| Russian Mathematical Surveys |
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This article is cited in 1 scientific paper (total in 1 paper) Brief communications Picard group of a connected affine algebraic group V. L. Popovab a Steklov Mathematical Institute of Russian Academy of Sciences, Moscow b National Research University Higher School of Economics, Moscow
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     All algebraic varieties considered below are defined over a basic algebraically closed field $k$. We follow the point of view on algebraic groups accepted in [1] and use the following notation. If $S$ is a connected semisimple algebraic group, then $\widehat S$ is its universal cover, and $\pi(S)$ is the kernel of the canonical isogeny $\widehat S\to S$. If $G$ is a connected affine algebraic group and $H$ is its closed subgroup, then $\varepsilon_{G, H}\colon \operatorname{Hom}_{\rm alg}(H,\mathbb G_m)\to \operatorname{Pic}(G/H)$ is the homomorphism that maps each character $\chi\colon H\to \mathbb G_m$ to the class of the one-dimensional homogeneous vector bundle over $G/H$ determined by $\chi$ (see [2], Theorem 4). If $\varphi\colon X\to Y$ is a morphism of smooth irreducible algebraic varieties, then $\varphi^*\colon {\rm Pic}(Y)\to {\rm Pic}(X)$ is the homomorphism of Picard groups associated with $\varphi$ (see [3], Chap. III, § 1.2). Recall that the derived subgroup of a connected reductive algebraic group is connected and semisimple (see [1], § § I.2.2 and II.14.2). The purpose of this note is to prove the following theorem. Theorem. Let $G$ be a connected affine algebraic group, let ${\mathcal R}_u(G)$ be its unipotent radical, let $\varrho\colon G\to G/{\mathcal R}_u(G)$ be the canonical homomorphism, let $S$ be the derived subgroup of the connected reductive group $G/{\mathcal R}_u(G)$, and let $\iota\colon S\hookrightarrow G/{\mathscr R}_u(G)$ be the identity embedding. Then the following canonical homomorphisms are isomorphisms: Corollary. The group $\operatorname{Pic}(G)$ is canonically isomorphic to the group $\operatorname{Hom}_{\rm alg}(\pi(S),\mathbb G_m)$ and is non-canonically isomorphic to the group $\pi(S)$. Example. Let $G=\operatorname{GL}_n$. Then the group ${\mathcal R}_u(G)$ is trivial, and the derived group of the group $G$ is the semisimple group $\operatorname{SL}_n$. The latter is simply connected, so the group $\pi(\operatorname{SL}_n)$ is trivial. Therefore, by the theorem, the group $\operatorname{Pic}(G)$ is trivial. This agrees with the fact that the group variety of the group $\operatorname{GL}_n$ is isomorphic to an open subset of $\mathbb A^{n^2}$. The following lemma is used in the proof of the theorem. Lemma. Let $X$ be an irreducible smooth algebraic variety, let $U$ be a nonempty open subset of $\mathbb A^{d}$, and let $u_0$ be a point of $U$. Then for the morphisms the homomorphisms $\alpha^*$ and $\beta^*$ are mutually inverse isomorphisms. Proof of the lemma. Consider the morphisms
$$
\begin{equation}
\gamma\colon X\times \mathbb A^d \to X,\ \ (x,a)\mapsto x,\quad\text{and}\quad \delta\colon X\times U \to X\times\mathbb A^d,\ \ (x, u)\mapsto (x, u).
\end{equation}
\tag{2}
$$
It follows from (1) and (2) that the following equalities hold:
$$
\begin{equation}
\begin{aligned} \, \gamma\circ\delta\circ\beta=\operatorname{id}_X\!\!\quad\text{and}\quad \alpha\circ\beta=\operatorname{id}_X\!. \end{aligned}
\end{equation}
\tag{3}
$$
As is known, $\gamma^*$ is an isomorphism (see [4], Chap. II, Proposition 6.6 and its proof) and $\delta^*$ is a surjection (see [4], Chap. II, Proposition 6.5, (a)). From this and the left-hand equality in (3) it follows that $\beta^*$ is an isomorphism. In view of the right-hand equality in (3), this shows that $\alpha^*$ also is an isomorphism, and $\alpha^*$ and $\beta^*$ are mutually inverse. $\Box$ Proof of the theorem and corollary. Since the connected affine algebraic group ${\mathcal R}_u(G)$ is unipotent, it follows from [5], Propositions 1 and 2, that
(a) the group variety of ${\mathcal R}_u(G)$ is isomorphic to an affine space; (b) the following commutative diagram exists: where $\tau$ is an isomorphism of group varieties (but, generally speaking, not of groups) and $\upsilon$ is the natural projection onto the second factor.In view of the lemma it follows from (a) and (4) that $\varrho^*$ is an isomorphism. According to [6], Theorem 1, the group $G/{\mathcal R}_u(G)$ contains a torus $Z$ such that the mapping
$$
\begin{equation*}
\mu\colon S\times Z\to G/{\mathcal R}_u(G),\quad (s,z)\mapsto sz
\end{equation*}
\notag
$$
is an isomorphism of group varieties (but not of groups in general). Consider the commutative diagram in which $\nu\colon S\to S\times Z$, $s\mapsto (s,e)$, where $e$ is the identity element. Since the group variety of the torus $Z$ is isomorphic to an open subset of the affine space, (5) and the lemma imply that $\iota^*$ is an isomorphism.
In view of the semisimplicity of the group $\widehat S$ the group $\operatorname{Hom}_{\rm alg}(\widehat S,\mathbb G_m)$ is trivial, and since the group $\widehat S$ is simply connected, the group $\operatorname{Pic}(\widehat S)$ is trivial (see [2], Proposition 1). According to [2], Theorem 4, it follows from this that $\varepsilon_{\widehat{S}, \pi(S)}$ is an isomorphism. This completes the proof of the theorem. The first part of the corollary follows directly from the theorem, while the second part follows from the fact that $\pi (S)$ is a finite abelian group. $\Box$ Remark. The above theorem corrects Theorem 6 in [2]. The latter asserts that the group $\operatorname{Pic}(G)$ is isomorphic to $\pi\bigl(G/{\mathcal R}(G)\bigr)$, where ${\mathcal R}(G)$ is the solvable radical of $G$. If the group extension $1\to {\mathcal R}(G)\to G\to G/{\mathcal R}(G)\to 1$ splits, then the group $G/{\mathcal R}(G)$ is isomorphic to the derived group of $G/{\mathcal R}_u(G)$, and so the formulated assertion is true in view of the above theorem. But in general this is not the case, as the example above shows: in it the group $G/{\mathcal R}(G)$ is isomorphic to $\operatorname{PGL}_n$, and $\pi(\operatorname{PGL}_n)$ is isomorphic to the group of all $n$th roots of $1$ in the field $k$. This latter group is non-trivial if $n$ is not a power of the characteristic of the field $k$ (however, the group $\operatorname{Pic}(G)$ is trivial for every $n$). I am grateful to Shuai Wang for bringing this example to my attention; this led to writing of this note. I am also indebted to S. O. Gorchinskiy, whose comments led to the above proof of the lemma and the emphasis on the canonical nature of the construction (the original proof of Lemma in the preprint [7] was more geometric).
Citation in format AMSBIB
\Bibitem{Pop23} |
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