Let $X\rightarrow B$ be a family of Kaehler manifolds with possibly singular fibers. Let $G$ be the monodromy group on $H^n(X_b,\mathbb{Z})$, where $n=\dim X_b$ with the smooth fiber $X_b$ over some $b \in B$. We say that an element $g \in G$ is neat if none of the eigenvalue of $g$ is a non-trivial root of unity. We say that $G$ is neat if all elements in $G$ are neat.

Here is my question: Is it true that if each local monodromy is neat then $G$ is neat? In other words, is it true that $G$ is neat if it is generated by neat elements?

I think my question is false if one removes the geometric picture that the group $G$ is the monodromy group of a family of Kahelr manifolds. We may use the fact that $G \subset Sp(k,\mathbb{Z})$ if $n$ is odd, for example.

**Edit**
As Jason pointed out below, I need to assume that the base space $B$ is simply connected and the global monodromy group is generated by the local monodromies.

differentquestions: the global monodromy group is not necessarily generated by local monodromies. There are geometric hypotheses that insure that the global monodromy group is generated by local monodromies, e.g., for the family of hyperplane sections of a fixed projective manifold obtained from aLefschetz pencilof hypereplane sections. Did you want to impose such a hypothesis? – Jason Starr Dec 26 '13 at 15:20