Question

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 ?

Answer 1

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 http://mathoverflow.net/questions/198/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.

Answer 2

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.