9

I heard that $\Bbb R$ and $\Bbb C$ are the only connected, locally compact fields. Does anyone know a proof for this result?

Dominik
  • 14,396
  • 4
    This MO question states the result and attributes it to Pontryagin.

    http://mathoverflow.net/questions/87967/is-the-reals-the-smallest-connected-ordered-topological-ring

     – Dylan Yott Mar 18 '13 at 19:51
  • @DylanYott: Thanks for the link. However, it doesn't offer any proofs. – Dominik Mar 18 '13 at 20:00
  • 4
    Another reference here shows that if you drop the connected assumption, your only options are finite extensions of : $\Bbb R, \ \Bbb Q_{p},$ or $\Bbb F_{p} ((t))$. I know that finite extensions of $\Bbb Q_{p}$ are disconnected, and I guess the same is true for $\Bbb F_{p} ((t))$, but I don't know much about that. The paper that proves the aforementioned result says the proof requires nontrivial facts about topological groups, including the existence of Haar measure. Here it is:

    http://math.uga.edu/~pete/8410Chapter5.pdf

     – Dylan Yott Mar 18 '13 at 20:11
  • It’s certainly true that the finite extensions of $\mathbb F_p((t))$ are totally disconnected: like the $p$-adic fields, their topology is defined by ideals (of the ring of local integers), and such topologies are always t.d. – Lubin Apr 12 '13 at 17:21
  • You might be interested in theorem 26.9 in Stroppel, Locally compact groups. – Watson Sep 28 '17 at 11:25

2 Answers2

3

The statement seems to be a corollary of Theorem 21 from Lev Pontrjagin book “Continuous groups”. Unfortunately, I have only Russian version of it. But this book is classical, so it should have an English translation.

Alex Ravsky
  • 90,434
3

We have to be careful, since fields with discrete topology are also locally compact! Conceptually, the proof that the only nondiscrete locally compact fields are of the type mentioned in the comments is not difficult, as I recall (but my recollection is not reliable). Use a Haar measure on the locally compact multiplicative group of the field. This measure is unique up to a multiplicative constant, let's call it $\mu$. Now take a compact set with nonzero measure, call it $X$. Let $\alpha$ be a nonzero element of the field, and compare $\mu(X)$ to $\mu'(X)=\mu(\alpha X)$. Here, $\mu$ and $\mu'$ both are Haar measures, and so must be related by a positive real constant, which we can call $|\alpha|$. This association, $\alpha\mapsto|\alpha|$, doesn't depend on the choice of $X$ nor on the original choice of $\mu$, and you have to verify that it is an absolute value (or almost), and now the remaining details escape me. But Weil starts his book “Basic Number Theory” with this topic, you can find it there.

Lubin
  • 62,818