Let $k$ be a field with absolute value $|\cdot |$ and let $S^1(k) := \lbrace x \in k \mid |x|=1\rbrace \le k^\times$. Since the absolute value definies a metric on $k$, $S^1(k)$ is a topological group. In particular, $S^1(k)$ has a classifying space $BS^1(k)$.

Question 1: What is known about $BS^1(k)$ ?

In case $k=\mathbb R$ (with the usual absolute value), $S^1(\mathbb R)=\lbrace \pm 1 \rbrace \cong \mathbb{Z}/2$ and $BS^1(\mathbb R) = P^\infty(\mathbb R)$.

Similarly, if $k=\mathbb C$ (with the usual absolute value), $S^1(\mathbb C)=S^1$ and $BS^1(\mathbb C) = P^\infty(\mathbb C)$. This motivates:

Question 2: Is it known under which conditions the relation $BS^1(k) = P^\infty(k)$ holds ?

  • $\begingroup$ If we know the homotopy groups of $S^1(k)$ then we know the homotopy groups of its classifying space. I could imagine that there are some valued fields with nasty metric topologies. $\endgroup$ – David Roberts May 17 '12 at 0:34
  • $\begingroup$ If you allow $k$ to be a noncommutative division ring, then you get the classifying space of $SU(2)$ from the quaternions. $\endgroup$ – S. Carnahan May 17 '12 at 8:17
  • $\begingroup$ S. Carnahan, thanks for the example (perhaps you can also include it there: mathoverflow.net/questions/56363/…). In fact, there's no reason to restrict to fields. $\endgroup$ – Ralph May 17 '12 at 9:49

Partially answering question 2:

If we give $k^\infty$ the topology of the limit of the inclusions $k^n \to k^{n+1}$ and consider the unit sphere $S^\infty(k)$ in there, then the sphere has a free action by $S^1(k)$, and the quotient is $P^\infty(k)$. As long as $S^\infty(k)$ is contractible, $S^\infty(k)\to P^\infty(k)$ is a universal $S^1(k)$-bundle, hence $P^\infty(k)$ is a classifying space for $S^1(k)$.

EDIT: Hmm, we can get a contraction $S^\infty(k) \times [0,1] \to S^\infty(k)$ if $k$ admits a contraction $k\times [0,1] \to k$ to 0, using How do you show that $S^{\infty}$ is contractible?.

How could we do this? Well, if $k$ is complete with respect to the norm, assuming we have a continuous map (not necessarily a ring map!) $\mathbb{Q} \to k$ (giving $\mathbb{Q}$ the usual norm topology) taking 0 to 0 and 1 to 1, then restricting to $0\le p/q \le 1$ we could argue by continuity and denseness of the rationals. Or perhaps we just try for a map $\mathbb{Q} \cap [0,1] \to k$. I don't know if this is doable for non-complete valued fields.

  • $\begingroup$ Actually, I think you can't find a contraction for non-complete valued fields, since you can't do it for $\mathbb{Q}$, for instance. The question is when can you find a continuous map $\mathbb{Q} \to k$ for a complete valued field. $\endgroup$ – David Roberts May 17 '12 at 1:12
  • $\begingroup$ If the norm is non-Archimedean, the topology is totally disconnected, so how could you have such a contraction? $\endgroup$ – Moshe May 17 '12 at 3:08
  • $\begingroup$ @Moshe That's the sort of thing I was thinking would obstruct the construction of a homotopy, but I couldn't think off the top of my head. $\endgroup$ – David Roberts May 17 '12 at 3:20
  • $\begingroup$ Thanks, David. That's an interesting approach that immediately unifies the cases k=R,C,H. Is my understanding right that whenever $k|\mathbb Q$ is complete field (or division ring) those valuation extends that of $\mathbb Q$ then we know that $BS^1(k)=P^\infty(k)$ ? (I think k=R,C,H are the only possibilities in the finite dimesional case, but maybe there are other infinite dim. examples). $\endgroup$ – Ralph May 17 '12 at 12:06
  • $\begingroup$ @Ralph - that's right. I don't have enough experience with valued fields to come up with other examples where this works. Quite possibly these are all the examples :) What about Lorentz series $k[[x]]$? $\endgroup$ – David Roberts May 18 '12 at 0:41

If $k=\mathbb{Q}_{p}^{\hat{}}$, I can say the following: $S^1 (k)$ is totally disconnected, since it is a subspace of the totally disconnected space $k$. Let $S^1 (k)^{\delta}$ be $S^1 (k)$, but equipped with the discrete topology. The natural map $S^1 (k)^{\delta} \to S^1 (k)$ is a weak homotopy equivalence, because a map from a standard sphere to $S^1 (k)$ is constant. It follows that the induced map $K(S^1 (k);1) = BS^1 (k)^{\delta} \to BS^1 (k) $ is a weak homotopy equivalence. I am reluctant to say that $BS^1 (k) $ is an Eilenberg Mac-Lane space since it is not a CW complex.

  • $\begingroup$ Johannes, thanks for your answer. I must admit that I don't know what $\mathbb{Q}_{p}^{\hat{}}$ means (the p-adic numbers are complete with respect to p-adic valuation, so $\hat{}$ doesn't mean a completion ?). Doesn't your argument show that $BS^1 (k)^{\delta} \to BS^1 (k)$ is a weak homotopy equivalence whenever $S^1(k)$ is totally disconnected ? $\endgroup$ – Ralph May 17 '12 at 11:40
  • $\begingroup$ $\mathbb{Q}^{\hat{}}_{p}$ is my (maybe not standard) notation for the $p$-adic rationals. My argument shows that for each totally disconnected group $G$, $BG^{\delta} \to BG$ is a weak equivalence. $\endgroup$ – Johannes Ebert May 17 '12 at 14:00

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