Infinitary braid groups

TL;DR

This paper explores the properties of infinitary braid groups, extending the concept of finite braid groups to infinitely many strings, particularly focusing on countable cases.

Contribution

It introduces and investigates the properties of more complex, infinitely many-string braid groups beyond the traditional finite cases.

Findings

Basic properties of countable infinitary braid groups analyzed

Extension of braid group concepts to infinite string scenarios

Foundations for further research on infinite braid structures

Abstract

An infinitary version of braid groups has been considered as a direct limit of n-braid groups. However, we can imagine more complicated braids with infinitely many strings. We invetisgate basic properties especially when the number of strings is countable.

Equations

$$\Phi (f)\#\Phi (g)$$

$$(1)$$

$$(2)$$

$$(3)$$

$$(4)$$

$$(1)$$

$$(2)$$

$$(3)$$

$$(4)$$

$$\psi_{\epsilon}(s)=\left\{\begin{array}[]{ll}(1-\frac{s}{\epsilon})(n-1)+\frac{s}{\epsilon}m,&\text{for }0\leq s\leq\epsilon;\\ m,&\qquad\qquad\text{for }\epsilon<s\leq 1-\epsilon;\\ \frac{s-1+\epsilon}{\epsilon}(n-1)+(1-\frac{s-1+\epsilon}{\epsilon})m,&\quad\text{for }1-\epsilon<s\leq 1\end{array}\right.$$

$$H^{\prime\prime}_{n-1}(s,t)=\left\{\begin{array}[]{ll}H^{\prime}((1-2t)(n-1)+2t\psi_{\epsilon}(s),s,1)&\text{for }0\leq t\leq 1/2;\\ (2-2t)g_{\epsilon}+(2t-1)(n-1)&\text{for }1/2<t\leq 1.\end{array}\right.$$

$$(1)$$

$$(2)$$

$$(3)$$

$$(4)$$

$$(1)$$

$$(2)$$

$$(3)$$

$$(4)$$

$$\left\{\begin{array}[]{lll}H(s,1)(n)&=&n;\\ H(s,t)(n)&=&H^{m}(n,s,2^{m+1}(t-\Sigma_{i=1}^{m}1/2^{i})),\\ &&\text{ if }\;\Sigma_{i=1}^{m}1/2^{i}\leq t\leq\Sigma_{i=1}^{m+1}1/2^{i}.\end{array}\right.$$

$$(†)$$

$$\text{such that}h:{\ast }_{i\in m}{\mathbb{Z}}_i\to {\ast }_{i\in E}{\mathbb{Z}}_i\text{ is an isomorphism },\sigma :m\to E$$

$$\text{is a bijection},p_{E}a=h{p}_m,h({\delta }_i)=U_i^{-1}{\delta }_{\sigma (i)}{U}_i\text{ for some }{U}_i,$$

$$h({\Pi }_{i\in m}{\delta }_i)={\Pi }_{i\in E}{\delta }_i,\text{where the products are taken under the}$$

$$\text{natural ordering on }\omega .$$

$$(1)$$

$$(2)$$

$$(3)$$

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Full text

Infinitary braid groups

Katsuya Eda

School of Science and Engineering, Waseda University, Tokyo 169, JAPAN

eda@waseda.jp

 and 

Takeshi Kaneto

Institute of Mathematics, University of Tsukuba, Tsukuba 305, JAPAN

The authors thank J. Morita, A. Tsuboi and K. Sakai for stimulating talks.

This preprint was written before 1993 when the first author was in Uviversity of Tsukuba. Now, according to the new result by W. Herfort and W. Hojka the conclusion of Lemma 2.11 and hence that of Theorem 2.8 becomes “cotorsion” instead of “complete mod-U”.

1. Introduction and definitions

An infinitary version of braid groups has been implicitly considered as a direct limit of $n$-braid groups (see Remark 1.5). However, as an intuitive object, we can imagine a more complicated braid with infinitely many strings (see Figure ($\ast$)).

In this paper we introduce an infinitary version of braid groups and study about fundamental properties of it especially in case the number of strings is countable. Our notation and notion are usual ones for braid groups and topology [2, 8, 11]. $\mathbb{Z},\mathbb{R}$ and $\mathbb{C}$ are the set of integers, real numbers and complex numbers, respectively. The unit interval $\mathbb{I}$ is $[0,1].$ For a path $f$ in a space $X,$ i.e. a continuous map $f:\mathbb{I}\to X,$ $f^{-}$ is the path defined by $f^{-}(t)=f(1-t).$ For paths $f,g:\mathbb{I}\to X$ with $f(1)=g(0),$ $f\cdot g$ is the path defined by: $f\cdot g(t)=f(2t)$ for $0\leq t\leq 1/2$ and $f\cdot g(t)=g(2t-1)$ for $1/2\leq t\leq 1.$ For paths $f$ and $g,$ $f\sim g$ if $f$ and $g$ are homotopic relative to $\{0,1\}.$ A constant path is denoted by the point as usual.

For a subset $X$ of $\mathbb{C}$ let $\mathbb{C}^{X}$ be the product space, regarding $X$ as an indexed set. For distinct $x,y\in X,$ let $\Delta_{xy}=\{u\in\mathbb{C}^{X}:u(x)=u(y)\}$ and $\Delta_{X}=\bigcup\{\Delta_{xy}:x\neq y,x,y\in X\}.$ The subspace $\mathbb{C}^{X}\setminus\Delta_{X}$ of $\mathbb{C}^{X}$ is denoted by $F_{X}.$ For a path $f$ in $F_{X}$ and $Y\subset X\subset\mathbb{C},$ the restriction $f|_{Y}$ is a path in $F_{Y},$ where $f|_{Y}(t)(y)=f(t)(y)$ for $y\in Y.$ Let $\Sigma_{X}$ be the group consisting of all permutations on the discrete set $X,$ which also acts freely on $F_{X}$ as coordinate transformations. For a path $f$ in $F_{X}$ and $\sigma\in\Sigma_{X},$ $f^{\sigma}$ is the path in $F_{X}$ defined by: $f^{\sigma}(t)(x)=f(t)(\sigma(x))$ for $t\in\mathbb{I},x\in X.$

Definition 1.1**.**

For a subset $X$ of $\mathbb{C},$ an $X$-braid $f:\mathbb{I}\to F_{X}$ is a path satisfying $f(0)=\operatorname{id}_{X}$ and $f(1)\in\Sigma_{X}.$ In case $f(1)=\operatorname{id}_{X},$ $f$ is called a pure $X$-braid. $X$-braids $f$ and $g$ are equivalent, if $f\sim g.$ Let $[f]$ be the equivalence class containing $f$ with respect to $\sim$ for an $X$-braid $f.$ Let $A_{X}$ be the set of $X$-braids, $B_{X}=\{[f]:f\in A_{X}\}$ and $P_{X}=\{[f]:f\text{ is a pure }X\text{-braid }\}.$ For $f,g\in A_{X},$ let $f\#g=f\cdot g^{f(1)}.$

In case $X$ is finite, the $X$-braid group is defined as the fundamental group of the orbit space $F_{X}/\Sigma_{X}$ in [6]. Then we can rewrite this definition in terms of paths in $F_{X},$ for $F_{X}$ is a covering space of $F_{X}/\Sigma_{X}$ (see Remark 1.5 (2)). Analogously, we define the $X$-braid group $B_{X}$ for a finite or infinite set $X$ in the sense of the next proposition, where the subgroup $P_{X}$ of $B_{X}$ is just the fundamental group of a space $F_{X}.$

Proposition 1.2**.**

Let $[f][g]=[f\#g]$ for $f,g\in A_{X}.$ Then, this operation is well-defined and $B_{X}$ becomes a group.

Proof.

Since the operation is clearly well-defined, we show the existence of identity and inverse. The equivalence class of the constant map $[\operatorname{id}_{X}]$ is clearly the left identity. For $f\in A_{X},$ let $\sigma=f(1)^{-1}\in\Sigma_{X}.$ Then, $(f^{-})^{\sigma}\in A_{X}$ and $[f][(f^{-})^{\sigma}]=[f\cdot(f^{-})^{\sigma f(1)}]=[f\cdot f^{-}]=e$ and hence $[(f^{-})^{\sigma}]$ is the left inverse of $[f].$ ∎

To simplify the notation, we adopt some convention in set theory, i.e. a non-negative integer $n$ is the set $\{0,\cdot\cdot\cdot,n-1\},$ [math] is the empty set and $\omega$ is the set of all non-negative integers. Hence, $m<n$ if and only if $m\in n$ for $m,n\in\omega.$ We denote the set of positive integers by $\omega_{+}.$ Then, an $n$-braid in the usual sense is the same as defined as above and we can see that the $X$-braid group $B_{X}$ coincides with the usual $n$-braid group if $X=n.$ We identify $f\in A_{X}$ with the indexed family of paths $(p_{x}:x\in X)$ in $\mathbb{C}$ satisfying $p_{x}(0)=x$ for $x\in X,$ $p_{x}(t)\neq p_{y}(t)$ for $x\neq y\in X$ and $\{p_{x}(1):x\in X\}=X,$ i.e. $p_{x}(t)=f(t)(x).$ Then, the restriction $f|_{Y}$ for $Y\subset X$ is the subfamily $(p_{x}:x\in Y)$ of $f=(p_{x}:x\in X).$ we call $p_{x}$ the $x$-string. Since a homotopy in a product space corresponds to homotopies in all components, we get the next proposition, which is obtained just by rewriting the definition of the equivalence $\sim.$

Proposition 1.3**.**

For $X$-braids $f=(p_{x}:x\in X),g=(q_{x}:x\in X),$ $f$ and $g$ are equivalent if and only if there exists an indexed family of homotopies $(H_{x}:x\in X)$ such that

1. (1)

$H_{x}:\mathbb{I}\times\mathbb{I}\to\mathbb{C}$* is continuous;* 2. (2)

$H_{x}(s,t)\neq H_{y}(s,t)$* for $s,t\in\mathbb{I},x\neq y$;* 3. (3)

$H_{x}(0,t)=x,H_{x}(1,t)=p_{x}(1)=q_{x}(1)$* for $t\in\mathbb{I}$;* 4. (4)

$H_{x}(s,0)=p_{x}(s)$* and $H_{x}(s,1)=q_{x}(s)$ for $s\in\mathbb{I}.$*

Now, we can see that $X$-braids are “strongly isotopic”in the sense of Artin [1] if and only if they are equivalent, in case $X$ is finite.

Definition 1.4**.**

Two $X$-braids $f$ and $g$ are strongly equivalent, if there exists a continuous map $H:\mathbb{C}\times\mathbb{I}\times\mathbb{I}\to\mathbb{C}$ such that

1. (1)

$H_{s,t}:\mathbb{C}\to\mathbb{C}$ is a homeomorphism, where $H_{s,t}(\alpha)=H(\alpha,s,t)$; 2. (2)

$H(\alpha,s,0)=H(\alpha,0,t)=H(\alpha,1,t)=\alpha$ for $\alpha\in\mathbb{C},s,t\in\mathbb{I}$; 3. (3)

$H(f(s)(x),s,1)=g(s)(x)$ for $s\in\mathbb{I},x\in X.$

In case $X$ is finite, E. Artin [1, Theorem 6] showed that $X$-braids $f,g$ are equivalent if and only if they are strongly equivalent. More precisely, he showed that for given $H_{x}\;(x\in X)$ in Proposition 1.3 there exists $H$ in Definition 1.4 such that $H(f(s)(x),s,t)=H_{x}(s,t)$ and $H(\alpha,s,t)=\alpha$ if $|\alpha|$ is large enough.

Remark 1.5**.**

(1) If we induce the box product topology to the space $\mathbb{C}^{\omega},$ the situation changes as follows. The path connected component of $\operatorname{id}_{\omega}$ consists of all $u\in F_{\omega}$ such that $u(n)=n$ for all but finite $n\in\omega.$ Therefore, the group obtained by the procedure in Definition 1.1 and Proposition 1.2 in this case is the sum $\bigcup\{B_{n}:n\in\omega\}$ under the natural identification of $B_{n}$ as a subgroup of $B_{n+1}$ and the homomorphic image of the canonical map to $\Sigma_{\omega}$ consists of permutations of finite support.

(2) One might suspect why we do not treat with the quotient space of $F_{X}$ using $\Sigma_{X}$ as in [6]. If we take such a quotient of $A_{X}$ in Definition 1.1 in case that $X$ is infinite, the quotient space does not become Hausdorff nor the quotient map is a regular cover. These are the reasons.

2. Basic results of $B_{X}$ and $\omega$-braids

To show some results of $B_{X}$ the next proposition is necessary. For its proof we debt A. Tsuboi.

Proposition 2.1**.**

For any $X\subset\mathbb{C},$ the space $F_{X}$ is path connected.

Proof.

Let $u\in F_{X}.$ We shall choose $(\alpha,1/2)\in\mathbb{I}\times\mathbb{I}$ for each $x\in X$ so that the dogleg segments from $(x,0)$ to $(u(x),1)$ via $(\alpha,1/2)$ do not intersect for distinct $x^{\prime}s.$ First, well-order $X$ so that $X=\{x_{\mu}:\mu<\kappa\}$ and the cardinality of $\{\nu:\nu<\mu\}$ is less than $2^{\aleph_{0}}$ for each $\mu<\kappa.$ Let $Q_{\mu}=(x_{\mu},0)$ and $S_{\mu}=(u(x_{\mu}),1)$ for $\mu<\kappa.$ Suppose that we have gotten $R_{\nu}\in\mathbb{C}\times\{1/2\}\;(\nu<\mu)$ so that $Q_{\nu}R_{\nu}S_{\nu}$ do not intersect mutually for distinct $\nu$’s. Since there are $2^{\aleph_{0}}$-many planes which contain $Q_{\mu}$ and $S_{\mu},$ there exists a plane which contains $Q_{\mu}$ and $S_{\mu}$ but does neither contain any $Q_{\nu}$ nor $S_{\nu}$ ($\nu<\mu).$ Take such a plane. Then, the intersection of the plane and $Q_{\nu}R_{\nu}S_{\nu}$ consists of at most two points for each $\nu<\mu.$ Therefore, there are less than $2^{\aleph_{0}}$-many points on the plane which are on some constructed $Q_{\nu}R_{\nu}S_{\nu}.$ Tracing $(\alpha,1/2)$’s in the plane, we get $2^{\aleph_{0}}$-many distinct dogleg segments in the plane which connect $x_{\mu}$ and $u(x_{\mu}).$ Therefore, we can choose $R_{\mu}$ so that $Q_{\mu}R_{\mu}S_{\mu}$ does not intersect with $Q_{\nu}R_{\nu}S_{\nu}$ for $\nu<\mu.$ For $t\in\mathbb{I},$ define $f(t)(x_{\mu})$ so that $(f(t)(x_{\mu}),t)$ is on $Q_{\mu}R_{\mu}S_{\mu}.$ Then, $f:\mathbb{I}\to F_{X}$ is a path from $\operatorname{id}_{X}$ to $u.$ ∎

Theorem 2.2**.**

If $X$ and $Y$ have the same cardinality for $X,Y\subset\mathbb{C},$ $B_{X}$ and $B_{Y}$ are isomorphic.

Proof.

Take a bijective function $\varphi:X\to Y.$ Then, $\varphi$ induces a homeomorphism between $F_{X}$ and $F_{Y}.$ Let $\sigma_{f}=f(1)$ for $f\in A_{X}\cup A_{Y}.$ For $f\in A_{X}$ and $g\in A_{Y},$ define paths $\varphi(f)$ in $F_{Y}$ and $\varphi^{-1}(g)$ in $F_{Y}$ by: $\varphi(f)(t)(y)=f(t)(\varphi^{-1}(y))$ and $\varphi^{-1}(g)(t)(x)=g(t)(\varphi(x))$ for $t\in\mathbb{I},x\in X,y\in Y.$ Then $\varphi(f)$ is a path from $\varphi^{-1}$ to $\sigma_{f}\varphi^{-1}.$ By Proposition 2.1, there exists a path $h$ from $\operatorname{id}_{Y}$ to $\varphi^{-1}\in F_{Y}.$ Define $\Phi:A_{X}\to A_{Y}$ by: $\Phi(f)=h\cdot\varphi(f)\cdot(h^{-})^{\varphi\sigma_{f}\varphi^{-1}}.$ It is easy to see $\Phi(f)\sim\Phi(g)$ for $f\sim g.$ On the other hand, $\varphi^{-1}(h\cdot\varphi(f)\cdot(h^{-})^{\varphi\sigma_{f}\varphi^{-1}})=\varphi^{-1}(h)\cdot f\cdot\varphi^{-1}((h^{-})^{\varphi\sigma_{f}\varphi^{-1}}=\varphi^{-1}(h)\cdot f\cdot(\varphi^{-1}(h^{-}))^{\sigma_{f}}.$ Therefore, $\varphi^{-1}(h^{-})\cdot\varphi^{-1}(\Phi(f))\cdot(\varphi^{-1}(h))^{\sigma_{f}}\in A_{X}$ is equivalent to $f,$ which shows that $\Phi(f)\sim\Phi(g)$ implies $f\sim g.$ Therefore, $\Phi$ induces an injection from $B_{X}$ to $B_{Y}.$ The surjectivity of the induced map can be proved similarly as above. The remaining thing to show is $\Phi(f\#g)\sim\Phi(f)\#\Phi(g).$

[TABLE]

∎

Next, we prove an exact sequence which shows a relationship between braid groups and permutation groups.

Proposition 2.3**.**

The following exact sequence holds for any subset $X$ of $\mathbb{C}:$

$0\;\;\to\;\;P_{X}\;\;\to\;\;B_{X}\;\;\to\;\;\Sigma_{X}\;\;\to\;\;0\;.$**

Proof.

Let $h:B_{X}\to\Sigma_{X}$ be the canonical homomorphism, i.e.        $h([f])(x)=f(1)(x)$ for $f\in A_{X},x\in X.$ Then, $\operatorname{Ker}(h)=P_{X}$ clearly. For $\sigma\in\Sigma_{X}$ there exists a path $f$ in $F_{X}$ from $\operatorname{id}_{X}$ to $\sigma$ by Proposition 2.1. Then, $f\in A_{X}$ and $h([f])=\sigma.$ ∎

It is well-known that $B_{n}$ are torsion-free for $n\in\omega$ [6, 3]. We shall show that $B_{\omega}$ is torsion-free. It follows from the next theorem, which shows a fundamental property of $\omega$-braids.

Theorem 2.4**.**

Let $f,g$ be $\omega$-braids. Then, $f\sim g$ if and only if $f|_{n}\sim g|_{n}$ for the restrictions $f|_{n},g|_{n}\;(n\in\omega).$

Corollary 2.5**.**

The $\omega$-braid group $B_{\omega}$ is torsion-free.

Proof.

Let $[f]^{m}=e$ for $f\in A_{\omega},m\neq 0$ and $\sigma=f(1).$ Then, $\sigma^{m}=e$ in $\Sigma_{\omega}$ and hence there exists a partition $\{E_{n}:n\in\omega\}$ of $\omega$ such that the cardinalities of $E_{n}$ are divisors of $m$ and each restriction of $\sigma$ to $E_{n}$ is a cyclic permutation on $E_{n}.$ The restriction $f|_{E_{n}}$ is an $E_{n}$-braid and $[f|_{E_{n}}]^{m}=e.$ Therefore, $[f|_{E_{n}}]=e$ and hence $\sigma|_{E_{n}}=e$ for each $n.$ Now, we have shown that $f$ is a pure $\omega$-braid.

Since $[f|_{n}]^{m}=e$ by the assumption, $f|_{n}\sim\operatorname{id}_{n}$ for every $n\in\omega.$ Hence, $[f]=e$ by Theorem 2.4. ∎

To show Theorem 2.4, some lemmas are necessary.

Lemma 2.6**.**

Let $f\in A_{n}$ be a pure $n$-braid for $n\in\omega$ such that $[f]=e$ and $f(s)(i)=i$ for $i\in k\leq n,s\in\mathbb{I}.$ Then, there exists a continuous map $H:\mathbb{C}\times\mathbb{I}\times\mathbb{I}\to\mathbb{C}$ such that

[TABLE]

Proof.

It suffices to show the case $k=n\!-\!1,$ since the conclusion can be obtained by repeated use of such a special case. By the assumption and [1, Theorem 6], $\operatorname{id}_{n}$ and $f$ are strongly equivalent, that is, there exists a continuous map $H^{\prime}:\mathbb{C}\times\mathbb{I}\times\mathbb{I}\to\mathbb{C}$ satisfying:

[TABLE]

For $\epsilon>0,$ define $\psi_{\epsilon}:\mathbb{I}\to[n-1,m]\;(\subset\mathbb{C})$ as follows:

[TABLE]

Let $g_{\epsilon}(s)=H^{\prime}(\psi_{\epsilon}(s),s,1).$ Since $H^{\prime}(\alpha,0,1)=H^{\prime}(\alpha,1,1)=\alpha$ for any $n\!-\!1\leq\alpha\leq m$ and $H^{\prime}(m,s,1)=m$ for any $s\in\mathbb{I},$ there exists $\epsilon>0$ such that the real part of $g_{\epsilon}(s)$ is greater than $n-1-1/2.$ Fix such an $\epsilon.$ Using $H^{\prime}((1-t)(n-1)+t\psi_{\epsilon}(s),s,1),$ we get a homotopy from $f$ to an $n$-braid whose $(n\!\!-\!\!1)$-string $g_{\epsilon}$ varies in $\{\alpha\in\mathbb{C}:{\mathbf{R}e}(\alpha)>n-1-1/2\}$ keeping the $i$-strings ($i\in n\!\!-\!\!1$) fixed. Then, we can easily make the $(n\!\!-\!\!1)$-string straight leaving the $i$-strings ($i\in n\!\!-\!\!1$) fixed. To perform the works altogether, let $H^{\prime\prime}_{i}(s,t)=i$ for $i\in n\!\!-\!\!1$ and

[TABLE]

To get the desired $H,$ let $H_{i}(s,t)=H^{\prime\prime}_{i}(s,1-t).$ Then, $H_{i}\;(i\in n)$ satisfy the following:

[TABLE]

Again by [1, Theorem 6], we can extend $H_{i}$’s to the desired $H.$ ∎

Lemma 2.7**.**

Let $f$ be a pure $\omega$-braid. Then, $f\sim\operatorname{id}_{\omega}$ if and only if $f|_{n}\sim\operatorname{id}_{n}$ for every $n\in\omega.$

Proof.

It suffices to show the one direction. Suppose that $f|_{n}\sim\operatorname{id}_{n}$ for every $n\in\omega.$ By induction we define pure $\omega$-braids $f_{m}$ and a continuous map $H^{m}:\mathbb{C}\times\mathbb{I}\times\mathbb{I}\to\mathbb{C}$ as the following:

[TABLE]

Using Lemma2.7and the facts $[f|_{m}]=e\;(m\in\omega)$ and $f_{m}(s)(k)=k\;(k\in m),$ we can define these above. Define $H:\mathbb{I}\times\mathbb{I}\to F_{\omega}$ as follows:

[TABLE]

Then, $H(0,t)=H(1,t)=\operatorname{id}_{\omega},$ $H(s,0)=f(s)$ and $H(s,1)=\operatorname{id}_{\omega}$ hold. The continuity of $H$ follows from the fact that $H^{m}(n,s,t)=n$ if $n<m$ and $\Sigma_{i=1}^{m}1/2^{i}\leq t.$ Therefore, $[f]=e.$ ∎

Proof of Theorem 2.4 It suffices to show the one direction. Suppose that $f|_{n}\sim g|_{n}$ for the restrictions $f|_{n},g|_{n}\;(n\in\omega).$ Then, $(f\cdot g^{-})|_{n}=f|_{n}\cdot g^{-}|_{n}\sim\operatorname{id}_{n}$ for every $n\in\omega.$ By Lemma 2.7, $f\cdot g^{-}\sim\operatorname{id}_{\omega}.$ Let $\sigma=f(1),$ then $g(1)=\sigma$ and $[g]^{-1}=[(g^{-})^{\sigma^{-1}}].$ Hence, $[f][g]^{-1}=[f\cdot(g^{-})^{\sigma^{-1}\sigma}]=[f\cdot g^{-}]=[\operatorname{id}_{\omega}]=e,$ i.e. $f\sim g.$

It is known that the abelianization of $B_{n},$ i.e. $B_{n}/(B_{n})^{\prime},$ is isomorphic to $\mathbb{Z}.$ In case of $B_{\omega}$ some difference should occur, since $\Sigma_{X}$ coincides with its commutator subgroup $(\Sigma_{X})^{\prime}$ for an infinite $X.$ We don’t know whether $B_{\omega}=(B_{\omega})^{\prime}$ or not. But, the abelianization of $B_{\omega}$ is not isomorphic to $\mathbb{Z},$ as we shall see in the following.

Some definition is necessary to state the next theorem. An abelian group $A$ is called complete modulo the Ulm subgroup (abbreviated by “complete mod-$U$”), if for any $x_{n}\in A(0<n\in\omega)$ with $n!\mid x_{n+1}-x_{n}$ there exists $x\in A$ such that $n!\mid x-x_{n}$ for all $0<n\in\omega.$ In other words, $A/U(A)$ is complete [7], where $U(A)=\bigcap_{n\in\omega_{+}}n!A.$ Since any homomorphic image of a complete mod-$U$ abelian group is also complete mod-$U,$ a complete mod-$U$ abelian group has no summand isomorphic to $\mathbb{Z}.$ (See [5] for further information about complete mod-$U$ groups.) This kind of group is related to the first integral singular homology groups of wild spaces [4, 5].

Theorem 2.8**.**

The abelianization of $B_{\omega}$ is complete modulo the Ulm subgroup.

It is well-known that $\Sigma_{X}=(\Sigma_{X})^{\prime}$ for infinite $X$ [10, p. 306]. Therefore, we get the following by Proposition 2.3.

Lemma 2.9**.**

For an infinite $X\subset\mathbb{C},$ $B_{X}=P_{X}(B_{X})^{\prime}.$

For $f\in A_{X},$ let $[f]_{a}$ be the element of the abelianization of $B_{X}$ corresponding to $f,$ i.e. the map $[f]\to[f]_{a}$ is the canonical homomorphism from $B_{X}$ to $B_{X}/(B_{X})^{\prime}.$ We shall treat with the cases $X=n$ for $n\in\omega$ and $X=\omega.$

Lemma 2.10**.**

For any pure $\omega$-braid $f$ and neighborhood $O$ of $\operatorname{id}_{\omega}$ in $F_{\omega},$ there exists a pure $\omega$-braid $g$ such that $[g]_{a}=[f]_{a}$ and $\operatorname{Im}(g)\subset O.$

Proof.

There exists $n\in\omega$ such that $O$ depends on the $i$-th co-ordinates ($i\in n$). Reminding the proof of the fact $B_{n+1}/(B_{n+1})^{\prime}\simeq\mathbb{Z},$ we conclude the existence of a pure $(n\!\!+\!\!1)$-braid $g^{\prime}$ such that $[g^{\prime}]_{a}=[f|_{n+1}]_{a}$ and $g^{\prime}(s)(i)=i$ for $i\in n$ and $s\in\mathbb{I}.$ Then, there exists an ($n$+1)-braid $h^{\prime}$ such that $[h^{\prime}]\in(B_{n+1})^{\prime}$ and $[g^{\prime}]=[f|_{n+1}][h^{\prime}].$ Since we may assume $|h^{\prime}(s)(i)|\leq n+1/2$ for $i\in n$ and $s\in\mathbb{I},$ extend $h^{\prime}$ to a pure $\omega$-braid $h$ so that $h(s)(i)=i$ for $n+1\leq i\in\omega,s\in\mathbb{I}.$ Then, $[h]\in(B_{\omega})^{\prime}$ and $(f\!\cdot\!h)|_{n+1}\sim g^{\prime}.$ Take $H$ in Definition 1.4 for $(f\cdot h)|_{n+1}$ and $g^{\prime}$ and define $g$ by: $g(s)(i)=H((f\cdot h)(s)(i),s,1).$ We have gotten a pure $\omega$-braid $g$ such that $g|_{n+1}=g^{\prime}$ and $[g]=[f][h],$ which imply $g\in O$ and $[g]_{a}=[f]_{a}.$ ∎

The next lemma is essentially included in [4, Theorem 1.1]. More precisely, what was necessary to prove it is that the one point union of cones $(CX,x)\vee(CY,y)$ satisfies the condition of the next lemma with the common point as base point and that the first integral singular homology group is an abelian group and a homomorphic image of the fundamental group. Therefore, we omit the proof.

Lemma 2.11**.**

Let $Y$ be a path-connected Hausdorff space and first countable at $y\in Y.$ Let $A$ be an abelian group which is a homomorphic image of $\pi_{1}(Y,y)$ and $h:\pi_{1}(Y,y)\to A$ be the homomorphism. Suppose that for any loop $f$ with base point $y$ there exist loops $f_{n}$ with base point $y$ $(n\in\omega)$ such that $h([f_{n}])=h([f])$ and $\operatorname{Im}(f_{n})$ converge to $y.$ Then, $A$ is complete mod-$U$.

Proof of Theorem 2.8. By Lemma 2.9, the abelianization of $B_{\omega}$ is a homomorphic image of $P_{\omega}.$ Now, the theorem follows from Lemmas 2.10 and 2.11.

Remark 2.12**.**

We have not succeeded to prove the torsion-freeness of $B_{X}$ for uncountable $X.$ As we remarked before Theorem 2.8, we don’t know whether $B_{\omega}=(B_{\omega})^{\prime}$ or not.

3. A representation of $B_{\omega}$ as an automorphism group

on the unrestricted free product

As is well-known, E. Artin [1] represented $B_{n}$ as an automorphism group on free groups of $n$-generators. On the other hand, G. Higman [9] introduced a notion “Unrestricted free product”. In this section, we represent $B_{\omega}$ as an automorphism group on the unrestricted free product of finitely generated free groups. Let $\mathbb{Z}_{i}\;(i\in\omega)$ be copies of the integer group $\mathbb{Z}$ and $p^{m}_{n}:\ast_{i\in m}\mathbb{Z}_{i}\to\ast_{i\in n}\mathbb{Z}_{i}$ the canonical projection for $n\leq m,$ where $\ast_{i\in m}\mathbb{Z}_{i}$ is the free product of $\mathbb{Z}_{i}$’s. We regard $\ast_{i\in m}\mathbb{Z}_{i}$ as the trivial group $\{e\}$ in case $m=0$ as usual. The unrestricted free product $\mathcal{G}_{\omega}$ is the inverse limit $\lim_{\leftarrow}(\ast_{i\in n}\mathbb{Z}_{i},p^{m}_{n}:n\leq m,m,n\in\omega).$ Let $p_{E}:\mathcal{G}_{\omega}\to\ast_{i\in E}\mathbb{Z}_{i}$ be the induced projection for a finite subset $E$ of $\omega.$ In the following, let $\delta_{i}$ be the generator of $\mathbb{Z}_{i}$ which corresponds to $1$ in $\mathbb{Z}.$ For a group $G,$ we denote the automorphism group of $G$ by $\operatorname{Aut}(G).$

Theorem 3.1**.**

Fix an inverse system $(\ast_{i\in n}\mathbb{Z}_{i},p^{m}_{n}:n\leq m,m,n\in\omega)$ for $\mathcal{G}_{\omega}.$ Let ($\dagger$) be a property of $a\in\operatorname{Aut}(\mathcal{G}_{\omega})$ as the following:

[TABLE]

Then, $B_{\omega}$ is naturally isomorphic to the subgroup of $\operatorname{Aut}(\mathcal{G}_{\omega})$ consisting of all automorphisms satisfying ($\dagger$).

To show the theorem, some lemmas are necessary.

Lemma 3.2**.**

[10, Theorem 1]** Let $h:\mathcal{G}_{\omega}\to\mathcal{F}$ be a homomorphism, where $\mathcal{F}$ is a free group. Then, there exist $m\in\omega$ and a homomorphism $\overline{h}:\ast_{i\in m}\mathbb{Z}_{i}\to\mathcal{F}$ such that $h=\overline{h}p_{m}.$

Lemma 3.3**.**

Suppose that $f_{m}:\mathbb{I}\to F_{m}\;(m\in\omega_{+})$ are satisfying $f_{m}(0)=\operatorname{id}_{m},$ $\{f_{m}(1)(i):i\in m\}\subset\omega$ and $f_{m}|_{n}\sim f_{n}$ for $n\leq m,m,n\in\omega_{+}.$ If $\bigcup_{m\in\omega_{+}}\{f_{m}(1)(i):i\in m\}=\omega,$ then there exists $f\in A_{\omega}$ such that $f|_{m}\sim f_{m}$ for every $m\in\omega_{+}.$

Proof.

By [1, Theorem 6], for each $m\in\omega_{+}$ there exists $H^{m}:\mathbb{C}\times\mathbb{I}\times\mathbb{I}\to\mathbb{C}$ such that

[TABLE]

Next, define $f^{k}_{m}:\mathbb{I}\to\mathbb{C}$ as follows: $f^{0}_{m}(s)=f_{m}(s)(m-1)$ and $f^{k}_{m}(s)=H^{m-k}(f^{k-1}_{m}(s),s,1)$ for $0<k\in m.$ Finally, let $f(s)(m)=f^{m}_{m+1}(s)$ for $m\in\omega.$ It is easy to check that $f(s)\in F_{\omega},$ $f(1)=\operatorname{id}_{\omega}$ and $f:\mathbb{I}\to F_{\omega}$ is continuous. By definition, $f|_{m}\sim f_{m}$ for every $m\in\omega_{+}.$ Now, $\bigcup_{m\in\omega_{+}}\{f_{m}(1)(i):i\in m\}=\omega$ implies $f(1)\in\Sigma_{\omega}$ and we have shown the lemma. ∎

Proof of Theorem 3.1. For $f\in A_{\omega},$ let $\sigma=f(1)$ and $n\in\omega_{+}.$ The restriction $f|_{n}$ does not always belong to $B_{n},$ but is still a braid in the sense of [1]. Therefore, $f|_{n}$ induces an isomorphism $e_{n}^{f}:\ast_{i\in n}\mathbb{Z}_{i}\to\ast_{i\in n}\mathbb{Z}_{\sigma(i)}.$ For $n\in\omega,$ denote the set $\{\sigma(i):i\in n\}$ by $\sigma[n].$ Then, the following diagram 3.4 commutes, i.e. $e_{n}^{f}p_{n}^{m}=p_{\sigma[n]}^{\sigma[m]}e_{m}^{f}$ for $n\leq m\;(m,n\in\omega).$

[TABLE]

Since $\sigma[n]\subset\sigma[m]$ for $n\leq m\;(m,n\in\omega)$ and $\omega=\bigcup_{n\in\omega}\sigma[n],$ $\mathcal{G}_{\omega}=\lim_{\leftarrow}(\ast_{i\in\sigma[n]}\mathbb{Z}_{i},p^{\sigma[m]}_{\sigma[n]}:n\leq m,m,n\in\omega).$ Therefore, $f$ induces an automorphism $a_{f}$ on $\mathcal{G}_{\omega},$ which satisfies the property ($\dagger$) by [1, Theorem 15]. It is easy to see that $a_{f}$ only depends on $[f]$ by Theorem 2.4 and the map $[f]\to a_{f}$ is a homomorphism from $B_{\omega}$ to $\operatorname{Aut}(\mathcal{G}_{\omega}).$ The injectivity of this homomorphism follows again from Theorem 2.4.

To show the surjectivity of this homomorphism, let $a\in\operatorname{Aut}(\mathcal{G}_{\omega})$ satisfy the property ($\dagger$). Then, there exist $h_{m}$ and a finite subset $E_{m}$ of $\omega$ which satisfy the properties in ($\dagger$) for each $m\in\omega.$ First, we show $E_{n}\subset E_{m}$ for $n\leq m.$ For $i\in E_{n},$ there is $g_{i}\in\mathcal{G}_{\omega}$ such that $a(g_{i})=\delta_{i},$ where we identify $\mathbb{Z}_{i}$ with the corresponding subgroup in $\mathcal{G}_{\omega}.$ Then, $\delta_{i}=p_{E_{n}}a(g_{i})=h_{n}p_{n}(g_{i})$ and hence $p_{n}(g_{i})\neq e,$ which implies $p_{m}(g_{i})\neq e.$ Therefore, $p_{E_{m}}(\delta_{i})=p_{E_{m}}a(g_{i})=h_{m}p_{m}(g_{i})\neq e,$ which implies $i\in E_{m}.$ For $i\in m,$ $p_{E_{m}}a(\delta_{i})=h_{m}p_{m}(\delta_{i})=h_{m}(\delta_{i})$ and hence $p_{E_{n}}^{E_{m}}h_{m}(\delta_{i})=p_{E_{n}}^{E_{m}}p_{E_{m}}a(\delta_{i})=p_{E_{n}}a(\delta_{i})=h_{n}p_{n}(\delta_{i})=h_{n}p_{n}^{m}p_{m}(\delta_{i})=h_{n}p_{n}^{m}(\delta_{i}).$ Therefore, $p_{E_{n}}^{E_{m}}h_{m}=h_{n}p_{n}^{m},$ i.e. the following diagram commutes.

[TABLE]

Since $h_{m}$ satisfies the conditions in ($\dagger$) and Diagram 3.5, by [1, Theorem 16] there exists a path $f_{m}:\mathbb{I}\to F_{m}$ such that $f_{m}(0)=\operatorname{id}_{m},$ $\{f_{m}(1)(i):i\in m\}=E_{m},$ $e^{m}_{f_{m}}=h_{m}$ and $f_{m}|n\sim f_{n}$ for $n\leq m,m,n\in\omega_{+}.$ If we can show that $f_{m}\;(m\in\omega_{+})$ satisfy the conditions of Lemma 3.3, we get $f\in A_{\omega}$ so that $a=a_{f}.$ Therefore, it suffices to show $\bigcup_{m\in\omega_{+}}E_{m}=\omega.$ For any $j\in\omega,$ there exist $m\in\omega_{+}$ and $\overline{h}:\ast_{i\in m}\mathbb{Z}_{i}\to\mathbb{Z}_{j}$ with $p_{\{j\}}a=\overline{h}p_{m}$ by Lemma 3.2. Take $g$ so that $a(g)=\delta_{j}.$ Then, $\overline{h}p_{m}(g)=p_{\{j\}}a(g)\neq e$ and hence $p_{m}(g)\neq e.$ Since $h_{m}$ is an isomorphism, $p_{E_{m}}(\delta_{j})=p_{E_{m}}a(g)=h_{m}p_{m}(g)\neq e,$ which implies $j\in E_{m}.$

Remark 3.4**.**

Instead of $\mathcal{G}_{\omega},$ there is another candidate to represent $B_{\omega}$ by its automorphism group. It is the fundamental group of the so-called Hawaiian ear ring, which is a subgroup of $\mathcal{G}_{\omega}$ and studied in [5]. But, a natural $\omega$-braid has no naturally corresponding automorphism on the fundamental group of the Hawaiian ear ring. For instance, think of a pure $\omega$-braid such that the first string goes straight, but the others go around the first string.

Supplementary remark In the case of a finite subset $X$ of $\mathbb{C},$ regarding $\mathbb{C}^{X}$ as the mapping space from the discrete space $X$ to $\mathbb{C},$ we can see natural correspondences among $f\in A_{X},$ a continuous maps $f^{\prime}:X\times\mathbb{I}\to\mathbb{C}$ $(f^{\prime}(x,t)=f(t)(x))$ and a level preserving embedding $f^{\prime\prime}:X\times\mathbb{I}\to\mathbb{C}\times\mathbb{I}$ $(f^{\prime\prime}(x,t)=(f(t)(x),t))$ whose image $f^{\prime\prime}(X\times\mathbb{I})\subset\mathbb{C}\times\mathbb{I}$ coincides with a finite braid in the intuitive sense. For $f,g\in A_{X},$ a homotopy between $f$ and $g$ in $F_{X}$ relative to $\{0,1\}$ corresponds to a level- preserving (ambient, in fact) isotopy between the two embeddings $f^{\prime\prime}$ and $g^{\prime\prime}$ keeping $X\times\{0,1\}$ fixed. In case $X$ is infinite, $f^{\prime\prime}$ may fail to be an embedding and is just a level preserving continuous injective map in general. And $f\sim g$ if and only if there exists a level preserving homotopy $H_{t}:X\times\mathbb{I}\to\mathbb{C}\times\mathbb{I}\;(t\in\mathbb{I})$ such that $H_{0}=f^{\prime\prime},$ $H_{1}=g^{\prime\prime}$ and $H_{t}$ is injective. This is our basic view point to infinitary braid groups $B_{X}$ in this paper.