I’m wondering whether there exist any continuous representations S1→GL(1,C)S^1\rightarrow GL(1,\mathbb{C}) with image not contained in U(1)U(1). I’m aware that any such representation is unitarisable, but wikipedia (https://en.wikipedia.org/wiki/Circle_group#Representations) makes the stronger claim that any such representation must take values in U(1)U(1). But I can’t think of an easy proof (or counter-example) of this.

Any help would be appreciated.

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S1S^1 is compact, so the image of a continuous representation must be a compact subgroup of C∗\mathbb{C}^{\ast}.

– Daniel Fischer♦

2 days ago

1

Is it obvious that the only infinite compact subgroup of C∗\mathbb{C}^* is S1S^1?

– ougoah

2 days ago

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Not utterly obvious. But it’s easy to see that a bounded subgroup of C∗\mathbb{C}^{\ast} must be contained in S1S^1.

– Daniel Fischer♦

2 days ago

Ah good point, thanks.

– ougoah

2 days ago

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1 Answer

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Any element of finite order must map to a root of unity. And such elements are dense in S1S^1.