TITLE: Quantum homology of $(S^2 \times S^2,\omega_{FS}\oplus \omega_{FS})$ and Poincare duality
QUESTION [7 upvotes]: I am having some issues computing Poincare duality for the quantum homology $QH(M)$ when $(M,\omega)=(S^2 \times S^2,\omega_{FS}\oplus \omega_{FS})$. I am using the simple novikov ring $\Lambda$ consisting of polynomials in variables $t$ and $t^{-1}$. I know that a basis for $QH(M)$ (over $\Lambda$) is given by the elements 
$$ 
[pt], \ [M], \ A:=[\{pt\}\times S^2], \ B:=[S^2 \times \{pt\}].
$$
Does someone know (or know a reference) how to determine which classes are Poincare dual to each other in this scenario? I imagine that $[pt]$ would be Poincare dual to $[M]$, but I am not sure.

REPLY [3 votes]: A bit late for this one, but I'll still post the answer for future visitors.
Poincaré duality on the quantum homology is just the same as Poincaré duality on normal homology, see for example the famous PSS paper [1, Section 2]. This means that for an element $\alpha= \sum_{A\in \Gamma} \alpha_A e^A$ with $\alpha_A\in H_{k+2c_1(A)}(M)$ and $\Gamma$ the image of $\pi_2(M)$ under the Hurewicz homomorphism, the Poincaré dual is given by
$$
PD(\alpha)=\sum_{A\in\Gamma}PD(\alpha_A)e^A.
$$
To get to your case with polynomials on $t,t^{-1}$ you just have to choose a basis for $H_2(M)$.
So $PD$ on your basis is just $PD$ on normal homology, which is uniquely characterised by the homology classes.
[1] "Symplectic Floer-Donaldson theory and quantum cohomology"; Piunikhin, Sergey, Dietmar Salamon, and Matthias Schwarz.