4
$\begingroup$

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.


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

$\endgroup$
6
  • 2
    $\begingroup$ I would guess that it's easier to prove that they aren't homotopy-equivalent. $\endgroup$ Apr 16, 2013 at 17:45
  • 1
    $\begingroup$ Is $\mathcal{U}(n)$ supposed to be the unitary group? (If so, I thought just plain $U(n)$ was much more standard notation.) $\endgroup$
    – Todd Trimble
    Apr 16, 2013 at 17:53
  • $\begingroup$ Yes, it denotes the unitary group, I've changed it now. $\endgroup$
    – Ollie
    Apr 16, 2013 at 18:15
  • 1
    $\begingroup$ Per Qiaochu's comment, aren't these obviously homomotopic to $U(n)$? $\endgroup$
    – Will Sawin
    Apr 16, 2013 at 18:24
  • 5
    $\begingroup$ I think that the $C(X,U(n))$'s can be distinguished by observing that the minimal degree of a unitary irreducible representation of $C(X,U(n))$, not of degree 1, must be $n$. $\endgroup$ Apr 16, 2013 at 18:52

2 Answers 2

6
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ sorry, I was overlooking Alain's comment which was folded. This is the same argument as his. $\endgroup$ Apr 18, 2013 at 7:21
  • $\begingroup$ Still, this is a neat proof, so I have accepted it. $\endgroup$
    – Ollie
    Apr 18, 2013 at 21:26
2
$\begingroup$

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 $n$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.