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I have recently read about induced representations and I have the following perhaps naive question about them.

Let $G$ be a finite or infinite (Lie) group. Can we construct all irreducible unitary representations of $G$ (except perhaps for 1-dimensional ones) via induction from some of its proper subgroups?

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  • $\begingroup$ Induction from the trivial subgroup is the regular representation. $\endgroup$
    – user148177
    Commented Sep 18, 2014 at 22:05
  • $\begingroup$ In the case of a complex semisimple Lie group, you can get the irreducible finite dimensional representations by induction from a special subgroup called a Borel subgroup (think upper triangular matrices in $GL_{n}$.) These representations are not the actual induced representations but they can be obtained from them in a nice way. The induction functor is slightly different than the standard case though because you use the universal enveloping algebra of the Lie algebra rather than the group algebra. Additionally, there is a geometric way to induce the irreducibles using cohomology. $\endgroup$
    – Siddharth Venkatesh
    Commented Sep 19, 2014 at 4:23
  • $\begingroup$ @SiddharthVenkatesh So does that mean that the answer to my question is negative in general? If yes, are there any conditions on the Lie group which make the statement valid? $\endgroup$
    – EPS
    Commented Sep 19, 2014 at 17:27
  • $\begingroup$ @Sam If your question is "is every irreducible representation of G exactly the induced representation from some representation of a proper subgroup H?", then I think that is never going to be the case. For example, any group has an irreducible trivial representation but the induced representation from a proper subgroup is never trivial. $\endgroup$
    – Siddharth Venkatesh
    Commented Sep 19, 2014 at 18:10
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    $\begingroup$ In fact, it is not too hard to see why any 1-dimensional irreducible representation of $G$ cannot be induced from a proper subgroup because those must have dimension divisible by $\frac{|G|}{|H|}.$ $\endgroup$
    – Siddharth Venkatesh
    Commented Sep 19, 2014 at 18:18

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No. For finite groups, the basic problem (as observed in the comments) is that a representation induced from a subgroup $H$ of a finite group $G$ has dimension divisible by $[G : H]$. In particular, if $H$ is a proper subgroup then an induced representation can't have dimension $1$, but there also exist finite groups $G$ such that the smallest index of a proper subgroup is larger than the smallest dimension of an irreducible representation which is not $1$-dimensional.

For example, take $G = A_5$. Since $A_5$ is simple, any homomorphism out of $A_5$ is either injective or trivial. In particular, the smallest $n$ such that there is a nontrivial homomorphism $A_5 \to S_n$ is $n = 5$, from which it follows that $5$ is the smallest possible index of a proper subgroup of $A_5$. But $A_5$ has irreducible representations of dimensions $3$ and $4$.

However, see Brauer's theorem on induced characters.

For Lie groups the problem is much worse: depending on what you mean by induced representations, almost all nontrivial induced representations are infinite-dimensional. But there are more sophisticated things one might mean by induction that can fix this; see the Borel-Weil theorem, which can be interpreted as a kind of "cohomological induction" (from a Borel subgroup).

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