INTERSECTION HOMOLOGY OF LINKAGE SPACES IN ODD DIMENSIONAL EUCLIDEAN SPACE

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1 INTERSECTION HOMOLOGY OF LINKAGE SPACES IN ODD DIMENSIONAL EUCLIDEAN SPACE DIRK SCHÜTZ Abstract. We consier the mouli spaces M (l) of a close linkage with n links an prescribe lengths l R n in -imensional Eucliean space. For > 3 these spaces are no longer manifols generically, but they have the structure of a pseuomanifol. We use intersection homology to assign a ring to these spaces that can be use to istinguish the homeomorphism types of M (l) for a large class of length vectors. These rings behave rather ifferently epening on whether is even or o, with the even case having been treate in an earlier paper. The main ifference in the o case comes from an extra generator in the ring which can be thought of as an Euler class of a stratifie bunle. 1. Introuction In this paper we continue our stuies of the mouli spaces of close n-gon linkages in high imensional Eucliean space. These spaces are etermine by a length vector l R n such that all entries are positive. More precisely, the mouli space we are intereste in is { }/ n M (l) = (x 1,..., x n ) (S 1 ) n l i x i = 0 SO() where SO() acts iagonally on the prouct of spheres. A natural question is how the topology of M (l) epens on l, an one of the first observations is that R n is ivie into finitely many chambers such that length vectors in the same chamber lea to homeomorphic mouli spaces. One may then ask whether length vectors from ifferent chambers (up to permutation of coorinates) have ifferent mouli spaces. In the planar case = 2 Walker [16] conjecture that the cohomology ring of these spaces is enough to istinguish them, which was then confirme by Farber, Hausmann an the author in [4, 13]. Furthermore, in [4] this was also shown for = 3, with the single exception of n = 4 where for two ifferent chambers the mouli space is the 2-sphere. Inee, it follows from the work of Schoenberg [12] that for n = + 1 each non-empty mouli space is a sphere, an for n the nonempty ones are iscs. The homology calculations of [14] inicate that for n + 2 the topology oes epen on l, but they also show that homology an cohomology are not enough to istinguish them. In [15] the author use intersection homology to istinguish mouli spaces of a large class of length vectors for 4 even. Here we show that this approach also works for 5 o. The main theorem we thus get for the topology of mouli spaces is the following. 1 i=1

2 2 DIRK SCHÜTZ Theorem 1.1. Let 2, l, l R n be generic, -normal length vectors. If M (l) an M (l ) are homeomorphic, then l an l are in the same chamber up to a permutation. For the precise efinition of generic an -normal we refer the reaer to Section 2. As mentione above, the remaining case is when 5 is o. For 4 the mouli spaces are no longer manifols, so in [15] a substitute for the cohomology ring using intersection homology was efine. For even specific generators were foun an it was shown that this ring is an exterior face ring similar to the situation when = 2. For o this ring has an extra generator, which makes the etermination of the ring more ifficult. However, this is the same situation as in the case = 3 where the cohomology ring was calculate by Hausmann an Knutson [11]. While we o not etermine the ring completely, we o obtain enough information to mimick the proof use in [4] to get the result for = 3, which relie on the cohomology escription of [11]. A crucial observation is that the extra generator can be thought of as an Euler class of a certain stratifie bunle over M (l) from which its multiplication with the other generators can be euce. Similar results to Theorem 1.1 have been obtaine in [5] an by Farber an Fromm in [3] for chain spaces an free polygon spaces, respectively. These spaces are close manifols for generic l an all 2, an the proofs o not rely on istinguishing between even an o. It may be possible to give a unifie proof of Theorem 1.1 by using Z/2 coefficients throughout. One coul ask whether the conition of -normality is necessary for n > + 1, as it is not necessary for = 2, 3. It seems unlikely though that this can be attacke using intersection homology. 2. Basic properties of linkage spaces We efine the chain space of a length vector l as { } n 1 C (l) = (x 1,..., x n 1 ) (S 1 ) n 1 l i x i = l n e 1 where e 1 = (1, 0,..., 0) R is the usual first coorinate vector. If we let SO( 1) act on S 1 by fixing the first coorinate, we see that SO( 1) acts iagonally on C (l) an M (l) = C (l)/so( 1). We also efine { n }/ N (l) = (x 1,..., x n ) (S 1 ) n l i x i = 0 O() so that N (l) = M (l)/(z/2). Definition 2.1. Let l R n be a length vector. A subset J {1,..., n} is calle l-short, if l j < l i. j J i / J i=1 i=1

3 INTERSECTION HOMOLOGY OF LINKAGE SPACES 3 It is calle l-long, if the complement is l-short. If every such subset is either l-short or l-long, the length vector is calle generic. A length vector l = (l 1,..., l n ) is calle orere, if l 1 l 2 l n. After permuting the coorinates we can always assume that l is orere. If l R n is orere an k n 3, we write S k (l) = {J {1,..., n} n J, J = k + 1, J is l-short}. The carinality of these sets is enote by a k (l) = S k (l), an we write S (l) for the simplicial complex which is the collection of all S k (l). If J {1,..., n}, we efine the hyperplane H J = (x 1,..., x n ) R n x j = j J j / J an let H = R n >0 J {1,...,n} where R n >0 = {(x 1,..., x n ) R n x i > 0}. Then H has finitely many components, which we call chambers. It is clear that a length vector l is generic if an only if l H. It is shown in [10] that if l an l are in the same chamber, then C (l) an C (l ) are O( 1)-equivariantly iffeomorphic. In particular, M (l) an M (l ) are homeomorphic. It is easy to see that two orere generic length vectors l, l are in the same chamber if an only if S k (l) = S k (l ) for all k = 0,..., n 3. Definition 2.2. Let l R n be a length vector an 2. Then l is calle -normal, if J J L (l) where L (l) are the subsets J {1,..., n} with 1 elements that are l-long. If L (l) =, we let the intersection above be {1,..., n}. We call a chamber -normal, if its length vectors are -normal. So for a length vector to be not -normal, we nee a long subset J {1,..., n} with 1 elements such that J oes not contain an element m with l m maximal. Because then we can replace any element of J with m to get another l-long subset with 1 elements, an the intersection of these sets will be empty. If l is orere, then l is -normal if an only if {n + 1, n + 2,..., n 1} is not l-long. For a generic length vector this is equivalent to S n (l) =. It follows from the efinition that every length vector with n 2 is 2-normal. Furthermore, there is only one chamber up to permutation which is not 3-normal, namely the one containing 1 l = (0,..., 0, 1, 1, 1). In [5], 4-normal was calle normal. 1 Technically, this l is not a length vector because of 0-entries. We interpret a 0-entry in a length vector as ε > 0 so small that ecreasing it oes not change the chamber. H J, x j

4 4 DIRK SCHÜTZ In the case = n 1, there are only two generic -normal length vectors l R n up to permutation, namely l = (1,..., 1, n 2) an l = (0,..., 0, 1). If n is large compare to, -normality gets more common, an we woul expect the ratio of all -normal chambers in R n by all chambers in R n to converge to 1. For = 2, 3 an l generic, the spaces M (l) are obtaine as a quotient space from C (l) using a free action, so they are close manifols. For 4 this is no longer the case an we get ifferent orbit types. Let x M (l) be represente by (x 1,..., x n ) (S 1 ) n. If we think of this as a ( n)-matrix, the rank of this matrix oes not epen on the representative of x. Definition 2.3. Let l R n >0 be a generic length vector an x M (l). Then the rank of x, rank(x), is the rank of a ( n)-matrix representing x. For k < we have an inclusion N k (l) M (l) an N k (l) are exactly those points with rank at most k. It was shown in [15] that these subsets form a stratification of M (l) an M (l) is a pseuomanifol for n + 1. Note that SO( 1) acts freely on points of rank 1 in C (l), so that the regular set is M (l) N 2 (l). The stratification we will look at is therefore given by = N 1 N 2 (l) N 3 (l) N 2 (l) M (l). The singular strata are therefore given by N k (l) N k 1 (l) for k = 2,..., 2 an they can easily seen to be connecte. The imension of M (l) (an N (l)) for n is given by n ( 2)( 3) = (n 3)( 1), 2 see [14]. We nee to recall a few basic facts about intersection homology. Definition 2.4. Let X be a stratifie pseuomanifol, a (general) perversity is a function p: {singular strata of X} Z. For our purposes, this mainly means functions p: {2,..., 2} Z where k {2,..., 2} correspons to the stratum N k (l) N k 1 (l). Also, we are mostly intereste in Goresky MacPherson perversities, which for us means that p(2) 2(n ) 1 p(k + 1) p(k) n + k, compare [15]. The top perversity t n = t is then given by t(k) = c n,k 2 for all k {2,..., 2}, where c n,k enotes the coimension of the stratum N k (l) N k 1 (l), an which is given by c n k(k 1),k = k(n ) +. 2 The perversities we nee for M (l) are given by p j for j = 0,..., n 1 via the formula (1) p j (k) = j k.

5 INTERSECTION HOMOLOGY OF LINKAGE SPACES 5 It is easy to check that these perversities are Goresky MacPherson, an that p i + p j = p i+j for i + j n 1. Given a perversity p, we enote the resulting intersection homology by I p H (X) where the efinition is as in [7]. Furthermore, in [7, 8] Goresky MacPherson efine the intersection pairing : I p H i (X) I q H j (X) I r H i+j n (X) where p, q an r are perversities such that p + q r, an show that it oes not epen on the stratification of X. Also, I 0 H n (X) contains a funamental class [X] which serves as a unit. Now let k, m > 0 an assume that p 0,..., p k is a sequence of perversities such that for all i, j 0 with i + j k we have p i + p j p i+j. We can then form the intersection ring of X with respect to p an m by IR (X) = k I pr H n rm (X), r=0 which is a grae ring with unit, where we treat proucts with graings r + s > k as zero. We are mainly intereste in the subring generate by the elements of I p1 H n m (X), calle the reuce intersection ring. We enote it by IR (X). This is a grae ring whose generators have egree m, an if the perversities are Goresky MacPherson, it is a homeomorphism invariant by [8]. The sequence of perversities we are usually intereste in is the one given by (1), an the intersection homology group we mainly nee is an the coefficient ring is given by Z. I pj H n j (M (l)), Note that p 0 is the zero perversity, so this group contains the funamental class of M (l) for j = 0. Similarly, if we form a length vector l J R n J from an orere length vector l using a subset J {1,..., n 1} by linking together the elements of J with n, the group I pj H n j (M (l)) contains the funamental class of M (l J ) for j = J. Note that we nee J n 1 in orer for M (l J ) to be a pseuomanifol. The conition of -normality of l implies that M (l J ) = for each J {1,..., n 1} with J n. The following result is proven in [15]. Proposition 2.5. Let 5 be o, n + 2 an l R n a generic length vector with M (l). Then I p1 H n 1(M (l)) = Z 1+a1(l). If for every j {1,..., n 1} with {j, n} l-short we set l j + = (l 1,..., ˆl j,..., l n + l j ) R n 1,

6 6 DIRK SCHÜTZ then the funamental classes [M (l j +)] I p1 H n 1(M (l)) are linearly inepenent (assuming that l is orere), as shown in [15]. The notation ˆl j inicates that this entry is omitte. It remains to fin one more generator for this group. 3. Stratifie bunles over linkage spaces Recall that M (l) can be viewe as C (l)/so( 1) where SO( 1) acts iagonally on the left of (S 1 ) n, fixing the first coorinate in R. We now form the space M (l) = C (l) SO( 1) R 1 which is the quotient space of C (l) R 1 using the equivalence relation given by (x, v) (Ax, Av) for A SO( 1), x C (l) an v R 1 with the stanar action of SO( 1) on R 1. There is a projection p: M (l) M (l) given by p([x, v]) = [x], which can be viewe as a stratifie fibre bunle, see Remark 3.2 below. For now it will be goo enough that M (l) is a pseuomanifol, so we begin by fining the right stratification. For k = 2,..., 2 let N k (l) = p 1 (N k (l)). Note that if [x, v] N k (l), then there is A SO( 1) with Ax C k (l) an Av R k, where R k R 1 so that the last k 1 coorinates are 0. Let Then for k = 3,..., 2 an N k 1 k (l) = {[x, v] M (l) x C k (l), v R k 1 }. N k 1 k (l) N k (l) N k+1(l) k N 3 2(l) N 2 (l) M (l). Also, p : M (l) N 2 (l) M (l) N 2 (l) is a vector bunle, because SO( 1) acts freely on points of rank 1 in C (l). Note that if [x, v] N k (l) N k 1 k (l) or [x, v] N k 1 k (l) N k 1 (l), then rankx = k. In the latter case v varies through R k 1, an in the former case v varies through R k R k 1. One now checks easily that M (l) is a pseuomanifol with the stratification N 1 2(l) N 2 (l) N 2 3(l) N 3 2(l) N 2 (l) M (l). The coimensions of the strata are for k = 2,..., 2. coim(n k (l) N k 1 k (l)) = c n,k + k 1 coim(n k 1 k (l) N k 1 (l)) = c n,k + k Recall the perversities p r for M (l) an 0 r n 1 given by p r (N k (l) N k 1 (l)) = k r,

7 INTERSECTION HOMOLOGY OF LINKAGE SPACES 7 k = 2,..., 2. Similarly, we efine the perversities q r for M (l) by q r (N k (l) N k 1 k (l)) = r k 1 q r (N k 1 k (l) N k 1 (l)) = r k. The inclusion i: M (l) M (l) given by i([x]) = [x, 0] is stratum preserving, as i(n k (l) N k 1 (l)) N k 1 k (l) N k 1 (l) an since p r (N k (l) N k 1 (l)) coim(n k (l) N k 1 (l)) = k r c n k = k(r + 1) c n,k k = q r+1 (N k 1 k (l) N k 1 (l)) coim(n k 1 k (l) N k 1 (l)) the inclusion inuces a map on intersection homology by [15, Lm.5.2]. i : I pr H (M (l)) I qr+1 H (M (l)) Lemma 3.1. The map i : I pr H (M (l)) I qr+1 H (M (l)) inuces an isomorphism with inverse coming from the projection p: M (l) M (l). Proof. The projection p: M (l) M (l) satisfies with p(n k 1 k (l) N k 1 (l)) N k (l) N k 1 (l) p(n k (l) N k 1 k (l)) = N k (l) N k 1 (l) q r+1 (N k (l) N k 1 k (l)) coim(n k (l) N k 1 k (l)) = (r + 1)k 1 c n,k k + 1 = r k c n,k = p r (N k (l) N k 1 (l)) coim(n k (l) N k 1 (l)), so projection inuces a homomorphism p : I qr+1 H (M (l)) I pr H (M (l)) which is seen to be the inverse of i as the straight-line homotopy between i p an the ientity on M (l) is stratum-preserving an therefore inuces the require chain homotopy. The pseuomanifol M (l) is non-compact, but we can form a similar compact pseuomanifol by letting an M (l) = C (l) SO( 1) D 1 M (l) = C (l) SO( 1) S 2 M (l) = M (l)/ M (l) with extra stratum corresponing to M (l). If we exten the perversity q r by efining q r ( ) = r( 2), we then have (2) I qr H n+1 r ( M (l)) = I qr H n+1 r(m (l))

8 8 DIRK SCHÜTZ for r > 1, since allowable k-chains in M (l) with k n+1 r + 1 cannot intersect the extra stratum. For r = 1 this oes not work, as the generator of I q1 H n (M (l)) bouns in M (l) in an allowable way. We can solve this by resorting to non-goresky MacPherson perversities, setting q 1 ( ) = 0. Now (2) also hols for r = 1, an we can form the intersection ring for M (l) given by IR ( M n (l)) = I qr H n+1 r( M (l)) r=0 with the intersection prouct coming from [6, Thm.5.3]. Note that proucts involving r 1 + r 2 > n are consiere 0. Define an element R I p1 H n 1(M (l)) as follows. The funamental class [M (l)] I p0 H n (M (l)) represents a generator X I q1 H n ( M (l)) which we can intersect with itself to obtain an element X 2 I q2 H n 1( M (l)). We then let R = p (X 2 ). We want to represent R more irectly. For this, efine for each j {1,..., n 1} by i j : M (l) M (l) i j ([x 1,..., x n 1 ]) = [x 1,..., x n 1, π(x j )] where π : S 1 R 1 is projection to the last 1 coorinates of S 1 R. Then i j (N k (l) N k 1 (l)) N k 1 k (l) N k 1 (l) an as i j is stratum-preserving homotopic to i, we get (i j ) = i : I pr H (M (l)) I qr+1 H (M (l)). Let l = (l 1,..., l n ) be a generic, orere length vector. For j = 1,..., n 1 recall the length vector l j + R n 1 given by l j + = (l 1,..., ˆl j,..., l n 1, l n + l j ). If we replace the last coorinate by l n l j, we get a length vector that we call l j R n 1. As in [15], we now get elements [M (l 1 +)],..., [M (l n 1 + )], [M (l 1 )],..., [M (l n 1 )] I p1 H n 1(M (l)). For the moment, there is still some ambiguity about the orientations of these elements. To resolve this, note that we can think of M (l j ±) as a subset of M (l) using the the stanar zero section. Then M (l) i j (M (l)) = {[x 1,..., x n 1, 0] x j S 0 } = M (l j +) M (l j ).

9 INTERSECTION HOMOLOGY OF LINKAGE SPACES 9 Choosing an orientation of M (l) then inuces an orientation on M (l) i j (M (l)) for all j = 1,..., n 1. In particular, we get (3) R = [M (l j +)] + [M (l j )] for all j = 1,..., n 1. As R = p (X 2 ) we see that 2([M (l j +)] + [M (l j )]) = 0 for even which iffers slightly from [15, Lm.7.5] because of ifferent orientation conventions. To simplify notation, let us write X j = [M (l j +)] I p1 H n 1(M (l)) = [M (l j )] I p1 H n 1(M (l)). X j Remark 3.2. The space M (l) is a stratifie bunle in the sense of [1]. To see this, let M be a compact smooth G-manifol, where G is a compact Lie group. If F enotes the orbit category of G, that is, the category with objects G/H for H a close subgroup of G, an whose morphisms are G-equivariant maps G/H G/H, then M M/G is an F-stratifie bunle by [1, Example 4.6], see also [2]. Now let V be a vector space an ρ: G GL(V ) a representation. Define the category V as a subcategory of topological spaces where the objects are the quotient spaces V/H for H a close subgroup of G. There is an obvious functor ϕ V : F V, an one can check that M G V agrees with the coen construction M F ϕ V escribe in [1, 6]. In particular, p: M G V M/G is a V-stratifie fibre bunle in the sense of [1]. Notice however that it is not a stratifie vector bunle in general. We can now think of the element R above as an Euler class, in that it represents an obstruction for the existence of a stratifie non-zero section σ : M (l) M (l). One woul expect that the above constructions can exten to M G V M/G, an one may ask how far this can be generalize to the setting of stratifie fibre bunles. 4. The reuce intersection ring of M (l) For the next lemma, we also use the notation for K {1,..., n 1}. X K = i K X i Lemma 4.1. Let 4 an l a generic, orere -normal length vector, an J {1,..., n 1} such that J {n} is l-short, an K {1,..., n 1} with K = J. Then there exists Y J I 0 H J ( 1) (M (l)) with { 1 K = J X K Y J = 0 else an R Y J = 0. Proof. In [15, Lm.8.1] explicit uals Y J for X J were constructe by efining appropriate embeings of (S 1 ) J into M (l). The relation Y J X K = 0 for K J was a consequence of being able to avoi letting the k-th coorinate x k of the element in M (l) point in the same irection as x n.

10 10 DIRK SCHÜTZ To o this the robot arm consisting of those links which were not part of J {n} ha to trace the area in R that coul be reache by the robot arm consisting of the links in J an which starte at l n e 1 R. To o this, the first robot arm has to trace a straight line, an then reach all other points using appropriate rotations, where not all links woul rotate the same way. For the first link (the one connecte to the origin), one can avoi completely this latter rotation, so one just has to avoi the points ±e 1 uring the trace of the straight line, which can easily be one. Note that even for = 4 we have n 6 to avoi trivial cases, so that the first robot arm has at least four links. Such a ual will then also satisfy Y J X k = 0, an therefore where we use (3). R Y j = (X k + X k ) Y j = 0, Let p 1 be the ual perversity to p 1, that is, the perversity with p 1 + p 1 = t. Lemma 4.2. Let 5 be o an l R n a generic, -normal, orere length vector with n + 3. Then there exists an element Y I p 1 H 1 (M (l)) with X n 1 Y = 0 X n 1 Y = 1. Remark 4.3. The proof of Lemma 4.2 relies on a elicate geometric construction that we postpone to Section 5. In the case = 5 this construction simplifies significantly, an we give this simplifie construction here as it alreay contains some of the ieas require in the general case. We will assume that l is orere. By the equivariant Morse-Bott function on C 5 (l) constructe in [14] we get an SO(4)- equivariant embeing C 5 (l n 1 ) D 4 into C 5 (l), where SO(4) acts iagonally on C 5 (l n 1 ) D 4, an C 5 (l n 1 ) {0} correspons to the obvious embeing C 5 (l n 1 ) C 5 (l). Pick an element p C 5 (l n 1 ) of rank 5 (or 4). Now efine f : S 3 C 5 (l n 1 ) by f(q) = q p where we think of S 3 as a subgroup of SO(4) via quaternion multiplication. Now observe that C 5 (l n 1 ) is 3-connecte. Firstly, the Morse-Bott function in [14, 3] can be moifie to a Morse function which has critical points only of inex 4(n 3 k) or 4(n 3 k) + 3 for k {0,..., n 3}, which makes C 5 (l n 1 ) simply connecte. Furthermore, the cohomology calculation in [5, Thm.2.1] shows that the first non-trivial homology group of C 5 (l n 1 ) has at least egree 4, which means this space is 3-connecte. Note that we require l n 1 to be 3-normal, which is implie by l being 5-normal, to ensure the vanishing of the thir homology group. We can therefore exten f to a map F : D 4 C 5 (l n 1 ) which can even be an embeing. Also, this embeing can be mae transverse to the map g : C 3 (l n 1 ) SO(4) C 5 (l n 1 ) given by g(x, A) = Ax. For imension reasons, this means that F misses g, so that all F (x) have rank at least 4. Finally, the map F : D 4 C 5 (l n 1 ) D 4 given by F (x) = (F (x), x) inuces a map F : S 4 M 5 (l) with F (S 4 ) M 5 (l n 1 ) = {[p]}. By letting F be constant in a neighborhoo of 0 (which lets F no longer be an embeing, but lets F remain an embeing) this intersection is transverse. Furthermore, F (S 4 ) M 5 (l n 1 ) =. Therefore F (S 4 ) represents

11 INTERSECTION HOMOLOGY OF LINKAGE SPACES 11 the require element Y, an since all points in F (S 4 ) have rank at least 4, we even get an element Y I 0 H 4 (M (l)). Also, note that we only require n + 2 = 7 here, if n = 6, the element X n 1 = 0. A similar construction can be one in the case = 9 using octonian multiplication, however, it is not clear how this construction coul generalize to the other cases of o. Proposition 4.4. Let 5 be o an l R n be a -normal, orere, generic length vector with n +2. Let k = a 1 (l). Then the intersection ring IR (M (l)) is generate by elements R, X 1,..., X k which satisfy the following relations: (1) RX i = X 2 i for all i = 1,..., k. (2) X i1 X im if {i 1,..., i m, n} is l-long. For n + 3 we can choose R to be the Euler class of the stratifie bunle M (l). This is not a complete list of relations, for example we have R m = 0 for m large enough simply by the construction of the intersection ring. Notice also that for n = + 2 we cannot have non-trivial proucts for egree reasons. Proof. Let n = + 2. Then we just choose the elements R, X 1,..., X k so that they form a basis of I p1 H n 1(M (l)), compare Proposition 2.5. Any proucts among these elements are zero for egree reasons, so the relations are trivially satisfie. Now let n > + 2. We now choose R an X i as in Section 3. By Lemma 4.1 an Lemma 4.2 these elements are linearly inepenent an form a basis of I p1 H n 1(M (l)) because of Proposition 2.5. We get R = X j + X j for all j = 1,..., n 1 by equation (3) of Section 3, so RX i = X i X i + X i X i. Now X i X i is represente by M (l i ) M (l i +) =, so (1) follows. Also, X i1 X im is represente by M (l i1 +) M (l im + ) which is empty by the conition that {i 1,..., i m, n} is l-long. Therefore (2) hols. Remark 4.5. We want to compare the previous result to the cohomology ring of M 3 (l) etermine in [11]. For a generic, orere length vector l R n their Theorem 6.4 states that H (M 3 (l)) = Z[R, V 1,..., V m 1 ]/I l where R an V i are of egree 2 an I l is the ieal generate by the three families S L,S S (l) Vi 2 + RV i for i = 1,..., n 1 for L {1,..., n 1} with L {n} l-long i L V i ( ) V i R L S 1 for L {1,..., n 1} l-long i S The first two families correspon to the relations in Proposition 4.4 after a change of sign. The thir family is more complicate, an we will not try to fin the corresponing relations for the intersection ring. We note however the following: If L {1,..., n 1} is l-long an S L is l-short, then either L S > 1 or S {n} is l-long. Therefore the relations in the thir family are of the form RW with W Z[R, V 1,..., V m 1 ]. This was alreay observe in [4, Lm.5].

12 12 DIRK SCHÜTZ In the case = 3 the stratifie bunle p: M 3 (l) M 3 (l) can be viewe as a complex line bunle, an it is shown in [11, Prop.7.3] that the negative Chern class agrees with R. Note that our R woul correspon to the positive Chern (Euler) class, which is consistent with the change of sign in the first relation of the next lemma below. Lemma 4.6. Let 5 be o an l R n be a -normal, orere, generic length vector with n + 2 an let k = a 1 (l). Let I l be the kernel of the surjection Φ: Z[R, X 1,..., X k ] IR (M (l)) inuce by Proposition 4.4. Then there exist elements W 1,..., W l Z[R, X 1,..., X k ] for some l 1 so that I l is generate by relations (1) RX i X 2 i for all i = 1,..., k. (2) X i1 X im if {i 1,..., i m, n} is l-long. (3) RW i for i = 1,..., l. Proof. By the Hilbert Basis Theorem we know that I l is finitely generate. Since the elements of the form (1) an (2) are in I l by Proposition 4.4, we can simply a them to any finite generating set. So these elements together with finitely many elements V 1,..., V l Z[R, X 1,..., X k ] form a generating set. Let us write each V i as a linear combination of monomials V i = j i j=1 a ijv ij with a ij Z {0}. We can first assume that no monomial V ij contains more than one factor of any X v, for we coul replace this monomial with the corresponing monomial having R u 1 X v in place of Xv u using a relation from (1). Now if a monomial V ij has no factor R, we can write it as V ij = X u1 X uv with J = {u 1,..., u v } an J = v. If J {n} is l-long, we can remove V ij using a relation of the form (2). If J {n} is l-short we get a Poincaré ual Y J to V ij from Lemma 4.1 with R Y J = 0. Then V i Y J = a ij 0 which contraicts V i I l. Therefore such a monomial cannot appear in V i. It follows that V i = RW i for some W i Z[R, X 1,..., X k ]. Let us now consier Z/2 coefficients. To simplify our iscussion, we will simply tensor the integral intersection ring with Z/2 to obtain a new ring that we enote by IR Z/2 (M (l)) = IR (M (l)) Z/2. Corollary 4.7. Let 5 be o an l R n be a -normal, generic length vector with + 3. Let R be the Euler class of the stratifie bunle M (l) with Z/2 coefficients. Then IR Z/2 (M (l))/ R = Λ Z/2 [S (l)]. Proof. We have the relations from Proposition 4.4 which reuce to Xi 2 for all i = 1,..., k an X i1 X im if {i 1,..., i m, n} is l-long. Over Z/2 this reuces to the exterior face ring of the short subsets. We now nee to fin a way to etect R in terms of intersection proucts. This is similar to the argument use in [4], however, since we have worse information about the intersection ring, the argument is a bit more involve. As a start, we nee the following result. Lemma 4.8. Let 5 be o an l R n be a -normal, generic length vector with n +3 an M (l). Let i, j {1,..., n 1} with i < j an let l ij + = (l j ) i +

13 R. Then I 0 H I p1 H INTERSECTION HOMOLOGY OF LINKAGE SPACES 13 (M (l ij + )) is a irect summan of I p1 H (M (l j )) an (M (l)). (M (l j )) is a irect summan of I p2 H Proof. Both statements have essentially the same proof, we will therefore focus on the secon statement. Note that inclusion M (l j ) M (l) inuces a map I p1 H (M (l j )) I p2 H (M (l)) by [15, Lm.5.2]. To see that this map is split-injective, we use the Morse argument use in [15, 7]. There is an SO( 1)- invariant Morse-Bott function on C (l) R whose absolute minimum is M (l j ) an whose other critical manifols are spheres of imension 2 of inex k( 1) for k {1,..., n 3}. This gives rise to a filtration of M (l) M 0 M 1 M m = M (l) where M 0 contains M (l j ) as a eformation retract in a stratification preserving way so that I p1 H (M (l j )) = I p2 H (M 0 ). We nee to show that I p2 H the long exact sequence (M l 1 ) I p2 H I p2 H (M l ) is split-injective for all l = 1,..., m. We have +1 (Ml, M l 1 ) I p2 H (M l 1 ) I p2 H (M l ) I p2 H (M l, M l 1 ) The proof is going to be along the lines of the proof of [15, Lm.7.1]. We remark that I p2 H (M 0 ) is torsion free by Proposition 2.5. We will see in the proof below that I p2 H (M l, M l 1 ) is also torsion-free, which will give rise to the splitting, an a torsion free group I p2 H (M l ) by inuction. Recall notation from [15, 6], namely for non-negative integers m, k with m k let N m,k = ((D 1 ) k (D 1 ) m k )/SO( 2), N m,k = ( ((D 1 ) k ) (D 1 ) m k )/SO( 2). Then I p2 H r (M l, M l 1 ) = I p2 H r (N n 3,k l, N n 3,k l ), where k l is the inex of the critical point containe in M l M l 1, that is, the corresponing critical sphere S 2 is of inex k l ( 1). Assume that k l n 5. Then I p2 H r (N n 3,k l, N n 3,k l ) = I 0 H r (N n 5,k l, N n 5,k l ) by [15, Lm.6.1]. For k l = n 5 this has only one non-vanishing group in egree r = by [15, 5], which is Z. For k l < n 5 we can use Poincaré uality (taking torsion into account, using [6, Cor.4.4.3]) I 0 H r (N n 5,k l, N n 5,k l ) = I t H r (N n 5,k l, + N n 5,k l ) an the latter is just orinary homology, which vanishes for r = 1 as + N n 5,k l for k l < n 5. Therefore the map I p2 H I p2 H (M l ) is split-injective if k l n 5. If k l = n 4, we get I p2 H r (N n 3,n 4, N n 3,n 4 ) = I p1 H r (N n 4,n 4, N n 4,n 4 ), + (M l 1 )

14 14 DIRK SCHÜTZ an the latter is Z/2 for r = + 1 an 0 for r = by [15, Prop.6.3]. Therefore our map is again split-injective. To calculate I p2 H r (N n 3,n 3, N n 3,n 3 ) a cellular chain complex is ientifie in [15, 6], an the lowest imensional cell is of imension + 2, compare [15, Lm.6.2]. Therefore the homology groups in egree are not affecte, an the result follows. Corollary 4.9. Let 5 be o an l R n be a -normal, generic length vector with n + 3. Let i, j {1,..., n 1} with i < j. If X i is a non-zero element of I p1 H n 1(M (l)), then X i X j is a non-zero element of I p2 H (M (l)) Z/2. Proof. Observe that X i X j is represente by M (l i +) M (l j ), an since X i 0 implies {i, n} is l-short, this set is non-empty. Therefore X i X j is the image of the funamental class in I 0 H (M (l i +) M (l j )). By Lemma 4.8 it follows that (M (l)) Z/2. 0 X i X j I p2 H Proof of Theorem 1.1. The cases n = 2, 3 were covere in [4, 13], while 4 even is in [15]. It remains to consier 5 o. That l is -normal implies n + 1. If n = + 1, there is only one chamber up to permutation, so there is nothing to show. Similarly, if n = + 2, then a 2 (l) = 0 an the chamber is etermine by a 1 (l), which is recovere from I p1 H n 1(M (l)), a homeomorphism invariant. We can therefore assume that n + 3. Assume also that l is orere. We want to say that R is the unique element of I p1 H n 1(M (l)) Z/2 such that multiplication by R inuces the squaring homomorphism Sq: I p1 H n 1 (M (l)) Z/2 I p2 H (M (l)) Z/2. By Proposition 4.4, R certainly has this property. To get uniqueness, let us also assume that a 2 (l) > 0, so that there exist i j with X i X j 0 (we can use i = 1, j = 2 since l is orere). Now let R = εr + X i1 + + X iu I p1 H n 1(M (l)) Z/2 satisfy R X = X 2 for all X I p1 H n 1(M (l)) Z/2, for some u 1. In particular X 2 j = εx 2 j + X i1 X j + + X iu X j for all j = 1,..., n 1. If any X iv X j were non-zero for j i v, then X 2 j Y i v,j 0 for the ual Y iv,j from Lemma 4.1. But since X 2 j Y i v,j = X j R Y iv,j = 0, we get X iv X j = 0 for all i v j. In particular i v 1 or 2 by the assumption X 1 X 2 0. By using j = i v (assuming that u 1) we also get X 2 i v = εx 2 i v + X 2 i v by multiplication with R, but we also have X 2 i v = (X 1 + X 1 )X i v = X 1 X i v

15 INTERSECTION HOMOLOGY OF LINKAGE SPACES 15 by multiplication with R = X 1 + X1. Therefore X2 i v means ε = 0 Z/2. Now 0 by Corollary 4.9 which X 2 1 = (X i1 + + X iu )X 1 = 0 contraicting X 2 1 = (X iu + X i u )X 1 = X i u X 1 0. So uner the extra conition that a 2 (l) > 0 we get that R is the only element in I p1 H n 1(M (l)) Z/2 such that multiplication by R gives Sq. Let l, l R n be generic, -normal length vectors with M (l) homeomorphic to M (l ). By [8] there is an isomorphism of intersection rings IR (M (l)) = IR (M (l )). If the Euler class R l of M (l) woul not be unique with the squaring property (after tensoring with Z/2), then neither woul be R l an we woul get a 2 (l) = 0 = a 2 (l ). But, up to permutation, the chamber of any l with a 2 (l) = 0 is etermine by a 1 (l) which we can obtain from the imension of I p1 H n 1(M l ) Z/2. So l an l are in the same chamber up to permutation. We can therefore assume that both R l an R l are unique with the squaring property, an the isomorphism of intersection rings inuces an isomorphism of exterior face rings Λ Z/2 [S (l)] = Λ Z/2 [S (l )] by Corollary 4.7, which inuces an isomorphism of simplicial complexes S (l) = S (l ) by [9]. This implies that l an l are in the same chamber up to permutation as in [4]. 5. Proof of Lemma 4.2 Let us begin with the strategy of the proof. From the equivariant Morse-Bott function in [14] we get an equivariant neighborhoo of C (l n 1 ) in C (l) of the form C (l n 1 ) D 1 where SO( 1) acts iagonally on the factors. As 5 is o, there is k 2 with 1 = 2k, an we can write R = R C k. We can let S 1 be the subgroup of U(k) SO(2k) = SO( 1) which rotates each complex coorinate. We now want to construct an S 1 -equivariant map f : S 2 C (l n 1 ) which extens (non-equivariantly) to f : D 1 C (l n 1 ) an which is constant near 0 D 1 an so that f(0) has rank at least 1. Then f inuces a map from complex projective space F : CP k M (l) via F (z) = q(f(z), z), where q : C (l n 1 ) D 1 M (l) is inclusion followe by the quotient map. Since f is constant near 0 an has rank 1 we get that F is an embeing near the point corresponing to 0, an F (CP k ) intersects M (l n 1 ) transversely in exactly one point, while F (CP k ) M (l n 1 + ) =. The require element is then Y = F [CP k ]. To get the S 1 -equivariant map f we will actually construct an (S 1 ) k -equivariant map, where each factor S 1 acts on its respective coorinate in C k. As a result, the map F will hit several singular strata in M (l). The next lemma gives a criterion so that F inuces an element of I p 1 H 1 (M (l)). Lemma 5.1. Let F : CP k M (l) be a stratum-preserving map for some stratification of CP k that has only even-imensional strata. Assume that for l = 0,..., k 2 the strata containe in F 1 (N 3+2l (l) N 2+2l (l)) have imension at most 2l, an that F 1 (N 2+2l (l) N 1+2l (l)) =. Then F inuces a homomorphism F : I 0 H 1 (CP k ) I p 1 H 1 (M (l)).

16 16 DIRK SCHÜTZ Proof. Note that the conition of the Lemma states that a stratum S which is sent to N 2(k l 1) (l) N 1 2(k l 1) (l) has coimension at least 2(k l). As p 1(N 2(k l 1) (l) N 1 2(k l 1) (l)) = c n,2(k l 1) 2(k l), we see that [15, Lm.5.2] applies. Let k 1 = {(t 1,..., t k ) R k t i [0, 1], t 1 + t k = 1} be the stanar (k 1)-simplex. Then k 1 = S 2 /(S 1 ) k an we can split the quotient map via j : k 1 S 2 given by j(t 1,..., t k ) = ( t 1,..., t k ) C k. So in orer to get an (S 1 ) k -equivariant map f : S 2 C (l n 1 ) it is enough to efine f on k 1. Let us start with a point x C (l n 1 ). Then x = (x 1,..., x n ) with x n = e 0, x n 1 = e 0, where we think of x j R = R C k an we think of the R-coorinate as the 0-th coorinate. Since l is -normal an orere, we get that {n ( 1),..., n 1} is l-short. Let K {1,..., n } be such that K {n ( 1),..., n 1} is long, while removing any element of K woul make it short. Define (x 1,..., x n ) C (l n 1 ) as follows. If j / K {n ( 1),..., n 1}, let x j = e 0. If j K {n ( 1),..., n 4, n 1}, let x j = e 0. Also, let x n 3 = (x n 3,0, x n 3,1, 0,..., 0), x = (x,0, x,1, 0,..., 0) with x n 3,0, x,0 ( 1, 0) R an x n 3,1, x,1 C be imaginary so that (x 1,..., x n ) C (l n 1 ). Note that these values can be chosen by the way K was efine, compare Figure 1. The values x n 3,1, x,1 are chosen imaginary, so that e 0, e 1, x have rank 3. Figure 1. The configuration x Note that e 1 S 2 C k correspons to x n 1, once C (l n 1 ) D 1 is embee in C (l). We start efining f by setting f(1, 0,..., 0) = (x 0,..., x n ). To exten f, for j = 2,... k 1 an t k 1 let 2 x j (t) = ( 1 (t j ε) 2 (t k ε) 2, 0,..., 0, t j ε, 0..., 0, t k ε) x j 1 (t) = ( 1 (t j ε) 2 (t k ε) 2, 0,..., 0, it j ε, 0..., 0, it k ε) x n ( 1) (t) = ( 1 (t k ε) 2, 0,..., 0, it k ε) where ε > 0 is small, an the non-zero entry is in the j-th complex coorinate of R C k. If ε is small enough, we can efine x n 3 (t) S 1 {0} S 1 an x (t) S 1 so that (x 1,..., x n, x n ( 1) (t),..., x n 4 (t), x n 3 (t), x (t), x n 1, x n ) C (l n 1 ) 2 The letter i now stans for the complex unit i C, it will no longer be use as an inex.

17 INTERSECTION HOMOLOGY OF LINKAGE SPACES 17 with t k 1. Note that x j an x j 1 epen on t j for j = 2,..., k 1 an t k. Basically, we let x compensate for the non-zero entries in the other complex variables, thus requiring that the absolute values of the 0-th an first coorinate in x n 3 an x will be slightly less. We can make this epen continuously on t, thus giving us a map k 1 C (l n 1 ). We still nee to slightly change this map. As j( k 1 ) is a contractible subset of S 2, there is a map A: k 1 U(k) with A(t) e 1 = j(t). Let f n 3 (t) = A(t) x n 3 (t), an f n j (t) = x n j (t) for j = 4,..., n 1. Choose f (t) so that (f 1,..., f, e 0, e 0 ) C (l n 1 ), that is, we have a map f : k 1 C (l n 1 ). In the next lemma we think of (f(t), j(t)) as an element of C (l) via the embeing C (l n 1 ) D 1. We also set n(t) to be the number of coorinates j with t j = 0 for any t k 1. Lemma 5.2. We have rank(f(t), j(t)) = 1 if n(t) = 0, an rank(f(t), j(t)) = 2n(t), if n(t) 1. Furthermore, if t = (t 1,..., t k ) satisfies t j = 0 for j {1,..., k}, then the j-th complex coorinate of each f m (t) is 0, m = 1,..., n. Proof. Let us write (f(t), j(t)) = (f 1,..., f, f n 1 ) C (l), wheref n 1 only epens on j(t). We ignore the n-th coorinate, as it is e 0 S 0. Since f j S 0 for j n an j n 1, an f is a linear combination of the other elements, the rank can be at most 1. Now assume that t j = 0 for j {1,..., k}. None of the elements f j has a non-zero entry in the j-th complex coorinate, so again the rank can be at most 2n(t). Let n(k) = 0. Then f n 1, f n 3 are the only elements (apart from f ) with nonzero entries in the first complex coorinate, an these two elements are linearly inepenent, as f n 1 is only in the real part, while f n 3 is only in the imaginary part. Apart from f n 1, f, f n 3, only f n 4, f n 5 have non-zero entry in the secon complex coorinate. Continuing, we see that f j, f j 1 increase the rank by two for each j = 2,..., k 1. Finally f n ( 1) an x n increase the rank to 1, as require. If n(k) > 0 let us istinguish the cases t 1 = 0 an t 1 > 0. If t 1 > 0, then f n 1, f n 3 are the only elements with non-zero entries in the first complex coorinate. If t k = 0, then each j {2,..., k 1} with t j > 0 prouces two elements f j, f j 1 with only the j-th complex coorinate non-zero. Together with the elements in S 0, we see the rank is 2n(t). If t 1 > 0 an t k > 0, there is a j {2,..., k 1} with t j = 0. Then f j, f j 1 have only something non-zero in the k-th complex coorinate, so they increase the rank by 2. All other j {2,..., k 1} with t j > 0 have f j, f j 1 as the only remaining elements with j-th complex coorinate non-zero, so they also increase the rank by 2. Together with the elements in S 0, the rank is again 2n(t). We thus have a map f : k 1 C (l) given by f(t) = (f(t), j(t)), which we can exten to an (S 1 ) k -equivariant map f : (S 1 ) k k 1 C (l), where (S 1 ) k acts on C k (on both sies) by coorinate-wise multiplication. Furthermore, there is a surjection p: (S 1 ) k k 1 S 2, which is also (S 1 ) k -equivariant, an which inuces the require (S 1 ) k -equivariant map f : S 2 C (l)

18 18 DIRK SCHÜTZ by Lemma 5.2. For every subset L {1,..., k} we get a subsphere S L S 2 of imension 2 L 1, whose entries are those points (z 1,..., z k ) S 2 with z j = 0 for j / L. It follows from Lemma 5.2 that f followe by the quotient map from C (l) to M (l) is a stratifie map, if we stratify S 2 by the S L. We nee to exten f to D 1. The basic iea is to stretch the robot arm mae up of the points f n 1,..., f n ( 1) into a straight line, pointing in the irection of e 0. Recall we write f(z 1,..., z k ) = (f 1,..., f, f n 1 ) C (l n 1 ) D 1 so that each f j S 1 for j = 1,..., n 2, with f n 1 D 1 S 1 in a small isc centere at e 0. In fact, f 1,..., f n S 0, an the 0-th coorinate (the real coorinate in R = R C k ) being negative for all f n 1,..., f n ( 1). We can ignore the last coorinate an simply think of (f 1,..., f ) C (l n 1 ). Denote by g the composition of f with the projection p: C (l) (S 1 ) 2 to the coorinates (f n ( 1),..., f ). Note that l n ( 1) f n ( 1) + + l f = ce 0 for some fixe c > 0. We can think of these 2 coorinates as a robot arm starting at the origin an ening at ce 0. We want to stretch out this robot arm until all coorinates point to e 0. To o this we use the flow Φ of the stanar graient of the height function on S 1 which has e 0 as its maximum an e 0 as its minimum. Consier G: S 2 [0, ) (S 1 ) 3 given by projecting g own to the coorinates (f n ( 1),..., f n 3 ) an then applying the flow Φ to each of the 3 coorinates. As we continue to flow, each coorinate approaches e 0. We thus get an inuce map Ḡ: D 1 (S 1 ) 3 such that Ḡ(0) = ( e 0,..., e 0 ). Denote by G j (z) S 1 the j-th coorinate of G(z) for z D 1. Then there is a unique element G 2 (z) S 1 whose 0-coorinate is negative, an which ensures that l n +1 G 1 (z) + + l n 3 G 3 (z) + l G 2 (z) = c( z )e 0 where c: [0, 1] (0, ) is a monotonely ecreasing function. Note that G 2 (z) exists by elementary geometry, an by the Implicit Function Theorem it epens smoothly on z. As the flow can be chosen to be invariant uner the SO( 1)-action on S 1, the function c only epens on z. We may think of the points (G 1 (z),..., G 2 (z)) (S 1 ) 2 as a robot arm epening on z D 1 which starts at the origin an has enpoint on the negative real axis in R C k. Also, on S 2 this agrees with g. We nee to exten this robot arm to a map f : D 1 C (l n 1 ). To o this, note that the coorinates (f 1,..., f n ) are all in S 0. Note that n 3 an there is at least one element f m with f m = e 0 (this is an element of the set K). Recall that K {n ( 1),..., n 1} is l-long, but removing any element of K makes this set l-short. Now let γ : [0, 1] (S 1 ) n with γ(0) = (f 1,..., f n ), so that in the m-th coorinate the point e 0 is rotate into e 0 along S 1 S 1, an so that l 1 γ 1 (t) + l n γ n (t) = (t)e 0 with : [0, 1] (0, ) a strictly monotone increasing map. Note that we only have to moify two extra coorinates besie m, so this is easily one. We can also o this so that p(γ 1 (t)),..., p(γ n (t))

19 INTERSECTION HOMOLOGY OF LINKAGE SPACES 19 have rank at least 3 for all t (0, 1) an for any projection p: R C k R C to one of the C-coorinates of C k (an keeping the R-coorinate). For this we shoul rotate the m-th coorinate iagonally through C k rather than through C {0}, an similarly with the other coorinates. We want to combine γ an G to obtain the map f : D 1 C (l n 1 ), using the formula f(z) = (γ(s( z )), G 1 (z),..., G 2 (z)) for some map s: [0, 1] [0, 1] with s(1) = 0. Note that both γ an G both en up on R {0} when aing up the coorinates, so we nee to choose s( z ) with l n l n 1 + c( z ) + (s( z )) = 0. Since is invertible on its image, an l n 1 c( z ) l n in this image by the choice of the set K, we can fin this s. Because of the (S 1 ) k -equivariance of f on S 2 (note that G is still equivariant on the interior of D 1, but γ is not), we get the inuce map F : CP k M (l) after restricting to the S 1 -iagonal-action. We clearly have F (CP k ) M (l n 1 ) = {F (0)}, where 0 CP k correspons to 0 D 1. Note however that ue to our construction, the rank of F (0) is 3, so it oes not represent a regular point of M (l). In orer to fix this, let us analyze ranks of images in more etail. Consier the restriction f : S 2 C (l n 1 ) which inuces a map F : CP k 1 M (l n 1 ). For any subset A {1,..., k} with A we have natural subspace CP A = CP A 1 consisting of those elements [z 1 : : z k ] with z i = 0 if i / A. These subspaces form a natural stratification of CP k 1 CP k 1 0 CP k 1 2 CP k 1 2k 4 CPk 1 2k 2 = CPk 1 an by the construction of f the restriction F is a stratifie map F : CP k 1 M (l n 1 ) with an F (CP k 1 2i CP k 1 2i 2 ) N 2i+3(l n 1 ) N 2i+2 (l n 1 ) for i k 2 F (CP k 1 CP k 1 2k 4 ) M (l n 1 ) N 2 (l n 1 ). Note that CP k has a similar stratification to CP k 1 using A {1,..., k + 1} with A, an the choice of γ ensures that we also have a stratifie map F : CP k M (l) with an F (CP k 2i CP k 2i 2) N 2i+3 (l n 1 ) N 2i+2 (l n 1 ) for i k 1 F (CP k CP k 2k 2) M (l n 1 ) N 2 (l n 1 ). Note that we think of CP k as a quotient space D 1 /, an the image of F is containe in a quotient space (D 1 C (l n 1 ))/ which is a subset of M (l). In terms of these quotient spaces, F is given by F ([y]) = [y, f(y)]. Now take a iffeomorphism ϕ of D 1 which sens 0 to a point y 0 near 0 that is sent to a point of rank 1 uner f, an is the ientity outsie a small neighborhoo of 0 D 1. Let us alter F to the map F ([y]) = [ϕ 1 (y), f(y)]. This map still has exactly one point in the intersection with M (l n 1 ), namely F ([y 0 ]), but this time the rank is 1. We can alter f to make it constant near y 0, thus ensuring

20 20 DIRK SCHÜTZ that the intersection is also transverse. To ensure that F is still strata-preserving, simply alter the stratification of CP k using ϕ. By Lemma 5.1 we get our element Y = F [CP k ] I p 1 H 1 (M (l)) an it satisfies the conitions require in Lemma 4.2. References [1] H.-J. Baues, D. Ferrario, Stratifie fibre bunles, Forum Math. 16 (2004), [2] M. Davis, Smooth G-manifols as collections of fiber bunles, Pacific J. Math. 77 (1978), [3] M. Farber, V. Fromm, The topology of spaces of polygons, Trans. Amer. Math. Soc. 365 (2013), [4] M. Farber, J.-Cl. Hausmann, D. Schütz, On the conjecture of Kevin Walker, J. Topol. Anal. 1 (2009), [5] M. Farber, J.-Cl. Hausmann, D. Schütz, The Walker conjecture for chains in R, Math. Proc. Cambrige Philos. Soc. 151 (2011), [6] G. Frieman, Intersection homology with general perversities, Geom. Deicata 148 (2010), [7] M. Goresky, R. MacPherson, Intersection homology theory, Topology 19 (1980), [8] M. Goresky, R. MacPherson, Intersection homology II, Invent. Math. 72 (1983), [9] J. Gubelaze, The isomorphism problem for commutative monoi rings, J. Pure Appl. Algebra 129 (1998), [10] J.-Cl. Hausmann, Sur la topologie es bras articulés, Algebraic topology Poznan 1989, , Lecture Notes in Math. 1474, Springer, Berlin, [11] J.-Cl. Hausmann, A. Knutson, The cohomology ring of polygon spaces, Ann. Inst. Fourier (Grenoble) 48 (1998), [12] I.J. Schoenberg, Linkages an istance geometry, I. Linkages, Inag. Math. 31 (1969), [13] D. Schütz, The isomorphism problem of planar polygon spaces, J. Topol. 3 (2010), [14] D. Schütz, Homology of mouli spaces of linkages in high-imensional Eucliean space, Algebr. Geom. Topol. 13 (2013), [15] D. Schütz, Intersection homology of linkage spaces, preprint, arxiv: , to appear in J. Topol. Anal. [16] K. Walker, Configuration spaces of linkages, Bachelor thesis, Princeton, Department of Mathematical Sciences, University of Durham, Science Laboratories, South R, Durham DH1 3LE, Unite Kingom aress: irk.schuetz@urham.ac.uk