Annals of Mathematics
Theory of Braids
Author(s): E. Artin
Source: Annals of Mathematics, Second Series, Vol. 48, No. 1 (Jan., 1947), pp. 101-126
Published by: Annals of Mathematics
Stable URL: http://www.jstor.org/stable/1969218 .
Accessed: 22/06/2013 05:14
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .
http://www.jstor.org/page/info/about/policies/terms.jsp
.
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of
content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms
of scholarship. For more information about JSTOR, please contact [email protected].
.
Annals of Mathematics is collaborating with JSTOR to digitize, preserve and extend access to Annals of
Mathematics.
http://www.jstor.org
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
ANNALS OF MATHEMATICS
Vol. 48, No. 1, January, 1947
THEORY OF BRAIDS
BY E.
ARTIN
(Received May 20, 1946)
A theoryof braids leading to a classificationwas givenin my paper "Theorie
der Zopfe" in vol. 4 of the HamburgerAbhandlungen(quoted as Z). Most of
the proofsare entirelyintuitive. That of the main theoremin ?7 is not even
convincing. It is possible to correctthe proofs. The difficulties
that one encountersif one triesto do so come fromthe fact that projectionof the braid,
whichis an excellenttool forintuitiveinvestigations,is a very clumsyone for
rigorousproofs. This has lead me to abandon projectionsaltogether. We shall
use the more powerfultool of braid coordinatesand obtain therebyfarther
reachingresultsof greatergenerality.
A fewwordsabout the initialdefinitions. The factthat we assume of a braid
stringthat it ends in a straightline is of course unimportant. It could be replaced by limit assumptionsor introductionof infinitepoints. The present
definition
was selectedbecause it makes some of the discussionseasier and may
be replaced any time by anotherone. I also wish to stress the fact that the
definitionof s-isotopyis of a provisionalcharacteronly and is replaced later
(Definition3) by a general notion of isotopy.
More than halfofthe paper is of a geometricnature. In thispart we develop
some resultsthat may escape an intuitiveinvestigation(Theorem 7 to 10).
We do not prove (as has been done in Z) that the relations(18) (19) are definingrelationsforthe braid group. We referthe readerto a paper by F. Bohnenblust1wherea proofof thisfactand of manyof our resultsis given by purely
group theoreticalmethods.
Later the proofsbecome more algebraic. With the developed tools we are
able to give a unique normalformfor every braid2(Theorem 17, fig.4 and remark followingTheorem 18). In Theorem 19 we determinethe centerof the
braid groupand finallywe give a characterisationof braids of braids.
I would like to mentionin this introductiona few of the more importantof
the unsolved problems:
1) Assumethat two braids can be deformedinto each otherby a deformation
of the most generalnatureincludingselfintersectionof each stringbut avoiding
intersectionof two different
strings. Are theyisotopic? One would be inclined
to doubt it. Theorem8 solves, however,a special case of this problem.
2) In Definition3, we introducea notion of isotopy that is already very
general. What conditionsmust be put on a many to many mappingso that
the result of Theorem 9 still holds?
I F. Bohnenblust, The algebraical braid group, Ann. Math., vol. 48, (1947), pp. 127-136.
2The freedom of the group of k-pure braids has been proved with other methods in:
W. FR6HLICH, Uber ein spezielle8 Transformationsproblembei einer besonderen Klasse von
Zopfen, Monatshefte fur Math. und Physik, vol. 44 (1936), p. 225.
101
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
102
E. ARTIN
3) Determine all automorphismsof the braid group.
4) With what braids is a given braid commutative?
5) Decide forany two givenbraidswhethertheycan be transformed
into each
otherby an innerautomorphismofthe group. Concerningapplicationsofbraid
theorythis is by far the most importantproblem.
The last threeof our questionsseem to requirean extensivestudyofthe automorphismsof freegroups.
We shall have to considernumerousfunctionsof several variables. All of
themare meantto be continuousin all the variables involvedso that the statementof continuityis always omitted. Only morestringentconditionsshall be
mentioned.
Let x, y, z be the Cartesian coordinatesof a 3-space. By a braid stringwe
mean a curvethat has preciselyone pointof intersectionwitheach plane z = a
so that z may be used as parameter. Denotingby X the two dimensionalvector
(x, y) we may thereforedescribethe stringby a vectorfunctionX = X(z). In
addition to that we assume the existenceof two constantsa, b such that X(z)
assumesa constantvalue X forall z ? a and a constantvalue X+ forall z _ b.
X- and X+ are called the ends of the string.
By an n-braidwe mean a set of n stringsXi(z) (i = 1, 2, ... n) withoutintersections(hence Xi(z) $ Xk(z) fori # k) wherethe numberingof the stringsis
consideredunessential.
Two n-braidsXi(z) and Yi(z) are called stronglyisotopic (s-isotopic)ifn vectorfunctionsXj(z, t) can be foundwiththefollowingproperties:They are defined
forall z and forall tofa certainintervalc < t < d, givean n-braidforeach special
t and are Xi(z) fort = c, Yj(z) fort = d. They are constantin z and t if z is
large enough and also constantif -z is large enough. We remarkat once that
the ends remainfixed.
THEOREM 1. s-isotopyis reflexive,
symmetric
and transitive.
This allows us to unite s-isotopicbraidsinto one class. We also obtain without difficulty:
THEOREM 2. Let g(z, t) be a numericalfunctiondefined
for all z and all t of a
certaint-interval.Assume thatit tendswithz to oo uniformly
in t (the sign is
unimportant). If Xi(z) is a braid thenthe new braids Xi(g(z, t)) for different
values of t are all s-isotopic. (They need not be s-isotopicto Xi(z) itself.)
COROLLARY 1. If thenumericalfunctiong(z) satisfieslim g(z) = + so and
lim g(z)
z-00
z=+oo
=
-
a, thenthebraidsXi(z) and Xi(g(z)) are s-isotopic.
PROOF: Put g(Z, t) = (1 - t) z + t g(z) for0 _ t < 1.
COROLLARY 2. Xi(z) is s-isotopicto any z-translation
Xi(z + t).
COROLLARY 3. The braids Xi( I z + t) are s-isotopicamong themselves
for
J
valuesof t. For largepositivevaluesof t all braidstringsare constant=
different
Xt. For largenegativevalues of t thepart of thebraid abovethexy-planelooks
like a z-translation
of Xj(z), thepart belowthisplane like its reflection
on thexyplane.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
THEORY OF BRAIDS
103
We mentiona fewtheoremsofwhichlittleuse shall be made in thispaper,but
that are usefulif braids are to be studied by means of theirprojectionon the
yz-planefromthe positivex-direction. By lexicographicalarrangementof vectors we mean, as usual, theirarrangementaccordingto the size of theiry-coordinateand, in case the y-coordinatesare equal, accordingto theirx-coordinate.
Two braids are said to have the same z-patternof projectionif the lexicographicalarrangementof the vectors Xi(z) is for each value of z the same as
that of the vectors Yi(z).
LEMMA.
Two braidswiththesame z-patternof projectionare s-isotopic.
PROOF: Put Xi(z, t) = (1 - t)Xi(z) + t Yi(z) for0 ? t ? 1. Assume we
could findvalues z, t and i $ k such that Xi(z, t) = Xk(Z, t). This means that
(1 - t)(Xi(Z) - Xk(Z)) + t(Yi(z) - Yk(z)) = 0. Let Xi(z) - Xk(z) = (a, b)
and Yi(z) - Yk(z) = (a', b'). The y-coordinateof our equation shows that b
and b' can not have the same sign hence b = b' = 0. Now the x-coordinateof
the equation leads to a = a' = 0. But thenXi(z) would not be a braid.
Two braids Xi(z) and Yj(z) are said to have the same patternof projection
if a monotonicallyincreasingfunctiong(z) with infinitelimitsexists such that
Xi(z) and Yi(g(z)) have the same z-patternof projection.
THEOREM 3. Two braids with the same patternof projectionare s-isotopic.
PROOF:Use the lemma and Corollary1 of Theorem2.
THEOREM 4. Let d be less than half theminimaldistancebetweentwoof the
stringsofXi(z). Let Yi(z) be a braidwiththesame endsas Xi(z) and assumethat
equallynumbered
stringsofthetwobraidshaveat each z levela distanceless thand.
Then thebraidsare s-isotopic.
The proofis done withthe same device as in the lemmaand is trivial.
DEFINITION: Two braids X,(z) and Yj(z) are called composable if:
1) They have the same number of strings.
2) Aftera suitable changein the numberingofthe stringswe have Y+l = X .
So the upper ends of Yj(z) mustfitthe lowerends of Xi(z).
If (aftera suitabletranslation)we join theseendswe obtain a new braid which
is said to be composedof X,(z) and Yi(z) (in this order). The formaldefinition
would be:
Select an a and a b such that Xi(z + a) is constantfornegativez and Yi(z + b)
is constantforpositivez. Put Zi(z) = Xi(z + a) forpositivez and = Yi(z + b)
fornegativez. This braidand any translationofit is called the composedbraid.
We still have a great freedomin the selectionof a, b and the translation. All
these braids are howevers-isotopic.
If we replaceboth braidsby othersthat are s-isotopicto themwe obtainbraids
that are s-isotopicto Zi(z). The prooffollowsfromthe fact that aftersuitable
selectionof a and b the necessarydeformationcan be carriedout independently
in bothsectionsof the composedbraid. This leads to:
DEFINITION 1. Two classes of braids A and B are called composableif the
braids in these classes are composable. The resultingbraids forma class denotedby AB. It is to be remarkedthat ifA is composablewithB then B may
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
104
E. ARTIN
not be composablewith A. And even if the classes are composablefromboth
sides then the commutativelaw may not hold. But the associative law does.
If A and B as well as AB and C are composablethenB and C as well as A and
BC are composable and we have A(BC) = (AB)C.
THEOREM 5. The classes of n-braidsforma groupoidunderour composition.
PROOF: The postulates that we must verifyare:
1) The kind of associative law we just have mentioned.
2) The existenceof two classes U and U' foreach given class A (dependent
on A) such that AU = U'A = A. U is obviouslythe class containinga constant braid with the same lower ends as A and U' the similarclass connected
withthe upper ends of A.
3) The existenceof a class A-' such that A-'A = U and AA-' = U'. Corollary 3 shows that the reflectionof a braid on the x, y plane gives such a class.
4) If A and B are givenclasses thereexistsa class C such that AC as well as
CB can be formed. This just means to constructan example of a braid with
given ends.
If U is one oftheunit classes call Gu theset ofall A that have U as leftas well
as rightunit. They forma group. If V is anotherunit and C a class such that
thusindicatedis an
UC = CV = C thenG, = C'1GuC and the transformation
isomorphism. The knowledgeof one of these groups reveals the structureof
all and as a matterof fact the structureof the whole groupoid. The braids in
sucha groupare thosewhoseupperendsare onlya permutationofthelowerends.
Next we prove that s-isotopycan be extendedto the wholespace:
an s-isotopy. Thenwe
THEOREM 6. Let Xi(z, t) be then functionsdescribing
forall X and z and all necessaryt,whosevalue
can finda functionF(X, z, t) defined
is a vector,
and thathas thefollowingproperties:
of theplane. That meansthatit is a one
1) For anyfixedz it is a deformation
toone correspondence
fort = a if that
oftheplane if t also is fixedand it is identity
is the beginningof the t-interval.
2) Shouldforany specialvalueofz theoriginalfunctionsXi(z, t) be independent
of t,thenF(X, z, t) = X forthatz and all X and t.
largedistancefromtheorigin
3) If a point(X, z) ofthe3-spacehas a sufficiently
thenF(X, z, t) = X for all t.
ofthespace movesthebraid4) F(Xi(z, a), z, t) = Xi(z, t). So thedeformation
stringspreciselyas the s-isotopydoes.
PROOF: 1) Select an r > 0 such that I Xi(z, t) - Xk(Z, t) I < 3r forall i # k
and all z and t. We firstconstructan auxiliaryfunctionG(X, Ps,, Q,) of X and
2n points Ps,, Q, (v = 1, 2, ***n) of the plane. The points P, are restricted
by the conditionthat theirmutual distanceshall always be greaterthan 3r, the
points Q, by the conditionthat Qi lies in the interiorof a circleCi of radius r
aroundPi . The value of G shall be X ifX is outsideof all the circlesor on the
peripheryof one. For X = Pi the value shall be Qi . If X is in the interior
of Ci bat different
fromPi draw a radius throughX and call R its intersection
with Ci. Define the functionvalueas that point on the straightline segment
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
THEORY
105
OF BRAIDS
R Qi that bisectsit in the same ratio as X bisectsR Pi. Our functionis continuousin all the variables,is a one to one correspondenceof the plane forfixed
P, and Q, and reducesto identityif P, = Q, forall v.
2) Divide the t intervalinto a finitenumberof parts tj < t < tail such that
the variationof everyXi(z, t) in that intervalis less than r forfixedz.
In a = to< t < ti we defineF(X, z, t) = G(X, X,(z, a), X,(z, t)). It has all
the necessaryproperties. Assume that we succeeded to defineF(X, z, t) for
all t of to< t < t, and to checkon the requiredproperties. For tm< t < tm+i
we define:
F(X, Z, t) = G(F(X, z,
tm), Xp(z, tm)
X,,(z,t)).
For t = tm we get the old value so it is a continuouscontinuation. The properties 1, 2, 3 follow immediately. For X = Xi(z, a) we get F(X, z, ti) =
Xi(z, ti) hence F(X, z, t) = Xi(z, t).
The extensionofan s-isotopyto thewholespace is not the onlyuse ofTheorem
6. We also use it to introducenew coordinatesfor the points of the space
called braid coordinates. They are much more flexiblein dealing with braids
and the principaltool in the proofsof most of the followingtheorems.
Let Xi(z) be a given braid, constantforz < a and forz > b. Considerthe
braids Xi(z + t) for0 _ t < b - a. They are all isotopic and let F(X, z, t)
be the extensionof this isotopy to the whole space. Then we have:
(1)
F(Xi(z),z,t)
= Xi(z+
t),
0<
t ? b - a.
With each point (x, z) of the space we associate now a 2 dimensionalvector
Y = Y(X, z) in the followingway:
For z ? a let Y = X.
For a ? z < b let Y be theunique solutionofF(Y, a, z - a) = X. For z = a
X = Y again.
we have F(X, a, 0) = X, therefore
For z > b put Y(X, z) = Y(X, b).
If z is fixedthen the mapping Y = Y(X, z) is a one to one correspondenceof
the plane that is certainlyidentity,outside a large circlewhose radius does not
depend on z. It is identityforall X if z < a. For z > b it is in generalnot
identitybut at least is the same mappingforall z > b.
F(Xi(a), a, z - a) = Xi(a + (z - a)) = Xi(z) if a < z < b because of (1).
This shows that the Y forthe point (X,(z), z) of the ith stringis Xi(a) = X,.
The same is trueforz < a and forz > b fortrivialreasons.
combination{ Y, z
We associate now withthe point (X, z) the corresponding
and call it the braid coordinatesof that point. They equal the ordinarycoordinatesforall z < a and also forall large I X I . All pointson the ith string
have the simple braid coordinates {X-, z}.
Anotherway to look upon the braid coordinatesis this: Interpretthemas
ordinarycoordinatesof a point of a 3-space. Then our 3-space is mapped by
onto thisnew one and the braid stringsare mapped
a one to one correspondence
onto verticallines with the same lower ends.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
106
E. ARTIN
Let now u be any real number. The mapping { Y, z}
{Y, z + u} in our
old space is a one to one correspondencethat has all the essentialfeaturesof
a translationand shall thereforebe called translationby u along the braid.
z I is
Each single stringremainsfixedas a whole. For large I X I and also if J
large in comparisonto I u I it is an ordinarytranslation. For the otherpoints
the z-coordinatedoes at least behave in the ordinaryway.
We can of course also findthe inversefunctionX = H(Y, z) that describes
the passage from Y back to X. Let now X(z) be any other braidstring(it
may intersectthe stringsof our braid). Apply to it a translationby u along
our braid. The braid coordinatesof (X(z), z) are { Y, z} whereY = Y(X(z), z).
The translationmoves it into { Y, z + u} . The ordinarycoordinatesare (X',
z + u) whereX' = H(Y(X(z), z), z + u). Lookingat the newstringas a whole,
we may replace z by z - u and obtain the new braidstring
X(z, u) = H(Y(X(z
-),
z
-
u), z)
such that (X(z, u), z) are the points of the new braidstring. Letting now the
u change,we see that we have beforeus an s-isotopicchangewhere the points
move only in horizontalplanes. Should the originalstringnot intersectthe
thenits translationdoes not eitherand thebraidformedout of the
braidstrings,
old braid by adding our new stringundergoesan s-isotopyunder translation.
We use this in the followingway: Let Xi(z) be an n-braidand replace its nth
stringby any otherstringX'(z) with the same ends. The new stringmay intersectXn(z) but shall not intersectthe otherstringsof our braid. This gives
a new n-braidand we apply now to our old braid a translationby a large u along
our new braid. What happens? The n - 1 firststringsremainfixed. If u
large then a verylow portionof the nth stringwill now be in the
is sufficiently
main part of the braid. In that very low portionthe stringXn(z) = X'n(z).
The stringX'(z) does not change under our translation. This shows that in
This is
the main portionof our braid X"(z) has moved into the positionX'(z).
of course compensatedby the fact that the nth stringis now entangledin the
otherstringsabove the main portionof the braid. But above the main section
of the braid,the firstn - 1 stringsare verysimple,namelyparallel lines. Rememberfinallythat we have shownin the precedingparagraphthat the translation can also be consideredas a horizontalmotion,as an s-isotopy.
movingonestringonly
THEOREM7. It is possibletoapplytoa braidan s-isotopy
intoany otherpositionprovidedthisis compenthisstringmay be brought
whereby
satedbya motionof thestringabovethemain sectionof thebraidwherethen - 1
otherstringsare parallel.
This suggeststhe definition:
DEFINITION 2. A braid withthe same upper and lower ends is called i-pure,
if all but the i th stringare constant. A class is called i-pure if it contains an
i-pare representative.
We see: if B and B' are braids or classes of braids having n - 1 stringsin
commonthen B = AB' whereA is i-pure.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
THEORY OF BRAIDS
107
Anotherusefulnotion connectedwith braid coordinatesis that of projection
along the braid. Considerthe plane z = zo . The mapping { Y, z} - { Y, zoI
is this projection. It carriesthe braidstringsinto theirintersectionwith the
plane and a pointnot on the braidinto a pointdifferent
fromtheseintersections.
Let now RB be the complementary
set to B in the 3-space. We introducethe
usual notion of homotopyof paths in RB. Two paths a = { Y(t), z(t)}, b =
I Y'(t), z'(t) }, 0 < t < 1 are called homotopicin RB: a -,,Bb if a functionof two
variables t and s (interval0, 1) { Y(t, s), z(t, s) } can be found,that is constant
for t = 0 and t = 1, gives the firstpath for s = 0 and the second fors = 1.
All pointsof the deformationhave to belong to RB whichmeans simplythat the
functionY avoids the values X7.
The compositionof homotopyclasses is introducedin the usual way and leads
to a groupoid. If we select a point P in RB and considerthe homotopyclasses
ofthose paths that have beginningas well as endpointat P, we obtain the Poincar6 group of RB. Let zo be the z coordinateof P and a = { Y(t), z(t)} any
elementof this group. The projectiona' = { Y(t), zo} is homotopicto a as the
function{ Y(t), zo + s(z(t)- zo)} shows. If two pathsin thisplane are B-homotropic,say by the function{ Y(t, s), z(t, s) }, thenthe function{ Y(t, s), zo} shows
that the paths are already homotopic in the plane. The Poincar6 group is
the same as that of a plane puncturedin the n pointsXi(zo). So it is
therefore
a freegroupwith n generators.
We must now carefullydescribethe generatorswe want to use. The plane
willbe eitherin the regionofz wherethe braidstringsassumethe constantvalues
X, and shall thenbe called a lowerplane or in the regionof the X+ whenwe call
it an upperplane. Take an upper plane and draw in it a ray that does not meet
any of the upper or lower ends. We intend to take the point P on that ray
far away. Each of the points X+ shall be connectedwith the besufficiently
ginningpoint Q of the ray by a brokenline withoutselfintersectionsuch that
two of the lines and also the ray have only the endpointQ in common. By 1i
we denotethe connectionthus establishedbetweenthe beginningpoint X+ and
the point P on the ray. By li(e) we mean the same path but startingwith
the parametervalue e. An orientationof the plane is selected. By ci(e) we
mean a curvewiththe windingnumber1 around X+ startingand endingat the
beginningpoint of li(e) that stays withina small neighborhoodof X+. It is
well known that the paths (for small e)
(2)
t = I(e)yc (e)l (e)
are freegeneratorsof the Poincar6groupof the puncturedplane. This pattern
to all otherupper planes by projection(in the ordiof paths is then transferred
nary sense) includingthe point P. We use the same names for the paths in
all the upper planes.
In a lower plane we firsttransferthe ray, the orientationand the point P to
it by ordinaryprojection. Since we now have to take eare of the lower ends,
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
108
E. ARTIN
new paths are selecteddenotedby 1', l'(e), c'(e). The corresponding
generators
shall be called t',.
What we intendto do with thissetup is roughlythis: If two braids are s-isoto consider
topic theymustby necessityhave thesame ends. It therefore
suffices
braids with given ends. For all these braids we use the same patternof paths
in the lower and upper planes. If B is a given braid then we project by braid
projectionthe generatorst' into the upper plane. We obtain paths 7i that are
now generatorsin the upper plane. Thereforethey can be expressedin terms
of the ti . It turnsout that these expressionsare a completeset of invariants
forthe isotopy classes.
As a firstindicationthat the studyof the homotopyclasses must give a solution of our problemlet us considerthe followingspecial case:
Let C be an (n - 1)-braid. Form two n-braidsby insertingin C an nthstring
in two ways but both time with the same ends. Select a zo such that the two
stringsare equal forz < zoand call P0 the pointon thenth stringsat thatzo-level.
In a similarfashionselect zi and P1 forthe upper end. Considernow the two
pieces of the nthstringsbetweenPo and Pi . If they are homotopicrelativeto
C we may call the two nthstringsC-homotopic. Then the followingrathersurprisingtheoremholds:
THEOREM 8. Let B and B' be obtained
fromC byinsertionofan nth stringwith
thenB and B' are s-isotopic. Every
gien ends. If thetwostringsare C-homotopic
class can be realizedby a braid. The converseis also truebut will be
homotopy
provedlater.
PROOF: 1) We use braid coordinatesof C and expressthe two insertedstrings
in termsof thesecoordinates:Yn(z), Y' (z). The factthat theydo not intersect
the stringsof C means just that the functionsavoid the values X, . By assumptionthereexists a function{ Y(t, s), z(t, s) } definedin zo < t < z1, 0 <
s ? 1 describingthe homotopyof the strings. So Y(t, s) will avoid the values
X,, have fixedbeginningand end pointsforall s and willbe Y"(t) fors = 0 and
Y'(t) for s = 1. The method consistsnow in forgettingabout the function
z(t, s) altogetherand to definea functionYn(z, s) as equal to Y(z, s) forzo <
z < zLand equal to X, forall otherz. The functionavoids X- and reducesto
0 and s = 1. So it gives the requireds-isotopy.
the givenstringsfors
2) If { Y(t), z(t) } is any curve definedin zo < z ? z1 that avoids the strings
of C and joins Po and P1, put as beforeYn(z) = Y(z) in that intervaland =
Xn forall otherz. The function{ Y(t), s *z(t) + (1 - s)t} shows that this nth
stringis homotopicto the given curve.
The s-isotopyof Theorem8 moves the nthstringonly.
REMARK.
Let us now returnto our upper plane. Joinone of the points X+ to P by a
curve h that avoids the braid with exceptionof its beginningpoint. Define
h(e) as beforeand let d(e) be a curve analogous to c,(e). Considerthe element
t = h(e)f'd(e) h(e). If we join the beginningpoint of li(e) to that of h(e) by a
path e that stays in a small neighborhoodof Xt then e lci(e)e is homotopicto
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
THEORY
109
OF BRAIDS
d(e). The path S' = li(e)-le h(e) is a certainelementof the group and may be
expressedin terms of the t. The element
SI- tiS' = h(E)-'elli(e)l(-)-'ci(e)li(e)li(-)-e
h (e)
is homotopicto t. Assume now that a similarexpressiont = S-1tiS is known
fromsome othersource and may even not be in a reducedform. We firstperformthe possible cancellationsin S only; a furthersimplification
of the expression S-1tiS is thenpossibleonly if S beginswitha powerof ti . Then S-1 ends
with the reciprocalof that power and we see that this term may indeed be
dropped. This shows that S is uniquely determinedbut fora power of ti and
we have thereforeS' = ttS or
(3)
li(e)le
h(e) = tS.
Let us now reinsertin our plane the one point X+ and considerhomotopies
in that new plane, puncturedin n - 1 points only. This homotopyshall be
denoted by -i . It amounts to put ti = 1 in all previousexpressions. The
resultingelementshall still be denoted by S. We obtain:
li(e))'e h(e)
-'
S.
The path lili(e)-le h(e)h-l is i-homotopicto a closed curvestartingat Xt and
remainingin a small neighborhoodof that point. It is thereforei-homotopic
to this point X+. This proves the formula
(4)
17lh'.'i S.
Let now B be a braid with the given ends. If we apply braid projectionto
a generatort' of a lower plane onto an upper plane and assume that the point
P has been selected sufficiently
far out, then the image 1iwill be an elementof
the Poincar6 group forP. If 1i is the projectionof l' we obtain equations of
the form:
(5)
t- = S'ltiSi
Si
1i(E)-le 1 (e)
1-11I Si
It is to be remarkedthat the propertiesof the braid coordinatesshow that the
formof the equation (5) does not depend on the preciselocation of the upper
and the lowerplane. It is also clear that it does not depend on e providedit is
only small enough. The positionof P plays also no role in it providedthat it
is far enough out. The equations (6) and (7) change of course theirmeaning
and the exponentr may depend on e and ei .
We may look upon this processin yet anotherway. Call g the straightline
segmentthat connectsP with its projectionin the lower plane and put ri =
gt g-1. They are elementsof the Poincar6 groupforP and as a matterof fact
a set of generators. If we subject them to braid projectionthey will go over
(6)
(7)
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
110
E. ARTIN
again into the Ti. But being elementswith the same beginningand end point
ri is, as we have seen before,homotopicto Ti. So equation (5) becomes
(8)
7'
SL1tiSi.
We are now ready fora generalizednotionof isotopy:
DEFINITION 3. A braid is called isotopicto anotherbraid if the space can
be mapped into itselfin such a way that pointson the firstbraid but no other
points of the space are mapped onto points of the second braid. In addition
to thiswe assume that the mappingis identityoutsideof a certainsphere. Inside that spherethe mappingmust of coursebe continuousbut need not be one
to one.
Considernow two isotopicbraids B and C. Locate the lowerand the upper
plane outsidethe sphereand select P so that g is also outsidethis sphere. The
two elementsri and its B-projectionti are B-homotopic. The surfaceconnecting the two paths does not meetB. Its image underour mappingwilltherefore
avoid C. This shows that the images of our paths are C-homotopic. But our
paths remain fixed. So ri and ti are C-homotopic. ri on the other hand is
C-homotopicto its C-projection. This C-projectionis thereforeC-homotopic
to Ti and this proves:
in formula(5) are
THEOREM 9. If B is isotopicto C thenthe exexpressions
thesamefor B and C.
The i-homotopyof formula(7) may be interpretedas homotopy'withrespect
to the braid resultingfroma cancellationof the ith string. Denote by Li the
piece of the ith stringthat startsat the upper plane and ends at the lowerplane.
The path ii' 2il'g-' is a closed path startingat P and as such i-homotopicto
its projectiononto the upper plane. But this projectionis obviouslythe left
side of (7). Computing2i out of the resultinghomotopywe get:
(9)
2 rIt ai liSigl/i_
and thisshows that Si determinesthe homotopyclass of the ith string.
(9) Interpretsthe ith stringbut it would not completelyexplain Si since it
is only an i-homotopy. Let 2' be any path connectingthe beginningpoints
of li(e) and 1(e) that stays in the immediatevicinityof the nth stringwithout
intersectingit. Its projectionis then a curve that may be used as ei . Now
the projectionof li(E)-'2zl'(E)g is the left side of (6):
(10)
Ii1
t$Si ,(e)r2,'(f)g
whichprovidesthefullgeometricmeaningof Si . The converseofTheorem8 is:
THEOREM 10. Let B and C be twobraidswiththesame endsand withthesame
firstn - 1.strings. AssumeeitherthatB and C are s-isotopicor thattheyare isotopicor thattheexpressions(5) are thesamefor bothbraids. Then thenth strings
and thereexistsan s-isotopymovingthenth stringonly.
are n-homotopic
PROOF: If they are s-isotopicthen they are isotopic because of Theorem 6.
If theyare isotopicthenthe expressions(5) are the same because of Theorem9.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
THEORY OF BRAIDS
111
If (5) is the same even fori = n only then the nth stringsare n-homotopicbecause offormula(9). The remarkto Theorem8 showstherestofthe contention.
THEOREM 11. If twobraidshave thesame ends and if theexpressions(5) are
thesamefor bothbraids thentheyare s-isotopic.
PROOF:For n = 1 our theoremis trivialsincethe two stringsare 1-homotopic.
Let it be provedforbraidswithn - 1 strings. If B and C are two n-braidsfor
whichthe assumptionholds,let B' and C' be the braids resultingfromcancellationof thenthstring. The expressions(5) forB' and C' are obtainedby putting
ThereforeB' and C' are s-isotn = 1. So (5) is also the same forB' and C'.
topic. Extend the s-isotopyto the whole space and apply this mappingto the
braid B. It carriesB into an s-isotopicbraid D having again the same expressions (5). D and C have now the firstn - 1 stringsin commonso that Theorem
10 shows that they are s-isotopic. This completesthe proof.
THEOREM 12. Isotopyand s-isotopyimplyeach other.
The prooffollowsfromTheorems6, 9 and 11.
THEOREM 13. The expressions(5) do not dependon thespecial braid-coordinates used. Theydependevenonlyon theclass and givetogether
withtheends a
theclass completely.
full systemofinvariantsof theclass thatdetermines
The proofis now obvious. It is to be remarked,however,that the expressions
dependon the selectionof generatorstiand t' . We mustnow develop methods
that allow the actual computationsof these invariantsand reveal the structure
of our groupoid.
To do so we have firstto change our notationslightly. Select in a plane n
pointsXi, a ray and paths li . Up to thispointthe numberingwas considered
unessential. Now we get a natural arrangementof our pointsby startingwith
our ray and goingaround Q in the positivesense of rotation(in a neighborhood
of Q). The firstpath that we meet shall be called 11, the next 12and so on.
The pointsXi are now numberedpreciselyas the paths leading to them. The
verysame patternis now used forthe upperplanes as well as forthelower ones.
The points Xi are now used as lower and upper ends of braids. We restrict
ourselvesto the investigationof braids whoselowerends are a subset of the Xi
and whose upper ends are anothersubset. These subsets may or may not be
thewholeset, no restrictionbeingput upon them. If B is such a braid and Xi,
Xi the lowerrespectivelyupper end of one of its stringswe write:
j = B(i).
Thus B maps a certainsubset of the numbers1, 2, * n onto anothersubset.
The numberingofthe generatorstiand t' so farwas connectedwiththe numbering of the string. Now we changethat and attach to themthe subscriptof the
point around which they run. We also drop the accent on the t' and write
uniformlytj for all the generatorsin the different
planes. This leads to the
followingsituation:
We have a groupF beforeus withthe n freegeneratorsti. For the Poincar6
group of a braid with the referencepoint in an upper or-a lower plane, not all
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
112
E. ARTIN
the generatorsare used; the Poincar6 group of such a plane is thereforeconsidereda subgroupof F generatedby a subset of the ti. Braid projectionof a
lower unto an upper plane will provideus with an isomorphicmappingof the
groupin the lower plane onto the groupin the upper one. If T is an element
of the group in the lower plane then its image under braid projectionshall be
denotedby B(T). In this new notation(5) takes on the form:
(11)
B(ti) = S7ltjSj
where j = B(i),
and wherethe numberingof all generatorshas been changed accordingto our
new convention. (11) alone already gives us the isomorphicmapping and in
this formcontainsalso the informationabout the upper and lower ends of the
stringsof the braid. In case all the pointsare used forthe lowerends,it will be
an automorphismof F. Otherwiseit maps a subgroupof F onto anothersubgroup.
Let A and B be two composablebraids and formthe composedbraid in such
a way that in AB the part B correspondsto negative,the part A to positivez.
of our projectionin termsof the
Returningfora momentto the interpretation
generatorsri whichallow to expressprojectionin termsof homotopieswe see:
AB(T) = A(B(T)).
We projectnamelya lowerplane ofAB firstonto the plane z = 0 and the result
onto an upper plane. Making use of the fact that (11) completelydetermines
the class we see:
THEOREM 14. The groupoidof braid classes whoselowerand upper ends are
subsetsof theXi is isomorphicto thegroupoidofmappingsin F indicatedby (11).
Consequentlywe expressthe braid class B in formof a substitution
(12)
B
=
t;si)
wheretirunsof courseonly throughcertainof the generators. It is convenient
to consideralso more generalsubstitutions
B
(ti
in the freegroupF wherecertainti are mapped onto powerproductsregardless
of whetherthe substitutionis derivedfroma braid or not. If the substitution
is derivedfroma braid class thenwe say brieflythat it is a braid. Also in this
generalcase we denote by B(T) the resultof applyingthe substitutionB onto
the power product T.
The braidsubstitutionshave one special propertythatwe mustderive. Draw
a hugecirclein a lowerplane startingat thereference
pointofthe Poincar6group
known
fromthe theoryof the homoand runningaround the braid. It is well
of
the
Poincar6groupis homotopic
topiesin a puncturedplane that thiselement
to the productof the generatorsti (of course only those that we need for our
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
THEORY
113
OF BRAIDS
braid) taken in the natural arrangementof the subscriptsaccordingto their
size. Braid projection onto an upper plane carriesthe circle into a similar
circle startingat P. This proves:
15. If B is a braid then
THEOREM
B (H ti) =t
(13)
ti
bothproductstakenin thenaturalarrangement
of theirsubscripts.
Select a subscripti < n and put X,(z) = Xv = constforv # i, i + 1. We
connect now Xi and Xi+, by a broken line startingat Xi and runningthen
parallelto li untilit comesnear the ray; thenit runsparallelto lib+untilit comes
close to Xi- withwhichit is then connected. If X(t) (O ? t ? 1) is the parametricrepresentationof this line we put
TXifor
X(z)
Xi(z)
z
?0
for 0 _ z _ 1
,Xi+, for z
1.
Then we draw a similarparallel curve betweenXi+, and Xi runningfarther
out and not intersectingthe previousone but at the ends. The stringXil+(z)
is explainedin a similarfashionthan Xi(z) but it has Xi+, as lower and Xi as
upperend. That this braid carriest,into itselfif v 0 i, i + 1 is seen by using
ordinaryprojectionwhichshows that T' is homotopicto t4. ti is mapped into a
transformof ti+1. To findthe transformer
we go back to (10). As parallel
curve we use one that will under ordinaryprojectionbecome the parts of li(e)
and li+1(e) up to the ray. Consider now the rightside of (10). Instead of li
we have to writeli 1in our new notation. The path projectsby ordinaryprojectionstillinto a homotopicpath. But thishomotopicpath is now obviously1.
So ti is carriedinto ti+1. The image of ti+1can now be found by a simpler
method,namely,by Theorem15. Since the productofthe timustremainfixed
we findby a simple computationthat till is mapped into ti-1titit+. The class
of this particularbraid shall be called oi and the corresponding
substitutionis:
(14)
(14)
a == ( ti, , ti+1
)
t4T~
titi+1)
K~~~~~~~~~~ti+1
with the understandingthat the generatorsthat are not mentionedare leftunchanged. For oj we have to compute the inverse substitutionand an easy
computationgives:
(15)
=
titi+1t7
si
To checkwhethera givenbraid is ai it suffices,
however,to check the following
properties:droppingthe ith and i + 1't stringwe must obtain a unit. In it
the i + 1't stringmust correspondto the unit homotopy. Afterreinserting
it
the ith stringmust have unit homotopy(always using the simplerformula(9)
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
114
E.
ARTIN
ratherthen (10)). Accordingto our theorythese checksalready determinethe
class.
From now on the nature of our proofswill be mostlyalgebraic.
Considera rathergeneralsubstitutionB that maps each generatorti onto a
transformQ j'tkQi = Ti (in reduced form)of some generatortk. When is B
a braid? The answeris given by the conditionof Theorem 15, namely that
(16)
TiT2... Tn= tlt2... tn.-
we assume each Qi written
The necessityis obvious. To prove the sufficiency
as a product of termst , e = 1= j . The numberof termsshall be called the
lengthof Qi and the sum of all theselengthsthe lengthofB. If the lengthof B
is 0 then (16) can hold onlyifeach Ti = tior B = 1 whichis the unitbraid. So
we may assume our contentionprovedforall braidswithsmallerlengththan B.
(16) can hold only if some cancellationstake place on the leftside. Since each
Ti is already reduced these cancellationsmust take place between adjacent
factorsof the leftside. Two cases are conceivable:
1) In a cancellation between two neighborsthe middle terms are never
affected. Carryingthem out in (16) therewill be a residue R, leftfromeach
Ti and this residuemust containthe middletermof Ti . We obtain:
RR2 ... Rn = tlt2... ta
and no furthercancellationis possible. This proves Ri = ti = middle term.
The termson the leftside of the middletermof T1 nevercould be cancelled at
all since no factoris on theirleft. So Q, = 1. Now thereis no furtherchance
forQ-1to be cancelled,so it mustbe 1 too. This shows Ti = tiso that this case
is settled.
2) Or else thereare two neighborsTi and Tj+1 = Q-1t8Qil4such that one or
bothofthemiddletermsare reachedin a cancellation. They cannot be affected
at thesame timesincetheirpositiveexponentpreventsit. Now twoalternatives
are forcedupon us:
first. ConsidertheproductTiTi+17T. Because ofthespecial
a) tkis affected
formof the T, a cancellationis now possible on both sides of Ti+1 . Carry it
l
out, termby term,on both sides until the middletermof Ti and Tj is reached
and stop the cancellationat that point even if it is possible to go on. More
than half of Ti and T7 will have been absorbed,the middle termof Ti+1 will
not yet be reachedand Ti+1will have lost as manyfactorsas the othertwo. A
few remnantsfromthese factorswill remainbut they will be shorterthan the
loss. It is of courseveryeasy to writethis down formally. What is important
Considernow
is, that the lengthof this productis shorterthan that of Tall.
find:
the substitutionBat71. We
Bo-i
)lTT
Bo
i = (T',
T,
T,
TiTi+1 T-1
ti+1)
Ti
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
v < i, i + 1.
115
THEORY OF BRAIDS
The product of the second line is T1T2 * T, = tit2
tn. It has still the
generalformofB but is shorter. So thissubstitutionA is a braid. Since B =
A vi, we see that B is a braid.
b) t8is affectedfirst. Then we findthat Ti-+TiTi~l is shorterthan Ti so that
Bi
ti,
tv
(sv
ti+1
)
v 4i,
+1
il
T+,
Ti+T
/s+ T+
is a braid. This proves that B = AaT1 is a braid.
Our proofalso shows that B can be expressedas a powerproductof the a, .
The proofwould also have workedif the subscriptsi ran througha subset of
all indices only. No conditionneed be put on the tkand Qi. (15) has to be
replaced by the conditionthat B leaves Dit (in the natural order) invariant.
The braids ai have of courseto be replacedby the corresponding
braids forthis
subset of ends.
The mostgeneralcase would be finallythis. A subset of the tj and mappings
ofthepreviouskindare given. B carries lttinto itj wherethe tj formanother
subset also in the natural order. To reducethis case to the previousone let B,
be a braid havingthe Xi as lowerends and the Xi as upper ones. The substitutionB B71 maps JjTtj
onto itselfand is thereforea braid. So B is a braid. Let
(i) and (j) be subsets of the indices both equal in number. Our resultshows
that
(17)
B(i)(J,-=
(Jis a braid.
(i, j natural arrangement).
Our resultsmay be expressedin the theorem:
THEOREM 16. A substitution
is a braidif, and onlyif, it has the generalform
and if it satisfiestheconditionof Theorem15. Thefull group
of a transformation
ofn-braidshas theai as generators. A generalbraidcan be expressedas a product
like theai but concerning
thelower
ofa braidoftheform(17) followedbygenerators
endsof thebraidonly.
A simplecomputationof substitutionsshows that the followingrelationshold
betweenthe a,:
(18)
(19)
aiak=akai
aia+iai
if I i-k
|I
2
= ar+1aifi+j.
In Z. I have shown that these relationsforma full set of definingrelations
forthe group. The method is geometricand can easily be made rigorousby
means of the tools developed in this paper. However a moreinterestingproof
shall be given in a paper by F. Bohnenblustwhichis essentiallyalgebraic and
leads deeper into the theoryof the group. I shall thereforeomit a proofhere
especiallysince no use shall be made of this factin this paper. All we shall use
is that these relationshold.
A simpleoperationis that of removinga string. What does it mean forthe
substitution? Let A be the braid and removethe stringwith Xr as lower and
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
116
E.
ARTIN
Xm as upper end. That means that we have to cancel the columnreferring
to
the image of tr. In addition to that we have to substituteeverywherein the
secondrowtm = 1 sincethat describestheshrinkagein thehomotopygenerators.
The inverseproblemin a somewhatsimplifiedformwould be;
Given a braid in the form(12), assume r does not occur among the i and m
not among thej. Form a new braid A by insertingone stringwithX, and Xm
will consistin thespecial
as respectivelowerand upperends. The simplification
positionwhichthe new stringis in, we shall namelyget A (tr) = ti . To do it,
enlargethe meaningof B by prescribingof a new substitutionC that it shall
have the same effecton the tias B and map tronto ti . This substitutionwill
in general not be a braid since the conditionconcerningthe product will be
violated. Definetwo new substitutionsa and ,3by theireffecton the tirespectivelytj and trand trresp tm
a(ti)
(20)Ltj
(20)
ti if i<r
=a(tr)
=
tr if i > r,
(
{:) =
t
[ji
1J
tMtjt;;
m
;X:
if j > m
13(tm)= tm.
Then A = ,3Ca is the desiredbraid. We firstprovethat it is a braid by showing that A carriesthe product
II ti * t * ][I ti into H
i<r
i>r
tI*t* tij>m t,
i<m
Indeed a carriesit into
II ti*try
tr1U ti*to-HI ti*tr
i>r
i<r
it into
C transforms
II tj tm= II
,<m
and #into
II
j<m
totmII to tm
*=
,>m
II
ti * j>m
II
j<m
tj tm
totm
I
m
ti.
It is seen immediatelythat cancellationof the rthstringleads back to B. So
A is the desiredbraid. ObviouslyA(tr) = ti .
We may now combineboth operations. If A is a braid, we may firstcancel
the rthstringand thenreinsertit withthe same ends so that it maps tr onto ti .
The new braid shall be denotedby A(r). Theorem7 shows that A may be obit fromtheleftby a uniquely determinedm-pure
tainedfromA(r) by multiplying
braid (m and not r-purebecause of our change of notation). That A(r) is
uniquelydeterminedby A followsfromTheorem8 since A(r)(tr) = tmdescribes
the homotopyclass of the rthstring. We may write:
A = UmA(r), Urnm-pure.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
117
THEORY OF BRAIDS
We know how to computeA(r) fromA and shall see a littlelater how U. can
be computed.
The elementsSi in theirdependencyon the braid shall be denotedby Sj(A).
In orderto have well definedelementsbeforeus, we must still make an agreement about the arbitrarypower of tj that still may be added as a leftfactor.
We chooseit in such a way that the sum of the exponentsof tj in Si is 0.
Let now A and B be two composablebraids. B maps t,into Sj(B)-ltjSj(B).
where k = A0j).
A maps this into A(Sj(B))-lYSk(A)-lFt.Sk(A).A(Sj(B)),
is determinedto withina power of tk we get to within
Since the transformer
such a power
(21)
Sk(AB) = Sk(A)A(Sj(B)),
k = A (i).
On the leftside and in the firstfactoron the rightthe sum is 0; in Sj(B) the
sum of the exponentsof tj is 0. A maps it into a powerproductwherethe sum
of the exponentsof tk is 0. So (21) is correctas it stands.
A ratherelementaryinvariantcan be derivedfrom(21). Calling Hj(A) the
sum of all exponentsin Sj(A) we obtain:
(22)
Hk(AB) = Hk(A) + Hj(B)
k = A(j).
Definingnow the "twiningnumber" T(A) as the sum of all Hk(A) we get:
(23)
T(AB) = T(A) + T(B).
Since T(Q) = e fore = i 1 thisinvariantcan in case ofthefullgroupofn-braids
also be explainedas the sum of all exponentsof the u, in any expressionof A by
the ai. This allows us to determinethe factor commutatorgroup without
makinguse ofthe factthat the systemofrelations(18), (19) is complete. Making all ai commutative(19) gives the equality of all the ai . In the factorcomj
~
So this group is infinite
A)
mutatorgroup a braid A shrinksthereforetotoT(A)
cyclicand T(A) gives the positionof A in it.
The -homologyclass of a stringis obtained fromS j(A) by the substitutionof
tj = 1. The resultmay be denoted by 3j(A). In orderto obtain a formula
similarto (21) we mustsee what effectthat substitutionhas on the second term
of tk . Aftersubstitutingtk=
on the rightof (21). A maps tj onto a transform
1 all termscomingfromtj will disappear. Hence we may substitutetj = 1 in
Sj(B). In additionto that,we mustalso substitutetk= 1 in theresultwherever
it appears fromthe rest of the substitution. The same effectis achieved if the
braidA is replacedby one wherethejth stringhas been dropped. Let us denote
this braid by A-j. Then we have:
(24)
Sk(AB) = Sk(A) .A-j(j(B)),
k = A(j).
Considerthe special case that A is k-pure. Then k = j, A_j a unit. Hence:
(25)
Ok(AB) = Ok(A)*Sk(B); if A is k-pure.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
118
E.
ARTIN
A stillmorespecial case is obtainedwhenboth A and B are k-pure. Theorem
8 tells that the homotopyclass of the kthstringtogetherwith the ends determines any k-purebraid completely. It also shows that everyhomotopyclass
is possible. (25) means thereforethat Sk(A) gives an isomorphicmappingof
the groupof k-purebraidswithgivenends onto the freegroupof the generators
tiwithi 5 k. If we denoteby Aik the k-purebraid that is mapped onto tithen
the Aik are the generatorsof our group and the mapping 3k(A) means just a
We prove
replacementof each Aik by ti . These braids satisfyAsk = Aki.
this by givingat the same time the fullsubstitutionof A k fori < k and e any
integer.
(26)
Ask
=
Aki
\r < i
trI
ti,
tr
tiCik,
trI
Cik trCik,
i
or > k)
tk
C ik tkc
/
r<k,
Cik
= (t, tkl) *(titk)E.
of generators.
Writingout the criticaltermswe see that all are transformations
The productpropertyholds so theyare braids. Ri(Ak) = tk, Sk(Atk) = t: and
thebraidreducesto a unitifwe dropeitherthe ith or the kthstring. This proves
all our contentions.
It is convenientto introducealso the inversemappingFk to Sk . It maps ti
onto A ik . Let it also have a meaningfortk whoseimage shall be 1.
Let A be a braid and assume k = ACj). We can writeA = U*A'( where U
=
Sk(U)
is k-pure. Making use of (25) we obtain Sk(A) = ;k(U)k(A(j))
since A(') maps by definitiontj onto tk, whence Sk(A(j)) = 1. Now applying
Fk gives U = Fk(3k(A)) or:
A(j) = k.
A = Fk(Sk(A)) A- ,
(27)
This is the algebraicformof Theorem7.
(27) also solvesthegeneralquestion: givena braid B; inserta new stringwith
the given homotopyclass Sk(A). We have learned to forma braid with the
homotopyclass 1, it is the braid A j. If we substitutethis and the givenSk(A)
in (27) we obtain A expressedby the A ik and A'j). Use now (26).
Anotherapplicationof (24) is this:let A be j-pureand assumeofB thatB(t ) =
Then Sk(B) = ;j(B 1) = 1. This leads to Sk(BAB-1) =
tk hence B l(tk) = tj.
BAB-1 is k-pureso we can apply Fk . This leads to:
B-j(j(A)).
BAB-1 = Fk(Bj(j(A))).
To get a still more generalformulalet B be now any braid. We firstreplace
in the previousformulaB by B(j). On the rightside B(W?appears. It is followed by the mappingFk whichanyhowmaps tk onto 1. So this braid may be
replaced by B itself. Use now (27) on B. We obtain
(28)
BAB-1 =
Fk(Sk(B))
.Fk(B(Sj(A)))
.
(Fk(S(B)))
1, B(j) = k.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
119
THEORY OF BRAIDS
B is herecompletelygeneralso thatwe have beforeus the generaltransformation
formulafora j-pure braid A. If A is givenas a power productof the Aij then
Si is a verytrivialmappingand so is Fk . The rightside is directlyexpressed
as a powerproductof the Aik . It is a verypowerfulformulathat allows us to
writedown transformation
formulaswhose direct computationwould be very
painful.
As an applicationlet A = Ai and B = ur. We give only the result of the
computationwhich is now very easy:
(29)
=A
urAiju'r
Aii
if r s i-1,ixj-1,j
A i+,,i
if r=i
,jAij_1A,7~
Ai-71A i-,si A
I
but
if r=jif r =i1
if r =j
tAilj+i
id
j-
1;
I
but i 5j . + 1;
A1,-1fori = j-1
Ai,j+1for i =j + 1.
It is to be remarkedthat the symmetryAi j = A ji gives otherexpressionsfor
the same transforms. We note the very special cases r = i, j. Since a very
we obtain fori < j the followingexplicit
simplecomputationgives Aii+ =
expressionsof the Ai in termsof the generatorsar
30 Aij =
(30
= 1-1
-1
...0
ai
ai+1
2
_2a ja_1 -2 ..
r
2 -1
=
(i-1
Tj-2
*-i+1G
iaii+1
...
-1
af-2
-1
a-.
As a second application we study the structureof the group I of n-braids
withidentitypermutation. We fistprove:
LEMMA: If A is an element
ofI thatmaps ti ontoitselffori _ j thentheSkfor
k > j do notdependon theseti .
PROOF: Droppingin the substitutionA the first
j columnswill give a substitution that still satisfiesthe product condition,so it is an (n - j)-braid. This
proves the lemma.
Now we use (27) forj = k = 1. On A (l) we use it fork = 2 and so on. Making use each time of the lemma we obtain a unique expressionof A:
THEOREM 17. The Aik are generators
of thegroupI. Everyelementcan be
expresseduniquelyin theform:
(31)
A
=
UlU2 ...Un-
whereeach Uj is a uniquelydetermined
powerproductof theAii using onlythose
withi > j. (Of courseA i = A ii).
The simple geometricmeaningof this normalformshall be givenlat,-rwhen
we interpretour resultsin termsofprojection.
What are the definingrelationsbetweenthese generators? Obviouslythose
that permitthe change of an arbitrarypowerproductof theminto the normal
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
120
E.
ARTIN
form. We musttherefore
findrulesforinterchanging
factorsof Uj withfactors
of Ui. For this purposewe derive all transformation
rules forthe expression
ArsAikA-E(e
t1). We use (28) with i instead of k forB = A"8, A = A a.
A simple computationyields:
THEOREM
18. The braidArsAikA-E(e = ?1) is i-pure. Thefollowingrules
giveits expressionas i-pure braid:
1) If i = r or s thenit has alreadythedesiredform. Since thei-purebraids
forma freegroupno otherexpressioncan be expected.
and if thepairs r, s and i, k do not separateeach
2) If all indicesare different
otherwe simplygetAik.
In all othercases we change,if necessary,
firstthenamesr, s in such a way that
thearrangement
i, r, s as comparedwiththenatural arrangement
of thesethree
numbersis a permutation
withthesame signas e.
3) If k = r we get:A76AirAa.
4) If k = s we obtain:
A7
Ai8A:A7rAiArA.
sur*Aie *A 'rA
8.
5) If finallythesubscriptsare all different
and if thepairs r, s and i, k separate
each othertheresultis:
A AYA ssAfir
*Ak *A AA
iArA !&
9
As definingrelationsthe ones wherei is the smallestindexof all are sufficient.
It also suffices
to have only e = +1 since e = -1 is just only the inverseautomorphism.
For braidswhosepermutationis not identity,a normalformis also easily obtained. Select to each of the n! permutationsir a braid B, with this permutation. Any braid can then be writtenas a product like that in Theorem 17
followedby a B,. This formis again unique.
The operationA() obviouslysatisfies:
(32)
(AB)(j)
= A(k)B(3),
k = B(j).
For the groupI, it is therefore
a homomorphicmappingand it sufficesto know
theresultforthegeneratorsAik. A (r) = 1 forr = i or k sinceAskis i- and k-pure.
It is Aik if r < i or > k as the substitutionshows. If r is betweeni and k we
apply (27) and Theorem 18. The result is:
1 for r = i or r = k
(33)
Ak =
A
if r not betweeni and k
1.AirAik'ir= A AkiA
N
if i < r < k.
We returnnow to the generalgroupofn-braids. Let A be a braid that leaves
tj fixedif eitherj < i or j > k. For the same reason as that in the previous
lemma,we findthat A maps ti, till ... tkonto expressionsdependingon these
variables alone. A can thereforebe expressedin termsof the generatorsa;,
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
121
THEORY OF BRAIDS
anal,I... O**k . These braids forma subgroupdenotedby Gik.
They behave
as iftheywerebraidsof k - i + 1 strings(fromi to k) only.
Put Cik = titian ... tk and considerthe followingsubstitutionCik
tr
(34)
(r
Cik =
ts
Cik
tCik
tr
< i or > k i < s < k
It is a braid of Gik since the productpropertyholds. It is obviouslycommutativewithany elementof Gik hencein the centerof thisgroup. If we drop the
kthstringin Cik and Ck- or the ith in Cik and Citl,k we always obtain identical
braids. Ci,k-1 leaves tk and Cilk leaves ti fixed. Hence:
(35)
Ck
=
Cik
Ci.k-1,
=
Ci+l,k
(Cii means 1).
Formula (27) gives:
(36)
Cik =
AikAiil,k
...
=
Ak-1kCik-1
Aii+]Aii-.2
... AikCi-4,k
This gives us the explicitexpressionsby the Aik .
Cik is closelyrelatedto a braid Dakof Gik whichmaps titi+*...
We see that
tk but tk onto CTik
tlCik .
ti+2
(37)
Cik =
tkl onto till,
(Dik)ki]l.
The product Oiail . . . ak-1 has the same effecton ti ti1,
-- tk-1 as Dik
4
Because of the product propertythis sufficesto establish equality. Hence:
(38)
Cik =
(aiai+1
...
ak- )k-i+l.
THEOREM 19. If n = 2 all braidsare commutative.If n _ 3 let k be one of
witheveryk-purebraid,thenB is a power
thenumbers< n. If B is commutative
This also determines
of C1.n.
of coursethecentersof the wholegroup,of I, Gi.
and ik = I n G-.
PROOF:If A is r-pureand B(r) = s, then BAB-' is s-pure. Put A = Aik.
It is pure only fori and k. Since BAikB' = Aik we see that B can at most
interchangei and k. If n > 3 theni may be replaced by anotherindex which
showsthat k remainsfixed. Consequentlyi remainsfixedtoo so B has identity
as permutation.
We now make use of (28) wherej = k. The braid B in the middletermon
the rightside came originallyfromBk which we introduce again. For A we
take any k-purebraid so that the left side is A again. If we then apply the
operationSk to both sides we get:
Sh(A) = ,Sk(B)-B-k (k(A)) *
(S.(B))1.
In this formulaSk(A) may be any power product T of the tj with i $ k.
This shows:
B-,k(T) = d-FTd, where d = Sh(B).
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
122
E. ARTIN
For T = To = Hik ti we have on the otherhand B4A(To) = To. So To is
commutativewith d. Since To occurs in a generatorsystemof the freegroup
d is a power of To say To . B-k transformsall T with To'. The same transformationis producedby (Cl,)1k ; B-k is thereforethis braid. Put now C =
B C'i.
We findCk = 1 so C is k-pure. But C is still commutativewith all
k-purebraids. They forma freegroupwithat least 2 generatorswhenceC = 1
or B = Cl,
This proves the contention.
For our next question we need a certainresultabout automorphismsof free
groups. Let F be a freegroup with the generatorst4. Divide the subscripts
into two classes p and q and in some otherway into the classes g and h. We
assume that thereare at least two tqand two th . Let x be a powerproductof
the tkthat appears amonga generatorsystemof F and y a similarpowerproduct
ofthe t4. Since we assumedthat thereare at least two th, x willnot be commutative with everyth . Define now the automorphismsC and D by:
(39)
C(tg) = tg, C(th) =
xathXa; D(tp)
=
tp,
D(tq)
=
y-btqyb
where a and b are positive integers. We ask for all automorphismsA that
satisfy:
D A = A C.
(40)
C leaves all tgas well as x invariant. Their image T under A will satisfy
(because of (40)?):
D(T) = T.
(41)
The powerproductsT = A(th) give D(T) = AC(th)= A(xathxG) hence:
(42)
D(T) = z' T zGwherez = A(x).
The equations (42) (41) exhaust(40) and onlythe conditionthatA is an automorphismwill have to be taken care of.
Denote by the letterP any power productof the t, alone, by Q one of the t,
and by R one of the t, and y. Any T can be writtenin the form:
T = P1QiP2Q2 ...
(43)
whereP1 may be absent. Then:
D(T)
(44)
=
P1lybQlybp2
...
Assume T satisfies(41). Each Qj must be commutativewith yb But y occursin a generatorsystemof F so Qj is a powerofy. Hence T is an R. We get
(45)
A(tg) = Rg.,
This takes care of (41).
(46)
ply-bQ1YbP2y
A(x) = z = Ro.
Assume now that T satisfies(42).
bQ2 ...
=
PQ1P2Qm * *
za.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
We get:
THEORY
OF
123
BRAIDS
If everyQ is a powerofy then T is an R and D(T) = T mustbe commutative
with za. But z is the image of a generatorso is itselfa generator. So T is a
power of z. Since one th at least will not be commutativewith x, its image T
will not be commutativewithz. So this case does not always happen.
Assumenow that Qi is the firstofthe Q in (46) that is not a powerofy. Then
the wholesegmentson theleftof thisfactorin (46) mustbe equal since z on the
rightside is also an R. We obtain: (the earlierQ are powersof y)
PAQ1P2..
= z PaQi *
a
R-lybR
PiY
or
Pi
Since y and z are generatorsthis is only possible if a
generatorand we get:
=
b. R-'yR is also a
z = R-'yR.
(47)
With this R put now T = R1'ToR. Because of D(R) = R (42) gives:
(48)
D(To)
=
Yb ToYb.
Writingnow To in the form(43) we get:
PlybQlybP2
...
=
y-bplQlp2
...
yb
The rightside shows that P1 mustbe absent. But also the presenceof P2 leads
to a contradiction. To is thereforea Q. Our resultsso farare:
(49)
A(t,) = R,, a = b, A(x) = R lyR, A(th) = R-1QhR.
A maps the group generatedby the t0and x into the group of the t, and y.
A-' satisfiesA-'D-' = C-1A-1wherethe roles of g and p are interchanged. It
maps thereforethe group of the t, and y into the group of the t, and x. The
mappingsare thereforeone to one and this showsthat the numberof subscripts
g is the same as that of the subscriptsp.
We go now back to the braids and ask what automorphismssatisfy:
(50)
CakA = ACa8X
a > 0.
Our conditionsare satisfied. The p are < i or > k, the q satisfyi < q ? k,
theg are < r or > s, the h satisfyingr ? h < s. x = Crs and y = cikare generators. We must have k - i = s - r.
Assume a little more about the automorphismA namely that it maps each
tj onto a transformof another. Then:
(51)
=
A(th) = R-1Qj1tQqR,
A(cr8) = R-lcikR
RltpRp,
A(tg)
whereeach Qq is a powerproductofthe tq, Rp and R powerproductsofthe tpand
of Cik.
Split A into two substitutions:
(52)
(53)
0 < j < k - i = s-r.
B(tr+4) = R-ltin+R,
B(tg) = R-ltpRv,
E(tp) = C,, E(ti-+) = Q, ltqQq where r + j corresponds to q.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
124
E. ARTIN
A maps on one hand onto R-'cikR; computeddirectlyonto IIJRi1Q-1tQQqR
c
whence fI Qqlt'Qq = C.
Now we see that E leaves Cik fixedand therefore
cr.
A = EB.
(54)
We see at the same time that E is always a braid. The conditionthat A be
an automorphismis thereforethat B is one and that E is a braid. Should A
be also a braid thenB is one and conversely.
In case A is a braidthe geometricsignificance
of (52), (53) and (54) is obvious.
B behaves as iftheith up to kthstringwerejust onlyone strand(onlythe product
ofthe thand ofthe tqplays a role). A is obtained by weavingthe patternE into
the ith up to the kth string,then, consideringthis partial braid as one string
only,interweavingit with the otherstringsaccordingto patternB.
1
i
1
n
i+
1
FIGURE
ri
+
n
FIGURE~ 2
The question whethera given braid A can be consideredas braid of braids
amountsto checkingrelationsof the type (50). It sufficesof course to consider
a = 1. Since we can decide whetheror not theyhold, this question is decided
too.
THEOREM20. The numbern is a groupinvariantof thegroupof braidswithn
strings.
PROOF:Theorem19 shows that Cl,, is a generatorof the centerand therefore
togetherwith its reciprocalcharacterizedby an inner propertyof the group.
The numberT(C1,,) = n(n - 1) is thereforealso an invariantsince it gives the
positionin the factorcommutatorgroup.
The structureof the groupdoes not depend on the positionof the ends. We
may thereforeput the ends in the special points with the coordinatesx = 0,
y = i, (i = 1, 2, *** n). As ray forthe Poincar6 group we may select x = 1,
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
THEORY
OF BRAIDS
125
y < 0 and as paths 1i the straightline segmentsformthe beginningpoint of
the ray to the ends. It is advisable to use as orientationof the plane the nega-
1
i
k- 1
FIGURE
k
3
FIGURE
4
tive one, so the sense of rotationfromthe positivey-axisto the positivex-axis.
This fixesall the necessarydata and we are now in a positionto interpretour
resultsin the projectionfromthe positivex-directiononto the yz-plane.
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions
126
E. ARTIN
Theorem 4 shows us that any braid is isotopic to anotherone whose strings
are brokenlines. It leaves us so muchfreedomthat we can assume in addition
that the projectionis freefromany triplepoints and that no two double points
occur at the same z-level.
Figures 1 and 2 show the generators as and at71 in their projection.
Corollary3 to Theorem2 shows indeed that they are reciprocal. The braid in
Fig. 1 maps t4obviouslyonto itselfif v $ i, i + 1. It maps ti onto ti+1and,
because of the product property,that is sufficientto establish its identity.
Theorem3 teaches how to read offfromthe projectionof a braid its expression
in termsof the generatorsao .
Formula (38) showsthat Cik is simplythe fulltwistof all the stringsfromthe
ith to the kthand that gives the geometricmeaningof Theorem 19.
Formula (30) gives now the projectionof the generatorAij. It is indicated
in Fig. 3.
The geometricmeaningofthe normalformofTheorem17 and that mentioned
afterTheorem18 is now also clear. Every braidis isotopicto anotherone whose
patternof projectionis especiallysimpleand is indicatedin Fig. 4 fora special
case. This patternis unique. The braid in Fig. 4 has the expression:
= A'A'A'4A
A A=A123
12A4A4r *A
-2 -A
A34A3-2*A-41
*A-4
-'A223A24A
Althoughit has been proved that everybraid can be deformedinto a similar
normalformthe writeris convincedthat any attemptto carrythis out on a
living person would only lead to violent protestsand discriminationagainst
mathematics. He would thereforediscouragesuch an experiment.
INDIANA
UNIVERSITY
This content downloaded from 129.215.149.97 on Sat, 22 Jun 2013 05:14:31 AM
All use subject to JSTOR Terms and Conditions