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Let $k$ be a field with characteristic zero and $G$ be a (connected or not) linear algebraic group defined over $k$. We know that $G$ has a Levi decomposition i.e., $G=R_u(G)\rtimes L$, where $R_u(G)$ is the unipotent radical of $G$ and $L$ is a Levi factor of $G$ which is a reductive complement to the unipotent radical.

Q: What does this mean? $G(k)=R_u(G)(k)\rtimes L(k)$ or $G(\bar{k})=R_u(G)(\bar{k})\rtimes L(\bar{k})$? where $\bar{k}$ is the algebraic closure of $k$.

I guess it follows from the fact that the first Galois cohomology of a unipotent group over fields with characteristic zero is trivial. Because I have learned the following: Suppose $1\rightarrow A \rightarrow B \rightarrow C \rightarrow1$ is a sequence of morhisms of algebraic groups and that the induced sequence $1\rightarrow A(l)\rightarrow B(l)\rightarrow C(l)\rightarrow 1$ is exact for some Galois extension $l/k$ finite or infinite. Then we get an exact sequence $1\rightarrow A(k)\rightarrow B(k)\rightarrow C(k)\rightarrow H^1(l/k,A)\rightarrow H^1(l/k,B) \rightarrow H^1(l/k,C)$.

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  • $\begingroup$ What definition do you have for $H_1 \ltimes H_2$ for $k$-groups $H_i$ such that it isn't immediate that its set of $k$-points is $H_1(k) \ltimes H_2(k)$? Or put another way, when you write "$G = R_u(G) \ltimes L$" are you abusing notation and referring to some isomorphism over $\overline{k}$? Please clarify the precise meaning you have in mind for "Levi decomposition" (it does work directly over $k$, but did you know that?). One doesn't need to bring in any Galois cohomology for describing the $k$-points of a semi-direct product of algebraic groups. $\endgroup$
    – nfdc23
    Commented Feb 26, 2017 at 1:20
  • $\begingroup$ Is it really true that one has such a splitting even for a disconnected group? In that setting, what is the definition of $\mathrm R_{\mathrm u}(G)$? (In particular, is it normal in $G$, or only in $G^\circ$?) $\endgroup$
    – LSpice
    Commented Feb 26, 2017 at 1:57
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    $\begingroup$ @LSpice: Over alg. closed fields $R_u(G^0)$ is characteristic in $G^0$, so it is normal in $G$ and hence is the maximal normal unipotent smooth connected subgroup of $G$. Thus, both definitions of $R_u(G)$ coincide. Over a field $k$, the splitting exists even if $G$ is disconnected when ${\rm{char}}(k)=0$: see section 5.4 (and in particular Prop. 5.4.1) in the article Reductive group schemes in Vol. I of "Autour Des Schemas en Groupes", No. 42-43 of the series Panoramas et Syntheses published by the SMF (2014). Counterexamples if ${\rm{char}}(k)>0$ are in A.6.4 of Pseudo-reductive Groups. $\endgroup$
    – nfdc23
    Commented Feb 26, 2017 at 2:55
  • $\begingroup$ @nfdc23: This is my point. I don't really understand Levi decomposition for a linear algebraic group $G$ over a field $k$ with characteristic zero. $G(k)=R_u(G)(k)\rtimes L(k)$ or $G(\bar{k})=R_u(G)(\bar{k})\rtimes L(\bar{k})$? It is not clear to me. Sorry for the stupied question. I understand from your comment that $G(k)=R_u(G)(k)\rtimes L(k)$, isn't? $\endgroup$
    – m07kl
    Commented Feb 26, 2017 at 9:36
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    $\begingroup$ Even over alg. closed fields, an algebraic group is not just a group: the variety structure must be remembered. If $G$ is any linear algebraic group over a perfect field $k$ then the unipotent radical of $G_{\overline{k}}$ is Galois-stable and hence (by perfectness of $k$) descends to a $k$-subgroup $R_u(G)\subset G$. A Levi $k$-subgroup of $G$ by definition is a smooth closed $k$-subgroup $L\subset G$ such that $R_u(G)\rtimes L\to G$ is a $k$-isomorphism. If char$(k)=0$, Mostow showed $L$ exists and is unique up to $R_u(G)(k)$-conjugacy. See the Prop. 5.4.1 reference in my previous comment. $\endgroup$
    – nfdc23
    Commented Feb 26, 2017 at 15:12

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