## Peter-Weyl theorem as proven in Cartier’s Primer

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I'm reading Pierre Cartier's A primer of Hopf algebras 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.]

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## 3 Answers

(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.

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Nice answer, thanks a lot! – darij grinberg Feb 24 2012 at 17:58
Ah-- the overlines denote complex conjugation (I always work over the complex numbers); the tilde means do the group inverse, and take complex conjugate (the latter not being needed if you are working over the reals). – Matthew Daws Feb 24 2012 at 18:13

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More generally, if $A$ is a bounded $G$-invariant operator acting on a unitary representation $(\rho,\mathcal{H})$ of $G$ and $A^*$ is it's adjoint (i.e. $(Ax,y) = (x,A^*y)$ for all $x,y\in \mathcal{H}$), then we have

$$(x,A^*\rho(g)y) = (Ax,\rho(g)y) = (\rho(g^{-1})Ax,y)$$

where in the last step we used unitarity of $\rho$. Now use $G$-invariance of $A$ and retrace the steps back

$$(\rho(g^{-1})Ax,y) = (A\rho(g^{-1})x,y) = (\rho(g^{-1})x,A^*y) = (x,\rho(g)A^*y).$$

QED

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Are you aware of the relations, $R_{f}^* = R_{f*}$ and $R_f R_h = R_{f \ast h}$? So if $R_f$ commutes with $L_g$ for all $f$, hence so does $R_{f^\ast}$.

So to say $R$ is a $*$ algebra representation by right convolutionn and $L$ are the left convolution. As an intuition why, these are equivalent to right and left translation by elements, which commute for every group. But you can also just check the integrals. The facts are all purely topological, no smooth structures are required and the proofs work perfectly fine for every compact group.

Actually conversely, every compact group is a projective limit of compact (possibly finite) Lie groups by the Peter-Weyl theorem, but that is another story (see the comments).

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Thanks - I'd have asked you for what $f^{\ast}$ means, but now I am sure it's what Matthew calls $\overline f$. – darij grinberg Feb 24 2012 at 17:58
Yes Matthew is correct. – Marc Palm Feb 24 2012 at 18:22
I'm now curious - Peter-Weyl gives an embedding of the group as a subgroup of a product of fin-dim unitary groups, but how does it give smooth structure on, say, a finite group? A profinite one? – Yemon Choi Feb 24 2012 at 18:56

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Yemon, pm may mean "every compact group with no small subgroups". – BR Feb 24 2012 at 20:19
No, I meant every compact group is a projective limit of lie groups. In the profinite case, you simply get a projective limit of finite groups. The article en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem explains it well – Marc Palm Feb 25 2012 at 10:49
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