Table Of ContentCONFIGURATION SPACES ARE NOT HOMOTOPY INVARIANT
4
RICCARDOLONGONIANDPAOLOSALVATORE
0
0
2 Abstract. Wepresentacounterexample tothe conjectureonthehomotopy
n invarianceofconfigurationspaces. Moreprecisely,weconsiderthelensspaces
a L7,1 and L7,2, and prove that their configuration spaces are not homotopy
J equivalent by showing that their universal coverings have different Massey
products.
8
]
T
A 1. Introduction
.
h The configurationspace F (M) ofpairwise distinct n-tuples of points in a man-
n
at ifold M has been much studied in the literature. Levitt reported in [4] as “long-
m standing” the following
[ Conjecture 1.1. The homotopy type of F (M), for M a closed compact smooth
n
1 manifold, depends only on the homotopy type of M.
v
There was some evidence in favor: Levitt proved that the loop space ΩF (M)
5 n
7 is a homotopy invariant of M. Recently Aouina and Klein [1] have proved that
0 a suitable iterated suspension of F (M) is a homotopy invariant. For example
n
1 the double suspension of F (M) is a homotopy invariant. Moreover F (M) is a
2 2
0
homotopyinvariantwhenM is2-connected(see[4]). Arationalhomotopytheoretic
4
version of this fact appears in [3]. On the other hand there is a similar situation
0
/ suggestingthattheconjecturemightfail: theEuclideanconfigurationspaceF3(Rn)
h has the homotopy type of a bundle over Sn−1 with fiber Sn−1∨Sn−1 but it does
t
a not split as a product in general[6]. Howeverthe loopspaces of F (Rn) and of the
3
m product Sn−1×(Sn−1∨Sn−1) are homotopy equivalent and also the suspensions
: of the two spaces are homotopic.
v
Lensspacesprovidehandyexamplesofmanifoldswhicharehomotopyequivalent
i
X
but not homeomorphic. The first of these examples are L and L . We show
7,1 7,2
r thattheir configurationspacesF (L )andF (L )arenothomotopyequivalent.
a 2 7,1 2 7,2
After recalling some definition, we will describe the universal coverings of these
configuration spaces. Such coverings can be written as bundles with same base
and fiber, but the first splits and the second does not. We will show that Massey
productsareallzerointhefirstcase,whilethereexistsanontrivialMasseyproduct
in the second case.
ThismeansthatF (L )isnothomotopyequivalenttoF (L ). Finallywewill
2 7,1 2 7,2
extendthisresultbyshowingthatF (L )isnothomotopyequivalenttoF (L )
n 7,1 n 7,2
for any n≥2. The same result holds for unordered configuration spaces.
2. Configuration spaces of lens spaces
The lens spaces are 3-dimensional oriented manifolds defined as
L :=S3/Z = (x ,x )∈C×C |x |2+|x |2 =1 /Z
m,n m 1 2 1 2 m
1
(cid:8) (cid:12) (cid:9)
(cid:12)
2 R.LONGONIANDP.SALVATORE
where the group action is defined by ζ((x ,x )) = (e2πi/mx ,e2πin/mx ), and ζ is
1 2 1 2
the generator of Z . It is known (see e.g. [7]) that L and L are homotopy
m 7,1 7,2
equivalent, though not homeomorphic.
ForanytopologicalspaceM,letF (M)betheconfigurationspaceofnpairwise
n
distinct points in M, namely F (M) := Mn \( ∆) where ∆ is the union of
n
all diagonals. We first want to compute the fundamental group of F (L ) and
2 7,1
S S
F (L ). Observe that S3 is the universal covering of L , for j = 1,2, and
2 7,2 7,j
thereforethefundamentalgroupofL isZ . Thenπ (F (L ))=Z ×Z because
7,j 7 1 2 7,j 7 7
π (L ×L ) = Z ×Z and removing the diagonal, which is a codimension 3
1 7,j 7,j 7 7
manifold, does not change the fundamental group.
The universal coverings F (L ) and F (L ) are the so-called “orbit configu-
2 7,1 2 7,2
ration spaces” and are given by pairs of points (x,y) of S3 which don’t lie on the
same orbit, i.e., x6=g(y) foreany g ∈Z . e
7
IntherestofthepaperweidentifyZ tothegroupof7thcomplexrootsofunity,
7
and we use the symbol ζt, t∈R, to denote the complex number e2πit/7.
Thefirstuniversalcoveringhasasimplestructure,namelywehavethefollowing
Proposition 2.1. F (L ) is homotopy equivalent to ∨ S2×S3.
2 7,1 6
Proof. Itis convenientto interpretS3 as the space ofquaternionsofunitary norm.
e
Then the action of Z on S3 = L is the left translation by the subgroup Z ⊂
7 7,1 7
C⊂H. We define a map F (L )→(S3\Z )×S3 by sending (x,y) to (xy−1,y).
2 7,1 7
Thisis ahomeomorphismsince xg6=ζk(y)=ζky isequivalenttoxy−1 6=ζk forany
7throotofunityζk,k ∈{0e,...,6}. Finallywe observethatS3 minus apointis R3
and hence S3\Z is homotopic to the wedge of six 2-dimensional spheres. (cid:3)
7
3. Massey products
WebrieflyrecallthedefinitionofMasseyproductsforatopologicalspaceX (see
∗
[5]). Letx,y,z ∈H (X)suchthat x∪y =y∪z =0. If we choosesingularcochain
∗
representatives x¯,y¯,z¯∈ C (X) then we have that x¯∪y¯= dZ and y¯∪z¯= dX for
some cochains Z and Y. Notice that
d(Z ∪z¯−(−1)deg(x)x¯∪X)=(x¯∪y¯∪z¯−x¯∪y¯∪z¯)=0,
andhencewecandefinehx,y,zitobethecohomologyclassofZ∪z¯−(−1)deg(x)x¯∪X.
Since the choice of Z and X is not unique, the Massey product hx,y,zi is well
∗
definedonlyinH (X)/hy,ziwherehy,ziistheidealgeneratedbyy andz. Clearly
Massey products are homotopy invariants. A rational homotopy theoretic version
of the following definition is in [2].
Definition 3.1. A space X is formal if the singular cochain complex C∗(X) is
∗
quasi-isomorphic to H (X) as augmented differential graded ring.
This means there is a zig-zag of homomorphisms inducing isomorphism in co-
∗ ∗
homology and connecting H (X) and C (X). It is easy to see that spheres are
formal. Moreover wedges and products of formal spaces are formal. By construc-
tionall Massey products on the cohomologyof a formalspace vanish. This in turn
implies the following result
Proposition 3.2. All Massey products in the cohomology of F (L ) are trivial.
2 7,1
e
CONFIGURATION SPACES ARE NOT HOMOTOPY INVARIANT 3
We deduce that in order to prove that F (L ) and F (L ) are not homotopy
2 7,1 2 7,2
equivalent, we only need to construct a nontrivial Massey product in the cohomol-
ogy F (L ). e e
2 7,2
e 4. Nontrivial Massey product for F (L )
2 7,2
The projectiononto the firstcoordinategivesF (L ) the structureof a bundle
2 7,2 e
over S3 with fiber S3\Z ≃∨ S2 that admits a section. It follows that the Serre
7 6
spectral sequence collapses and the cohomology reing splits as a tensor product, so
that it does not detect the nontriviality of the bundle. In particular we have that
H2(F (L )) ∼= Z6 and H4(F (L )) = 0. This in turn implies that the Massey
2 7,2 2 7,2
product of any triple in H2 is well defined.
WeewanttocomputeMasseeyproducts“geometrically”,namelyusingintersection
theory on the Poincar´edual cycles as in [5].
Let us define the embedded “diagonal”3-spheres ∆ ⊂S3×S3, for k =0,...6,
k
by ∆ := {(x,ζk(x))|x ∈ S3}. Clearly ∆ is the standard diagonal. The space
k 0
F (L ) is the complement of the union of the diagonals
2 7,2
6
e F (L )=(S3×S3)\ ∆ .
2 7,2 k
!
k=0
a
By Poincar´eduality we heave the isomorphism
6 6
Hp (S3×S3)\ ∆k ∼=H6−p S3×S3, ∆k .
!! !!
k=0 k=0
a a
Under this identification the cup product in cohomology corresponds to the inter-
section product in homology.
We observe that there exists an isotopy H : S3×[0,1] → S3×S3 (where k is
k
considered mod 7) defined by H ((x ,x ),t) = ((x ,x ),(ζk−1+tx ,ζ2(k−1+t)x )).
k 1 2 1 2 1 2
The images of Hk at times 0 and 1 are respectively ∆k−1 and ∆k, and the full
image of H is a submanifold A ⊂ S3 × S3 which represents an element in
k k
H S3×S3, 6 ∆ Poincar´edualtoaclassa ∈H2(F (L )).Byusingthe
4 k=0 k k 2 7,2
May(cid:16)er-Vietori(cid:16)s`sequence(cid:17)o(cid:17)ne can easily see that the classes ak span H2(F2(L7,2))
e
under the relation 6 a =0. The main result of the paper is the following
k=0 k
e
Theorem 4.1. ThPe Massey product ha ,a ,a +a i contains the class a ∪ι and
4 1 2 6 2
hence is nontrivial.
Proof. It is easy to check that A intersects only A and A where again k is
k k+3 k+4
consideredmod 7. Hence inthe computationofha ,a ,a +a i wemustcheckthe
4 1 2 6
following
Lemma 4.2. The submanifolds A and A intersect transversally and
1 4
S1×[0,1]∼=A ∩A = (0,x ),(0,ζλx ) |x |=1, λ∈[0,1] .
1 4 2 2 2
Proof. We only need to verify th(cid:8)a(cid:0)t the tangent spa(cid:1)c(cid:12)es to A and A at(cid:9)the point
(cid:12) 1 4
(0,x ),(0,ζλx ) span a six dimensional vector space. Recall that we are repre-
2 2
senting points in S3 as elements (x ,x ) in C×C such that |x |2+|x |2 =1, and
(cid:0) (cid:1) 1 2 1 2
hence tangent vectors at (0,x ) are real linear combinations of the vectors (1,0),
2
4 R.LONGONIANDP.SALVATORE
(i,0) and (0,ix ). These immediately give rise to the following tangent vectors to
2
A at (0,x ),(0,ζλx ) :
1 2 2
(cid:0) (1,0),(ζλ/2,(cid:1)0) , (i,0),(iζλ/2,0) , (0,ix ),(0,iζλx )
2 2
and to the(cid:16)following tange(cid:17)nt ve(cid:16)ctors to A at th(cid:17)e sam(cid:0)e point: (cid:1)
4
(1,0),(−ζλ/2,0) , (i,0),(−iζλ/2,0) , (0,ix ),(0,iζλx ) .
2 2
Finally c(cid:16)onsider the path(cid:17)in A(cid:16)∩A given by (cid:17) (cid:0) (cid:1)
1 4
s7→ (0,x ),(0,ζλ+sx .
2 2
Its derivative for s = 0 gives, up(cid:0)to a scalar factor,(cid:1)the vector (0,0),(0,iζλx2) .
By a simple inspection one sees that the linear space spanned by these vectors is
six dimensional. (cid:0) (cid:3)(cid:1)
Let us consider the closed 2-disc
D ={(r,x)|0≤r≤1,r2+|x|2 =1, x∈C}⊂S3.
2
Lemma 4.3. The intersection A ∩A is the relative boundary of the 3-manifold
1 4
D ×[0,1]∼=X := (r,x),(ζ4tr,ζtx) (r,x)∈D , 0≤t≤1 .
2 14 2
Proof. The pieces of the bou(cid:8)nd(cid:0)ary of X14 corre(cid:1)s(cid:12)(cid:12)pond to r = 0, t = 0 a(cid:9)nd t = 1.
Clearly ∂ X =A ∩A . If we now show that the other pieces belong to one of
r=0 14 1 4
the diagonals ∆ , the Lemma is proved. Since ζk =ζk+7 we have
k
∂ X ={((r,x),(r,x))}⊂∆
t=0 14 0
∂ X = (r,x),(ζ4r,ζx) ⊂∆ .
t=1 14 4
(cid:3)
(cid:8)(cid:0) (cid:1)(cid:9)
The next step is to find the intersection of X with A and A . We observe
14 2 6
that the inclusion S3 → S3 ×S3 sending x to (1,x) represents the generator of
H S3×S3, 6 ∆ ∼=Z Poincar´edual to a class ι∈H3(F (L ))∼=Z.
3 k=0 k 2 7,2
(cid:16) (cid:17)
Lemma 4.4.`The manifolds X and A do not intersect. Moreover X and A
14 6 e 14 2
intersect transversally and X ∩A =A ∩S3 is Poincar´e dual to the class a ∪ι.
14 2 2 2
Proof. The intersection of X with A is given by the solution to the system of
14 6
equations
ζ4tr =ζ5+sr
(ζtx2 =ζ10+2sx2
for 0 ≤ r ≤ 1, r2+|x|2 = 1, 0≤ t ≤ 1 and 0 ≤ s ≤ 1. If we equate the exponents
ofthe ζ’s in the firstand in the secondequationwe immediately see that there are
no solutions for 0≤t≤1.
TheintersectionofX withA isgivenbythesolutiontothesystemofequations
14 2
ζ4tr=ζ1+sr
(ζtx=ζ2+2sx
which has solutions (1,0),(ζ1+s,0) where 0 ≤ s ≤ 1. In fact, from the second
equationwegettheequationt=2+2s(mod7),whichhasnosolutionfor0≤t≤1.
(cid:0) (cid:1)
Therefore we must have x = 0 and r = 1. From the first equation we have that
CONFIGURATION SPACES ARE NOT HOMOTOPY INVARIANT 5
ζ4t = ζ1+s which implies t = (1+s)/4. Therefore X ∩A is a path connecting
14 2
∆ with ∆ which equals A ∩S3.
1 2 2
Finally we have to check transversality for X and A . By repeating the argu-
14 2
mentsofLemma4.2,wededucethatthetangentspacetoA at((1,0),(ζ1+s,0))=
2
((1,0),(ζ4t,0))isspannedby((i,0)(iζ1+s,0)),((0,1),(0,ζ2+2s)),((0,i),(0,iζ2+2s))
and((0,0),(iζ1+s,0)) while the tangentspace to X at the same point is spanned
14
by ((0,1),(0,ζt)), ((0,i),(0,iζt)) and ((0,0),(iζ4t,0)). These vectors clearly span
a six dimensional space. (cid:3)
This concludes the proof since a ∪ι does not belong to the subspace generated
2
by a ∪ι and (a +a )∪ι in
4 2 6
6
H5(F (L ))=ha ∪ι|k =0,...,6i a ∪ι.
2 7,2 k k
,
k=0
X
e (cid:3)
5. Generalizations
We extend our result to the n points configuration space, namely we have that
F (L )isnothomotopictoF (L ). TheuniversalcoveringF (L )istheorbit
n 7,1 n 7,2 n 7,j
configuration space of n-tuples of points in S3 lying in pairwise distinct Z -orbits.
7
The forgetful map (x ,...,x ) 7→ (x ,x ) defines a bundle Fe(L ) → F (L )
1 n 1 2 n 7,j 2 7,j
whichadmitsasection. Forexamplethevaluesx ,...,x ofthesectionarepairwise
3 n
distinct points very close to 1 multiplied by x . e e
1
By naturality we deduce that F (L ) has a nontrivial Massey product on H2.
n 7,2
On the other hand right multiplication by x−1 induces a product decomposition
1
Fn(L7,1)=S3×Yn−1, where Yn−e1 is the n−1 points orbit configuration space of
theZ -spaceS3\Z . Theforgetfulmappickingthefirstcoordinatedefinesabundle
7 7
Ye →S3\Z havingasfiberS3with14pointsremoved. Byiteratingthisprocedure
2 7
wefindatoweroffibrationsexpressingYn−1 astwistedproduct,uptohomotopy,of
thewedgesofspheres∨6S2,∨13S2,andsoon. TheadditivehomologyofYn−1splits
as tensor product of the homology of the factors, by the Serre spectral sequence.
In particular there is a map ∨(n−1)(7n−2)/2S2 → Yn−1 inducing isomorphism on
H2. The product map S3 × ∨(n−1)(7n−2)/2S2 → Fn(L7,1) induces isomorphism
on the cohomology groups H2,H3,H5. Thus all Massey products on elements of
H2(F (L )) must vanish. e
n 7,1
The unorderedconfigurationspace C (L )=F (L )/Σ has as fundamental
n 7,j n 7,j n
grouep the wreath product Σ ≀Z and has the same universal cover as the ordered
n 7
configurationspace. It follows that also all unordered configurationspaces are not
homotopy invariant.
Our approachshows that infinite other pairs of homotopic lens spaces have non
homotopic configuration spaces. It might be interesting to study whether the ho-
motopy type of configuration spaces distinguishes up to homeomorphism all lens
spaces.
References
[1] M. Aouina, J. R. Klein “On the homotopy invariance of configuration spaces”,
math.AT/0310483
6 R.LONGONIANDP.SALVATORE
[2] P.A.Griffiths,J.W.Morgan“RationalHomotopyTheoryandDifferentialForms”Progress
inMathematics 16,Birkha¨user(1981)
[3] P. Lambrechts, D. Stanley “The rational homotopy type of configuration spaces of two
points”,http://gauss.math.ucl.ac.be/ lambrech/publications.html
[4] N.Levitt“Spacesofarcsandconfigurationspacesofmanifolds”Topology34,217-230(1995)
[5] W.S.Massey“Higherorder linkingnumeber”Conferenceon AlgebraicTopology, University
ofChicagoCircle,174-205(1969) reprintedinJour.ofKnotTheoryanditsRamification 7,
no.3,393-414(1998)
[6] W.S.Massey,“Thehomotopytypeofcertainconfiguration spaces”Bol.Soc.Mat.Mex.37
(1992), 355-365
[7] A.Ranicki“Notes onReidemeisterTorsion”,http://www.maths.ed.ac.uk/ aar/papers/
DipartimentodiMatematica“G.Castelnuovo” —Universita` di Roma “La Sapienza”
e
E-mail address: [email protected]
DipartimentodiMatematica—Universita` diRoma “TorVergata”
E-mail address: [email protected]
