I hope this is a simple question...

For the trefoil knot 3_1, whose knot group is given by a presentation of the fundamental group, $\pi_1(M) = \langle a,b: aba = bab \rangle$, where the meridian and longitude cycles can be identified as,

$$m = a,$$

$$l = ba^2ba^{-4}$$

I understand the presentation, and have actually worked them out with Wirtinger(spelling?), but the thing I don't understand is how these cycles are identified.

I know that a and b refer to $\rho(a)$ and $\rho(b)$, where these are commuting (brought into Jordan form) complex 2x2 matrices in $SL_2(C)$ representing the cycles.

I guess I just don't understand how these cycles are 'identified'...

  • $\begingroup$ Are you asking how you prove that $m=a$ and $l=ba^2ba^{-4}$? Where does $SL_2(\mathbb C)$ come into the picture? Are you thinking of a representation of the knot group into $SL_2(\mathbb C)$? $\endgroup$ Jun 15, 2013 at 3:47
  • $\begingroup$ Yes, I am asking how $m=a$ and $l=ba^2ba^{-4}$. Here $m$ is the contractible loop. I guess the other information was to try and help context... $\endgroup$
    – nate
    Jun 15, 2013 at 3:55

1 Answer 1


enter image description here Okay, here's my picture of the trefoil and derivation of the wirtinger presentation that you already worked out. Any curve in the knot complement starting at the basepoint (near the X in my picture) will pick up a Wirtinger generator every time the curve passes under the appropriate arc. So the meridian, regarded as a small loop around the arc with an X, picks up an $a$, giving you the $m=a$. The longitude is defined to be a parallel copy of the knot with trivial linking number. If you run an obvious parallel arc to the knot, it has linking number $\pm3$ the way I've drawn it, so I added three twists to cancel out this linking number. So I read off $\ell=baca^{-3}$ as I travel around the red curve, which is $\ell=ba(aba^{-1})a^{-3}=ba^2ba^{-4}$, as desired.

  • $\begingroup$ So two things I notice: 1.) choice of starting point X (being at "a") is arbitrary, and 2.) leaving X in the direction shown, one passes through the "b" loop first, then the "a" loop, etc., picking up a "ba...", as opposed to "ac..." - which was by `reading off the sign posts along the road and not counting the signs on the loops.... (Does that make sense?) Very good explanation though and great effort! $\endgroup$
    – nate
    Jun 15, 2013 at 18:28
  • $\begingroup$ @Nate: yes the basepoint is chosen based on the fact that you said $a$ was the meridian. Also, I don't understand what you are saying in 2). Are you saying my calculation is wrong? As you travel along the red arc from the basepoint you pass under b followed by a followed by c, giving you $bac=baaba^{-1}$ followed by $a^{-3}$ for the twists, so I think it is correct as stated. $\endgroup$ Jun 15, 2013 at 19:03
  • $\begingroup$ Sorry, no, I was just wrong in my thinking of #2, I agree and it also concurs with what I have read about this knot. $\endgroup$
    – nate
    Jun 15, 2013 at 21:45

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