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Let $K$ be a compact or a finite group with a closed subgroup $H$. Let $C(K)$ be the convolution algebra of continuous functions on $K$.

The Peter Weyl theorem asserts that the $*$ algebra $C(K)$ and the subalgebra $C(K)^H$ of functions with $$ f(h^{-1}k h) = f(k) \qquad h\in H$$ are Morita equivalent (wrong!). The algebra $C(K)^H$ can have additional modules.

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I guess that if you understand the finite case, the compact case will be very similar. So let us suppose K finite. Here is a candidate : let S be any multiplicity one left-submodule of C(K) containing a copy of each simple left module. Then take $M=N:=S. C(K)^H$. I am much too lazy to check anything at the moment, but I'm pretty sure it works for H=K or H=\{e\}$ ! On the other hand this is maybe not as nice a construction as what you would hope for. –  Jef Mar 17 '12 at 13:09
Thx. What is $S.C(K)^H$? –  Marc Palm Mar 17 '12 at 13:41
Well, the product of these two subsets of $C(K)$, or if you prefer, the $(C(K)^H)^{opp}$ submodule of $C(K)$ for the right multiplcation that is spanned by S (which is a $C(K)$-submodule for the left multiplication). –  Jef Mar 17 '12 at 14:48
Argh, I don't know how to edit the previous comment. I meant the product of these two subvectorspaces, not subsets. I.e. linear combinations of products. –  Jef Mar 17 '12 at 14:49

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I don't believe this statement. In the finite case $C(K)$ is the usual group algebra. Let $K$ be nonabelian of order $6$ and let $H\subset K$ have order $2$. Then $C(K)^H$ is not Morita equivalent to $C(K)$: It is semisimple, but it has four simple modules up to isomorphism whereas $C(K)$ has three.

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I think pm might be confused with the algebra of bi-invariant functions which also happens to be a commutant in the spirit of Morita's theorem: $Hom_K(L(X),L(X))=L(H\K/H)$ where $X=K/H$. –  Reimundo Heluani Mar 17 '12 at 18:54
Okay, thank you for this counter example. So you consider $A_3 = <d>$ inside $S_3$. The conjugation orbits are $1$, $<d> - 1$ and $S_3 -<d>$, but that is how far I can go. @Reimundo Heluani: No, I was not confused with this example. My proof goes as follows. I identify invariant functionals with irreducible representation,and observe that the invariant functionals on $C(K)$ are isomorphic to invariant functionals on $C(K)^H$. No I guess, I am simply mistaken that modules over $C(K)^H$ are not associated to representations anymore;( –  Marc Palm Mar 17 '12 at 19:48

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