# why does the regular action of the structure group not imply triviality of a fibre system?

Let $P$ and $X$ be algebraic varieties, $\pi:P\to X$ a morphism and $G$ an algebraic group acting on $P$. Serre calls the triple $(G,P,X)$ a fibre system an proves that if it is locally isotrivial (i.e. trivial in the etale topology) then the action of $G$ must be sharply transitive on each fibre of $\pi$.

Now my question is: Conversely if we assume that the the action of $G$ is sharply transitive on each fibre why it does not imply that $P$ is trivial in the Zariski topology? (It certainly cannot be true, but I don't see why).

Let $\mathbb{G}_m=\operatorname{Spec}k[t,t^{-1}]$ be the multiplicative group. Consider the map $\pi:\mathbb{G}_m\rightarrow\mathbb{G}_m, t\mapsto t^2$.
The finite group $\mathbb{Z}/2\mathbb{Z}$ acts on the fiber by $t\mapsto -t$ and the action is sharply transitive. However the map is not locally trivial for the Zariski topology. In fact the preimage of any non empty open subset is connected.
It is locally trivial for the étale topology though. Indeed, $\pi$ is étale so you can consider it as an étale covering of $\mathbb{G}_m$. Then the pullback of $\pi$ by $\pi$ is isomorphic to $\mathbb{G}_m\coprod\mathbb{G}_m$ with $\mathbb{Z/2Z}$ permuting the two components and so it is trivial.
• Thanks for the answer @Roland. It helps to clarify. However I still feel a bit uneasy about the paragraph where you show although the action of $\mathbb Z/2\mathbb Z$ Is sharply transitive, the cover is not trivial. Can you elaborate a bit more on that please. – ugur efem Apr 28 '16 at 10:26
• @ugurefem Assume it is locally trivial for the Zariski topology, then there exists an open subset $U\subset\mathbb{G}_m$ whose preimage is isomorphic to $\mathbb{Z/2Z}\times U$. But this is not connected while every open subset of $\mathbb{G}_m$ is connected since $\mathbb{G}_m$ is irreducible. – Roland Apr 28 '16 at 13:55