Link homotopic but not isotopic

0.49 MB·

by  Bakul Sathaye

LINK HOMOTOPIC BUT NOT ISOTOPIC TO THE UNLINK
BAKUL SATHAYE
Abstract. We provide a family of links with n components (n≥3) that are link homotopic but
not link isotopic to the unlink. Further, every sublink of these links is link isotopic to the unlink,
6 and all the Milnor invariants of these links vanish.
1
0
2

b 1. Introduction
e
F An n-component link L in S3 is a collection of piecewise linear maps (l ,...,l ) : S1 → S3,
1 n
3 where the images l1(S1),...,ln(S1) are pairwise disjoint. A link with one component is a knot.
2 Two links in S3, L and L , are said to be isotopic if there is an orientation preserving homeo-
1 2
morphism h : S3 → S3 such that h(L ) = L and h is isotopic to the identity map.
] 1 2
T The notion of link homotopy was introduced by Milnor in [4]. Two links L and L(cid:48) are said
G to be link homotopic if there exist homotopies h , between the maps l and the maps l(cid:48) so that
i,t i i
. the sets h (S1),...,h (S1) are disjoint for each value of t. In particular, a link is said to be
h 1,t n,t
t link homotopically trivial if it is link homotopic to the unlink. Notice that this equivalence allows
a
self-crossings, that is, crossing changes involving two strands of the same component.
m
The question that arises now is how different are these two notions of link equivalence.
[
Fromthedefinitionitisclearthatalinkisotopyisalinkhomotopyaswell. Linkhomotopyallows

2
self-crossings,whichdramaticallysimplifiestheequivalencebetweenlinks. Underlinkhomotopy,all
v
knots become trivial knots, despite the fact that they are non-trivial under isotopy. Link homotopy
2
9 in some sense measures the linking between different components of a link. So the question reduces
2 to whether two link homotopic links are also isotopic. We have the example of the Whitehead link,
5
which is a non-trivial link that becomes trivial under link homotopy. This gives us an example of
0
. a 2-component link that is link homotopic, but not isotopic, to the unlink, answering our question
1
for links with two components.
0
6 In this article, we provide examples of links with more than 2 components. In particular, we
1 prove the following.
:
v
i MainTheorem. Foreverym ≥ 3thereisanm-componentlink, L , withthefollowingproperties:
X m
r (1) Lm is link homotopic to the unlink
a
(2) L is not link isotopic to the unlink
m
(3) Every proper sublink of L is link isotopic to the unlink.
m
Moreover, examples exist with the additional property:
(4) All Milnor’s µ-invariants, µ(i ...i ), are 0.
1 s
In a subsequent paper, these links will be used to construct a family of new examples of 4-
manifolds which have singular metrics with nonpositive curvature, but do not support a smooth
Riemannian structure with nonpositive sectional curvature.
Acknowledgements. I am grateful to my advisor, Jean-Fran¸cois Lafont, for posing the question
addressedinthispaperandforhissupportandguidancethroughtheprocess. Iwouldliketothank
J.-B. Meilhan for his inputs.
1
2
2. Jones polynomial
The Jones polynomial gives an invariant of an oriented link, that assigns to each link a Laurent
polynomial in the variable t1/2 with integer coefficients. If two links are link isotopic then they
have the same Jones polynomial.
The Jones polynomial is a function
V : {Oriented links in S3} → Z[t−1/2,t1/2]
with the following properties.
(1) V(unknot) = 1.
(2) V(L∪(cid:13)) = (−t−1/2−t1/2)V(L).
(3) V satisfies the Skein relation: whenever three oriented links L ,L and L are the same,
+ − 0
except in the neighbourhood of a point where they are as shown in Figure 1, then
t−1V(L )−tV(L )+(t−1/2−t1/2)V(L ) = 0.
+ − 0
Figure 1. Skein relation
The Jones polynomial is discussed in detail in the book [2].
3. Whitehead link
The Whitehead link is given in Figure 2. It can be seen to be link homotopic to the unlink by
making a crossing change at the central crossing, A. However, its Jones polynomial is computed to
be t−3/2(−1+t−2t2+t3−2t4+t5), while the Jones polynomial of the unlink is (−t−1/2−t1/2).
Figure 2. Whitehead link
This shows that the Whitehead link is not link isotopic to the unlink. Moreover, removing any
component from the Whitehead link leaves us with the unknot. Thus, it satisfies the conclusion of
the Main Theorem for m = 2.
4. New Examples
A Brunnian link is a non-trivial link such that each sublink is link isotopic to an unlink. An
example of a Brunnian link with 3 components are the Borromean rings (Figure 3). For m ≥ 3,
a family of Brunnian links, B , was given by Milnor [4] as shown in Figure 4. Notice that these
m
links satisfy properties (2) and (3) of our theorem, but not property (1).
3
Figure 3. Borromean rings
Figure 4. Brunnian link with m components, B
m
Definition (Whitehead double). Let K be a component of a link L in S3, regarded as h({0}×S1)
for some embedding h : D2×S1 → S3\(L\K), such that K and h((0,1)×S1) have linking number
zero. Let n be a non-zero integer. Consider in the solid torus T = D2 ×S1 the knot W with n
n
crossings, as depicted in Figure 5. The knot h(W ) is called the Whitehead n-double of K, and is
n
denoted by W (K).
n
(a) n negative twists (b) n positive twists
Figure 5. Whitehead n twists, W
n
If the components of the link L are numbered, L = K ∪...∪K , then the link obtained by
1 m
considering the Whitehead n-double on the i-th component of L will be denoted by Wi(L). We
n
are now ready to prove our Main Theorem.
Case m = 3: We first consider the case of 3-component links. Let B be the Brunnian link with
3
3 components, with its components numbered as shown in Figure 6. Consider the Whitehead n-
double (with negative twists) of the third component of B to obtain the new link, W3(B ). To
3 n 3
simplify notation, we will denote this by W (B ) hereafter (Figure 7).
n 3
Observe that this link is link homotopic to the unlink. Even a single crossing change in W (B )
n 3
unwindsthelinktogiveanunlink. NotethatW (B )satisfiesproperties(1)and(3)ofourtheorem.
n 3
4
Figure 6. Brunnian link with 3 components, B
3
Figure 7. Whitehead double of B
3
We claim that there is an integer n > 0 for which W (B ) is not link isotopic to the unlink,
n 3
that is, it satisfies property (2). In fact, such an n occurs in every pair of consecutive odd, or even,
integers.
We show this by showing that the Jones polynomial of W (B ) for some n is not the same as
n 3
that of the unlink. Let us first consider the case when n is even.
Consider the skein relation at the crossing indicated in Figure 8.
Figure 8
t−1V(W (B ))−tV(W (B ))+(t−1/2−t1/2)V(L ) = 0. (1)
n−2 3 n 3 3
If our claim is false, then both W (B ) and W (B ) must be link isotopic to the unlink and
n−2 3 n 3
must have the same Jones polynomial as the unlink with 3 components, namely, (−t−1/2−t1/2)2.
In that case, the above equation becomes
t−1(−t−1/2−t1/2)2−t(−t−1/2−t1/2)2+(t−1/2−t1/2)V(L ) = 0. (2)
3
Therefore,
(t−1/2−t1/2)V(L ) = (t−t−1)(−t−1/2−t1/2)2. (3)
3
5
That is,
V(L ) = (−t−1/2−t1/2)3. (4)
3
NowwecomputetheJonespolynomialforL throughmultipleapplicationsoftheskeinrelations.
3
5. Computations
Consider the link L and focus on the part of the link indicated in Figure 9.
3
Figure 9
We apply the skein relation at the crossings indicated to get the following equations.
t−1V(L ) = tV(L1)−(t−1/2−t1/2)V(L2)
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tV(L1) = t−1V(L3)+(t−1/2−t1/2)V(L4)
tV(L2) = t−1V(L5)+(t−1/2−t1/2)V(L6)
6
Under isotopy the links L3,L4,L5 and L6 can be drawn as shown in Figure 10.
Figure 10
Now define the link A as shown.
Notice that the links L3,L4,L5 and L6 can be written, up to isotopy, using the links A and the
Brunnian link, B . Indeed, we have L3 = B ∪(cid:13), L4 = B , L5 = B , L6 = A∪(cid:13).
3 3 3 3
SonowourcomputationisreducedtofindingtheJonespolynomialforB andA. Afterapplying
3
further Skein relations, we can calculate:
V(B ) = −t3+3t2−2t+4−2t−1+3t−2−t−3
3
V(A) = t2+2+t−2.
Using the Skein relations above, this allows us the calculate:
V(L ) = t9/2−2t7/2+t5/2−2t3/2−2t1/2−2t−1/2−2t−3/2+t−5/2−2t−7/2+t−9/2. (5)
3
However, this is different from the expression we obtained in (4) for V(L ), proving our assumption
3
to be incorrect. Hence, both W and W cannot be unlinks. This shows that at least one of any
n n−2
pair of consecutive even numbers gives a link with properties as required in the Main Theorem.
We have thus proved our Main Theorem for links with m = 3 components.
If n is odd, the orientation on the last component of L changes. However, this does not change
3
our final expression for V(L ), giving us the same result for any pair of consecutive odd integers.
3
Case m > 3: Now consider the case when the number of components is greater than 3. Consider
them-componentBrunnianlink, B (m > 3), andtheWhiteheadn-doubleofthem-thcomponent
m
to obtain the link W (B ). Using the Skein relations, similar to the ones in Figure 8, we obtain
n m
the following link L (see Figure 11).
m
If W (B ) and W (B ) were both isotopic to the unlink, then computations similar to the
n m n−2 m
equations (1)−(4) show that the Jones polynomial for L will be
m
V(L ) = (−t−1/2−t1/2)m. (6)
m
On the other hand, we compute the Jones polynomial of L by application of Skein relations at
m
the crossing (A) and then at (B). This reduces the link L to the links L1 ,...,L6 , where
m m m
L3 = B ∪(cid:13), L4 = B , L5 = B and L6 = L ∪(cid:13).
m m m m m m m m−1
7
Figure 11
Then similar computations give us the following recurrence relation for the Jones polynomial for
L , for every m > 3:
m
V(L ) = V(L )(t5/2−t3/2−t−3/2+t−5/2)+(−t1/2−t−1/2)m(t2−2t+3−2t−1+t−2) (7)
m m−1
We put z = t1/2 to turn V(L ) into a Laurent polynomial in z with integer powers.
m
V(L ) = V(L )(z5−z3−z−3+z−5)+(−z−z−1)m(z4−2z2+3−2z−2+z−4) (8)
m m−1
Let deg(V(L )) denote the degree of V(L ) in z. We claim that deg(V(L )) > m for every
m m m
m ≥ 3. From equation (5), we know that deg(V(L )) = 9 > 3, hence the result is true for m = 3.
3
Assume that this is true for m−1. From the relation (8) we see that the highest power of z
in V(L ) comes from one of the terms, z5V(L ) and (−z − 1)mz4. By induction hypothesis,
m m−1 z
we know that, deg(V(L )) > m − 1. Hence, deg(z5V(L )) > 5 + m − 1 = m + 4. This
m−1 m−1
shows that degree of V(L ) must be the degree of the term z5V(L ), that is, deg(V(L )) =
m m−1 m
deg(V(L ))+5. Also, deg(V(L )) > m+4 > m, which proves our claim.
m−1 m
It also gives us a recurrence relation for the degree of V(L ), namely,
m
deg(V(L )) = deg(V(L ))+5.
m m−1
This in turn gives us that, for m ≥ 3, deg(V(L )) = 5(m−3)+9.
m
This shows that
degree of V(L ) > degree of (−z− 1)m, for m ≥ 3,
m z
showingthattheexpressionobtainedin(6)mustbedifferentfromtheoneintherecurrencerelation
(7).
Thusoneofthelinks,W (B )orW (B ),mustbenon-isotopictotheunlink. Thiscompletes
n m n−2 m
the proof for the Main Theorem for m ≥ 3.
Remark 1. Consider the Seifert matrix of the link W−(B ), the Whitehead double of B with n
n 3 3
negative twists, where n ≥ 4 is even.
 0 
.
.
 S . 
 
 0 
 
Sn =  0 
 
 1 
 
 −1 
0···0 −1 0 −1 n−4
2
where S is nothing but the Seifert matrix for the link L from Figure 9.
3
8
On looking at the form of the matrix, one is tempted to compare the signatures of the two links.
To do this, consider the symmetric bilinear form given by A = S +ST. The signature of the link
n n n
W−(B ), σ(W−(B )), is by definition, the signature of A .
n 3 n 3 n
 0 
.
.
 A . 
 
 0 
 
An =  −1 ,
 
 1 
 
 −2 
0···0 −1 1 −2 n−4
where A gives the bilinear form corresponding to the link L . (Detailed discussion on Seifert form
3
and signature of links can be found in [2], Chapters 6 and 8.)
Suppose A defines a bilinear form on the vector space V, and A on V ⊕ R. Let w be the
n n
eigenvector for the largest eigenvalue, λ , of A , and H be the subspace orthogonal to w . The
n n n n
form of the matrix A allows us to conclude that w → (0,...,0,1) and λ → ∞, as n → ∞. This
n n n
in turn gives us that subspace H approaches V as n → ∞. Hence, the form A | approaches
n n Hn
A | . However, A | is same as the form A.
n V n V
This shows that the eigenvalues of A | must also approach those of A. Thus, as long as none
n Hn
of the eigenvalues of A are zero, for a large enough value of n, we would obtain σ(A | ) = σ(A),
n Hn
which would force σ(A ) = σ(A)+1. Alternatively, if W+(B ) is a link having n positive twists
n n 3
with n ≥ 4 being even, then a similar analysis shows that for large n, σ(A ) = σ(A)−1.
n
Thus, for large enough n even, σ(W+(B )) = σ(L )−1 and σ(W−(B )) = σ(L )+1. Hence at
n 3 3 n 3 3
least one of the links, W+(B ) or W−(B ), must have signature different from that of the unlink,
n 3 n 3
proving the existence of a link as in the Main Theorem for m = 3. The same method would work
for links W (B ) for m > 3 as well.
n m
Unfortunately, the following observation shows that det(A) = 0, so A will have zero as an
eigenvalue, showing that the above method will not work for these examples of links.
A coloring (modulo n) of a link L is a function c from the set of arcs in its link diagram to Z/nZ
if at every crossing we have 2c(x ) ≡ c(x )+c(x ) mod n, where x ,x ,x are arcs at a crossing as
i j k i j k
shown.
Figure 12
A system of these equations gives us the coloring matrix. Deleting any one column and one row
yields a new matrix. We know that det(A) is equal to the determinant of this new matrix for link
L . And this determinant is 0 if and only if the link has a coloring mod n for every n ∈ Z, and in
m
particular, if it has a coloring over Z ([1], Chapter 3).
For every m ≥ 3, we have the following coloring of L over Z, which forces det(A) = 0.
m
9
Figure 13
Remark 2. As per the comments by an anonymous referee, one can use 3-manifold theory to
show that the Whitehead double of (any non-split) link is isotopically non-trivial, and in fact, non-
split. The complement of a Brunnian link in S3 gives an irreducible 3-manifold with incompressible
boundary, and so does the complement of the Whitehead n-twists in the solid torus.
Let W be the Whitehead n twists in the solid torus D2×S1 (Figure 5). Let M := S3\N(B )
n 1 m
and M := (D2 × S1) \ N(W ), where N(B ) and N(W ) denote the tubular neighbourhoods.
2 n m n
Let T2 be the torus boundary of the tubular neighborhood of the m-th component of B , N(Bm).
m m
Then we have S3\N(W (B )) = M ∪ M . Since M and M are irreducible manifolds, gluing
n m 1 T2 2 1 2
them along their incompressible boundary gives us another irreducible manifold. This shows that
W (B ) cannot be a trivial link, since otherwise we get a 2-sphere that separates S3\N(W (B )).
n m n m
To see that gluing M and M along their incompressible boundary gives us another irreducible
1 2
manifold, say M, let us first assume that M is not irreducible. Then there must be a copy of
α : S2 (cid:44)→ M which separates M. Also, α(S2) must intersect T2 , since otherwise it would lie
completely in either M or M , which are irreducible. We may assume that α is transverse to T2.
1 2
Then α(S2)∩T2 is a codimension 1 submanifold in T2. Then α−1(T2) (cid:44)→ S2 also has codimension
1, and hence must be a union of copies of S1 in S2.
ConsideracopyofS1, sayS1, inα(S2)∩T2 thatboundsadiskinM . SinceT2 isincompressible
β 1
in M, it must bound a disk in T2. This gives a copy of S2 in M . Being the complement of the
1
link B , M is aspherical. Hence the 2-sphere must bound a 2-ball in M . Then the map α can be
m 1 1
perturbed so that α(S2)∩T2 no longer contains S1. Doing this for all the copies of S1 in S2∩T2
β
that bound a disk in M will eventually show that there is a copy of S2 in M that separates M
1 2 2
which contradicts the irreducibility of M .
2
Remark 3. J.-B. Meilhan pointed out that a result of Meilhan-Yasuhara (Theorem 1, [3]) can
also be used to show that W (B ) satisfy (2) for every n > 0. The result shows that these links
n m
havenon-zeroMilnorisotopyinvariants[5], showingthattheyarelinkisotopicallynon-trivial. And
hence they also fail to satisfy property (4) of the Main Theorem. In Section 7, we give examples
satisfying properties (1)−(4).
6. Non-isotopic links
Inthissectionwepointoutthatthemethoddescribedintheprevioussectionsisusefultofurther
show that each of the links, W (B ), are in fact non-isotopic links. To see this, observe that one
n 3
can obtain the Jones polynomial of W (B ) as a function of W (B ), using the Skein relation in
n 3 n−2 3
equation (1).
Define Φ : Z[t−1/2,t1/2] → Z[t−1/2,t1/2], given by Φ(p) = t−2p+t−1(t−1/2 −t1/2)V(L ), where
3
V(L ) is the Jones polynomial of the link L as given in equation (5).
3 3
10
Then V(W (B )) = Φ(V(W (B ))), for every n ≥ 2. Hence we only need to analyse the
n 3 n−2 3
function Φ in order to understand the Jones polynomials of the Whitehead doubles.
One can see that the degree of Φ(p) depends on that of the polynomial p. Let us denote
by deg (p) the degree of the lowest degree term of p. Then, deg (Φ(p)) = min {deg (p) − 2,
L L L
deg (V(L ))−3/2}. We know from our previous computations that deg (V(L )) = −9/2. Hence,
L 3 L 3
deg (Φ(p)) = min {deg (p)−2, −6}.
L L
For n = 2, we have that V(W (B )) = Φ(V(W (B ))). Since W (B ) is nothing but the unlink
2 3 0 3 0 3
with 3 components, V(W (B )) = (−t−1/2 − t1/2)2. Hence deg (V(W (B ))) = −1, and hence
0 3 L 0 3
deg (V(W (B ))) = deg (Φ(V(W (B )))) = −6. Continuing in a similar fashion, we get that
L 2 3 L 0 3
deg (V(W (B ))) = −8, deg (V(W (B ))) = −10 and so on.
L 4 3 L 6 3
Similarly,computationsshowthatV(W (B )) = −t4+2t3−t2+2t+1+t−2−t−3+2t−4−t−5,which
1 3
in turn gives us that deg (V(W (B ))) = deg (Φ(V(W (B )))) = −7, deg (V(W (B ))) = −9 and
L 3 3 L 1 3 L 5 3
so on.
We can generalize this to obtain that deg (V(W (B ))) = −n−4 for n ≥ 1. This shows that
L n 3
each of these links are neither isotopic to the unlink, nor to each other.
Further computations show that for links with more than 3 components,
5−5m
deg (V(W (B ))) = −(n−1) for n ≥ 1 and m > 3.
L n m
2
This shows that the links are all non-isotopic links.
7. Milnor invariants
Let L be an m-component link in S3. Let F(L) be the fundamental group, π (S3 − L), and
1
F (L) be its q-th lower central subgroup. We now define Milnor’s µ-invariants [5].
q
We have a presentation of the group F(L)/F (L) with m generators, α ,...,α , where α
q 1 m i
represents the i-th meridian of the link L. For each i = 1,...,m, the longitude l of the i-th
i
component of L is represented by a word in α ,...,α .
1 m
On substituting α = 1 + κ , and α−1 = 1 − κ + κ2 − κ3 + ... in the expression for l , we
j j j j j j i
can represent l by a formal power series in κ ,...,κ . Let µ(j ...j i) denote the coefficient of
i 1 m 1 s
κ ...κ in this series, for every s ≥ 1.
j1 js
Let ∆(i ...i ) denote the greatest common divisor of µ(j ...j ), where j ...j (2 ≤ s < r)
1 r 1 s 1 s
ranges over all sequences obtained by cancelling at least one of the indices i ...i , and permuting
1 r
the remaining indices cyclically. And now µ(i ...i ) is defined as the residue class of µ(i ...i )
1 r 1 r
modulo ∆(i ...i ). These are Milnor’s isotopy invariants of links.
1 r
Consider the link obtained from the Brunnian link B by taking the Whitehead double twice,
3
on the first and last component (Figure 14). We denote this link by W1,3 (B ), where k and n are
n,m 3
the number of twists on the first and third components respectively.
Figure 14. W1,3(B )
k,n 3