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Milnor mu-bar invariants

Contents

Idea

In Link Groups , John Milnor introduced the notion of the Link Group? as a way to study links. The notion of equivalence of links that Milnor used is slightly different to that obtained by extending the usual notion of equivalence of knots. In Milnor’s paper, the crucial aspect of links was the interactions between distinct components. Thus for Milnor, a link in a manifold MM is a map nS 1M\coprod_n S^1 \to M such that the components have disjoint images. Similarly, two links are homotopic if there is a homotopy between the maps which is a link at every time. Thus links can be deformed in such a manner that individual components can pass through themselves, but not through other components. Also link components can have self-intersections or the map on a component can be a constant map. Milnor uses the term proper link to refer to a link in which the map is a homeomorphism onto its image.

The Whitehead link is a simple example of a link that is not trivial under ambient isotopy but is trivial under Milnor’s notion of homotopy.

The μ\mu-invariants come from explicit descriptions of the link groups of particular links. Specifically, Milnor calls a link almost trivial if every proper sublink is trivial (see Brunnian link). Such a link corresponds to an element in a particular link group which can be completely described by certain numbers.

Let us begin by describing the link group. Milnor’s alternative description is as follows. Consider the complement of a link LL in an open 33-manifold MM. We choose a basepoint in this complement and so have the fundamental group. We define a relation on this group as follows: two loops α\alpha, β\beta are equivalent if the link Lα 1βL \cup \alpha^{-1} \beta is homotopic in MM to one of the form L1L' \cup 1 (where 11 is the constant loop at the basepoint). The link group is the group of equivalence classes of such loops.

A more practical description is the following.

Definition

Let LL be a link in an open 33-manifold MM. Let G(L)G(L) be the fundamental group of the complement of LL. Let L iL^i denote the sublink obtained by deleting the iith component of LL. Let A i(L)A_i(L) be the kernel of the natural inclusion G(L)G(L i)G(L) \to G(L^i) and [A i][A_i] its commutator subgroup. Let E(L)=[A 1][A 2][A n]E(L) = [A_1] [A_2] \cdots [A_n]. This is a normal subgroup of G(L)G(L). The quotient, 𝒢(L)G(L)/E(L)\mathcal{G}(L) \coloneqq G(L)/E(L) is the link group of LL.

Milnor’s first theorem on this group was to show that this group is an invariant of the homotopy class of the link, at least for proper links.

Theorem (Milnor, Theorem 1)

If two proper links are homotopic, then their link groups are isomorphic.

To study this group for a particular link, we need to find some particular elements in it. These are the meridians and the parallels. Basically, a meridian goes around one component of the link once, in a specific direction, whilst a parallel goes along it. Technically, the parallels of a link are not elements of its link group, but cosets.

Choose a component L iL_i of the link LL. Choose orientations of the ambient manifold, MM, and of the circle. To define the meridian and parallel of L iL_i we need to choose a path from the basepoint, x 0x_0, to a point on the image of L iL_i which does not intersect the image of LL at any other time. Let pp be such a path, so then p(1)p(1) is a point on the image of L iL_i.

Definition

The iith meridian of LL is the element α i𝒢(L)\alpha_i \in \mathcal{G}(L) defined as follows. Choose a small neighbourhood NN of p(1)p(1). Define a path by going along pp until we are inside NN, then go around a closed loop in NN which has linking number +1+1 with the part of the image of L iL_i inside NN. Then return to x 0x_0 along pp.

The iith parallel of LL is the coset β i𝒜 i𝒢(L i)\beta_i \mathcal{A}_i \in \mathcal{G}(L^i) defined as follows. The subgroup 𝒜 i\mathcal{A}_i is the kernel of the homomorphism 𝒢(L)𝒢(L i)\mathcal{G}(L) \to \mathcal{G}(L^i). Go along pp to its end. Then go around the image of L iL_i according to the orientation of the circle. Finally return to x 0x_0 along pp. The preimage of this element defines a coset in 𝒢(L)\mathcal{G}(L) which we write as β i𝒜 i\beta_i \mathcal{A}_i.

The basic method of studying a link via link groups is to consider a link as an element of the link group of the link obtained by removing one of its components. To show that this is a reasonable thing to do, Milnor proved the following theorem.

Theorem (Milnor, Theorem 3)

Let LL be a proper link with nn components. Let ff, ff' be closed loops in the complement of LL. If they represent conjugate elements of 𝒢(L)\mathcal{G}(L) then the links (L,f)(L,f) and (L,f)(L,f') are homotopic.

μ\mu-Invariants

For Brunnian links, which Milnor calls almost trivial links, the classification question reduces to looking at elements of the link group of trivial links. It is important to note that the ambient space here is Euclidean space, 3\mathbb{R}^3.

Let LL be an nn-component Brunnian link. Then we consider the element β n𝒢(L n)\beta'_n \in \mathcal{G}(L^n) corresponding to the nnth parallel. Upon removing a further component, say the n1n-1st, this element becomes trivial since we are then looking at L n1L^{n-1} which is trivial. Thue β n𝒜 n1(L n)\beta'_n \in \mathcal{A}_{n-1}(L^n), the kernel of 𝒢(L n)𝒢(L n1,n)\mathcal{G}(L^n) \to \mathcal{G}(L^{n-1,n}) (here L n1,nL^{n-1,n} is LL with both the n1n-1st and nnth components removed). Now 𝒜 n1(L n)\mathcal{A}_{n-1}(L^n) is the smallest normal subgroup containing the meridian α n1\alpha_{n-1} (since removing the n1n-1st component is the same thing as allowing the meridian α n1\alpha_{n-1} to collapse) and so every element of 𝒜 n1\mathcal{A}_{n-1} can be written as a word in alphabet of powers and conjugates of α n1\alpha_{n-1}. Milnor uses the notation

α n1 σ,σ(L n1,n) \alpha_{n-1}^\sigma, \quad \sigma \in \mathcal{R}(L^{n-1,n})

to write this. The exponent σ\sigma itself decomposes as

(1)σ=μ(i 1i n2,n1n)k i 1k i n2 \sigma = \sum \mu(i_1 \cdots i_{n-2}, n-1 n) k_{i_1} \cdots k_{i_{n-2}}

where the summation is over all permutations i 1i n2i_1 \cdots i_{n-2} of 11, 22, …, n2n-2.

Theorem (Milnor, Section 5)

The integers μ(i 1i n2,n 1n)\mu(i_1 \cdots i_{n-2}, n_1 n) are homotopy invariants of LL. The homotopy class of LL is completely specified by these integers.

There was nothing special about the choice of components. A similar procedure works for any pair of components. The resulting integers obey the following rules:

(2)μ(i 1i 2i n2,i n1i n) =μ(i ni 1i 2i n3,i n2i n2) μ(i 1i νrj 1j nν2s) =(1) nνμ(h 1h n2rs) \begin{aligned} \mu(i_1 i_2 \cdots i_{n-2}, i_{n-1} i_n) &= \mu(i_n i_1 i_2 \cdots i_{n-3}, i_{n-2} i_{n-2}) \\ \mu(i_1 \cdots i_{\nu} r j_1 \cdots j_{n-\nu-2} s) &= (-1)^{n-\nu} \sum \mu(h_1 \cdots h_{n-2} r s) \end{aligned}

In the second identity, the summation is over all shuffle products of (i 1i ν)(i_1 \cdots i_{\nu}) with (j 1j nν2)(j_1 \cdots j_{n - \nu - 2}).

Let us expand on the definition of the μ\mu-invariants. We start with the exponential notation. The following holds for an arbitrary proper link, LL, embedded in an open 33-manifold MM.

Let J𝒢(L)J \mathcal{G}(L) be the integral group ring of 𝒢(L)\mathcal{G}(L). As mentioned above, any element of 𝒜 i(L)\mathcal{A}_i(L) is a product of powers of conjugates of α i\alpha_i. We can write such an element in the form α i s\alpha_i^s for sJ𝒢(L)s \in J \mathcal{G}(L) by interpreting:

α i x+y =α i xα i y α i kx =(α i x) k α i β =βαβ 1 \begin{aligned} \alpha_i^{x + y} &= \alpha_i^x \alpha_i^y \\ \alpha_i^{k x} &= (\alpha_i^x)^k \\ \alpha_i^\beta &= \beta \alpha \beta^{-1} \end{aligned}

where x,yJ𝒢(L)x, y \in J\mathcal{G}(L), kk \in \mathbb{Z}, and β𝒢(L)\beta \in \mathcal{G}(L).

We write the kernel of J𝒢(L)J𝒢(L i)J \mathcal{G}(L) \to J \mathcal{G}(L^i) as 𝒦 i(L)\mathcal{K}_i(L). Using these, we define:

(L)J𝒢(L)/𝒦 1(L) 2++𝒦 n(L) 2 \mathcal{R}(L) \coloneqq J \mathcal{G}(L) / \mathcal{K}_1(L)^2 + \cdots + \mathcal{K}_n(L)^2

Now the notation α i s\alpha_i^s for an element of 𝒜 i(L)\mathcal{A}_i(L) does not provide an injective map from J𝒢(L)J\mathcal{G}(L) to 𝒜 i(L)\mathcal{A}_i(L). The kernel is the ideal 𝒦 i(L)+(𝒦 1(L) 2++𝒦 n(L) 2)\mathcal{K}_i(L) + (\mathcal{K}_1(L)^2 + \cdots + \mathcal{K}_n(L)^2) which is naturally isomorphic to (L i)\mathcal{R}(L^i).

Now let us suppose that LL is trivial. Then G(L)G(L) is the free product of the fundamental group of MM with the infinite cyclic groups generated by the (elements representing the) meridians of LL. Let these be a 1a_1, …, a na_n and let k i=a i1k_i = a_i - 1 in JG(L)J G(L). Milnor defines a canonical word to be a product of the form ϕ 0k j 1ϕ 1k j 2ϕ 2k j pϕ p\phi_0 k_{j_1} \phi_1 k_{j_2} \phi_2 \cdots k_{j_p} \phi_p with p0p \ge 0, ϕ iπ 1(M)\phi_i \in \pi_1(M), and 1j in1 \le j_i \le n. A canonical sentence is a sum or difference of any number of canonical words. It turns out (Milnor, Theorem 7) that each element of (L)\mathcal{R}(L) is represented by a unique canonical sentence.

Now let us return to the case of the almost trivial link in Euclidean space. From above, we have the element β n𝒜 i(L n)\beta'_n \in \mathcal{A}_i(L^n) corresponding to the nnth parallel. Removing any other component allows us to trivialise β n\beta'_n since removing, say, the iith component leaves us with L iL^i which is homotopic to the trivial link on n1n-1 components. Removing the ii component corresponds to setting a ia_i to 11 in JG(L n)J G(L^n), equivalently to setting k i=0k_i = 0. So upon setting k i=0k_i = 0 we must have that β n1\beta'_n \mapsto 1 and thus (by uniqueness) σ0\sigma \mapsto 0. Hence k ik_i divides σ\sigma, and so every canonical word in σ\sigma is of the form k i 1k i n2k_{i_1} \cdots k_{i_{n-2}} for some permutation of 11, 22, …, n2n-2. Sorting them out by permutation, we get the expression in (1).

Now, how do we interpret or calculate these invariants? We need to work out what an expression of the form in (1) is saying. Consider a canonical word, k i 1k i n2k_{i_1} \cdots k_{i_{n-2}}. The corresponding element is:

α n1 k i 1k i n2 \alpha_{n-1}^{k_{i_1} \cdots k_{i_{n-2}}}

Let us write α=α n1\alpha = \alpha_{n-1}. Now α k 1\alpha^{k_1} is α a 11=a 1αa 1 1α 1\alpha^{a_1 - 1} = a_1 \alpha a_1^{-1} \alpha^{-1}. Thus this tells us to go around L 1L_1, then L n1L_{n-1}, back around L 1L_1, and finally back around L n1L_{n-1}. Each time we introduce a new power, we do the same except that we replace the loop around L n1L_{n-1} with the loop so far constructed.

So the general method is as follows: choose two components of the link. Write one of them as a word in the meridians of the others. Then simplify this word using the other chosen link as the “base”: namely, write everything in terms of conjugates of that base. This will then separate out into the desired form and, hopefully, the link invariants can be read off.

References

  • John Milnor (1954). Link groups. Ann. of Math. (2), 59, 177–195. MR
Revised on March 1, 2013 17:54:15 by Andrew Stacey (92.21.167.146)