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\begin{document}

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\section*{Milnor mu-bar invariants}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{knot_theory}{}\paragraph*{{Knot theory}}\label{knot_theory}

[[!include knot theory - contents]]

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{link_group}{Link Group}\dotfill \pageref*{link_group} \linebreak
\noindent\hyperlink{invariants}{$\mu$-Invariants}\dotfill \pageref*{invariants} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

In \emph{\hyperlink{jmLinkGroups}{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 \textbf{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 \textbf{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 \emph{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.

\hypertarget{link_group}{}\subsection*{{Link Group}}\label{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 $L' \cup 1$ (where $1$ is the constant loop at the basepoint). The \textbf{link group} is the group of equivalence classes of such loops.

A more practical description is the following.

\begin{defn}
\label{linkgroup}\hypertarget{linkgroup}{}
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, $\mathcal{G}(L) \coloneqq G(L)/E(L)$ is the \textbf{link group} of $L$.

\end{defn}
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.

\begin{theorem}
\label{htyinv}\hypertarget{htyinv}{}
If two proper links are homotopic, then their link groups are isomorphic.

\end{theorem}
To study this group for a particular link, we need to find some particular elements in it. These are the \textbf{meridians} and the \textbf{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 \emph{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 $L_i$.

\begin{defn}
\label{meridian}\hypertarget{meridian}{}
The \textbf{$i$th meridian} of $L$ is the element $\alpha_i \in \mathcal{G}(L)$ 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 $p$.

\end{defn}
\begin{defn}
\label{parallel}\hypertarget{parallel}{}
\end{defn}
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.

\begin{theorem}
\label{conjelts}\hypertarget{conjelts}{}
Let $L$ be a proper link with $n$ components. Let $f$, $f'$ be closed loops in the complement of $L$. If they represent conjugate elements of $\mathcal{G}(L)$ then the links $(L,f)$ and $(L,f')$ are homotopic.

\end{theorem}
\hypertarget{invariants}{}\subsection*{{$\mu$-Invariants}}\label{invariants}

For [[Brunnian links]], which Milnor calls \textbf{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, $\mathbb{R}^3$.

Let $L$ be an $n$-component Brunnian link. Then we consider the element $\beta'_n \in \mathcal{G}(L^n)$ 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 $\beta'_n \in \mathcal{A}_{n-1}(L^n)$, the kernel of $\mathcal{G}(L^n) \to \mathcal{G}(L^{n-1,n})$ (here $L^{n-1,n}$ is $L$ with both the $n-1$st and $n$th components removed). Now $\mathcal{A}_{n-1}(L^n)$ 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 $\mathcal{A}_{n-1}$ can be written as a word in alphabet of powers and conjugates of $\alpha_{n-1}$. Milnor uses the notation

\begin{displaymath}
\alpha_{n-1}^\sigma, \quad \sigma \in \mathcal{R}(L^{n-1,n})
\end{displaymath}
to write this. The exponent $\sigma$ itself decomposes as

\begin{equation}
\sigma = \sum \mu(i_1 \cdots i_{n-2}, n-1 n) k_{i_1} \cdots k_{i_{n-2}}
\label{muinvs}\end{equation}
where the summation is over all permutations $i_1 \cdots i_{n-2}$ of $1$, $2$, \ldots{}, $n-2$.

\begin{theorem}
\label{muinvs}\hypertarget{muinvs}{}
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.

\end{theorem}
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:

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

\begin{displaymath}
\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}
\end{displaymath}
where $x, y \in J\mathcal{G}(L)$, $k \in \mathbb{Z}$, and $\beta \in \mathcal{G}(L)$.

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

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

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

\end{theorem}
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 $a_1$, \ldots{}, $a_n$ and let $k_i = a_i - 1$ in $J G(L)$. Milnor defines a \textbf{canonical word} to be a product of the form $\phi_0 k_{j_1} \phi_1 k_{j_2} \phi_2 \cdots k_{j_p} \phi_p$ with $p \ge 0$, $\phi_i \in \pi_1(M)$, and $1 \le j_i \le n$. A \textbf{canonical sentence} is a sum or difference of any number of canonical words. It turns out (\hyperlink{jmLinkGroups}{Milnor, Theorem 7}) that each element of $\mathcal{R}(L)$ 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 $\beta'_n \in \mathcal{A}_i(L^n)$ corresponding to the $n$th parallel. Removing any other component allows us to trivialise $\beta'_n$ 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 $J G(L^n)$, equivalently to setting $k_i = 0$. So upon setting $k_i = 0$ we must have that $\beta'_n \mapsto 1$ 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 $1$, $2$, \ldots{}, $n-2$. Sorting them out by permutation, we get the expression in \eqref{muinvs}.

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

\begin{displaymath}
\alpha_{n-1}^{k_{i_1} \cdots k_{i_{n-2}}}
\end{displaymath}
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.

\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item [[John Milnor]] (1954). Link groups. \emph{Ann. of Math. (2)}, \emph{59}, 177--195. \href{http://www.ams.org/mathscinet-getitem?mr=71020}{MR}

\end{itemize}


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