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Let $G$ be locally compact group. Define group algebra as $$L^1(G)=\{f\colon G\to\Bbb{C}\mid\int\lvert f(x)\rvert\, dx<\infty\}$$ with convolution product. When is the group algebra $L^1(G)$ semisimple?

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up vote 10 down vote accepted

Always, according to Naĭmark, Normed Algebras, VII p. 380.

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Thanks a lot! Is there a simple proof? – user51514 Jun 2 '14 at 5:53
Naĭmark's proof is pretty simple: if $f$ is in the reducing ideal of $L^1(G)$, i.e. $m(f^*\cdot f)=0$ for every positive linear functional $m$ on $L^1(G)$, then $\|f\cdot\varphi\|_2^2=(\varphi,f^*\cdot f\cdot\varphi)=0$ for all $\varphi\in L^2(G)$. So $\int f(g)\varphi(g^{-1}h)dg=0$ for all $\varphi\in L^2(G)$ and almost all $h\in G$. Taking for $\varphi$ the characteristic function of an arbitrary summable set, one concludes that $f(g)=0$ almost everywhere on $G$, qed. – Francois Ziegler Jun 2 '14 at 6:13

I found my answer also in corollary 4.34 page 103 of a book which was named "A course in abstract harmonic analysis" By Folland. In there it was proven that for any locally compact commutative group $G$ the group algebra $L^1(G)$ and measure algebra $M(G)$ are semi-simple.

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Your question did not assume G is commutative. The result you state here only applies to commutative groups – Yemon Choi Oct 17 '14 at 3:35
Unless you have misquoted Folland? – Yemon Choi Oct 18 '14 at 20:40

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