Infinitary braid groups
Katsuya Eda
School of Science and Engineering,
Waseda University, Tokyo 169, JAPAN
[email protected]
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 (∗)).
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].
Z,R and C are the set of integers, real numbers and
complex numbers, respectively. The unit interval I is [0,1].
For a path f in a space X, i.e. a continuous map f:I→X, f−
is the path defined by f−(t)=f(1−t).
For paths f,g:I→X with f(1)=g(0), f⋅g is the path
defined by: f⋅g(t)=f(2t) for 0≤t≤1/2 and f⋅g(t)=g(2t−1)
for 1/2≤t≤1.
For paths f and g,
f∼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 C let CX be the product space, regarding X as an indexed set. For distinct x,y∈X, let
Δxy={u∈CX:u(x)=u(y)} and
ΔX=⋃{Δxy:x=y,x,y∈X}.
The subspace CX∖ΔX of CX is denoted by FX.
For a path f in FX and Y⊂X⊂C,
the restriction f∣Y is a path in FY,
where f∣Y(t)(y)=f(t)(y) for y∈Y.
Let ΣX be the group consisting of all permutations on the discrete
set X, which also acts freely on FX as coordinate transformations.
For a path f in FX and σ∈ΣX, fσ is the path in FX defined by:
fσ(t)(x)=f(t)(σ(x)) for t∈I,x∈X.
Definition 1.1**.**
For a subset X of C, an X-braid
f:I→FX is a path satisfying f(0)=idX
and f(1)∈ΣX.
In case f(1)=idX, f is called a pure X-braid.
X-braids f and g are equivalent, if f∼g.
Let [f] be the equivalence class containing f with respect to ∼
for an X-braid f.
Let AX be the set of X-braids,
BX={[f]:f∈AX} and PX={[f]:f is a pure X-braid }.
For f,g∈AX, let f#g=f⋅gf(1).
In case X is finite, the X-braid group is defined as the fundamental
group of the orbit space FX/ΣX in [6]. Then we can rewrite
this definition in terms of paths in FX, for FX is a covering space
of FX/ΣX (see Remark 1.5 (2)). Analogously, we define the X-braid
group BX for a finite or infinite set X in the sense of the next
proposition, where the subgroup PX of BX is
just the fundamental group of a space FX.
Proposition 1.2**.**
Let [f][g]=[f#g] for f,g∈AX.
Then, this operation is well-defined and BX 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
[idX] is clearly the left identity.
For f∈AX, let
σ=f(1)−1∈ΣX.
Then, (f−)σ∈AX and [f][(f−)σ]=[f⋅(f−)σf(1)]=[f⋅f−]=e and hence
[(f−)σ] 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,⋅⋅⋅,n−1}, [math] is
the empty set and ω is the set of all non-negative integers.
Hence, m<n if and only if m∈n for m,n∈ω.
We denote the set of positive integers by ω+.
Then, an n-braid in the usual sense is the same as defined as above and
we can see that the X-braid group BX coincides with the usual n-braid
group if X=n.
We identify f∈AX with the indexed family of paths
(px:x∈X) in C
satisfying px(0)=x
for x∈X, px(t)=py(t) for x=y∈X
and {px(1):x∈X}=X,
i.e. px(t)=f(t)(x).
Then, the restriction f∣Y for Y⊂X is the subfamily
(px:x∈Y) of f=(px:x∈X).
we call px 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 ∼.
Proposition 1.3**.**
For X-braids f=(px:x∈X),g=(qx:x∈X),
f and g are equivalent if and only if there exists an indexed family of
homotopies (Hx:x∈X) such that
- (1)
Hx:I×I→C* is continuous;*
2. (2)
Hx(s,t)=Hy(s,t)* for s,t∈I,x=y;*
3. (3)
Hx(0,t)=x,Hx(1,t)=px(1)=qx(1)* for t∈I;*
4. (4)
Hx(s,0)=px(s)* and Hx(s,1)=qx(s) for s∈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:C×I×I→C such that
- (1)
Hs,t:C→C is a homeomorphism, where
Hs,t(α)=H(α,s,t);
2. (2)
H(α,s,0)=H(α,0,t)=H(α,1,t)=α for
α∈C,s,t∈I;
3. (3)
H(f(s)(x),s,1)=g(s)(x) for s∈I,x∈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
Hx(x∈X) in Proposition 1.3 there exists H in Definition 1.4
such that H(f(s)(x),s,t)=Hx(s,t) and H(α,s,t)=α if ∣α∣
is large enough.
Remark 1.5**.**
(1) If we induce the box product topology to the space Cω,
the situation changes as follows. The path connected component of
idω consists of all u∈Fω such that
u(n)=n for all but finite n∈ω.
Therefore, the group obtained by the procedure in Definition 1.1 and
Proposition 1.2 in this case is the sum ⋃{Bn:n∈ω}
under the natural identification of Bn as a subgroup of Bn+1
and the homomorphic image of the canonical
map to Σω
consists of permutations of finite support.
(2) One might suspect why we do not treat with the quotient space of
FX using ΣX as in [6].
If we take such a quotient of AX 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 BX and ω-braids
To show some results of BX the next proposition is necessary. For
its proof we debt A. Tsuboi.
Proposition 2.1**.**
For any X⊂C, the space FX
is path connected.
Proof.
Let u∈FX.
We shall choose (α,1/2)∈I×I for each x∈X so that the dogleg segments
from (x,0) to (u(x),1) via (α,1/2) do not intersect for distinct
x′s.
First, well-order X so that X={xμ:μ<κ} and
the cardinality of {ν:ν<μ} is less than 2ℵ0 for each
μ<κ.
Let Qμ=(xμ,0) and Sμ=(u(xμ),1) for μ<κ.
Suppose that we have gotten
Rν∈C×{1/2}(ν<μ) so that
QνRνSν do not intersect mutually for distinct ν’s.
Since there are 2ℵ0-many planes
which contain Qμ and Sμ, there exists a plane which
contains Qμ and Sμ but does neither contain any
Qν nor Sν (ν<μ).
Take such a plane. Then, the intersection of the plane and
QνRνSν consists of at most two points for
each ν<μ.
Therefore, there are less
than 2ℵ0-many points on the plane which are on some
constructed QνRνSν.
Tracing (α,1/2)’s in the plane, we get 2ℵ0-many
distinct dogleg segments in the plane which connect xμ and u(xμ).
Therefore, we can choose Rμ so that
QμRμSμ does not intersect with
QνRνSν for ν<μ.
For t∈I, define f(t)(xμ) so that
(f(t)(xμ),t) is on QμRμSμ.
Then, f:I→FX is a path from idX to u.
∎
Theorem 2.2**.**
If X and Y have the same cardinality for
X,Y⊂C, BX and BY are isomorphic.
Proof.
Take a bijective function φ:X→Y.
Then, φ induces a homeomorphism between FX and FY.
Let σf=f(1) for f∈AX∪AY.
For f∈AX and g∈AY,
define paths φ(f) in FY and φ−1(g) in FY by:
φ(f)(t)(y)=f(t)(φ−1(y)) and
φ−1(g)(t)(x)=g(t)(φ(x)) for t∈I,x∈X,y∈Y.
Then φ(f) is a path from φ−1 to σfφ−1.
By Proposition 2.1, there exists a path h from idY
to φ−1∈FY. Define Φ:AX→AY by:
Φ(f)=h⋅φ(f)⋅(h−)φσfφ−1.
It is easy to see Φ(f)∼Φ(g) for f∼g.
On the other hand, φ−1(h⋅φ(f)⋅(h−)φσfφ−1)=φ−1(h)⋅f⋅φ−1((h−)φσfφ−1=φ−1(h)⋅f⋅(φ−1(h−))σf.
Therefore, φ−1(h−)⋅φ−1(Φ(f))⋅(φ−1(h))σf∈AX is equivalent to f, which shows that
Φ(f)∼Φ(g) implies f∼g.
Therefore, Φ induces an injection from BX to BY.
The surjectivity of the induced map can be proved similarly as above.
The remaining thing to show is Φ(f#g)∼Φ(f)#Φ(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 C:
0→PX→BX→ΣX→0.**
Proof.
Let h:BX→ΣX be the canonical homomorphism, i.e.
h([f])(x)=f(1)(x) for f∈AX,x∈X.
Then, Ker(h)=PX clearly. For σ∈ΣX there exists a path f
in FX from idX to σ by Proposition 2.1.
Then, f∈AX and h([f])=σ.
∎
It is well-known that
Bn are torsion-free for n∈ω [6, 3].
We shall show that Bω is torsion-free. It follows from the next theorem,
which shows a fundamental property of ω-braids.
Theorem 2.4**.**
Let f,g be ω-braids. Then,
f∼g if and only
if f∣n∼g∣n for the restrictions f∣n,g∣n(n∈ω).
Corollary 2.5**.**
The ω-braid group Bω is torsion-free.
Proof.
Let [f]m=e for f∈Aω,m=0 and σ=f(1). Then, σm=e in Σω and hence there exists a partition
{En:n∈ω} of ω such that
the cardinalities of En are divisors of m and each restriction of σ
to En is a cyclic permutation on En. The restriction f∣En is an En-braid and [f∣En]m=e. Therefore, [f∣En]=e and hence
σ∣En=e for each n. Now, we have shown that f is a pure
ω-braid.
Since [f∣n]m=e by the assumption,
f∣n∼idn for every n∈ω. Hence, [f]=e by
Theorem 2.4.
∎
To show Theorem 2.4, some lemmas are necessary.
Lemma 2.6**.**
Let f∈An be a pure n-braid for n∈ω such that
[f]=e and f(s)(i)=i for i∈k≤n,s∈I.
Then, there exists a continuous map
H:C×I×I→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],
idn and f are strongly equivalent, that is,
there exists a continuous map
H′:C×I×I→C
satisfying:
[TABLE]
For ϵ>0, define ψϵ:I→[n−1,m](⊂C) as follows:
[TABLE]
Let gϵ(s)=H′(ψϵ(s),s,1).
Since H′(α,0,1)=H′(α,1,1)=α for any n−1≤α≤m and
H′(m,s,1)=m for any s∈I,
there exists ϵ>0 such that the real part of gϵ(s) is greater
than n−1−1/2. Fix such an ϵ.
Using H′((1−t)(n−1)+tψϵ(s),s,1), we get a homotopy from
f to an n-braid whose (n−1)-string gϵ varies in
{α∈C:Re(α)>n−1−1/2} keeping the i-strings
(i∈n−1) fixed. Then, we can easily make
the (n−1)-string
straight leaving the i-strings (i∈n−1) fixed.
To perform the works altogether, let Hi′′(s,t)=i for i∈n−1
and
[TABLE]
To get the desired H, let Hi(s,t)=Hi′′(s,1−t).
Then, Hi(i∈n) satisfy the following:
[TABLE]
Again by [1, Theorem 6], we can extend Hi’s to the desired H.
∎
Lemma 2.7**.**
Let f be a pure ω-braid.
Then, f∼idω if and only if f∣n∼idn for every n∈ω.
Proof.
It suffices to show the one direction. Suppose that
f∣n∼idn for every n∈ω.
By induction we define pure ω-braids fm and a continuous map
Hm:C×I×I→C as the following:
[TABLE]
Using Lemma2.7and the facts [f∣m]=e(m∈ω) and
fm(s)(k)=k(k∈m), we can define these above.
Define H:I×I→Fω as follows:
[TABLE]
Then, H(0,t)=H(1,t)=idω, H(s,0)=f(s) and
H(s,1)=idω hold. The continuity of H follows from the fact
that Hm(n,s,t)=n if n<m and Σi=1m1/2i≤t.
Therefore, [f]=e.
∎
Proof of Theorem 2.4
It suffices to show the one direction. Suppose that
f∣n∼g∣n for the restrictions f∣n,g∣n(n∈ω).
Then, (f⋅g−)∣n=f∣n⋅g−∣n∼idn for every n∈ω.
By Lemma 2.7, f⋅g−∼idω.
Let σ=f(1), then g(1)=σ and [g]−1=[(g−)σ−1].
Hence, [f][g]−1=[f⋅(g−)σ−1σ]=[f⋅g−]=[idω]=e, i.e. f∼g.
It is known that the abelianization of Bn, i.e. Bn/(Bn)′, is isomorphic
to Z.
In case of Bω some difference should occur, since ΣX
coincides with its commutator subgroup (ΣX)′ for an infinite X.
We don’t know whether Bω=(Bω)′ or not.
But, the abelianization of Bω is not isomorphic to 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 xn∈A(0<n∈ω) with n!∣xn+1−xn there
exists x∈A such that n!∣x−xn for all 0<n∈ω.
In other words, A/U(A) is complete [7],
where U(A)=⋂n∈ω+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 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 Bω is complete modulo the Ulm subgroup.
It is well-known that ΣX=(ΣX)′ for infinite X
[10, p. 306].
Therefore, we get the following by Proposition 2.3.
Lemma 2.9**.**
For an infinite X⊂C, BX=PX(BX)′.
For f∈AX, let [f]a be the element of the
abelianization of BX corresponding to f, i.e. the map [f]→[f]a
is the canonical homomorphism from BX to BX/(BX)′.
We shall treat with the cases X=n for n∈ω and X=ω.
Lemma 2.10**.**
For any pure ω-braid f and neighborhood O of idω in
Fω, there
exists a pure ω-braid g such that
[g]a=[f]a and Im(g)⊂O.
Proof.
There exists n∈ω such that O depends on the i-th co-ordinates (i∈n).
Reminding the proof of the fact Bn+1/(Bn+1)′≃Z, we
conclude the existence of a pure (n+1)-braid g′ such that
[g′]a=[f∣n+1]a and g′(s)(i)=i for i∈n and s∈I.
Then, there exists an (n+1)-braid h′ such that [h′]∈(Bn+1)′ and
[g′]=[f∣n+1][h′]. Since we may assume ∣h′(s)(i)∣≤n+1/2 for
i∈n and s∈I, extend h′ to a pure ω-braid h so that
h(s)(i)=i for
n+1≤i∈ω,s∈I.
Then,
[h]∈(Bω)′ and (f⋅h)∣n+1∼g′.
Take H in Definition 1.4 for (f⋅h)∣n+1 and g′ and
define g by: g(s)(i)=H((f⋅h)(s)(i),s,1).
We have gotten a pure ω-braid g such that g∣n+1=g′ and
[g]=[f][h], which imply g∈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)∨(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∈Y.
Let A be an abelian group which is a homomorphic image of π1(Y,y) and
h:π1(Y,y)→A be the homomorphism.
Suppose that for any loop f with base point
y there exist loops fn with base point y (n∈ω)
such that h([fn])=h([f]) and Im(fn) converge to y.
Then, A is complete mod-U.
Proof of Theorem 2.8.
By Lemma 2.9, the abelianization of Bω is a homomorphic image
of Pω.
Now, the theorem follows from
Lemmas 2.10 and 2.11.
Remark 2.12**.**
We have not succeeded to
prove the torsion-freeness of BX for uncountable X.
As we remarked before Theorem 2.8, we don’t know whether Bω=(Bω)′
or not.
3. A representation of Bω as an automorphism group
on the unrestricted free product
As is well-known, E. Artin [1] represented Bn
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 Bω as an automorphism group on
the unrestricted free product of finitely generated free groups.
Let Zi(i∈ω) be copies of the integer group Z and
pnm:∗i∈mZi→∗i∈nZi the canonical
projection for n≤m, where
∗i∈mZi is the free product of Zi’s.
We regard ∗i∈mZi as the trivial
group {e} in case m=0 as usual.
The unrestricted free product Gω is the inverse limit
lim←(∗i∈nZi,pnm:n≤m,m,n∈ω).
Let pE:Gω→∗i∈EZi be the induced projection for
a finite subset E of ω. In the following, let δi be the
generator of Zi which corresponds to 1 in Z.
For a group G, we denote the automorphism group of G by Aut(G).
Theorem 3.1**.**
Fix an inverse system
(∗i∈nZi,pnm:n≤m,m,n∈ω)
for Gω.
Let (†) be a property of a∈Aut(Gω)
as the following:
[TABLE]
Then, Bω is naturally isomorphic to the subgroup
of Aut(Gω)
consisting of all automorphisms satisfying (†).
To show the theorem, some lemmas are necessary.
Lemma 3.2**.**
[10, Theorem 1]**
Let h:Gω→F be a homomorphism, where
F is a free group.
Then, there exist m∈ω and a homomorphism
h:∗i∈mZi→F such that h=hpm.
Lemma 3.3**.**
Suppose that fm:I→Fm(m∈ω+) are satisfying
fm(0)=idm, {fm(1)(i):i∈m}⊂ω and fm∣n∼fn for n≤m,m,n∈ω+.
If ⋃m∈ω+{fm(1)(i):i∈m}=ω,
then there exists f∈Aω such that f∣m∼fm for every
m∈ω+.
Proof.
By [1, Theorem 6], for each m∈ω+ there exists
Hm:C×I×I→C such that
[TABLE]
Next, define fmk:I→C as follows:
fm0(s)=fm(s)(m−1) and fmk(s)=Hm−k(fmk−1(s),s,1)
for 0<k∈m.
Finally, let f(s)(m)=fm+1m(s) for m∈ω.
It is easy to check that f(s)∈Fω, f(1)=idω
and f:I→Fω is continuous.
By definition, f∣m∼fm for every m∈ω+.
Now, ⋃m∈ω+{fm(1)(i):i∈m}=ω implies
f(1)∈Σω and we have shown the lemma.
∎
Proof of Theorem 3.1.
For f∈Aω, let σ=f(1) and n∈ω+.
The restriction f∣n does not always belong to Bn, but is still a braid
in the sense of [1]. Therefore,
f∣n induces an isomorphism
enf:∗i∈nZi→∗i∈nZσ(i).
For n∈ω, denote the set {σ(i):i∈n} by σ[n].
Then, the following diagram 3.4 commutes, i.e. enfpnm=pσ[n]σ[m]emf
for n≤m(m,n∈ω).
[TABLE]
Since σ[n]⊂σ[m]
for n≤m(m,n∈ω) and ω=⋃n∈ωσ[n],
Gω=lim←(∗i∈σ[n]Zi,pσ[n]σ[m]:n≤m,m,n∈ω). Therefore, f induces an automorphism af on Gω,
which satisfies the property (†) by [1, Theorem 15].
It is easy to see that af only depends on [f] by Theorem 2.4 and
the map [f]→af is a homomorphism from Bω to
Aut(Gω). The injectivity of this homomorphism follows again
from Theorem 2.4.
To show the surjectivity of this homomorphism, let a∈Aut(Gω) satisfy the property (†).
Then, there exist hm and a finite subset Em of ω which satisfy the properties in (†) for each m∈ω.
First, we show En⊂Em for n≤m.
For i∈En, there is gi∈Gω such that a(gi)=δi,
where we identify Zi with the corresponding subgroup
in Gω.
Then, δi=pEna(gi)=hnpn(gi) and hence pn(gi)=e, which implies pm(gi)=e.
Therefore, pEm(δi)=pEma(gi)=hmpm(gi)=e, which implies i∈Em.
For i∈m, pEma(δi)=hmpm(δi)=hm(δi) and hence
pEnEmhm(δi)=pEnEmpEma(δi)=pEna(δi)=hnpn(δi)=hnpnmpm(δi)=hnpnm(δi).
Therefore, pEnEmhm=hnpnm, i.e. the following diagram commutes.
[TABLE]
Since hm satisfies the conditions in (†) and Diagram 3.5, by
[1, Theorem 16]
there exists a path fm:I→Fm such that fm(0)=idm,
{fm(1)(i):i∈m}=Em,
efmm=hm and fm∣n∼fn for n≤m,m,n∈ω+.
If we can show that fm(m∈ω+) satisfy the conditions of
Lemma 3.3, we get f∈Aω so that a=af. Therefore, it suffices
to show ⋃m∈ω+Em=ω.
For any j∈ω, there exist m∈ω+ and
h:∗i∈mZi→Zj with p{j}a=hpm
by Lemma 3.2. Take g so that a(g)=δj.
Then, hpm(g)=p{j}a(g)=e and hence pm(g)=e.
Since hm is an isomorphism,
pEm(δj)=pEma(g)=hmpm(g)=e, which implies
j∈Em.
Remark 3.4**.**
Instead of Gω, there is another candidate to represent Bω
by its
automorphism group. It is the fundamental group of the so-called Hawaiian
ear ring, which is a subgroup of Gω and studied in [5].
But, a natural ω-braid has no naturally corresponding
automorphism on the fundamental group of the Hawaiian ear ring.
For instance, think of a pure ω-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 C, regarding CX as the mapping space
from the discrete space X to C, we can see natural correspondences among
f∈AX, a continuous maps f′:X×I→C
(f′(x,t)=f(t)(x)) and a level preserving embedding
f′′:X×I→C×I (f′′(x,t)=(f(t)(x),t))
whose image f′′(X×I)⊂C×I coincides
with a finite braid in the intuitive sense. For f,g∈AX, a homotopy
between f and g in FX relative to {0,1} corresponds to a level-
preserving (ambient, in fact) isotopy between
the two embeddings f′′ and g′′ keeping X×{0,1} fixed.
In case X is infinite,
f′′ may fail to be an embedding and is just a level preserving continuous injective map in general. And f∼g if and only if there exists a level preserving homotopy Ht:X×I→C×I(t∈I) such that H0=f′′, H1=g′′ and Ht is injective. This is our basic view point to infinitary braid groups BX in this paper.