Date: Fri, 27 Feb 2004 11:06:33 -0500
From: Tom Goodwillie <tomg@math.brown.edu>
>
>If EG_+ smash_G E = E_{hG} is contractible, then so is its functional
>dual
>D(E_{hG}) = F(EG_+ smash_G E, S) = F(EG_+, F(E, S))^G = D(E)^{hG}, i.e.,
>
>the homotopy fixed point spectrum of D(E).  So from the norm sequence
>
>         D(E)_{hG} -N-> D(E)^{hG} --> D(E)^{tG}
>
>the contractibility of D(E)_{hG} is equivalent to the contractibility of
>
>the Tate spectrum D(E)^{tG}. (Sometimes it is denoted t_G D(E)^G or
>hat H(G, D(E).)
>
>If D(E) is a G-equivariant ring spectrum, e.g. if E is the suspension
>spectrum of a finite G-space, then the Tate spectrum is an algebra over
>the homotopy fixed point spectrum, so if the latter is contractible,
>surely so is the Tate spectrum.  Thus in this case D(E)_{hG} will be
>contractible.

Well, that's very nice. So in that case we can say that if E has
trivial homotopy orbits then both E and D(E) have trivial homotopy
fixed points and homotopy orbits.

This raises a question: If E is (nonequivariantly equivalent to) a
finite spectrum and G (finite) acts on E, is E equivalent (in the
weakest possible sense: there is a G-map which is nonequivariantly an
equivalence) to the suspension spectrum of a finite G-CW complex,
after suspending by some (orientable?) representation?

I was trying a different approach to Greg's question. Reduce it to an
analogous algebraic question: Let G act on a finite chain complex V
over a field. If the (hyper)homology H(G;V) is trivial, does it
follow that H(G;D(V)) is also trivial? The only interesting case, of
course, is characteristic p where p divides the order of G. I can do
it for p-groups (and more generally nilpotent groups), then I get
stuck.

The key point is that under Greg's hypotheses if G is a p-group then
E must be trivial mod p. In fact, the first nontrivial mod p
cohomology group of E would have to survive to E_{hG} because when a
p-group acts on a nontrivial finite p-group the group of fixed
elements is nontrivial.

  -- Tom

Subject: Greg's question
Date: Fri, 27 Feb 2004 16:36:49 -0500
From: Tom Goodwillie <tomg@math.brown.edu>

For the list:

Greg, the answer is yes. The statement even generalizes to compact
Lie groups as long as the adjoint representation is orientable.

The key is to start with the case when G is connected. If the
connected group G acts on the spectrum E and E is bounded below then
the only way E_{hG} can be contractible is if E is contractible (The
first nontrivial homotopy group of E must be the same as that of
E_{hG}.)

Now let G be any compact Lie group and embed it as a subgroup of a
compact connected Lie group U. If G acts on E (a spectrum
nonequivariantly equivalent to a finite one), then consider the
induced U-spectrum Ind(E) = U_+ \smash_G E. The homotopy orbit
spectrum Ind(E)_{hU} is the same as E_{hG}, so if the latter is
contractible we find that Ind(E) is contractible. (That's actually if
and only if.)

The same applies to the G-action on the dual spectrum D(E). And

D(Ind(D(E))) = Coind(E) = Map^G(U_+,E),

so Greg's question becomes:

If the induced U-spectrum Ind(E) is contractible, must the coinduced
U-spectrum Coind(E) be contractible, too?

Well, they are not quite the same, but they differ by a little twist
that doesn't matter much for these purposes. I believe that Coind(E)
is Ind(S^{-V}E) where V is a representation of G, the difference
between the adjoint rep of U (restricted to G) and that of G. And
that means that the homology of Coind(E) and of  Ind(E) are related
by a Thom isomorphism as long as that representation is orientable,
so if one is contractible then the other has no homology and
therefore (being bounded below) is contractible.

In the nonorientable case (e.g. G=O(2)) I believe the statement is false.

  -- Tom