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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2 Answers
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Always, according to Naĭmark, Normed Algebras, VII p. 380.
answered Jun 2, 2014 at 5:35
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1$\begingroup$ 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. $\endgroup$ Jun 2, 2014 at 6:13
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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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1$\begingroup$ Your question did not assume G is commutative. The result you state here only applies to commutative groups $\endgroup$ Oct 17, 2014 at 3:35
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$\begingroup$ When is the group $C$-star of $G$ ,$C^*(G)$, semi-simple? Does it depend on the group $G$? $\endgroup$ Nov 16, 2018 at 13:44