Problem. Let $ \Gamma $ be a discrete group. Denote its full group $ C^{*} $-algebra by $ {C^{*}}(\Gamma) $. If $ \Gamma $ is a finite group, then is it true that $ {C^{*}}(\Gamma) \odot {C^{*}}(\Gamma) = {C^{*}}(\Gamma) \otimes_{\sigma} {C^{*}}(\Gamma) $?
Notation:
- If $ A $ and $ B $ are $ C^{*} $-algebras, then $ A \otimes_{\sigma} B $ denotes their spatial tensor product, which is defined as the completion of the algebraic tensor product $ A \odot B $ with respect to the spatial norm $ \| \cdot \|_{\sigma} $.
- $ {C^{*}}(\Gamma) $ is the completion of $ \mathbb{C}[\Gamma] $ with respect to the norm $$ \| f \| \stackrel{\text{def}}{=} \sup \! \left( \left\{ \| \pi(f) \|_{\mathcal{B}(\mathcal{H})} ~ \Big| ~ \text{ $ (\pi,\mathcal{H}) $ is a non-degenerate $ * $-representation of $ \mathbb{C}[\Gamma] $ } \right\} \right). $$
- $ \mathbb{C}[\Gamma] $ denotes the group ring of $ \Gamma $, i.e., the ring of all finitely supported functions $ f: \Gamma \to \mathbb{C} $, where the sum of two functions is defined component-wise and the product of two functions is their convolution. (Note: If $ \Gamma $ is already finite, then every function $ f: \Gamma \to \mathbb{C} $ is automatically finitely supported.)
If $\Gamma$ is finite then the group ring is finite-dimensional, and so it is already closed. Thus $C^*(\Gamma) $ is finite-dimensional. The same happens with the algebraic tensor product: it is finite-dimensional, so closed.