$C(G) \otimes C(G) \simeq C(G \times G)$

Question

Let be $G$ a compact group and $C(G)$ the vector space of continue complex functions defined over $G$, $C(G):=\{f:G \to \mathbb{C}\}$, where $f$ is a continue function. How can I prove that $$ C(G) \otimes C(G) \simeq C(G \times G) ?$$

Answer 1

In general, if it is the ordinary vector space tensor product we are talking about, the canonical map $C(G) \otimes C(G) \to C(G \times G)$ that takes $\phi \otimes \psi$ to the function $(g, h) \mapsto \phi(g)\psi(h)$ is not an isomorphism. It does however give an isomorphism onto a dense subspace of $C(G \times G)$ (equipped as usual with the topology of uniform convergence, given by the sup norm). None of this depends on the group structure of $G$, incidentally.

This is not a subject I am intimately acquainted with (I am a category theorist, not a specialist in Banach space theory or TVS theory), but some useful information can be gleaned from this online text. Proposition 5.29 (page 35) seems to be the result most directly relevant to the OP. In brief, for compact (Hausdorff) spaces $K, K'$, the Banach space $C(K \times K')$ is isomorphic to the injective tensor product $C(K) \otimes_\varepsilon C(K')$ of the Banach spaces $C(K)$, $C(K')$. As for how the ordinary tensor product of vector spaces figures in, there is a canonical isomorphism $C(K \times K') \cong C(K, C(K'))$, and what we can say is that image of the composite

$$C(K) \otimes C(K') \to C(K \times K') \cong C(K, C(K'))$$

is the subspace consisting of continuous functions $K \to C(K')$ with finite-dimensional image (in the infinite-dimensional Banach space $C(K')$). See example 1.11 (page 8); see also example 1.13 for the density statement.

Once all this is squared away, then one can begin contemplating how the group structure on $G$ manifests itself. Undoubtedly one will have a Banach space version of a Hopf algebra structure on $C(G)$, notably involving a comultiplication

$$\delta: C(G) \to C(G) \otimes_\varepsilon C(G)$$

that is a Banach algebra map. But this is beyond the scope of the actual question.