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54-00-InvariantFormsOnRepresentationsOfCompactGroups.tex
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54-00-InvariantFormsOnRepresentationsOfCompactGroups.tex
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\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{InvariantFormsOnRepresentationsOfCompactGroups}
\pmcreated{2013-03-22 13:23:40}
\pmmodified{2013-03-22 13:23:40}
\pmowner{bwebste}{988}
\pmmodifier{bwebste}{988}
\pmtitle{invariant forms on representations of compact groups}
\pmrecord{11}{33933}
\pmprivacy{1}
\pmauthor{bwebste}{988}
\pmtype{Theorem}
\pmcomment{trigger rebuild}
\pmclassification{msc}{54-00}
\endmetadata
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\begin{document}
Let $G$ be a real Lie group. TFAE:
\begin{enumerate}
\item Every real representation of $G$ has an invariant positive definite form, and $G$ has at least one faithful representation.
\item One faithful representation of $G$ has an invariant positive definite form.
\item $G$ is compact.
\end{enumerate}
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).
\begin{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 $\R^{\dim V}$ 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 $\int_Gdg=1$). 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 $\rho'$ of $K$, with $\rho$ a $T$-subrepresentation of $\rho'$. 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 $\fr g$, its Lie algebra,
we find that $\fr g$ in the direct sum of simple algebras $\fr g_1,\ldots,\fr g_n$, in the sense that $\fr g_i$ has no proper nontrivial ideals, meaning that $\fr g_i$ is simple in the usual sense or it is abelian.
\end{proof}
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\end{document}