Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What kind of examples of groups and representations should I keep in mind, which do not uniquely decompose into irreducible representations? I am mostly interested in characteristic zero representation, but I am also curiuous about positive characteristic.

What about the traceability of such representation? What algebraic structure give these issues? For which groups, we do not have such issues?

share|cite|improve this question
Think about how the plane can be specified by axes. Even if one demands a right angle ... With representations of finite groups, for example, the unique decomposition into subspaces is not into irrecdivcible – Mark Bennet Jul 27 '11 at 21:24
up vote 5 down vote accepted

You have to be careful about the difference between irreducibles and indecomposibles. The represenation $\mathbb{Z}\to GL_2$ defined by $1 \mapsto \left( \begin{array}{cc} 1& 1\\ 0&1\end{array}\right)$ is the basic example of a represenation that is reducible but doesn't decompose into irreducibles.

On the other hand if you're asking when things split up into indecomposibles, there is the Krull-Schmidt theorem which says that if a module satisfies both the ascending chain condition and the descending chain condition, then it is a direct sum of indecomposibles. So in particular this would include any finite dimensional representation $G\to GL_n(\mathbb C)$

For counterexamples, maybe you could look at this book, I just now found it by google searching 'krull-schmidt counterexample'... the first example they give is that for $R= \mathbb{Z}[\sqrt{-5}]$ you have $\langle 3,2+\sqrt{-5}\rangle \oplus \langle 3,2-\sqrt{-5}\rangle \cong R\oplus \langle 3\rangle$, so the decompositions aren't unique in this case in the sense that the summands on the LHS and RHS aren't isomorphic as $R$-modules. But this is a statement about rings rather than group representations, so it might not be quite what you're after...

share|cite|improve this answer
It is sort of an example of group representations. You run into something similar looking at integral (as in Z) representations of a cyclic group of order 5. While you mostly stick with finitely generated free abelian groups to build your modules over, as representations they are only finitely generated torsion-free. In particular, they can be projective without being free (as expected), but over rings like R you don't have Krull-Schmidt even for projectives. The failure of this is measured by the ideal class group and is an old part of algebraic number theory. – Jack Schmidt Jul 28 '11 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.