4
$\begingroup$

Let $V$ be a chain complex, which is either $Z$ or $Z/2$ graded. A circle action on $V$ is by definition an action of the dga $H_\ast(S^1)$. This consists of a map $D : V → V$ , which is of square zero, commutes with $d$, and in the $Z$ graded case is of degree 1. the homotopy invariant is $V^{hS^1}= V [[t]]$ with differential $d+tD$. That is if $vf(t)$ is an element $V [[t]]$, then $d(vf(t)) = (dv)f(t) + (Dv)tf(t)$

The Tate complex is $V_{Tate} = V ((t))$ again with differential $d+tD.$

The homotopy coinvariants is the space $V_{hS^1} = t^−1V [t^−1] = V_{Tate}/V^{hS^1}$ with differential induced from that on $V_{Tate}$

Could anyone explain to me how this works? why homotopy (co)invariant is that? I am not quite familiar with homotopy (co)invariant.(but I know what it is)

$\endgroup$

2 Answers 2

5
$\begingroup$

These chain complexes are modeling a standard construction in equivariant homotopy theory. Let $S^\infty$ have the standard $S^1$ action (this is a model for what is called $ES^1$) and add a disjoint basepoint, denoting the result $S^\infty_+$. There is a cofiber sequence $S^\infty_+ \to S^0 \to \widetilde{S^\infty}$, which defines the last space. Now smash that cofiber sequence with the space ${\rm Map}(S^\infty_+, X)$ and take $S^1$ fixed sets. That is consider the (stable - we'll need that later) $S^1$ fixed sets of $$S^\infty_+ \wedge {\rm Map}(S^\infty_+, X) \to S^0 \wedge {\rm Map}(S^\infty_+, X) \to \widetilde{S^\infty} \wedge {\rm Map}(S^\infty_+, X). $$ In the middle one of course has the space of $S^1$-equivariant maps from $ES^1$ to $X$, which are the "homotopy fixed sets." In general the fixed sets $X^{S^1}$ can be viewed as the $S^1$-equivariant maps from a point to $X$ (or in the based case here a "pointed point", that is $S^0$), and since $ES^1$ is contractible but with a free $S^1$-action this ends up being a "more homotopy invariant" analogue to fixed sets. The chain complex version of this is to take equivariant homomorphisms from the chains on $S^\infty_+$ to a given $V$, which yields $V[[t]]$.

Next, smashing with $\widetilde{S^\infty}$ is the colimit of smashing with $\widetilde{S^{2n + 1}}$ and in homology each map in this colimit is multiplication by the Euler class of the standard representation of $S^1$ (viewed as a bundle over a point) so this is a localization by inverting that Euler class, yielding $V((t))$.

Finally, smashing with $S^\infty_+$ yields an equivalence when one smashes with an equivariant map which is a non-equivariant homotopy equivalence (sometimes called weak equivariant equivalences - but I might prefer "really weak"). Thus since $S^\infty_+ \to S^0$ is a really weak equivariant equivalence so is $X \cong {\rm Map}(S^0, X) \to {\rm Map}(S^\infty_+, X)$ and thus the left hand term is (strongly) equivariantly homotopy equivalent to $S^\infty_+ \wedge X$. Stably (sorry I do not have time/space to elaborate), the fixed sets of this are equivalent to the quotient $(ES^1_+ \wedge X)/S^1$. This is exactly what one should mean by "homotopy coinvariants" - smashing with $ES^1$ and then taking the quotient (that is, the coinvariants). In the world of chain complexes since this is a cofiber sequence one gets the model you describe.

In summary, if you are familiar with a more general story in equivariant homotopy theory then this construction on chain complexes makes sense. If you're strongly interested in $S^1$-equivariant homotopy theory, I would recommend Greenlees's book "Rational $S^1$-equivariant stable homotopy theory", though you really need to know some stable homotopy theory to begin with. You still might want to know where these constructions all come from, for which the short answer is that Tate first noticed that one could splice together group homology and cohomology to get a coherent theory. Later people (tom Dieck? Segal?) noticed that theory had to do with localization and then folks like Greenlees and May developed space-level constructions (which yield versions of Tate theory in any generalized cohomology theory, including K-theory which is a lot of fun).

$\endgroup$
2
  • $\begingroup$ @Sinha, Could you explain why taking equivariant homomorphisms from chains on $S^\inf_{+}$ to a given $V$ yields $V[[t]]$? Thanks! $\endgroup$
    – HYYY
    Commented Nov 14, 2010 at 10:43
  • $\begingroup$ If $V$ has trivial action, then these equivariant homomorphisms coincide with homomorphisms from the chains on $CP^\infty$. $\endgroup$
    – Dev Sinha
    Commented Nov 20, 2010 at 6:47
4
$\begingroup$

I have a different way to approach to this sort of question which appears in the paper:

Klein, John R. The dualizing spectrum of a topological group. Math. Ann. 319 (2001), no. 3, 421–456

The above was written using the language of naive $G$-spectra, but one can get the construction in the differential graded setting by means of a very simple idea.

The idea is this: if $R$ is an algebra augmented over a commutative ring $k$, then we have a composition pairing $$ \hom_{R\text{-mod}}(k,R) \otimes_{R} \hom_{R\text{-mod}}(R,M) \to \hom_{R\text{-mod}}(k,M) $$ that can be defined for any left $R$-module $M$. We are using here the observation that $R$ is a bimodule over itself to give $\hom_{R\text{-mod}}(-,R)$ and $\hom_{R\text{-mod}}(R,-)$
the structure of $R$-modules (the first of these is a right module and the second one is a left module).

Notice that $M = \hom_{R\text{-mod}}(R,M)$ canonically, so we can rewrite the pairing as $$ \hom_{R\text{-mod}}(k,R) \otimes_{R} M \to \hom_{R\text{-mod}}(k,M) $$

There is a derived version of this construction too: we can replace $k$ by a free resolution $\hat k$ over $R$, and $M$ by any chain complex over $R$ to get a pairing $$ D_R \otimes_R M \to \hom_{R\text{-mod}}(\hat k,M) $$ where $D_R = \hom_{R\text{-mod}}(\hat k,R)$ and where $\otimes_R$ now means derived tensor product.

The target is the derived invariants of $R$ acting on $M$, and I tend to denote it by $M^{hR}$. So with these changes, we have a map $$ D_R \otimes_R M \to M^{hR} . $$ I called this the norm map. Tate cohomology is defined to be homotopy cofiber of this map (algebraic mapping cone).

When $R = C_*(G)$ is the dga given by the chains on a $d$ dimensional compact Lie group, it turns out that is not so hard to identify $D_R$: it is quasi-isomorphic over $R$ to the chains on $S^{\text{ad}} =$ the one point compactification of the adjoint representation.

When $G=S^1$ one gets in this case $D_R = C_\ast(S^1)$ with trivial action, and the norm map in this case is a map $$ C_*(S^1) \otimes_{k} M_{hS^1} \to M^{hS^1} $$ which is the same as writing a map $$ \Sigma M_{hS^1} \to M^{hS^1} . $$ Yet another approach appears in the paper:

Klein, John R. Axioms for generalized Farrell-Tate cohomology. J. Pure Appl. Algebra 172 (2002), no. 2-3, 225–238.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .