Are the groups $C( \mathbb{R} ; U(n) )$ isomorphic? 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.

$_{^1\text{ e.g. looking at Murray-von Neumann equivalence classes of projections will distinguish them.}}$
 A: 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.
A: Here is another proof. The elements of $C(\mathbb{R};U(n)) $ satisfying $f^2=1$ are functions whose values (under the standard representation of U(n)) are self-adjoint unitaries. There are $n+1$ conjugacy classes of such unitaries (each self-adjoint unitary can be represented as a diagonal matrix of 1s and -1s and counting the 1s gives the conjugacy class). Moreover the continuous map $tr:U(n)\to\mathbb{C}$ induced by the standard representation takes a different integer value on each conjugacy class, so there cannot be a function $f\in C(\mathbb{R};U(n))$ taking values in two such classes.
This shows that there are $n+1$ conjugacy classes of elements $f\in C(\mathbb{R};U(n))$ satisfying $f^2=1$. Hence these groups are non-isomorphic for different $n$.
