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.



Enveloping $C^*$-algebra
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](https://en.wikipedia.org/wiki/Representation_theory_of_finite_groups "Wikipedia") 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](https://mathscinet.ams.org/mathscinet-getitem?mr=2304316) for instance.


