I know that any finite-dimensional complex representation of a finite group $G$ is determined by its characters. This is immediate, in view of the complete reducibility of this category modules.

My question is, do we need complete reducibility when we work in a category of modules over complex finite-dimensional semisimple Lie algebras in order to objects are characterized by their characters?

Thank you.

  • 6
    $\begingroup$ Assume a module $M$ is determined up to isomorphism by its character $\chi$. Write $\chi = \chi_1 + \cdots + \chi_n$ as a sum of irreducibles. Let $M_i$ be a simple module affording $\chi_i$. The direct sum $M_1 \oplus \cdots \oplus M_n$ affords $\chi$ and by assumption is isomorphic to $M$. Hence $M$ is a direct sum of simple modules, so semisimple. Complete reducibility holds. $\endgroup$
    – Jay Taylor
    Dec 28, 2019 at 7:41
  • $\begingroup$ @JayTaylor please explain, why is it always possible to write a character (of a module over a Lie algebra) as a sum of characters of irreducibles? $\chi = \chi_1 + \cdots + \chi_n$? $\endgroup$
    – GA316
    Dec 28, 2019 at 8:02
  • 4
    $\begingroup$ Characters are additive in short exact sequences, so the character can't tell the difference between a direct sum and a nontrivial extension. In particular, the character of any module can be written as a sum of characters of simple modules (namely those of its Jordan-Hölder composition series). $\endgroup$ Dec 28, 2019 at 10:43
  • $\begingroup$ @BertramArnold Such series exists for infinite-dimensional modules as well? because I remember for arbitrary highest weight modules of Kac-Moody algebras such series doesn't exist. Kindly explain to me more. thanks. $\endgroup$
    – GA316
    Dec 28, 2019 at 13:29
  • $\begingroup$ For finite dimensional algebras over algebraically closed fields, simple modules are characterized up to isomorphism by their characters. Probably algebraically closed is not needed. $\endgroup$ Dec 28, 2019 at 15:37

1 Answer 1


Doc, I am not sure what your question is, but the answer is yes. Whatever definition of character you are using, any two extensions of $M$ by $N$ will have the same character. Thus, a non-trivial extension has the character as $M\oplus N$. Bingo: non-isomorphic modules will have the same characters...

  • $\begingroup$ Thanks. can you please tell me what is the meaning of extensions? $\endgroup$
    – GA316
    Jan 6, 2020 at 7:51
  • 2
    $\begingroup$ @GA316 In this context, I believe it means a short exact sequence of the form $0\to M\to M'\to N\to 0$ With a split short exact sequence, $M'\cong M\oplus N$, but if you do not have complete reducibility, then there will exist non-split short exact sequences. $\endgroup$
    – Aaron
    Jan 6, 2020 at 8:17
  • 1
    $\begingroup$ @Aaron Exactically, Aaron explained it well. A character is always a homomorphism from the Grothenideck group to some other group. Since $M^\prime$ and $M\oplus N$ give the same element of the Grothendieck group, they cannot be distinguished by a character... $\endgroup$
    – Bugs Bunny
    Jan 6, 2020 at 12:11
  • $\begingroup$ Thank you. Can you suggest some references regarding these? $\endgroup$
    – GA316
    Jan 6, 2020 at 18:12
  • $\begingroup$ What are "these"? $\endgroup$
    – Bugs Bunny
    Jan 8, 2020 at 9:49

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