Poincaré duality and quantum (co)homology for $S^2 \times S^2$

Question

Consider the symplectic manifold $(M,\omega):=(S^2 \times S^2, \omega_{FS}\oplus \omega_{FS})$. The homology $H_*(M;\mathbb{C})$ has as a basis the 4 non-trivial classes: $[pt],[M],A$ and $B$, where $$ A=[pt \times S^2] \quad \text{and}\quad B=[S^2 \times pt]. $$ Denoting by $\mathcal{F}$ the field of Laurent polynomials $\mathcal{F}:=\mathbb{C}[[s]$ in a formal variable $s$, the quantum homology $QH(M):=H(M;\mathbb{C})\otimes_{\mathbb{C}}\mathcal{F}$ is equipped with the quantum product $\ast$. An element of $\mathcal{F}$ is written $\sum_{j\in \mathbb{Z}}z_js^j$ and $z_j=0$ for all $j$ large enough. I know that $$ A\ast B=[pt] \quad \text{and} \quad A\ast A=B\ast B=[M]s^2. $$ I would like to compute Poincaré duality in this picture, but for some reason I keep getting it wrong. I would be very grateful if someone could explain to me how it works. What I would like to know is that are the dual classes to each of the classes $[pt],[M],A$ and $B$?

Answer 1

To my understanding, the quantum homology and the quantum cohomology are basically the same thing. So the Poincare duals of $$[pt] \in QH_0(M), \quad [M] \in QH_4(M), \quad A \in QH_2(M), \quad B \in QH_2(M)$$ are just $$PD[pt] \in QH^4(M), \quad PD[M] = 1 \in QH^0(M), \quad PD(A) \in QH^2(M), \quad PD(B) \in QH^2(M)$$ where $PD$ denotes the Poincare dual for the ordinary homology class.

By the way, the structure of the quantum homology highly depends on choice of the coefficient ring $\mathcal{F}$. According to your expression, you seem to take $\mathcal{F} = \mathbb{C}[s]$ with $s = e^{-A/2} = e^{-B/2}$. Since $M$ is monotone, we don't need to worry about infinite sums. The variable $s$ has degree $-2$, so if $C$ is a homology class of degree $c$, the Poincare dual of $C \otimes s^k \in QH_{c -2k}(M)$ is $PD(C) \otimes s^{-k} \in QH^{(4-c) + 2k}(M)$.

I recommend the book $J$-holomorphic Curves and Symplectic Topology by McDuff and Salamon. In the 2nd edition, Example 11.1.13 explains exactly the same example and the Poincare duality is explained in Remark 11.1.20.