MathOverflow is a question and answer site for professional mathematicians. 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

Let $G$ be a Lie group, and $(\pi,V)$ is a continuous representation of $G$ which has finite composition series. A question I have which might be somehow naive is that: for any irreducible representation $(\sigma,W)$ of $G$, is it true that $(\sigma,W)$ occurs as one composition factor if and only if the set $Hom_G(V,W)$ is nonzero?

I have no idea how difficult or how easy this question might be, and any reference or answer is appreciated.

Edit: Thanks a lot for all of your answers, comments and examples. Now if $G$ is real reductive, $(\pi,V)$ is smooth admissible. Is there a way to determine all of the composition factors of $V$?

share|cite|improve this question
I'm not too familiar with the etiquette on MO myself, but since your original question was already answered, you might want to make a separate question about real reductive groups. – Kimball Mar 21 '11 at 14:48

No. This holds only if $V$ is semisimple. Consider the case when $V$ has two composition factors. This means that one is an invariant subspace and the other is the quotient of $V$ by this invariant subspace. If there is also an invariant subspace isomorphic to the quotient then $V$ is the direct sum of these two representations and so $V$ is decomposable.

share|cite|improve this answer

Bruce's answer is perfectly satisfactory, but you might want to see an explicit example. Let $B$ be the group of all upper triangular matrices in $GL_2({\mathbb R})$ and let $\pi:B\to GL_2({\mathbb R})$ be the inclusion map, which you might consider as a representation on $\mathbb R^2$. Let $\chi_1,\chi_2:B\to GL_1({\mathbb R})$ be the representations given by $$ \chi_j\left(^{a_1}\ ^x_{a_2} \right)=a_j. $$ Then $\chi_1$ is a subrepresentation and $\chi_2$ is a quotient of $\pi$. So both are subquotients, but $Hom_B(\pi,\chi_1)$ is zero.

share|cite|improve this answer

The answers given by Bruce and Anton are sensible, but I'd add that it's important in Lie theory to distinguish the behavior of reductive groups from that of arbitrary Lie groups: finite dimensional representations of the former are semisimple (motivating the label "reductive") but often not in general. It's also important to distinguish finite and infinite dimensional representations, since you use the term "continuous". Even for a well-behaved simple Lie group, typical infinite dimensional representations of finite length are often not semisimple: those in the "principal series", etc. This is mirrored in a more elementary way in related Lie algebra representations, starting with Verma modules: these have finite composition series but are only rarely semisimple (then only when they are in fact simple).

share|cite|improve this answer

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.