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Let $G$ be a group, $K$ an algebraically closed field of characteristic zero and $\rho_1,\rho_2:G\to \mathrm{GL}_n(K)$ be two semi-simple representations. What I would like to be able to determine is if $\rho_1$ is a subquotient of $\rho_2$. Since the character determines the isomorphism-classes of representations uniquely, and being a subquotient is independent of the choice of isomorphism-class, I would like to know if I can determine/test "being a subquotient" via the character?

That is to say, is there a partial ordering $\leq$ on functions $G\to K$ such that $$\rho_1\leq \rho_2 \iff \chi_{\rho_1}\leq \chi_{\rho_2}.$$

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  • $\begingroup$ There is a reductive algebraic group scheme $H$ over $K$ and a homomorphism $G\to H$, such that every semi-simple representation of $G$ factors through $H$. Your question has a positive answer for representations of $H$, so theoretically there is such a partial order. $\endgroup$
    – Kapil
    Feb 6, 2022 at 4:20
  • $\begingroup$ @Kapil How can we read of "being a subquotient" of the character for reductive algebraic groups? Also, do you happen to know a reference for the existence of $H$? $\endgroup$
    – curious math guy
    Feb 6, 2022 at 4:40
  • $\begingroup$ The standard representation theory for reductive groups (see e.g. the book of Springer in Linear Algebraic Groups) should give the answer to the first question. About the existence of $H$, the basic result is that if $\rho:G\to\mathrm{GL}_n$ is a semi-simple representation, then the Zariski closure of the image is a reductive group. Now one has to take a projective limit of such images to get $H$. $\endgroup$
    – Kapil
    Feb 6, 2022 at 9:53
  • $\begingroup$ If G is very large (e.g. an uncountable free group), then H constructed as this projective limit will be too large to be of finite type. $\endgroup$
    – Peter McNamara
    Feb 6, 2022 at 10:30

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