invariant forms on representations of compact groups

Let $G$ be a real Lie group. TFAE:

1. Every real representation of $G$ has an invariant positive definite form, and $G$ has at least one faithful representation.

2. One faithful representation of $G$ has an invariant positive definite form.

3. $G$ is compact.

Also, any group satisfying these criteria is reductive, and its Lie algebra is the direct sum of simple algebras and an abelian algebra (such an algebra is often called reductive).

Proof.

$(1)\Rightarrow (2)$: Obvious.

$(2)\Rightarrow (3)$: Let $\mathrm{\Omega }$ be the invariant form on a faithful representation $V$. Let then representation gives an embedding$\rho :G\to \mathrm{SO}(V,\mathrm{\Omega })$, the group of automorphisms of $V$ preserving $\mathrm{\Omega }$. Thus, $G$ is homeomorphic to a closed subgroup of $\mathrm{SO}(V,\mathrm{\Omega })$. Since this group is compact, $G$ must be compact as well.

(Proof that $\mathrm{SO}(V,\mathrm{\Omega })$ is compact: By induction on $dimV$. Let $v\in V$ be an arbitrary vector. Then there is a map, evaluation on $v$, from$\mathrm{SO}(V,\mathrm{\Omega })\to S^{dimV-1}\subset V$ (this is topologically a sphere, since $(V,\omega )$ is isometric to ${ℝ}^{dimV}$ with the standard norm). This is a a fiber bundle, and the fiber over any point is a copy of $\mathrm{SO}(v^{⟂},\mathrm{\Omega })$, which is compact by the inductive hypothesis. Any fiber bundle over a compact base with compact fiber has compact total space. Thus $\mathrm{SO}(V,\mathrm{\Omega })$ is compact).

$(3)\Rightarrow (1)$: Let $V$ be an arbitrary representation of $G$. Choose an arbitrary positive definite form $\mathrm{\Omega }$ on $V$.Then define

$$\tilde{\mathrm{\Omega }}(v,w)={\int }_G\mathrm{\Omega }(gv,gw)𝑑g,$$

where $dg$ is Haar measure (normalized so that ${\int }_G𝑑g=1$). Since $K$ is compact, this gives a well defined form. It is obviously bilinear, b$\mathrm{SO}(V,\mathrm{\Omega })$y the linearity of integration, and positive definite since

$$\tilde{\mathrm{\Omega }}(gv,gv)={\int }_G\mathrm{\Omega }(gv,gv)𝑑g\ge \underset{g\in G}{inf}\mathrm{\Omega }(gv,gv)>0.$$

Furthermore, $\tilde{\mathrm{\Omega }}$ is invariant, since

$$\tilde{\mathrm{\Omega }}(hv,hw)={\int }_G\mathrm{\Omega }(ghv,ghw)𝑑g={\int }_G\mathrm{\Omega }(ghv,ghw)d(gh)=\tilde{\mathrm{\Omega }}(v,w).$$

For representation $\rho :T\to \mathrm{GL}(V)$ of the maximal torus $T\subset K$, there exists a representation ${\rho }^{\prime }$ of $K$, with $\rho$ a $T$-subrepresentation of ${\rho }^{\prime }$. Also, since every conjugacy class of $K$ intersects any maximal torus, a representation of $K$ is faithful if and only if it restricts to a faithful representation of $T$. Since any torus has a faithful representation, $K$ must have one as well.

Given that these criteria hold, let $V$ be a representation of $G$, $\mathrm{\Omega }$ is positive definite real form, and $W$ a subrepresentation. Now consider

$$W^{⟂}=\{v\in V|\mathrm{\Omega }(v,w)=0\forall w\in W\}.$$

By the positive definiteness of $\mathrm{\Omega }$, $V=W\oplus W^{⟂}$. By induction, $V$ is completely reducible.

Applying this to the adjoint representation of $G$ on $𝔤$, its Lie algebra,we find that $𝔤$ in the direct sum of simple algebras ${𝔤}_1,\mathrm{\dots },{𝔤}_n$, in the sense that ${𝔤}_i$ has no proper nontrivial ideals, meaning that ${𝔤}_i$ is simple in the usual sense or it is abelian.∎