In knot theory, an area of mathematics, the link group of a link is an analog of the knot group of a knot. They were described by John Milnor in his Bachelor's thesis, Template:Harv.

## Definition

It is not the fundamental group of the link complement, since the components of the link are allowed to move through themselves, though not each other, but thus is a quotient group of the link complement, since one can start with this, and then by knotting or unknotting components, some of these elements may become equivalent to each other.

## Examples

The link group of the n-component unlink is the free group on n generators, $F_{n}$ , as the link group of a single link is the knot group of the unknot, which is the integers, and the link group of an unlinked union is the free product of the link groups of the components.

The link group of the Hopf link, the simplest non-trivial link – two circles, linked once – is the free abelian group on two generators, ${\mathbf {Z} }^{2}.$ Note that the link group of two unlinked circles is the free nonabelian group on two generators, of which the free abelian group on two generators is a quotient. In this case the link group is the fundamental group of the link complement, as the link complement deformation retracts onto a torus.

## Milnor invariants

Milnor defined invariants of a link (functions on the link group) in Template:Harv, using the character ${\bar {\mu }},$ which have thus come to be called "Milnor's μ-bar invariants", or simply the "Milnor invariants". For each k, there is an k-ary function ${\bar {\mu }},$ which defines invariants according to which k of the links one selects, in which order.

Milnor's invariants can be related to Massey products on the link complement (the complement of the link); this was suggested in Template:Harv, and made precise in Template:Harv and Template:Harv.

As with Massey products, the Milnor invariants of length k + 1 are defined if all Milnor invariants of length less than or equal to k vanish. The first (2-fold) Milnor invariant is simply the linking number (just as the 2-fold Massey product is the cup product, which is dual to intersection), while the 3-fold Milnor invariant measures whether 3 pairwise unlinked circles are Borromean rings, and if so, in some sense, how many times (i.e., Borromean rings have a Milnor 3-fold invariant of 1 or –1, depending on order, but other 3-element links can have an invariant of 2 or more, just as linking numbers can be greater than 1).

Another definition is the following: let's consider a link $L=L_{1}\cup L_{2}\cup L_{3}$ . Suppose that $lk(L_{i},L_{j})=0;i,j=1,2,3;i . Find any Seifert surfaces for link components- $F_{1},F_{2},F_{3}$ correspondingly, such that $F_{i}\cap L_{j}=\emptyset ,i\neq j$ . Then the Milnor 3-fold invariant equals minus the number of triple points in $F_{1}\cap F_{2},\cap F_{3}$ counting with signs; Template:Harv.

Milnor invariants can also be defined if the lower order invariants do not vanish, but then there is an indeterminacy, which depends on the values of the lower order invariants. This indeterminacy can be understood geometrically as the indeterminacy in expressing a link as a closed string link, as discussed below (it can also be seen algebraically as the indeterminacy of Massey products if lower order Massey products do not vanish).

Milnor invariants can be considered as invariants of string links, in which case they are universally defined, and the indeterminacy of the Milnor invariant of a link is precisely due to the multiple ways that a given links can be cut into a string link; this allows the classification of links up to link homotopy, as in Template:Harv. Viewed from this point of view, Milnor invariants are finite type invariants, and in fact they (and their products) are the only rational finite type concordance invariants of string links; Template:Harv.

The number of linearly independent Milnor invariants of length k+1 is $mN_{k}-N_{k+1},$ where $N_{k}$ is the number of basic commutators of length k in the free Lie algebra, namely:

$N_{k}={\frac {1}{k}}\sum _{d|m}\phi (d)\left(m^{k/d}\right),$ 