I'm looking for simple examples of calculations of equivariant homology and of equivariant bordism.

I have a finite group G acting on an CW-complex X. I would like to calculate the equivariant homology $H_\ast^G(X)= H_\ast(X\times_G EG)$. In all of the examples I know, either the calculation is degenerate, for example because action of $G$ is taken to be free in which case $H_\ast^G(X)\simeq H_\ast(X/G)$, or else the setting is quite complex (string topology for instance) and I have trouble following. What I need is a good "second example".

I am looking first for a useful example of a simple non-degenerate equivariant homology calculation. One example I am specifically interested in is to calculate $H_\ast^{C_p}(C_q)$, where $C_p\ltimes C_q$ is a metacyclic group, so that $C_p$ is a cyclic group of order p acting on $C_q$ (with addition as the group operation) by multiplication by a $p$th root of unity in $C_q$. For example, $C_2$ would act on $C_7$ by multiplication by $6$, and $C_3$ would act on $C_7$ by multiplication by $2$. The action is not free (it fixes $0\in C_q$) so this example is not degenerate, yet the groups are small and finite.

I have the same question with regard to equivariant bordism. In the above example for instance, consider smooth maps from CW-complexes to the Eilenberg-Maclane space $K(C_q,1)$, where two such maps are considered equivalent if their image differs by a smooth $C_p$-action on $K(C_q,1)$. How to calculate $\Omega_\ast^{C_p}(C_q)$?

  • $\begingroup$ There's the case $X = \{*\}$, of course. Finite groups acting on surfaces would be a good source of examples in the case $G$ finite. If you want a continuous group, Seifert-fibred 3-manifolds with orientable fibres all have fixed-point free $SO_2$-actions but relatively diverse orbit decompositions. These could all be done with equivariant decompositions, together with knowledge of the homology of $EG/H$ for $H$ subgroups of $G$. I imagine other people will have far better suggestions but, not knowing any better that's where I'd start. $\endgroup$ Dec 8, 2010 at 1:28
  • $\begingroup$ note that $EG/H$ is just $BH$. $\endgroup$ Dec 22, 2010 at 14:51
  • $\begingroup$ @Ryan, could you expand on what you mean by an "equivariant decomposition"? $\endgroup$
    – Mark Grant
    Feb 4, 2011 at 15:26

2 Answers 2


Sorry this got too long for a comment, so I post it as an answer.

A CW-complex equipped with a $G$- action is not the right thing to consider. A $G$-CW complex is a space, that can be built using blocks of the form $G/H \times (D^n,S^{n-1})$ (in the same way you build a CW-complex). A $G$-CW complex has better properties than a CW-complex, for example the fixed point set of any subgroup is a subcomplex. However most CW-complexes with a $G$ action can also be given the structure of a CW-complex.

In the example of $G=\{1,t\}=\mathbb{Z}/2$ acting on $S^1\times S^1$ by flipping the components, one could for example take the following $G$-CW- structure with 1 $0$-cell $P$ of type $G/G$, two 1-cells $A,B$ of type $G/1,G/G$ and 1 2-cell $C$ of type $G/1$:

image of G-CW structure http://www.freeimagehosting.net/uploads/d0acae4eb9.jpg

Then the cellular chain complex may be considered as $\mathbb{Z}[G]$ chain complex. Here it is:

$\mathbb{Z}[P]\leftarrow\mathbb{Z}[A,tA,B] \leftarrow \mathbb{Z}[C]$

, where the differentials are given by $A,B\mapsto 0, C\mapsto A+tA+B$. Forgetting the names, we can write the chain complex as

$\mathbb{Z}[G/G]\leftarrow\mathbb{Z}[G/1]\oplus \mathbb{Z}[G/G] \leftarrow \mathbb{Z}[G/1]$

These maps are $G$-equivariant. If one would apply $\otimes_{\mathbb{Z}[G]}\mathbb{Z}$ to this chain complex and take the homology, one would just get the cellular homology of the quotient $H_*(X/G)$.

I think, that if one takes a projective resolution of this chain complex first, one should get $H_*(X\times_GEG)$. (The cellular chain complex of $X\times EG$ is a free resolution of the cellular chain complex of $X$).

EDIT:I am not sure, whether I understand the second paragraph correctly. Group homology can be defined as the homology opf the classifying space, which reads $H_*(G):=H_*(BG)$. So this causes confusion, if it is not clear, whether one considers the group homology or the homology of the discrete space given by forgetting the group structure.

If the second was the case: Then the answer would be: write the discrete space as a union of orbits $\amalg_{i\in I}G/H_i$ and the result would be (like above) $\bigoplus_{i\in I} H_*(H_i)$ (denoting the group homology).

Or (more likely) it asks for some sort of "equivariant group homology", which I would read as $H_*^C_p(BC_q)$. Using a nice functorial construction for $BC_q$, the $C_p$ action on the group also gives a $C_p$ action on the classifying space $BC_q$ and also on $EC_q$. However in this case I would guess, that $EC_q\times EC_p$ is a free,contractible $C_p\ltimes C_q$-CW-complex. Then it would be a model for $EC_p\ltimes C_q$ and hence:

$(BC_q \times EC_p)/C_p=(EC_q/C_q\times EC_p)/C_p=(EC_q\times EC_p)/(C_p\ltimes C_q)$. Then $H_*^{C_p}(C_q)$ would just be the group homology of $C_p\ltimes C_q$.


How about $\mathbb{Z}_2$ acts on $S^n\times S^n$:

  • by transposition of factors
  • by a reflection in one factor.
  • $\begingroup$ How is this action not free? $\endgroup$ Dec 8, 2010 at 12:46
  • $\begingroup$ If $\tau$ is the nontrivial group element, then $^\tau (x,x)=(x,x)$ for all $x\in S^n$ so the first action is not free. The second action is free. $\endgroup$ Dec 8, 2010 at 13:07
  • $\begingroup$ I mean to reflect in an equatorial plane (in say, the first factor). Then $\{equator\}\times S^n$ is left fixed. $\endgroup$
    – Mark Grant
    Dec 8, 2010 at 14:23
  • 1
    $\begingroup$ Okay... then how would the calculation go? $\endgroup$ Dec 8, 2010 at 15:08
  • $\begingroup$ The only way I know of computing $H_*^G(X)=H_*(EG\times_G X)$ is by looking at the Leray-Serre spectral sequence of the Borel fibration $X\to EG\times_G X\to BG$. I think in both these cases all differentials are trivial, so it's just a case of computing $E^2_{*,*}=H^*(G, H^*(X))$. $\endgroup$
    – Mark Grant
    Dec 8, 2010 at 16:43

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