order to have a diagram like (\[eq:diag1\]) we need $k>2\dim(\Sigma_X)+1$, but this is the only difference with the previous case.

Alexander polynomial and line arrangements
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By the works of Milnor [@Mi] and Lê [@Le] we know in particular that if $f\in{\mathbb{C}}[x_0,\ldots,x_n]$ is a homogeneous polynomial then the map $f:{\mathbb{C}}^{n+1}\setminus f^{-1}(0)\rightarrow{\mathbb{C}}^*$ is a smooth locally trivial fibration; its generic fibre, usually denoted by $F$, is called *Milnor fibre*. To $F$ we can associate the geometric monodromy operator $h:F\rightarrow F$ and the induced algebraic monodromy operators $T_i:H^i(F,{\mathbb{C}})\rightarrow H^i(F,{\mathbb{C}})$.

Let $C=V(f)\subset{\mathbb{P}}^2$ be a reduced curve. The *Alexander polynomial* of $C$ is the characteristic polynomial of $T_1$, and is denoted by $\Delta_C$.

If $f$ has degree $d$ then $h$ is given by $\underline{x}\mapsto\eta_d\cdot\underline{x}$, where $\zeta_d$ is a primitive $d$-th root of unity; hence both $h$ and $T$ have order $d$, so $T$ is diagonalisable with roots of unity of order $d$ as eigenvalues. Moreover, the Milnor fibre of $C$ is a $d$-fold cover of $U:={\mathbb{P}}^2\setminus C$ and the geometric monodromy $h$ is a generator of the group of deck transformations of $F$; this implies that

$$\Delta_C(t)=(t-1)^{r-1}\prod_{1<k|d}\Phi_k(t)=(t-1)^{r-1}q(t).$$

where $r$ is the number of irreducible components of $C$. We call $q(t)$ the non-trivial part of 