A translation of A.A.Vinogradov's "On the free product of ordered groups"

TL;DR

This paper translates Vinogradov's 1949 work demonstrating that the free product of two ordered groups can be endowed with an order, extending the understanding of ordered group constructions.

Contribution

It provides the first English translation of Vinogradov's proof that free products of ordered groups are themselves orderable.

Findings

Proves free products of ordered groups are orderable

Includes minor corrections to the original proof

Highlights significance for ordered group theory

Abstract

This is a translation from Russian of the article by A.A.Vinogradov: On the free product of ordered groups, Mat. Sb. (N.S.), 1949, Volume 25(67), Number 1, 163-168. The main result is that the free product of two ordered groups is orderable. Minor corrections were made and highlighted in footnotes.

Figures

Equations

$$(a>0\text{ and }x>y)⟹(ax>ay\text{ and }xa>ya).$$

$$A=\sum _{i=1}^n{\alpha }_i{a}_i,A^{\prime }=\sum _{i=1}^n{\alpha }_i^{\prime }{a}_i,$$

$${\alpha }_1={\alpha }_1^{\prime },\dots ,{\alpha }_{r-1}={\alpha }_{r-1}^{\prime },{\alpha }_r>{\alpha }_r^{\prime }.$$

$$\begin{array}[]{lccl}n-m\leq k-i&\text{ when }&m<i,&\text{ and }\\ n-m<k-i&\text{ when }&m\geq i.\end{array}$$

$$X=\left|\left|\begin{array}[]{ccccccc}1&x_{12}&x_{13}&x_{14}&\cdot&\cdot&\cdot\\ &1&x_{23}&x_{24}&\cdot&\cdot&\cdot\\ &&1&x_{34}&\cdot&\cdot&\cdot\\ &&&1&\cdot&\cdot&\cdot\\ &&&&\cdot&\\ &&&&&\cdot\\ &&&&&&\cdot\\ \end{array}\right|\right|$$

$$A=\left|\left|\begin{array}[]{ccccccc}1\\ &a\\ &&1\\ &&&a\\ &&&&\cdot\\ &&&&&\cdot\\ &&&&&&\cdot\\ \end{array}\right|\right|$$

$$y_{in}=-x_{in}+\sum _{i<{\alpha }_1<n}{x}_{i{\alpha }_1}{x}_{{\alpha }_{1}n}-\sum _{i<{\alpha }_1<{\alpha }_2<n}{x}_{i{\alpha }_1}{x}_{{\alpha }_1{\alpha }_2}{x}_{{\alpha }_{2}n}+\cdots +$$

$$+(-1{)}^{n-i}{x}_{i,i+1}{x}_{i+1,i+2}\dots x_{n-1,n}$$

$$b_{ik}=1(y_{i1}{x}_{1k}+y_{i3}{x}_{3k}+\cdots +y_{i,2l+1}{x}_{2l+1,k})+$$

$$a(y_{i2}{x}_{2k}+y_{i4}{x}_{4k}+\cdots +y_{i,2r}{x}_{2r,k}).$$

$$y_{i1}{x}_{1k}+y_{i3}{x}_{3k}+\cdots +y_{i,2l+1}{x}_{2l+1,k}=\text{vskip}6.0ptplus2.0ptminus2.0pt$$

$$-\sum x_{in}{x}_{nk}+\sum \sum _{i<{\alpha }_1<n}{x}_{i{\alpha }_1}{x}_{{\alpha }_{1}n}{x}_{nk}-\sum \sum _{i<{\alpha }_1<{\alpha }_2<n}{x}_{i{\alpha }_1}{x}_{{\alpha }_1{\alpha }_2}{x}_{{\alpha }_{2}n}{x}_{nk}+\cdots ,$$

$$Y=\left|\left|\begin{array}[]{ccccccc}1&y_{12}&y_{13}&y_{14}&\cdot&\cdot&\cdot\\ &1&y_{23}&y_{24}&\cdot&\cdot&\cdot\\ &&1&y_{34}&\cdot&\cdot&\cdot\\ &&&1&\cdot&\cdot&\cdot\\ &&&&\cdot&\\ &&&&&\cdot\\ &&&&&&\cdot\\ \end{array}\right|\right|,$$

$$U=\left|\left|\begin{array}[]{ccccccc}u_{1}\\ &u_{2}\\ &&u_{3}\\ &&&\cdot\\ &&&&\cdot\\ &&&&&\cdot\\ \end{array}\right|\right|,\qquad V=\left|\left|\begin{array}[]{ccccccc}v_{1}\\ &v_{2}\\ &&v_{3}\\ &&&\cdot\\ &&&&\cdot\\ &&&&&\cdot\\ \end{array}\right|\right|.$$

$$\overline{A_{a}}=\left|\left|\begin{array}[]{ccccccc}1\\ &a\\ &&1\\ &&&a\\ &&&&\cdot\\ &&&&&\cdot\\ &&&&&&\cdot\\ \end{array}\right|\right|,$$

$$\overline{B_{b}}=\left|\left|\begin{array}[]{ccccccc}1\\ &b\\ &&1\\ &&&b\\ &&&&\cdot\\ &&&&&\cdot\\ &&&&&&\cdot\\ \end{array}\right|\right|.$$

$$r_1=\prod _1^n{a}_i{b}_i,r_2=\left(\prod _1^n{a}_i{b}_i\right)a_k,r_3=b_k\prod _1^n{a}_i{b}_i,r_4=\prod _1^n{b}_i{a}_i,$$

$$R_1=\prod _1^n\overline{\overline{A_i}}\overline{\overline{B_i}},R_2=\left(\prod _1^n\overline{\overline{A_i}}\overline{\overline{B_i}}\right)\overline{\overline{A_k}},R_3=\overline{\overline{B_k}}\prod _1^n\overline{\overline{A_i}}\overline{\overline{B_i}},R_4=\prod _1^n\overline{\overline{B_i}}\overline{\overline{A_i}}$$

$$r_{ik}=\sum _{i\le i_2\le i_3\le \dots \le i_{2n}\le k}{\overline{\overline{a}}}_{i{i}_2}^{(1)}{\overline{\overline{b}}}_{i_2{i}_3}^{(1)}{\overline{\overline{a}}}_{i_3{i}_4}^{(2)}{\overline{\overline{b}}}_{i_4{i}_5}^{(2)}\cdots {\overline{\overline{a}}}_{i_{2n-1},i_{2n}}^{(n)}{\overline{\overline{b}}}_{i_{2n},k}^{(n)}.$$

$$r_i\mapsto R_i.$$

Full text

For any elements $x$ and $y$ of the group either $x>y$, or $y>x$, or $x=y$. 2. 2.

If $x>y$ and $y>z$, then $x>z$. 3. 3.

If $x>y$, then $axb>ayb$ for any elements $a$ and $b$ of the group.

the additive group of the ring (field) is ordered, and 2. 2.

for any elements $a$, $x$, $y$ of the ring (field),

[TABLE]

[TABLE]

where some of the $\alpha_{i}$ and $\alpha^{\prime}_{i}$ might be zero, and $a_{1}>\ldots>a_{n}$. We set $A>A^{\prime}$ if for some $r\in\left\{1,\ldots,n\right\}$,

[TABLE]

It is easy to check that the conditions from Definition 2 hold. ∎

We call a triangular matrix any matrix, finite or infinite, with zeroes under the main diagonal.

[TABLE]

Say that $X>Y$ if either of the following conditions holds:

• •

$x_{ii}=y_{ii}$ for $i=1,\ldots,k-1$, and $x_{kk}>y_{kk}$ for some $k$,

• •

$x_{ik}>y_{ik}$ for some $k>i$, and their preceding entries coincide.

One easily checks that the conditions of Definition 1 are satisfied. ∎

We denote by $\mathfrak{M}$ the direct product of two ordered groups $\mathfrak{A}$ and $\mathfrak{B}$. A pair of the form $(a,e_{1})$ where $e_{1}$ is the identity of $\mathfrak{B}$ will be denoted simply by $a$, and a pair of the form $(e,b)$ where $e$ is the identity of $\mathfrak{A}$ will be denoted by $b$.

Consider now the following transcendental triangular matrix:

[TABLE]

We denote by $\mathfrak{G}$ the free abelian group generated by the entries $x_{ij}$ of $X$. This group is orderable (see [2] and references therein). By Lemma 1, the group algebra $\mathfrak{K}=\mathbb{Q}\mathfrak{G}$ is orderable, and thus has no zero divisors. The field of fractions $\operatorname{Frac}(\mathfrak{K})$ of this algebra is also orderable [3]. Consider the group algebra $\mathfrak{L}=\operatorname{Frac}(\mathfrak{K})\mathfrak{M}$, where $\mathfrak{M}=\mathfrak{A}\times\mathfrak{B}$ as above. According to Lemmas 1 and 3, the algebra $\mathfrak{L}$ is orderable.

[TABLE]

where $1$ is the unit of $\mathfrak{L}$ and $a\in\mathfrak{L}$ is neither [math] nor $1$. Then every entry of the matrix $B=X^{-1}AX$ located to the right of or on the main diagonal is non-zero.

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

From this follows:

[TABLE]

[TABLE]

where the external sums are over all odd integers $n$ between $i$ and $k$. This equality shows that the coefficient of $1$ in $b_{ik}$ is non-zero, and so $b_{ik}\neq 0$. ∎

[TABLE]

[TABLE]

Let $\mathfrak{A}$ and $\mathfrak{B}$ be ordered groups. As before, we construct an algebra $\mathfrak{L}=\operatorname{Frac}(\mathbb{Q}\mathfrak{G})\mathfrak{M}$ with $\mathfrak{M}=\mathfrak{A}\times\mathfrak{B}$, where now the free abelian group $\mathfrak{G}$ is generated by the set of all formal entries not only of $X$, but also of $Y$, $U$, and $V$. To every $a=(a,e_{1})\in\mathfrak{M}$ we associate the diagonal matrix

[TABLE]

and to every $b=(e,b)\in\mathfrak{M}$ the diagonal matrix

[TABLE]

Clearly the two sets of matrices $\left.\overline{\mathfrak{A}}=\left\{\overline{A_{a}}\,\,\right|a\in\mathfrak{A}\right\}$ and $\left.\overline{\mathfrak{B}}=\left\{\overline{B_{b}}\,\,\right|b\in\mathfrak{B}\right\}$ form groups naturally isomorphic to $\mathfrak{A}$ and $\mathfrak{B}$ respectively.

Put $\overline{\overline{\mathfrak{A}}}=U^{-1}X^{-1}\overline{\mathfrak{A}}XU$ and $\overline{\overline{\mathfrak{B}}}=V^{-1}Y^{-1}\overline{\mathfrak{B}}YV$. We are going to show that the representations of $\mathfrak{A}$ and $\mathfrak{B}$ given by $a\mapsto\overline{\overline{A_{a}}}$ and $b\mapsto\overline{\overline{B_{b}}}$ induce a faithful representation of the free product $\mathfrak{A}\ast\mathfrak{B}$, that is, given elements of $\mathfrak{A}\ast\mathfrak{B}$ of type

[TABLE]

the corresponding matrices

[TABLE]

are not the identity matrix. We will write down the proof for $R_{1}$ only, as the three remaining cases are similar.

Every entry $\overline{\overline{a}}_{kl}^{\,i}$ of the matrix $\overline{\overline{A_{i}}}$ is equal to $u_{k}^{-1}a_{kl}^{\prime\,i}u_{l}$, where $a_{kl}^{\prime\,i}$ is an entry of $A_{i}^{\prime}=X^{-1}\overline{A_{i}}X$, and $u_{k}^{-1}$ and $u_{l}$ are diagonal entries of the matrices $U^{-1}$ and $U$. Similarly, $\overline{\overline{b}}_{kl}^{\,i}=v_{k}^{-1}b_{kl}^{\prime\,i}v_{l}$, where $b_{kl}^{\prime\,i}$ is an entry of $B_{i}^{\prime}=X^{-1}\overline{B_{i}}X$, and $v_{k}^{-1}$ and $v_{l}$ are diagonal entries of the matrices $V^{-1}$ and $V$.

By Lemma 4, every matrix in the groups $\mathfrak{A}^{\prime}=X^{-1}\overline{\mathfrak{A}}X$ and $\mathfrak{B}^{\prime}=Y^{-1}\overline{\mathfrak{B}}Y$ different from the identity matrix has only non-zero entries to the right of or on the main diagonal. The entries of the matrix $R_{1}$ are given by

[TABLE]

Here $i\leq k$. This sum can be regarded as a polynomial in the diagonal entries of $U$, $V$ and of their inverses. The coefficients of this polynomial are products of entries of the matrices $A^{\prime}_{1},B^{\prime}_{1},A^{\prime}_{2},B^{\prime}_{2},\ldots$. Observe that no monomial occurs twice in the sum as it is given. Moreover, every coefficient is non-zero, since it is a product of non-zero elements of the algebra $\mathfrak{L}$, which has no zero divisors.

Therefore, we have a faithful representation of the free product $\mathfrak{A}\ast\mathfrak{B}$, given by

[TABLE]

Every diagonal entry of $R_{i}$ is either the unit of $\mathfrak{L}$ or a positive invertible element of $\mathfrak{L}$ distinct from the unit. It follows then from Lemma 2 that all matrices of all four types $R_{i}$ together form an orderable group. Therefore, the free product $\mathfrak{A}\ast\mathfrak{B}$ is orderable. ∎

The proof presented here for two factors obviously works for any number of factors.