I am reading Brian Hall's Lie Groups, Lie Algebras, and Representation Theory. His Prop. 4.36 reads

If $G$ is a compact matrix Lie group, $G$ has the complete reducibility property.

The "complete reducibility property" means that every finite representation (of $G$) is completely reducible (i.e., is equivalent to a direct sum of irreducible representations). His proof invokes Haar measure, which he (admittedly) does not prove the existence of. He also uses (again without proof) that the Haar measure of a compact group is finite.

Is there a proof that does not require Haar measure or integration on manifolds?

In particular, the fact that we are only interested in compact matrix Lie groups should simplify things. I found some sources that suggest that existence of Haar measure is "trivial" to establish on Lie groups, but this requires familiarity with volume forms on manifolds, and I would prefer to even avoid this.


Question: "Is there a proof that does not require Haar measure or integration on manifolds?"

Answer: If your group $G$ is algebraic and semi-simple there is an elementary and algebraic proof in "Fulton/Harris - Representation theory a first course", Proposition C.15 using semi simple Lie algebras.

Hence if your Lie group is algebraic there is for every finite dimensional $G$-module $W$ a decomposition

$$W \cong W_1 \oplus \cdots \oplus W_k$$

where $W_i$ are irreducible $G$-modules.

When the book speaks about a "complementary $G$-module" $W'$ this means $V \cong W \oplus W'$. Hence by induction you get the above result.


Note: The book of FU speaks of complex representations. It may be the proof in FU can be adapted to your situation: That every finite dimensional real representation $W$ of $G$ decompose into a direct sum of irreducible representations.

  • $\begingroup$ Here is the statement of the theorem you cited: "Let $V$ be a representation of the semisimple Lie algebra $\mathfrak g$ and $W \subset V$ a submodule. Then there exists a submodule $W' \subset V$ complementary to $W$. $\endgroup$ – WillG May 5 at 20:24
  • $\begingroup$ Could you explain how this implies the complete reducibility property for compact matrix Lie groups? $\endgroup$ – WillG May 5 at 20:24
  • $\begingroup$ @WillG: It would appear that the usual induction on the dimension works. Every nonzero module has an irreducible submodule; take one and find a complementary submodule, and if the latter is nonzero restrict to it and apply the induction hypothesis. $\endgroup$ – Marc van Leeuwen May 6 at 8:14

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