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.

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.

Let $X\rightarrow B$ be a family of Kahelr 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.

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.