nLab
Milnor mu-bar invariants

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 $M$ is a map ${\coprod }_n{S}^1\to M$ 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 $\mu$-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.

Link Group

Let us begin by describing the link group. Milnor’s alternative description is as follows. Consider the complement of a link $L$ in an open $3$-manifold $M$. We choose a basepoint in this complement and so have the fundamental group. We define a relation on this group as follows: two loops $\alpha$, $\beta$ are equivalent if the link $L\cup {\alpha }^{-1}\beta$ is homotopic in $M$ to one of the form $L\prime \cup 1$ (where $1$ 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 $L$ be a link in an open $3$-manifold $M$. Let $G(L)$ be the fundamental group of the complement of $L$. Let $L^i$ denote the sublink obtained by deleting the $i$th component of $L$. Let $A_i(L)$ be the kernel of the natural inclusion $G(L)\to G(L^i)$ and $[A_i]$ its commutator subgroup. Let $E(L)=[A_1][A_2]\cdots [A_n]$. This is a normal subgroup of $G(L)$. The quotient, $𝒢(L)≔G(L)/E(L)$ is the link group of $L$.

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

Definition

The $i$th meridian of $L$ is the element ${\alpha }_i\in 𝒢(L)$ defined as follows. Choose a small neighbourhood $N$ of $p(1)$. Define a path by going along $p$ until we are inside $N$, then go around a closed loop in $N$ which has linking number $+1$ with the part of the image of $L_i$ inside $N$. Then return to $x_0$ along $p$.

The $i$th parallel of $L$ is the coset ${\beta }_i{𝒜}_i\in 𝒢(L^i)$ defined as follows. The subgroup ${𝒜}_i$ is the kernel of the homomorphism $𝒢(L)\to 𝒢(L^i)$. Go along $p$ to its end. Then go around the image of $L_i$ according to the orientation of the circle. Finally return to $x_0$ along $p$. The preimage of this element defines a coset in $𝒢(L)$ which we write as ${\beta }_i{𝒜}_i$.

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 $L$ be a proper link with $n$ components. Let $f$, $f\prime$ be closed loops in the complement of $L$. If they represent conjugate elements of $𝒢(L)$ then the links $(L,f)$ and $(L,f\prime )$ are homotopic.

$\mu$-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, ${ℝ}^3$.

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

$${\alpha }_{n-1}^{\sigma },\sigma \in ℛ(L^{n-1,n})$$\alpha_{n-1}^\sigma, \quad \sigma \in \mathcal{R}(L^{n-1,n})

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

(1)$$\sigma =\sum \mu (i_1\cdots i_{n-2},n-1n)k_{i_1}\cdots k_{i_{n-2}}$$\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 $i_1\cdots i_{n-2}$ of $1$, $2$, …, $n-2$.

Theorem (Milnor, Section 5)

The integers $\mu (i_1\cdots i_{n-2},n_{1}n)$ are homotopy invariants of $L$. The homotopy class of $L$ 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)$$\begin{array}{rl}\mu (i_1{i}_2\cdots i_{n-2},i_{n-1}{i}_n)& =\mu (i_n{i}_1{i}_2\cdots i_{n-3},i_{n-2}{i}_{n-2})\\ \mu (i_1\cdots i_{\nu }r{j}_1\cdots j_{n-\nu -2}s)& =(-1{)}^{n-\nu }\sum \mu (h_1\cdots h_{n-2}rs)\end{array}$$\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 $(i_1\cdots i_{\nu })$ with $(j_1\cdots j_{n-\nu -2})$.

Let us expand on the definition of the $\mu$-invariants. We start with the exponential notation. The following holds for an arbitrary proper link, $L$, embedded in an open $3$-manifold $M$.

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

$$\begin{array}{rl}{\alpha }_i^{x+y}& ={\alpha }_i^x{\alpha }_i^y\\ {\alpha }_i^{kx}& =({\alpha }_i^x{)}^k\\ {\alpha }_i^{\beta }& =\beta \alpha {\beta }^{-1}\end{array}$$\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 $x,y\in J𝒢(L)$, $k\in \mathbb{Z}$, and $\beta \in 𝒢(L)$.

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

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

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

Theorem (Milnor, Theorem 5)

Let $L$ be a link which is homotopic to one in with the $i$th component is constant. Then every element of ${𝒜}_i(L)$ can be expressed uniquely in the form ${\alpha }_i^{\sigma }$ with $\sigma \in ℛ(L^i)$.

Now let us suppose that $L$ is trivial. Then $G(L)$ is the free product of the fundamental group of $M$ with the infinite cyclic groups generated by the (elements representing the) meridians of $L$. Let these be $a_1$, …, $a_n$ and let $k_i=a_i-1$ in $JG(L)$. Milnor defines a canonical word to be a product of the form ${\varphi }_0{k}_{j_1}{\varphi }_1{k}_{j_2}{\varphi }_2\cdots k_{j_p}{\varphi }_p$ with $p\ge 0$, ${\varphi }_i\in {\pi }_1(M)$, and $1\le j_i\le n$. A canonical sentence is a sum or difference of any number of canonical words. It turns out (Milnor, Theorem 7) that each element of $ℛ(L)$ 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 $\beta {\prime }_n\in {𝒜}_i(L^n)$ corresponding to the $n$th parallel. Removing any other component allows us to trivialise $\beta {\prime }_n$ since removing, say, the $i$th component leaves us with $L^i$ which is homotopic to the trivial link on $n-1$ components. Removing the $i$ component corresponds to setting $a_i$ to $1$ in $JG(L^n)$, equivalently to setting $k_i=0$. So upon setting $k_i=0$ we must have that $\beta {\prime }_n\mapsto 1$ and thus (by uniqueness) $\sigma \mapsto 0$. Hence $k_i$ divides $\sigma$, and so every canonical word in $\sigma$ is of the form $k_{i_1}\cdots k_{i_{n-2}}$ for some permutation of $1$, $2$, …, $n-2$. 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, $k_{i_1}\cdots k_{i_{n-2}}$. The corresponding element is:

$${\alpha }_{n-1}^{k_{i_1}\cdots k_{i_{n-2}}}$$\alpha_{n-1}^{k_{i_1} \cdots k_{i_{n-2}}}

Let us write $\alpha ={\alpha }_{n-1}$. Now ${\alpha }^{k_1}$ is ${\alpha }^{a_1-1}=a_1\alpha a_1^{-1}{\alpha }^{-1}$. Thus this tells us to go around $L_1$, then $L_{n-1}$, back around $L_1$, and finally back around $L_{n-1}$. Each time we introduce a new power, we do the same except that we replace the loop around $L_{n-1}$ 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