I'm trying to understand Slide 35 of Joan Birman's presentation on Lorenz knots, available here: https://www.math.columbia.edu/~jb/Lorenz-general-audience.pdf

I'm struggling with this particular bit. Let $G=\pi_1(S^3\smallsetminus \mathcal{T}) = \langle U,V: U^2=V^3\rangle$ be the fundamental group of the trefoil knot complement. She says that every free homotopy class of $G$ is represented by the cyclic word in $U$ and $V$ of the form $W=C^kUV^{\epsilon_1}....UV^{\epsilon_r}$ where $\epsilon_i=\pm1$ and $C=U^2=V^3$ generates the center of $G$.

My questions:

1) I don't see where she gets this particular presentation of the free homotopy class from. My guess is that this might have something to do with the fact that the fundamental group of the trefoil knot complement is the same as the braid group on 3 strands, $B_3$, and braid groups have a solvable word problem. So maybe that's where the presentation comes from...? I can't find the explicit justification for it though.

2) Just a clarification: isn't the trefoil knot complement a connected space? If so, isn't every element of the fundamental group a free homotopy class since all fundamental groups of a connected space is isomorphic, regardless of the base-point? If so,then Birman's presentation is true for all elements of the fundamental group and not just a special class of them, right?

  • $\begingroup$ For number 2, a change in basepoint corresponds to conjugation by a word. Which does not give every element, just all the conjugates of the word $W$. $\endgroup$
    – N. Owad
    Aug 28, 2017 at 0:27
  • $\begingroup$ @N.Owad thanks, that helped. Any clue about how the free homotopy classes of the fundamental group can be represented in the manner Birman suggested though? Or where I might try go looking for answers? $\endgroup$
    – asldjk
    Aug 28, 2017 at 9:05
  • $\begingroup$ I think all she does is (ignoring the $C^k$ for the moment) look at reduced words in the braid group. $V^{-1}=V^2$ and if you had $V^3$, you could replace it with the identity. Similarly for $U$. So, $W$ is all the reduced words. The $C$ is in the center, so she just shoves all them to the front of the word. $\endgroup$
    – N. Owad
    Aug 29, 2017 at 0:17
  • $\begingroup$ @N.Owad Yes, but suppose I had $W=UV^3$. Then this would give me $W=CU$, which is not in the form she described. My thought then is that there must be some other element in the same free homotopy class as $CU$ that does in fact have the form she described, but I'm not sure how to show this. $\endgroup$
    – asldjk
    Aug 29, 2017 at 10:00
  • $\begingroup$ I should add that the presentation has to have a sequence "$UV$"'s with appropriate powers after the $C$, so that's why $W=CU$ is not in the form she describes. $\endgroup$
    – asldjk
    Aug 29, 2017 at 10:04

1 Answer 1


This is what I have. It's just a sketch, but I think it's right. If it's wrong, feel free to point it out!

Note that the generic element $W\in G$ is of the form $W=U^{a_1}V^{b_1}U^{a_2}V^{b_1}...U^{a_n}V^{b_n}$, where $a_i,b_i\in\mathbb{N}$, for all $1\leq i\leq n$. By using the fact that $C=U^2=V^3$, we can reduce it to the following form: $W=C^{k}U^{a'_1}V^{b'_1}UV^{\pm1}UV^{\pm1}...U^{a'_m}V^{b'_m}$

where $a'_1,a'_m \in \{0,1\}$, where $b'_1,b'_m \in \{0,1,-1\}$.

Suppose that none of the following exponents are zero: $a'_1,a'_m,b'_1, b'_m$. Then we are done because the reduced form of $W$ is exactly in the form we want. Now, suppose that $a'_1=0,b'_1=\pm1$ and that $a'_m=1$, $b'_m=\pm1$. Then, $W= C^{k}V^{\pm1}...UV^{\pm1}$ is free homotopic to $(UV^{\pm1})W(UV^{\pm1})^{-1}=C^{k}U(V^{\pm1})^2...UV^{\pm1}(UV^{\pm1})^{-1} = C^{k}UV^{\mp1}...UV^{\pm1}$,

which is exactly in the form we want.

The other cases (i.e. when at least one of the exponents $a'_1,a'_m,b'_1, b'_m$ are zero) can be solved through similar methods and exploiting the fact that $C$ commutes with every element $g\in G$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .