3

I have a sort of follow-up question to Representation theory of $\mathbb{R}$?

If we have a finite field $F$, then $F^{\times}$ is a cyclic group. And so I believe that I understand how the representations (homomorphisms $G \to GL(V)$) work out.

My question is how this works when $F$ is an infinite field?

I get from the other answer, that considering representations of fields (considered as groups) is a bit tricky, but I am wondering what can be said when the one takes the units.

If this is too complicated or too broad, then I would welcome a reference.

For example, what is the representation theory of $\mathbb{R}^\times$?

Community
  • 1
John Doe
  • 3,529
  • 5
  • 50
  • 89
  • 2
    If I have time tomorrow, I can elaborate on my comment from the other question about considering such a group as an algebraic group (which is probably one of the best view points for an arbitrary field). The basic idea is that this group can be given a scheme structure as spec of $F[t,t^{-1}]$. – Tobias Kildetoft Sep 11 '16 at 19:27
  • @TobiasKildetoft: Thank you. I will look forward to that. – John Doe Sep 11 '16 at 19:28
  • $F^{\times}$ can be a complicated group in general, and the assumption that it's the multiplicative group of a field isn't super helpful other than knowing that it's commutative. $\mathbb{R}^{\times}$ is just ${ \pm 1 } \times \mathbb{R}$, so it's about as hard to understand as $\mathbb{R}$. – Qiaochu Yuan Sep 11 '16 at 20:15
  • @TobiasKildetoft: I offered a bounty on the question. – John Doe Nov 14 '16 at 15:11
  • @QiaochuYuan: I offered a bounty on the question. – John Doe Nov 14 '16 at 15:12
  • @QiaochuYuan: Why should $F ^\times$ be commutative simply because $F$ is infinite? What about $F$ being the field of quaternions? – Alex M. Nov 14 '16 at 15:12
  • 1
    @Alex: the standard convention in English is that fields are commutative. Noncommutative or skew fields are usually called division rings. I think the convention might be different in French. – Qiaochu Yuan Nov 14 '16 at 16:00
  • Yes, I meant that the field is commutative. – John Doe Nov 14 '16 at 18:06
  • In what category? I have no idea what it would look like if you treat $R^$ as a discrete group. Maybe its OK, and I am just a chicken. However it seems to me, that $R^$ is not even countably generated, and there is a chance to do some elementary set theory. – Charlie Frohman Nov 21 '16 at 13:45
  • However, if you are only interested in continuous finite dimensional representations, this is an exercise in canonical forms. – Charlie Frohman Nov 21 '16 at 13:46

0 Answers0