Algebraic $K$-theory of the infinite place

Abstract

We show that the algebraic $K$-theory of generalized archimedean valuation rings occurring in Durov’s compactification of the spectrum of a number ring is given by stable homotopy groups of certain classifying spaces. We also show that the “residue field at infinity” is badly behaved from a $K$-theoretic point of view.

1 Introduction

In number theory, it is a universal principle that the spectrum of $\mathbb {Z}$ should be completed with an infinite prime. This is corroborated, for example, by Ostrowski’s theorem, the product formula

$$\begin{aligned} \prod _{p \le \infty } |x|_p = 1, \ \ x \in \mathbb {Q}^{\times }, \end{aligned}$$

the Hasse principle, Artin–Verdier duality, and functional equations of $L$-functions.

This “compactification” ${\mathrm {Spec} \; \! } {\widehat{\mathbb {Z}}}:= {\mathrm {Spec} \; \! } {\mathbb {Z}}\cup \{ \infty \}$ was just a philosophical device until recently: Durov has proposed a rigorous framework which allows for a discussion of, say, $\mathbb {Z}_{(\infty )}$, the local ring of ${\mathrm {Spec} \; \! } {\widehat{\mathbb {Z}}}$ at $p=\infty$ [1]. The purpose of this work is to study the $K$-theory of the so-called generalized rings intervening at the infinite place.

Algebraic $K$-theory is a well-established, if difficult, invariant of arithmetical schemes. For example, the pole orders of the Dedekind $\zeta$-function $\zeta _F(s)$ of a number field $F$ are expressible by the ranks of the $K$-theory groups of ${\mathcal {O}_F}$, the ring of integers. By definition, $K$-theory only depends on the category of projective modules over a ring. Therefore, this interacts nicely with Durov’s theory of generalized rings which describes (actually: defines) such a ring $R$ by defining its free modules. For example, the free $\mathbb {Z}_{(\infty )}$-module of rank $n$ is defined as the $n$-dimensional octahedron, i.e.,

$$\begin{aligned} \mathbb {Z}_{(\infty )}(n) := \left\{ (x_1, \dots , x_n) \in \mathbb {Q}^n, \sum _i |x_i| \le 1 \right\} . \end{aligned}$$

The abstract theory of such modules is a priori more complicated than in the classical case since $\mathbb {Z}_{(\infty )}$-modules fail to build an abelian category. Nonetheless, using Waldhausen’s $S_\bullet$-construction it is possible to study the algebraic K-theory of $\mathbb {Z}_{(\infty )}$ and similar rings occurring for other number fields (Theorem 3.10, Definition 3.12).

Theorem 3.14. The $K$-groups of $\mathbb {Z}_{(\infty )}$ are given by

$$\begin{aligned} K_i(\mathbb {Z}_{(\infty )}) = \pi _{i}^\mathrm {s} (B \mu _2 \sqcup \{*\}, *) = {\left\{ \begin{array}{ll} \mathbb {Z}&{} i = 0 \ (Durov [Dur, 10.4.19]) \\ \mathbb {Z}/ 2 \oplus \mu _2 &{} i=1 \\ \mathrm {a\,finite\,group} &{} i > 1. \end{array}\right. } \end{aligned}$$

The $\mathbb {Z}/2$-part in $K_1$ stems from the first stable homotopy group $\pi _1^s$, while $\mu _2 = \{\pm 1 \}$ arises as the subgroup of $\mathbb {Z}_{(\infty )}$ of elements of norm $1$, i.e., the subgroup of (multiplicative) units of $\mathbb {Z}_{(\infty )}$. The finite $K$-group for $i>1$ is the abutment of an Atiyah–Hirzebruch spectral sequence.

This theorem is proven for more general generalized valuation rings including ${\mathcal {O}_F}_{(\sigma )}$, the ring corresponding to an infinite place $\sigma$ of a number field $F$. In this case the group $\mu _2$ above is replaced by the group $\{x \in F, |\sigma (x)| = 1\}$. The basic point is this: the only admissible monomorphisms (i.e., the ones occurring in the $S_\bullet$-construction of $K$-theory)

$$\begin{aligned} \mathbb {Z}_{(\infty )}(1) = [-1, 1] \cap \mathbb {Q}\rightarrow \mathbb {Z}_{(\infty )}(2) \end{aligned}$$

are given by mapping the interval to one of the two diagonals of the lozenge. Thereby, the Waldhausen category structure on free $\mathbb {Z}_{(\infty )}$-modules turns out to be equivalent to the one of finitely generated pointed $\{\pm 1\}$-sets, whose $K$-theory is well-known. In the course of the proof we also show that other plausible definitions, such as the $S^{-1} S$-construction, the $Q$-construction, and the $+$-construction yield the same $K$-groups.

We finish this note by pointing out two $K$-theoretic differences of the infinite place: we show that $K_0(\mathbb {F}_\infty ) = 0$ (Proposition 4.2), as opposed to $K_0(\mathbb {F}_p) = \mathbb {Z}$. Also, the completions at infinity are not well-behaved from a $K$-theoretic viewpoint. These remarks raise the question whether the “local” ring $\mathbb {Z}_{(\infty )}$ should be considered regular or, more precisely, whether

$$\begin{aligned} K_0(\mathbb {Z}_{(\infty )}) \rightarrow K'_0(\mathbb {Z}_{(\infty )}) \!:=\! \mathbb {Z}[\text {finitely presented }\mathbb {Z}_{(\infty )}\!-\!\mathbf {Mod}] / \text {short exact sequences} \end{aligned}$$

is an isomorphism. Unlike in the classical case, there does not seem to be an easy resolution argument in the context of Waldhausen categories. Another natural question is whether there is a Mayer–Vietoris sequence of the form

$$\begin{aligned} K_i(\widehat{\mathbb {Z}}) \rightarrow K_i(\mathbb {Z}) \oplus K_i(\mathbb {Z}_{(\infty )}) \rightarrow K_i(\mathbb {Q}) \rightarrow K_{i-1}(\widehat{\mathbb {Z}}), \end{aligned}$$

where $\widehat{\mathbb {Z}}$ is a generalized scheme obtained by glueing ${\mathrm {Spec} \; \! } {\mathbb {Z}}$ and $\mathrm {Spec} \; \! \mathbb {Z}_{(\infty )}$ along $\mathrm {Spec} \; \! \mathbb {Q}$. The usual proof of this sequence proceeds by the localization sequence, which is not available in our context.

Throughout the paper, we use the following notation: $F$ is a number field with ring of integers ${\mathcal {O}_F}$. Finite primes of ${\mathcal {O}_F}$ are denoted by $\mathfrak {p}$. We write $\Sigma _F$ for the set of real and pairs of complex embeddings of $F$. The letter $\sigma$ usually denotes an element of $\Sigma _F$. It is referred to as an infinite prime of ${\mathcal {O}_F}$.

2 Generalized rings

In a few brushstrokes, we recall the definition of generalized rings and their modules and some basic properties. Everything in this section is due to Durov. All references in brackets refer to [1], where a much more detailed discussion is found.

A monad in the category of sets is a functor $R : \mathbf {Sets}\rightarrow \mathbf {Sets}$ together with natural transformations $\mu : R \circ R \rightarrow R$ and $\epsilon : \mathrm {Id}\rightarrow R$ required to satisfy an associativity and unitality axiom akin to the case of monoids. We will write $R(n) := R(\{1, \dots , n\})$. An $R$-module is a set $X$ together with a morphism of monads $R \rightarrow \mathrm {End}(X)$, where the endomorphism monad $\mathrm {End}(X)$ satisfies $\mathrm {End}(X)(n) = \mathrm {Hom}_{\mathbf {Sets}}(X^n, X)$. In other words, $X$ is endowed with an action

$$\begin{aligned} R(n) {\times }X^n \rightarrow X \end{aligned}$$

satisfying the usual associativity conditions. Thus, $R(n)$ can be thought of as the $n$-ary operations (acting on any $R$-module).

Definition 2.1

(Durov [5.1.6]) A generalized ring is a monad $R$ in the category of sets satisfying two additional properties:

• $R$ is algebraic, i.e., it commutes with filtered colimits. Since every set is the filtered colimit of its finite subsets, this implies that $R$ is determined by $R(n)$ for $n \ge 0$ [4.1.3].

• $R$ is commutative, i.e., for any $t \in R(n)$, $t' \in R(n')$, any $R$-module $X$ (it suffices to take $X = R(n {\times }n')$) and $A \in X^{n{\times }n'}$, we have

  $$\begin{aligned} t(t'(A)) = t'(t(A)), \end{aligned}$$

  where on the left hand side $t'(A) \in X^n$ is obtained by letting act $t'$ on all rows of $A$ and similarly (with columns) on the right hand side.

For a unital associative ring $R$ (in the sense of usual abstract algebra), let

$$\begin{aligned} R(S) := \oplus _{s \in S} R \end{aligned}$$

be the free $R$-module of rank $\sharp S$, where $S$ is any set. The addition and multiplication on $R$ turn this into an (algebraic) monad which is commutative iff $R = R(1)$ is [3.4.8]. Indeed, the required map

$$\begin{aligned} R(1) {\times }R(1) \rightarrow R(1) \end{aligned}$$

(1)

is just the multiplication in $R$, while the addition is reformulated as

$$\begin{aligned} R(2) {\times }(R(1) {\times }R(1)) \rightarrow R(1), ((x_1, x_2), (y_1, y_2)) \mapsto \sum x_i y_i. \end{aligned}$$

Note that (1) is required to exist for any monad, so multiplication is in a sense more fundamental than addition, which requires the particular element $(1, 1) \in R(2)$ [3.4.9].

Reinterpreting a ring as a monad in this way defines a functor from commutative rings to generalized rings, which is easily seen to be fully faithful: given two classical rings $R$, $R'$, and a map of monads, i.e., a collection of maps $R(n) = R^n \rightarrow R'(n) = R'^n$, one checks that the maps for $n \ge 2$ are determined by $R \rightarrow R'$. In the same vein, $R$-modules in the classical sense are equivalent to $R$-modules (in the generalized sense). Henceforth, we will therefore not distinguish between classical commutative rings and their associated generalized rings.

The initial generalized ring is the monad $\mathbb {F}_0: \mathbf {Sets}\rightarrow \mathbf {Sets}$, $M \mapsto M$. Its modules are just the same as sets. The monad $\mathbf {Sets}\ni M \mapsto M \sqcup \{ * \}$ is denoted $\mathbb {F}_1$. Neither of these two generalized rings is induced by a classical ring. See Definition 3.2 for our main example of a non-classical ring.

Given a morphism $\phi : R \rightarrow S$ of generalized rings, the forgetful functor $\mathbf {Mod}(S) \rightarrow \mathbf {Mod}(R)$ between the module categories has a left adjoint $\phi ^*: \mathbf {Mod}(R) \rightarrow \mathbf {Mod}(S)$ called base change. We also denote it by $-\otimes _R S$. Being a left adjoint, this functor preserves colimits [4.6.19]. For example, for a generalized ring $R$, the unique map $\mathbb {F}_0\rightarrow R$ of generalized rings induces an adjunction

$$\begin{aligned} \mathbf {Sets}= \mathbf {Mod}(\mathbb {F}_0) \leftrightarrows \mathbf {Mod}(R): \text {forget} \end{aligned}$$

Its left adjoint is explicitly given by $X \mapsto R(X)$, the so-called free R-module on some set $X$. That is,

$$\begin{aligned} \mathrm {Hom}_{\mathbf {Mod}(R)}(R(X), M) = \mathrm {Hom}_{\mathbf {Sets}}(X, M), \end{aligned}$$

as in the classical case.

Coequalizers and arbitrary coproducts exist in $\mathbf {Mod}(R)$, for any generalized ring $R$ [4.6.17]. Therefore, arbitrary colimits exist. Base change functors $\phi ^*$ commute with coequalizers. Moreover, arbitrary limits exist in $\mathbf {Mod}(R)$, and commute with the forgetful functor $\mathbf {Mod}(R) \rightarrow \mathbf {Sets}$ [4.6.1].

An $R$-module $M$ is called finitely generated if there is a surjection $R(n) \twoheadrightarrow M$ for some $0 \le n < \infty$ [4.6.9]. Unless the contrary is explicitly mentioned, all our modules are supposed to be finitely generated over the ground generalized ring in question. An $R$-module $M$ is projective iff it is a retract of a free module, i.e., if there are maps $M \mathop {\rightarrow }\limits ^{i}R(n) \mathop {\rightarrow }\limits ^{p}M$ with $p i = \mathrm {id}_M$. As in the classical case this is equivalent to the property that for any surjection of $R$-modules $N \twoheadrightarrow N'$, $\mathrm {Hom}_{\mathbf {Mod}(R)}(M, N)$ maps onto $\mathrm {Hom}_{\mathbf {Mod}(R)}(M, N')$ [4.6.23]. The categories of (finitely generated) free and projective $R$-modules are denoted $\mathbf {Free}(R)$ and $\mathbf {Proj}(R)$, respectively.

As usual, an ideal $I$ of $R$ is a submodule of $R(1)$. A proper ideal $I \subsetneq R(1)$ is called prime if $R(1) \backslash I$ is multiplicatively closed [6.2.2].

3 Archimedean valuation rings

3.1 Definitions

Let $K$ be an integral domain equipped with a norm $|-|: K \rightarrow \mathbb {R}^{\ge 0}$. We will write $Q$ for the quotient field of $K$. We put $E := \{ x \in K, |x| = 1 \}$. We also write $|x|$ for the $L^1$-norm on $K^n$, i.e., $|x| = \sum _i |x_i|$. Throughout, we assume:

Assumption 3.1

• (A) $|K^{\times }| = \{|k|, k \in K^{\times }\} \subset \mathbb {R}^{\ge 0}$ is dense.

• (B) $E \subset K^{\times }$.

Definition 3.2

The (generalized) valuation ring associated to $(K, |-|)$ is the submonad $\mathcal {O}$ of $K$ given by

$$\begin{aligned} \mathcal {O}(S) := \left\{ x = (x_s) \in \bigoplus _{s \in S} K, |x| := \sum _{s \in S} |x_s| \le 1 \right\} . \end{aligned}$$

This is clearly algebraic. Moreover, the multiplication of the monad, i.e., $\mathcal {O}\circ \mathcal {O}\rightarrow \mathcal {O}$ is well-defined by restricting the one of $K$ (and is therefore commutative):

$$\begin{aligned} \mathcal {O}(\mathcal {O}(n)) = \left\{ (y_x) \in \bigoplus _{x \in \mathcal {O}(n)} K, \sum _x |y_x| \le 1 \right\} \rightarrow \mathcal {O}(n) \end{aligned}$$

sends $(y_x)$ to (the finite sum) $\sum _x y_x \cdot x$. A priori, this expression is an element of $K^n$, only, but is actually contained in $\mathcal {O}(n)$ since

$$\begin{aligned} \left| \sum _x y_x \cdot x \right| \le \left( \sum _x |y_x| \right) \cdot \sup |x| \le 1. \end{aligned}$$

In the case of an archimedean valuation, this definition of $\mathcal {O}$ is the one of Durov [1, 5.7.13]. For non-archimedean valuations, Durov’s original definition gives back the (generalized ring corresponding to the) ordinary ring $\{x \in K, |x| \le 1\}$ which is different from Definition 3.2 (see Example 3.4).

By definition, an $\mathcal {O}$-module $M$ is therefore a set such that an expression $\sum _{i=1}^n \lambda _i m_i$ is defined for $n \ge 0$, $m_i \in M$, $\lambda _i \in K$ such that $\sum |\lambda _i| \le 1$, obeying the usual laws of commutativity, associativity and distributivity. Maps $f: M \rightarrow N$ of $\mathcal {O}$-modules are described similarly: they satisfy $f(\sum _i \lambda _i m_i) = \sum _i \lambda _i f(m_i)$. The set $\{0\}$, with its obvious $\mathcal {O}$-module structure is both an initial and terminal $\mathcal {O}$-module. Given a map $f: M' \rightarrow M$ of $\mathcal {O}$-modules, the (co)kernel is defined to be the (co)equalizer of the two morphisms $f$ and $M' \rightarrow 0 \rightarrow M$. As was noted above, the forgetful functor $\mathcal {O}-\mathbf {Mod}\rightarrow \mathbf {Sets}$ preserves limits, so the kernel $\ker f$ is just $f^{-1}(0)$. The cokernel is described by the following proposition. Also see Remark 3.11 for an explicit example of a cokernel computation.

Proposition 3.3

Given a map $f: M' \rightarrow M$ of $\mathcal {O}$-modules, the cokernel is given by

$$\begin{aligned} \hbox {coker}(f) = M / \sim , \end{aligned}$$

(2)

where $\sim$ is the equivalence relation generated by $\sum _{i \in I} \lambda _i m_i \sim \sum _{i \in I} \lambda _i \tilde{m}_i$, where $I$ is any finite set, $\lambda = (\lambda _i) \in \mathcal {O}(\sharp I)$ and $m_i, \tilde{m}_i \in M$ are such that either $m_i = \tilde{m}_i$ or both $m_i$, $\tilde{m}_i \in f(M') \subset M$. This set is endowed with the $\mathcal {O}$-action via the natural projection $\pi : M \rightarrow \hbox {coker}(f)$.

Proof

This follows from the description of cokernels given in [1, 4.6.13]. It is also easy to check the universal property directly: we clearly have $\pi \circ f = 0$. Given a map $t: M \rightarrow T$ of $\mathcal {O}$-modules such that $t f = 0$, we need to see that $t$ factors uniquely through $\hbox {coker}f$. The unicity of the factorization is clear since $M \rightarrow \hbox {coker}f$ is onto. The existence is equivalent to $t(m_1) = t(m_2)$ whenever $\pi (m_1) = \pi (m_2)$. This is obvious from the definition of the equivalence relation $\sim$ above. $\square$

The base change functor resulting from the monomorphism $\mathcal {O}\subset K$ of generalized rings is denoted

$$\begin{aligned} (-)_K : \mathbf {Mod}(\mathcal {O}) \rightarrow \mathbf {Mod}(K). \end{aligned}$$

Actually, using Assumption 3.1, we may pick $t \in K^{\times }$ such that $|t|<1$. Then, $K$ is the unary localization $K = \mathcal {O}[1/t]$. This is shown in [1, 6.1.23] for $K = \mathbb {R}$. The proof for a general domain is the same. Therefore $K$ is flat over $\mathcal {O}$, so $(-)_K$ preserves finite limits, in particular kernels [1, 6.1.2, 6.1.8]. Recall from p. 4 that $(-)_K$ also preserves colimits, such as cokernels.

Let $E(n) := \{ x \in K(n) = K^n, |x| = 1 \}$ be the “boundary” of $\mathcal {O}(n)$. (This is merely a collection of sets, not a monad.) We write $\mathcal {O}$ for $\mathcal {O}(1)$ and $E$ for $E(1)$, if no confusion arises. In particular, $x \in \mathcal {O}$ means $x \in \mathcal {O}(1)$. The $i$-th standard coordinate vector $e_i = (0, ..., 1, ..., 0)$ is called a basis vector of $\mathcal {O}(n)$ ($1 \le i \le n$).

Example 3.4

Let $F$ be a number field with ring of integers ${\mathcal {O}_F}$. We fix a complex embedding $\sigma : F \rightarrow \mathbb {C}$ and take the norm $|-|$ induced by $\sigma$. Let $K$ be either $\mathcal {O}_F[1/N]$ where $N \in \mathbb {Z}$ has at least two distinct prime divisors, or $F$, or $\widehat{F}^{\sigma }$, the completion of $F$ with respect to $\sigma$. The respective generalized valuation rings will be denoted $\mathcal {O}_{F, 1/N, (\sigma )}$, $\mathcal {O}_{F, (\sigma )}$, and $\mathcal {O}_{F,\sigma }$, respectively. For example, $\mathcal {O}_{F, (\sigma )} = \mathcal {O}_{F, (\overline{\sigma })}$. Assumption 3.1(A) is satisfied: for $\mathcal {O}_F[1/N]$, pick two distinct prime divisors $p_1 \ne p_2$ of $N$. The elements $p_1^{n_1} p_2^{n_2} \in K$ are invertible for any $n_1$, $n_2 \in \mathbb {Z}$. The subgroup $\{ \log (| p_1^{n_1} p_2^{n_2} |), n_i \in \mathbb {Z}\} \subset \mathbb {R}$ is dense: otherwise it was cyclic, in contradiction to the $\mathbb {Q}$-linear independence of $\log p_1$ and $\log p_2$ (Gelfand’s theorem).

As for Assumption 3.1(B), let $x \in K$ with $|x|=1$. If $\sigma$ is a real embedding, $x = \pm |x| = \pm 1$. If $\sigma$ is a complex embedding, let $\overline{\sigma }$ be its complex conjugate and $\overline{x} \in K$ be such that $\sigma (\overline{x})=\overline{\sigma }(x)$. Then $\sigma (x) \sigma (\overline{x})=\sigma (x) \overline{\sigma (x)} = |\sigma (x)|^2 = 1$ implies $x \in K^{\times }$.

According to Durov, $\mathcal {O}_{F, (\sigma )}$ is the replacement for infinite places of the local rings ${\mathcal {O}_F}_{(\mathfrak {p})}$ at finite places. However, the analogy is relatively loose, as is shown by the following two remarks: first, for $p < \infty$, let $|x|_p := p^{-v_p(x)}$ for $x \in \mathbb {Q}^{\times }$. Then the generalized ring $\mathbb {Z}_{|-|_p}$ (in the sense of Definition 3.2) maps injectively to the localization $\mathbb {Z}_{(p)}$ of $\mathbb {Z}$ at the prime ideal $p$, but the map is a bijection only in degrees $\le p$. (Less importantly, Assumption 3.1(A) is not satisfied for $\mathbb {Z}_{|-|_p}$.)

Secondly, recall that the semilocalization ${\mathcal {O}_F}_{(\mathfrak {p}_1, \mathfrak {p}_2)} = {\mathcal {O}_F}_{(\mathfrak {p}_1)} \cap {\mathcal {O}_F}_{(\mathfrak {p}_2)}$ at two finite primes is one-dimensional. In analogy, pick two $\sigma _1, \sigma _2 \in \Sigma _F$ and consider $\mathcal {O}:= \mathcal {O}_{(\sigma _1)} \cap \mathcal {O}_{(\sigma _2)} \subset F$, i.e.,

$$\begin{aligned} \mathcal {O}(n) := \left\{ (x_1, \dots , x_n) \in F^n, \sum _k |\sigma _i(x_k)| \le 1 \ \ \text { for }i=1, 2\right\} . \end{aligned}$$

Let $\mathfrak p_i = \{ x \in \mathcal {O}, |\sigma _i(x)| < 1 \}$ and $\mathfrak p := \{ x \in \mathcal {O}, |\sigma _1(x) \sigma _2(x)| < 1 \}$. These are ideals: for example, for $x = (x_j) \in \mathcal {O}(n)$, $s_1, \dots , s_n \in \mathfrak p$, we need to check $\sum s_j x_j \in \mathfrak p$: if, say, $|\sigma _1(s_1)| < 1$ then

$$\begin{aligned} \left| \sigma _1 \left( \sum _j s_j x_j\right) \right| \le \sum |\sigma _1 (s_j)| |\sigma _1(x_j)| < \sum |\sigma _1(x_j)| \le 1. \end{aligned}$$

The complement $\mathcal {O}\backslash \mathfrak p = \{ x, |\sigma _1(x)| = |\sigma _2(x)| = 1 \}$ is multiplicatively closed (and contains $1$). We get a chain of prime ideals

$$\begin{aligned} 0 \subsetneq \mathfrak p_1 \subset \mathfrak p \subsetneq \mathcal {O}. \end{aligned}$$

The middle inclusion is, in general, strict, namely when $F = \mathbb {Q}[t] / p(t)$ with some irreducible polynomial $p(t)$ having zeros $a_1, a_2 \in \mathbb {C}$ with $|a_1| = 1$, $|a_2| < 1$. That is, $\mathrm {Spec} \; \! \mathcal {O}$ is not one-dimensional.

3.2 Projective and free $\mathcal {O}$-modules

In this section we gather a few facts about projective and free $\mathcal {O}$-modules. We begin with a handy criterion for monomorphisms of certain $\mathcal {O}$-modules (Lemma 3.5). Lemma 3.6 concerns a particular unicity property of the basis vectors $e_i = (0, \dots , 0, 1, 0, \dots , 0) \in \mathcal {O}(n)$. This is used to prove Theorem 3.7: every projective $\mathcal {O}$-module is free, provided that the norm is archimedean. This improves a result of Durov which treats only the cases where $\mathcal {O}$ is either the “unclompeted local ring” of a number ring at an infinite place $\sigma$, $\mathcal {O}_{F, (\sigma )}$, in the case where $\sigma$ is a real embedding or the “completed local ring” $\mathcal {O}_{F, \sigma }$ for both real and complex places. Therefore, we only study the $K$-theory of free $\mathcal {O}$-modules in this paper (but see Remark 3.18). We also use Lemma 3.6 to establish a highly combinatorial flavor of automorphisms of free $\mathcal {O}$-modules (Proposition 3.9), which will later give rise to the computation of higher $K$-theory of $\mathcal {O}$.

Lemma 3.5

(compare [1, 2.8.3.]) Let $f : M' \rightarrow M$ be a map of $\mathcal {O}$-modules. We suppose both $M'$ and $M$ are submodules of free $\mathcal {O}$-modules. (For example, they might be projective.) Then the following are equivalent:

• a) $f_Q: M'_Q \rightarrow M_Q$ is injective, where $Q$ is the quotient field of $K$,

• b) $f_K: M'_K \rightarrow M_K$ is injective,

• c) $f$ is injective (as a map of sets),

• d) $f$ is a monomorphism of $\mathcal {O}$-modules,

Proof

Consider the diagram

Its horizontal maps are injective since both modules are submodules of free modules and, for these, $\mathcal {O}(n) \subset K(n) = K^n \subset Q(n) = Q^n$. This shows (a) $\Rightarrow$ (b) $\Rightarrow$ (c). (c) implies (d) since the forgetful functor $\mathbf {Mod}(\mathcal {O}) \rightarrow \mathbf {Sets}$ is faithful. (d) $\Rightarrow$ (b): by Assumption 3.1, we may pick $t \in K^{\times }$ with $|t|<1$. Any two element of $M'_K$ are of the form $m'_1 / t^n$, $m'_2 / t^n$, where $m'_1, m'_2 \in M'$ and $n \ge 0$. Suppose that $f_K(m'_1 / t^n) = f(m'_1) / t^n$ agrees with $f_K(m'_2/t^n)$. The multiplication with $t^{-n}$ is injective on $M'_K$, since $M'$ ($M'_K$) is a submodule of a free $\mathcal {O}$- ($K$-, respectively) module. Thus $f(m'_1) = f(m'_2)$ so the assumption (d) implies our claim. Finally (b) $\Rightarrow$ (a) follows from the flatness of $Q$ over $K$. $\square$

The following lemma can be paraphrased by saying that the basis vectors $e_i = (0, \dots , 1, \dots 0) \in \mathcal {O}(n)$ cannot be generated as a nontrivial $\mathcal {O}$-linear combination of other elements of $\mathcal {O}(n)$.

Lemma 3.6

Suppose that $K$ is a field (as opposed to a domain). Suppose further that

$$\begin{aligned} e_i = \sum _{j=1}^m \lambda _j f_j \end{aligned}$$

(3)

with $f_j \in \mathcal {O}(n)$ and $(\lambda _j)_j \in \mathcal {O}(m)$, $\lambda _j \ne 0$. Then for each $j$, $f_j = \mu _j \cdot e_i$ with $\mu _j \in E$.

Proof

The proof proceeds by induction on $m$, the case $m=1$ being trivial.

Each $f_j$ can be written as $f_j = \sum _{l=1}^n \kappa _{jl} e_l$ with $(\kappa _{jl})_l \in \mathcal {O}(n)$. We get

$$\begin{aligned} 1 = |e_i| \mathop = \limits ^{{(3)}} |\sum \lambda _j f_j| \le \sum |\lambda _j| |f_j| \le \sum |\lambda _j| \le 1. \end{aligned}$$

(4)

Therefore equality holds throughout. We have $e_i = \sum _{j,l} \lambda _j \kappa _{jl} e_l$. This $K$-linear relation between the basis vectors of $K^n$ yields $1 = \sum _j \lambda _j \kappa _{ji}$. Hence

$$\begin{aligned} 1 \le \sum _j |\lambda _j \kappa _{ji}| \le \underbrace{\left( \sum |\lambda _j|\right) }_{{\mathop = \limits ^{(4)}}1} \cdot \max _j |\kappa _{ji}|. \end{aligned}$$

On the other hand, $|\kappa _{ji}| \le 1$, so there is some $j_0$ such that $|\kappa _{j_0i}|=1$. Using $\sum _l |\kappa _{j_0l}| \le 1$ we see $\kappa _{j_0l} = 0$ for all $l\ne i$, thus $f_{j_0} = \kappa _{j_0i} e_i.$ Put $\mu _{j_0} := \kappa _{j_0i} (\in E)$, so

$$\begin{aligned} (1-\lambda _{j_0} \mu _{j_0}) e_i = \sum _{j \ne j_0} \lambda _{j} f_j \end{aligned}$$

holds. If $|\lambda _{j_0} \mu _{j_0}| = 1$, we are done since all other $\lambda _j$, $j \ne j_0$ must vanish in this case. If $|\lambda _{j_0} \mu _{j_0}| < 1$, then

$$\begin{aligned} e_i = \sum _{j \ne j_0} \frac{\lambda _j}{1 - \lambda _{j_0} \mu _{j_0}} f_j. \end{aligned}$$

This finishes the induction step since the right hand side is actually an $\mathcal {O}$-linear combination of the $f_j$, for

$$\begin{aligned} \sum _{j \ne j_0} |\lambda _{j} | \mathop = \limits ^{(4)} 1 - |\lambda _{j_0}| = 1 - |\lambda _{j_0} \mu _{j_0}| \le |1 - \lambda _{j_0} \mu _{j_0}|. \end{aligned}$$

$\square$

Theorem 3.7

Suppose that the norm $|-|$ giving rise to the generalized valuation ring $\mathcal {O}$ is archimedean. Then every projective $\mathcal {O}$-module $M$ is free.

Proof

Let $K'$ be the completion (with respect to the norm $|-|$) of $Q$, the quotient field of $K$. By Ostrowski’s theorem, we have either $K' = \mathbb {R}$ or $K' = \mathbb {C}$ (with their usual norms). Let us write $-' := - \otimes _\mathcal {O}\mathcal {O}'$, where $\mathcal {O}' := \mathcal {O}_{K'}$ is the generalized valuation ring belonging to $K'$. We consider the following maps of $\mathcal {O}'$-modules, where $O_i$ are certain free $\mathcal {O}$-modules that are defined in the course of the proof:

$$\begin{aligned} O_3' \rightarrow O_2' \rightarrow O_1' \mathop {{\longrightarrow }}\limits ^{p'}M' \mathop {{\longrightarrow }}\limits ^{\phi , \cong }O_0'. \end{aligned}$$

First, $M'$ is a projective $\mathcal {O}'$-module: given a projector $p : O_1 := \mathcal {O}(n_1) \rightarrow \mathcal {O}(n_1)$ with $M = \hbox {im}p$, we get $M' = \hbox {im}p'$. By the afore-mentioned result of Durov [1, 10.4.2], there is an isomorphism of $\mathcal {O}'$-modules, $\phi : M' \mathop {\rightarrow }\limits ^{\cong }O_0' := \mathcal {O}'(n_0)$. The composition $\phi \circ p'$ is surjective, so for any basis vector $e_i \in O_0'$ ($1 \le i \le n_0$), there is some $\mathcal {O}'$-linear combination $\sum _{j \le n_1} \lambda _{ij} e_j$ mapping to $e_i$ under $\phi p'$. Thus, $\sum _j \lambda _{ij} \phi p' (e_j) = e_i$. Therefore, by Lemma 3.6, $\phi p' (e_j) \in E' \cdot e_i$ for each $j$. Here $E' = \{ x \in \mathcal {O}', |x| = 1\}$ (which is $S^1 \subset \mathbb {C}$ or $\{\pm 1\} \subset \mathbb {R}$ depending on $K'$). We put $O_2 := \sqcup _{j_2 \in J_2} e_{j_2} \mathcal {O}= \mathcal {O}(J_2)$, where the coproduct runs over

$$\begin{aligned} J_2 := \{1 \le j_2 \le n_1, \phi p'(e_{j_2}) \in E' e_i \ \ \text{ for } \text{ some } i \le n_0\}. \end{aligned}$$

The inclusion $J_2 \subset \{1, \dots , n_1\}$ induces a ($\mathcal {O}$-linear!) injection $f_{21}: O_2 \rightarrow O_1$. According to the previous remark, $O'_2 \mathop {{\longrightarrow }}\limits ^{\phi p' f_{21}'}O'_1$ is surjective. Consider the map $J_2 \rightarrow \{1, \dots , n_0 \}$ which maps $j_2$ to the (unique) $i$ with $e_i \in E' \phi p'(e_{j_2})$. This map is onto. By Assumption 3.1, we may pick some $J_3 \subset J_2$ on which it is a bijection. Let $f_{32}: O_3 := \sqcup _{j_3 \in J_3} e_{j_3} \mathcal {O}= \mathcal {O}(J_3) \rightarrow O_2 = \mathcal {O}(J_2)$ be the map induced by $J_3 \subset J_2$. Set $f_{31} = f_{21} \circ f_{32}$. Then the composition $O'_3 \mathop {\rightarrowtail }\limits ^{f_{31}'}O'_1 \mathop {\rightarrow }\limits ^{p'}M' \mathop {{\longrightarrow }}\limits ^{\phi , \cong }O'_0$ is an isomorphism of $\mathcal {O}'$-modules. Note that $f_{31}$ and $p$ are $\mathcal {O}$-linear maps, but $\phi$ is defined over $\mathcal {O}'$, only. Writing $v := p \circ f_{31}$, we must show the implication

$$\begin{aligned} v' \text { isomorphism} \Rightarrow v \text{ isomorphism }. \end{aligned}$$

The elements $m_j := p(e_j) \in M$, $j \le n_1$, generate $M$. The map $v' \otimes _{\mathcal {O}'} K' = v_{Q} \otimes _{Q} K'$ is an isomorphism of $K'$-vector spaces. The inclusion of the quotient field $Q \rightarrow K'$ is fully faithful, so that $v_{Q}$ is also an isomorphism. Hence there is some $k_j = a_j / b_j \in Q \backslash \{ 0 \}$ such that $k_j m_j \in \hbox {im}v$. According to Assumption 3.1, we can pick some $N \in K^{\times }$ such that $|a_j / N|, |b_j / N| \le 1$ for all $j$. Then $m_j a_j / N \in \hbox {im}v$. Similarly, pick some $t \in \mathcal {O}$ with $0 < |t| \le \min _j |a_j / N|$. Then $t M \subset \hbox {im}v$.

To show the surjectivity of $v$, we fix $m \in M$ and pick some $o_3 \in O_3$ with $tm = v(o_3)$. Since $M \subset M'$ and $v'$ is an isomorphism, there is a unique $\tilde{o}'_3 \in O'_3$ with $v'(\tilde{o}'_3) = m$. Hence $v(o_3) = v'(o_3) = v'(t\tilde{o}'_3)$, so that $t \tilde{o}'_3 = o_3$. In other words, $o'_3 = t^{-1} o_3 \in O'_3 \cap (O_3)_K = O_3$. This shows the surjectivity of $v$. The injectivity of $v$ is clear, since $O_3 \subset O'_3$ and $v'$ is injective. Consequently, $v$ is an isomorphism. $\square$

Definition 3.8

Recall that $\mathbf {Free}(\mathcal {O})$ is the category of (finitely generated) free $\mathcal {O}$-modules. In $\mathbf {Free}(\mathcal {O})$ let cofibrations ($\rightarrowtail$) be the monomorphisms whose cokernel (in the category of all $\mathcal {O}$-modules) lies in $\mathbf {Free}(\mathcal {O})$. Morphisms which are obtained as cokernels of cofibrations are called fibrations and denoted $\twoheadrightarrow$. Let weak equivalences $\mathop {\rightarrow }\limits ^{\sim }$ be the isomorphisms.

Proposition 3.9

Let $f: M' \rightarrow M$ be a monomorphism of free $\mathcal {O}$-modules with projective cokernel $M''$ (for example, a cofibration). Then there is a unique isomorphism $\phi : M \cong M' \sqcup M''$ such that the following diagram is commutative

(5)

Proof

Let $M' = \mathcal {O}(n')$, $M = \mathcal {O}(n)$ and let $f_i := f(e_i) \in M$, $1 \le i \le n'$ be the images of the basis vectors.

We claim that $f$ factors through $\sqcup _{i \le n, e_i \in f(M')} e_i \mathcal {O}= \mathcal {O}(\tilde{n}') \subset M = \mathcal {O}(n)$, where $\tilde{n}' := \sharp \{i \le n, e_i \in f(M') \}$. To show this, write $f(M') \ni m' = \sum _{i \in I} \lambda _i e_i$, where all $\lambda _i \ne 0$ and the $e_i$ are the basis vectors of $M$. Put

$$\begin{aligned} m' = \underbrace{\sum _{e_i \notin f(M')} \lambda _i e_i}_{=:m'_1} + \underbrace{\sum _{e_i \in f(M')} \lambda _i e_i}_{=:m'_2}. \end{aligned}$$

By Assumption 3.1, we can pick some $t \in K^{\times }$ such that $|t| \le 1/2$. Then $t m'_1 = t m' - t m'_2 \in f(M')$. Let $i$ be such that $e_i \notin f(M')$. We need to see $\lambda _i = 0$.

We write $(-)_Q$ for the functor $- \otimes _\mathcal {O}\mathcal {O}_Q$, where $\mathcal {O}_Q$ is the generalized valuation ring associated to the unique extension of the norm $|-|$ in $K$ to the quotient field $Q$ of $K$. The functor $(-)_Q$ preserves colimits, in particular $\hbox {coker}(f_Q) = (\hbox {coker}f)_Q$. In addition, $f_Q$ is a monomorphism by Lemma 3.5. The assumption $e_i \notin f(M')$ implies $e_i \notin f_Q(M'_Q)$: suppose that $e_i = \sum _{i' \le n'} \kappa _{i'} f_{i'}$ where $(\kappa _{i'}) \in \mathcal {O}_Q(n')$ and $f_{i'} := f(e_{i'})$ are the images of the basis vectors of $M'$. By Lemma 3.6, we have $f_{i'} = \epsilon _{i'} e_i$ for all $i'$, with some $\epsilon _{i'} \in \mathcal {O}_Q$, $|\epsilon _{i'}|=1$. But $f_{i'}$ also lies in $M$ (as opposed to $M_Q$). Thus, $\epsilon _{i'}$ must lie in $\mathcal {O}$, that is, $e_i \in f(M')$. Therefore, to prove the claim we may assume $K$ is a field.

Now, by Lemma 3.6, $e_i$ is not a non-trivial $\mathcal {O}$-linear combination of other elements of $M$. As $e_i \notin f(M')$, Proposition 3.3 implies

$$\begin{aligned} \pi ^{-1}(\pi (e_i)) = \{ e_i \}. \end{aligned}$$

(6)

Fix a section $\sigma : M'' \rightarrow M$ of $\pi$, which exists by the assumption that $M''$ be projective. We obtain $\sigma (\pi (e_i)) = e_i$. Hence,

$$\begin{aligned} 0 = \sigma (0_{M''}) = \sigma (\pi (t m'_1)) = \sum _{e_i \notin f(M')} t \lambda _i \sigma (\pi (e_i)) = \sum _{e_i \notin f(M')} t \lambda _i e_i, \end{aligned}$$

so that $\lambda _i = 0$. The claim is shown.

By the claim, $f$ induces a bijection $\tilde{f}: M' = \mathcal {O}(n') \rightarrow \mathcal {O}(\tilde{n}')$, which gives rise to a bijection $K^{n'} \rightarrow K^{\tilde{n}'}$. This shows $\tilde{n}' = n'$. We conclude that the basis vectors $e_i \in M'$ get mapped under $f$ to $\epsilon _i e_{J(i)}$ where $\epsilon _i \in E$ and $J : \{1, \dots , n'\} \rightarrow \{1, \dots , n\}$ is an injective set map. In fact, suppose $\tilde{f}^{-1}(e_i) = \sum _{j \in J} \lambda _{ij} e_j$ with $(\lambda _{ij}) \in \mathcal {O}(J)$ with all $\lambda _{ij} \ne 0$. Equivalently, $\sum \lambda _{ij} \tilde{f}(e_j) = e_i$. Therefore, by Lemma 3.6 (applied with $Q$ instead of $K$), $\tilde{f}_Q(e_j) \in E_Q \cdot e_i$ for all $j$, where $E_Q = \{ q \in Q, |q| = 1\}$. Since $\tilde{f}$ and therefore, by Lemma 3.5, $\tilde{f}_Q$ is injective, this implies that only one summand appears in this sum, i.e., $\tilde{f}(e_j) = \lambda _{ij}^{-1} e_i$ for some $j \in J$. A priori, $\lambda _{ij}^{-1}$ only lies in $Q$, but $\tilde{f}(e_j) \in \mathcal {O}(n')$ shows that $\epsilon _i := \lambda _{ij}^{-1} \in \mathcal {O}$, hence in $E$.

By Assumption 3.1, $\epsilon _i \in E$ is a unit in $K$. We can therefore define $\phi ': \mathcal {O}(n') \rightarrow M'$ by mapping the basis vectors $e_{i}$ of $\mathcal {O}(n')$ (which correspond, in the above notation, to the basis vectors $e_{J(i)}$ of $M$) to $\epsilon _i^{-1} e_i$. Also, let $\phi '': \mathcal {O}(n-n') \subset M \rightarrow M''$ be the map which sends the remaining basis vectors $e_{j'}$ for $j' \notin \hbox {im}J$ to $\pi (e_{j'})$. Put

$$\begin{aligned} \phi := \phi ' \sqcup \phi '': M = \mathcal {O}(n) = \mathcal {O}(n') \sqcup \mathcal {O}(n-n') \rightarrow M' \sqcup M''. \end{aligned}$$

Both $\phi '$ and $\phi ''$ are onto, hence so is $\phi$. This follows from the construction of coproducts of modules over generalized rings [1, 4.6.15]. (Also see [1, 10.4.7] for an explicit description of the coproduct for modules over archimedean valuation rings.) Alternatively, the surjective maps $\phi '$ and $\phi ''$ are epimorphisms of $\mathcal {O}$-modules. Hence their coproduct $\phi$ is an epimorphism. As $M' \sqcup M''$ is projective, $\phi$ has a section, so it is also surjective. The map $\phi$ is injective, as can be seen by checking the definition or using Lemma 3.5(b) $\Rightarrow$ (c). Hence $\phi$ is an isomorphism.

We finally show the unicity of $\phi$ or, in other words, that there are no non-trivial automorphism of cofiber sequences

$$\begin{aligned} 0 \rightarrow M' \rightarrowtail M \twoheadrightarrow M'' \rightarrow 0. \end{aligned}$$

Suppose $\tilde{\phi }$ is another isomorphism fitting into (5). We replace $\phi$ by $\tilde{\phi }\phi ^{-1}$ and $\tilde{\phi }$ by $\mathrm {id}_M$ and assume $f$ is the standard inclusion $M' \rightarrow M = M' \sqcup M''$ and $\pi$ is the standard projection onto $M''$. Applying the base change functor $(-)_Q$ (see above), we may assume that $K$ is a field. Then $M''_K$ is a free $K$-module, so the endomorphism $\phi _K: M_K \rightarrow M_K$ is given by a matrix

$$\begin{aligned} B = \left( \begin{array}{l@{\quad }l} \mathrm {Id}_{M'} &{} A \\ 0 &{} \mathrm {Id}_{M''} \end{array} \right) , \end{aligned}$$

where $A$ is the matrix corresponding to the map $M''_K \rightarrow M'_K$ (of free $K$-modules). On the other hand, $\phi$ is a map of free $\mathcal {O}$-modules, so every column in $B$ is in $\mathcal {O}(n)$. This forces $A = 0$, so that $\phi = \mathrm {id}_M$. $\square$

Theorem 3.10

The category $(\mathbf {Free}(\mathcal {O}), \rightarrowtail , \mathop {\rightarrow }\limits ^{\sim })$ defined in 3.8 is a Waldhausen category.

Proof

The only non-trivial thing to show is the stability of cofibrations under cobase-change. By Proposition 3.9, a cofibration sequence $M' \mathop {\rightarrowtail }\limits ^{\iota }M \mathop {\twoheadrightarrow }\limits ^{\pi }M''$ in $\mathbf {Free}(\mathcal {O})$ is isomorphic to $M' \rightarrowtail M' \sqcup M'' \twoheadrightarrow M''$. Hence, given any map $f: M' \rightarrow \tilde{M}'$, the pushout of $\iota$ along $f$, $\tilde{M}' \rightarrow \tilde{M}' \sqcup _{M'} M$ is isomorphic to $\tilde{M}' \rightarrow \tilde{M}' \sqcup M''$ which is a monomorphism with cokernel $M''$. $\square$

Remark 3.11

Mahanta uses split monomorphisms as cofibrations in the category of finitely generated modules over a fixed $\mathbb {F}_1$-algebra (i.e., pointed monoid) to define $G$- (a.k.a. $K'$-)theory of such algebras [3]. In $\mathbf {Free}(\mathcal {O})$, we have seen that all cofibrations are split, but not conversely: the cokernel of the split monomorphism $\varphi : \mathbb {Z}_\infty (1) \rightarrow \mathbb {Z}_\infty (2)$, $e_1 \mapsto \frac{e_1}{2} + \frac{e_2}{2}$ is not free. This follows either from Proposition 3.9 or by an explicit computation, using Proposition 3.3. Indeed, two elements $x_i e_1 + y_i e_2 \in \mathbb {Z}_\infty (2)$ ($i=1,2$) are identified in $\hbox {coker}\varphi$ iff $|y_1 - x_1| = |y_2 - x_2| < 1$. On $\hbox {coker}\varphi$, multiplication with $1/2$ is therefore not injective. Thus $\hbox {coker}\varphi$ is not a submodule of a free $\mathbb {Z}_\infty$-module, in particular it is not projective.

3.3 $K$-theory

In this subsection, we compute the $K$-theory of the generalized valuation ring $\mathcal {O}$ (Definition 3.2) or, more precisely, of the category of free $\mathcal {O}$-modules. By Theorem 3.7, every projective $\mathcal {O}$-module is free, provided that the norm is archimedean.

We define the $K$-theory using Waldhausen’s $S_\bullet$-construction, which has the advantage of being immediately applicable (Theorem 3.10). Other constructions, such as Quillen’s $Q$-construction can also be applied (slightly modified, since $\mathcal {O}$-modules do not form an exact category). The resulting $K$-groups do not depend on the choice of the construction.

Recall the definition of $K$-theory of a Waldhausen category $\mathcal {C}$ (see e.g. [7, Section IV.8] for more details). We always assume that the weak equivalences of $\mathcal {C}$ are its isomorphisms. The category $S_n \mathcal {C}$ consists of diagrams

(7)

such that $A_{i,j} \rightarrowtail A_{i,k} \twoheadrightarrow A_{j,k}$ is a cofibration sequence. Varying $n$ yields a simplicial category $S_\bullet \mathcal {C}$. The subcategory of isomorphisms is denoted $w S_\bullet \mathcal {C}$. Applying the classifying space construction of a category yields a pointed bisimplicial set $S(\mathcal {C})_{n, m} := B_m w S_n \mathcal {C}$. For example, $S(\mathcal {C})_{n,0} = \text {Obj} (S_n \mathcal {C})$. The $K$-theory of $\mathcal {C}$ is defined as

$$\begin{aligned} K_i(\mathcal {C}) := \pi _{i+1} d (B_* w S_\bullet \mathcal {C}), \end{aligned}$$

where $d(-)$ is the diagonal of a bisimplical set.

By Theorem 3.10, we are ready to define the algebraic K-theory of $\mathcal {O}$. More precisely, we consider the Waldhausen category of (finitely generated) free $\mathcal {O}$-modules, which is the same as projective $\mathcal {O}$-modules in all cases of interest by Theorem 3.7.

Definition 3.12

$$\begin{aligned} K_i (\mathcal {O})&:= K_i (\mathbf {Free}(\mathcal {O})) = \pi _{i+1} (d B w S_\bullet \mathbf {Free}(\mathcal {O})), \ \ i \ge 0. \end{aligned}$$

Lemma 3.13

Given two normed domains and a ring homomorphism $f: K \rightarrow K'$ between them satisfying $|f(x)| = |x|$ (so that $f$ restricts to a map $f: \mathcal {O}\rightarrow \mathcal {O}'$), the functor $f^*: \mathbf {Free}(\mathcal {O}) \rightarrow \mathbf {Free}(\mathcal {O}')$, $M \mapsto M \otimes _{\mathcal {O}} \mathcal {O}'$ is (Waldhausen-)exact and therefore induces a functorial map

$$\begin{aligned} f^* : K_i (\mathcal {O}) \rightarrow K_i (\mathcal {O}'). \end{aligned}$$

Proof

As pointed out at p. 4, $f^*: \mathbf {Mod}(\mathcal {O}) \rightarrow \mathbf {Mod}(\mathcal {O}')$ preserves cokernels. Secondly, tensoring with $\mathcal {O}'$ preserves cofibrations since a map $M \rightarrow M'$ of free (or projective) $\mathcal {O}$-modules is a monomorphism iff $M_Q \rightarrow M'_Q$ is one (where $Q$ is the quotient field of $K$, Lemma 3.5) and the statement is true for $Q$-modules: the map $Q \rightarrow Q'$ is injective since $|f(1)| = |1| = 1$ and therefore flat. $\square$

The group $K_0(\mathcal {O})$ is the free abelian group generated by the isomorphisms classes of free $\mathcal {O}$-modules modulo the relations

$$\begin{aligned}{}[\mathcal {O}(n') \sqcup \mathcal {O}(n'')] = [\mathcal {O}(n')] + [\mathcal {O}(n'')]. \end{aligned}$$

Indeed, any cofiber sequence satisfies additivity of the ranks of the involved free modules, as one sees by tensoring the sequence with the quotient field $Q$ of $K$. Therefore, $K_0(\mathcal {O}) = \mathbb {Z}$.

We now turn to higher $K$-theory of $\mathcal {O}$. Recall that $E : = \{x \in \mathcal {O}, |x| = 1 \}$ is the subgroup of norm one elements. Let us write $\mathrm {GL}_n (\mathcal {O}) := \mathrm {Aut}_{\mathcal {O}}(\mathcal {O}(n))$. According to Proposition 3.9,

$$\begin{aligned} \mathrm {GL}_n(\mathcal {O}) = E \wr S_n = E^n \rtimes S_n, \end{aligned}$$

(8)

where the symmetric group $S_n$ acts on $E^n$ by permutations. For $E = \mu _2 = \{ \pm 1 \}$, this group is known as the hyperoctahedral group. As usual, we write

$$\begin{aligned} \mathrm {GL}(\mathcal {O}) := \mathop {\mathop {\lim }\limits _{\longrightarrow }}\limits _{n}\mathrm {GL}_{n}(\mathcal {O}) \end{aligned}$$

for the infinite linear group, where the transition maps are induced by $\mathrm {GL}_n(\mathcal {O}(n) \ni f \mapsto f\sqcup \mathrm {id}_{\mathcal {O}}$. For any group $G$, let $G_\mathrm {ab}= G / [G, G]$ be its abelianization. We write $\pi ^\mathrm {s}_{i}(-)$ for the stable homotopy groups of a space and abbreviate $\pi ^\mathrm {s}_{i} := \pi ^\mathrm {s}_{i}(S^0)$.

Theorem 3.14

Let $\mathcal {O}$ be a generalized valuation ring as defined in 3.2. Then for $i\ge 0$, there is an isomorphism

$$\begin{aligned} K_i(\mathcal {O}) \cong \pi ^\mathrm {s}_{i} (BE_+, *), \end{aligned}$$

where the right hand side denotes the $i$-th stable homotopy group of the classifying space of $E$ (viewed as a discrete group), with a disjoint base point $*$. For a map $f$ as in Lemma 3.13, this isomorphism identifies $f^*$ in $K$-theory with the map on stable homotopy groups induced by $E(\mathcal O) \rightarrow E(\mathcal O')$.

For $i=1$, $2$ we get

$$\begin{aligned} K_1(\mathcal {O})&= \mathrm {GL}(\mathcal {O})_\mathrm {ab}= E {\times }\mathbb {Z}/2 \nonumber \\ K_2 (\mathcal {O})&= \mathop {\mathop {\lim }\limits _{\longrightarrow }}\limits _{n} \mathrm {H}_2([\mathrm {GL}_n(\mathcal {O}), \mathrm {GL}_n(\mathcal {O})], \mathbb {Z}) \end{aligned}$$

(9)

where the right hand side in (9) is group homology with $\mathbb {Z}$-coefficients.

Before proving the theorem, we first discuss our main example, when $\mathcal {O}$ comes from an infinite place of a number field, as in Example 3.4. Then, we prove a preliminary lemma.

Example 3.15

Let us consider a number field $F$ with the norm induced by some complex embedding $\sigma \in \Sigma _F$ (see p. 3 for notation). The torsion subgroup $E_\text {tor}$ of $E := \{x \in F^{\times }, |x| = 1\}$ agrees with the finite group $\mu _F$ of roots of unity. The exact localization sequence involving all finite primes of ${\mathcal {O}_F}$,

$$\begin{aligned} 1 \rightarrow \mathcal {O}_F^{\times }\rightarrow F^{\times }\rightarrow L := \ker (\oplus _{\mathfrak {p}< \infty } \mathbb {Z}\rightarrow \text {cl}(F)) \rightarrow 0, \end{aligned}$$

shows $F^{\times }/ \mu _F \cong \mathcal {O}_F^{\times }/ \mu _F \oplus L$. Hence it is free abelian by Dirichlet’s unit theorem. Thus

$$\begin{aligned} E \subset \mu _F \oplus \mathbb {Z}^{r_1 + r_2 - 1} \oplus L, \end{aligned}$$

where $r_1$ and $r_2$ are the numbers of real and pairs of complex embeddings. Therefore, $E = \mu _F \oplus \mathbb {Z}^S$, where $S := \hbox {rk}E$ is at most countably infinite. Of course, $E = \{ \pm 1 \}$ whenever $\sigma$ is a real embedding, but also, for example, for any complex embedding of $F = \mathbb {Q}[\root 3 \of {2}]$. For $F = \mathbb {Q}[\sqrt{-1}]$, $E$ is the (countably) infinitely generated group of pythagorean triples [2] (see also [8] for a description of the group structure of pythagorean triples in more general number fields).

The group $\mu _F$ is cyclic of order $w$, so the long exact sequence of group homology,

$$\begin{aligned} \mathrm {H}_i(\mu _F, \mathbb {Z}) \mathop {{\longrightarrow }}\limits ^{\cdot n}\mathrm {H}_i(\mu _F, \mathbb {Z}) \rightarrow \mathrm {H}_i(\mu _F, \mathbb {Z}/n) \rightarrow \mathrm {H}_{i-1}(\mu _F, \mathbb {Z}), \end{aligned}$$

together with the Atiyah–Hirzebruch spectral sequence

$$\begin{aligned} \mathrm {H}_p (\mu _F, \pi ^\mathrm {s}_{q}) = \mathrm {H}_p (B \mu _F, \pi ^\mathrm {s}_{q}) \Rightarrow \pi ^\mathrm {s}_{p+q} (B \mu _F) = \pi ^\mathrm {s}_{p+q} ((B \mu _F)_+, *) \end{aligned}$$

yield at least for small $p$ and $q$ explicit bounds on $\pi ^\mathrm {s}_{p+q}((B \mu _F)_+, *)$: the $E^2$-page reads

$$\begin{aligned} \begin{array}{l|l@{\quad }l@{\quad }l@{\quad }l} q \uparrow &{}&{}&{}&{}\\ 2 &{} \pi ^\mathrm {s}_{2} = \mathbb {Z}/2 &{} \mathbb {Z}/w' &{} \mathbb {Z}/w'&{}\\ 1 &{} \pi ^\mathrm {s}_{1} = \mathbb {Z}/2 &{} \mu _F / 2 = \mathbb {Z}/w' &{} \mathbb {Z}/w'&{}\\ 0 &{} \mathbb {Z}&{} \mu _F = \mathbb {Z}/w &{} 0 &{}\\ \hline &{} 0 &{} 1 &{} 2 &{} p \rightarrow \end{array} \end{aligned}$$

where $w'=(2,w)$. In general, $\pi ^\mathrm {s}_{p+q}((B \mu _F)_+, *)$ is finite for $p+q>0$. For $i > 0$,

$$\begin{aligned} K_i({\mathcal {O}_F}_\sigma )&= K_i({\mathcal {O}_F}_{(\sigma )}) \\&= \pi ^\mathrm {s}_{i} (B (\mu _F \oplus \mathbb {Z}^{\oplus S})_+, *) \\&= \pi ^\mathrm {s}_{i}\left( (B\mu _F)_+ \vee \bigvee _S S^1, *\right) \\&= \pi ^\mathrm {s}_{i} (B\mu _F) \oplus \bigoplus _S \pi ^\mathrm {s}_{i-1}. \end{aligned}$$

In particular

$$\begin{aligned} K_1({\mathcal {O}_F}_{(\sigma )})&= \mathbb {Z}/2 \oplus \mu _F \oplus \mathbb {Z}^{\oplus S}, \\ K_2({\mathcal {O}_F}_{(\sigma )})&= G \oplus (\mathbb {Z}/2)^{\oplus S}, \end{aligned}$$

where $G$ is a finite (abelian) group which is filtered by a filtration whose graded pieces are subquotients of $\mathbb {Z}/2$ and $\mathbb {Z}/w'$. (Determining $G$ would require studying the differentials of the spectral sequence).

Lemma 3.16

The map

$$\begin{aligned} \mathrm {GL}(\mathcal {O})_\mathrm {ab}\rightarrow E {\times }\mathbb {Z}/2, (\epsilon , \sigma ) \mapsto \left( \prod _{i=1}^\infty \epsilon _i, \text {parity}(\sigma )\right) \end{aligned}$$

is an isomorphism. Here the representation of elements of $\mathrm {GL}(\mathcal {O})$ is as in (8). The group $[\mathrm {GL}(\mathcal {O}), \mathrm {GL}(\mathcal {O})]$ is perfect.

Proof

For $i \ge 1$ and $\epsilon \in E$, let $\epsilon _i = (1, \dots , 1, \epsilon , 1, \dots ) \in E {\times }E {\times }\dots$ be the vector with $\epsilon$ at the $i$-th spot. Let $\sigma _i = (i, i+1) \in S_n$ be the permutation swapping the $i$-th and $i+1$-st letter. The $\epsilon _i$ and $\sigma _i$, for $i \ge 1$ and $\epsilon \in E$, generate $G := \mathrm {GL}(\mathcal {O})$ as we have seen in the proof of Proposition 3.9. In $G$, we have relations $\sigma _i \sigma _{i+1} \sigma _i = \sigma _{i+1} \sigma _i \sigma _{i+1}$, which implies $\sigma _i = \sigma _{i+1}$ in $G_\mathrm {ab}$. Moreover, in $G$ we have the relation $\epsilon _i \sigma _i = \sigma _{i+1} \epsilon _{i+1}$, so that we get $\epsilon _i = \epsilon _{i+1}$ in $G_\mathrm {ab}$. This shows the first claim.

The perfectness of $[\mathrm {GL}(\mathcal {O}), \mathrm {GL}(\mathcal {O})]$ is a special case of [6, Prop. 3], for example. Alternatively, the above implies that $H := [\mathrm {Aut}(\mathcal {O}(n)), \mathrm {Aut}(\mathcal {O}(n))]$ is given by $H = L \rtimes A_n$, where the alternating group $A_n$ acts on $L := \ker (\prod _{i=1}^n E \rightarrow E, (\epsilon ^1, \dots , \epsilon ^n) \mapsto \prod \epsilon ^i) (\cong E^{n-1})$ by restricting the $S_n$-action on $E^n$. Now, the perfectness of $A_n$ for $n \ge 5$ and a simple explicit computation shows $H_\mathrm {ab}=1$ for $n \ge 5$. $\square$

We now prove Theorem 3.14. This theorem is actually an immediate consequence of Proposition 3.9, together with well-known facts about $K$-theory of $G$-sets, where $G$ is some group [7, Ex. IV.8.9]. For example, the $K$-theory of the Waldhausen category of finite pointed sets (which would correspond to the impossible case $E=1$) is

$$\begin{aligned} K_i(\mathbb {F}_1) := K_i ((\text {finite pointed sets}, \text { injections, bijections})) = \pi ^\mathrm {s}_{i}, \end{aligned}$$

the stable homotopy groups of spheres. More generally, for some (discrete) group $G$, the $K$-theory of the category $\mathbf {Free}(G)$ of finitely generated (i.e., only finitely many orbits) pointed $G$-sets on which the $G$-action is fixed-point free, together with bijections as weak equivalences and injections as cofibrations, is known to be the stable homotopy group of $(BG)_+$. By Proposition 3.9, the canonical functor

$$\begin{aligned} \mathbf {Free}(E) \rightarrow \mathbf {Free}(\mathcal {O}), (E^X) \sqcup \{ * \} \mapsto \mathcal {O}(X) \end{aligned}$$

induces an equivalence of the categories of cofibrations and therefore an isomorphism of $K$-theory. For the convenience of the reader, we recall the necessary arguments, which also includes showing that other definitions of higher $K$-theory (of free $\mathcal {O}$-modules) yield the same $K$-groups.

Proof

Let $Q \mathbf {Free}(\mathcal {O})$ be Quillen’s $Q$-construction, i.e., the category whose objects are the ones of $\mathbf {Free}(\mathcal {O})$ and

$$\begin{aligned} \mathrm {Hom}_{Q \mathbf {Free}(\mathcal {O})} (A, B) := \{ A \twoheadleftarrow A' \rightarrowtail B \} / \sim , \end{aligned}$$

where two such roofs are identified if there is an isomorphism between them which is the identity on $A$ and $B$. It forms a category whose composition is given by the composite roof defined by the cartesian diagram

Here, we use that $A''$ exists (in $\mathbf {Free}(\mathcal {O})$) since it is the kernel of the composite $B' \twoheadrightarrow B \twoheadrightarrow B / A'$, which is split by Proposition 3.9. The subcategory $S := \text {Iso} (\mathbf {Free}(\mathcal {O}))$ of $\mathbf {Free}(\mathcal {O})$ consisting of isomorphisms only is a monoidal category under the coproduct. Hence $S^{-1} S$ is defined. We claim

$$\begin{aligned} \Omega B Q \mathbf {Free}(\mathcal {O}) = B (S^{-1} S). \end{aligned}$$

Indeed, the proof of [7, Theorem IV.7.1] carries over: the extension category $\mathcal E \mathbf {Free}(\mathcal {O})$ is defined as in loc. cit. and comes with a functor $t: \mathcal E \mathbf {Free}(\mathcal {O}) \rightarrow Q \mathbf {Free}(\mathcal {O}), (A \rightarrowtail B \twoheadrightarrow C) \mapsto C$. The fiber $\mathcal E_C := t^{-1} C$ ($C \in \mathbf {Free}(\mathcal {O})$) consists of sequences $A \rightarrowtail B \twoheadrightarrow C$. The functor

$$\begin{aligned} \phi : S \rightarrow \mathcal E_C, \ \ A \mapsto A \rightarrowtail A \sqcup C \twoheadrightarrow C \end{aligned}$$

induces a homotopy equivalence $B(S^{-1} S) \rightarrow B(S^{-1} \mathcal E_C)$ in the classical case of an exact category (instead of $\mathbf {Free}(\mathcal {O})$). In our situation, $\phi$ is an equivalence of categories since any extension in $\mathbf {Free}(\mathcal {O})$ splits uniquely (Proposition 3.9). Thus [7, Theorem IV.4.10] gives

$$\begin{aligned} B Q \mathbf {Free}(\mathcal {O}) = K_0(S) {\times }B \mathrm {GL}(\mathcal {O})^+, \end{aligned}$$

where the right hand side is the $+$-construction with respect to the perfect normal subgroup $[\mathrm {GL}(\mathcal {O}), \mathrm {GL}(\mathcal {O})]$ (Lemma 3.16). In the same vein, Waldhausen’s comparison of the $Q$-construction and his $S_\bullet$-construction carries over: $d (B w S_\bullet \mathbf {Free}(\mathcal {O}))$ is weakly equivalent to $B Q \mathbf {Free}(\mathcal {O})$.

Finally, by the Barratt–Priddy theorem (see e.g. [5, Th. 3.6])

$$\begin{aligned} \pi _i (B \mathrm {GL}(\mathcal {O})^+) \cong \pi ^\mathrm {s}_{i} (B E_+, *). \end{aligned}$$

The identification of the low-degree $K$-groups is the standard calculation of the $S^{-1}S$-construction [7, IV.4.8.1, IV.4.10]. $\square$

Remark 3.17

The calculation of $K_1(\mathcal {O})$ could also be done using the description of $K_1$ of a Waldhausen category due to Muro and Tonks [4].

Remark 3.18

Recall that for an (ordinary) ring $R$ the following two properties of an $R$-module $M$ are equivalent: (i) it is projective, (ii) there is another projective module $M'$ such that $M \sqcup M'$ is free. I have not been able to show the corresponding statement for projective $\mathcal {O}$-modules. For example, for a projector $p: \mathcal {O}(n) \rightarrow \mathcal {O}(n)$ with $M = \hbox {im}p$, it is not true that the canonical map

$$\begin{aligned} \phi : M \sqcup \ker p \rightarrow \mathcal {O}(n) \end{aligned}$$

is an isomorphism of $\mathcal {O}$-modules: for $n=2$ and the projector $p$ given by the matrix

$$\begin{aligned} \left( \begin{array}{l@{\quad }l} 1/2 &{} 1/2 \\ 1/2 &{} 1/2 \end{array} \right) , \end{aligned}$$

$\ker p$ is the free $\mathcal {O}$-module of rank $1$, generated by $(e_1 - e_2) /2 \in \mathcal {O}(2)$. In this case, $\phi$ induces an isomorphism of $M \sqcup \ker p$ with the free $\mathcal {O}$-module of rank $2$ generated by $(e_1 \pm e_2)/2$, but not with $\mathcal {O}(2) = (e_1, e_2)$. The analogous statement of Proposition 3.9 for cofibrations of projective $\mathcal {O}$-modules, as well as the computation of $K_i(\mathbf {Proj}(\mathcal {O}))$ for $i > 0$ (using Waldhausen’s cofinality theorem) would carry over verbatim if the above statement about projective $\mathcal {O}$-modules holds. However, the distinction between projective and free modules is only relevant for non-archimedean valuations, by Theorem 3.7.

4 The residue field at infinity

We finish this work by noting two differences (as far as $K$-theory is concerned) to the case of classical rings, namely the $K$-theory of the residue “field” at infinity, and the behavior with respect to completion. For simplicity, we restrict our attention to the case $F = \mathbb {Q}$.

Let $p < \infty$ be a (rational) prime with residue field $\mathbb {F}_p$. There is a long exact sequence

$$\begin{aligned} K_n (\mathbb {F}_p) \rightarrow K_n (\mathbb {Z}_{(p)}) \rightarrow K_n(\mathbb {Q}) \mathop {\rightarrow }\limits ^{\delta }K_{n-1} (\mathbb {F}_p) \end{aligned}$$

which stems from the fact that $\mathbb {Z}_{(p)}$ (the localization of $\mathbb {Z}$ at the prime ideal $(p)$) is a Noetherian regular local ring of dimension one. Moreover, for $n=1$ the map $\delta$ is the $p$-adic valuation $v_p : \mathbb {Q}^{\times }\rightarrow \mathbb {Z}$. The situation is less formidable at the infinite places, as we will now see. The (generalized) valuation ring $\mathbb {Z}_{(\infty )}$ (Definition 3.2) is not Noetherian: ascending chains of ideals need not terminate. Indeed, consider a finitely generated ideal $I = (m_1, \ldots , m_n) \subset \mathbb {Z}_{(\infty )}$. Then $|I| = \{ |m|, m \in I \} = [0, \max _i |m_i|] \cap |\mathbb {Z}_{(\infty )}|$. In particular, an ideal of the form $\{x \in \mathbb {Z}_{(\infty )}, |x| < \lambda \}$, $\lambda \le 1$ is not finitely generated, since $|\mathbb {Z}_{(\infty )}|$ is dense in $[0,1]$. This should be compared with the well-known fact that the valuation ring of a non-archimedian field is noetherian iff the field is trivially or discretely valued.

Definition 4.1

[1, 4.8.13] Put $\mathbb {F}_\infty := \mathbb {Z}_{(\infty )}/ \widetilde{\mathbb {Z}_{(\infty )}}$, where $\widetilde{\mathbb {Z}_{(\infty )}}$ is the submonad given by

$$\begin{aligned} \widetilde{\mathbb {Z}_{(\infty )}}(n) = \{ x \in \mathbb {Q}^n, |x| < 1\}. \end{aligned}$$

We refer to loc. cit. for the general definition of strict quotients of generalized rings by appropriate relations. For us, it is enough to note that every element of $\mathbb {Z}_{(\infty )}(n)$ is uniquely represented by $z = \sum _{i \in I} \lambda _i \epsilon _i e_i$, where $I \subset \{1, \dots , n\}$, $0 < \lambda _i \le 1$, $\sum \lambda _i \le 1$, $\epsilon _i \in E_{\mathbb {Z}_{(\infty )}} = \{ \pm 1 \}$, and $e_i$ is the standard basis vector. Two elements $z, z' \in \mathbb {Z}_{(\infty )}(n)$ get identified in $\mathbb {F}_\infty (n)$ (Notation: $z \equiv z'$) iff

$$\begin{aligned} |z| < 1\quad \text { and }\quad |z'|<1 \end{aligned}$$

(10)

or

$$\begin{aligned} |z| = |z'|=1,\ I_z=I_{z'}, \quad \text{ and } \quad \epsilon _{i, z}=\epsilon _{i, z'}\quad \text { for all }i \in I_z. \end{aligned}$$

(11)

That is, as a set $\mathbb {F}_\infty (n)$ consists of the faces of the $n$-dimensional octahedron. Again, $0$ is the initial and terminal $\mathbb {F}_\infty$-module, so we can speak about (co)kernels.

As usual, we put

$$\begin{aligned} K_0({\mathbb {F}_\infty }) := \left( \bigoplus _{M \in {\mathbf {Free}}({\mathbb {F}_\infty }) / Iso} {\mathbb {Z}} \right) / [M] = [M^{\prime }] + [M^{\prime \prime }], \end{aligned}$$

with a relation for each monomorphism $M' \rightarrow M$ in $\mathbf {Free}(\mathbb {F}_\infty )$ such that its cokernel $M''$ (computed in $\mathbf {Mod}(\mathbb {F}_\infty )$) lies in $\mathbf {Free}(\mathbb {F}_\infty )$. Similarly, we define $K_0^{\mathbf {Proj}}(\mathbb {F}_\infty )$ using projective $\mathbb {F}_\infty$-modules. Using the above, one sees that $\mathbb {F}_\infty$ is not finitely presented as a $\mathbb {Z}_{(\infty )}$-module. Thus, one should not expect a natural map $i_*: K_0(\mathbb {F}_\infty ) \rightarrow K_0(\mathbb {Z}_{(\infty )})$. Actually, $K$-theory of $\mathbb {F}_\infty$-modules behaves badly in the sense of the following proposition:

Proposition 4.2

$K_0^\mathbf {Proj}(\mathbb {F}_\infty ) = 0$, $K_0(\mathbb {F}_\infty ) = \mathbb {Z}$. In particular, there is no exact localization sequence (regardless of the maps involved)

$$\begin{aligned} K_1(\mathbb {Z}_{(\infty )})&= \mathbb {Z}/2 {\times }\{ \pm 1 \} \rightarrow K_1(\mathbb {Q}) = \mathbb {Q}^{\times }\rightarrow K_0(\mathbb {F}_\infty ) \rightarrow K_0(\mathbb {Z}_{(\infty )}) \\&= \mathbb {Z}\rightarrow K_0(\mathbb {Q}) = \mathbb {Z}, \end{aligned}$$

or similarly with $K_0^\mathbf {Proj}(\mathbb {F}_\infty )$ instead.

Proof

We first show that any projective $\mathbb {F}_\infty$-module $M$ which is generated by $n$ elements contains $\mathbb {F}_\infty$ as a submodule, such that the cokernel is a projective $\mathbb {F}_\infty$-module generated by $n-1$ elements. This implies that $K^\mathbf {Proj}_0(\mathbb {F}_\infty )$ is generated by $[\mathbb {F}_\infty ]$ (which is obvious for $K_0(\mathbb {F}_\infty )$).

The projective module $M$ is specified by a projector $\pi : \mathbb {F}_\infty (n) \rightarrow \mathbb {F}_\infty (n)$ with $M = \pi (\mathbb {F}_\infty (n))$. Let $a_i := \pi (e_i) \in \mathbb {F}_\infty (n)$. We pick $a_{ij} \in [-1, 1] \subset \mathbb {R}$ such that $a_i \equiv \sum _{j \in J_i} a_{ij} e_j$ with $a_{ij} \ne 0$ for all $j \in J_i$. Set $A := (a_{ij}) \in \mathbb {R}^{n{\times }n}$. We may assume that the number $n$ of generators of $M$ is minimal, i.e., there is no surjection $p': \mathbb {F}_\infty (n') \rightarrow M$ with $n' < n$. Indeed, if there is such a surjection, it has a section $\sigma '$ since $M$ is projective, and $\pi ' := \sigma ' p'$ would again be a projector.

The minimality of $n$ implies that $a_i \not \equiv a_j$ for all $i \ne j$. Otherwise, the restriction of $\pi$ to $\mathbb {F}_\infty (n \backslash \{ i \}) \subset \mathbb {F}_\infty (n)$ would be surjective. Similarly, the minimality implies $a_i \not \equiv 0 \in \mathbb {F}_\infty (n)$ for all $i$. Also, put $B = (b_{ij}) := A^2 \in \mathbb {R}^{n {\times }n}$. Using $(b_{ij})_j \equiv \pi (a_i) \equiv a_i \not \equiv 0 \in \mathbb {F}_\infty (n)$, we obtain $\sum _{j} |b_{ij}| = 1$ and $\sum _j |a_{ij}| = 1$ by (10).

The minimality of $n$ implies $i \in J_i$ or equivalently, $a_{ii} \ne 0$: otherwise $a_i \equiv \pi (a_i) \equiv \sum _{j \in J_i \backslash \{ i\}} a_{ij} a_j$ would be an $\mathbb {F}_\infty$-linear combination of the remaining columns of $A$.

For every $i \le n$,

$$\begin{aligned} 1&= \sum _j |b_{ij}| = \sum _j | \sum _k a_{ik} a_{kj}| \\&\le \sum _j \sum _k |a_{ik}| |a_{kj}| = \sum _k |a_{ik}| \underbrace{\left( \sum _j |a_{kj}|\right) }_{=1} \\&= 1, \end{aligned}$$

so equality holds. In particular, the terms $\hbox {sgn}(a_{ik} a_{kj})$ are either all (for arbitrary $i, j, k \le n$) non-negative or non-positive. Picking $k = j := i$, we see that they are non-negative, since $\hbox {sgn}(a_{ii}^2) > 0$, for $a_{ii} \ne 0$.

Let $I^> := \{i, a_{ii} > 0\}$ and likewise with $I^<$. Then $I^> \sqcup I^- = \{ 1, \dots , n \}$. Moreover, for $i \in I^>$ and $j \in I^<$, $a_{ii} a_{ij} \ge 0$ and $a_{ij} a_{jj} \ge 0$ imply $a_{ij} = 0$. In other words, the matrix $A$ decomposes as a direct sum matrix $A^> \sqcup A^<$, where $A^>$ and $A^<$ are the submatrices of $A$ consisting of the rows and columns with indices in $I^>$ and $I^<$, respectively. We may therefore assume $A = A^>$, say. For $i (\in I^>)$, and any $j$, $a_{ii} a_{ij} \ge 0$ implies $a_{ij} \ge 0$, i.e., the entries of $A$ are all non-negative.

Fix some $i \le n$. As $\pi$ is a projector, $a_i \equiv \pi (a_i)$, i.e.,

$$\begin{aligned} a_i \equiv \sum _{j \in J_i} a_{ij} e_j \equiv \sum a_{ij} \pi (e_j) \equiv \sum _{j \in J_i, k \in J_j} a_{ij} a_{jk} e_k \in \mathbb {F}_\infty (n). \end{aligned}$$

By (10), (11), this implies $\hbox {sgn}(a_{ik}) = \hbox {sgn}(\sum _j a_{ij} a_{jk})$, which gives

$$\begin{aligned} J_i = \cup _{j \in J_i} J_j. \end{aligned}$$

(12)

Indeed, “$\subset$” is easy to see without using the non-negativity of the entries. Conversely, for $k \notin J_i$, $\sum _j a_{ij} a_{jk} = 0$. Since all $a_{**} \ge 0$, this implies $a_{jk} = 0$ for all $j \in J_i$, i.e., $k \notin \cup _{j \in J_i} J_j$.

Now, pick some $i \le n$ such that $J_i$ is maximal, i.e., not contained in any other $J_j$, $i \ne j$. Then $i \notin J_j$ for any $i \ne j$ by (12). In other words, the $i$-th row only contains a single non-zero entry. For simplicity of notation, we may suppose $i = 1$.

Consider the diagram

where $\rho$ is the projection onto the last $n-1$ coordinates, $\iota$ is the injection in the first coordinate. The lower left-hand map is a monomorphism since the first row of $A$ is nonzero. Its cokernel $M'$ is the projective module determined by the matrix $(a_{ij})_{2 \le i,j \le n}$. This exact sequence shows that $K^\mathbf {Proj}_0(\mathbb {F}_\infty )$ is generated by $[\mathbb {F}_\infty ]$.

On the other hand, consider the projective $\mathbb {F}_\infty$-module $P$ defined by the projector $\left( \begin{array}{cc} 1/2 &{} 0 \\ 1/2 &{} 1 \end{array} \right)$ [1, 10.4.20]. It consists of $5$ elements and can be visualized as

The composition $\mathbb {F}_\infty \mathop {{\longrightarrow }}\limits ^{(1/2, 1/2)}\mathbb {F}_\infty (2) \twoheadrightarrow P$ is a monomorphism with cokernel $\mathbb {F}_\infty$. The pictured inclusion $P \rightarrow \mathbb {F}_\infty (2)$ has cokernel $\mathbb {F}_\infty$, spanned by $e_1$. This shows that $[\mathbb {F}_\infty (2)] = 2 [\mathbb {F}_\infty ] = [P] + [\mathbb {F}_\infty ] = 3 [\mathbb {F}_\infty ]$. Hence $K^\mathbf {Proj}_0(\mathbb {F}_\infty ) = 0$.

Finally, we have to show $K_0(\mathbb {F}_\infty ) = \mathbb {Z}$. For this, consider a cofiber sequence

$$\begin{aligned} \mathbb {F}_\infty (n^{\prime }) \mathop {\rightarrowtail }\limits ^{i}\mathbb {F}_\infty (n) \mathop {\twoheadrightarrow }\limits ^{p}\mathbb {F}_\infty (n^{\prime \prime }). \end{aligned}$$

We have to show $n = n' + n''$. Pick a section $\sigma$ of $p$. The natural map $i \sqcup \sigma : \mathbb {F}_\infty (n') \sqcup \mathbb {F}_\infty (n'') \rightarrow \mathbb {F}_\infty (n)$ is injective, as one easily shows. Thus $n' + n'' \le n$ for cardinality reasons. Conversely, for any basis vector $e_i \in \mathbb {F}_\infty (n) \backslash \hbox {im}i$, $p^{-1} (p(e_i)) = \{ e_i \}$, as one shows in the same way as for $\mathbb {Z}_\infty$-modules, cf. (6). Thus $\sigma (p(e_i)) = e_i$, so there are at most $n''$ such basis vectors by the injectivity of $\sigma$. Moreover, at most $n'$ of the basis vectors $e_i$ of $\mathbb {F}_\infty (n)$ are in $\hbox {im}i$ by the injectivity of $i$. This shows $n' + n'' \ge n$. $\square$

Remark 4.3

For $p \le \infty$, let $Fib$ be the homotopy fiber of $\Omega K(\mathbb {Z}_{(p)}) \rightarrow \Omega K(\mathbb {Q})$ and $\widehat{Fib}$ the one of $\Omega K(\mathbb {Z}_p) \rightarrow \Omega K(\mathbb {Q}_p)$. The localization sequence for $K$-theory shows in case $p<\infty$ that $Fib$ and $\widehat{Fib}$ are homotopy equivalent (and given by $K(\mathbb {F}_p)$). Here $\Omega$ is the loop space and $K(-)$ is a space (or spectrum) computing $K$-theory, for example the $S_\bullet$-construction. However, for $p=\infty$, we have

so that $\pi _0(Fib) \subsetneq \pi _0(\widehat{Fib})$.