The third homology of $\mathrm{SL}_2(\mathbb{Q})$

TL;DR

This paper computes the third homology group of SL_2 over the rationals with half-integral coefficients, revealing a structure linked to prime-specific involutions and the indecomposable K_3 of Q.

Contribution

It provides an explicit calculation of the third homology of SL_2(Q) and describes the eigenspace decomposition related to prime operators, connecting algebraic K-theory and homology.

Findings

The third homology group is described as a direct sum over primes.

Each prime p induces an involution with eigenspaces of order related to p+1.

The kernel of the homology map relates to the indecomposable K_3 of Q.

Abstract

We calculate the third homology of $\mathrm{SL}_2(\mathbb{Q})$ with half-integral coefficients. Corresponding to each prime $p$ there is an operator on this group with square the identity. The kernel of the (split surjective) homomorphism to the indecomposable $K_3$ of $\mathbb{Q}$ is a the direct sum over all primes of the $(-1)$-eigenspaces of these operators. The $(-1)$-eigenspace of the operator corresponding to the prime $p$ is cyclic of order the odd part of $p+1$.

Equations

$${\mathrm{H}}_3\left({\mathrm{SL}}_2\left(\mathbb{Q}\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)\cong K_3^{\mathrm{ind}}(\mathbb{Q})\left[\frac{1}{2}\right]\oplus \left(\underset{p}{⨁}\mathcal{P}({\mathbb{F}}_p)\left[\frac{1}{2}\right]\right)\cong \mathbb{Z}/3\oplus \left(\underset{p}{⨁}\mathbb{Z}/{\left(p+1\right)}_{\text{mbox}odd}\right)$$

$${\mathrm{H}}_3{\left({\mathrm{SL}}_2\left(\mathbb{Q}\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)}_0\cong \underset{p}{⨁}{\mathrm{H}}_3{\left({\mathrm{SL}}_2\left({\mathbb{Q}}_p\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)}_0.$$

$${\mathrm{H}}_3{\left({\mathrm{SL}}_2\left({\mathbb{Q}}_p\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)}_0\cong \mathcal{P}({\mathbb{F}}_p)\left[\frac{1}{2}\right].)$$

$${\mathrm{H}}_3{\left({\mathrm{SL}}_2\left(\mathbb{C}(x)\right),\mathbb{Z}\right)}_0\to \underset{p\in {\mathbb{P}}^1(\mathbb{C})}{⨁}\mathcal{P}(\mathbb{C})$$

$$\frac{G{\otimes }_{\mathbb{Z}}G}{<x\otimes y+y\otimes x|x,y\in G>}$$

$${\mathrm{H}}_3\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\right)\to K_3^{\mathrm{ind}}(F).$$

$$1\to {\mathrm{SL}}_2\left(F\right)\to {\mathrm{GL}}_2(F)\to F^{\times }\to 1$$

$${\mathrm{H}}_3{\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)}_{F^{\times }/(F^{\times }{)}^2}\cong K_3^{\mathrm{ind}}(F)\left[\frac{1}{2}\right]$$

$${\mathrm{H}}_3{\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)}_0={\mathcal{I}}_F{\mathrm{H}}_3\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\left[\frac{1}{2}\right]\right).$$

$${\mathrm{H}}_3\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\right)\cong K_3^{\mathrm{ind}}(F)\oplus {\mathrm{H}}_3{\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\right)}_0$$

$$R_{x,y}:0=\left[x\right]-\left[y\right]+\left[\frac{y}{x}\right]-\left[\frac{1-x^{-1}}{1-y^{-1}}\right]+\left[\frac{1-x}{1-y}\right]x,y\ne 1.$$

$$\lambda :\mathcal{P}(F)\to {\mathrm{S}}_{\mathbb{Z}}^2(F^{\times }),[x]\mapsto \left(1-x\right)\circ x$$

$$0\to \widetilde{{\mathrm{Tor}}_1^{\mathbb{Z}}({\mu }_F,{\mu }_F)}\to K_3^{\mathrm{ind}}(F)\to \mathcal{B}(F)\to 0$$

$$S_{x,y}:0=\left[x\right]-\left[y\right]+⟨x⟩\left[\frac{y}{x}\right]-⟨x^{-1}-1⟩\left[\frac{1-x^{-1}}{1-y^{-1}}\right]+⟨1-x⟩\left[\frac{1-x}{1-y}\right],x,y\ne 1.$$

$$\mathcal{P}(F)=(\mathcal{RP}(F){)}_{F^{\times }/(F^{\times }{)}^2}={\mathrm{H}}_0\left(F^{\times }/(F^{\times }{)}^2,\mathcal{RP}(F)\right).$$

$$\mathcal{RP}(F)\to {\mathcal{I}}_F^2\oplus {\mathrm{S}}_{\mathbb{Z}}^2(F^{\times })$$

$$\mathcal{RP}(F)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\mathcal{R}\mathcal{B}(F):$$

$$0=2\cdot (\left[-1\right]+⟨-1⟩\left[-1\right]).$$

$$0\to {\mathrm{Tor}}_1^{\mathbb{Z}}({\mu }_F,{\mu }_F)\to {\mathrm{H}}_3\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\right)\to \mathcal{R}\mathcal{B}(F)\to 0$$

$$0\to {\mathrm{Tor}}_1^{\mathbb{Z}}({\mu }_F,{\mu }_F)\left[\frac{1}{2}\right]\to {\mathrm{H}}_3\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)\to \mathcal{R}\mathcal{B}(F)\left[\frac{1}{2}\right]\to 0.$$

$$\left\{xy\right\}=\left\{x\right\}+\left\{y\right\}\text{mbox}and2\left\{x\right\}=0\text{mbox}forallx,y\in F^{\times }.$$

$${\psi }_1\left(x\right):=[x]+⟨-1⟩[x^{-1}]$$

$$\psi_{2}\left(x\right):=\left\{\begin{array}[]{ll}\left\langle{1-x}\right\rangle\left(\left\langle{x}\right\rangle[x]+[x^{-1}]\right),&x\not=1\\ 0,&x=1\end{array}\right.$$

$${\mathrm{H}}_3{\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)}_0={\mathcal{I}}_F{\mathrm{H}}_3\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)\cong {\mathcal{I}}_F{\mathcal{RP}}_1(F)\left[\frac{1}{2}\right]$$

$${\mathcal{RP}}_1(F)\left[\frac{1}{2}\right]={\widetilde{\mathcal{RP}}}_1(F)\left[\frac{1}{2}\right]={\mathrm{e}}_{+}^{-1}\widetilde{\mathcal{RP}}(F)\left[\frac{1}{2}\right]$$

$${\mathrm{H}}_3{\left({\mathrm{SL}}_2\left(F\right),\mathbb{Z}\left[\frac{1}{2}\right]\right)}_0\cong {\mathcal{I}}_F{\mathcal{RP}}_{+}(F)\left[\frac{1}{2}\right].$$

$$⟨⟨x⟩⟩C_F={\psi }_2\left(x\right)=⟨x-1⟩⟨⟨-x⟩⟩\left[x\right]\text{mbox}in\widetilde{\mathcal{RP}}(F)$$

$$0={\psi }_1\left(x\right)=\left[x\right]+⟨-1⟩\left[x^{-1}\right]⟹\left[x^{-1}\right]=-⟨-1⟩\left[x\right]$$

$${\psi }_2\left(x\right)$$

Full text

1. Introduction

Many years ago, in an article on the homology of Lie groups made discrete, Chi-Han Sah, quoting S. Lichtenbaum, cited our lack of any precise knowledge of the structure of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$ as an example of the poor state of our understanding of the homology of linear groups of general fields (see [13, pp 307-8]). Where such understanding does exist, even now, it tends often to come from connections with algebraic $K$-theory or Lie group theory where a bigger suite of mathematical tools is available. For example, we know the structure of $\mathrm{H}_{3}\left(\operatorname{SL}_{3}\left(\mathbb{Q}\right),\mathbb{Z}\right)$ because homology stability theorems tell us that it is isomorphic to $\mathrm{H}_{3}\left(\operatorname{SL}_{n}\left(\mathbb{Q}\right),\mathbb{Z}\right)$ for all larger $n$ ([9]) and this stable homology group is in turn isomorphic, via a Hurewicz homomorphism, to $K_{3}(\mathbb{Q})/\{-1\}\cdot K_{2}(\mathbb{Q})=K^{\mathrm{\small ind}}_{3}(\mathbb{Q})$ (indecomposable $K_{3}$) by [15, Lemma 5.2], which is known to be cyclic of order $24$ by the result of Lee and Szczarba ([10]).

For any field $F$, the natural map $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to\mathrm{H}_{3}\left(\operatorname{SL}_{3}\left(F\right),\mathbb{Z}\right)\cong K_{3}(F)/\{-1\}\cdot K_{2}(F)\to K^{\mathrm{\small ind}}_{3}(F)$ can be shown to be surjective ([9]). When $F=\mathbb{C}$, or more generally when $F$ is algebraically closed, it has long been known, thanks to the work of Sah and his co-authors, that this map is an isomorphism. Note that when $F$ is a number field, or a global function field, the map $\mathrm{H}_{3}\left(\operatorname{SL}_{3}\left(F\right),\mathbb{Z}\right)\to K^{\mathrm{\small ind}}_{3}(F)$ is an isomorphism, since $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\cong\mathrm{H}_{3}\left(\operatorname{SL}_{\infty}\left(F\right),\mathbb{Z}\right)$ by stability (see [9]), $\mathrm{H}_{3}\left(\operatorname{SL}_{\infty}\left(F\right),\mathbb{Z}\right)\cong K_{3}(F)/\{-1\}\cdot K_{2}(F)$ by [15, Lemma 5.2] and furthermore $\{-1\}\cdot K_{2}(F)=K_{3}^{M}(F)$ (Milnor $K_{3}$) by the calculations of [1]. Thus, for any number field $F$, the kernel of the map $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to K^{\mathrm{\small ind}}_{3}(F)$ is just the kernel of the stability homomorphism $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to\mathrm{H}_{3}\left(\operatorname{SL}_{3}\left(F\right),\mathbb{Z}\right)$.

One natural obstruction to the injectivity or surjectivity of the stability homomorphisms

$\mathrm{H}_{\bullet}\left(\operatorname{SL}_{n}\left(F\right),\mathbb{Z}\right)\to\mathrm{H}_{\bullet}\left(\operatorname{SL}_{n+1}\left(F\right),\mathbb{Z}\right)$ lies in the action of the multiplicative group $F^{\times}$: For any $a\in F^{\times}$ conjugation on $\operatorname{SL}_{n}\left(F\right)$ by a matrix $M$ of determinant $a$ induces an automorphism of $\mathrm{H}_{\bullet}\left(\operatorname{SL}_{n}\left(F\right),\mathbb{Z}\right)$ which depends only on $a$. In particular, $a^{n}=\det(\mathrm{diag}(a,\ldots,a))$ acts trivially. Since the stability homomorphism is a map of $\mathbb{Z}[F^{\times}]$ modules, both $a^{n}$ and $a^{n+1}$ act trivially on its image, and so the action of $F^{\times}$ on the image of this map is trivial. It follows that the stability homomorphism factors through the coinvariants of $F^{\times}$ on $\mathrm{H}_{\bullet}\left(\operatorname{SL}_{n}\left(F\right),\mathbb{Z}\right)$ and has image lying in the invariants of $F^{\times}$ on $\mathrm{H}_{\bullet}\left(\operatorname{SL}_{n+1}\left(F\right),\mathbb{Z}\right)$. In particular, when $F^{\times}$ acts nontrivially on $\mathrm{H}_{\bullet}\left(\operatorname{SL}_{n}\left(F\right),\mathbb{Z}\right)$, the stability homomorphism has a nontrivial kernel, since it contains $\mathcal{I}_{F}\mathrm{H}_{\bullet}\left(\operatorname{SL}_{n}\left(F\right),\mathbb{Z}\right)$, where $\mathcal{I}_{F}$ denotes the augmentation ideal of the group ring $\mathbb{Z}[F^{\times}]$.

For example, the calculations of Suslin in [14] tell us that for any infinite (or sufficiently large) field $F$ the map $\mathrm{H}_{2}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to\mathrm{H}_{2}\left(\operatorname{SL}_{3}\left(F\right),\mathbb{Z}\right)\cong K_{2}(F)$ is surjective with kernel $\mathcal{I}_{F}\mathrm{H}_{2}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ isomorphic to $I(F)^{3}$ where $I(F)$ denotes the fundamental ideal in the Grothendieck-Witt ring of the field $F$. In the case $F=\mathbb{Q}$, this kernel is isomorphic to the $\mathbb{Z}[\mathbb{Q}^{\times}]$-module $\mathbb{Z}$ on which $-1$ acts by negation and all primes act trivially.

B. Mirzaii has shown ([11]) for infinite fields $F$ that the kernel of the stability homomorphism $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to\mathrm{H}_{3}\left(\operatorname{SL}_{3}\left(F\right),\mathbb{Z}\right)=K^{\mathrm{\small ind}}_{3}(F)$, when tensored with ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$, is $\mathcal{I}_{F}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$; i.e., it is again the case that the only obstruction to injective stability is the nontriviality of the action of the multiplicative group. He subsequently ([12]) generalised this result to rings with many units (including local rings with infinite residue fields).

The main theorem of this article (Theorem 4.3) describes the structure of $\mathcal{I}_{\mathbb{Q}}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ as a $\mathbb{Z}[\mathbb{Q}^{\times}]$-module. $-1\in\mathbb{Q}^{\times}$ acts trivially, but each prime acts nontrivially. Since the the squares of rational numbers act trivially, each prime induces a decomposition into $(+1)$- and $(-1)$-eigenspace. The theorem states that this module is the direct sum over all primes of these $(-1)$-eigenspaces. The $(-1)$-eigenspace of the prime $p$ is isomorphic, via a natural residue homomorphism $S_{p}$, to ${\mathcal{P}(\mathbb{F}_{p})}\left[\tfrac{1}{2}\right]$, the scissors congruence group of the field $\mathbb{F}_{p}$. It follows that as an abelian group

[TABLE]

where $\left(m\right)_{\mbox{ \tiny odd}}$ denotes the odd part of $m\in\mathbb{Q}^{\times}$; i.e. $\left(m\right)_{\mbox{ \tiny odd}}=2^{-v_{2}(m)}m$.

As explained in Section 5 below, this theorem can be stated equivalently as follows: For any field $F$, let $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)_{0}$ denote the kernel of the surjective homomorphism $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to K^{\mathrm{\small ind}}_{3}(F)$. Then the map $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)\to\prod_{p}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}_{p}\right),\mathbb{Z}\right)$ (product over all primes) induces an isomorphism

[TABLE]

(In Section 3 we give a new more streamlined proof of the identification

[TABLE]

The main tool we use is the description of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ in terms of refined scissors congruence groups. The scissors congruence group $\mathcal{P}(F)$ of a field $F$ was introduced by Dupont and Sah in [3]. It is an abelian group defined by a presentation in terms of generators and relations and it was shown by the authors to be closely related to $K^{\mathrm{\small ind}}_{3}(F)=\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ when $F$ is algebraically closed. Soon afterwards Suslin proved ([15, Theorem 5.2]) that the connection to $K^{\mathrm{\small ind}}_{3}(F)$ persists for all infinite fields $F$ (see Theorem 2.4 below). However, to derive an analagous result for $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ for general fields it is necessary to factor in the action of the multiplicative group of the field. The refined scissors congruence group ${\mathcal{RP}}(F)$ of the field $F$ – introduced in [5] – is defined by generators and relations analagously to the scissors congruence group but as a module over $\mathbb{Z}[F^{\times}]$ and not merely an abelian group. It can then be shown to bear approximately the same relation to $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ as $\mathcal{P}(F)$ has to $K^{\mathrm{\small ind}}_{3}(F)$. (For a precise statement, see Theorem 2.5 below.) Using some later results of the author about refined scissors congruence groups, our starting point in this article is essentially a presentation of $\mathcal{I}_{\mathbb{Q}}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ as a module over the group ring $\mathbb{Z}[\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^{2}]$ as well as the existence of module homomorphisms $S_{p}:\mathcal{I}_{\mathbb{Q}}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\right)\to\mathcal{P}(\mathbb{F}_{p})$ (where the target is a module via $a\cdot x=(-1)^{v_{p}(a)}x$ for $a\in\mathbb{Q}^{\times}$), one for each prime $p$.

Remark 1.1**.**

In our main theorem, we prove that the module homomorphism

$\mathcal{I}_{\mathbb{Q}}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)\to\bigoplus_{p}\mathcal{P}(\mathbb{F}_{p})$ induced by the maps $S_{p}$, ranging over all primes $p$, becomes an isomorphism after tensoring with ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$. It is natural to ask whether the original homomorphism is an isomorphism over $\mathbb{Z}$.

I do not know. Our methods of proof and $2$-torsion ambiguities in existing results require us to work over ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$. However, it is not hard to show even over $\mathbb{Z}$ that the cokernel of this map is annihilated by $4$.

Remark 1.2**.**

It is to be expected that some version of the main result should hold for general number fields and even global fields. In order to arrive at such a result it would appear necessary first to determine whether the action of the (square classes of) the global units is trivial on the groups $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$. There is some mild evidence suggesting that this is so: (i) for any field the square class $\left\langle{-1}\right\rangle$ acts trivially (see Theorem 2.6 below) and (ii) for local fields with finite residue field, the units act trivially. (This follows, for example, from Corollary 3.13 below.) We hope to examine these questions elsewhere.

Remark 1.3**.**

There ought also to be analogous results for geometric function fields, at least over algebraically closed, or quadratically closed, fields.

For example, let $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{C}(x)\right),\mathbb{Z}\right)_{0}$ denote the kernel of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{C}(x)\right),\mathbb{Z}\right)\to K^{\mathrm{\small ind}}_{3}(\mathbb{C}(x))$. There is a natural surjective homomorphism of $\mathbb{Z}[\mathbb{C}(x)^{\times}]$-modules

[TABLE]

where the action of $\mathbb{C}(x)^{\times}$ on the component $\mathcal{P}(\mathbb{C})$ indexed by a given $p$ on the right is given by $f\cdot x:=(-1)^{v_{p}(f)}x$. By analogy with our main theorem below, it is natural to ask whether this map is an isomorphism. (The group $\mathcal{P}(\mathbb{C})$ is known to be a $\mathbb{Q}$-vector space and one would expect the left-hand side also to be uniquely divisible, so that the result should hold without the need to invert $2$.)

Acknowledgements. I thank the referee for a very careful and thorough reading of the article, and in particular for identifying a gap (now filled) in the proof of Theorem 3.11.

1.1. Layout of the article

In Section 2 we review some of the relevant known results about scissors congruence groups and their relation to the third homology of $\mathrm{SL}_{2}$ of fields. We introduce here the module $\mathcal{RP}_{+}(F)$ associated to a field $F$, which coincides with module $\mathcal{RP}_{1}(F)$ on tensoring with ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$, but has the advantage of being a quotient rather than a submodule of ${\mathcal{RP}}(F)$, and thus is defined by a presentation. Our main results in the article depend on computations in $\mathcal{RP}_{+}(F)$.

In Section 3, we use the algebraic properties of the refined scissors congruence groups to calculate $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ for fields $F$ which are complete with respect to a discrete valuation. The results of this section give an update and a strengthening of the main results of [7].

In Section 4 we specialize to the case of the field $\mathbb{Q}$ and state the main theorem.

Section 5 contains the proof of the main theorem (Theorem 4.3) using the results and methods outlined in Sections 3 and 4.

In Section 6, we describe some further applications of the main theorem; for example, the calculation of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left({\mathbb{Q}}[{t},{t}^{-1}]\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ and an explicit description of a basis for the $\mathbb{F}_{3}$-vector space elements of order dividing $3$ in $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$.

1.2. Notation

For a commutative unital ring $R$, ${R}^{\times}$ denotes the group of units of $R$.

For any abelian group $A$, we denote $A\otimes{\mathbb{Z}}\left[\frac{1}{n}\right]$ by ${A}\left[\frac{1}{n}\right]$. For any prime $p$, ${A}_{(p)}$ denotes the vector space $\{a\in A|\ pa=0\}$, of elements of order dividing $p$ in $A$.

If $q$ is a prime power, $\mathbb{F}_{q}$ will denote the finite field with $q$ elements.

For a group $G$ and a $\mathbb{Z}[G]$-module $M$, $M_{G}$ will denote the module of coinvariants; $M_{G}=\mathrm{H}_{0}\left(G,M\right)=M/\mathcal{I}_{G}{M}$, where $\mathcal{I}_{G}$ is the augmentation ideal of $\mathbb{Z}[G]$.

Given an abelian group $G$ we let $\mathrm{S}^{2}_{\mathbb{Z}}(G)$ denote the group

[TABLE]

and, for $x,y\in G$, we denote by $x\circ y$ the image of $x\otimes y$ in $\mathrm{S}^{2}_{\mathbb{Z}}(G)$.

For any rational prime $p$, $v_{p}:\mathbb{Q}^{\times}\to\mathbb{Z}$ denotes the corresponding discrete valuation, determined by $a=p^{v_{p}(a)}\cdot(m/n)$ with $m,n$ not divisible by $p$.

For a field $F$, we let $\mathrm{R}_{F}$ denote the group ring $\mathbb{Z}[{F}^{\times}/({F}^{\times})^{2}]$ of the group of square classes of $F$ and we let $\mathcal{I}_{F}$ denote the augmentation ideal of $\mathrm{R}_{F}$. If $x\in F^{\times}$, we denote the corresponding square-class, considered as an element of $\mathrm{R}_{F}$, by $\left\langle{x}\right\rangle$. The generators $\left\langle{x}\right\rangle-1$ of $\mathcal{I}_{F}$ will be denoted $\left\langle\!\left\langle{x}\right\rangle\!\right\rangle$.

2. Refined scissors congruence groups

In this section we review some of the relevant known facts about the third homology of $\mathrm{SL}_{2}$ of fields and its description in terms of refined scissors congruence groups.

2.1. Indecomposable $K_{3}$

For any field $F$ there is a natural surjective homomorphism

[TABLE]

When $F$ is quadratically closed (i.e. when $F^{\times}=(F^{\times})^{2}$) this map is an isomorphism. However, in general, the group extension

[TABLE]

induces an action – by conjugation – of $F^{\times}$ on $\mathrm{H}_{\bullet}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ which factors through ${F}^{\times}/({F}^{\times})^{2}$. It can be shown that the map (1) is a homomorphism of $\mathrm{R}_{F}$-modules (where ${F}^{\times}/({F}^{\times})^{2}$ acts trivially on $K^{\mathrm{\small ind}}_{3}(F)$) and induces an isomorphism

[TABLE]

(see [11, Proposition 6.4]), but – as our calculations in [4] show – the action of ${F}^{\times}/({F}^{\times})^{2}$ on $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ is in general non-trivial.

Let $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)_{0}$ denote the kernel of the surjective homomorphism ${\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to K^{\mathrm{\small ind}}_{3}(F)}$. This is an $\mathrm{R}_{F}$-submodule of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$. Note that the isomorphism (2) implies that

[TABLE]

Remark 2.1**.**

When $F$ is a number field the surjective homomorphism $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to K^{\mathrm{\small ind}}_{3}(F)$ is split as a map of $\mathbb{Z}$-modules. In fact, $K^{\mathrm{\small ind}}_{3}(F)$ is a finitely generated abelian group and it is enough to show that there is a torsion subgroup of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ mapping isomorphically to the (cyclic) torsion subgroup of $K^{\mathrm{\small ind}}_{3}(F)$. But this latter statement follows from the explicit calculations of C. Zickert in [16, Section 8]. It follows that, as an abelian group,

[TABLE]

for any number field $F$.

However, there is no such decomposition of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ as an $\mathrm{R}_{F}$-module. For details, see Remark 4.7 below.

2.2. Scissors Congruence Groups

For a field $F$, with at least $4$ elements, the scissors congruence group (also called the pre-Bloch group), $\mathcal{P}(F)$, is the group generated by the elements $\left[x\right]$, $x\in F^{\times}$, subject to the relations

[TABLE]

The map

[TABLE]

is well-defined, and the Bloch group of $F$, $\mathcal{B}(F)\subset\mathcal{P}(F)$, is defined to be the kernel of $\lambda$.

For the fields with $2$ and $3$ elements the following definitions allow us to include these fields in the statements of some of our results:

$\mathcal{P}(\mathbb{F}_{2})=\mathcal{B}(\mathbb{F}_{2})$ is a cyclic group of order $3$ with generator denoted $C_{\mathbb{F}_{2}}$. We let $\left[1\right]:=0$ in $\mathcal{P}(\mathbb{F}_{2})$.

$\mathcal{P}(\mathbb{F}_{3})$ is cyclic of order $4$ with generator $\left[-1\right]$. We have $\left[1\right]:=0$ in $\mathcal{P}(\mathbb{F}_{3})$. $\mathcal{B}(\mathbb{F}_{3})$ is the subgroup generated by $2\left[-1\right]$.

We recall (see, for example, [5, Lemma 7.4]):

Lemma 2.2**.**

If $q$ is a prime power then $\mathcal{B}(\mathbb{F}_{q})$ is cyclic of order $(q+1)/2$ when $q$ is odd and $q+1$ when $q$ is even.

The following corollary is particularly relevant to this article:

Corollary 2.3**.**

If $q$ is a prime power then ${\mathcal{P}(\mathbb{F}_{q})}\left[\tfrac{1}{2}\right]$ is cyclic of order $\left(q+1\right)_{\mbox{ \tiny odd}}$.

The Bloch group is closely related to the indecomposable $K_{3}$ of the field $F$:

Theorem 2.4**.**

For any field $F$ there is a natural exact sequence

[TABLE]

where $\widetilde{\mathrm{Tor}^{\mathbb{Z}}_{1}(\mu_{F},\mu_{F})}$ is the unique nontrivial extension of $\mathrm{Tor}^{\mathbb{Z}}_{1}(\mu_{F},\mu_{F})$ by $\mathbb{Z}/2$.

(See Suslin [15] for infinite fields and [5] for finite fields.)

2.3. The refined scissors congruence group

For a field $F$ with at least $4$ elements, ${\mathcal{RP}}(F)$ is defined to be the $\mathrm{R}_{F}$-module with generators $\left[x\right]$, $x\in F^{\times}$ subject to the relations

[TABLE]

Of course, from the definition it follows immediately that

[TABLE]

Let $\Lambda=(\lambda_{1},\lambda_{2})$ be the $\mathrm{R}_{F}$-module homomorphism

[TABLE]

where $\lambda_{1}:{\mathcal{RP}}(F)\to\mathcal{I}_{F}^{2}$ is the map $\left[x\right]\mapsto\left\langle\!\left\langle{1-x}\right\rangle\!\right\rangle\left\langle\!\left\langle{x}\right\rangle\!\right\rangle$, and $\lambda_{2}$ is the composite

[TABLE]

It can be shown that $\Lambda$ is well-defined.

The refined scissors congruence group of $F$ (when $F$ has at least $4$ elements) is the $\mathrm{R}_{F}$-module $\mathcal{RP}_{1}(F):=\mathrm{Ker}(\lambda_{1})$.

The refined Bloch group of the field $F$ (with at least $4$ elements) to be the $\mathrm{R}_{F}$-module

[TABLE]

We can also define appropriate notions for the fields with $2$ and $3$ elements as follows:

$\mathcal{P}(\mathbb{F}_{2})={\mathcal{RP}}(\mathbb{F}_{2})={\mathcal{RB}}(\mathbb{F}_{2})$ is simply an additive group of order $3$ with distinguished generator, denoted $C_{\mathbb{F}_{2}}$.

${\mathcal{RP}}(\mathbb{F}_{3})$ is the cyclic $\mathrm{R}_{\mathbb{F}_{3}}$-module generated by the symbol $\left[-1\right]$ and subject to the one relation

[TABLE]

$\mathcal{P}(\mathbb{F}_{3})=\mathrm{H}_{0}\left(\mathbb{F}_{3}^{\times},{\mathcal{RP}}(\mathbb{F}_{3})\right)$ is then cyclic of order $4$ generated by the symbol $\left[-1\right]$. ${\mathcal{RB}}(\mathbb{F}_{3})$ is the submodule of order $2$ in ${\mathcal{RP}}(\mathbb{F}_{3})$ generated by $\left[-1\right]+\left\langle{-1}\right\rangle\left[-1\right]$.

The symbol $\left[1\right]$ continues to denote [math] in ${\mathcal{RP}}(\mathbb{F}_{2})$ and ${\mathcal{RP}}(\mathbb{F}_{3})$.

We recall some results from [5]: The main result there is

Theorem 2.5**.**

Let $F$ be any field.

There is a natural complex

[TABLE]

which is exact everywhere except possibly at the middle term. The middle homology is annihilated by $4$.

In particular, for any field there is a natural short exact sequence

[TABLE]

2.4. Scissors congruence groups and $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)_{0}$

In [15] Suslin defines the elements $\left\{x\right\}:=[x]+[x^{-1}]\in\mathcal{P}(F)$ and shows that they satisfy

[TABLE]

In particular, $\left\{x\right\}=0$ in ${\mathcal{P}(F)}\left[\tfrac{1}{2}\right]$.

There are two natural liftings of these elements to ${\mathcal{RP}}(F)$: given $x\in F^{\times}$ we define

[TABLE]

and

[TABLE]

(If $F=\mathbb{F}_{2}$, we interpret this as $\psi_{i}\left(1\right)=0$ for $i=1,2$. For $F=\mathbb{F}_{3}$, we have $\psi_{1}\left(-1\right)=\psi_{2}\left(-1\right)=\left[-1\right]+\left\langle{-1}\right\rangle\left[-1\right]$. )

The maps $F^{\times}\to{\mathcal{RP}}(F),x\mapsto\psi_{i}\left(x\right)$ are $1$-cocyles: $\psi_{i}\left(xy\right)=\left\langle{x}\right\rangle\psi_{i}\left(y\right)+\psi_{i}\left(x\right)$ for all $x,y\in F^{\times}$. (See [4, Section 3]). In general, the elements $\psi_{i}(x)$ have infinite order however.

We define $\widetilde{\mathcal{RP}}(F)$ to be ${\mathcal{RP}}(F)$ modulo the submodule generated by the elements $\psi_{1}\left(x\right)$, $x\in F^{\times}$. Likewise, $\widetilde{\mathcal{P}}(F)$ is the group $\mathcal{P}(F)$ modulo the subgroup generated by the elements $\left\{x\right\}$, $x\in F^{\times}$. Note that since the elements $\left\{x\right\}$ are annihilated by $2$, we have ${\mathcal{P}(F)}\left[\tfrac{1}{2}\right]={\widetilde{\mathcal{P}}(F)}\left[\tfrac{1}{2}\right]$.

For any field there is natural homomorphism of $\mathrm{R}_{F}$-modules $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to{\mathcal{RP}}(F)$ and we have ([7, Corollary 2.8, Corollary 4.4]):

Theorem 2.6**.**

For any field $F$, the map $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to{\mathcal{RP}}(F)$ induces an isomorphism of $\mathrm{R}_{F}$-modules

[TABLE]

and furthermore

[TABLE]

where $\mathrm{e}^{-1}_{+}$ denotes the idempotent $(1+\left\langle{-1}\right\rangle)/2\in{\mathrm{R}_{F}}\left[\tfrac{1}{2}\right]$.

Note that it follows that the square class $\left\langle{-1}\right\rangle$ acts trivially on $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$.

To simplify the right-hand side we define the module $\mathcal{RP}_{+}(F)$ to be $\widetilde{\mathcal{RP}}(F)$ modulo the submodule generated by the elements $(1-\left\langle{-1}\right\rangle)\left[x\right]$, $x\in F^{\times}$. Thus $\mathcal{RP}_{+}(F)$ is the $\mathrm{R}_{F}$-module generated by the the elements $\left[x\right],x\in F^{\times}$ subject to the relations

1. (1)

$\left[1\right]=0$ 2. (2)

$S_{x,y}=0$ for $x,y\not=1$ 3. (3)

$\left\langle{-1}\right\rangle\left[x\right]=\left[x\right]$ for all $x$. 4. (4)

$[x^{-1}]=-\left[x\right]$ for all $x$

The theorem then says that the map $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to{\mathcal{RP}}(F)$ induces an isomorphism

[TABLE]

Note that the natural map $\mathcal{RP}_{+}(F)\to\widetilde{\mathcal{P}}(F)$ induces an isomorphism $\mathcal{RP}_{+}(F)_{F^{\times}}\cong\widetilde{\mathcal{P}}(F)$. Furthermore, the results of [5, Section 7] immediately imply that $k^{\times}$ acts trivially on $\mathcal{RP}_{+}(k)$ when $k$ is a finite field. Thus $\mathcal{RP}_{+}(k)=\widetilde{\mathcal{P}}(k)$ for a finite field $k$.

2.5. Some algebra in ${\mathcal{RP}}(F)$

For any field $F$ the elements $C(x):=\left[x\right]+\left\langle{-1}\right\rangle\left[1-x\right]+\left\langle\!\left\langle{1-x}\right\rangle\!\right\rangle\psi_{1}\left(x\right)\in{\mathcal{RP}}(F)$, $x\in F^{\times}\setminus\{1\}$ can be shown to be independent of $x$ (see [4, Section 3.2]). We denote this constant, as well as its image in any quotient of ${\mathcal{RP}}(F)$, by $C_{F}$.

We review some of the fundamental properties of the element $C_{F}\in{\mathcal{RP}}(F)$ (for proofs see [4, Section 3.2]).

Proposition 2.7**.**

The element $C_{F}\in{\mathcal{RP}}(F)$ has the following properties:

1. (1)

$3\cdot C_{F}=\psi_{1}\left(-1\right)$* and $6\cdot C_{F}=0$.* 2. (2)

$2\left\langle\!\left\langle{x}\right\rangle\!\right\rangle C_{F}=\psi_{1}\left(x\right)-\psi_{2}\left(x\right)$* for all $x\in F^{\times}$.* 3. (3)

$2\cdot C_{F}=0$* if and only if $T^{2}-T+1$ splits in $F$. *

Corollary 2.8**.**

For any field $F$, we have $3\cdot C_{F}=0$ in $\widetilde{\mathcal{RP}}(F)$ and

[TABLE]

for all $x\in F^{\times}\setminus\{1\}$.

Proof.

$3\cdot C_{F}=0$ in $\widetilde{\mathcal{RP}}(F)$ since $\psi_{1}\left(-1\right)=0$ in $\widetilde{\mathcal{RP}}(F)$. Thus $-\left\langle\!\left\langle{x}\right\rangle\!\right\rangle C_{F}=2\cdot\left\langle\!\left\langle{x}\right\rangle\!\right\rangle C_{F}=-\psi_{2}\left(x\right)$ since $\psi_{1}\left(x\right)=0$ in $\widetilde{\mathcal{RP}}(F)$. Furthermore, in $\widetilde{\mathcal{RP}}(F)$ we have

[TABLE]

and hence

[TABLE]

∎

Observe that in $\widetilde{\mathcal{RP}}(F)$ we have $C_{F}=\left[x\right]+\left\langle{-1}\right\rangle\left[1-x\right]$ since $\psi_{1}\left(x\right)=0$, and in $\mathcal{RP}_{+}(F)$ we have $C_{F}=\left[x\right]+\left[1-x\right]$ since $\left\langle{-1}\right\rangle$ acts trivially by definition on $\mathcal{RP}_{+}(F)$.

It will be convenient below to introduce the following additional notation in $\mathcal{RP}_{+}(F)$:

[TABLE]

With this notation, we then have

[TABLE]

2.6. A character-theoretic local-global principle

We will use the following character-theoretic principles:

Let $G$ be an abelian group satisfying $g^{2}=1$ for all $g\in G$. Let $\mathcal{R}$ denote the group ring ${\mathbb{Z}}[G]$.

For a character $\chi\in\widehat{G}:=\mathrm{Hom}(G,\mu_{2})$, let ${\mathcal{R}}^{\chi}$ be the ideal of $\mathcal{R}$ generated by the elements $\{g-\chi(g)\ |\ g\in G\}$. In other words ${\mathcal{R}}^{\chi}$ is the kernel of the ring homomorphism $\rho(\chi):\mathcal{R}\to\mathbb{Z}$ sending $g$ to $\chi(g)$ for any $g\in G$. We let ${\mathcal{R}}_{\chi}$ denote the associated $\mathcal{R}$-algebra structure on $\mathbb{Z}$; i.e. ${\mathcal{R}}_{\chi}:=\mathcal{R}/{\mathcal{R}}^{\chi}$.

If $M$ is an $\mathcal{R}$-module, we let ${M}^{\chi}={\mathcal{R}}^{\chi}M$ and we let

[TABLE]

Thus ${M}_{\chi}$ is the largest quotient module of $M$ with the property that $g\cdot m=\chi(g)\cdot m$ for all $g\in G$.

In particular, if $\chi=\chi_{0}$, the trivial character, then ${\mathcal{R}}^{\chi_{0}}$ is the augmentation ideal $\mathcal{I}_{G}$, ${M}^{\chi_{0}}=\mathcal{I}_{G}M$ and ${M}_{\chi_{0}}=M_{G}$.

Given $m\in M$, $\chi\in\widehat{G}$, we denote the image of $m$ in ${M}_{\chi}$ by ${m}_{\chi}$. For example, for any character $\chi\in\widehat{F^{\times}/(F^{\times})^{2}}$, we can give a presentation of the $\mathrm{R}_{F}$-module ${\mathcal{RP}_{+}(F)}_{\chi}$, which is our main object of study below, as follows: It is the $\mathrm{R}_{F}$-module with generators ${\left[x\right]}_{\chi}$, $x\in F^{\times}$, subject to the relations

1. (1)

$\left\langle{a}\right\rangle\cdot{\left[x\right]}_{\chi}:=\chi(a){\left[x\right]}_{\chi}$ for all $a,x\in F^{\times}$ 2. (2)

${\left[1\right]}_{\chi}=0$ 3. (3)

[TABLE]

for all $x,y\not=1$ 4. (4)

$\chi(-1){\left[x\right]}_{\chi}={\left[x\right]}_{\chi}$ for all $x$, and 5. (5)

${\left[x\right]}_{\chi}=-{\left[x^{-1}\right]}_{\chi}$ for all $x$.

We will need the following result ([7, Section 3])

Proposition 2.9**.**

1. (1)

For any $\chi\in\widehat{G}$, $M\to{M}_{\chi}$ is an exact functor on the category of ${\mathcal{R}}\left[\tfrac{1}{2}\right]$-modules. 2. (2)

Let $f:M\to N$ be an ${\mathcal{R}}\left[\tfrac{1}{2}\right]$-module homomorphism. For any $\chi\in\widehat{G}$, let ${f}_{\chi}:{M}_{\chi}\to{N}_{\chi}$. Then $f$ is bijective (resp. injective, surjective) if and only if ${f}_{\chi}$ is bijective (resp. injective, surjective) for all $\chi\in\widehat{G}$.

Corollary 2.10**.**

For any field $F$ and any $\chi_{0}\not=\chi\in\widehat{F^{\times}/(F^{\times})^{2}}$, the natural $\mathrm{R}_{F}$-homomorphism $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to\mathcal{RP}_{+}(F)$ induces an isomorphism

[TABLE]

Proof.

Since $\chi\not=\chi_{0}$, there exists $x\in F^{\times}$ with $\chi(x)=-1$ and hence for any $\mathrm{R}_{F}$-module $M$ we have ${\left({M_{F^{\times}}}\left[\tfrac{1}{2}\right]\right)}_{\chi}=0$. Applying the functor ${(-)}_{\chi}$ to the exact sequence $0\to\mathcal{I}_{F}M\to M\to M_{F^{\times}}\to 0$ thus shows that ${{M}\left[\tfrac{1}{2}\right]}_{\chi}={\left(\mathcal{I}_{F}{M}\left[\tfrac{1}{2}\right]\right)}_{\chi}$. The stated result thus follows from the isomorphism of $\mathrm{R}_{F}$-modules

[TABLE]

(Theorem 2.6). ∎

The following lemma will play a central role in all that follows:

Lemma 2.11**.**

Let $\chi\in\widehat{F^{\times}/(F^{\times})^{2}}$. If $x\in F^{\times}$ with $\chi(x)=-1$, then

[TABLE]

Proof.

From Corollary 2.8, we have

[TABLE]

However, we can suppose that $\chi(-1)=1$, since otherwise ${\mathcal{RP}_{+}(F)}_{\chi}=0$, and hence $\chi(-x)=\chi(x)=-1$. Thus, we obtain

[TABLE]

∎

3. Fields with a valuation

3.1. Valuations and the modules $\mathcal{L}_{v}$

Given a field $F$ and a (surjective) valuation $v:F^{\times}\to\Gamma$, where $\Gamma$ is a totally ordered additive abelian group, we let $\mathcal{O}_{v}:=\{x\in F^{\times}\ |\ v(x)\geq 0\}\cup\{0\}$ be the associated valuation ring, with maximal ideal $\mathcal{M}_{v}=\{x\in\mathcal{O}_{v}\ |\ v(x)\not=0\}$, group of units $U_{v}:=\mathcal{O}_{v}\setminus\mathcal{M}_{v}$ and residue field $k=k(v):=\mathcal{O}_{v}/\mathcal{M}_{v}$.

Given an $\mathrm{R}_{k}$-module $M$, we denote by $\mathrm{Ind}^{F}_{k}M$ the $\mathrm{R}_{F}$-module $\mathrm{R}_{F}\otimes_{\mathbb{Z}[U_{v}]}M$ (noting that the ring $\mathbb{Z}[U_{v}]$ surjects naturally onto $\mathrm{R}_{k}$ and maps naturally to $\mathrm{R}_{F})$.

We recall the following result (see [7, Section 5]):

Lemma 3.1**.**

There is a natural homomorphism of $\mathrm{R}_{F}$-modules $S_{v}:\widetilde{\mathcal{RP}}(F)\to\mathrm{Ind}^{F}_{k}\widetilde{\mathcal{RP}}(k)$ given by

[TABLE]

Now let $\mathcal{L}_{v}\subset\widetilde{\mathcal{RP}}(F)$ be the $\mathrm{R}_{F}$-submodule generated by $\{\left[u\right]\ |\ u\in U_{1}=U_{1,v}:=1+\mathcal{M}_{v}\subset U_{v}\}$. (Caution: This is a slightly different definition from that given in [7].)

The following is a refinement of [7, Lemma 5.2]:

Proposition 3.2**.**

Given a valuation $v$ on the field $F$, there is a natural short exact sequence of $\mathrm{R}_{F}$-modules

[TABLE]

Proof.

Certainly, $\mathcal{L}_{v}\subset\mathrm{Ker}(S_{v})$ and, since $S_{v}$ is clearly surjective, it induces a surjective homomorphism of $\mathrm{R}_{F}$-modules

[TABLE]

To prove the Proposition it will thus suffice to construct an $\mathrm{R}_{F}$-module homomorphism $T_{v}:\mathrm{Ind}^{F}_{k}\widetilde{\mathcal{RP}}(k)\to{\widetilde{\mathcal{RP}}(F)}_{v}$ satisfying $T_{v}\circ S_{v}=\mathrm{Id}_{{\widetilde{\mathcal{RP}}(F)}_{v}}$.

We will require the following three lemmas:

Lemma 3.3**.**

If $v(x)\not=0$, then

[TABLE]

Proof of Lemma 3.3: If $v(x)>0$, then $C_{F}=\left[x\right]+\left\langle{-1}\right\rangle\left[1-x\right]=\left[x\right]$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$ since $1-x\in U_{1}$.

If $v(x)<0$, then $v(x^{-1})>0$ and $C_{F}=\left\langle{-1}\right\rangle C_{F}=\left\langle{-1}\right\rangle\left[x^{-1}\right]=-\left[x\right]$, since $0=\psi_{1}\left(x\right)=\left[x\right]+\left\langle{-1}\right\rangle\left[x^{-1}\right]$ in $\widetilde{\mathcal{RP}}(F)$.

Lemma 3.4**.**

For all $x\in F^{\times}$, $u\in U_{1}$, $\left[x\right]=\left[xu\right]$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$.

Proof of Lemma 3.4: Since $\left[u\right]=0$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$ for all $u\in U_{1}$, we may assume $x\not\in U_{1}$. Then

[TABLE]

But this implies $\left[x\right]=\left[xu\right]$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$ since $u,(1-x)/(1-xu)\in U_{1}$.

Lemma 3.5**.**

For all $x\in F^{\times}$, $u\in U_{1}$ we have $\left\langle{u}\right\rangle\left[x\right]=\left[x\right]$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$.

Proof of Lemma 3.5: Let $u\in U_{1}$, $u\not=1$. By Corollary 2.8, $\left\langle\!\left\langle{u}\right\rangle\!\right\rangle C_{F}=\left\langle{u-1}\right\rangle\left\langle\!\left\langle{-u}\right\rangle\!\right\rangle\left[u\right]=0$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$; i.e., $\left\langle{u}\right\rangle C_{F}=C_{F}$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$ for all $u\in U_{1}$.

Now for $x\in F^{\times}$, $u\in U_{1}$ (and $x\not=1$, $xu\not=1$) we have the following sequence of identities in ${\widetilde{\mathcal{RP}}(F)}_{v}$:

[TABLE]

and hence

[TABLE]

proving the Lemma.

By Lemma 3.5, the $\mathbb{Z}[U_{v}]$-action on ${\widetilde{\mathcal{RP}}(F)}_{v}$ descends to an $\mathrm{R}_{k}$-module structure. Combining this with Lemma 3.4, there is a well-defined $\mathrm{R}_{k}$-module homomorphism

[TABLE]

Thus there is an induced $\mathrm{R}_{F}$-module homomorphism

[TABLE]

Now, by choosing $u\in U_{v}\setminus U_{1}$, and noting that then $C_{k}=\left[\bar{u}\right]+\left\langle{-1}\right\rangle\left[1-\bar{u}\right]$, we see that $T_{v}(1\otimes C_{k})=C_{F}$. Hence if $v(x)\not=0$ we have $T_{v}(S_{v}(\left[x\right]))=\left[x\right]$ in ${\widetilde{\mathcal{RP}}(F)}_{v}$ by Lemma 3.3. On, the other hand, if $u\in U_{v}$, then $T_{v}(S_{v}(\left[u\right]))=\left[u\right]$ (using Lemma 3.4 again), so that $T_{v}\circ S_{v}=\mathrm{id}_{{\widetilde{\mathcal{RP}}(F)}_{v}}$ as required. ∎

Tensoring with ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$, taking the $\mathrm{e}^{-1}_{+}$-component and using Theorem 2.6 we deduce

Corollary 3.6**.**

There is a natural short exact sequence of $\mathrm{R}_{F}$-modules

[TABLE]

3.2. Discrete valuations and the specialization homomorphism

Suppose that $v:F^{\times}\to\mathbb{Z}$ is a discrete valuation on the field $F$ with residue field $k=k(v)$. Let $\chi_{v}:F^{\times}/(F^{\times})^{2}\to\mu_{2}$ denote the associated character defined by $\chi_{v}(a)=(-1)^{v(a)}$.

For an abelian group $M$, we let ${M}\left\{v\right\}$ denote the $\mathrm{R}_{F}$-module $\mathrm{R}_{\chi_{v}}\otimes_{\mathbb{Z}}M$. Equivalently, we equip $M$ with the $\mathrm{R}_{F}$-module structure $\left\langle{a}\right\rangle m:=(-1)^{v(a)}m$.

Theorem 3.7**.**

Let $F$ be a field with discrete valuation $v:F^{\times}\to\mathbb{Z}$ and residue field $k$. Then we have natural isomorphisms

[TABLE]

Proof.

By Corollary 3.6, to prove the left-hand isomorphism, we must prove that ${\left(\mathrm{e}^{-1}_{+}{\mathcal{L}_{v}}\left[\tfrac{1}{2}\right]\right)}_{\chi_{v}}=0$; i.e., we must prove that ${\left[u\right]}_{\chi_{v}}=0$ in ${\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]$ for all $u\in U_{1}$. This, in turn, follows from

Lemma 3.8**.**

${\left[x\right]}_{\chi_{v}}=C_{F}$* in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi_{v}}$ whenever $v(x)>0$.*

For, given this lemma, if $u\in U_{1}$ then $v(1-u)>0$ and hence

[TABLE]

Proof of Lemma 3.8: Suppose that $x\in F^{\times}$ with $v(x)>0$. If $v(x)$ is odd then $\chi_{v}(x)=-1$ and hence

[TABLE]

by Lemma 2.11.

Suppose, on the other hand, that $v(x)=2k$ with $k\geq 1$. Let $\pi\in F^{\times}$ with $v(\pi)=1$. So $x=\pi^{2k}u$ for some $u\in U_{v}$. Let $y=\pi u$ and $z=\pi^{1-2k}\in F$. Then in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi_{v}}$ we have

[TABLE]

We can identify all of the terms ocurring (except for ${\left[x\right]}_{\chi_{v}}$) using the case of odd valuation:

$v(y)=1\implies{\left[y\right]}_{\chi_{v}}=C_{F}$.

$v(z^{-1})=2k-1\implies{\left[z^{-1}\right]}_{\chi_{v}}=C_{F}\implies{\left[z\right]}_{\chi_{v}}=-C_{F}$ since $\left[a\right]=-\left[a^{-1}\right]$ in $\mathcal{RP}_{+}(F)$.

[TABLE]

and thus

[TABLE]

Futhermore

[TABLE]

Since $\chi_{v}(z)=-1=\chi_{v}(1-z)$ and $\chi_{v}(1-z^{-1})=1$, we therefore deduce

[TABLE]

proving Lemma 3.8 (since $3\cdot C_{F}=0$ in $\mathcal{RP}_{+}(F)$).

The second isomorphism of the theorem follows from the general calculation for any $\mathrm{R}_{k}$-module $M$:

[TABLE]

∎

Remark 3.9**.**

The isomorphism ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi_{v}}\cong{{\mathcal{P}(k)}\left[\tfrac{1}{2}\right]}\left\{v\right\}$ of Theorem 3.7 is induced by the map of $\mathrm{R}_{F}$-modules

[TABLE]

Remark 3.10**.**

Observe that this map makes sense when $k(v)=\mathbb{F}_{2}$ or $\mathbb{F}_{3}$.

3.3. Fields complete with respect to a discrete valuation

Let $F$ be a field with discrete valuation $v:F^{\times}\to\mathbb{Z}$, maximal ideal $\mathcal{M}_{v}$ and residue field $k$. For $n\geq 1$, let $U_{n}$ denote the subgroup $1+\mathcal{M}_{v}^{n}$ of $U=U_{v}$.

Observe that if $F$ is complete with respect to the valuation $v$ and if muliplication by $m$ is invertible on $k$, then $U_{1}^{m}=U_{1}$ since $U_{n}/U_{n+1}\cong k$ for all $n$. Taking $m=2$, we deduce that $U_{1}=U_{1}^{2}$ whenever $F$ is complete with residue characteristic not equal to $2$. On the other hand, for the field $\mathbb{Q}_{2}$, one has $U_{1}^{2}=U_{3}$. More generally, if $F$ is any finite degree extension of $\mathbb{Q}_{2}$ it is easily seen that there exists some $n>0$ such that $U_{n}\subset U_{1}^{2}$. On the other hand, for the complete field $F=\mathbb{F}_{2}((x))$, $U_{1}^{2}$ has infinite index in $U_{1}$ and hence $U_{n}\not\subset U_{1}^{2}$ for all $n>0$.

The following significantly improves [7, Theorem 6.1]:

Theorem 3.11**.**

Let $v:F^{\times}\to\mathbb{Z}$ be a discrete valuation on the field $F$. Suppose that there exists $n>0$ such that $U_{n}\subset U_{1}^{2}$. Then the homomorphism $S_{v}:\mathcal{RP}_{+}(F)\to\mathrm{Ind}^{F}_{k}{\mathcal{RP}_{+}(k)}$ induces an isomorphism of $\mathrm{R}_{F}$-modules

[TABLE]

Proof.

By Proposition 2.9 and Corollary 3.6, this is equivalent to the statement ${{\mathcal{L}_{v}}\left[\tfrac{1}{2}\right]}_{\chi}=0$ for all characters $\chi\in\widehat{F^{\times}/(F^{\times})^{2}}$ satisfying $\chi(-1)=1$ and $\chi\not=\chi_{0}$ (the trivial character). To see this, apply the exact functor ${(\ )}_{\chi}$ to the exact sequence of Corollary 3.6 and observe that for any $\chi$ and for any ${\mathrm{R}_{F}}\left[\tfrac{1}{2}\right]$-module $M$ we have

[TABLE]

Now, if $\chi\not=\chi_{0}$ and $\chi|_{U}$ is trivial then necessarily $\chi=\chi_{v}$ and this is Theorem 3.7.

So we can suppose that there exists $u\in U$ with $\chi(u)=-1$. Since $U_{n}\subset U_{1}^{2}$ we have $\chi(u)=1$ for all $u\in U_{n}$. Let $n_{0}\geq 1$ be minimal such that $\chi(u)=1$ for all $u\in U_{n_{0}}$. Thus there exists $u\in U_{n_{0}-1}$ with $\chi(u)=-1$ (where $U_{0}:=U$). Since

[TABLE]

replacing $u$ by $u^{-1}$ if necessary, we can suppose that $\chi(1-u)=-1=\chi(u)$. Let $p:=1-u$. So $\chi(p)=-1$ and $v(p)=n_{0}-1\geq 0$.

Let $a\in F^{\times}$ with $v(a)>0$. We will prove that ${\left[a\right]}_{\chi}=C_{F}$ (from which the required result follows as in the proof of Theorem 3.7):

First consider the case $\chi(a)=1$. In ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$ we have

[TABLE]

where $w:=(1-p)/(1-ap)$ (and using $\chi(p)=-1=\chi(1-p)$, $\chi(1-p^{-1})=1$).

Since $\chi(p)=\chi(1-p)=-1$, we have ${\left[p\right]}_{\chi}=\chi(1-p)C_{F}=-C_{F}$ by Lemma 2.11. Similarly, $\chi(ap)=\chi(a)\chi(p)=-1$ while $\chi(1-ap)=1$, since $1-ap\in U_{n_{0}}$, so ${\left[ap\right]}_{\chi}=C_{F}$.

We have $\chi(w)=\chi(1-p)\chi(1-ap)=-1$. So ${\left[w\right]}_{\chi}=\chi(1-w)C_{F}$. But

[TABLE]

So ${\left[w\right]}_{\chi}=-\chi(1-a)C_{F}$.

Finally, $\chi(aw)=\chi(a)\chi(w)=-1$. So ${\left[aw\right]}_{\chi}=\chi(1-aw)C_{F}$. But $1-aw=(1-a)/(1-ap)$ so that $\chi(1-aw)=\chi(1-a)$. We deduce

[TABLE]

which forces ${\left[a\right]}_{\chi}=C_{F}$, as required.

We now comsider the case $\chi(a)=-1$. Then ${\left[a\right]}_{\chi}=\chi(1-a)C_{F}$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$ by Lemma 2.11. If $\chi(1-a)=1$ this gives the required conclusion.

This leaves us with the case that $\chi(a)=-1=\chi(1-a)$. We have ${\left[a\right]}_{\chi}=-C_{F}$ by Lemma 2.11. Consider again the identity

[TABLE]

in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$. We have ${\left[a\right]}_{\chi}={\left[p\right]}_{\chi}=-C_{F}$. Since $\chi(w)=\chi(1-p)\chi(1-ap)=-1$. $\chi(1-w)=-\chi(1-a)=1$ we have ${\left[w\right]}_{\chi}=C_{F}$ by Lemma 2.11. Furthermore, $\chi(aw)=1$ and $\chi(1-aw)=\chi(1-a)\chi(1-ap)=-1$ gives ${\left[1-aw\right]}_{\chi}=C_{F}$ and hence ${\left[aw\right]}_{\chi}=0$. We conclude that $0=-{\left[ap\right]}_{\chi}-C_{F}$ and hence

[TABLE]

On the other hand, note that $\chi(ap)=1$ and $v(ap)>0$ so that

[TABLE]

by the case $\chi(a)=1$ above. Comparing (4) and (5), we conclude that $C_{F}=0$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$ and so the required identity ${\left[a\right]}_{\chi}=C_{F}$ holds in this case also. ∎

Corollary 3.12**.**

Let $v:F^{\times}\to\mathbb{Z}$ be a discrete valuation on the field $F$. Let $\chi\not=\chi_{0}\in\widehat{{F}^{\times}/({F}^{\times})^{2}}$. Suppose that $\chi(-1)=1$ and that there exists $n>0$ such that $\chi|_{U_{n}}=1$. Then

1. (1)

[TABLE] 2. (2)

Suppose further that $\chi|_{U_{1}}\not=1$. Then $C_{F}=0$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$.

Proof.

1. (1)

This follows from the proof of Theorem 3.11 since the condition $U_{n}\subset U_{1}^{2}$ is only used to ensure that $\chi|_{U_{n}}=1$ for any given $\chi$. 2. (2)

Suppose that there exists $u\in U_{1}$ such that $\chi(u)=-1$. Let $a=1-u$. replacing $u$ by $u^{-1}$ if necessary, we can suppose that $\chi(a)=-1$. Thus $v(a)>0$ and $\chi(a)=\chi(1-a)=-1$. By the last case considered in the proof of Theorem 3.11, it follows that $C_{F}=0$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$.

∎

Corollary 3.13**.**

Let $v:F^{\times}\to\mathbb{Z}$ be a discrete valuation on the field $F$ with residue field $k$. Suppose that there exists $n>0$ such that $U_{n}\subset U_{1}^{2}$. Then there is an isomorphism of $\mathrm{R}_{k}$-modules

[TABLE]

Proof.

By Theorem 3.11

[TABLE]

(as $\mathrm{R}_{F}$-modules).

By [7, Lemma 5.4], there is an $\mathrm{R}_{k}$-module isomorphism

[TABLE]

∎

Corollary 3.14**.**

Let $F$ be a field with discrete valuation $v$ satisfying

1. (1)

the residue field $k=k(v)$ is either finite or quadratically closed or real closed, and 2. (2)

there exists $n\geq 1$ such that $U_{n}\subset U_{1}^{2}$.

Then we have natural isomorphisms of $\mathrm{R}_{F}$-modules

[TABLE]

and thus there is a (split) exact sequence of $\mathrm{R}_{F}$-modules

[TABLE]

Proof.

Recall first that $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\operatorname{SL}_{2}\left(F\right)\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)_{0}\cong\mathcal{I}_{F}{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]$ as $\mathrm{R}_{F}$-module by Theorem 2.6.

On the one hand, by Theorem 3.7, the map $S_{v}$ induces an isomorphism of $\mathrm{R}_{F}$-modules.

[TABLE]

On the other hand, Theorem 3.11 gives an isomorphism of $\mathrm{R}_{F}$-modules

[TABLE]

The conditions on the residue field $k$ imply that $\mathcal{I}_{k}{\mathcal{RP}_{+}(k)}\left[\tfrac{1}{2}\right]=0$ and hence that ${\mathcal{RP}_{+}(k)}\left[\tfrac{1}{2}\right]={\mathcal{P}(k)}\left[\tfrac{1}{2}\right]$ with trivial $\mathrm{R}_{k}$-module structure. Thus the result follows from:

Lemma 3.15**.**

Let $M$ be an $\mathrm{R}_{k}$-module with trivial action of $k^{\times}$. Then

[TABLE]

Proof of Lemma 3.15: By Proposition 2.9, to prove that the natural homomorphism, $S$ say,

[TABLE]

is an isomorphism, it is enough to prove that ${S}_{\chi}$ is an isomorphism for all $\chi\in\widehat{F^{\times}/(F^{\times})^{2}}$.

When $\chi=\chi_{0}$ both domain and target of ${S}_{\chi}$ are [math]. Likewise, if there exists $u\in U_{v}$ for which $\chi(u)=-1$, then (since $M$ has trivial $\mathbb{Z}[U]$-module structure by hypothesis), $\left\langle{u}\right\rangle$ acts as multiplication by both $1$ and $-1$ on the target and domain, so that they vanish. This leaves only $\chi=\chi_{v}$, and ${S}_{\chi_{v}}$ is tautologically an isomorphism. ∎

Remark 3.16**.**

Note that the $\mathrm{R}_{F}$-module direct sum decomposition in Corollary 3.14 is just the decomposition into $+1$ and $-1$-eigenspaces for the action of $\left\langle{\pi}\right\rangle$ where $\pi$ is any element of $F$ with $v(\pi)=1$.

Example 3.17**.**

Let $F$ be a local field with finite residue field. Suppose that either $F$ has characteristic [math] or $\mathrm{char}(k)\not=2$. Then

[TABLE]

as $\mathrm{R}_{F}$-modules. In particular, for all primes $p$ we have

[TABLE]

Example 3.18**.**

Consider the case $F={\mathbb{C}}\left(\!\left(x\right)\!\right)$. Then we have

[TABLE]

(since $\mathcal{P}(\mathbb{C})$ is a $\mathbb{Q}$-vector space).

4. The field $\mathbb{Q}$

For a field $F$ with discrete valuation $v$, we let $\bar{S}_{v}$ denote the composite $\mathrm{R}_{F}$-module homomorphism

[TABLE]

(See remark 3.10 above.) By abuse of notation, we will use the same symbol for the $\bar{S}_{v}$ restricted to $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)_{0}$.

Theorem 4.1**.**

[4*, Theorem 5.1]**

Let $F$ be a field and let $\mathcal{V}$ be a family of discrete valuations on $F$ satisfying*

1. (1)

For any $x\in F^{\times}$, $v(x)=0$ for all but finitely many $v\in\mathcal{V}$. 2. (2)

The map

[TABLE]

is surjective.

Then the maps $\{\bar{S}_{v}\}_{v\in\mathcal{V}}$ induce a natural surjective homomorphism

[TABLE]

Remark 4.2**.**

On the face of it, the collection of maps $\{\bar{S}_{v}\}_{v\in\mathcal{V}}$ as above induces an $\mathrm{R}_{F}$-module homomorphism with target the product – rather than the direct sum – of the scissors congruence groups:

[TABLE]

However, when we restrict to $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)_{0}$ and tensor with ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$ the image lies in the direct sum instead, in view of the isomorphism $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)_{0}\cong\mathcal{I}_{F}{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]$ and the fact that $S_{v}(\left\langle\!\left\langle{a}\right\rangle\!\right\rangle\left[b\right])=\left\langle\!\left\langle{a}\right\rangle\!\right\rangle S_{v}(\left[b\right])=0$ whenever $v(a)$ is even.

Specializing to the case $F=\mathbb{Q}$ and $\mathcal{V}=\mathrm{Primes}$, the set of all primes, we obtain a surjective homomorphism $\bar{S}=\{\bar{S}_{p}\}_{p\in\mathrm{Primes}}$ of $\mathrm{R}_{\mathbb{Q}}$-modules

[TABLE]

In the next section we will prove the following main theorem:

Theorem 4.3**.**

The map

[TABLE]

is an isomorphism of $\mathrm{R}^{+}_{\mathbb{Q}}$-modules.

Remark 4.4**.**

Since $\left\langle{-1}\right\rangle\in\mathrm{R}_{\mathbb{Q}}$ acts trivially on both of the modules in (6), this is a map of $\mathrm{R}^{+}_{\mathbb{Q}}$-modules where

[TABLE]

Remark 4.5**.**

We note that the image of the map

[TABLE]

does not lie in the direct sum $\bigoplus_{p\in\mathrm{Primes}}{{\mathcal{P}(\mathbb{F}_{p})}\left[\tfrac{1}{2}\right]}\left\{p\right\}$:

Let $t\in\operatorname{SL}_{2}\left(\mathbb{Z}\right)$ be the element of order $3$

[TABLE]

Denote (also) by $C_{\mathbb{Q}}\in\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$ the image of $1\in\mathbb{Z}/3=\mathrm{H}_{3}\left(\left\langle{t}\right\rangle,\mathbb{Z}\right)$ under the map induced by the inclusion $\left\langle{t}\right\rangle\to\operatorname{SL}_{2}\left(\mathbb{Q}\right)$. Then $C_{\mathbb{Q}}\in\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$ maps to $C_{\mathbb{Q}}\in\mathcal{RP}_{+}(\mathbb{Q})$ ([4, Remark 3.14]). Note that $\bar{S}_{p}(C_{\mathbb{Q}})=C_{\mathbb{F}_{p}}\in\widetilde{\mathcal{P}}(\mathbb{F}_{p})$ for all primes $p$. Furthermore, $C_{\mathbb{F}_{p}}\not=0$ precisely when $p\equiv 2\mod{3}$ (i.e., precisely when $3|p+1$), by [5, Lemma 7.11].

In particular, the image of $C_{\mathbb{Q}}$ under the map $\{\bar{S}_{p}\}_{p}$ lies in the product, but not the direct sum, of the scissors congruence groups of the residue fields.

Remark 4.6**.**

In view of Corollary 3.14 above, Theorem 4.3 can be stated equivalently as follows: The natural homomorphism $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)\to\prod_{p}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}_{p}\right),\mathbb{Z}\right)$ induces an isomorphism

[TABLE]

Remark 4.7**.**

We observe that – unlike in the local case – the short exact sequence of $\mathrm{R}_{\mathbb{Q}}$-modules

[TABLE]

has no $\mathrm{R}_{\mathbb{Q}}$-splitting (it is $\mathbb{Z}$-split, however):

Suslin’s map gives a canonical isomorphism ${K^{\mathrm{\small ind}}_{3}(\mathbb{Q})}_{(3)}\cong{\mathcal{B}(\mathbb{Q})}_{(3)}=\mathbb{Z}/3\cdot C_{\mathbb{Q}}\subset\widetilde{\mathcal{P}}(\mathbb{Q})$ and we can let $C_{\mathbb{Q}}$ also denote the corresponding element of $K^{\mathrm{\small ind}}_{3}(\mathbb{Q})$.

Recall that $\mathrm{R}_{\mathbb{Q}}$ acts trivially on $K^{\mathrm{\small ind}}_{3}(\mathbb{Q})$. Suppose that there were an $\mathrm{R}_{\mathbb{Q}}$-module splitting $j:{K^{\mathrm{\small ind}}_{3}(\mathbb{Q})}\left[\tfrac{1}{2}\right]\to\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$. Then we would have $j(C_{\mathbb{Q}})=C_{\mathbb{Q}}+h$ for some $h\in\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)_{0}$. We must have $\mathrm{R}_{\mathbb{Q}}$ acts trivially on $j(C_{\mathbb{Q}})$ and hence $\left\langle\!\left\langle{p}\right\rangle\!\right\rangle j(C_{\mathbb{Q}})=\left\langle\!\left\langle{p}\right\rangle\!\right\rangle(C_{\mathbb{Q}}+h)=0$ for all primes $p$. However, we can choose a prime $p$ such that $\left\langle\!\left\langle{p}\right\rangle\!\right\rangle h=0$ in $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)_{0}$ and $p\equiv 2\pmod{3}$. Then $\bar{S}_{p}(\left\langle\!\left\langle{p}\right\rangle\!\right\rangle(C_{\mathbb{Q}}+h))=\left\langle\!\left\langle{p}\right\rangle\!\right\rangle C_{\mathbb{F}_{p}}=-2C_{\mathbb{F}_{p}}\not=0$, giving us a contradiction. So no such $\mathrm{R}_{\mathbb{Q}}$-splitting $j$ can exist.

5. Proof of the Main Theorem

In this section we prove Theorem 4.3.

Recall now that $\mathrm{R}^{+}_{\mathbb{Q}}=\mathbb{Z}[G]$ where $G=\mathbb{Q}_{+}/\mathbb{Q}_{+}^{2}=\mathbb{Q}^{\times}/\pm(\mathbb{Q}^{\times})^{2}$. As a multiplicative $\mathbb{F}_{2}$-space, the set of all primes form a (number-theoretically) natural basis of $\mathbb{Q}_{+}/\mathbb{Q}_{+}^{2}$. Thus the space of characters $\widehat{\mathbb{Q}_{+}/\mathbb{Q}_{+}^{2}}$ is naturally parametrised by the subsets of the set $\mathrm{Primes}$ of positive prime numbers: if $S\subset\mathrm{Primes}$ then the corresponding character $\chi_{S}$ is defined by

[TABLE]

for all $p\in\mathrm{Primes}$ or, equivalently,

[TABLE]

for all $x\in\mathbb{Q}^{\times}$. Conversely, the character $\chi$ corresponds to the subset

[TABLE]

(Thus, for a prime number $p$, $\chi_{p}$ is the unique character satisfying $\mathrm{supp}(\chi_{p})=\{p\}$.)

The following lemma is immediate from the definition of the $\mathrm{R}_{\mathbb{Q}}$-module structure on ${\mathcal{P}(\mathbb{F}_{p})}\left\{p\right\}$.

Lemma 5.1**.**

Let $\chi\in\widehat{\mathbb{Q}_{+}/\mathbb{Q}_{+}^{2}}$. Let $p$ be a prime. Then

[TABLE]

Corollary 5.2**.**

For $\chi\in\widehat{\mathbb{Q}_{+}/\mathbb{Q}_{+}^{2}}$ we have

[TABLE]

It thus follows from Proposition 2.9 that to prove Theorem 4.3 it is enough to prove that, for any prime $p$, $\bar{S}_{p}$ induces an isomorphism

[TABLE]

for any prime $p$, while

[TABLE]

whenever $\mathrm{supp}(\chi)$ contains at least two distinct primes. The first of these statements is an immediate consequence of Theorem 3.7 above. The second is Corollary 5.8 below.

Lemma 5.3**.**

Let $F$ be a field. Let $\chi\in\widehat{{F}^{\times}/({F}^{\times})^{2}}$. Suppose that $a\in F^{\times}$ satisfies $\chi(1-a)=-1$. Then ${\left[a\right]}_{\chi}=0$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$.

Proof.

If $\chi(-1)=-1$ we have ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}=0$. So we can suppose $\chi(-1)=1$. In this case ${\left[1-a\right]}_{\chi}=C_{F}$ by Lemma 2.11. But $C_{F}={\left[a\right]}_{\chi}+{\left[1-a\right]}_{\chi}$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$. ∎

Lemma 5.4**.**

Let $F$ be a field. Let $\chi\in\widehat{{F}^{\times}/({F}^{\times})^{2}}$ with $\chi(-1)=1$. Let $\ell\in F^{\times}$ satisfy $\chi(\ell)=-1$ and $\chi(1-\ell)=1$. Then

[TABLE]

in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$ for all $a\in\mathbb{P}^{1}(F)$.

Proof.

Observe that ${\left[1-\ell\right]}_{\chi}=0$ by Lemma 5.3. In particular, the result holds for $a\in\{0,1,\infty\}$.

For all $a\in F^{\times}\setminus\{1\}$ we have in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$

[TABLE]

and hence

[TABLE]

where

[TABLE]

Thus

[TABLE]

We consider now the four possible values of $(\chi(a),\chi(1-a))$:

1. (1)

$\chi(a)=-1$ and $\chi(1-a)=1$.

Then $\chi(a^{-1})=-1=\chi(1-a^{-1})$. Furthermore ${\left[a\right]}_{\chi}=C_{F}$ and ${\left[a^{-1}\right]}_{\chi}=-C_{F}$ by Lemma 2.11. By (7) we thus have

[TABLE]

where $\chi(y)=-1=\chi(z)$ and $\chi(w)=1$.

We divide further into sub-cases according to the value of $\chi(1-y)$:

1. (a)

$\chi(1-y)=1$: Then ${\left[y\right]}_{\chi}=C_{F}$ by Lemma 2.11 and hence ${\left[y\right]}_{\chi}={\left[a\right]}_{\chi}$ as required. 2. (b)

$\chi(1-y)=-1$: Then ${\left[y\right]}_{\chi}=-C_{F}$ by Lemma 2.11. However, by (8), $\chi(1-z)=\chi(1-y)\chi(\ell)=1$ and $\chi(1-w)=\chi(1-y)\chi(a\ell)=-1$ so that ${\left[z\right]}_{\chi}=C_{F}$ and ${\left[w\right]}_{\chi}=0$ by Lemmas 2.11 and 5.3. Hence, by (7), we now have $0=-C_{F}+C_{F}-C_{F}-0$ and hence $C_{F}=0$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$. Thus ${\left[y\right]}_{\chi}=0={\left[a\right]}_{\chi}$ as required, in this case also. 2. (2)

$\chi(a)=-1$ and $\chi(1-a)=-1$.

Then $\chi(a^{-1})=-1$ and $\chi(1-a^{-1})=1$. Thus ${\left[a\right]}_{\chi}=-C_{F}$ and ${\left[a^{-1}\right]}_{\chi}=C_{F}$. This gives $0=C_{F}-{\left[y\right]}_{\chi}+{\left[z\right]}_{\chi}+{\left[w\right]}_{\chi}$ where $\chi(y)=-1=\chi(w)$ and $\chi(z)=1$.

1. (a)

$\chi(1-y)=1$: Then ${\left[y\right]}_{\chi}=C_{F}=-{\left[a\right]}_{\chi}$. However, by (8) again, $\chi(1-z)=-1$ and $\chi(1-w)=1$ so that ${\left[z\right]}_{\chi}=0$ and ${\left[w\right]}_{\chi}=C_{F}$. From (7) we have $0=C_{F}-C_{F}+0+C_{F}$ and hence $C_{F}=0$ in ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$ as required. 2. (b)

$\chi(1-y)=-1$: Then ${\left[y\right]}_{\chi}=-C_{F}={\left[a\right]}_{\chi}$ again as required. 3. (3)

$\chi(a)=1$ and $\chi(1-a)=-1$.

Then ${\left[a\right]}_{\chi}=0={\left[a^{-1}\right]}_{\chi}$ by Lemma 5.3. Thus from (7) we have $0={\left[y\right]}_{\chi}+{\left[z\right]}_{\chi}-{\left[w\right]}_{\chi}$ where $\chi(y)=\chi(z)=\chi(w)=1$.

1. (a)

$\chi(1-y)=1$: Then $\chi(1-z)=-1=\chi(1-w)$. Hence ${\left[z\right]}_{\chi}=0={\left[w\right]}_{\chi}.$ Thus ${\left[y\right]}_{\chi}=0={\left[a\right]}_{\chi}$ as required. 2. (b)

$\chi(1-y)=-1$: Then ${\left[y\right]}_{\chi}=0={\left[a\right]}_{\chi}$ by Lemma 5.3. 4. (4)

$\chi(a)=1=\chi(1-a)$.

Then $\chi(a^{-1})=1=\chi(1-a^{-1})$ also. Equation (7) thus gives $0={\left[a^{-1}\right]}_{\chi}+{\left[y\right]}_{\chi}-{\left[z\right]}_{\chi}+{\left[w\right]}_{\chi}$ with $\chi(z)=-1=\chi(w)$. Furthermore $\chi(1-z)=-\chi(1-y)=\chi(1-w)$. Hence ${\left[z\right]}_{\chi}={\left[w\right]}_{\chi}=-\chi(1-y)C_{F}$ by Lemma 2.11. This gives

[TABLE]

as required.

∎

A straightforward induction gives:

Corollary 5.5**.**

Let $F$ be a field. Let $\chi\in\widehat{{F}^{\times}/({F}^{\times})^{2}}$ with $\chi(-1)=1$. Let $\ell\in F^{\times}$ satisfy $\chi(\ell)=-1$ and $\chi(1-\ell)=1$. Then

[TABLE]

for all $a\in\mathbb{P}^{1}(F)$ and all $m\in\mathbb{Z}$.

Corollary 5.6**.**

Let $F$ be a field. Let $\chi\in\widehat{{F}^{\times}/({F}^{\times})^{2}}$ with $\chi(-1)=1$. Let $\ell\in F^{\times}$ satisfy $\chi(\ell)=-1$ and $\chi(1-\ell)=1$. Then

[TABLE]

for all $a\in F$ and all $t\in\mathbb{Z}$.

Proof.

In ${{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]}_{\chi}$ we have

[TABLE]

for any $a\in F$. ∎

Proposition 5.7**.**

Let $\chi\in\widehat{\mathbb{Q}_{+}/\mathbb{Q}_{+}^{2}}$. If $\left|\mathrm{supp}(\chi)\right|\geq 2$ then

[TABLE]

in ${{\mathcal{RP}_{+}(\mathbb{Q})}\left[\tfrac{1}{2}\right]}_{\chi}$ for all $t\in\mathbb{Z}$ and $a\in\mathbb{Q}$.

Proof.

Let $p=\mathrm{min}(\mathrm{supp}(\chi))$. Then $\chi(p)=-1$ and $\chi(1-p)=\chi(p-1)=1$. So

[TABLE]

for all $a\in\mathbb{Q}$ and $t\in\mathbb{Z}$ by Corollary 5.6.

Now let $q=\mathrm{min}(\mathrm{supp}(\chi)\setminus\{p\})$.

Suppose first that $p>2$. The either $q-1$ or $q+1$ is not divisble by $p$. If $p$ does not divide $q-1$ take $\ell=q$. Otherwise take $\ell=-q$. Then $\chi(\ell)=-1$ and $\chi(1-\ell)=1$ so that for all $a\in\mathbb{Q}$ ${\left[a\right]}_{\chi}={\left[a+t\ell\right]}_{\chi}$ for all $t\in\mathbb{Z}$ and hence ${\left[a\right]}_{\chi}={\left[a+tq\right]}_{\chi}$ for all $t\in\mathbb{Z}$.

Thus for all $a\in\mathbb{Q}$ we have

[TABLE]

proving the proposition in this case.

Suppose now that $p=2$.

If $q\equiv 5\pmod{8}$ then $v_{2}(1-q)=2$ so that $\chi(1-q)=1$ and we can take $\ell=q$ and argue as above.

If $q\equiv 3\pmod{8}$ the corresponding argument applies with $\ell=-q$.

If $q\equiv-1\pmod{8}$ we can take $\ell=3q$. Then $\chi(\ell)=-1$ (since $q\not=3$). Furthermore we have $\ell-1\equiv 4\pmod{8}$ and

[TABLE]

This implies $\chi(\ell-1)=\chi(1-\ell)=1$ and we can conclude as before.

Finally, if $q\equiv 1\pmod{8}$ we take $\ell=-3q$ and argue as in the previous case. ∎

Corollary 5.8**.**

Let $\chi\in\widehat{\mathbb{Q}_{+}/\mathbb{Q}_{+}^{2}}$ and suppose that $\left|\mathrm{supp}(\chi)\right|\geq 2$. Then ${{\mathcal{RP}_{+}(\mathbb{Q})}\left[\tfrac{1}{2}\right]}_{\chi}=0$.

Proof.

We will show that ${\left[a\right]}_{\chi}=0$ for all $a\in\mathbb{Q}$. By Proposition 5.7 we have ${\left[a\right]}_{\chi}={\left[a+t\right]}_{\chi}$ for all $t\in\mathbb{Z}$, $a\in\mathbb{Q}^{\times}$. It follows that ${\left[a\right]}_{\chi}={\left[1\right]}_{\chi}=0$ if $a\in\mathbb{Z}$. Thus also ${\left[1/a\right]}_{\chi}=0$ for all $a\in\mathbb{Z}\setminus\{0\}$.

Note that it is enough to prove ${\left[a\right]}_{\chi}=0$ for all $a>0$ (if necessary replacing $a$ by $a+t$ with $t\in\mathbb{Z}$ large). So let $a=r/s$ with $0<r,s\in\mathbb{Z}$. We proceed by induction on $h:=\mathrm{min}(r,s)$. The case $h=1$ has already been proved. Suppose now that $n\geq 1$ and the statement is known for $h\leq n$. Consider the case $h=n+1$. Replacing $a$ by $1/a$ if necessary we can suppose $s<r$ and $s=n+1$. Then there exists $t\in\mathbb{Z}$ such that $0<r^{\prime}:=r-ts<s$. So

[TABLE]

where now $h=r^{\prime}\leq n$ and we are done. ∎

6. Some related calculations

6.1. The module $\mathcal{RP}_{+}(\mathbb{Q})$

The module $\mathcal{RP}_{1}(F)$ arises inevitably in the calculation of the third homology of $\operatorname{SL}_{2}\left(A\right)$ for various rings $A$. For example, if $F$ is any infinite field we have ([6, Theorem 8.1])

[TABLE]

and there is a natural short exact sequence of $\mathrm{R}_{F}$-modules ([8, Theorem 7.4, Example 7.9])

[TABLE]

Furthermore, there is a natural short exact sequence

[TABLE]

As noted above, for any field $F$ the natural $\mathrm{R}_{F}$-homomorphism $\mathcal{RP}_{1}(F)\to\mathcal{RP}_{+}(F)$ induces an isomorphism

[TABLE]

In Theorem 4.3 above we have calculated $\mathcal{I}_{\mathbb{Q}}{\mathcal{RP}_{+}(\mathbb{Q})}\left[\tfrac{1}{2}\right]$. This easily gives a computation of ${\mathcal{RP}_{+}(\mathbb{Q})}\left[\tfrac{1}{2}\right]$. Namely, for any field $F$ there is a short exact sequence of $\mathrm{R}_{F}$-modules (see, for example, [7, Lemma 2.7])

[TABLE]

(where, $\mathrm{R}_{F}$ acts trivially on $\mathcal{P}(F)$). Now, by definition, there is an exact sequence of abelian groups

[TABLE]

Tensoring with ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$ and using the fact that $K_{2}(\mathbb{Q})$ is a torsion $\mathbb{Z}$-module, we deduce that

[TABLE]

where $V={\mathrm{S}^{2}_{\mathbb{Z}}(\mathbb{Q}^{\times})}\left[\tfrac{1}{2}\right]$ is a free ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$-module of countable rank. Furthermore the exact sequence

[TABLE]

splits as a sequence of ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$-modules since the subgroup $\mathbb{Z}/3\cdot C_{\mathbb{Q}}\subset{\mathcal{RP}_{+}(\mathbb{Q})}\left[\tfrac{1}{2}\right]$ maps isomorphically to ${\mathcal{B}(\mathbb{Q})}\left[\tfrac{1}{2}\right]$. Thus, in view of Theorem 4.3 we have:

Lemma 6.1**.**

As a ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$-module, ${\mathcal{RP}_{+}(\mathbb{Q})}\left[\tfrac{1}{2}\right]$ is a direct sum of an infinite torsion group and a free ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$-module $V$ of countable rank. More particularly:

[TABLE]

Corollary 6.2**.**

As an abelian group we have

[TABLE]

6.2. The module $\mathcal{D}_{\mathbb{Q}}$ and the $3$-torsion in $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$

We let $\mathcal{D}_{F}$ denote the $\mathrm{R}_{F}$-submodule of $\mathcal{RP}_{+}(F)$ generated by $C_{F}$. Note that $3\cdot\mathcal{D}_{F}=0$; $\mathcal{D}_{F}$ is an $\mathbb{F}_{3}$-vector space.

For any field $F$, let $\mathcal{H}=\mathcal{H}_{F}$ denote the $\mathrm{R}_{F}$-submodule of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ generated by the image of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Z}\right),\mathbb{Z}\right)$.

Remark 6.3**.**

Since the $\mathrm{R}_{F}$-module structure on $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ is induced from the action of $\mathrm{GL}_{2}(F)$ by conjugation on $\operatorname{SL}_{2}\left(F\right)$, $\mathcal{H}=\mathcal{H}_{F}$ is just the subgroup $\sum_{g\in\mathrm{GL}_{2}(F)}\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Z}\right)^{g},\mathbb{Z}\right)$ in $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$; i.e. it is the subgroup of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)$ generated by $\operatorname{SL}_{2}\left(\mathbb{Z}\right)$ and its $\mathrm{GL}_{2}(F)$-conjugates.

Proposition 6.4**.**

Suppose that $\mathrm{char}(F)\not=3$ and $\zeta_{3}\not\in F$.

Then the map $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to\mathcal{RP}_{+}(F)$ induces an isomorphism ${\mathcal{H}}\left[\tfrac{1}{2}\right]\cong{\mathcal{H}}_{(3)}\cong\mathcal{D}_{F}$.

Proof.

As above, let

[TABLE]

and let $G$ be the cyclic subgroup of order $3$ generated by $t$. By [4, Remark 3.14], the composite map $\mathbb{Z}/3=\mathrm{H}_{3}\left(G,\mathbb{Z}\right)\to\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),\mathbb{Z}\right)\to\widetilde{\mathcal{RP}}(F)$ sends $1$ to $C_{F}$ for any field $F$.

We recall that $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Z}\right),\mathbb{Z}\right)\cong\mathbb{Z}/12$. Furthermore, the inclusion $G\to\operatorname{SL}_{2}\left(\mathbb{Z}\right)$ induces an isomorphism

[TABLE]

Thus ${\mathcal{H}}_{(3)}\cong{\mathcal{H}}\left[\tfrac{1}{2}\right]$ maps onto $\mathcal{D}_{F}$.

On the other hand, the kernel of the map $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(F\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)\to{\mathcal{RP}_{+}(F)}\left[\tfrac{1}{2}\right]$ is isomorphic to ${\mu_{F}}\left[\tfrac{1}{2}\right]$. In particular, if $\zeta_{3}\not\in F$, the induced map ${\mathcal{H}}_{(3)}\to\mathcal{D}_{F}$ is also injective. ∎

Lemma 6.5**.**

We have $\mathcal{D}_{\mathbb{Q}}={\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)}_{(3)}$ and

[TABLE]

Proof.

Since ${K^{\mathrm{\small ind}}_{3}(\mathbb{Q})}\left[\tfrac{1}{2}\right]\cong{\mathcal{B}(\mathbb{Q})}\left[\tfrac{1}{2}\right]=\mathbb{Z}/3\cdot C_{\mathbb{Q}}$ we have a ($\mathbb{Z}$-split) short exact sequence of $\mathrm{R}_{\mathbb{Q}}$-modules

[TABLE]

Consider the composite map

[TABLE]

where the right-hand arrow is an isomorphism by Theorem 4.3. We finish by observing that ${\mathcal{P}(\mathbb{F}_{p})}\left[\tfrac{1}{2}\right]\cong\mathbb{Z}/\left(p+1\right)_{\mbox{ \tiny odd}}$ has no $3$-torsion except when $p\equiv-1\mod{3}$ and when $p\equiv-1\mod{3}$ the element $C_{\mathbb{F}_{p}}=\bar{S}_{p}(C_{\mathbb{Q}})$ has order $3$ by [5, Lemma 7.11]. ∎

Remark 6.6**.**

By our main theorem, $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ has (odd) torsion of every possible size. However, the elements of order $3$ in this group all come from the obvious source: the torsion of order $3$ in $\operatorname{SL}_{2}\left(\mathbb{Z}\right)$ and its $\mathrm{GL}_{2}(\mathbb{Q})$- conjugates in $\operatorname{SL}_{2}\left(\mathbb{Q}\right)$. More precisely, a basis for the $\mathbb{F}_{3}$-vector space ${}_{3}{\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)}$ is $\{\tau\}\cup\{\tau_{p}\ |p\equiv-1\mod{3}\}$ where $\tau$ is the image of $1\in\mathbb{Z}/3=\mathrm{H}_{3}\left(\left\langle{t}\right\rangle,\mathbb{Z}\right)\to\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ and $\tau_{p}$ is the image of $1\in\mathbb{Z}/3=\mathrm{H}_{3}\left(\left\langle{t}\right\rangle^{D_{p}},\mathbb{Z}\right)\to\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),{\mathbb{Z}}\left[\tfrac{1}{2}\right]\right)$ with $D_{p}:=\mathrm{diag}(p,1)\in\mathrm{GL}_{2}(\mathbb{Q})$.

Remark 6.7**.**

Although our main results are over the coefficient ring ${\mathbb{Z}}\left[\tfrac{1}{2}\right]$, it is possible to say something about the $2$-torsion structure of $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$. Theorem 4.3 implies that $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$ is a torsion group. (This is already known from the rank calculations in [2]). For any global field $F$ there is a well-defined homomorphism (induced by the maps $\bar{S}_{v}$)

[TABLE]

where $v$ ranges over the discrete valuations. Our main theorem tells us that when $F=\mathbb{Q}$ the kernel and cokernel of this homomorphism are $2$-torsion groups. In fact it can be shown that the cokernel is annihilated by $4$ (since the cokernel of each of the maps $\bar{S}_{v}$ is annihilated by $4$). It follows that $\mathrm{H}_{3}\left(\operatorname{SL}_{2}\left(\mathbb{Q}\right),\mathbb{Z}\right)$ contains elements of order $2^{n}$ for all $n$; i.e. it also contains $2$-torsion of all possible orders.