I'm reading [Pierre Cartier's *A primer of Hopf algebras*](http://www.math.osu.edu/~kerler.2/VIGRE/InvResPres-Sp07/Cartier-IHES.pdf) to educate myself. In its subsection 3.3 (which doesn't need any Hopf algebra theory), he sketches a proof why compact Lie groups are algebraic. One step in this proof is the Peter-Weyl theorem. Here is something I don't understand in the proof of this theorem: Why is the space $C\_{\lambda, f}$ invariant under left translations $L\_g$? It is clear to me that $R\_f$ commutes with $L\_g$, but I don't see why $R\_f^{\ast}$ should also commute with $L\_g$, and I would need this assumption to prove the $L\_g$-invariance of $C\_{\lambda,f}$.


[**Disclosure:** I am neither an analyst nor a Lie group theorist, so this might be a trivial question.]



Peter-Weyl theorem as proven in Cartier's Primer
(Same as pm's answer, with details.)


Well, if $$ R\_f(\varphi)(h) = \int\_G \varphi(g) f(g^{-1}h) \ dg $$
then the adjoint satisfies
\begin{align\*}
(R\_f^\*(\varphi)|\psi) &= (\varphi|R\_f(\psi))
= \int\_{G\times G} \varphi(h) \overline{ \psi(g) f(g^{-1}h) } \ dg \ dh \\
&= \int\_{G\times G} \varphi(h) \tilde f(h^{-1}g) \overline{\psi(g)} \ dh \ dg
= (R\_{\tilde f}(\varphi) | \psi) \end{align\*}
where $\tilde f(g) = \overline{ f(g^{-1}) }$. So $R\_f^\* = R\_{\tilde f}$ is again a right translation operator.


