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Let $U=\mathbb{P}^1-\{p_1, \ldots, p_n\}$ be a Zariski open subset of the projective line. Consider a rank $r$ local system of complex vector spaces $V$ on $U$ and assume that the monodromy representation

$$ \rho: \pi_1(U, t) \to GL(r, \mathbb{C}) $$

attached to $V$ is trivial (that is, $\rho(\gamma)$=identity matrix for each $\gamma$ in $\pi_1(U, t)$). I'm interested in two questions:

1) Is it true that $V$ is isomorphic to its dual $V^\vee$?

2) What are the cohomology groups $H^i(U, V)=0$ for $i=0, 1, 2$?

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    $\begingroup$ Is this an exercise? I mean, 1) is pretty obvious, since you assume that the action of the fundamental group on $V$ is trivial, then you're only looking at the underlying vector space, and any f.d. vector space is isomorphic to the linear dual. $\endgroup$
    – Fernando Muro
    Commented Jun 25, 2013 at 14:45
  • $\begingroup$ No, Fernando, it's not an exercise. Thanks for explaining 1). What about 2)? Is it true that $H^i(U, V)=0$ for $i=1$ and $2$? $\endgroup$
    – local
    Commented Jun 25, 2013 at 14:47
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    $\begingroup$ For 2: $U$ is affine and $V$ is a coherent $\mathcal{O}_V$-module, so all the higher cohomology groups vanish. $\endgroup$
    – Alicia Garcia-Raboso
    Commented Jun 25, 2013 at 15:04
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    $\begingroup$ Isn't V a locally constant sheaf of $\mathbb{C}$-modules, rather than a coherent module? To answer 2), as Fernando points out, triviality of the monodromy representation means that $V$ is globally constant, and hence $H^i(U,V)\cong H^i(U,\mathbb{C})\otimes_{\mathbb{C}}V$, and $H^i(U,\mathbb{C})$ is easy to calculate. $\endgroup$
    – ChrisLazda
    Commented Jun 25, 2013 at 15:14
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    $\begingroup$ I don't think so. If $n=0$, then $U$ is just $\mathbb{P}^1$, so the cohomology is just that of the sphere, and so we have $h^0=1$, $h^1=0$, $h^2=1$. But if $n\geq 1$ then $U$ is homotopic to the wedge product of $n-1$ spheres, and so we will have $h^0=1$, $h^1=n-1$ and $h^2=0$. $\endgroup$
    – ChrisLazda
    Commented Jun 25, 2013 at 16:37

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