Relative equivariant cohomology

Question

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

Best,

AB

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 $H^0_G(S^1, \pm 1;R) = R$.