Let us assume that $X=\mathbb{R}\times S^1$ is given with a $G=\mathbb{Z}_2$ action that corresponds to the symmetry $(x,e^{i\theta})\mapsto(-x,e^{-i\theta})$. I want to compute the equivariant cohomology of $X$ relative to the fixed point set $X^G=\{(0,-1),(0,1)\}$. Since the subspace $X^G$ is $G$-invariant, there is a natural generalization of Borel's construction to the relative equivariant cohomology $$ H^*_G(X, X^G) = H^*(X\times_G EG,X^G \times_G EG). $$ And we know that this has an associated long exact sequence $$ \cdots \rightarrow H^n_G(X, X^G) \rightarrow H^n_G(X) \rightarrow H^n_G(X^G) \rightarrow H^{n+1}_G(X, X^G)\rightarrow \cdots. $$ How do we really calculate $H^n_G(X)$? When we take a field like $F_2$ with two elements, one can at least take advantage of Smith theory to conclude that since the cochain $C^*(X, X^G)$ is $G$-free, the relative equivariant cohomology can be identified with the cohomology of the subcomplex of invariants, which vanishes above the dimension of $X$ [1]. Can anybody help me with the lower relative cohomology groups?

[1] R.B. Sher, R.J. Daverman, Handbook of Geometric Topology, North Holland, 1st ed., p. 13



  • 3
    $\begingroup$ Your pair is equivariantly homotopy equivalent to $S^1$ with the reflection map and the same fixed point set, so you will get the same relative cohomology. Now by hand you can see that the relative Borel construction gives you in this case the suspension of $EG$. All you have is something in degree 0. $\endgroup$
    – mme
    Nov 14, 2018 at 18:50
  • $\begingroup$ @MikeMiller So you're saying that all I have is the equivariant cohomology of the suspension of $S^\infty$ that due to being contractible is again that of $S^\infty$? $\endgroup$ Nov 14, 2018 at 19:56

1 Answer 1


The relative Borel homology of a pair $(X,A)$ of $G$-spaces is well-defined up to equivariant homotopy equivalence (actually, up to equivariant maps which are nonequivariant homotopy equivalences of pairs). So we may reduce your example to $(S^1, \pm 1)$ with the reflection action.

Now $(S^1, \pm 1)$ has two fixed points and two free arcs that are sent to one another by the involution. The Borel construction $S^1 \times_{\Bbb Z/2} E(\Bbb Z/2)$ looks like a copy of $B(\Bbb Z/2)$ above the two fixed points and an arc's worth of copies of $E(\Bbb Z/2)$; collapsing that arc, this Borel construction is homotopy equivalent to $B(\Bbb Z/2) \vee B(\Bbb Z/2)$.

The relative Borel construction just says "collapse those two fibers that look like $B(\Bbb Z/2)$". So what you're left with is the suspension of $E(\Bbb Z/2)$. So $H^*_G(S^1, \pm 1;R)$ is a copy of the coefficient ring $R$ concentrated in degree zero.

This is consistent with the relative long exact sequence you mention: $(\pm 1) \times_{\Bbb Z/2} E(\Bbb Z/2)$ gives you $B(\Bbb Z/2) \sqcup B(\Bbb Z/2)$, and the map $$(\pm 1) \times_{\Bbb Z/2} E(\Bbb Z/2) \to S^1 \times_{\Bbb Z/2} E(\Bbb Z/2)$$ is just wedging the two copies of $B(\Bbb Z/2)$ together. In particular, the map $H^*_G(\pm 1) \to H^*_G(S^1)$ is an isomorphism in all degrees greater than 2, and is the map $R \oplus R \to R$ given by addition in degree zero; its cokernel is $H^0_G(S^1, \pm 1;R) = R$.

  • $\begingroup$ This is what I was looking for. Thank you! $\endgroup$ Nov 14, 2018 at 22:51

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