Third homology of ${\mathrm{SL}}_2$ and the indecomposable $K_3$

Abstract

It is known that, for an infinite field $F$, the indecomposable part of $K_3(F)$ and the third homology of ${\mathrm{SL}}_2(F)$ are closely related. In fact, there is a canonical map $\alpha : H_3({\mathrm{SL}}_2(F),\mathbb {Z})_{F^*}\rightarrow K_3(F)^\mathrm{ind}$. Suslin has raised the question: Is $\alpha$ an isomorphism? Recently Hutchinson and Tao have shown that this map is surjective. In this article, we show that $\alpha$ is bijective if and only if the natural maps $H_3({\mathrm{GL}}_2(F), \mathbb {Z})\rightarrow H_3({\mathrm{GL}}_3(F), \mathbb {Z})$ and $H_3({\mathrm{SL}}_2(F), \mathbb {Z})_{F^*}\rightarrow H_3({\mathrm{GL}}_2(F), \mathbb {Z})$ are injective.

1 Introduction

For an infinite field $F$, Suslin has proved that the Hurewicz homomorphism

$$\begin{aligned} h_3: K_3(F)=\pi _3(B{\mathrm{SL}}(F)^+) \longrightarrow H_3(B{\mathrm{SL}}(F)^+, \mathbb {Z})\simeq H_3({\mathrm{SL}}(F), \mathbb {Z}) \end{aligned}$$

is surjective with 2-torsion kernel. In fact, he has shown that $h_3$ sits in the exact sequence

$$\begin{aligned} K_2(F) \overset{l(-1)}{\longrightarrow } K_3(F) \longrightarrow H_3({\mathrm{SL}}(F), \mathbb {Z}) \longrightarrow 0, \end{aligned}$$

where the homomorphism ${l(-1)}: K_2(F) {\rightarrow } K_3(F)$ coincides with multiplication by $l(-1) \in K_1(\mathbb {Z})$ [10, Lemma 5.2, Corollary 5.2]. Let

$$\begin{aligned} \alpha : H_0({F^*},H_3({\mathrm{SL}}_2(F),\mathbb {Z}))\rightarrow K_3(F)^\mathrm{ind}\end{aligned}$$

be the composition of the following sequence of homomorphisms

$$\begin{aligned}&H_0({F^*}, H_3({\mathrm{SL}}_2(F), \mathbb {Z})) \overset{\mathrm{inc}_*}{\longrightarrow } H_3({\mathrm{SL}}(F), \mathbb {Z}) \overset{{\bar{h}}_3^{-1}}{\overset{\simeq }{\longrightarrow }} K_3(F)/l(-1)K_2(F)\\&\qquad \qquad \qquad \qquad \qquad \overset{p}{\longrightarrow } K_3(F)^\mathrm{ind}:= K_3(F)/K_3^M(F), \end{aligned}$$

where $\mathrm{inc}_*$ is induced by the inclusion $\mathrm{inc}: {\mathrm{SL}}_2(F) \rightarrow {\mathrm{SL}}(F)$, and $p$ is induced by the inclusion $l(-1)K_2(F) \subseteq \mathrm{im}(K_3^M(F)\rightarrow K_3(F))$. For algebraically closed fields, it was known that $\alpha$ is an isomorphism [1, 9]. Following this, Suslin raised the following question:

Question (Suslin). Is it true that $H_0({F^*}, H_3({\mathrm{SL}}_2(F), \mathbb {Z}))$ coincides with $K_3(F)^\mathrm{ind}$? (See [9], Question 4.4]).

In other words, is $\alpha$ bijective for an arbitrary infinite field $F$? This question is true after killing 2-power torsion elements (i.e. after tensoring the both sides of this map with $\mathbb {Z}[1/2]$) or when ${F^*}={F^*}^2=\{a^2|a \in {F^*}\}$ [6, Proposition 6.4].

Recently Hutchinson and Tao have proved that $\alpha$ is surjective [4, Lemma 5.1]. The following theorem is our main result, which improves an argument of Hutchinson and Tao in [4].

Theorem

Let $F$ be an infinite field. The following conditions are equivalent.

• (i) The homomorphism $\alpha : H_0({F^*}, H_3({\mathrm{SL}}_2(F), \mathbb {Z})) \rightarrow K_3(F)^\mathrm{ind}$ is bijective.

• (ii) The natural homomorphisms $H_3({\mathrm{GL}}_2(F), \mathbb {Z})\rightarrow H_3({\mathrm{GL}}_3(F), \mathbb {Z})$ and $H_0({F^*}, H_3({\mathrm{SL}}_2(F), \mathbb {Z})) \rightarrow H_3({\mathrm{GL}}_2(F), \mathbb {Z})$ are injective.

In the mean time we also establish that the kernel of the homomorphism

$$\begin{aligned} H_3(\mathrm{inc}): H_3({\mathrm{GL}}_2(F), \mathbb {Z})\rightarrow H_3({\mathrm{GL}}_3(F), \mathbb {Z}) \end{aligned}$$

is equal to

$$\begin{aligned} \mathrm{im}(H_3({\mathrm{SL}}_2(F), \mathbb {Z}) \rightarrow H_3({\mathrm{GL}}_2(F), \mathbb {Z})) \cap {F^*}\cup H_2({\mathrm{GL}}_1(F), \mathbb {Z})), \end{aligned}$$

where the cup product is induced by the natural diagonal inclusion $\mathrm{inc}: {F^*}\times {\mathrm{GL}}_1(F) \rightarrow {\mathrm{GL}}_2(F)$. It seems that, for an arbitrary field, not much is known about the kernel of

$$\begin{aligned} H_0\left( {F^*}, H_3({\mathrm{SL}}_2(F), \mathbb {Z})\right) \rightarrow H_3({\mathrm{GL}}_2(F), \mathbb {Z}), \end{aligned}$$

except that it is a 2-power torsion group (see proof of Theorem 6.1 in [6]).

1.1 Notation

In this article by $H_i(G)$ we mean the homology of group $G$ with integral coefficients, namely $H_i(G, \mathbb {Z})$. By ${\mathrm{GL}}_n$ (resp. ${\mathrm{SL}}_n$) we mean the general (resp. special) linear group ${\mathrm{GL}}_n(F)$ (resp. ${\mathrm{SL}}_n(F)$), where $F$ is an infinite field. If $A \rightarrow A'$ is a homomorphism of abelian groups, by $A'/A$ we mean $\mathrm{coker}(A \rightarrow A')$ and we take other liberties of this kind. Here by $\Sigma _n$ we mean the symmetric group of rank $n$.

2 The group $H_1\left( {F^*}, H_2({\mathrm{SL}}_2)\right)$

We start this section by looking at the corresponding Lyndon/Hochschild-Serre spectral sequence of the commutative diagram of extensions

So we get a morphism of spectral sequences

By an easy analysis of this spectral sequence we obtain the following commutative diagram with exact rows

The following theorem is due to Hutchinson and Tao [4, Theorem 3.2], which is very fundamental in their proof of the surjectivity of $\alpha$.

Theorem 2.1

The inclusion ${\mathrm{SL}}_2 \longrightarrow {\mathrm{SL}}_3$ induces a short exact sequence

$$\begin{aligned} 0 \longrightarrow H_1\left( {F^*}, H_2({\mathrm{SL}}_2)\right) \longrightarrow H_1\left( {F^*}, H_2({\mathrm{SL}}_3)\right) \longrightarrow k_3^M(F) \longrightarrow 0, \end{aligned}$$

where $k_3^M(F):=K_3^M(F)/2$.

Since the action of ${F^*}$ on $H_2({\mathrm{SL}}_3)$ is trivial,

$$\begin{aligned} H_1\left( {F^*}, H_2({\mathrm{SL}}_3)\right) \simeq {F^*}\otimes K_2^M(F). \end{aligned}$$

So we consider $H_1({F^*}, H_2({\mathrm{SL}}_2))$ as a subgroup of ${F^*}\otimes K_2^M(F)$. It is easy to see that the map

$$\begin{aligned} H_1\left( {F^*}, H_2({\mathrm{SL}}_3)\right) \longrightarrow k_3^M(F) \end{aligned}$$

is induced by the natural product map ${F^*}\otimes K_2^M(F) \longrightarrow K_3^M(F)$. Since the $n$-th Milnor $K$-group, $K_n^M(F)$, is naturally isomorphic to the $n$-th tensor of ${F^*}$ modulo the two families of relations

$$\begin{aligned}&a_1 \otimes \dots \otimes a_{n-1} \otimes (1-a_{n-1}), \qquad a_i \in {F^*}, a_{n-1} \ne 1,\\&a_1 \otimes \cdots a_i \otimes a_{i+1} \cdots \otimes a_n + a_1 \otimes \cdots a_{i+1} \otimes a_{i} \cdots \otimes a_n, \, \, a_i \in {F^*}, \end{aligned}$$

it easily follows that the kernel of the product map ${F^*}\otimes K_2^M(F)\longrightarrow K_3^M(F)$ is generated by elements $a \otimes \{b,c\} + b \otimes \{a, c\}$. This proves the following lemma.

Lemma 2.2

As a subgroup of $H_1({F^*}, H_2({\mathrm{SL}}_3))={F^*}\otimes K_2^M(F)$, the group $H_1({F^*}, H_2({\mathrm{SL}}_2))$ is generated by elements $a \otimes \{b,c\} + b \otimes \{a, c\}$ and $2d \otimes \{e,f\}$.

To go further, we need to introduce some notations. Let $G$ be a group and set

$$\begin{aligned} {\mathbf {c}}({g}_1, {g}_2,\ldots , {g}_n):=\sum _{\sigma \in S_n} {\mathrm{sign}(\sigma )}[{g}_{\sigma (1)}| {g}_{\sigma (2)}|\ldots |{g}_{\sigma (n)}] \in H_n(G), \end{aligned}$$

where ${g}_i \in G$ pairwise commute and $S_n$ is the symmetric group of degree $n$. Here we use the bar resolution of $G$ [2, Chapter I, Section 5] to define the homology of $G$.

Lemma 2.3

Let $G$ and $G'$ be two groups.

• (i) If $h_1\in G$ commutes with all the elements $g_1, \dots , g_n \in G$, then

  $$\begin{aligned} {\mathbf {c}}(g_1h_1, g_2,\ldots , g_n)= {\mathbf {c}}(g_1, g_2,\ldots , g_n)+{\mathbf {c}}(h_1, g_2,\ldots , g_n). \end{aligned}$$

• (ii) For every $\sigma \in S_n$, ${\mathbf {c}}(g_{\sigma (1)},\ldots , g_{\sigma (n)})=\mathrm{sign(\sigma )} {\mathbf {c}}(g_1,\ldots , g_n)$.

• (iii) The cup product of ${\mathbf {c}}(g_1,\ldots , g_p)\in H_p(G)$ and ${\mathbf {c}}(g_1',\ldots , g_q') \in H_q(G')$ is ${\mathbf {c}}((g_1, 1), \dots , (g_p,1),(1,g_1'), \ldots , (1,g_q')) \in H_{p+q}(G \times G')$.

Proof

The proof follows from direct computations, so we leave it to the interested readers. $\square$

3 The kernel of $H_3({\mathrm{GL}}_2)\longrightarrow H_3({\mathrm{GL}}_3)$

For simplicity, in the rest of this article, we use the following notation

$$\begin{aligned} k_{a,b,c}:={\mathbf {c}} (\mathrm{diag}(a,1), \mathrm{diag}(1,b),\mathrm{diag}(1,c)) \in H_3({\mathrm{GL}}_2). \end{aligned}$$

The following theorem has been proved in [7, Theorem 3.1].

Theorem 3.1

The kernel of ${\mathrm{inc}_1}_*:H_3({\mathrm{GL}}_2) \longrightarrow H_3({\mathrm{GL}}_3)$ consists of elements of the form $\sum k_{a,b,c}+k_{b,a,c}$ such that

$$\begin{aligned} \sum a\otimes \{b, c\}+b\otimes \{a, c\}=0 \in {F^*}\otimes K_2^M(F). \end{aligned}$$

In particular $\mathrm{ker}({\mathrm{inc}_1}_*) \subseteq {F^*}\cup H_2({\mathrm{GL}}_1) \subseteq H_3({\mathrm{GL}}_2)$, where the cup product is induced by the natural diagonal inclusion ${F^*}\times {\mathrm{GL}}_1 \longrightarrow {\mathrm{GL}}_2$. Moreover $\mathrm{ker}({\mathrm{inc}_1}_*)$ is a $2$-torsion group.

Let $\Psi$ and $\Phi$ be the following compositions,

$$\begin{aligned} {F^*}&\otimes&K_2^M(F) \overset{\mathrm{id}_{F^*}\otimes \iota }{\longrightarrow } {F^*}\otimes H_2({\mathrm{GL}}_2) \overset{\cup }{\longrightarrow } H_3\left( {F^*}\times {\mathrm{GL}}_2\right) \overset{\mathrm{inc}_*}{\longrightarrow } H_3({\mathrm{GL}}_3),\\ {F^*}&\otimes&K_2^M(F) \overset{\mathrm{id}_{F^*}\otimes \iota }{\longrightarrow } {F^*}\otimes H_2({\mathrm{GL}}_2) \overset{\cup }{\longrightarrow } H_3\left( {F^*}\times {\mathrm{GL}}_2\right) \overset{\beta _*}{\longrightarrow } H_3({\mathrm{GL}}_2), \end{aligned}$$

respectively, where $\iota :K_2^M(F)\simeq H_2({\mathrm{SL}}_2)_{F^*}\longrightarrow H_2({\mathrm{GL}}_2)$ is the natural inclusion given by the formula $\{a,b\} \mapsto {\mathbf {c}} (\mathrm{diag}(a,1),\mathrm{diag}(b,b^{-1}))$ [3, Proposition A.11] and $\beta : {F^*}\times {\mathrm{GL}}_2 \longrightarrow {\mathrm{GL}}_2$ is given by $(a, A) \mapsto aA$. It is easy to see that

$$\begin{aligned} \Psi (a \otimes \{b,c\})&= {\mathbf {c}} \left( \mathrm{diag}(a,1,1), \mathrm{diag}(1,b,1),\mathrm{diag}\left( 1,c,c^{-1}\right) \right) ,\\ \Phi (a \otimes \{b, c\})&= {\mathbf {c}} \left( \mathrm{diag}(a,a), \mathrm{diag}(b,1),\mathrm{diag}\left( c,c^{-1}\right) \right) . \end{aligned}$$

Lemma 3.2

Let $\Theta$ be the composition

$$\begin{aligned} H_3({\mathrm{GL}}_2)/H_3({\mathrm{GL}}_1) \overset{\varphi }{\longrightarrow } H_1({F^*}, H_2({\mathrm{SL}}_2)) \hookrightarrow {F^*}\otimes K_2^M(F). \end{aligned}$$

Then

• (i) $\Theta (k_{a,b,c}+k_{b,a,c})=a\otimes \{b,c\}+b \otimes \{a,c\}$,

• (ii) $\Theta ({\mathbf {c}}(\mathrm{diag}(a, a),\mathrm{diag}(b,1), \mathrm{diag}(c, c^{-1})))= 2a \otimes \{b,c\}$,

• (iii) $\Theta (k_{c,a,b})= b\otimes \{a,c\}-a \otimes \{b,c\}$.

Proof

(i) It is easy to see that the exact sequence

$$\begin{aligned} 0 \longrightarrow H_3({\mathrm{SL}}_3)_{F^*}\longrightarrow H_3({\mathrm{GL}}_3)/H_3({\mathrm{GL}}_1) \overset{\psi }{\longrightarrow } H_1\left( {F^*}, H_2({\mathrm{SL}}_3)\right) \longrightarrow 0 \end{aligned}$$

splits canonically by the map

$$\begin{aligned} {F^*}\otimes K_2^M(F)=H_1\left( {F^*}, H_2({\mathrm{SL}}_3)\right) \longrightarrow H_3({\mathrm{GL}}_3)/H_3({\mathrm{GL}}_1) \end{aligned}$$

defined by $\Psi$. Now consider the commutative diagram

We have

$$\begin{aligned} {\mathrm{inc}_1}_*(k_{a,b,c})&= {\mathbf {c}}(\mathrm{diag}(a,1,1),\mathrm{diag}(1,b,1), \mathrm{diag}(1,c, 1))\\&= {\mathbf {c}}\left( \mathrm{diag}(a,1,1),\mathrm{diag}(1,b,1), \mathrm{diag}\left( 1,c, c^{-1}\right) \right) \\&+{\mathbf {c}}(\mathrm{diag}(a,1,1),\mathrm{diag}(1,b,1), \mathrm{diag}(1,1,c))\\&= {\mathbf {c}}\left( \mathrm{diag}(a,1,1),\mathrm{diag}(1,b,1), \mathrm{diag}\left( 1,c, c^{-1}\right) \right) \\&-{\mathbf {c}}(\mathrm{diag}(b,1,1),\mathrm{diag}(1,a,1), \mathrm{diag}(1,1,c))\\&= {\mathbf {c}}\left( \mathrm{diag}(a,1,1),\mathrm{diag}(1,b,1), \mathrm{diag}\left( 1,c, c^{-1}\right) \right) \\&-{\mathbf {c}}\left( \mathrm{diag}(b,1,1),\mathrm{diag}(1,a,1), \mathrm{diag}\left( 1,c^{-1},c\right) \right) \\&-{\mathbf {c}}(\mathrm{diag}(b,1,1),\mathrm{diag}(1,a,1), \mathrm{diag}(1,c, 1))\\&= \Psi \left( a \otimes \{b,c\}+b \otimes \{a,c\}\right) -{\mathrm{inc}_1}_*(k_{b,a,c}). \end{aligned}$$

Hence ${\mathrm{inc}_1}_*(k_{a,b,c}+k_{b,a,c})=\Psi (a \otimes \{b,c\}+ b \otimes \{a,c\})$. Therefore

$$\begin{aligned} \Theta (k_{a,b,c}+k_{b,a,c})&= \psi \circ {\mathrm{inc}_1}_*(k_{a,b,c}+k_{b,a,c})\\&=\psi \circ \Psi \left( a \otimes \{b,c\}+b \otimes \{a,c\}\right) \\&= a \otimes \{b,c\}+b \otimes \{a,c\}. \end{aligned}$$

(ii) Consider the composition

$$\begin{aligned} {F^*}\otimes K_2^M(F) \overset{\Phi }{\longrightarrow } H_3({\mathrm{GL}}_2)/H_3({\mathrm{GL}}_1) \overset{\Theta }{\longrightarrow } {F^*}\otimes K_2^M(F). \end{aligned}$$

The image of $\Phi (a\otimes \{b,c\})= {\mathbf {c}}(\mathrm{diag}(a, a),\mathrm{diag}(b,1), \mathrm{diag}(c, c^{-1}))$ in the group $H_3({\mathrm{GL}}_3)/H_3({\mathrm{GL}}_1)=H_3({\mathrm{SL}}_3)_{F^*}\oplus {F^*}\otimes K_2^M(F)$ is equal to

$$\begin{aligned} {\mathrm{inc}_1}_*\circ \Phi (a\otimes \{b,c\})&={\mathbf {c}}\left( \mathrm{diag}(a, a,1),\mathrm{diag}(b,1,1), \mathrm{diag}\left( c, c^{-1},1\right) \right) \\&={\mathbf {c}}\left( \mathrm{diag}(a, a, a^{-2}),\mathrm{diag}(b,1,1), \mathrm{diag}\left( c, c^{-1},1\right) \right) \\&\,\quad +{\mathbf {c}}\left( \mathrm{diag}(1, 1,a^2),\mathrm{diag}(b,1,1), \mathrm{diag}\left( c, c^{-1},1\right) \right) \\&={\mathbf {c}}\left( \mathrm{diag}(a, 1, a^{-1}), \mathrm{diag}(b, 1, b^{-1}),\mathrm{diag}\left( c, c^{-1},1\right) \right) \\&\,\quad +{\mathbf {c}}\left( \mathrm{diag}(a^2, 1, 1), \mathrm{diag}(1, b, 1),\mathrm{diag}\left( 1, c, c^{-1}\right) \right) . \end{aligned}$$

Therefore $\Theta \circ \Phi (a\otimes \{b,c\})=\psi \circ {\mathrm{inc}_1}_*\circ \Phi (a\otimes \{b,c\}) =2 a\otimes \{b,c\}$.

(iii) First note that

$$\begin{aligned} \Phi (a \otimes \{b,c\})&= {\mathbf {c}}\left( \mathrm{diag}(a,a),\mathrm{diag}(b,1), \mathrm{diag}\left( c, c^{-1}\right) \right) \\&= {\mathbf {c}}(\mathrm{diag}(a,1),\mathrm{diag}(b,1), \mathrm{diag}(c,1))\\&-k_{c,a,b}+ k_{a,b,c}+k_{b,a,c}. \end{aligned}$$

Therefore

$$\begin{aligned} \Theta (k_{c,a,b})&=\Theta (k_{a,b,c}+k_{b,a,c})-\Theta (\Phi (a \otimes \{b,c\}))\\&=b\otimes \{a,c\}-a \otimes \{b,c\}. \end{aligned}$$

$\square$

Proposition 3.3

Let ${\mathrm{inc}_2}_*: H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow H_3({\mathrm{GL}}_2)$ be induced by the natural map $\mathrm{inc}_2:{\mathrm{SL}}_2 \longrightarrow {\mathrm{GL}}_2$. Then

$$\begin{aligned} \mathrm{im}({\mathrm{inc}_2}_*)\cap \Big ({F^*}\cup H_2({\mathrm{GL}}_1)\Big )= \mathrm{ker}({\mathrm{inc}_1}_*). \end{aligned}$$

Proof

By Theorem 3.1, the kernel of ${\mathrm{inc}_1}_*: H_3({\mathrm{GL}}_2) \longrightarrow H_3({\mathrm{GL}}_3)$ consists of elements of the form $\sum k_{a,b,c}+k_{b,a,c}$ such that

$$\begin{aligned} \sum a\otimes \{b, c\} +b\otimes \{a, c\}=0 \in {F^*}\otimes K_2^M(F). \end{aligned}$$

By Lemma 3.2, we see that

$$\begin{aligned} \Theta \left( \sum k_{a,b,c}+k_{b,a,c}\right) = \sum a\otimes \{b, c\} +b\otimes \{a, c\}=0. \end{aligned}$$

Since the sequence

$$\begin{aligned} H_3({\mathrm{SL}}_2)_{F^*}\overset{{\mathrm{inc}_2}_*}{\longrightarrow } H_3({\mathrm{GL}}_2)/H_3({\mathrm{GL}}_1) \longrightarrow H_1\left( {F^*}, H_2({\mathrm{SL}}_2)\right) \longrightarrow 0, \end{aligned}$$

is exact, $\sum k_{a,b,c}+k_{b,a,c} \in \mathrm{im}({\mathrm{inc}_2}_*)$. Therefore

$$\begin{aligned} \mathrm{ker}({\mathrm{inc}_1}_*) \subseteq \mathrm{im}({\mathrm{inc}_2}_*)\cap \Big ({F^*}\cup H_2({\mathrm{GL}}_1)\Big ). \end{aligned}$$

Now let $x \in \mathrm{im}({\mathrm{inc}_2}_*)\cap ({F^*}\cup H_2({\mathrm{GL}}_1))$. Then $x$ is of the following form

$$\begin{aligned} x=\sum {\mathbf {c}} (\mathrm{diag}(a_i,1), \mathrm{diag}(1,b_i),\mathrm{diag}(1,c_i)). \end{aligned}$$

Thus $\det _*(x)=\sum {\mathbf {c}} (a_i, b_i,c_i)=0$, where $\det _*: H_3({\mathrm{GL}}_2) \longrightarrow H_3({F^*})$ is induced by the determinant. By the inclusion $\bigwedge _\mathbb {Z}^3 {F^*}\hookrightarrow H_3({F^*})$, we have $a\wedge b \wedge c \mapsto {\mathbf {c}}(a, b,c)$ (see for example [10, Lemma 5.5]). Thus

$$\begin{aligned} \sum a_i \otimes b_i\otimes c_i=\sum a' \otimes a'\otimes b'+ \sum a'' \otimes b''\otimes a'' +\sum b''' \otimes a'''\otimes a'''. \end{aligned}$$

Under the composition ${F^*}^{\otimes 3} \longrightarrow {F^*}\otimes H_2({F^*}) \longrightarrow H_3({\mathrm{GL}}_2)$ defined by

$$\begin{aligned} a \otimes b \otimes c \mapsto a \otimes {\mathbf {c}} (b,c) \mapsto {\mathbf {c}} (\mathrm{diag}(a,1), \mathrm{diag}(1,b),\mathrm{diag}(1,c))=k_{a,b,c}, \end{aligned}$$

we see that $x$ has the following form

$$\begin{aligned} x=\sum {\mathbf {c}} (\mathrm{diag}(a,1), \mathrm{diag}(1,a),\mathrm{diag}(1,b)). \end{aligned}$$

For simplicity, we assume that $x={\mathbf {c}} (\mathrm{diag}(a,1), \mathrm{diag}(1,a),\mathrm{diag}(1,b))$. By Lemma 3.2, $\Theta (x)= a\otimes \{a, b\} - b \otimes \{a, a\}=0$. Thus

$$\begin{aligned}&{\mathbf {c}} \left( \mathrm{diag}(a,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}\left( 1,b, b^{-1}\right) \right) \\&\quad =\Psi \left( a\otimes \{a, b\}\right) =\Psi \left( b\otimes \{a, a\}\right) \\&\quad ={\mathbf {c}} \left( \mathrm{diag}(b,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}\left( 1,a,a^{-1}\right) \right) \!, \end{aligned}$$

and so

$$\begin{aligned} \begin{array}{c} +{\mathbf {c}} (\mathrm{diag}(a,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}(1,b, 1))\\ -{\mathbf {c}} (\mathrm{diag}(a,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}(1,1, b))\\ =\\ -{\mathbf {c}} (\mathrm{diag}(b,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}(1,1,a)). \end{array} \end{aligned}$$

Hence in $H_3({\mathrm{GL}}_3)$ we have

$$\begin{aligned} {\mathrm{inc}_1}_*(x)&= {\mathbf {c}} (\mathrm{diag}(a,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}(1,b,1))\\&= {\mathbf {c}} (\mathrm{diag}(a,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}(1,1,b))\\&-{\mathbf {c}} (\mathrm{diag}(b,1,1), \mathrm{diag}(1,a,1),\mathrm{diag}(1,1, a))\\&= 0 \end{aligned}$$

Therefore $x \in \mathrm{ker}({\mathrm{inc}_1}_*)$ and this completes the proof of the proposition. $\square$

4 The indecomposable part of the third K-group

Define the pre-Bloch group $\mathfrak {p}(F)$ of $F$ as the quotient of the free abelian group $Q(F)$ generated by symbols $[a]$, $a \in {F^*}-\{1\}$, by the subgroup generated by elements of the form

$$\begin{aligned}{}[a] -[b]+\bigg [\frac{b}{a}\bigg ]-\bigg [\frac{1- a^{-1}}{1- b^{-1}}\bigg ] + \bigg [\frac{1-a}{1-b}\bigg ], \end{aligned}$$

where $a, b \in {F^*}-\{1\}$, $a \ne b$. Define

$$\begin{aligned} \lambda ': Q(F) \longrightarrow {F^*}\otimes {F^*}, \, \, \, \, [a] \mapsto a \otimes (1-a). \end{aligned}$$

By a direct computation, we have

$$\begin{aligned} \lambda '\left( [a] -[b]+\left[ \frac{b}{a}\right] -\left[ \frac{1- a^{-1}}{1- b^{-1}}\right] + \left[ \frac{1-a}{1-b}\right] \right) =a \otimes \left( \frac{1-a}{1-b}\right) +\left( \frac{1-a}{1-b}\right) \otimes a. \end{aligned}$$

Let $({F^*}\otimes {F^*})_\sigma :={F^*}\otimes {F^*}/ \langle a\otimes b + b\otimes a: a, b \in {F^*}\rangle$. We denote the elements of $\mathfrak {p}(F)$ and $({F^*}\otimes {F^*})_\sigma$ represented by $[a]$ and $a\otimes b$ again by $[a]$ and $a\otimes b$, respectively. Thus we have a well-defined map

$$\begin{aligned} \lambda : \mathfrak {p}(F) \longrightarrow \left( {F^*}\otimes {F^*}\right) _\sigma \!, \, \, \, [a] \mapsto a \otimes (1-a). \end{aligned}$$

The kernel of $\lambda$ is called the Bloch group of $F$ and is denoted by $B(F)$. Therefore we obtain the exact sequence

$$\begin{aligned} 0 \longrightarrow B(F) \longrightarrow \mathfrak {p}(F) \longrightarrow \left( {F^*}\otimes {F^*}\right) _\sigma \longrightarrow K_2^M(F) \longrightarrow 0. \end{aligned}$$

The following remarkable theorem is due to Suslin [10, Theorem 5.2].

Theorem 4.1

Let $F$ be an infinite field. Then we have the exact sequence

$$\begin{aligned} 0 \longrightarrow {\mathrm{Tor}_1^{\mathbb {Z}}}(\mu (F), \mu (F))^\sim \longrightarrow K_3(F)^\mathrm{ind}\longrightarrow B(F) \longrightarrow 0, \end{aligned}$$

where ${\mathrm{Tor}_1^{\mathbb {Z}}}(\mu (F), \mu (F))^\sim$ is the unique nontrivial extension of the group ${\mathrm{Tor}_1^{\mathbb {Z}}}(\mu (F), \mu (F))$ by $\mathbb {Z}/2$ if $\mathrm{char}(F)\ne 2$ and is equal to ${\mathrm{Tor}_1^{\mathbb {Z}}}(\mu (F), \mu (F))$ if $\mathrm{char}(F)= 2$.

The following theorem has been proved in [8, Theorem 4.4].

Theorem 4.2

Let $F$ be an infinite field. Then we have the exact sequence

$$\begin{aligned} 0 \longrightarrow {\mathrm{Tor}_1^{\mathbb {Z}}}(\mu (F), \mu (F))^\sim \longrightarrow \tilde{H}_3({\mathrm{SL}}_2(F)) \longrightarrow B(F) \longrightarrow 0, \end{aligned}$$

where $\tilde{H}_3({\mathrm{SL}}_2(F)):=H_3({\mathrm{GL}}_2)/(H_3({\mathrm{GL}}_1) + {F^*}\cup H_2({\mathrm{GL}}_1))$.

These two theorems suggest that $K_3(F)^\mathrm{ind}$ and $\tilde{H}_3({\mathrm{SL}}_2(F))$ should be isomorphism. But there is no natural homomorphism from one of these groups to the other one! But there is a natural map from $H_3({\mathrm{SL}}_2)_{F^*}$ to both of them. Hutchinson and Tao have proved that $H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow K_3(F)^\mathrm{ind}$ is surjective [4, Lemma 5.1]. The next lemma claims that this is also true for the other map.

Lemma 4.3

The map $\varsigma :H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow \tilde{H}_3({\mathrm{SL}}_2)$, induced by the natural map ${\mathrm{SL}}_2 \longrightarrow {\mathrm{GL}}_2$, is surjective.

Proof

Consider the exact sequence

$$\begin{aligned} H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow H_3({\mathrm{GL}}_2)/H_3({\mathrm{GL}}_1) \overset{\varphi }{\longrightarrow } H_1({F^*}, H_2({\mathrm{SL}}_2)) \longrightarrow 0. \end{aligned}$$

By Lemma 3.2, we have

$$\begin{aligned}&\Theta (k_{a,b,c}+k_{b,a,c}) =a\otimes \{b,c\} +b \otimes \{a,c\}, \\&\Theta ({\mathbf {c}}(\mathrm{diag}(a, a),\mathrm{diag}(b,1), \mathrm{diag}(c, c^{-1})))=2a\otimes \{b,c\}. \end{aligned}$$

Since $H_1({F^*}, H_2({\mathrm{SL}}_2))$ as a subgroup of $H_1({F^*}, H_2({\mathrm{SL}}_3))={F^*}\otimes K_2^M(F)$ is generated by elements $a\otimes \{b,c\} +b \otimes \{a,c\}$ and $2d\otimes \{e,f\}$ and since the elements $a\cup {\mathbf {c}}(b, c)=k_{a,b,c}$ vanish in $\tilde{H}_3({\mathrm{SL}}_2)$, $H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow \tilde{H}_3({\mathrm{SL}}_2)$ must be surjective. $\square$

Now we are ready to prove our main theorem.

Theorem 4.4

Let $F$ be an infinite field. The following conditions are equivalent.

• (i) The homomorphism $\alpha : H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow K_3(F)^\mathrm{ind}$ is bijective.

• (ii) The natural homomorphisms ${\mathrm{inc}_1}_*: H_3({\mathrm{GL}}_2) \longrightarrow H_3({\mathrm{GL}}_3)$ and ${\mathrm{inc}_2}_*:H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow H_3({\mathrm{GL}}_2)$ are injective.

Proof

(ii) $\Rightarrow$ (i) Consider the surjective map $\varsigma : H_3({\mathrm{SL}}_2)_{F^*}\longrightarrow \tilde{H}_3({\mathrm{SL}}_2)$ from Lemma 4.3. Let $\varsigma (x)=0$. Then ${\mathrm{inc}_2}_*(x) \in \mathrm{im}({\mathrm{inc}_2}_*) \cap {F^*}\cup H_3({\mathrm{GL}}_1)$. But by Proposition 3.3 and the assumptions

$$\begin{aligned} \mathrm{im}({\mathrm{inc}_2}_*) \cap {F^*}\cup H_3({\mathrm{GL}}_1)=\mathrm{ker}({\mathrm{inc}_1}_*)=0. \end{aligned}$$

From this we have ${\mathrm{inc}_2}_*(x)=0$ and hence $x=0$. Therefore $\varsigma$ is an isomorphism. Now the claim follows by comparing the exact sequence of Theorem 4.2 and Suslin’s Bloch-Wigner exact sequence in Theorem 4.1.

(i) $\Rightarrow$ (ii) Let $\bar{F}$ be the algebraic closure of $F$. By a theorem of Merkurjev and Suslin, $K_3(F)^\mathrm{ind}\longrightarrow K_3(\bar{F})^\mathrm{ind}$ is injective [5, Proposition 11.3]. Thus from the commutative diagram

and the injectivity of $\alpha$, we deduce the injectivity of the map $H_3({\mathrm{SL}}_2)_{F^*}\!\!\longrightarrow \!\! H_3({\mathrm{SL}}_2(\bar{F}))$. Now the injectivity of $H_3({\mathrm{SL}}_2)_{F^*}\! \longrightarrow \! H_3({\mathrm{GL}}_2)$ follows from the injectivity of $H_3({\mathrm{SL}}_2(\bar{F}))\! \longrightarrow \!\!H_3({\mathrm{GL}}_2(\bar{F}))$ [6, Theorem 6.1] and commutativity of the diagram

On the other hand, by Proposition 3.3, $\mathrm{ker}({\mathrm{inc}_1}_*) \subseteq H_3({\mathrm{SL}}_2)_{F^*}\subseteq H_3({\mathrm{GL}}_2)$. Let ${\mathrm{inc}_1}_*(x)=0$. It easily follows from the commutative diagram

that $x \in \mathrm{ker}(H_3({\mathrm{GL}}_2(\bar{F})) \longrightarrow H_3({\mathrm{GL}}_3(\bar{F})))=0$. Therefore $x=0$ [6, Theorem 5.4(iii)]. $\square$

Remark 4.5

(i) From the spectral sequence $\mathcal {E}_{p,q}^2$, one gets the exact sequence

$$\begin{aligned} H_4({\mathrm{GL}}_2)/H_4({\mathrm{GL}}_1)&\longrightarrow H_2\left( {F^*}, H_2({\mathrm{SL}}_2)\right) \overset{{\mathfrak {d}}_{2,2}^2}{\longrightarrow } H_3({\mathrm{SL}}_2)_{F^*}\\&\longrightarrow H_3({\mathrm{GL}}_2)/ H_3({\mathrm{GL}}_1) \longrightarrow H_1\left( {F^*}, H_2({\mathrm{SL}}_2)\right) \longrightarrow 0. \end{aligned}$$

Thus the injectivity of ${\mathrm{inc}_2}_*$ is equivalent to triviality of the differential ${\mathfrak {d}}_{2,2}^2$.

(ii) Theorem 3.1 gives a clear description of elements of the kernel of ${\mathrm{inc}_1}_*$. But there is no such information about the kernel of ${\mathrm{inc}_2}_*$. It is easy to see that $s_{a,b,c}:={\mathbf {c}} (\mathrm{diag}(a,a^{-1}), \mathrm{diag}(b,b^{-1}),\mathrm{diag}(c,c^{-1}))$ is in the kernel of ${\mathrm{inc}_2}_*$ and is 2-torsion:

$$\begin{aligned} s_{a,b,c}&={\mathbf {c}} \left( \mathrm{diag}\left( a,a^{-1}\right) \!, \mathrm{diag}\left( b,b^{-1}\right) \!,\mathrm{diag}\left( c,c^{-1}\right) \right) \\&={\mathbf {c}} \left( w.\mathrm{diag}\left( a,a^{-1}\right) .w^{-1}, w.\mathrm{diag}\left( b,b^{-1}\right) .w^{-1}, w.\mathrm{diag}\left( c,c^{-1}\right) .w^{-1}\right) \\&={\mathbf {c}} \left( \mathrm{diag}\left( a^{-1},a\right) \!, \mathrm{diag}\left( b^{-1},b\right) \!,\mathrm{diag}\left( c^{-1},c\right) \right) \\&=-s_{a,b,c}, \end{aligned}$$

where $w:=\left( \begin{array}{l@{\quad }l} 0&{}-1\\ 1&{}0\\ \end{array}\right)$. But it is not clear to us why it should be zero. It is not difficult to see that $\mathrm{ker}({\mathrm{inc}_2}_*)$ is a 2-power torsion group (see for example the proof of Theorem 6.1 in [6]).