8
PETER HILTON
4. Properties of LS-category
We will be content here to mention specifically two properties of LS-category,
one general and one special.
THEOREM 4.1. Let LS-catX ::::; n. Then all n-fold cup-products of positive-
dimensional cohomology classes in X are zero.
PROOF. Consider the ordered sequence o:1,
0:2, · · · ,
O:n,
with o:i E Hm; (X, Gi),
i
= 1, 2, · · · , n. Then their cup-product is an element in Hm(X; Gi
® · · · ®
Gn),
m =
E
mi, given by the composite map
(4.1) X~ xn ~ K(Gl, ml) X ... X K(Gn, mn) ~ K(Gl
® ... ®
Gn, m),
where p represents the identity map
Hm(K(Gbml)
X··· X
K(Gn,mn))
~
G1
® · · · ®
Gn.
Obviously we may define the fat wedge
T
for any product of n spaces, so we embed
( 4.1) in the homotopy commutative diagram
X
xn
i
ix
K(G1,m1)
X··· X
K(Gn,mn)
j
i
rn X
~i
T (K(Gl,
ml)
X ... X
K(Gn, mn))
Certainly, pj
~
0. Thus if
ixJ.L
~
Ll, then p(llo:;).Ll
~
0, proving the theorem. D
THEOREM 4.2. Let LS-catX = 2. Then rr1X is free.
5
We first prove a key group-theoretical lemma.
LEMMA 4.3. Let G be any group. Then if
j:
G*G
~
G x G is the canonical
map from the free product to the direct product and if Ll: G
~
G x G is the
diagonal, the group
r
1
{LlG) is free.
PROOF OF LEMMA. We write, forgE
G,
g' for the element g appearing in the
first factor of G*G, and g" for the same element appearing in the second factor.
We then claim that
j-1(.LlG)
is free on the setS of elements g'g", g-=/:- e E G.
First it is obvious that S
~
j-
1
(LlG). Second, we claim that S is a free set,
that is, that it freely generates the subgroup of
j-
1
(
LlG) it generates. For let
(4.2)
be a reduced word in the elements of
S,
Ei
= ±1. For there to be any cancellation
in (4.2) there must exist
i
with
fi+l
=
-Ei;
but then we would only get cancellation
if gi = gi+l• that is, if (4.2) were not reduced. Thus S generates a free subgroup
F(S) of
r
1(.LlG)
and we must prove that, in fact, F(S) =
j-1(.LlG).
Notice that
we may writeS= S U (e), F(S) = F(S).
Now an element x of
j-1
(.LlG), just by virtue of being an element of G*G, has
a unique expression as
5
We include a proof of this theorem since the sketched proof given in
[J
a] is not quite correct.
The theorem itself is due to Fox [Fa].
Previous Page Next Page