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](https://en.wikipedia.org/wiki/Special_group_(algebraic_group_theory))), but how about the general case?



Quotient of a reductive group by a Levi subgroup and locally triviality
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 $G \rightarrow G/L$.


