Let $C_\ast$ be any chain complex of $R$-modules. Then for any $k\in\mathbb{Z}$ we obtain a $R$-bilinear map $$\langle-,-\rangle:H^k\!C_\ast\times H_kC_\ast\longrightarrow R, ~~~~~~~(\alpha,a)\longmapsto \alpha(a),$$ where $H_kC_\ast=\frac{Ker\partial}{Im\partial}$ and $H^k\!C_\ast=\frac{Ker\delta}{Im\delta}$ and $\delta:Hom_R(C_\ast,R)\to Hom_R(C_{\ast+1},R)$, correct?
If $H^k\!C_\ast$ and $H_kC_\ast$ are free of rank $m$ and $n$ with basis $e_i$ and $e_j$, then this bilinear form is uniquely given by a matrix $M$, such that $\langle x,y\rangle\mapsto x^t M y$, where $M_{i,j}=\langle e_i,e_j\rangle$, correct?
If $X$ is a Seifert surface with genus $g$ of an $n$-component link $L\subseteq S^3$, which looks like 
with the bands possibly tiwisted, linked and knotted, what is the matrix of the bilinear form $$H^1\!X\!\times\!H_1X\cong \mathbb{Z}^{2g+n-1}\!\times\mathbb{Z}^{2g+n-1} \longrightarrow\mathbb{Z}, ~~~~~~~(\alpha,a)\longmapsto \alpha(a),?$$
Is this just the identity matrix or is it $[lk(L_i,L_j)]_{i,j}$? Do we need to take $H_1\!(S^3\!\setminus\!X)\!\times\!H_1X\longrightarrow\mathbb{Z}$?
See Lickorish Knot theory p.51 Proposition 6.3.