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Let $X$ be a smooth projective variety over $\mathbb{C}$. Let $\rho: \pi_1(X,x)\rightarrow Gl(n,\mathbb{C})$ be a semisimple representation of fundamental group of $X$. The monodromy group $M(\rho, x)$ of the representation $\rho$ is defined to be the zariski closure of the image $\rho(\pi_1(X,x))$ in $Gl(n,\mathbb{C})$.

Q: Why is the above monodromy group reductive?

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    $\begingroup$ You don't show it's semisimple first, but rather use the Lie-Kolchin theorem. Stability of a subspace under $\rho(\pi_1)$ is the same as under $M$ (see this using a basis beginning with one of the subspace), so semisimplicity of $\rho$ implies each $M$-stable subspace has an $M$-stable complement. For any normal Zariski closed subgroup $N \subset M$, an $N$-stable subspace is also $M$-stable by normality and so admits an $N$-stable (even $M$-stable) complement; i.e., $N$ acts semisimply. By Lie-Kolchin, if $N$ is unipotent this forces $N=1$. Using $N=\mathscr{R}_u(M)$ implies $M$ is reductive. $\endgroup$
    – nfdc23
    Jun 26 '17 at 21:49
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    $\begingroup$ In fact it is impossible to show semisimplicity in this situation. Consider the case $n=1$. Then as long as the image of $\rho$ is infinite, its Zariski closure is $\mathbb G_m$, which is not semisimple. $\endgroup$
    – Will Sawin
    Jun 26 '17 at 21:57
  • $\begingroup$ @WillSawin Thank you for pointing out the trivial mistake. I have modifed it. $\endgroup$
    – Feng Hao
    Jun 26 '17 at 22:01
  • $\begingroup$ @nfdc23 It is crystal clear. Thank you $\endgroup$
    – Feng Hao
    Jun 26 '17 at 22:08
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    $\begingroup$ It's essentially the same argument, nothing to do with $\pi_1$'s: the image of $\rho$ has the same stable subspaces as its Zariski closure, and being semisimple is a property of the collection of stable subspaces (i.e., each has another that is a linear complement), so the substance in the converse is that a (possibly disconnected!) reductive linear algebraic group $G$ in char. 0 has all algebraic representations completely reducible. Tori are easy, the derived group of $G^0$ goes via semisimple Lie algebras in char. 0 (real content!), so $G^0$ is OK, and then average (again: char. 0). $\endgroup$
    – nfdc23
    Jun 27 '17 at 3:41

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