# invariant forms on representations of compact groups

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

###### Proof.

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

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

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

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

 $\tilde{\Omega}(v,w)=\int_{G}\Omega(gv,gw)dg,$

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

 $\tilde{\Omega}(gv,gv)=\int_{G}\Omega(gv,gv)dg\geq\inf_{g\in G}\Omega(gv,gv)>0.$

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

 $\tilde{\Omega}(hv,hw)=\int_{G}\Omega(ghv,ghw)dg=\int_{G}\Omega(ghv,ghw)d(gh)=% \tilde{\Omega}(v,w).$

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 $G$, $\Omega$ is positive definite real form, and $W$ a subrepresentation. Now consider

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

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

Applying this to the adjoint representation  of $G$ on $\mathfrak{g}$, its Lie algebra, we find that $\mathfrak{g}$ in the direct sum of simple algebras $\mathfrak{g}_{1},\ldots,\mathfrak{g}_{n}$, in the sense that $\mathfrak{g}_{i}$ has no proper nontrivial ideals, meaning that $\mathfrak{g}_{i}$ is simple in the usual sense or it is abelian. ∎

Title invariant forms on representations of compact groups InvariantFormsOnRepresentationsOfCompactGroups 2013-03-22 13:23:40 2013-03-22 13:23:40 bwebste (988) bwebste (988) 11 bwebste (988) Theorem msc 54-00