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[It turns out what was wanted were groups with H^k(G,Z)=0 for many k;
the following shows H^k(G,Z/2Z)=0 is possible for many k -- a weaker statement.
--djr]
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Date: Sun, 25 Feb 96 23:38:31 CST
From: rusin (Dave Rusin)
To: adem@math.wisc.edu
Subject: Groups with lots of vanishing cohomology

Alex,

Did I understand you to ask for an example of the following type:

WANTED: a group  G_k  such that  H^n(G_k, F_2) = 0  for  n = 1, 2, ..., k
but not for  n=k+1.

(You have such a group  G_k  for  k= 0, 1, 2, 3  and want examples with 
higher  k.)

If that's what you want, you're in luck  --- such examples exist with large
k, and in all likelihood exist for arbitrarily many different values of  k.

One need only consider groups which are semidirect products of the form  
G_p = (Z/2)^p . C,  where  C  is cyclic of order  q = 2^p - 1, and this number
is a (Mersenne) prime. The action I intend for  C  on  (Z/2)^p  is the
action of the  Sylow  q-subgroup  of  GL_n(F_2). A generator of such a 
(cyclic)  C  can thus be represented by a  p x p  matrix over  F_2. It's
not diagonalizable, but over the Galois field  K=F_(2^p)  it can be diagonalized
as  diag(w, w^2, w^4, w^8, ..., w^(2^(p-1)) ), where  w  is a primitive
q-th root of  1.  In other words, if you extend the action of  C  on  
V = (F_2)^p  to an action on  W = V \tensor K, a basis of  W
may be chosen so that  C  acts diagonally. This then makes it easy to compute
the action of  C  on the tensor powers of  V:  we know  T^n(V) \tensor  K =
T^n(W), and the action of  C  on the latter is of course diagonalizable --
the eigenvalues are the products of  n  eigenvalues of the action
of  C  on  W.  Could  1  be among the eigenvalues? Yes, surely, but since
the order of  w  is  2^p-1,  Prod (w^(2^i))^e_i  is equal to  1  iff
Sum (2^i*e_i) = 0 mod  2^p - 1. The solution to this diophantine equation
for which  Sum(e_i)  is smallest is the one with  e_0=e_1=...=e_(p-1)=1,
in which  Sum(e_i) =  p. In other words, there is an eigenvector in
T^n(W)  with eigenvalue  1  only if  n >= p.  It follows in particular
that  T^n(V)  has no fixed points under the action of  C  for  n<p.

Since  H^n(G_p, F_2) = [ H^n((Z/2)^p, F_2) ] ^ C = T^n(V)^C  (oops, really I
should throw in a dual space somewhere), it follows that  
H^n(G_p, F_2) = 0   for  n = 1, 2, ..., p-1.  I don't think it's hard
to show that  H^p(G_p, F_2)  isn't zero, but we don't need that if
our only goal is to show the existence of groups with arbitarily large
initial cohomology groups zero, with a non-zero group following.

Of course it isn't proven that there are infinitely many Mersenne primes,
but I think most people expect that there are. In any event, the cases
p=5 and p=7  give you moderately small groups with larger vanishing
of cohomology than you implied  you had in other examples.  Taking
p=11,213 (or whatever the current record is) gives an impressive example.

Or did I misunderstand your question? This is too easy.

Enjoyed your visit.

dave

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