# Semisimplicity for representations of a compact group

I was reading the first chapter of the book Representation Theory by Joe Harris and William Fulton, and it states the following proposition:

Proposition 1.5. If $$W$$ is a subrepresentation of a representation $$V$$ of a finite group $$G$$, then there is a complementary invariant subspace $$W'$$ of $$V$$, so that $$V = W \oplus W'$$.

Then, it is asserted that this property is called complete reducibility, or semisimplicity, and that for continuous representations of compact groups, this property still holds.

Is there any self-contained material proving this proposition under these assumptions? I have searched in the book, but have not found anything "elementary".

Thanks.

• I suppose, the answer depends on what you call "elementary". If you are familiar with invariant measures, the proof is indeed elementary and, as far as I know, can be found in almost every book on representation theory. Commented Jul 31 at 21:20
• Welcome to mse! I've edited your question to use mathjax (which is searchable) rather than an image (which isn't) so that other users will have an easier time finding this. In the future you should do the same ^_^ Commented Jul 31 at 21:37

For arbitrary compact (Hausdorff) groups $$G$$ this requires the construction of Haar measure on $$G$$. If $$G$$ is a compact Lie group there are some tricks that can be used but the general case requires a measure-theoretic argument.

Given the existence of Haar measure the argument is to use Weyl's unitary trick. Namely, if $$V$$ is a f.d. representation of a compact group $$G$$, we pick an arbitrary inner product $$\langle \cdot, \cdot \rangle$$ on $$V$$, then use Haar measure to average this inner product over $$G$$ to produce a new inner product

$$\langle x, y \rangle_G = \int_G \langle gx, gy \rangle \, dg$$

which, by the invariance of Haar measure, is now $$G$$-invariant. Now we can show by a straightforward calculation that the orthogonal complement of an invariant subspace with respect to $$\langle \cdot, \cdot \rangle_G$$ is another invariant subspace. This argument is easiest to understand in the case that $$G$$ is a finite group, in which case Haar measure is just normalized counting measure on $$G$$ and the integral above is a finite sum

$$\langle x, y \rangle_G = \frac{1}{|G|} \sum_{g \in G} \langle gx, gy \rangle.$$

The two halves of this argument are more or less completely independent from each other. Personally I have never looked at the proof of the existence of Haar measure in detail. For the purposes of understanding Fulton and Harris you only, at most, need this result for compact Lie groups and, as mentioned, here there are tricks that can be used to avoid the general measure-theoretic argument.

If the $$G$$ group you are interested in is a compact Lie group -- that is, a group which is also a smooth compact manifold with the two structures being compatible -- then the construction of a left-invariant measure on $$G$$ is much easier than the general construction of Haar measure for a topological group: the tangent bundle of $$TG$$ of $$G$$ is trivialized by (say) the left action of $$G$$ on itself, giving an isomorphism $$TG \cong G\times T_eG$$. A choice of a nonzero top-form $$\omega_e \in \bigwedge^n(T_eG)^*$$ corresponds, via this isomorphism, to a left-invariant top form $$\omega$$ on $$G$$. If we scale $$\omega$$ so that $$\int_G \omega =1$$, then if we define $$\int_G f = \int_{G} f\omega,$$ the operation $$f \mapsto \int_{G} f$$ plays the role that the averaging $$f\mapsto Av(f):=|G|^{-1}\sum_{g \in G} f(g)$$ operation does for finite groups.

One pretty good reference for this material is Part III of Brian C Hall's book "Lie groups, Lie algebras, and representations", which is in Springer's Graduate Texts in Mathematics series.