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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.


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

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2  
I would guess that it's easier to prove that they aren't homotopy-equivalent. –  Qiaochu Yuan Apr 16 '13 at 17:45
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Is $\mathcal{U}(n)$ supposed to be the unitary group? (If so, I thought just plain $U(n)$ was much more standard notation.) –  Todd Trimble Apr 16 '13 at 17:53
    
Yes, it denotes the unitary group, I've changed it now. –  Ollie Margetts Apr 16 '13 at 18:15
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Per Qiaochu's comment, aren't these obviously homomotopic to $U(n)$? –  Will Sawin Apr 16 '13 at 18:24
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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 $n$. –  Alain Valette Apr 16 '13 at 18:52

2 Answers 2

up vote 6 down vote accepted

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.

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sorry, I was overlooking Alain's comment which was folded. This is the same argument as his. –  Makoto Yamashita Apr 18 '13 at 7:21
    
Still, this is a neat proof, so I have accepted it. –  Ollie Margetts Apr 18 '13 at 21:26

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 $n$.

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