Figure 1. The left-handed and right-handed trefoils

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1 The Knot Group A knot is n emedding of the irle into R 3 (or S 3 ), k : S 1 R 3. We shll ssume our knots re tme, mening the emedding n e extended to solid torus, K : S 1 D 2 R 3. The imge is lled tuulr neighorhood of the originl knot. There exist knots tht re not tme, these re lled wild knots; see knot. Note: If the emedding is smooth or piee-wise liner (PL) it n e shown tht the knot is tme. Two knots re regrded s equivlent if there is n mient isotopy 1 of R 3 tht tkes the imge of one onto the other. A knot with no rossings when projeted onto plne is lled n unknot. The next simplest knot is the trefoil whih hs plne projetion with three rossings. (Any knot with just one or two rossings n e deformed into n unknot.) In ft, there re tully two trefoil knots tht re mirror imges of eh other. They re lled the left-hnded nd righthnded trefoils. See Figure 1. Figure 1. The left-hnded nd right-hnded trefoils Exerise. Plot the system of prmetri equtions elow with omputer for t [0, 2π]. Whih trefoil is it? Find equtions for the other. Write formul for n extension to tuulr neighorhood. x(t) = sin t + 2 sin2t y(t) = ost 2 os 2t z(t) = sin 3t The knot group of knot is the fundmentl group of the omplement on the knot. It is invrint under mient isotopy. First we do the unknot. Rell tht S 3 n e formed y gluing two unknotted solid tori together, with the meridin of eh going to the longitude of the other. Thus, in S 3 the omplement of n unknot n e homotoped to nother to solid torus. In R 3 we just get solid torus minus point. Thus the fundmentl group of the omplement of n unknot is n infinite yli group. 1 An isotopy is homotopy where eh level is n homeomorphism. An isotopy of the whole spe is lled n mient isotopy. 1

2 2 Next we do trefoil knot. (Both versions give isomorphi groups.) The method generlizes to ny knot. We will use PL version of the trefoil nd tke s tuulr neighorhood, N, one with squre ross setions, 1/3 1/3, tht fits into R 2 [0, 1] s in Figure 2. Sine omplement of tuulr neighorhood is deformtion retrt of the knot omplement, the fundmentl groups re isomorphi. A loseup view of one rossing is shown in Figure 3. We prtition N into six (esy!) piees. The lower prt is N {z 1/3} nd hs three omponents tht we ll B 1, B 2 nd B 3. The upper prt is N {z 1/3} lso hs three omponents tht we ll A 1, A 2 nd A 3. The A i s ll meet the z = 1 plne, while the B i s rest on the z = 0 plne. A 1 B 1 A 2 B 3 A 3 B 2 Figure 2. PL trefoil squre tuulr neighorhood A 1 A 2 A 3 B 1 Figure 3. Closeup of rossing We wish then to find the fundmentl group of R 3 N. We shll deompose this spe into two open sets nd pply the Seifert-vn Kmpen theorem. Let nd V 1 = {(x, y, z) z > 0} N V 2 = {(x, y, z) z < 1/6} N.

3 Let V 3 = V 1 V 2. Now we ompute the fundmentl groups of these three spes. We let p = (0, 0, 1/12) e the ommon se point, ut leve this out of the nottion for onveniene. The ovious inlusion mps indue the following ommuttive digrm. i π 1 (V 1 ) k π 1 (V 3 ) π 1 (R 3 N) j π 1 (V 2 ) We strt with V 2 euse it is simplest. It is the {z < 1/6} open lower hlf spe with three trenhes of depth 1/6 dug out, tht is, l V 2 = {z < 1/6} (B 1 B 2 B 3 ). There is deformtion retrtion to the hlf spe {z < 0}. This is homeomorphi to R 3. Thus, π 1 (V 2 ) is the trivil group. Next look t V 3. It is equl to {0 < z < 1/6} (B 1 B 2 B 3 ). This hs deformtion retrtion to plne, sy z = 1/12, with three disjoint retngles ut out. This in turn hs deformtion retrtion to wedge of three irles. (Prove this!) Thus, π 1 (V 3 ) = F 3. Figure 4 illustrtes hoie of genertors, α, β, nd γ for V 3. In the figure B 1, B 2 nd B 3 re holes. 3 B 1 α γ p B 2 B 3 β Figure 4. V 3 with hoie of genertors Finlly, onsider V 1. It n e thought of s the open hlf spe {z > 0} with three tunnels dug out. This is homeomorphi to n open retngulr ox with three tunnels dug. We n use n isotopy to strighten the tunnels s shown in Figure 5. This spe n e homotoped to n open retngle with three disks removed, whih in turn n e homotoped to wedge of three irles. Thus, π 1 (V 1 ) = F 3.

4 4 We will ll the genertors,, nd. Eh loops one round one of the tunnels. So, now we hve the following ommuttive digrm. i,, k α, β, γ π 1 (R 3 N) j Sine π 1 (V 2 ) is trivil, it follows tht π 1 (R 3 N) is generted y the imges of the genertors of π 1 (V 1 ) under the homomorphism indued y inlusion. Now the triky prt is to find the reltions. l Figure 5. V 1 deformed to retngle with three holes To do this we first study one rossing. We hve drwn four loops in Figure 6(left). Eh is homotopi to genertor. Two of them re tully the sme genertor. We lel them,, nd. We will write for nd so on nd use the sme symols for the loops nd the orresponding group elements. Study the loop 1. Notie tht it is homotopi to the α genertor of π 1 (V 3 ). Tht might e esier to see in Figure 6(right). But this genertor is mpped to the identity in π 1 (V 2 ). Thus, we hve the reltion 1 = 1 in π 1 (R 3 N). Eh rossing indues similr reltion. By the Seifert-vn Kmpen theorem, we rrive t presenttion for π 1 (R 3 N). We use the stylized digrm in Figure 7 to do the omputtion for our trefoil knot. This gives π 1 (R 3 N) =,, 1 = 1, 1 = 1, = 1.

5 5 p p Figure 6. A reltion t rossing Figure 7. Digrm for group reltions This is lled the Wirtinger presenttion of the knot group. The method n e pplied to ny knot. Often it is useful to derive different presenttions. For the trefoil group we show one of the two most ommonly used presenttions.,, 1 = 1, 1 = 1, =,, 1 = 1, 1 = 1, = 1 1 T1, T2, T1, T2 =,, 1 = 1, = 1 1 T2 =,, 1 = 1, = 1 1, 1 = 1 T1 =,, = 1 1, 1 = 1 T1 =, 1 = 1 T4 =, = T1, T2

6 6 We n elinize this group y dding the reltion =. You n hek this implies = nd thus the eliniztion gives n infinite yli group. It turns out tht the eliniztion of ny knot group is infinite yli. Finlly, we prove tht the trefoil group itself is not infinite yli. From this we n onlude tht the trefoil is not equivlent to the unknot. From n erlier hndout we know, = = x, y x 2 = y 3. Cll this group G. Let Q = x 2. There is homomorphism from G onto G/Q. But G/Q = x, y x 2 = y 3, x 2 = 1 = x, y x 2 = 1, y 3 = 1 = Z/2 Z/3, whih is not elin. But ny homomorphi imge of n infinite yli group must e elin. Exerise. Find presenttion for the knot group of the figure-8 knot; see Figure 8(left). Show tht it is equivlent to p, q p 1 qpq 1 pq = qp 1 qp. Show tht it is not infinite yli nd tht it is not isomorphi to the trefoil group. Show tht its eliniztion is infinite yli. Figure 8. The figure-8 knot nd the Hopf link Exerise. By the wy, the figure-8 knot is mient isotopi to its mirror imge. See if you n show this y drwing the mirror imge nd then grdully deforming it into the originl. Exerise. These ides n e extended to links. Figure 8 lso shows the Hopf link. Find the fundmentl group of its omplement. Show tht the result is isomorphi to Z 2. Referenes. (1) Knots nd Links, y Dle Rolfsen, AMS Chelse Pulishing. (2) Knots nd Surfes, y N. D. Gilert nd T. Porter, Oxford Siene Pulitions. Mihel Sullivn, My e e used for nonprofit edutionl purposes.