Erratum to: Bloch–Wigner theorem over rings with many units

1 Erratum to: Math. Z. (2011) 268:329–346 DOI 10.1007/s00209-010-0674-9

2 Introduction

In this erratum, we correct a mistake that I made in [1]. The formulation of our main theorem [1, Theorem 5.1] is not correct. The correct form of the theorem, which is sufficient for our applications, is as follows:

Let $R$ be a commutative ring with many units and define

$$\begin{aligned} \tilde{H}_3(\mathrm{SL}_2(R), \mathbb Z ):=H_3(\mathrm{GL}_2(R), \mathbb Z )/(H_3(\mathrm{GL}_1(R), \mathbb Z ) + {R^{*}}\cup H_2(\mathrm{GL}_1(R), \mathbb Z ), \end{aligned}$$

which is induced by the diagonal inclusion of $R^*\times \mathrm{GL}_1(R)$ in $\mathrm{GL}_2(R)$. Then, there is a quotient $M$ of $H_1(\Sigma _2, {R^{*}}\otimes {R^{*}})$ which fits into exact sequences

$$\begin{aligned}&0 \longrightarrow T_R \longrightarrow \tilde{H}_3(\mathrm{SL}_2(R), \mathbb Z ) \longrightarrow \mathfrak p (R) \longrightarrow (R^*\otimes _\mathbb Z R^*)_\sigma \longrightarrow K_2(R) \longrightarrow 0,\\&\quad {\mathrm{Tor}_1^\mathbb{Z }}(\mu (R),\mu (R)) \longrightarrow T_R \longrightarrow M \longrightarrow 0. \end{aligned}$$

When $R$ is an integral domain, the left-hand side map in the second exact sequence is injective.

3 The main theorem

Lemma 3.1 in [1] is not correct. In fact, in the proof of the lemma, the map

$$\begin{aligned} F_q \otimes _{\mathrm{Stab}_{\mathrm{GL}_2}(\infty )} \mathbb Z \rightarrow F_q \otimes _{\mathrm{GL}_2} C_1(R^2), \quad s\otimes 1 \mapsto s \otimes (\infty ,0) \end{aligned}$$

is not well defined. Because of this mistake, we made few uncorrect claims in [1]. Here, we give a correct formulations of the lemma and these claims that are sufficient for our applications. We follow the notations in [1].

Lemma 3.1

The groups $E_{1,0}^2$ and $E_{1,1}^2$ are trivial. Also, there is a surjective map $H_1(\Sigma _2, {R^{*}}\otimes {R^{*}}) \twoheadrightarrow E_{1,2}^2$, where the action of $\Sigma _2=\{1, \sigma \}$ on ${R^{*}}\otimes {R^{*}}$ is defined by $\sigma (a \otimes b)=-b \otimes a$. In particular, $E_{1,2}^2$ is a $2$-torsion group.

Proof

In [1] on page 335, we have shown that $H_0(\mathrm{GL}_2,H_1(X))\simeq \mathbb Z$ and $d_{2,0}^1=\mathrm{id}_\mathbb Z$. Consider the differential $d_{2,1}^1:E_{2,1}^1=H_1(\mathrm{GL}_2,H_1(X)) \rightarrow \mathrm{ker}(d_{1,1}^1)\simeq {R^{*}}$. Let $\varphi : {R^{*}}\simeq H_1(\mathrm{GL}_2,C_2(R^2)) \rightarrow H_1(\mathrm{GL}_2,H_1(X))$. It is easy to see that $d_{2,1}^1\circ \varphi =\mathrm{id}_{R^{*}}$. These facts immediately imply the triviality of $E_{1,0}^2$ and $E_{1,1}^2$. To compute $E_{1,2}^2$, first note that $\mathrm{ker}(d_{1,2}^1)\simeq H_2({R^{*}}) \oplus ({R^{*}}\otimes {R^{*}})^\sigma$. Again, one can easily see that the composition

$$\begin{aligned} H_2({R^{*}})\!\simeq \! H_2\left( \mathrm{GL}_2,C_1\left( R^2\right) \right) \!\rightarrow \! H_2(\mathrm{GL}_2,H_1(X))\! \rightarrow \! H_2({R^{*}}) \!\oplus \!({R^{*}}\!\otimes \!{R^{*}})^\sigma \end{aligned}$$

is given by $x \mapsto (x, 0)$. Thus, $E_{1,2}^2\simeq ({R^{*}}\otimes {R^{*}})^\sigma /A$. By an easy analysis of our main spectral sequence, we have

$$\begin{aligned} E_{1,2}^2=({R^{*}}\otimes {R^{*}})^\sigma /A \hookrightarrow H_3(\mathrm{GL}_2)/H_3({R^{*}}\times {R^{*}}). \end{aligned}$$

We call this map $\beta$. We know that

$$\begin{aligned} H_1(\Sigma _2, {R^{*}}\otimes {R^{*}})= \frac{({R^{*}}\otimes {R^{*}})^\sigma }{(1+\sigma )({R^{*}}\otimes {R^{*}})}= \frac{({R^{*}}\otimes {R^{*}})^\sigma }{\langle a \otimes b-b\otimes a: a, b \in {R^{*}}\rangle }. \end{aligned}$$

Let $x_{a,b}=a \otimes b-b\otimes a$. Then,

$$\begin{aligned} h=[(a,1)|(1,b)]-[(1,b)|(a,1)] -[(b,1)|(1,a)]+[(1,a)|(b,1)] \end{aligned}$$

is the cycle that represents the element $x_{a,b} \in ({R^{*}}\otimes {R^{*}})^\sigma \subseteq H_2({R^{*}}\times {R^{*}})^\sigma$. Let $\tau$ be the automorphism of transposition of terms. Then, $\tau (h)-h=0$. Now, by [2, Lemma 2.5], the image of the class $\overline{h}$ under $\beta$ is given by $-\overline{\rho _s(h)}$, where $s=\left( \begin{array}{c@{\quad }c} 0 &{} 1 \\ 1 &{} 0 \end{array} \right)$ and

$$\begin{aligned} \rho _s(h){:=}&+[s|(1,a)|(b,1)]-[(a,1)|s|(b,1)]+[(a,1)|(1,b)|s]\\&-[s|(b,1)|(1,a)]+[(1,b)|s|(1,a)]-[(1,b)|(1,1)|s]\\&-[s|(1,b)|(a,1)]+[(b,1)|s|(a,1)]-[(b,1)|(1,a)|s]\\&+[s|(a,1)|(1,b)]-[(1,a)|s|(1,b)]+[(1,a)|(b,1)|s]. \end{aligned}$$

Now, by a direct computation, one can see that $\overline{\rho _s(h)}=0$. Thus, $x_{a,b} \in A$ and therefore $(1+\sigma )({R^{*}}\otimes {R^{*}}) \subseteq A$. This implies the surjectivity that we are looking for. $\square$

Now, we are ready to correct Theorem 5.1 in [1].

Theorem 5.1

Let $R$ be a commutative ring with many units and define $\tilde{H}_3(\mathrm{SL}_2(R)):=H_3(\mathrm{GL}_2)/(H_3(\mathrm{GL}_1) + {R^{*}}\cup H_2(\mathrm{GL}_1))$. There is a quotient $M$ of $H_1(\Sigma _2, {R^{*}}\otimes {R^{*}})$ which fits into exact sequences

$$\begin{aligned}&0 \longrightarrow T_R \longrightarrow \tilde{H}_3(\mathrm{SL}_2(R)) \longrightarrow B(R) \longrightarrow 0,\\&\quad {\mathrm{Tor}_1^\mathbb{Z }}(\mu (R),\mu (R)) \longrightarrow T_R \longrightarrow M \longrightarrow 0. \end{aligned}$$

When $R$ is an integral domain, the left-hand side map in the second exact sequence is injective.

Proof

The proof is very similar to the proof of Theorem 5.1 in [1]. $\square$

Remark 0.1

Here, we further make some minor corrections.

• (i) In Proposition 2.1 and Remark 5.2 in [1], we have to assume that either the coefficient group is $\mathbb Z [1/2]$ or the ring $R$ has the property ${R^{*}}={R^{*}}^2$ ($K_1(R)=K_1(R)^2$ for Remark 5.2).

• (ii) The claim made in Remark 2.2 in [1] is not correct, and Suslin’s claim in [2, Remark 2.2] remains true.

The rest of our claims in [1] remains true.