Let $L_n(p)$ be the $2n+1$ dimensional Lens space $$ S^{2n+1}/\mathbb{Z}_p $$ where the action is given as $z_i\rightarrow e^{\frac{2\pi}{p}}z_i$, $i=1,...,n+1$, with $z_i$ the coordinates of $\mathbb{C}^{n+1}$ such that $S^{2n+1}$ is $|z_1|^2+...+|z_{n+1}|^2=1$. For $k\neq 0,2n+1$ the homology groups with coefficients in a commutative ring $R$ are $$ H_k(L_n(p),R)=\left\{\begin{array}{cc} R/pR & \text{if $k$ is odd}\\ T_p(R) & \text{if $k$ is even } \end{array}\right. $$ where $T_p(R)\subset R$ is the $p-$torsion subgroup: $$ T_p(R)=\left\{x\in R \ | \ px=0\right\} \ . $$ Since $R$ is an abelian group, we can consider $R^{\vee}$, its Pontryagin. Fix an extension $\Gamma$ of $R$ by $R^{\vee}$: $$ 1\rightarrow R^{\vee}\rightarrow \Gamma \rightarrow R\rightarrow 1 \ . $$ This induces a long exact sequence of homology and cohomology groups, with connecting homomorphisms given by the Bockstein maps. For instance in cohomology $\beta :H^k(L_n(p),R)\rightarrow H^{k+1}(L_n(p),R^{\vee})$.
Given $\Sigma \in H_{2n}(L_n(p),R)$ we consider its Poincare' dual cocycle $A=PD(\Sigma)\in H^1(L_n(p),R)$, and its image under Bockstein $\beta(A)\in H^2(L_n(p),R^{\vee})$. Since $R^{\vee}$ is also a ring, the cup product associated with the product $R^{\vee}\times R^{\vee}\rightarrow R^{\vee}$ allows to construct a class $\beta(A)^n\in H^{2n}(L_n(p),R^{\vee})$, and the natural pairing $R\times R^{\vee}\rightarrow \mathbb{R}/\mathbb{Z}$ allows the construction of the self-linking invariant: $$ lk(\Sigma)=\int _{L_n(p)}A\cup \beta(A)^n \in \mathbb{R}/\mathbb{Z} $$
Preliminary question: how this self-linking invariant is related with the more standard linking form which only uses the sequence $\mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}$?
Let us first fix $n=1$, $p=2$ so that $L_1(p)=\mathbb{RP}^3$, and $R=\mathbb{Z}_2$. Since $T_2(\mathbb{Z}_2)=\mathbb{Z}_2$ there is precisely one 2-cycle $\Sigma \in H_2(L_1(2),\mathbb{Z}_2)$, and one 1-cycle $\gamma\in H_1(L_1(2),\mathbb{Z}_2)$. How can I compute $$ lk(\Sigma)=\int _{L_1(2)}A\cup \beta(A) $$ explicitely? Moreover I have the intuition that $\beta(A)$ should be the Poincare' dual of $\gamma$, but I don't know how to prove this.
For general $n$ and $p$ fix $R=\mathbb{Z}_p$, so that $R/pR=T_p(R)=\mathbb{Z}_p$, and again choose one generator $\Sigma $ of $H_{2n}(L_n(p),\mathbb{Z}_p)$. How can I compute $lk(\Sigma)$? By extending the intuition before I also guess that, fixing a generator $\gamma$ of $H_1(L_n(p),\mathbb{Z}_p)$, $\beta(A)$ is the Poincare' dual of $\beta(A)^n$. Is this correct?