character $\rho=(1, \dots, 1)$ gives rise to the usual cohomology with (untwisted) complex coefficients, that is, $H^*(M_{{{\mathcal A}}}, \;_{\rho}{\mathbb{C}}) = H^*(M_{{{\mathcal A}}}, {\mathbb{C}})$, for $\rho =1$.

\[points\] Let ${{\mathcal A}}$ be an arrangement of $n$ points in ${\mathbb{C}}$. The complement has the homotopy type of a bouquet of $n$ circles, a $K(\pi,1)$ space (see Definition \[high\]), with fundamental group the free group on $n$ generators. Then $H^*(M_{{{\mathcal A}}}, \;_{\rho}{\mathbb{C}})=0$, for $n=1$ and $\rho \ne 1$.

[*Having Theorem in mind, we may ask: is there an analog of the Orlik-Solomon algebra for twisted coefficients, in other words, is the twisted cohomology combinatorially determined, at least for rank $1$ complex local systems? This appears to be the major open problem in the topology of arrangements.*]{} See Section \[sect:VR\] for more on this.

Homotopy groups {#sect:pi1}
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We discuss first the fundamental group of the complement of an arrangement ${{\mathcal A}}$. As before, we are interested in combinatorial aspects of $\pi_1(M_{{{\mathcal A}}})$.

For instance, the abelianity of the fundamental group of the complement is equivalent to its nilpotency, and equivalent to the property of the lattice of being in general position in codimension $2$. See [@Mac] for more equivalences.

\[fiber:pi1\]

1.  \[Ftype\] The 