Let $C(\mathbb{R};{U}(n))$ denote the topological group of continuous functions $\mathbb{R}\to {U}(n)$ with pointwise multiplication and compact-open topology. My question is:


**Are these groups isomorphic for different values of $n$?**


I suspect the answer is no (it feels like it should be obvious), but proving this in the category of topological groups seems difficult. I would *like* to associate to $C(\mathbb{R}; {U}(n))$ the enveloping C\*-algebra $C\_b(\mathbb{R};M\_n)$, since these are much easier to distinguish$^1$. So a second question would be:


**Does there exist a functor $TopGrp\to C^\*Alg$ taking $C(\mathbb{R};{U}(n))$ to $C\_b(\mathbb{R};M\_n)$?**


The same question could be asked of the measure theoretic versions of these groups $\mathcal{M}(\mathbb{R};{U}(n))$. These are called current groups, although the literature seems unhelpful for the isomorphism problem.




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$\_{^1\text{ e.g. looking at Murray-von Neumann equivalence classes of projections will distinguish them.}}$



Are the groups $C( \mathbb{R} ; U(n) )$ isomorphic?
How about this argument? If I remember correctly, the irreducible representations of $U(n)$ are either 1-dimensional or at least $n$-dimensional. Suppose that there was an isomorphism $\phi\colon C(\mathbb{R}; U(m)) \rightarrow C(\mathbb{R}; U(n))$ for $n < m$. We have the embedding $i\colon U(m) \rightarrow C(\mathbb{R}, U(m))$ as the constant functions, and the evaluation $e\_t\colon C(\mathbb{R}, U(n)) \rightarrow U(n)$ at $t$ for any $t \in \mathbb{R}$. By the above remark, $e\_t \circ \phi \circ i$ has a commutative image and has a kernel containing $SU(n)$. On the other hand, $\prod\_{t \in \mathbb{R}} e\_t$ is an injective homomorphism. Thus, we get a contradiction.


