Equivariant cohomology of three copies of $\mathbb{CP}^1$ joined at the fixed points.

Question

In Goresky, Kottwitz and MacPherson's classic paper, they give the following example of a space whose equivariant cohomology is not free as an $H^*(BS^1)$-module:

Let $K = S^1$ act on $\mathbb{CP}^1 \cong S^2$ by rotation with fixed points at the North and South poles. Let $X$ be three copies of $\mathbb{CP}^1$ joined at these fixed points so as to form a “ring”. Then $X$ is a projective algebraic variety but the equivariant cohomology $H^*_K(X)$ is not a free module, and the restriction map $\gamma$ (of (6.3.1)) fails to be an injection.

I want to compute the $S^1$-equivariant cohomology of $X$:

I would like to use the Mayer-Vietoris sequence. So I decompose the space $X$ into $A$ and $B$ as follows,

which gives me the following pieces,

The Mayer-Vietoris sequence then looks like: $$ \cdots \to H^*_{S^1}(X) \to H^*_{S^1}(A)\oplus H^*_{S^1}(B) \to H^*_{S^1}(A\cap B) \to \cdots $$ Clearly, both $A$ and $B$ equivariantly deformation retract to $S^2$ whose $S^1$-equivariant cohomology is given by pairs of polynomials which have the same constant coefficient (I only care about the $H^*(BS^1)$-module structure, not the product structure). And, $A\cap B$ is a disjoint union of a point and a circle (on which $S^1$ acts freely). So, $$ H^*_{S^1}(A) = \{(f, g) \in k[t]^2 : f(0) = g(0)\} $$ $$ H^*_{S^1}(B) = \{(h, i) \in k[t]^2 : h(0) = i(0)\} $$ $$ H^*_{S^1}(A \cap B) = H^*_{S^1}(\{\text{pt}\} \sqcup S^1) = k[t] \oplus k $$ (Note that I am working with coefficients in a field $k$.)

The inclusion induced map $H^*_{S^1}(A) \to H^*_{S^1}(A \cap B)$, is $(f, g)\mapsto (f, g(0))$. So the map in the Mayer-Vietoris sequence, $$ H^*_{S^1}(A)\oplus H^*_{S^1}(B) \to H^*_{S^1}(A\cap B) $$ is, $$ (f, g, h, i) \mapsto (f - h, g(0) - i(0)). $$ Clearly, this map is surjective, so the Mayer-Vietoris sequence splits into short exact sequences: $$0 \to H^*_{S^1}(X) \to H^*_{S^1}(A)\oplus H^*_{S^1}(B) \to H^*_{S^1}(A\cap B) \to 0 $$ So, $H^*_{S^1}(X)$ is the kernel of the map described above, which is, $$ \{((f, g), (h, i)) \in k[t]^2 \oplus k[t]^2 : f(0) = g(0), h(0) = i(0), f - h = 0, g(0) - i(0) = 0 \} $$ or equivalently $$ \{(f, g, i) \in k[t]^3 : f(0) = g(0) = i(0) \} $$ which does seem to be a free $H^*(BS^1) \cong k[t]$-module, and it indeed injects into the equivariant cohomology of the fixed point set which is $k[t]^3$.

Is there an error in my computation?

Answer 1

Actually the map $H^*_{S^1}(A) \oplus H^*_{S^1}(B) \to H^*_{S^1}(A\cap B)$ in the Mayer-Vietoris sequence is not surjective.

Recall, $H^*_{S^1}(A) \cong H^*_{S^1}(B) \cong \{(f, g) \in k[t]^2 : f(0) = g(0) \}$ and $H^*_{S^1}(A\cap B) \cong k[t] \oplus k$.

The map $j^* : H^*_{S^1}(A) \oplus H^*_{S^1}(B) \to H^*_{S^1}(A \cap B)$ induced by inclusions $A\cap B \hookrightarrow A$ and $A \cap B \hookrightarrow$, is given by $((f, g), (h, i)) \mapsto (f-h,g(0)-i(0))$. Now in the domain we have that $f(0) = g(0)$ and $h(0) = i(0)$, which forces $(f-h)(0) = g(0)-i(0)$. Thus the image of $j^*$ consists of only those pairs $(p, q) \in k[t] \oplus k$, satisfying $p(0) = q$ so the map is not surjective. In particular, in degree $0$ the map is, $k\oplus k \to k\oplus k$, $(a, b) \mapsto (a-b, a-b)$ and the cokernel is, $$ H^1_{S^1}(X) = \text{coker } j^0 \cong k. $$ Since $H^*(BS^1) = k[t]$ is evenly graded, any free $H^*(BS^1)$-module must also be evenly graded. Since $H^{\text{odd}}_{S^1}(X) \ne 0$ we conclude that $H^*_{S^1}(X)$ is not a free $H^*(BS^1)$-module.

In fact we can compute the $S^1$-equivariant cohomology. Notice that in degrees $\ge 2$, the map $j^*$ is indeed surjective. Explicitly, for $n \ge 1$, $j^{2n} : k^2 \oplus k^2 \to k$, $((a, b), (c, d)) \mapsto a - c$. Also, we have $H^{\text{odd}}_{S_1}(A\cap B) = 0$. So for we have the following short exact sequences, $$ 0 \to H^{2n}_{S^1}(X) \to H^{2n}_{S^1}(A) \oplus H^{2n}_{S_1}(B) \xrightarrow{j^{2n}} H^{2n}_{S^1}(A\cap B) \to 0 $$

for $n \ge 1$ and so $H^{2n}_{S^1}(X) = \ker j^{2n} \cong k^3$. Thus, $$ H^0_{S^1}(X) = k $$ $$ H^1_{S^1}(X) = k $$ $$ H^{2n}_{S^1}(X) = k^3 $$ $$ H^{2n+1}_{S^1}(X) = 0. $$.

So k[t]-module structure is given by $$ H^*_{S^1}(X) = k[t] \oplus k\langle u \rangle \oplus k[t][-2]^2 $$ where $u$ is of degree $1$, and $t\cdot u = 0$ (so $u$ is a torsion element).