Let $f:X \to Y$ be a smooth morphism between projective varieties. Suppose $Y$ is a homogeneous space. Under what additional condition on $f$, can we conclude that every fibers of $f$ are isomorphic?

## 1 Answer

For your particular case, there is a sufficient hypothesis that the morphism is étale locally a product. Let $k$ be a field. Let $G$ be a smooth $k$-group scheme. Let $$\Psi_X:G\times_{\text{Spec}(k)} X \to X\times_{\text{Spec}(k)} X, \ \ \Psi_X(g,x) = (g\cdot x,x), $$ $$\Psi_Y:G\times_{\text{Spec}(k)} Y \to Y\times_{\text{Spec}(k)} Y, \ \ \Psi_Y(g,y) = (g\cdot y,y)$$ be $k$-actions of $G$ on $X$, resp. $Y$. Assume that $f$ is $G$-equivariant, $f(g\cdot x) = g\cdot f(x)$. Finally, assume that the action $\Psi_Y$ is smooth and surjective, i.e., $Y$ is a $G$-homogeneous space whose stabilizer subgroup schemes are $k$-smooth. Then the morphism $f$ is étale locally a product.

It suffices to prove this after étale base change, thus assume that $Y$ has a $k$-point $y$. Then $\Psi_Y$ induces a smooth $k$-morphism, $$\psi_{Y,y}:G\to Y,\ \ \psi_{Y,y}(g) = g\cdot y.$$ Denote $X\times_{f,Y,y} \text{Spec}(k)$ by $X_y$. Then there is a commutative diagram. $$ \begin{array}{ccc} G\times_{\text{Spec}(k)} X_y & \xrightarrow{\text{pr}_1\circ \Psi_X} & Y \\ \ \ \ \ \ \ \downarrow{\text{pr}_G} & & \downarrow{f} \\ G & \xrightarrow{\psi_{Y,y}} & X \end{array}$$ It is not too hard to check that this is actually a Cartesian diagram. Thus, after the smooth base change $\psi_{Y,y}$, the morphism $f$ becomes a product. Of course there are étale local sections of $\psi_{Y,y}$. Thus, after étale base change of $Y$, the morphism $f$ becomes a product.

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