Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Currently I am reading Milnor's paper "Link Groups". In the paper he defines the link group, for every $n$ component link $L$, as a certain quotient of the fundamental group $\pi_1 (S^3 \setminus L)$. On wikipedia in the article titled "link group" one can read that for trivial links this link group is isomorphic to the free group. However I think that this is impossible because the normal closure of the group generated by every meridian is abelian. So are the link groups of the trivial links free?

share|improve this question
4  
Presumably the term "link group" is being used in different ways. Typically it just means $\pi_1(S^3\setminus L)$, and for the trivial link this is free on the meridians. I don't recall Milnor's definition, but if he is taking a quotient, his "link group" means something different. –  Paul May 29 '12 at 13:05
1  
Indeed, read the wikipedia article and the mathscinet review of Milnor's paper with more care and you will see that they are using the term "link group" in two different ways. –  Lee Mosher May 29 '12 at 13:12
1  
Milnor's link group is not the same as $\pi_1(S^3 \setminus L)$, but nonetheless the article and the review use the term differently. –  Lee Mosher May 29 '12 at 13:18
2  
Milnor's link group is invariant under "link homotopy", which allows strands to pass through themselves but not through other strands (this isn't obvious -- it's one of the main theorems of his paper). In particular, under Milnor's definition the group of any knot is $\mathbb{Z}$ (since there are no other strands to get in the way of making the knot trivial). This is very different from the fundamental group of the link complement. –  Andy Putman May 29 '12 at 15:16
    
Today I realized that the term link group means something else in both articles. Thanks for your comments anyway. –  W. Politarczyk May 30 '12 at 19:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.