Is there a non-unitary continuous representation of S1S^1?

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.

=================

1

 

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

1

 

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

=================

1 Answer
1

=================

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