# Form of an algebraic group

Jump to: navigation, search

2010 Mathematics Subject Classification: Primary: 14L [MSN][ZBL]

A form of an algebraic group $G$ defined over a field $k$ is an algebraic group $G'$ defined over $k$ and isomorphic to $G$ over some extension $L$ of $k$. In this case $G'$ is called an $L/k$-form of $G$. If $K_s$ is the separable closure of $k$ in a fixed algebraically closed ground field $K$ (a universal domain), then $k_s/k$-forms are simply called $k$-forms of $G$. Two $L/k$-forms of a group are said to be equivalent if they are isomorphic over $k$. The set of equivalence classes of $L/k$-forms of $G$ is denoted by $E(L/k,G)$ (in the case $L=k_s$ by $E(k,G)$) (see [Vo], [Ti], [Sp]).

Example. Let $k=\R$, $K=\C$. Then

$$G' = \Big\{ \begin{pmatrix} x & y\\ -y & x\end{pmatrix} : x^2+y^2 = 1 \Big\}$$ and

$$G = \{\def\diag{ {\rm diag}}\diag (x.y) : xy=1\}$$ are two subgroups of the general linear group $\def\GL{ {\rm GL}}\GL$ defined over $k$, and $G'$ is a $k$-form of $G$ (the isomorphism $\def\phi{\varphi} : G' \to G$, defined over $K$, is given by the formula

$$\phi \Big(\begin{pmatrix} x & y\\ -y & x\end{pmatrix}\Big) = \diag(x+iy,x-iy).$$ This $k$-form is not equivalent to $G$ (if one regards $G$ as a $k$-form of itself relative to the identity isomorphism $G\to G$). In this example, the set $E(k,G)$ consists of the two elements represented by the two $k$-forms above.

The problem of classifying forms of algebraic groups can be naturally reformulated in the language of Galois cohomology, [Se], [Vo]. Namely, suppose that $L/k$ is a Galois extension with Galois group $\def\G{\Gamma}\G_{L/k}$ (equipped with the Krull topology). The group $\G_{L/k}$ acts naturally on the group $\def\Aut{ {\rm Aut}}\Aut_L(G)$ of all $L$-automorphisms of $G$, and also on the set of all $L$-isomorphisms from $G'$ to $G$ (in coordinates, these actions reduce to applying the automorphisms in $\G_{L/k}$ to the coefficients of the rational functions defining the respective mappings). Let $\phi: G'\to G$ be some $L$-isomorphism, let $\def\s{\sigma}\s\in\G_{L/k}$ and let $\phi^\s$ be the image of $\phi$ under the action of $\s$. Then the mapping $\G_{L/k}\to\Aut_L G$, $\s \mapsto c_\s= \phi^\s\circ \phi^{-1}$, is a continuous $1$-cocycle of $\G_{L/k}$ with values in the discrete group $\Aut_L G$. When replacing $\phi$ by another $L$-isomorphism $G'\to G$, this cocycle changes to a cocycle in the same cohomology class. Thus arises a mapping $E(L/k, G)\to {\rm H}^1(\G_{L/k},\Aut_L G)$. The main importance of the cohomological interpretation of the forms of $G$ consists in the fact that this mapping is bijective. In the case when all automorphisms $c_\s$ are inner, $G'$ is called an inner form of $G$, and otherwise an outer form.

For connected reductive groups there is a thoroughly developed theory of forms, where relative versions of the structure theory of linear algebraic groups over an algebraically closed field are established: $k$-roots, the $k$-Weyl group, the Bruhat decomposition over $k$, etc. Here the role of maximal tori is played by maximal $k$-split tori, and that of Borel subgroups by minimal $k$-parabolic subgroups [Bo], [Hu], , [Ti]. This theory enables one to reduce the question of classifying forms to that of classifying anisotropic reductive groups over $k$ (see Anisotropic group; Anisotropic kernel). The question of classifying the latter depends essentially on the properties of the field $k$. If $k=\R$ and $K=\C$, then the characterization of forms of semi-simple algebraic groups is the same as that of real forms of complex semi-simple algebraic groups (see Complexification of a Lie group).

How to Cite This Entry:
Form of an algebraic group. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Form_of_an_algebraic_group&oldid=42615
This article was adapted from an original article by V.L. Popov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article