# trefoil knot and meridian/longitudinal cycles

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'...

• 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 $SL_2(\mathbb C)$? Jun 15, 2013 at 3:47
• 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...
– nate
Jun 15, 2013 at 3:55

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.
• @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. Jun 15, 2013 at 19:03