n=0 n=0
where hn is dual to the inclusion En Vn for each n 0. Let t = hg and tn = h*
*ngn
for n 0. Then we show that
Proposition 3.6. ker(t) is an ideal generated by ker(t2).
Assuming this proposition and that g is a ring homomorphism, then ker(g) is
an ideal contained in ker(t) and hence, E2 = V2 will imply that ker(g) = ker(t).
Because g and t are epimorphisms as direct sums of maps dual to inclusions, it
follows that h is an isomorphism. Therefore, En = Vn for all n 0.
Remark 3.7. A particular case of the Proposition 3.6 appeared as [1, Lemma 3.3]
but the proof given in [1, p.12] is obviously inaccurate.
3.5. The proof of 3.6. The strategy we are pursuing in proving 3.6 is the follo*
*w-
ing. We first study the map t in terms of additive bases and show that ker(t) i*
*s an
ideal. Then we define a map ff : S(n) ! S(n) which controls the way t maps basis
elements. By using ff, we show that if ker(t) is not generated by ker(t2), then*
* we
AN ELEMENTARY INVARIANT PROBLEM 7
have an infinite ascent phenomenon inside a certain subset S of S(n). Because S
will be finite, our assumption is false, proving 3.6.
Let us start with an additive basis for En given by the monomials
Yr
cI = ck11...cknn effli,1i,1...effli,ni,n
i=1
where I 2 S(n) is of the form (3.4)and let (vI) be the dual basis in E0n. Then *
*we
have the following duality equations
X
(3.5) cK = [K : I]xI forK 2 S(n)
I2S(n)
X
(3.6) t(uI) = [K : I]vK forI 2 S(n)
K2S(n)
where the coefficients [K : I] 2 F are zero almost everywhere in each sum.
L 1
Lemma 3.8. There is a ring structure on the direct sum n=0E0nsuch that t is
a ring homomorphism with respect to 3.2.
L 1
Proof.Let us define a ring structure on the direct sum n=0E0nby the following
equations which are similar to 3.2:
vK vL = ffivKL , cKL = fficKO cOL
where K 2 S(n), L 2 S(m), and ffi 2 {-1, 1}is chosen such that the second
equation is an identity in En+m . Then we can use (3.5), (3.6), and 3.2 to show*
* that
t is a ring homomorphism.
Next, we need a way to bound indices I by natural numbers (see 3.9). Also, we
use the following lexicographical order: two sequences of integers a = (a1, ...*
*, am )
and a0= (a01, ..., a0m) of the same length m 0 are in the relation a > a0if a*
*1 > a01
or there is 1 s < m such that aj = a0jfor 1 j s and as+1 > a0s+1.
Definition 3.9. Let n 0 and I 2 S(n) of the form (3.4). Then we define the
degree of I by
Xn Xr
deg(I) = (kj+ ffli,j)
j=1 i=1
Lemma 3.10. Let n 0. Then there is an injective map ff : S(n) ! S(n) such
that for any I 2 S(n) and K 2 S(n) the following conditions hold:
(1) [K : I] = 1 if I = ff(K)
(2) [K : I] = 0 if I > ff(K)
(3) If I is of the form (3.4), then I is in the image of ff if and only if
Xr
kj- kj+1 ffli,j+1for all1 j < n
i=1
(4) [K : I] = 0 if deg(I) 6= deg(ff(K)).
Proof.Let us assume that I 2 S(n) has the form (3.4)and K the form
K = (a1, ..., an; OE1,1, ..., OE1,n; ...; OEr,1, ..., OEr,n)
8 MARIAN F. ANTON
Then xI is lexicographically with respect to I the leading term of the expansio*
*n of
cK as in (3.5)if and only if:
Xr Xr
kj = aj+ ... + an + OEi,j+1+ ... + OEi,n for1 j < n
i=1 i=1
kn = an
ffli,j= OEi,j for1 j n and 1 i r
Given I 2 S(n), this system of equations has a solution K 2 S(n) if and only if
I satisfies the condition (3) of the lemma. Moreover, if this condition is sati*
*sfied,
then the solution K is unique. Hence, if we define ff(K) to be the only element*
* in
S(n) such that xff(K)is the leading term of cK , then the map ff is injective a*
*nd
satisfies (1) to (3). In order to prove (4) we only need to observe that all mo*
*nomials
xI occurring with non-zero coefficients in the expansion of cK have the same va*
*lue
for deg(I) and xff(K)is one of these monomials.
With these preparations and notations, we can prove 3.6 as follows. By 3.8,
t is a ring homomorphism and therefore, ker(t) contains the ideal generated by
ker(t2), denoted (ker(t2)). Now, assume that ker(t) is bigger than (ker(t2)). F*
*or
each n 0 and d 0, let M be the set of all elements x 2 A0nsuch that x 2 ker*
*(t),
x 62 (ker(t2)), and x can be written in the form
X
(3.7) x = (x : I)uI
I2S(n), deg(I)=d
where we define (x : I) 2 F for all I 2 S(n) by setting (x : I) = 0 if deg(I) 6*
*= d.
Let us fix n and d such that the set M is not empty (we can do this according to
our assumption, (3.6), and 3.10 (4)). Now, let S be the set all I 2 S(n) satisf*
*ying
the following conditions: deg(I) = d and there is x 2 M such that (x : J) = 0 f*
*or
all J < I. This set is finite and not empty. Let I0 2 S be lexicographically *
*the
maximal element of S.
We claim that we can find an element I002 S which contradicts the maximality
property of I0. More specifically, we have the following
Proposition 3.11. Let x0 2 M such that (x0 : I) = 0 for all I < I0. Then there *
*is
u 2 (ker(t2)) such that u can be written in the form (3.7)with (u : I0) = (x0 :*
* I0)
and (u : I) = 0 for I < I0.
Now, the element x00= x0 - u belongs to M and there is I002 S(n) such that
deg(I00) = d and (x00: I) = 0 for all I < I00but (x00: I00) 6= 0. Therefore, I0*
*02 S and
by the properties of u and x0, (x00: I) = 0 for all I I0. It follows that I00*
*> I0,
proving the claim and 3.6.
3.6. The proof of 3.11. We need to make use of the map ff and its properties
given in 3.10.
Case 1: Assume that there is K 2 S(n) such that I0 = ff(K). By 3.10 (1) (2)
we have [K : I0] = 1 and [K : I] = 0 for all I > I0. Combining these with (3.6)*
*and
the properties of x0 and I0, we have
K ff X
0 = c , t(x0)= [K : I](x0 : I) = (x0 : I0)
I2S(n), deg(I)=d
where <, >is the duality pairing. Hence, u = 0 has the desired properties.
AN ELEMENTARY INVARIANT PROBLEM 9
Case 2: Assume that there is no K 2 S(n) such that I0 = ff(K). Then by 3.10
(3) there is 1 j < n such that I0 = I1I2I3 where I1 2 S(j - 1), I2 2 S(2), and
I3 2 S(n - j - 1) are of the form
I1 = (k1, ..., kj-1; ffl1,1, ..., ffl1,j-1; ...; fflr,1, ..., fflr,j-1)
*
*Xr
I2 = (kj, kj+1; ffl1,j, ffl1,j+1; ...; fflr,j, fflr,j+1) withkj- kj+1 < *
* ffli,j+1
*
*i=1
I3 = (kj+2, ..., kn; ffl1,j+2, ..., ffl1,n; ...; fflr,j+2, ..., fflr,n)
Subcase 2(a): Assume that t(uI2) = 0. Then uI0 = uI1uI2uI3 2 (ker(t2))
and u = (x0 : I0)uI0has the desired properties.
Subcase 2(b): Assume that t(uI2) 6= 0. Let d0= deg(I2) and define T to be
the set of all K 2 S(2) such that deg(ff(K)) = d0. Because ff is injective, T i*
*s a set
of finite order, say m. Let us label the elements of T by K1, K2,..., Km such t*
*hat
ff(K1) < ff(K2) < ... < ff(Km ).
Lemma 3.12. There is 1 s m such that I2 < ff(Ks) and [Ki: I2] = 0 for all
1 i < s.
Proof.By (3.6)and 3.10 (4), we have
mX
t(uI2) = [Ki: I2]vKi
i=1
Because t(uI2) 6= 0, there is 1 s m such that [Ks : I2] 6= 0 and [Ki: I2] =*
* 0 for
1 i < s. By 3.10 (3) there is no K 2 S(2) such that I2 = ff(K) and combining
this with 3.10 (2), we get I2 < ff(Ks).
Corollary 3.13. With s as in 3.12, the following system of equations
mX
0 = [Kj : I2] + [Kj : ff(Ki)]~i for all1 j m
i=s
has a solution ~s, ~s+1,..., ~m 2 F .
Proof.By 3.10 (1) (2), we have [Ki : ff(Ki)] = 1 for all s i m and [Kj :
ff(Ki)] = 0 for all 1 j < i m. Combining these with 3.12, we deduce that the
system has indeed a solution.
Let ~i2 F for s i m be a solution of the system 3.13 and define
mX
w = uI2+ ~iuff(Ki)
i=s
Then we have
mX mX
t(w) = ([Kj : I2] + [Kj : ff(Ki)]~i)vKj by (3.6)and 3.10 (4)
j=1 i=s
= 0 by 3.13
Therefore, w 2 ker(t2) and u = (x0 : I0)uI1wuI3 has the desired properties,
proving 3.11.
10 MARIAN F. ANTON
4. Etale obstruction classes
In this section we keep the hypotheses and notations from x2 and show that the
condition M2 = I2 in 1.1 is satisfied if certain homology classes vanish.
Namely, let SL2(R) be the special linear group of 2 x 2 matrices over R with
determinant 1 and consider the following map
` -1 '
(4.1) ' : D1(R) ! SL2(R), '(a) = a0 0a fora 2 D1(R).
By 2.2, the mod p group cohomology H*D1(R) can be identified with the ring
Or p + 1
(4.2) Fp[x] (yi), r = _____, |x|= 2, |yi|= 1 for1 i r.
i=1 2
For each I 2 S(1) of the form I = (k, ffl1, ffl2, ..., fflr) as in x3.2, let
rY 1
xI = xk yffliiand a(I) = _(ffl1 + ffl2 + ... + fflr - k)
i=1 2
where we recall that ffli2 {0, 1}for 1 i r. Then H*D1(R) has an additive ba*
*sis
given by the monomials xI with I 2 S(1). Let (uI) be the dual basis in H*D1(R).
Then the map '* induced by ' on mod p homology sends each basis element uI with
I 2 S(1) to a homology class '*(uI) in H*SL2(R).
Definition 4.1. Let fi be the Bockstein derivation on mod p cohomology. We call
an identification of H*D1(R) with the ring (4.2)admissible if x 2 fiH1D1(R).
Definition 4.2. Given an admissible identification, we say that '*(uI) 2 H*SL2(*
*R)
where I 2 S(1) is an etale obstruction class if a(I) is a positive integer.
Proposition 4.3. Let p an odd regular prime and fix an admissible identification
of H*D1(R) with (4.2). Then the linear subspace of H*SL2(R) spanned by the etale
obstruction classes is trivial if and only if the condition M2 = I2 in 1.1 is s*
*atisfied.
Remark 4.4. For p = 3, there is only one sequence I 2 S(1) satisfying 4.2, name*
*ly
I = (0, 1, 1). Therefore, given an admissible identification, there is only on*
*e ob-
struction class living in H2SL2(R). By [1, (5.1)], this homology group is zero,
reproving 1.2 as a consequence of 1.1 and 4.3. For p > 3, it is an open problem
whether M2 = I2 or not.
4.1. The proof of 4.3. The idea is to use the following commutative diagram of
groups and group homomorphisms
D1(R) x D1(R) _'xId//_SL2(R) x D1(R)
|ø| |ß|
|fflffl incl. fflffl|
D2(R) ____________//_GL2(R)
where ' is the map (4.1), Id is the identity map of D1(R), incl. is the inclusi*
*on map
D2(R) GL2(R), and the other two maps are defined by the formulae
` '
ø(a, b)= '(a) b0 0b
` '
ß(A, b)= A b0 0b
AN ELEMENTARY INVARIANT PROBLEM 11
for a, b 2 D1(R) and A 2 SL2(R) (on the right hand side we use matrix multipli-
cation). This diagram induces a commutative diagram on mod p cohomology
* Id
(4.3) H*D1(R) OH*D1(R)O 'oo__H*SL2(R) OH*D1(R)O
ø*|| ß* ||
| res2 | f*2
H*D2(R) oo_____________H*GL2(R) oo________H*X2(R)
where we have identified H*GL2(R) with H*BGL2(R). According to 2.1, 2.2, and
their proofs, there are identifications
`0 Or
H*D1(R) Fp[x] (yi)
i=1
Or
H*D2(R) Fp[x1, x2] (yi,1, yi,2)
i=1
Or
H*X2(R) Fp[c1, c2] (ei,1, ei,2)
i=1
such that x = fi(yi), x1 = fi(yi,1), x2 = fi(yi,2), c1 = fi(ei,1), 2c2 = fi(ei,*
*2), res2Of*2
is the inclusion determined by ei,1= yi,1+ yi,2, ei,2= x1yi,2+ x2yi,1, and ø* is
the map determined by ø*(yi,1) = 1 yi- yi 1, ø*(yi,2) = yi 1 + 1 yi for
all 1 i r. Assuming these identifications and notations, we have the follow*
*ing
lemmas which imply 4.3:
Lemma 4.5. The map ø* is a ring isomorphism mapping the image M2 of res2Of*2
to the following subring inside H*D1(R) H*D1(R)
Or
ø*(M2) = Fp[x2 1, 1 x] (xyi 1, 1 yi)
i=1
Proof.By applying fi to ø*(yi,1) and ø*(yi,2) for any 1 i r, we obtain ø*(x*
*1) =
1 x - x 1 and ø*(x2) = 1 x + x 1. Hence, we easily see that ø* is surje*
*ctive
and therefore, an isomorphism, because 2 is invertible (p is odd) and the source
and the target of ø* are graded vector spaces of the same finite dimension in e*
*ach
degree. Also, ø*(ei,1) = 2(1 yi), ø*(ei,2) = 2(1 xyi- xyi 1) for 1 i r*
* and
by applying fi, ø*(c1) = 2(1 x), ø*(c2) = 2(1 x2 - x2 1). Therefore, 1 *
*yi,
1 x, xyi 1, and x2 1 for 1 i r generate ø*(M2) as a ring.
Lemma 4.6. The image of '* is invariant with respect to the ring automorphism
of H*D1(R) given by x 7! -x, y 7! -yi for 1 i r.
Proof.The group automorphism of D1(R) given by a 7! a-1 for a 2 D1(R) induces
the multiplication by (-1) map on H1D1(R). The map '(a) ! '(a-1) for a 2 D1(R)
is a conjugation inside SL2(R) which induces the trivial map on H*SL2(R). Hence,
we conclude that the image of '* is invariant with respect to the map described*
* in
the lemma.
Lemma 4.7. If all etale obstruction classes associated to `0 in H*SL2(R) vanish,
then ø*(I2) = ø*(M2).
12 MARIAN F. ANTON
Proof.The idea is to use the additive basis of H*D1(R) H*D1(R) given by
the monomials xI xJ with I, J 2 S(1). By 4.5, the monomials xI xJ where
I, J 2 S(1) with a(I) a non-positive integer form an additive basis for ø*(M2).*
* By
4.6, the monomials xI xJ where I, J 2 S(1) with a(I) an integer span a linear
subspace which contains the image of '* Id. Finally, by our hypothesis, given *
*any
z 2 H*SL2(R), z0 2 H*D1(R), and I, J 2 S(1) with a(I) a positive integer, the
following equation is satisfied
<'*(z) z0, uI =uJ>0
Here, (uL) denotes the dual basis in H*D1(R) of the basis (xL) in H*D1(R) where
L 2 S(1) and <, >the duality pairing. Hence, by putting together these observa-
tions we conclude that Im('* Id) ø*(M2). According to the diagram (4.3), we
have ø*(M2) ø*(I2) Im('* Id) and the conclusion follows.
Lemma 4.8. If there is an etale obstruction class associated to `0 in H*SL2(R)
which is not zero, then ø*(I2) 6= ø*(M2).
Proof.By a spectral sequence argument, Im(ß*) agrees with the invariants of
H*SL2(R) H*D1(R) with respect to the action of the the cokernel of ß which is
D1(R)=D1(R)2 (Z=2)r. Because this group acts trivially on H*D1(R) H*D1(R)
and 2 is invertible, we conclude that
(4.4) Im('* Id) = ('* Id)(Im(ß*)) = ø*(I2).
With the same notations as in the proof of 4.7, if there is I0 2 S(1) with a(I0*
*) a pos-
itive integer and '*(uI0) 6= 0, then there is z 2 H*SL2(R) such that 6= 0.
Consequently, '*(z) 1 belongs to ø*(I2) by (4.4)but not to ø*(M2) by 4.5.
Lemma 4.9. The linear subspace of H*SL2(R) spanned by the etale obstruction
classes is independent of the admissible identification chosen.
Proof.Let (x; y1, y2, ..., yr) and (~x; ~y1, ~y2, ..., ~yr) be two sets of ring*
* generators for
H*D1(R) with x, ~x2 fiH1D1(R) and yi, ~yi2 H1D1(R) for 1 i r. Because
fiH1D1(R) is a one-dimensional vector space, ~x= ~x for some non-zero scalar
~ 2 Fp. Because H1D1(R) is r-dimensional, there is (`i,j) 2 GLr(Fp) such that
Xr
~yi= `i,jyj for1 i r.
j=1
Then, for each I 2 S(1) of the form I = (k; ffl1, ffl2, ..., fflr) the monomial
Yr
x~I= ~xk ~yfflii
i=1
can be written as a linear combination of monomials of the form xIoewhere oe 2 *
* r
and Ioe= (k; ffloe(1), ffloe(2), ..., ffloe(r)) and vice versa. Observe that a(*
*I) = a(Ioe) for any
I 2 S(1) and oe 2 r. Let (uI) be the dual basis of (xI) and (~uI) the dual of *
*(~xI)
where I 2 S(1). Hence, we conclude that each etale obstruction class '*(uJ) with
J 2 S(1) is a linear combination of '*(~uI) where I 2 S(1) with a(I) a positive
integer, and vice versa.
AN ELEMENTARY INVARIANT PROBLEM 13
4.2. More general rings of integers. According to [2], we can generalize 1.1 as
follows. Suppose that F is a finite field extension of Q and OF the ring of alg*
*ebraic
integers in F . Assume that OF satisfies the following conditions:
(1) In OF there is only one prime ideal above p
(2) The Picard group of OF has no p-torsion
(3) OF contains a primitive p-th root of unity
Let R = OF [1=p] and r = r2+1 where r2 is half the number of complex embeddings
of F . Then we can define the etale obstruction classes associated to an admiss*
*ible
identification of H*D1(R) with a ring of the form (4.2)exactly as in 4.2. With
these adjustments, the proofs go through word by word and we have the following
overall result:
Theorem 4.10. Let R = OF [1=p] with p an odd prime and OF a ring of algebraic
integers satisfying (1) to (3) above. Then the linear subspace of H*SL2(R) span*
*ned
by the etale obstruction classes associated to an admissible identification is *
*trivial
if and only if the maps resn and resn O f*nas in 1.1 have the same image for all
n 0, including in the stable range.
Remark 4.11. The statement that resn and resn O f*nhave the same image in the
stable range was proved in a paper of S. Mitchell [7]. It should be noted that
H*BDn(R) and H*Xn(R) are completely stable in the sense that they are images
of the n = 1 versions, but this is not true for H*BGLn(R). Hence the case n < 1
does not follow from Mitchell's result.
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[9]Voevodsky, V.: The Milnor conjecture, preprint MPI, 1997
Department of Pure Mathematics, University of Sheffield, Hicks Building, Shef*
*field
S3 7RH, U.K. and IMAR, P.O. Box 1-764, Bucharest, RO 70700
Current address: University of Kentucky, Department of Mathematics, 715 POT, *
*Lexington,
KY 40506-0027, U.S.A.
E-mail address: Marian.Anton@imar.ro