Suppose $K=\mathbb{Q}_p$ ist the p-adic field and $G$ a finite group with a normal subgroup $S$. I would like to get all the irreducible representations of $G$ over $K$ .

I tried to solve this by using induction from the irreducible representations of $S$ and then checked with Mackey if they are irreducible or not.

What can one do if the induced representations decomposes into two irreducibles? Is there some theorem which tells, how many of them are different and how hey decomposes?


1 Answer 1


For finite groups and a field $F$ of characteristic zero, you can identify the $G$-endomorphisms of two induced representations $$ Hom_G( I_H^G \pi, I_K^G \sigma) $$ with the space of functions $f: H \backslash G / K \rightarrow Hom_F(V_\pi, V_\sigma*)$ with $f(hgk)= \pi(h) f(g) \sigma(k)$. If $H=G$ and $\pi = \sigma$, i.e., the case you are interested in, everything becomes a convolution algebra. If you find a basis, they give you projections to irreducible components, and you can decompose explicitly.

Equivalently, you could use Frobenius reciprocity $$ Hom_H( \pi, R_H I_K^G \sigma) $$ with Mackey induction restriction formula giving you a decomposition $$R_H I_K^G \sigma$$ in terms of $H\backslash G/K$.

All this is very combinatorial. There is no general algorithm for general $G$ even in the case of complex representation, e.g., nobody knows how to decompose the parabolic induction in $GL_3(\mathbb{Z} / p^N)$. So, one might say decomposing induced representations into irreducibles is an art rather then a theory.

The situation is slightly different in the special case, where $H$ is normal. Then induction exhaust all irreducible representations and it is easy to parametrize them. However, also here you encounter the same difficulties only later:( To understand this special case, I suggest you should learn Clifford theory.

Edit in response to a comment: Let $N$ be a normal subgroup. $G$ acts on the representation $\pi$ of $N$ via $g: \pi \mapsto \pi^g(x) = \pi(gxg^{-1})$.

  1. $Ind_N^G \pi$ decomposes with single multiplicity iff $End_G( I_N^G \pi)$ is abelian.

  2. $Ind_N^G \pi$ is irreducible, if the cardinality of the $G$ orbit $\pi$ has the same cardinality as $G/N$.

  3. More generally, the cardinality of representations contained in $I_N^G \pi$ is given by $C(\pi) /N$ for the centralizers $C(\pi) = \{ g \in G : \pi^g =\pi\}$.

1 is obvious by Schur's lemma: Assume $I_N^G \pi = \bigoplus \sigma^{\oplus m_\sigma}$ then as algebra isomorphisms $$ End_G( I_N^G \pi) = \bigoplus End_G( \sigma^{\oplus m_\sigma}) = \bigoplus M(m(\sigma), F).$$

Proof of 2+3 via Mackey formula, Schur's lemma and Frobenius reciprocity: As vectorspace isomorphisms $$Endo_G(I_N^G \pi) = Hom_N(\pi , Res\, I_N^G \pi) = \bigoplus_{g \in G/N} Hom_N(\pi, \pi^g).$$

  • $\begingroup$ Thank you, I am interested only in the case, where $H$ is normal, as the other part seems way to hard. In my expample, which I try to solve, $H$ has two irreducible representations, which both decomposes. How could Clifford theory tell me, if 3 or 4 of them are not isomorphic? $\endgroup$
    – user33046
    Apr 12, 2013 at 11:25
  • $\begingroup$ None of my claim uses Clifford theory. Clifford theory is a theory which describes the irreducible representations of G in terms of G orbits of reps pi of N and reps of C(pi), so doesn't really help for this particular question. $\endgroup$
    – Marc Palm
    Apr 13, 2013 at 18:01

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