Let $G$ be a linear algebraic group over an algebraically closed field $\mathbb C$ of characteristic zero and $U$ its unipotent radical, then $H:=G/U$ is a reductive group. Assume that I have a one parameter subgroup $\lambda:\mathbb C^\times\to H$. I would like to lift this to a one parameter subgroup $\tilde\lambda:\mathbb C^\times\to G$ such that $\lambda = \pi\circ \tilde\lambda$, where $\pi:G\to H$ is the canonical projection. If $\mathbb C$ are the complex numbers, then this is always possible by the two theorems in this paper.

My question is: Is it true for any algebraically closed field? Do I even need the characteristic zero assumption? I am basically looking for references of such a lifting statement in the context of linear algebraic groups instead of lie groups.


1 Answer 1


This is true for quotients by smooth unipotent normal subgroups over any field whatsoever: if $f:G \rightarrow G'$ is a surjective homomorphism between group schemes of finite type over a field $k$ such that $\ker f$ is a unipotent smooth group (in particular, affine) then for any $k$-torus $T' \subset G'$ there exists a $k$-torus $T \subset f^{-1}(T')$ such that $f:T \rightarrow T'$ is an isomorphism. The main content of the proof is the structure of connected solvable groups over algebraically closed fields.

For a proof, we can replace $G'$ with $T'$ and replace $G$ with $f^{-1}(T')$ so that $G'=T'$ is a torus. Since $\ker f$ is smooth and affine over $k$, it follows by descent theory that $f$ is smooth and affine (as $G \times_{G'} G = G \times (\ker f)$). Thus, since $G'$ is now smooth and affine, so is $G$. We can even replace $G$ with $G^0$, so $G$ is a connected solvable linear algebraic group. If $T$ is a maximal $k$-torus in $G$ then $G_{\overline{k}} = T_{\overline{k}} \ltimes R_u(G_{\overline{k}})$ by the structure theorem for connected solvable groups over algebraically closed fields.

Under the quotient map $f:G \twoheadrightarrow G'=T'$ necessarily $R_u(G_{\overline{k}})$ has to be killed by $f_{\overline{k}}$. But $G_{\overline{k}}/R_u(G_{\overline{k}})$ is a torus (in fact naturally isomorphic to $T_{\overline{k}}$), so the image of the smooth unipotent $(\ker f)_{\overline{k}} = \ker(f_{\overline{k}})$ in this quotient of $G_{\overline{k}}$ is trivial. Hence, $\ker(f_{\overline{k}}) = R_u(G_{\overline{k}})$. Thus, $f_{\overline{k}}$ induces an isomorphism $T_{\overline{k}} \rightarrow T'_{\overline{k}}$, so likewise $f$ induces an isomorphism $T \rightarrow T'$. QED

The smoothness of the kernel is essential; e.g., if $\ker f$ is a non-smooth unipotent group scheme then the conclusion is false. The most famous class of examples are the so-called unipotent isogenies between connected semisimple groups, associated to non-degenerate quadratic forms in characteristic 2. More specifically, if $(V,q)$ is any non-degenerate quadratic space of dimension $2n+1 \ge 3$ over a field $k$ of characteristic 2 then the defect space $V^{\perp}$ for the associated symmetric bilinear form $B_q$ on $V$ is a line and $B_q$ induces a non-degenerate bilinear form $\overline{B}_q$ on $\overline{V} = V/V^{\perp}$ that is symplectic (since in characteristic 2), with ${\rm{SO}}(q) \rightarrow {\rm{Sp}}(\overline{B}_q)$ an isogeny from adjoint type B$_n$ to split simply connected type C$_n$ whose kernel is a unipotent group scheme (in fact a non-central but commutative direct product of copies of $\alpha_2$). If $q$ is not homothetic to a split form (e.g., if it is anisotropic) then ${\rm{SO}}(q)$ is $k$-anisotropic, so none of the non-trivial split maximal $k$-tori in the quotient can be lifted (not even up to isogeny).

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    $\begingroup$ Wonderful, I'll count your answer among my most cherished christmas presents. Cheers! $\endgroup$ Dec 24, 2014 at 21:20

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