Consider a finite dimensional $C^*$-algebra $\cal{A}$. Is there any enveloping $C^*$-algebra $\cal{C^*(G)}$ such that $\cal{A}\cong C^*(G)$ for some locally compact group $\cal{G}$?
(Note that "$\cong$" is the $C^*$-algebra isomorphism.
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2$\begingroup$ If $\mathcal{A}$ is finite-dimensional, then $\mathcal{G}$ then will have to be finite, no? So then you are equivalently asking which finite-dimensional semisimple algebras over $\mathbb{C}$ are group algebras. Or yet equivalently, under what conditions a multiset of positive integers coincides with the degrees of irreducible representations (groupprops.subwiki.org/wiki/…) of a finite group. $\endgroup$– Tobias FritzCommented Jan 11, 2018 at 8:29
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$\begingroup$ @ Tobias Fritz .By the [corollary 3.3.27, Banach Algebras and Automatic continuity H. G. Dales]( Let $I$ be a closed ideal of finite co-dimension in $L^l(G)$. Then $I$ is a *-ideal and $L^l(G)/I$ is *-isomorphic to a finite-dimensional $C^*$-algebra), $L^l(G)/I$ is a finite-dimensional $C^*$-algebra. My goal, is find a Banach algebra\ $C^*$-algebra which it is related to a locally compact group, and it is isomorph to $L^l(G)/I$. In the best case, I want to find some group algebra such that it is isomorph to $L^l(G)/I$. $\endgroup$– Albert haroldCommented Jan 11, 2018 at 9:01
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$\begingroup$ I mean, whether there exist a locally compact group $H$, such that $L^1(G)/I\cong L^1(H)$ or $L^1(G)/I\cong C^*(H)$? $\endgroup$– Albert haroldCommented Jan 11, 2018 at 9:23
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$\begingroup$ So you're meaning to ask whether there exist $G$ and $I$ such that $\mathcal{A}\cong L^1(G)/I$? This is a different question than the one that you've asked. In fact, now the answer will be positive, and you can still take $G$ to be finite. So if this is your actual question, then I can write an answer later (or somebody else does it before me). $\endgroup$– Tobias FritzCommented Jan 11, 2018 at 9:23
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2$\begingroup$ I see, so you're interested in the case $\mathcal{A} = L^1(G)/I$ and want to find an isomorphism to some $C^*(H)$. Then my original comment and similarly the accepted answer apply. In fact, there will be $G$ and $I$ such that $L^1(G)/I$ is not isomorphic to a group algebra. E.g. take any nonabelian finite simple group $G$ and let $I$ be the ideal supported on the trivial representation. Then $\mathbb{C}[G]/I$ is a direct sum of matrix algebras with all direct summands of dimension $>1$, which cannot be a group algebra as per the accepted answer. $\endgroup$– Tobias FritzCommented Jan 11, 2018 at 9:29
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1 Answer
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No, in general.
If $C^\ast(G)$ is to be finite-dimensional then $G$ must be a finite group, and so $C^*(G)= \mathbb{C}(G)$ is complex group algebra of $G$. Basic results from the representation theory of finite groups identify this algebra, up to $*$-isomorphism, as follows:
$$ C^*(G) \cong \bigoplus_{[\pi]\in \widehat{G}} M_{d_\pi}(\mathbb C),$$
where $\widehat{G}$ is the set of isomorphism classes of irreducible representations of $G$, and $d_\pi$ is the dimension of the vector space underlying the representation $\pi$. Since every group admits a one-dimensional representation (the trivial representation), $C^*(G)$ will always have a one-dimensional summand.
The problem of characterising the group algebras of finite groups among all finite-dimensional $C^*$-algebras---that is, the problem of characterising the multisets $\{ d_\pi \ |\ [\pi]\in \widehat{G}\}$ among all finite multisets of positive integers---is still open; see here for instance.
