\magnification=\magstep1
\input amstex
\documentstyle{imappt}
\def\bz{{\Bbb Z}}
\def\br{{\Bbb R}}
\def\tr{\operatorname{tr}}
\def\deg{\operatorname{deg}}
\def\os{\operatorname{os}}
\NoBlackBoxes
\centerline{{\bf GROUP ACTIONS AND THE SINGULAR SET}} 
\smallskip
\centerline{by}
\smallskip
\centerline{DANIEL H. GOTTLIEB {\smc and} MURAD OZAYDIN} 
\medskip
\noindent{\bf 1. Introduction}

Suppose a compact Lie group $G$ is acting on a $G$--CW complex $X$.  The
{\it singular set} $X^{S}$ consists of all points in $X$ with
non-trivial isotropy subgroups.  If the action were free, then the singular
set would be empty, otherwise $X^{S}$ is non-empty.  The main purpose of this
paper is to show that the {\it trace} of the action, denoted  $\tr(G,X)$,
is precisely equal to the trace of the singular set $\tr(G,X^{S})$ when the
action is not free.

The trace of an action was introduced in \cite{{\bf G}} .  It is closely
related to the exponent of Browder and Adem, studied in \cite{{\bf B1}},
\cite{{\bf B2}}, and \cite{{\bf A}}.  It is an
integer invariant of the action which characterizes a free action of a
finite group on a finite complex, and also equals the number of 
points are in the
smallest orbit of an action when $G$ is an elementary abelian $p$-group.

We first recall briefly the definition of the trace of an action as defined
in \cite{{\bf G}}.  Then we note that the main results of \cite{{\bf G}}
may be expressed in the category of $G$--CW complexes.  The key lemma is
established.  It gives an obstruction to adjoining an equivariant cell
without changing the trace.  This implies the main theorem that $\tr(G,X)$
equals $\tr(G,X^{S})$.  

We apply the key lemma to investigate the relationship between the trace
of an action, and the orbit size of the action which we define to be the
greatest common divisor of the cardinalities of the orbits. While these
two numbers are always equal when an elementary abelian $p$-group acts on
a finite complex (indeed we will show here that these numbers are equal if
the group has elementary abelian Sylow $p$-subgroups for all $p$),
we will show that for cyclic subgroups examples can be found
in which the trace and orbit size assume any arbitrary integers subject
only to the restriction that the first integer divides the second.

Finally in the last section, we compare the trace and the exponent. We
observe they are equal when  $G$  is a finite group and  $X$  is a finite
complex. It is an open question if they agree when  $X$  is only finite
dimensional. While the trace is defined for every action, the exponent
as defined in \cite{{\bf A}} is only defined for finite group actions on
finite dimensional complexes. We propose an extension of the definition of the
exponent to compact
Lie groups.

Many of these results comprise chapter III of \cite{{\bf O}},
the Ph.D thesis of Murad
\"Ozayd\i n , Purdue University.
\medskip
\noindent{\bf 2. The trace of an action}

Given a map $f: Y\to X$ between two topological spaces, a
{\it transfer} $\tau$ of trace $k$ ($k\in\bz$) for $f_{*}$ is a graded
homomorphism
$$
\tau: H_{*}(X;\bz)\to H_{*}(Y;\bz)
$$
so that $f_{*}\circ\tau(x) = kx$ for all $x$ in $H_{*}(X;\bz)$.  The set of
all traces associated to transfers for $f_{*}$ form a subgroup of $\bz$,
and we define $\deg(f)$ (degree of $f$) to be the non-negative
generator  of this subgroup.

Let $F\to E\to B$ be a fiber bundle.  The {\it trace} of this bundle
is the least common multiple of the degrees of projection maps $q$ of the
pullback bundles $F\to f^{*}(E)\overset q\to\longrightarrow Y$ taken over
all maps  $f: Y\to B$.  The {\it trace of the action} of a group $G$
on a space $X$ is the trace of the Borel construction $X\to X_{G}\to
B_{G}$, denoted $\tr(G,X)$.

Now a key fact about traces is Proposition 6.6 of \cite{{\bf G}}:
\proclaim{Proposition 1} Let $(G,M)$ be an action on a compact manifold and
suppose $B_{G}$ has finite type.  Then $\tr(G,M) = \tr(p)$ where $p: E\to
B$ is a pullback of the universal fibration $p_{G}:M_{G}\to B_{G}$ by any
map $f: B\to B_{G}$ which is a $(\dim M +2)$-homotopy equivalence.  If
$B$ is a closed oriented manifold, then $\tr(G,M) = \deg(p)$.
\endproclaim
Now this proposition applies when $G$ is a compact Lie group,
since $B_{G}$ has finite type in this case.  We want to extend this result
for manifolds $M$ replaced by a finite $G$--CW complex $X$.  Recall that a
$G$--CW complex is built from equivariant cells of the form $G/K\times e$
 (where $K$,
a closed subgroup of $G$, is the isotropy subgroup of this cell $e$ ) with
equivariant attaching maps (\cite{{\bf I1}}).  A smooth compact manifold
with a smooth $G$ action has a $G$--CW structure \cite{{\bf I2}}.
\proclaim{Lemma 2} Let $G$ be a compact Lie group and let $X$ be a finite
$G$--CW complex.  Then $X$ is a equivariant retract of a compact oriented
$G$-manifold.
\endproclaim

\underbar{Proof:} We may equivariantly imbed $X$ as a subspace of a
representation space of $G$.  Then Schultz, \cite{{\bf S}}, states that $X$
is an equivariant retract of some open invariant neighborhood $V$.  Using
the Haar measure, we can find a smooth invariant map from $V$ to $\br$
which is zero on $X$ and $1$ near the frontier of $V$.  The inverse image
of a suitably chosen interval gives the compact oriented invariant manifold
which we are seeking. $\square$

Hence Proposition 1 is true when $M$ is a finite $G$--CW complex, since traces
are preserved by equivariant retractions.

If $f: B\to B_{G}$ pulls back a fibre bundle $p: E\to B$ so that $\tr(p)
= \tr(G,X)$, then we say that $p$ {\it realizes} $\tr(G,X)$.  When $B$
is a closed oriented $m$-manifold, then $\tr(G,X) = \deg(p)$.  That is the
same as saying that the image of $p_{*}: H_{m}(E)\to H_{m}(B)\cong\bz$
is generated by $\tr(G,X)[B]\in H_{m}(B;\bz)$.

We can easily find a closed compact oriented manifold $B$ so that $f: B\to
B_{G}$ is highly connected for $G$ a compact Lie group.   It suffices to
find a highly connected, closed, oriented manifold $E$ with a free, orientation preserving $G$-action.  Then $B = E/G$ and $f$ is a classifying map for the
principal bundle $G\to E\to B$.  Any compact Lie group $G$ is isomorphic to
a subgroup of a special orthogonal group $SO(n)$.  But $SO(n)$ acts freely
(and preserves the orientation because $SO(n)$ is connected) on the highly
connected Stiefel manifold $SO(n+N)/SO(N) = E$.
\medskip
\noindent {\bf 3. The obstruction to preserving the trace}

We investigate the effect of adjoining an equivariant cell to a $G$--CW
complex on the trace of the action. Let $X$ be $G$--CW complex.  The case to
understand is the adjunction of an equivariant cell, because any finite
$G$--CW complex is built in finitely many such steps.  Recall that an
equivariant $n$-cell $G/K\times e^{n}$,  (where $K$ is a closed subgroup, $G$ acts on
the first factor) is adjoined by extending  a map $f: S^{n-1}\to X^{K}$
equivariantly to $\widetilde f: G/K\times S^{n-1}\to X$, the boundary of
our $n$-cell.  The next result establishes an obstruction to removing an
equivariant cell and preserving the trace.
\proclaim{Key Lemma} Let $G$ be a compact Lie group, $X$ a finite $G$--CW
complex and $Y = (G/K\times e^{n}) \underset{\widetilde f}\to\cup X$ as
above.  Then there is an $\alpha\in H^{n+d}(B_{K};\bz)$ such that
$$
\tr(G,Y)\big\vert\tr(G,X)\big\vert\tr(G,Y) \ |\alpha|
$$
where $d = \dim G/K$ and $|\alpha|$ is the (additive) order of $\alpha$.
\endproclaim

\underbar{Proof:}  We know that $\tr(G,Y)\big\vert\tr(G,X)$ because there is a
$G$-map from $X$ to $Y$ (\cite{{\bf G}}, 6.2a).  To show that
$\tr(G,X)\big\vert\tr(G,Y) \ |\alpha|$, we first choose a closed oriented manifold
$E$ of connectivity $N > n+d$, where $G$ acts freely on $E$.  Then, as
above, we have manifolds, $E/G = B$ and $E/K$.  Now $E/K$ is $N$-homotopy
equivalent to $B_{K}$, so $H^{n+d}(B_{K})\cong H^{n+d}(E/K)$.  So the
$\alpha$ we are seeking in $H^{n+d}(B_{K})$, we may regard as in
$H^{n+d}(E/K)$.  Now $E/K$ is a closed oriented manifold of dimension $\dim
E/K = \dim E - \dim K = \dim B + \dim G - \dim K = \dim B + \dim G/K =
m+d$, where $m$ is the dimension of $B$.  Now by Poincare duality
$H^{n+d}(E/K)\cong H_{m-n}(E/K)$.  Hence we can imagine the $\alpha$ in the
theorem as an element in $H_{m-n}(E/K)$.

We construct $\alpha$ as follows.  There is an element $\omega\in
H_{m}((Y\times E)/G)$ so that $p_{*}(\omega) = \tr(G,Y) [B]$.  Consider the
Meyer-Vietoris exact sequence of the union $(Y\times E)/G = ((X\times
E)/G)\cup ((G/K \times e^{n}\times E)/G)$, noting that $(G/K \times
e^{n}\times E)/G =  e^{n}\times E/K$  and also the
intersection of the two subspaces is equal to $(G/K \times S^{n-1}\times
E)/G$:
$$
\aligned
&\rightarrow H_{m}((X\times E)/G)\oplus H_{m}(e^{n}\times(E/K))\overset
{i_{*}+ j_{*}}\to\longrightarrow H_{m}((Y\times
E)/G)\overset\delta\to\longrightarrow\\
&H_{m-1}(S^{n-1}\times (E/K))|\overset{(k,\ell)_{*}}\to\longrightarrow
H_{m-1}((X\times E)/G)\oplus H_{m-1}(e^{n}\times(E/K))\endaligned
$$
Then $\delta(\omega) = [S^{n-1}]\times\alpha + 1\times\beta\in
H_{m-1}(S^{m-1}\times E/K)$ where $\alpha$ is some element in
$H_{m-n}(E/K)$ and $\beta$ is some element in $H_{m-1}(E/K)$.  But note that
$(k,\ell)_{*} (1\times\beta) = 1\times\beta\ \oplus\ g $ where $g$ is 
some other term
contained in the $H_{m-1}((X\times E)/G)$-summand.  Hence, by the exactness
of the Meyer-Vietoris exact sequence, $\beta$ must be zero .  Hence
$\delta(\omega) = [S^{n-1}]\times\alpha$.

Now $|\alpha|\omega$ is in the kernel of $\delta$, hence it is in the image of
$i_{*}+ j_{*}$ where $i$ and $j$ are the inclusions of $(X\times E)/G$ and
$(e^{n}\times (E/K))$ into $(Y\times E)/G$ respectively.  So there are $u$ and $v$
so that $i_{*}(u) + j_{*}(v) = |\alpha| \omega$. 
So $p_{*}(i_{*}(u) + j_{*}(v)) =p_{*}(|\alpha|\omega)$.
 We have $|\alpha|\tr(G,Y)[B] =
p_{*}(|\alpha|\omega) = p_{*}(i_{*}(u) + j_{*}(v)) = a[B] + b[B] = (a+b)[B]$ 
for some
integers $a$ and $b$.  Now $\tr(G,X)$ divides $a$ and $\tr(G,G/K)$ divides
$b$, so
$(a+b)$ is a multiple of the greatest common
divisor of $\tr(G,X)$ and $\tr(G,G/K)$.  But $G/K$ maps equivariantly into
$X$, so $\tr(G,X)$ divides $\tr(G,G/K)$.  Hence $\tr(G,X)$ divides $a+b$
and hence divides $|\alpha| \ \tr(G,Y)$ as was to be shown. $\square$

In order to apply the key lemma we need some control over the
cohomology of the isotropy subgroup $K$.  The easiest case is when $K$ is
trivial, so all cohomology in positive dimensions vanish.  We call two $G$
spaces $X$ and $Y$ $G$--{\it related} if there are $G$-maps $f: X\to Y$ and
$f': Y\to X$.  The {\it singular set} of a $G$-space $X$, denoted
$X^{S}$, consists of all points in $X$ with nontrivial isotropy subgroups.
\proclaim{Theorem 3} Let $G$ be a compact Lie group and let $X$ be a $G$-
space which is $G$-related to a finite $G$--CW complex. Then $\tr(G,X) =
\tr(G,X^{S})$, when $X^{S}$ is not empty.
\endproclaim

\underbar{Proof:} If $X$ is $G$-related to $Y$ then the singular sets
$X^{S}$ and $Y^{S}$ are also $G$-related.  Hence $\tr(G,X) = \tr(G,Y)$ and
$\tr(G,X^{S}) = \tr(G,Y^{S})$, so we only need to prove $\tr(G,Y) =
\tr(G,Y^{S})$.  The finite $G$--CW complex $Y$ is built from the subcomplex
$Y^{S}$ by adding equivariant cells with trivial isotropy subgroup.  But
the trace does not change when we add free cells, by the key lemma.
$\square$

\noindent {\bf 4. Trace and orbit size}

When $G$ is a finite group, a naive invariant measuring the nontriviality of
the $G$-action on $X$ is the {\it orbit size}, denoted $\os(G,X)$. This
is defined as the greatest common divisor of the cardinalities of all
the orbits. That is
$$
\os(G,X) = \gcd \{[G:G_{x}]: x\in X\}.
$$
 
In this section we compare the properties of the trace and the orbit
size. 

\noindent (i) $\tr(G,X)\big\vert \os(G,X)$ if $X$ is an arbitrary space
(\cite{{\bf G}}, 6.2a and
6.7b).

\noindent (ii) Suppose that $G_{p}$ is a Sylow $p$--subgroup of $G$.
Then
$\tr(G,X) = \Pi\tr(G_{p},X)$ where the product is over
all primes $p$ dividing the order of $G$. (\cite{{\bf G--O}}, theorem
14;
\cite{{\bf O}}, theorem 2.5)

\noindent (iii) $\os(G,X) = \Pi\os(G_{p},X)$ where the product
is taken over all primes $p$
dividing the order of $G$.

\noindent (iv) $\tr(G,X) = \os(G,X)$.
If all Sylow $p$--subgroups of $G$ are elementary abelian and $X$ is a
finite complex.
This follows from (ii) and (iii)
and (\cite{{\bf G}}, theorem 7.4) or (\cite{{\bf B2}}, theorem 1.1).

\noindent (v) $\os(G,X)\big\vert \Lambda_{f}$. 
where $X$ is a finite complex and $f:X \rightarrow X$ is
a $G$--map and the Lefschetz number of $f$ is denoted by $\Lambda_{f}$.
(\cite{{\bf O}},appendix)



Now we show that the trace equals the orbit size when $X$ is low
dimensional. On the other hand for cyclic groups acting on
closed oriented manifolds, we can find examples where the trace and
orbit size are any pair of positive integers subject to condition (i).
These results will follow from the key lemma and theorem 3.

\proclaim{Corollary 4} Let $G$ be a finite group and let $X$ be
$G$-related to a finite, one dimensional $G$--CW complex (a $G$-graph).
Then $\tr(G,X) = \os(G,X)$.
\endproclaim

\underbar{Proof:} Assume $X^{S}$ is not empty.
Since $\tr(G,X)$ and $\os(G,X)$ are invariants of
$G$-relatedness, it suffices to prove this for a finite, one dimensional
$G$--CW complex $X$.  For any finite (isotropy) group $K$, we know 
 $H^{1}(K;\bz)$ is
trivial, hence by the key lemma
the trace of the zero-skeleton is equal to $\tr(G,X)$.  But
for a finite discrete set $X^{\circ}, \ \tr(G,X^{\circ}) = \os(G,X^{\circ})$. 
When the singular set is
empty, i.e., when the action is free then $\tr(G,X) = |G|= \os(G,X)$
($G$ a
finite group and $X$ a finite $G$--CW complex).  The first equality
follows
from the key lemma by peeling off free equivariant cells till we get to
the zero skeleton.
 $\square$

\proclaim{Corollary 5} Let $G$ be a finite group acting smoothly on a
connected
manifold $M$.  Assume that there is a free orbit and either: a) $M$ is a
compact surface; or b) $M$ is an oriented 3-manifold with an orientation
preserving action.
Then $\tr(G,M) = \os(G,M)$.
\endproclaim

\underbar{Proof:} $M$ is a finite $G$--CW complex \cite{{\bf I2}}.  Since
there is a free orbit the hypothesis implies that the singular set $M^{S}$
is at most one dimensional,
(because the action is orientation preserving in the three manifold case
the codimension of the singular set must be greater than 1).
Therefore Theorem 3 and Corollary 4 imply
$\tr(G,M) = \os(G,M)$. $\square$  

\proclaim{Lemma 6} Let $\phi: \bz_{p^{n+1}}\to\bz_{p^{n}}$ be the cononical
epimorphism.
Thinking of $\bz_{p^{n}}$ as a subgroup of $S^{1}$, we have
the standard action on $S^{2k+1}$ given by complex multiplication on the
unit sphere in complex $(k+1)$--space. Let $\bz_{p^{n+1}}$ act on $S^{2k+1}$ via
$\phi$.  Then $\os(\bz_{p^{n+1}}, S^{2k+1}) = p^{n}$ and
$\tr(\bz_{p^{n+1}}, S^{2k+1}) =\max (p^{n-k},1)$.
\endproclaim

\underbar{Proof:}  Since the action is orientation preserving, the trace
equals the fiber number (\cite{{\bf G}}, 6.4) The fiber number is given by
the exponent of the transgression of the fundamental class $[\overline
S^{2k+1}]$ in the cohomology Serre spectral sequence of the Borel
construction:
$$
H^{2k+1} (S^{2k+1}) \cong E^{0,2k+1}_{2}\to 
E^{2k+2,0}_{2} \cong H^{2k+2}(\bz_{p^{n+1}})
$$
(because there are only 2 nonzero rows at the $E^{2}$ level).  The action
being induced from the action of $\bz_{p^{n}}$, the transgression map factors
through $H^{2k+2}(\bz_{p^{n}})\to H^{2k+2}(\bz_{p^{n+1}})$, by the naturality
of the Serre spectral sequence. $H^{*}(\bz_{p^{n}})$ is generated by the
powers of $\alpha$ in $H^{2}(\bz_{p^{n}}) \cong \bz_{p^{n}}$, and
$\phi^{*}(\alpha) = p\beta$ where $\beta$ is an analogous generator for
$H^{2}(\bz_{p^{n+1}})$. So $\phi^{*}(\alpha^{k+1}) = p^{k+1}\beta^{k+1}$. The
image of $[\overline
S^{2k+1}]$ under the transgression in the Borel construction of the
action of $\bz_{p^{n}}$ is a generator of $H^{2k+2}(\bz_{p^{n}})$ because
$\bz_{p^{n}}$ is acting freely. Thus the exponent of the image of $[\overline
S^{2k+1}]$ in $H^{2k+2}(\bz_{p^{n+1}}) \cong \bz_{p^{n+1}} = <\beta^{k+1}>$ is
$p^{n+1}/p^{k+1} = p^{n-k}$ if $k \leq n$ and 1 otherwise. Also 
$\os(\bz_{p^{n+1}}, S^{2k+1}) = p^{n}$ because every orbit has $p^{n+1}/p =
p^{n}$ elements.
$\square$

Let $G$ and $H$ act on spaces $X$ and $Y$ respectively. We say that $G
\times H$ acts on $X \times Y$ with the {\it product action} if $g
\times h(x \times y) = g(x) \times h(x)$ for all $g \in G$, $h \in H$, $x
\in X$, $y \in Y$.

\proclaim{Lemma 7. }  If $(G \times H)$ is a product action, and
if the order of  $G$ is relatively prime to the order of $H$, then

\noindent a) $\tr(G \times H, X \times Y) = \tr(G,X) \cdot \tr(H,Y)$

\noindent b) $\os(G \times H, X \times Y) = \os(G,X) \cdot \os(H,Y)$ 
\endproclaim

\underbar{Proof:} We identify $G$ with the subgroup $G \times 1 \subset
G \times H$ and $H$ with the subgroup $1 \times H$. Then the composition
$X \rightarrow X \times Y \rightarrow X$ given by $x \mapsto (x \times
y_{0}) \mapsto x$ is a composition of $G$--maps. Thus $X$ and $X \times
Y$ are
$G$--related. 
Both trace and orbit size are preserved under $G$--relatedness, hence
$\tr(G,X)
= \tr(G,X \times Y)$ and $\os(G,X) = \os(G,X \times Y)$. Similarly
$\tr(H,Y) =
\tr(H,X \times Y)$ and $\os(H,Y) = \os(H,X \times Y)$. Now it follows
from (ii)
and the fact that $|G|$ and $|H|$ are relatively prime that $\tr(G
\times H, X
\times Y) = \tr(G, X \times Y) \cdot \tr(H, X \times Y)$ which equals
$\tr(G,
X) \cdot \tr(H,Y)$ as required. Similarly, using (iii) and the relative
primeness of $|G|$ and $|H|$, we get $\os(G \times H, X \times Y) =
\os(G,X)
\cdot
\os(H,Y)$.

\proclaim{Theorem 8} Let $m$ and $n$ be positive integers  with $m$
dividing
$n$. There exists an action $(G,M)$ where $G$ is cyclic, $M$ is a closed
oriented manifold and the action is orientation preserving and
effective,
such
that $\tr(G,M) = m$ and $\os(G,M) = n$.
\endproclaim

\underbar{Proof:} We factor $m$ and $n$ into prime factors. Then 
$m = p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{k}^{e_{k}}$ and 
$n = p_{1}^{e_{1}+f_{1}}p_{2}^{e_{2}+f_{2}}\ldots p_{k}^{e_{k}+f_{k}}$,
where $0 \leq e_{i} , f_{i}$ for all $i = 1, \ldots , k$. For each $i$
consider
the action $(\bz _{p_{i}}^{e_{i}+f_{i}+1}, S^{2f_{i}+1})$ as described in
Lemma
6. So the trace of the action is $p_{i}^{e_{i}}$ and the orbit size is 
${ p_{i}}^{e_{i}+f_{i}}$. Now let $G$ be the product of the 
$ \bz _{p_{i}}^{e_{i}+f_{i}+1}$, so $G$ is a cyclic group.
Let $X$ be the product of the $S^{2f_{i}+1}$. Then by lemma 7, the
product
action $(G,X)$ has trace equal to $m$ and orbit size equal to $n$ . This
proves
the theorem except for the conclusion that the action should be
effective. We
attach a fee $G$--cell to $X$ to get the action $(G,Y)$ which is
effective. By
the key lemma, $\tr(G,Y)$ equals $m$; and $\os(G,X)$ obviously equals
$n$. But
now $Y$ is not a manifold. Using lemma 2 we can find a compact oriented
manifold
on which G acts effectively, preserving orientation, and which
retracts
equivariantly onto $Y$ . Then doubling the manifold and the action we
get a compact oriented
manifold
$M$ on which G acts effectively, preserving orientation, and which
retracts
equivariantly onto $Y$. 
Thus the trace and orbit size of $(G,M)$ still
equal
$m$ and $n$ . $\square$

Finally we show that corollary 4 is true only for dimension of $X$ equal
to one, and corollary 5 needs the
hypothesis
that there is a free orbit, by considering the following example which
also gives a non-trivial use of the key lemma.


\proclaim{Example 9} Let $\bz_{4}$ act on $S^{2}$ through the epimorphism
$\bz_{4}\to \bz_{2}$ and then via the antipodal action.  Then
$\tr(\bz_{4},S^{2}) = 1$, and $\os(\bz_{4},S^{2}) = 2$ .
\endproclaim

\underbar{Proof:} By lemma 6, $\tr(\bz_{4},S^{3}) = 1$.  But $S^{3}$ is
obtained from $S^{2}$ by adding an equivariant 3 cell of isotropy type
$\bz_{2}$.  Since $H^{3}(\bz_{2}) = 0$ we get $\tr(\bz_{4},S^{2}) =
\tr(\bz_{4},S^{3}) = 1$ by the key lemma.  $\square$

\noindent {\bf 5. The trace and the exponent}

We have already mentioned that the trace is closely related to the
exponent  of
(\cite{{\bf A}}, definition 1.4 and 3.1) . The trace is defined for any
action
while the exponent is defined only for actions of finite groups acting
on
connected finite dimensional CW--complexes. We will show that the trace
equals
the exponent, but only for finite complexes. Whether they are equal for
actions
on non-compact, finite dimensional complexes is open. We will suggest
an extension
of the definition of the exponent to the case of compact Lie groups
acting on
possibly disconnected $G$--CW complexes.

\proclaim{Proposition 10} If $G$ is a finite group and $X$ is a
connected compact
$G$--CW complex, then $\tr(G,X) =e_{G}(X)$, i.e. the trace equals the
exponent.
\endproclaim

\underbar{Proof:} If $X$ is a connected orientable manifold and the
finite
group $G$ acts preserving orientation, then the trace and the exponent
both
equal the fibre number (\cite{{\bf G}}, theorem 6.4; \cite{{\bf A}},
corollary
3.13). Now lemma 2 states that a compact $G$--CW complex is an
equivariant
retract of of a closed oriented $G$--manifold. Since both the trace and
the
exponent are invariant under equivariant retracts, they are
equal. $\square$

It would be useful to know if the trace equaled the exponent for
non-compact
finite dimensional $G$--CW complexes since the exponent works just as
well in
that case as in the compact case. For example $e_{G}(X) = e_{G}(X^S)$ .
So if
trace equals exponent, then our main theorem, theorem 3, could be
extended to
non-compact situations for finite $G$.

We propose to define $e_{G}(X)$ in the case where $X$ is a possibly
disconnected
finite dimensional $G$--CW complex and $G$ is a connected Lie group.
First if
$G$ is finite and acts transitively on the set of connected components,
where
$e_{G}(X)$ is the subgroup of $G$ which fixes a component $G_{0}$, we
define
$$ e_{G}(X) = [G:G_{0}] \cdot e_{G_{0}}(X)$$
Then if $X$ is  generally disconnected, we note that $X$ is the union
of spaces
$X_i$ on which $G$ acts transitively and we define
$$ e_{G}(X) = \operatorname{ gcd}\{e_{G}(X_i)\}$$
Finally if $G$ is a compact Lie group we propose the definition
$$ e_{G}(X) = \operatorname { lcm} \{e_{H}(X)\}\  
\hbox{for all finite subgroups $H$ of $G$}.$$

The new exponent $e_{G}(X)$  satisfies all the relevant properties listed
in the
introduction of \cite{{\bf A}} when G is finite. Also, for G finite and X
compact, the
trace equals the exponent. This follows because the trace satisfies the
first
two equations above (\cite{{\bf O}}, proposition 3.3; 
\cite{{\bf G}},p. 397). In the case of G compact and X compact we no
longer know
if they are equal, but $e_{G}(X)$ divides $\tr(G,X)$ since $\tr(H,X)$
divides
$\tr(G,X)$ by (\cite{{\bf G}}, 6.2b). (It is unusual for a useful invariant
to
divide the trace, it is usually the other way.) The trace would be equal
to the
exponent if the answer to the following question is  yes. Does 
$$  \tr(G,X) = \hbox{lcm}\ \{\tr(H,X)\} \ 
\hbox{for all finite subgroups $H$ of $G$ ?} $$

A similar question, whose affirmative answer would imply that trace
equals exponents when $G$ is finite and $X$ is finite dimensional, is the
following. Does
$$  \tr(G,X) = \hbox{gcd}\ \{\tr(G,Y)\} \
\hbox{for all finite $G$-subcomplexes $Y$ of $X$ ?} $$   

%\vfill\eject
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\vskip1truein
\noindent Purdue University

\noindent Oklahoma University
\enddocument

 

