Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.]

share|improve this question

3 Answers 3

up vote 4 down vote accepted

(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.

share|improve this answer
Nice answer, thanks a lot! –  darij grinberg Feb 24 '12 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 '12 at 18:13

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

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


share|improve this answer

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).

share|improve this answer
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 '12 at 17:58
Yes Matthew is correct. –  plusepsilon.de Feb 24 '12 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 '12 at 18:56
Yemon, pm may mean "every compact group with no small subgroups". –  B R Feb 24 '12 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 –  plusepsilon.de Feb 25 '12 at 10:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.