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Let $G$ be an affine algebraic group (i.e. a $k$-variety which is also a group and the group multiplication and inversion are morphisms of varieties). A character of $G$ is a morphism of algebraic groups $\chi:G\to k^\times$.

You may assume for the following that $k^\times$ acts algebraically on $G$, but I do not know if that is required at all.

It is well-known that I can always embed $G$ as a closed, algebraic subgroup of some $\mathrm{Gl}_n$. Now given any (nonzero) character $\chi$, can I embed $G$ in such a way that $\chi=\det_n|_G$, the restriction to $G$ of the determinant on $\mathrm{Gl}_n$? If no, can you characterize the characters that do satisfy this property? Because certainly, some characters of $G$ arise in this manner.

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If $\chi$ is such a character, and $\theta$ is any linear character, then you can show $\chi\theta$ is such a character by using the direct sum of the two representations. I think that finishes it: yes they all arise this way. –  Jack Schmidt Aug 16 '12 at 20:15

1 Answer 1

up vote 5 down vote accepted

Let $G$ be a closed subgroup of $\newcommand{\GL}{\operatorname{GL}}\GL_n$ and let $\chi : G \to k^\times$ be a character. Then consider the subgroup $$G_\chi = \left\{ ~ \begin{bmatrix} g & . & . \\ . & \det(g)^{-1} & . \\ . & . & \chi(g) \end{bmatrix} ~:~ g \in G ~\right\} \leq \GL_{n+2}$$ By definition of character, this is a closed subgroup of $\GL_{n+2}$ and clearly the determinant of the “$g$” matrix is now $\chi(g)$. $G_\chi \cong G$ by definition of homomorphism. Hence every (linear) character is the determinant of a faithful representation.

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That. Is sweet. And elementary. Thanks a bunch! =) –  Jesko Hüttenhain Aug 18 '12 at 0:45

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