Suppose $G$ is a connected reductive algebraic group over an algebraically closed field $k$, and suppose $L$ is a Levi subgroup (of some parabolic subgroup of $G$), is it always true that the principal fibration $G\rightarrow G/L$ is locally trivial in Zariski topology?

Note that this is true if $G=GL_n$ or $L$ is a maximal torus, as products of general linear groups are special algebraic groups (in the sense here), but how about the general case?

  • $\begingroup$ Well, I misunderstood, since some data are missing in the question. A Levi subgroup means a Levi subgroup of some a parabolic subgroup? $\endgroup$
    – YCor
    Feb 18, 2016 at 22:28
  • $\begingroup$ @ YCor Yes, I edit it. $\endgroup$
    – user148212
    Feb 18, 2016 at 22:30
  • $\begingroup$ Note that your quotient $G/L$ is always an affine variety, in any characteristic. $\endgroup$ Feb 18, 2016 at 23:15

1 Answer 1


Yes, this is also true over an arbitrary infinite field $k$. The technique to be used is exactly the same as the one which shows that the natural map $G(k) \rightarrow (G/P)(k)$ is surjective for any parabolic $k$-subgroup $P \subset G$ for any infinite field $k$.

Say $L$ is a Levi factor of a parabolic $k$-subgroup $P$ of $G$. There is always a compatible dynamic description of $P$ and $L$: a $k$-homomorphism $\lambda: {\rm{GL}}_1 \rightarrow G$ such that $L = Z_G(\lambda)$ is the centralizer of $\lambda$ and $P = P_G(\lambda) = Z_G(\lambda) \ltimes U_G(\lambda)$ represents the functor of points $g \in G(R)$ (for a $k$-algebra $R$) such that the orbit morphism $f_g:{\rm{GL}}_1 \rightarrow G_R$ over $R$ defined by $t \mapsto \lambda(t)g\lambda(t)^{-1}$ extends (necessarily uniquely) to $\mathbf{A}^1_R \rightarrow G_R$. Here, $U_G(\lambda)$ is the subgroup of points $g \in P_G(\lambda)$ such that $f_g(0)=1$.

The multiplication map $U_G(\lambda^{-1}) \times P_G(\lambda) \rightarrow G$ is an open immersion (since it is an etale monomorphism, for example), so $U_G(\lambda^{-1}) \times U_G(\lambda)$ is a locally closed subvariety of $G$ mapping isomorphically onto a dense open subvariety $\Omega \subset G/L$. Since $G(k)$ is Zariski-dense in $G$ (this is where we use that $k$ is infinite), the image of $G(k)$ in $G/L$ is Zariski-dense, so the image of $G(k)$ in $(G/L)_{\overline{k}}$ is Zariski-dense. Hence, the translates of $\Omega$ by the left-action of $G(k)$ on $G/L$ constitute an open cover of $G/L$ (as we may check on $\overline{k}$-points!) over which there are sections. That establishes the Zariski-local triviality of $G \rightarrow G/L$.


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