What is the $S^1$-equivariant cup product on $S^2$? Consider the sphere $S^2 = \mathbb{CP}^1$ with the $S^1 = \{ \tau \in \mathbb{C} \mid |\tau| = 1 \}$ action given by
$$ \tau \cdot [z_1, z_2] = [\tau ^ k \cdot z_1, z_2] $$
The corresponding $S^1$-equivariant cohomology is $H_{S^1}^\ast(\mathbb{P}^1) = \mathbb{Z}[u] \cdot 1 \oplus \mathbb{Z}[u]\cdot y$, where $1$ and $y$ correspond to the generators of the ordinary cohomology of $\mathbb{P}^1$ in degrees 0 and 2 respectively, and $\mathbb{Z}[u] = H^\ast (\mathbb{CP}^\infty)$ is the $S^1$-equivariant cohomology of a point, with $u$ in degree 2.
My question is, "what is the equivariant cup product in this case?"
My progress: since the cup product will make $H_{S^1}^\ast(\mathbb{P}^1)$ a unital $\mathbb{Z}[u]$-algebra, the product is determined by the value of $y \smile y$, which for degree reasons will be a linear combination of $u \cdot y$ and $u^2 \cdot 1$. Using Morse theory, with the $S^1$-invariant Morse function $f : [z_1, z_2] \mapsto |z_1|^2$, we can deduce that the $u^2 \cdot 1$ component vanishes. I expect the coefficient of the $u \cdot y$ component to be nonzero (when $k \ne 0$) and to depend on $k$. I would be grateful for any assistance.
 A: The coefficient is $k$, so $y \smile y = kuy$.
A result on page 536 of [1] states the $k=1$ case, and the proof provided in the references of [1] uses Chern character theory.
From the $k=1$ case, we can deduce all further cases as follows.
Given a map $f : G \to H$, there is a natural map $f_\ast$ from $G$-equivariant cohomology to $H$-equivariant cohomology of any space $X$ with a $H$-action, where the $G$ action on $X$ is induced by the map $f$.
We will use the map $f: S^1 \to S^1 / (\mathbb{Z}/k) \cong S^1$, and look at the induced map $f_\ast$.
We are able to calculate the map for the action of rotation of a sphere, and we can use this to deduce what the cup product will be on the codomain of $f_\ast$.
First, $f_\ast(y) = y$.
This follows becaues the induced map $f_\ast$ will be an isomorphism on the non-equivariant part of the equivariant cohomologies (which is found by setting $u=0$), as can be seen by looking at the fibre of the bundle 
$$ES^1 \times_{S^1} X \to ES^1 / S^1 = \mathbb{CP}^\infty,$$
which will be isomorphic to $X$.
Next, $f_\ast(u) = ku$. We set $X$ equal to a point, since by the functoriality of $f_\ast$ with respect to the space $X$, this case is sufficient to deduce the result for all $X$.
On the fundamental groups, $f_\ast : \pi_1(S^1) \to \pi_1(S^1)$ is multiplication by $k$. The map $f$ yields a natural map between the long exact sequences of homotopy groups associated to the principal bundle $S^1 \to ES^1 \to BS^1$. Since $ES^1$ has trivial homotopy groups, this long exact sequence has isomorphisms $\pi_2(BS^1) \to \pi_1(S^1)$. Thus the naturality of the map between the long exact sequences and our knowledge of the maps behaviour on the fibres force the induced map $f_\ast : \pi_2(BS^1) \to \pi_2(BS^1)$ to also be multiplication by $k$. Since, for $BS^1$, the Hurwitz homomorphism $\pi_2 \to H_2$ is an isomorphism, we deduce that the corresponding map in degree 2 on both homology and cohomology is multiplication by $k$, as desired.
Finally, the cup product is calculated using the fact that we know that the map $f_\ast$ on equivariant cohomology is a ring homomorphism which takes $y \mapsto y$ and $u \mapsto ku$. In particular,
$$y \smile_k y = f_\ast(y) \smile_k f_\ast(y) = f_\ast(y \smile y) = f_\ast(uy) = kuy.$$
[1] Hori, Kentaro; Katz, Sheldon; Klemm, Albrecht; Pandharipande, Rahul; Thomas, Richard; Vafa, Cumrun; Vakil, Ravi; Zaslow, Eric, Mirror symmetry, Clay Mathematics Monographs 1. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-2955-6/hbk). xx, 929 p. (2003). ZBL1044.14018.
