Part 1 : Preliminaries

Save this PDF as:

Size: px
Start display at page:

Download "Part 1 : Preliminaries"


1 Singularity Knots of Minimal Surfaces in R 4 revised version, October 2008 Abstract We study knots in S 3 obtained by the intersection of a minimal surface in R 4 with a small 3-sphere centered at a branch point. We construct examples of new minimal knots. In particular we show the existence of non-fibered minimal knots. We show that simple minimal knots are either reversible or fully amphicheiral; this yields an obstruction for a given knot to be a simple minimal knot. Properties and invariants of these knots such as the algebraic crossing number of a braid representative and the Alexander polynomial are studied. Part 1 : Preliminaries 1 Introduction We study here knots associated to singularities of minimal surfaces in R 4 ; we call them minimal knots. They are derived from minimal surfaces in the same way that algebraic knots are derived from algebraic curves. We remind the reader that algebraic curves are a special case of minimal surfaces; thus minimal knots are a generalization of algebraic knots. How much of a generalization? This is the question that we want to address here. We need to recall how minimal knots are constructed. We let D be a disk centered at 0 and endowed with a Riemann (complex) structure; if X : D R 4 is a conformal harmonic map (in other words a minimal map) and dx(0) = 0, we say that p = X(0) is a singularity or branch point of X. The topology of X(D) around p is entirely determined by the (possibly 1

2 singular) knot or link obtained by intersecting the minimal disk X(D) with a small 3-sphere S 3 ɛ of R 4 around p of radius ɛ. We denote this intersection K ɛ ; if p has only one preimage in D, then K ɛ has only one connected component for ɛ small enough. If, moreover, X(D) is embedded in a neighborhood of p, then the K ɛ s are smoothly embedded curves all isotopic to fixed knot K and X(D) B(0, ɛ) is topologically a cone over K. In their groundbreaking study of singularities of minimal disks in 4- manifolds ([11]), M. Micallef and B. White showed how the square knot and the figure eight knot can be realized as minimal knots. These two knots are fibered, like the algebraic knots and this led H. Rosenberg and others to ask if all minimal knots are fibered. We answer this question negatively in the last section where we show several new examples of minimal knots: one of them, namely 9 46 is not fibered. Thus the class of minimal knots is much wider than the class of algebraic knots. This leads us to wonder if every knot is a minimal one. We ask: is there an obstruction for a knot to be a minimal one? We answer this question for simple minimal knots. These are the simplest type of minimal knots; roughly speaking, more general minimal knots can be built by a finite sequence of cablings starting with a simple minimal knot, possibly with singularities occurring along the way. We show that an amphicheiral knot such as 8 17 cannot be simple minimal. The general question: find a knot which is not a minimal knot remains open for the moment. Hopefully we will discuss it soon in another paper. Let us now say a word about our techniques. Minimal knots come from branch points which can be expressed by an analytic expansion in z and z. In the case of simple minimal knots, which are the focus of our study, this expression is not complicated and easily translates into a braid representation for the knot. Most of our work, for example the proof of the symmetry of the simple minimal knots, is done on this braid. When looking for examples of minimal knots, we feed the data of the braids into the KnotPlot. The software computes the Alexander and Jones polynomials of the knots; if the crossing number is not too large, we can then identify these knots in the Rolfsen or Hoste-Thislewaite tables. 2

3 1.0.1 Minimal Knots Singularities of minimal maps - which define minimal knots - are also called branch points. We shall presently see the reason for that name when we recall the definition of minimal maps from the disk. A map X : D C 2 from the unit disk of C into C 2 is minimal if and only if X is harmonic with respect to the metric induced on D by X. Since a harmonic map from a surface remains harmonic if we conformally change the metric on the surface, we put Definition 1 Let D be the unit disk in C and let z = x + iy be a complex coordinate on C. A map X : D C 2 is minimal if it is conformal and harmonic with respect to the flat metric on D; in other words if it verifies the following two conditions D X = 4 z z X = 0. (H) X x = X y, < X x, X y When the derivative of X vanishes we have >= 0. (C) Proposition 1 Let D be the unit disk in C and let X : D R 4 be minimal and suppose that X(0) = 0 and dx(0) = 0. Then there is a holomorphic coordinate in D and a coordinate system on R 4 such that X writes in a neighbourhood of 0 z ( Re(z ) + o( z ), Im(z ) + o( z ), o( z ), o( z ) ). We now assume that X is injective. We denote by S ɛ (resp. B ɛ ) the sphere (resp. ball) in R 4 centered at 0 and of radius ɛ. We put K ɛ = S ɛ X(D). This is how Milnor derives knots from algebraic curves; he proves (his proof also works in our more general context) that for ɛ small enough 1) (B ɛ, X(D)) is a cone over (S ɛ, K ɛ ), 2) all knots K ɛ = S ɛ X(D) with η ɛ are isotopic. The ensuing knot type is said to be associated to the singularity of X at 0. It follows from the implicit function theorem that there exists a real function r ɛ such that K ɛ is parametrized by X(r ɛ (θ)e iθ ), with e iθ going through S 1. A natural question arises: which knot types occur via this construction? If 3

4 X is a holomorphic map to C 2, it is a classical result that the associated knots are iterated torus knots (cf. [2] or [9]). Holomorphic curves are a special case of area-minimizing surfaces (Wirtinger inequality) which, in turn, are a special case of minimal surfaces. Micallef and White ([11]) proved that the knots of area minimizing surfaces are iterated torus knots, as are algebraic knots. But in the case of general minimal surfaces, other knot types can occur, for example they showed that ( D C C z (z 3 + o ( z 3 ), z 4 z 4 + z 5 + z 5 + o ( z 5 )) is a branch point of a minimal surface and the corresponding minimal knot is the square knot (see fig.3). Micallef and White left open the following question: can every knot isotopy type can be realized as the knot of a minimal branch point? 1.1 Simple minimal knots We investigate in the present paper a specific class of knots of branched points of minimal disks. They are given by the following Proposition which is inspired by a remark in [11] Proposition 2 Let, p and q be integers, p >, q > and let φ be a real number. We assume that the map X 0 : z ( Re(z ), Im(z ), Re(e iφ z p ), Im(z q ) ) is injective and we denote by K(, p, q, φ) the associated knot type. Then K(, p, q, φ) is associated to a branch point of a minimal disk. Proof. Proposition 2 means that, although X 0 itself may not be minimal, there is a minimal map which is close enough to X 0 in the sense that its associated knot is K(, p, q, φ). The proof consists in finding a deformation X t of X 0, t [0, 1] where i) X 1 is minimal, ii) there exists an η > 0 such that for every t [0, 1], X t is injective on the disk D(0, η). We look for a map X 1 of the form X 1 : z ( z + f(z), g(z) + h (z) ) where f, g and h are holomorphic functions. Such a map is harmonic; it is moreover ) (1) 4

5 conformal if and only if it satisfies the equations (C) above. This translates into f (z) = 1 z 1 g (z)h (z). If g(z) = z p + z q and h(z) = z p z q, such a function f exists and verifies f(z) = o( z ). In a neighbourhood of 0, there exists another function φ(z) such that φ(z) = 1 + f(z) z. We put w = zφ(z); we have w = z + o( z ). ote that w = z φ(z) so we can rewrite X 0 as X 0 (z) = ( Re(w ), Im(w ), Re[(1 + a(w))e iφ w p ], Im[(1 + b(w))w q ] ) where a(w) and b(w) are o(1). We introduce a parameter t and deform X 1 into X 0 by setting X t (w) = ( Re(w ), Im(w ), Re[(1 + ta(w))e iφ w p ], Im[(1 + tb(w))(w q )] ). We now prove the existence of an η > 0 such that all the X t s are injective on D(0, η). We start by noticing the following trivial Lemma 1 Let w and w be complex numbers, w w, which verify X t (w) = X t (w ) for some t [0, 1]. Then there exists ν 1 with ν = 1, such that w = νw. In particular w = w. Thus, if we let w = re iθ we will know that an X t is injective on a D(0, η) if and only the following map is injective X t : D(0, η) \ {0} R 4 X t : w = re iθ (cos θ, sin θ, Re[(1 + ta(w))e iφ e piθ ], Im[(1 + tb(w))e qiθ ]. We have assumed X 0 to be injective, hence X 0 is also injective. ote that, for r 0 and θ [0, 2π], the value X 0 (re iθ ) only depends on θ. So there exists a positive real number C > 0 such that for every w, ν, with w 0, ν = 1 and ν 1, X 0 (w) X 0 (νw) > C. 5

6 ext we pick an η > 0 small enough such that for every w with w < η, a(w) < C 10, b(w) < C 10. It follows that, for t [0, 1], w D(0, η) \ {0}, ν = 1 and ν 1, X t (w) X t (νw) > C 2. Thus the X t s and hence the X t s are injective on D(0, η). This, together with Lemma 1 above, proves that the X t s are injective on a small disk around 0. Moreover for ɛ small enough, X t (D(0, η)) S ɛ constitutes an isotopy between the knots associated to the singularities of X 0 and X 1. This concludes the proof of Proposition 2. ote that if q = p (and φ = 0 but we will see later that this condition is not necessary), then K(, p, q, φ) is a (, p) torus knot. We call a knot of the type K(, p, q, φ) simple minimal. The word simple comes from the fact that general knots of minimal surfaces can be seen as iterated versions of possibly singular K(, q, p, φ) s. We plan to devote another paper to the general case and focus here on simple minimal knots. We point out that simple minimal knots are similar in their expression to Lissajous knots [1], [6], ( see also KnotPlot [8] for a generator of Lissajous knots ), and can help understand them. evertheless we shall see that the properties of these two classes of knots are quite different. 1.2 Outline of the paper. In Part 1, we give a general description of the properties of simple minimal knots K(, p, q, φ). We describe the braid representation- refered to as the minial braid- that is naturally attached to K(, p, q, φ). The detailed study of this minimal braid is the purpose of Part 2 where we show that simple knots are invariant by a change of phase. From then on we drop the φ and denote a simple minimal knot by K(, p, q). We also prove an estimate for the crossing number of the braid. 6

7 Part 3 contains our main results about simple minimal knots. On the one hand we show that they all satisfy one symmetry or another, hence the knot 8 17 for example is not simple minimal. On the other hand, we give several new examples of simple minimal knots. Among them is the knot 9 46 which is not fibered, thus settling Harold Rosenberg s question negatively. 1.3 Main results Let us first recall that there are two natural symmetries among knots that are involutions : the mirror symmetry s m, (symmetry of a knot with respect to a orientation reversing symmetry of S 3 ) and the inversion of a knot s i which maps a knot to the same knot but with the reverse orientation. K is invertible if it is invariant by s i and amphicheiral if it is invariant with respect to s m up to inversion. In Part 3 we prove that the knot type K(, p, q, φ) is independent of the phase φ; we easily derive Theorem 1 A simple minimal knot is either reversible or fully amphicheiral. More precisely : 1. All knots K(, p, q, φ) are invertible. 2. If p + q is odd, then (, p, q) is positive amphicheiral. 3. If is even and p + q is even or if is odd and p and q are even, then K(, p, q, φ) is periodic of order two. As in [1], where Lissajous knots are studied, the existence of symmetries together with results of [5] and [10] yield properties on the Arf invariant and Alexander polynomial. It also gives obstructions on knots to be realized as simple minimal knots. Corollary 1 A negative amphicheiral or chiral knot cannot be the knot of a simple minimal knot. Thus the knot 8 17 is the first negative amphicheiral knot in the Rolfsen classification that can not be a simple minimal knot ( 9 32 is the first chiral knot that cannot be a simple minimal knot ). Though these knots cannot be simple minimal knots, we do not yet know if they can be realized as general minimal knots, for example cable knots of minimal singular knots. ext we investigate if minimal knots are fibered. 7

8 Definition 2 A knot K is fibered if there is a differentiable mapping φ : S 3 \ K S 1 which defines a fiber bundle; for every t S 1, the fiber φ 1 (e it ) is the interior of a compact orientable differentiable surface with boundary K. ote that fibered knots have an algebraic characterization: the commutator subgroup of the knot group π 1 (S 3 \ K) is finitely generated ( cf. [3]). Knots of holomorphic curves singularities are always fibered; if the surface is given locally by the equation F (z, w) = 0, then the (Milnor) fibration is simply given by F (z, w) φ(z, w) = F (z, w) restricted to a sphere of sufficiently small radius around the singularity. The fibration when it exists, yields a monodromy mapping h and a gluing map θ such that S 3 I F \ K, where F is the fiber of the fibration, (x,0) (θ(x),1) i.e. the Seifert surface spanning K. This monodromy map provides a way to compute the Alexander polynomial of the knot : P (x) = det(h xid) where h : H 1 (M, R) H 1 (M, R) is the linear map induced by h. In particular, the coefficient of the highest order term of P is ±1; we will use this fact to find non-fibered minimal knots. In the last section we construct examples of minimal braids with three, four or five strands and identify them. We show : Theorem 2 The knot 9 46 (see fig. 13 ) is a prime knot which is not fibered but is a minimal knot; the minimal surface is locally given by z ( z 4 + o( z 4 ), z 13 + z 13 + z 5 z 5 + o( z 13 ) ). Computer calculations seem too indicate that the simple knots K(, p, q) with odd are fibered. We give also a description of simple minimal knots for a small number of strands and small p ( or q). Theorem 3 Simple minimal knots of type K(3, p, 4), p 4 (respectively K(3, p, 5), K(3, p, 7), K(4, p, 5)) consist of only five (resp. four, 8

9 six, five) different knots that appear periodically with respect to p. precisely (we indicate only the smallest p for which they appear) K(3, 4, 4) = T (3, 4) (torus knot) K(3, 5, 4) = 3 1 # 3 1 K(3, p, 4), p 4 : K(3, 7, 4) = 1 K(3, 8, 4) = 4 1 #(3 1 # 3 1 ) K(3, 10, 4) = 4 1 K(3, 5, 5) = T (3, 5) K(3, 7, 5) = K(3, p, 5), p 5 : K(3, 8, 5) = 1 K(3, 8, 4) = 4 1 #(3 1 # 3 1 ) K(3, 10, 5) = 6 2 # 6 2 K(3, 7, 7) = T (3, 7) K(3, 8, 7) = 5 1 # 5 1 K(3, 10, 7) = 1 K(3, p, 7), p 7 : K(3, 11, 7) = K(3, 14, 7) = # K(3, 19, 7) = K(4, 5, 5) = T (4, 5) K(4, 7, 5) = 5 2 # 5 2 K(4, p, 5), p 5 : K(4, 9, 5) = 1 K(4, 11, 5) = K(4, 13, 5) = 9 46 More ACKOWLEDGMETS.We are grateful to Harold Rosenberg for asking us about the fibration of minimal singularities. We thank Joan Birman for her suggestions and Dror bar atan for his helpful advice. 2 Definition of a minimal knot Germ of a minimal singularity We consider a map X : D R 4 with a branch point as in Prop. 1; we put M = X(D) and m = X(0). The associated knot K is defined by intersecting M with a small sphere S ɛ (m) centered at m. For our present purpose, we 9

10 now describe an equivalent but slightly different point of view where we see K as sitting inside a cylinder around m. The minimal surface M is not a submanifold around m; nevertheless there exists a 2-plane in C 2 denoted by T m M which we can define as tangent to M at m ( with the notatons of Proposition 1. T m M is generated by the first two coordinates). By taking complements we also have a normal plane m M. The ambient space is thus split as C 2 = T m M m M. We let S ɛ be a small circle of radius ɛ in m M centered at the origin and we intersect M with the 3-cylinder Cyl ɛ := S ɛ m M T m M m M. We denote this intersection by K ɛ. With ɛ small, this knot stays in a neighbourhood of the equator S ɛ = S 3 (m, ɛ) T m M (in fact it runs times along S ɛ ) hence K ɛ can be equally viewed as a knot of the 3-sphere. We now drop the ɛ and parametrize K in terms of trigonometric functions as follows: ( K(, p, q, φ) : [0, 2π] R 4 θ (cos θ, sin θ, cos (pθ + φ p ), sin (qθ + φ q )) By changing θ into θ + α if necessary, we may assume that one of the two phases is zero. We thus define Definition 3 A simple minimal knot is a knot isotopic to the curve given by the one-to-one parametrization ( K(, p, q, φ) : [0, 1] Cyl R 4 t (cos 2πt, sin 2πt, cos(2πpt + φ), sin 2πqt) ) These knots are similar to another type of knots parametrized by trigonometrical functions and hence known as Lissajous knots (cf. [1] or [6]) Lissajous Knots Definition 4 A Lissajous knot is a curve parametrized one-to-one by L : [0, 1] R 3 t (cos 2πt, cos(2πpt + φ 1 ), cos(2πqt + φ 2 )). ) (2) 10

11 otice that these knots are parametrized by five quantities instead of four for simple minimal knots. It is a direct consequence of this definition to show that Lemma 2 The projection into a vertical 3-plane of a minimal knot is a Lissajoux knot. If the minimal knot is of type K(, p, q) then the corresponding Lissajous knot has a braid representation of 2 strands induced by the braid representation of K(, p, q). Part 2 : Minimal Braids 3 The minimal braid representation of a simple minimal knot The braid B(, p, q, φ) that we use to represent K(, p, q, φ) comes naturally from the way this knot is a -fold covering of S ɛ. We cut K(, p, q, φ) into pieces {K(θ) 0 θ < 2π 2lπ },..., {K(θ) 2(l + 1)π θ < },... for 0 l 1. Each of these pieces projects 1-to-1 onto S ɛ and we take it to be a strand of B(, p, q, φ). In other words, each strand of B(, p, q, φ) is the data of a map from one of the intervals [ 2lπ, 2(l+1)π [ into mm. Identifying each of these intervals with [0, 1[ and m M with C, we write these maps as functions B k, k = 0,..., 1, where B k : [0, 1[ C = m M ( ) ( ) 2πp 2πq B k (t) = cos (t + k + φ + i sin (t + k). A REMARK ABOUT THE OTATIO. The careful reader will have noticed that the phase φ above is different from the phase φ of the previous paragraphs. Indeed by writing 2π(t + k + φ) instead of 2π (t + k) + φ, we have changed the notation for the phase: the old φ is the new φ multiplied by 2π. 11

12 Figure 1: braid and braid diagram of K(3,4,4) ow that we are dealing with the braid, our new notation for the phase will make computations less cumbersome. The B k s define a braid B(, p, q, φ) which represents K(, p, q, φ) and which we call the minimal braid. ote that we reconstruct the knot K(, p, q, φ) by closing this braid : since B k (1) = B k+1 (0) for k = 0,..., 1, we can connect the k-th strand to the k + 1-th strand. By forgetting one of the two components of the B k s, we get the braid diagram. More precisely we write a z in C = m M as z = x + iy and define Definition 5 The braid diagram K (, q) is the union of the graphs of the k = 0,, 1. y k = ImB k : [0, 1] R, ote that K (, q) is always identical to the braid diagram of the (, q)- torus knot. To recover the braid from the braid diagram, we need to know which strand is above and which one is below at an intersection of any two strands y k and y l. This is the goal of the next section. First let us clarify which coordinates we use. The braid diagram sits in a plane and is made up of graphs of functions y k of t, with t [0, 1]: so we will use will use t (resp. y) for the horizontal coordinate (resp. vertical 12

13 coordinate). Going from the braid diagram to the braid itself is done by adding a third component, where the ReB k (t) s live. We will use the coordinate x for this component. 3.1 Crossing locus of the braid diagram K (, q) We will first find the values of the t-coordinates of the crossing points of K (, q) i.e. the solutions of y k (t) = y l (t) for k, l = 0,..., 1 with k l. Then we will determine which strand is above at each crossing i.e. what is the sign of ReB k (t) ReB l (t). There is a small problem we need to address before we start. It turns out that for an even number of strands the parameter value t = 0 corresponds to a crossing point; namely if = 2b, we have h 0 (0) = h b (0). We want to avoid having a crossing point on the boundary of the interval so we introduce a small positive ɛ such that ɛ does not parametrize an intersection of two strands; we study the braid diagram on the new interval [ɛ, 1 + ɛ[ instead of [0, 1[. Lemma 3 Let K(, p, q, φ) be a simple minimal knot. Its braid diagram K (, q) has q( 1) crossing points. Furthermore, for any k < l, k = 0,..., 1, the strands y k and y l, meet at points (t(n, k, l), y k (t(n, k, l))) where (2n + 1) t(n, k, l) := k + l, t(n, k, l) [ɛ, 1 + ɛ[, 4q 2 and where n is any integer that verifies 2qɛ y k (t(n, k, l)) = ( 1) n cos π q(k l). + q(k + l) 1 2 n < 2qɛ Proof. Suppose that t satisfies y k (t) y l (t) = q(k + l) q.

14 That is Using we obtain : ImB k (θ) ImB l (θ) = sin 2π q (t + k) sin 2π q (t + l). sin a sin b = 2 cos( a + b 2 ) sin(a b 2 ), ImB k (θ) ImB l (θ) = 2 cos ( 2π qt ) q(k + l) q(k l) + π sin π. (3) It follows from (3), that t parametrizes a double point if the first cosine factor is zero, that is if there exists some integer n such that that is 2π qt t = + π q(k + l) (2n + 1) 4q = π 2n k + l 2. If q and are not coprime then the second term is the RHS of equation 3 may be zero; then the knot becomes singular. In the regular case, the integer n has to verify ɛ (2n + 1) 4q k + l 2 < 1 + ɛ i.e. i.e. i.e. ɛ + k + l 2 4q 2q(k + l) ɛ + (2n + 1) 4q (2n + 1) < 4q < 1 + k + l 2 + ɛ + 2q(k + l) + 4q ɛ 2q q(k + l) ɛ n < 2q q(k + l) q ɛ. 14

15 We can now compute the y-coordinate of the corresponding point of the braid diagram, i.e. the height of the crossing point on the braid diagram; it is given by y k (t(n, k, l)) = sin 2π q(t + k) = ( 1)n sin ( π q (k l) + π 2 = ( 1) n cos π q(k l) ) Example 1 (braid with two strands (=2)) There are as many crossing points as integers n such that q (2n + 1) < 3q; that is q 1 n < q 1 + q. 2 2 Hence in that case the braid diagram has q crossing points Regularity of K(, p, q, φ) We return to the minimal knot and its naturally associated braid. First we rule out, by reparametrization, the case where, p and q have a common divisor: the knot would be covered more than once. The braid may be singular in the sense that some strands intersect : if this is the case, B k (t) = B l (t) for some k, l and t. To solve this equation, we need a lemma complementary to Lemma 3 Lemma 4 1) The projections ReB k and ReB l, k < l, for K(, p, q, φ) meet at points parametrized by t(m, k, l) := m 2p φ k + l 2, where m is any integer such that p (k + l + 2φ + 2ɛ) m < p (k + l + 2φ ɛ). (4) 2) The braid B(, p, q, φ) is singular if there are integers m and n such that φ = m 2p Such a φ is called a critical phase. (2n + 1). 4q 15

16 Proof. We have cos (2π p ) (t + k + φ) if and only if sin ( 2π p ( t + k + l 2 2π p cos (2π p ) (t + l + φ) = 0. )) + φ. sin (2π p ) (k l) = 0 ( t + k + l + φ 2 ) = πm t = m 2p φ k + l 2. Given these values of t, the inequalities ɛ < t < 1 + ɛ translate easily into the inequalities (4) of Lemma 4. Hence K(, p, q, φ) has a selfintersection if there is a t [ɛ, 1 + ɛ[ and m (resp. n) that satisfy the assumptions of Lemma 3 (resp. Lemma 4). Part 2) of the Lemma follows. We derive: Proposition 3 Let, q and p be integers such that p = 1 or q = 1; then for almost all φ s, K(, p, q, φ) is regular. K(, p, q, φ) is singular for a finite number of φ s which are all rational. Example 2 Knots K(, p, p, φ) are the torus knots T (, p). These knots project on T m M C 2 onto a circle. If φ (in our old notation π), 4 2 K(, p, p, φ) projects on the second component p M of C 2 onto an ellipse. But if φ =, the projection is a line segment, hence the knot becomes singular. We will show in section 4 that, surprisingly, a minimal knot does not 4 change its type as the phase varies. This is a striking difference with the Lissajous knots where a suitable combined variation of the two phases may change the knot (cf. [1]). 3.2 Sign of the crossing points The sign S(t, k, l, φ) of the crossing point is by definition given by the combination of the following two pieces of data (see figure 2): 1. which strand is in front of the other one 16

17 -1 +1 Figure 2: Sign of a crossing point 2. which strand goes upwards. 1. The answer is given by the sign of the difference ReB k (t) ReB l (t), that is the sign of the difference in the x-coordinates cos 2π 2π p(t + k + φ) cos p(t + l + φ) 2. We notice that the k-th strand is going upwards if the derivative of the function x sin 2π qx at t + k is positive (in which case its derivative at t + l is negative). It follows from 1) and 2) that the sign is given by the product S(t, k, l, φ) := [cos 2π q(t+k) cos 2π 2π 2π q(t+l)][cos p(t+k+φ) cos p(t+l+φ)]. We remind the reader that a difference of cosines can be written as a product of sines and derive S(t, k, l, φ) = 4 sin 2π q(t+k + l 2π ) sin 2 p(t+k + l 2 +φ) sin π p(k l) sin π q(k l). We now suppose that t is of the form t = t(m, k, l) as given by Lemma 3. We derive sin 2π q(t + k + l 2 ) = sin(π 2 sin 2π p(t + k + l + φ) = sin 2π 2 p + nπ) = ( 1)n (φ + 4q ) (1 + 2n). otations. If x is a real number we denote its integral part by [x]; if n is an integer, we denote by P (n) Z 2 its congruence modulo 2. With these notations and those of Lemma 3, the previous computations yield the following Lemma 5 [ Crossing sign formula] Let K(, p, q, φ) be a simple minimal knot; then the crossing points of its braid diagram K(, q) are parametrized by values t(m, k, l) [ɛ, 1 + ɛ[ with k < l, k, l = 0,..., 1 and m [ 2qɛ + q(k + l) 1 2, 2qɛ 17 + q(k + l) q [,

18 The sign at the crossing point parametrized by t(m, k, l) is given by the formula S(k, l, m, φ) = 1 + T (k, l) + P (m) + R(m, φ) (5) where T (k, l) = P ([ q ( )]) k l + P ([ p ( )]) k l (the sum is taken modulo Z 2 ) and where ([ 2 R(m, φ) = P p(φ + ]) (1 + 2m)). 4q 3.3 A word in the braid group We recall that the braid group B is generated by 1 elements, σ 1,..., σ 1 where σ i exchanges the i-th strand with the i+1-th one. The σ i s are subject to the following relations i = 0, 1, σ i σ i+1 σ i = σ i+1 σ i σ i+1 i, j = 0, 1 with i j 2, σ i σ j = σ j σ i. The goal of the present paragraph is to write a specific expression in the σ i s, σ 1 i s which represents the braid B(, p, q, φ). In other words we will specify one representative of a preimage of B(, p, q, φ) in the free group F 1 on the 1 generators σ i s. To derive a word in the braid group from the braid, we go from t = 0 (resp. t = ɛ) to t = 1 (resp. t = 1 + ɛ) and list the crossing points we encounter. The y-coordinate of each crossing point will give us the σ i it corresponds to; the sign of the crossing point will tell us whether to write σ i or in our word in the braid group. There is one problem, though. There can be several crossing points with the same t-coordinate; in that case they correspond to σ i s which commute with one another (see above the relations defining B ). So in order to specify our word in the σ i s we have to order the crossing points in a specific way. σ 1 i We list the crossing points as C i, 1 i q( 1). For each i, we denote by t i (resp. by k i, l i, k i < l i ) the t-coordinate of (resp. the pair of strands which meet at) C i. We will sometimes also use the notation {k, l} i for this 18

19 pair of strands. The ordering of the C i s is defined by the conditions (O1) If i < j then t i t j. There are, in general, several pairs of strands intersecting for the same t- coordinate; these different pairs of strands correspond to commuting σ i s or. We order these by setting (O2) If i < j and t i = t j, then k i + l i < k j + l j. (O3) If i < j, t i = t j and k i + l i = k j + l j, then k i < k j. σ 1 i ow that the C i s are defined without ambiguity, we can write the braid word. We recall (cf. Lemma 3) that the y-coordinates of the C i s can take 1 different values y 1 > y 2 >... > y 1. For an i, 1 i ( 1)q, we define an integer n(i) by setting n(i) = s if and only if y(c i ) = y s and we let ɛ(i) { 1, +1} be the sign of the crossing point C i. We end up with the following word in F 1 b(, q, p, φ) = ( 1)q i=1 σ ɛ(i) n(i). Its image in B is a representative of B(, p, q, φ). Since we will be dealing with words of length ( 1)q we introduce, for an integer m, the sets m = {n 1 n m}. Thus the word b(, q, p, φ) is given by the data of the maps n : q( 1) 1, ɛ : q( 1) Z 2. 4 (on) dependence of the knot type on the phase φ Proposition 4 Let, p, q be integers as above and let φ and φ be two elements of [0, 2π]. Then the knots K(, p, q, φ) and K(, p, q, φ ) - or K(, q, p, φ) and the mirror image of K(, q, p, φ )- can be represented by conjugate braids. 19

20 Corollary 2 Up to taking a mirror image, the isotopy type of the knot K(, q, p, φ) does not depend on the phase φ. We first prove the easy Lemma 6 It is enough to prove Proposition 4 in the case where φ = φ + ( A 2 p + B ) q for two integers A and B. To prove Lemma 3 we first need to deal with the critical phases, that is the φ s such that K(, p, q, φ) has (a) self-intersection(s) (cf. Lemma 4) We derive from Lemma 4 2) that Lemma 7 If φ 1 and φ 2 are two critical phases, there exist two integers A and B such that φ 1 φ 2 = ( A 2 p + B ). q We also have Lemma 8 Let φ 1, φ 2 be two numbers, φ 1 < φ 2 and suppose that there is no critical phase in the interval [φ 1, φ 2 ]. Then the braids B(, q, p, φ 1 ) and B(, q, p, φ 2 ) are identical. We order the critical phases between 0 and 2π and denote them respectively φ 0, φ 1,, φ M. We assume that φ u < φ < φ u+1 and φ v < φ < φ v+1 for some u, v, with 0 u, v M. It follows from Lemma 7 that φ u φ v = X with X = ( A + B ) for some 2 p q integers A, B. Thus φ + X belongs to the interval [φ v, φ v+1 ] (like φ does); hence the braids B(, q, p, φ ) and B(, q, p, φ + ( A + B )) are the same. 2 p q This proves Lemma 86. We now prove Proposition 4 for two phases φ and φ as in Lemma 7: keeping A fixed, we work by induction on B. We ask the reader to go back to the formula for the signs of the crossing points ( Lemma 5). (S)he will notice that it is a sum of three terms; only one of these three terms, namely R, involves the phase φ. Moreover a straightforward computation on R yields 20

21 Lemma 9 R (m, φ + 2 (Ap + Bq ) ) = P (A) + R(m + B, φ). If B = 0, Proposition 4 follows from Lemma 9 : if A is even (resp. odd) B(, q, p, φ) is the same as (resp. the mirror image of) of B, p, q, φ + A. 2p We prove Proposition 4 by induction on B: it is enough to show that K(, q, p, φ) and the mirror image of K(, q, p, φ + ) can be represented 2q by conjugate braids. We are going to write two words in the braid group B which represent respectively the word B(, p, q, φ + ) and a braid conjugate to the mirror 2q image of B(, p, q, φ)). For reasons that will appear later, we extend the definition of the braid strands B k s to the interval [0, 2] instead of [0, 1]. We keep the notation C i for the crossing points, as well as their ordering. There will be twice as many crossing points on [0, 2] as when the braid was defined over [0, 1] so the index i of the C i s will go from 1 to 2( 1)q. We put b(, p, q, φ + q( 1) 2q ) = i=1 σ η(i) n(i). To avoid confusion, we point out that the exponents of the σ n s in b(, p, q, φ) (resp. b(, p, q, φ + )) are denoted by ɛ (resp. η). 2q To describe the mirror image of B(, p, q, φ), we start with a word representing B(, p, q, φ) which is slightly different from the word we used before. amely we take the word q( 1)+ ˆb(, p, q, φ) = ( 1) 2 i=1+ ( 1) 2 σ ɛ(i) n(i) which represents B(, p, q, φ) (now the reader understands why we wanted to extend the interval over which the B k s are defined). To derive from ˆb(, p, q, φ) a word representing the mirror image of B(, p, q, φ), we change the parity of every exponent into its inverse and replace every σ i by the corresponding σ i ; we get the word q( 1)+ b(, q, p, φ) = ( 1) 2 i=1+ ( 1) 2 21 σ 1+ɛ(i) n(i).

22 Proposition 4 will follow immediately from Lemma 10 b(, q, p, φ) = b(, p, q, φ + 2q ). To prove this lemma, we introduce the map ( γ : i i + ( 1) 2 ) and write q( 1) b(, q, p, φ) = i=1 It is clear that Lemma 10 will follow from Sublemma 1 For every i, 1 i ( 1)q, 1) n(i) = n(γ(i)) 2) η(i) = 1 + ɛ(γ(i)). σ 1+ɛ(γ(i)) n(γ(i)) Proof of the Sublemma. We introduce a notation: for an i, 1 i ( 1)q, we define the integer m i by t i = k i + l i 2 + 4q (1 + 2m i). The formula in Lemma 3 tells us that two consecutive crossing points of a pair {k, l} differ by. Thus, if we go from t to t +, for a t [0, 1], we 2q 2q encounter one and only one crossing point for each of the ( 1) pairs {k, l}. 2 It follows that, for i, 1 i ( 1)q, (6) t i + 2q = t i+ ( 1) 2 = t γ(i) (7) Hence y(c i ) + y(c γ(i) ) = sin 2π q(t i) + sin 2π q(t i + 2q ) = 0 It follows that n(γ(i)) = n(i) This proves 1) of the Sublemma. To prove 2), we first derive from (7) that m γ(i) = m i

23 The number η(i) gives us the sign of the crossing point C i B(, p, q, φ + ). It follows from (5) that 2q of the braid η(i) = 1 + T (k i, l i ) + P (m i ) + R(m i, φ + 2q ) = 1 + T (k i, l i ) + P (m i ) + R(m i + 1, φ) (Lemma 9) = T (k i, l i ) + P (m i + 1) + R(m i + 1, φ) = 1 + ɛ(γ(i)). This proves 2) of Sublemma and concludes the proof of Proposition 4. 5 The algebraic crossing number 5.1 Definition - Background We remind the reader that the algebraic crossing number of B(, q, p, φ) is the sum of the signs (i.e. +1 or 1) of its crossing points. We will denote it e(b, q, p, φ). It is an invariant of the conjugacy class of the braid but it is not an isotopy invariant of the knot K, q, p, φ. In the present case, where the braid comes from a branched immersion in 4-space, its algebraic crossing number can be seen as the number of double points which concentrate at the branch point: here is the result of [13] specialized to the context of the present paper: Theorem 4 Let Σ be a closed Riemann surface without boundary, let M be an orientable 4-manifold and let f : Σ M be a (not necessarily minimal) topological embedding which has one branch point p. Suppose that in a neighbourhood of p, f is parametrized as in Prop. 1. Then the degree of the normal bundle f is [f(σ)].[f(σ)] e(b(, q, p, φ)) where [f(σ)].[f(σ)] denotes the self-intersection number of f(σ). This adds geometric significance to the algebraic crossing number and makes it worthwhile listing some of its properties. 23

24 5.2 A few estimates We derive from Proposition 4 that Proposition 5 Up to sign, the algebraic crossing number e(b(, p, q, φ)) does not depend on the phase φ. This allows us to drop the φ in the next two results which are about the absolute value of the crossing number. Proposition 6 Suppose that q and p are of different parities. Then e(b(, q, p)) = 0. Proposition 7 Let, p, q be as above and suppose that p and q are mutually prime. Then 1) If = 2, e(b(2, q, p)) = 2 2) If = 3, e(b(3, q, p)) 6 3) If 4, e(b(, q, p)) Remark. This estimate does not depend on p and q: notice the sharp contrast with the case of the (, q)-torus knot (i.e. when p = q) where every crossing number is positive and we have e(b(, q, q) = q( 1). Proof of Proposition 7 We begin by choosing a phase φ which will make our proof work. We go back to the expression of R(., φ) of Lemma 5 and we see that we can choose φ such that for every m, we have ([ ]) p R(m, φ) = P q m. We first investigate the m s for which p m is an integer: q Lemma 11 If m belongs to M(k, l), p m is an integer if and only if both q conditions below are satisfied i) m = q ii) k + l { 2, 1, }. Moreover, if condition ii) is verified, then q belongs to M(k, l). 24

25 Proof. We assume that m M(k, l) and that p m is an integer; since p q and q are mutually prime, then m is of the form m = aq. If we plug in aq in Lemma 3, we get k + l 1 2q a k + l q. Since 1 k + l 2 3, we derive that a = 1. Thus k + l satisfies k + l 1 2q 1 k + l q. Since < q the left hand-side yields k + l and the right-hand side yields k + l 2. We leave it to the reader to check the rest of the lemma. Lemma 12 If m q, we have R(2q m, φ) = R(m, φ) + 1. Proof Since p m is not an integer, q R(2q m, φ) = P ([2p p q m]) = 1 + P ([p m]) = 1 + R(m, φ). q If k and l are different integers, 1 k, l 1, we denote by M(k, l) the set of integers m such that the t(m, k, l) given in lemma 1 (2.1) above is a crossing point of the k-th and l-th strands. Lemma 13 There is a 1-to-1 correspondence between M(k, l) and M( 1 k, 1 l) given by t(m, k, l) t(2q m 1, 1 k, 1 l). Proof. We notice that m verifies 3? w.r.t. the integers k and l if and only if 2q m 1 verifies q (2 (k + l) 2) 1 2 2q m 1 q (2 (k + l) 2) q. 25

26 ote that we have taken ɛ = 0; this does not change the proof but simplifies the notations. We rewrite the inequalities above as q (( k 1)+( l 1)) 1 2 2q m 1 q ( 1 k)+( 1 l)) q. Our goal is now to cancel out the signs of most points in M(k, l) with points in M( 1 k, 1 l). We denote by σ(k, l) the signed number of intersection points of the k-th and l-th strands. We put together Lemma 11 and the two identities and derive T (k, l) = T ( k 1, l 1) P (m) = P (2q m 1) σ(k, l) + σ( k 1, l 1) = ( 1) T (k,l) Σ m M(k,l) ( 1) P (m) ( ( 1) R(m,φ) ( 1) R(2q m 1,φ)). At this point we introduce the slightly unorthodox notation M(k, l) + 1 = {m + 1 m M(k, l)} and rewrite the sum above (we drop the ( 1) T (k,l) factor) Σ m M(k,l) ( 1) P (m) ( 1) R(m,φ) + Σ n M(k,l)+1 ( 1) P (n) ( 1) R(2q n,φ) (8) It follows from Lemma 12 that, if k + l is not equal to either 2, 1 or, all n s in ( 8) verify R(2q n, φ) = R(n, φ) + 1 and the terms in m and in n cancel out two by two except maybe for the smallest and the largest ones. Thus σ(k, l) + σ( k 1, l 1) 2. If k + l = (resp. k + l = 2), then ( k 1) + ( l 1) = 2 (resp. ( k 1) + ( l 1) = ). Both pairs of strands {k, l} and { k 1, l 1} contain a crossing point for which m = q: the signs of these two points do not cancel out. Thus we can only claim σ(k, l) + σ( k 1, l 1) 4. 26

27 Finally if k + l = 1, then {k, l} = { k 1, l 1}. We can write in this case 2 σ(k, l) = σ(k, l) + σ( k 1, l 1) 4. To put together these estimates we write e(b(, p, q, φ)) Σ 2 s=1 Σ k+l=s σ(k, l) + σ( k 1, l 1) +Σ k+l= 1 σ(k, l). We immediately derive the estimates for = 2 and = 3. For the general case, we conclude, using the following remark: if s is an integer, the number of pairs of strands {k, l} such that k + l = s is not larger than s ( s ) ( ) ( ) 2 1 e(b(, p, q, φ)) 2Σ 3 s= = Part 3 : Simple Minimal Knots 6 Symmetries of simple minimal knots We use here the same terminology as in the first Part Section 1.3 ; we only recall that a knot is said to be strongly symmetric if it is symmetric with respect to an ambient isometry of S 3. Results of Part 2 and easy computations yields Theorem 1 of Part 1, or in full details : Theorem 5 A simple minimal knot is always invertible : it is either reversible or fully amphicheiral. More precisely : 1. All knots K(, p, q, φ) are invertible. 2. Furthermore, if p + q is odd, then K(, p, q, φ) is fully amphicheiral: 27

28 3. if is even and p + q is even or if is odd and p and q are even, then K(, p, q, φ) is periodic of order two : it is invariant by a rotation of angle π and the linking number of the corresponding axis of rotation in S 3 with the knot is equal to. These symmetries yields properties of the Arf invariant and Alexander polynomial of these knots which are similar to those of Lissajous knots ( cf. [1]). Let us first prove that a simple minimal knot is invertible. Proof. The change of parametrization t t changes the orientation of the knot. Suppose that φ = 0, π; then we see from definition 3 that the same change of parametrization is induced by the ambient symmetry (x, y, z, t) (x, y, z, t) which reverses the orientation when restricted to S 3. It follows in particular that knots K(, p, q, 0) and K(, p, q, π) - which are mirror images of one another - are invertible. This fact, together with Proposition 4 -on the independance of the knot type with respect to the phase- finishes the proof of the theorem. We may also consider the change of parametrization t t+ 1 which does 2 not change the knot as a whole. Let us remind the reader that if θ is a real number and m is an integer, then sin m(θ + π) = ( 1) m sin(mθ), cos m(θ + π) = ( 1) m cos(mθ), where θ is any real number and m is an integer. We conclude as before that this change of parametrization is induced by the following ambient symmetries. Lemma 14 The knot K(, p, q, φ) is invariant under the diffeomorphism Φ,p,q of S 3 defined by (x, y, z, w) (( 1) x, ( 1) y, ( 1) q z, ( 1) p w). According to the parities of the integers involved we obtain different symmetries of the knot. We notice that Φ,q,p is an involution in all cases. 1. If is even, then (a) if p+q is even, Φ,q,p is an orientation preserving symmetry. From the nondegeneracy condition p and q cannot be both even hence p and q are odd; then Φ,q,p is a rotation of angle π and K(, p, q, φ) is periodic of order two ( in a weak sense since the linking number of the rotation axis with K(, p, q, φ) may be zero). 28

29 (b) if p + q is odd Φ,q,p is orientation reversing. and K is strongly fully amphichireal. (c) Using the braid description of the knots, symmetries s m and s i induce symmetries of the braid. Let us denote by (t, k) a point in the braid that lies on the k-th strand. We notice that 2π 2π (x + k) + π = (x + k + 2 ) (9) Thus, if p and q have the same (resp. a different) parity, B(, q, p, 0) is preserved (resp. transformed into its mirror image) by the following transformation : if k 2, (t, k) (t, k + 2 ) if k 2, (t, k) (t, k 2 ). This symmetry switches the strands, while keeping the parameter t fixed. 2. If is odd, then (a) if p+q is even, Φ,q,p is orientation preserving. If p and q are both even, then Φ,q,p is a rotation of angle π around a vertical axis. The linking number of this axis with K(, p, q, φ) is. Hence K(, p, q, φ) is periodic of order two. If p and q are both odd, then Φ,q,p is orientation preserving and has no fixed points. (b) if p + q is odd Φ,q,p is orientation reversing and K(, p, q, φ) is strongly fully amphichireal (c) For the minimal braid, we notice that 2π 2π (x+k)+π = (x k + 1 ) = 2π 2 (x 1 2 +k ). 2 If p and q have the same (resp. a different) parity, B(, q, p, φ) is preserved (resp. transformed into its mirror image) by the transformation if k 1 and t 1 then (t, k) (t + 1, k + 1) if k 1 and t 1 then (t, k) (t 1, k + 1) if k 1 and t 1 then (t, k) (t + 1, k +1) if k 1 and t 1 then (t, k) (t 1, k +1) This symmetry switches the first half of a strand with the second half of another. 29

30 Figure 3: A knot that can not be simple minimal : 8 17 These facts finish the proof of Theorem 4. We can also deduce from these considerations and from results in [10] that Corollary 3 The Alexander polynomial of the simple minimal knot K(, p, q, φ) is a square modulo two, if p and q are not both odd; the Arf invariant of K(, p, q, φ) is then zero. 7 A knot that cannot be a minimal knot We consider simple minimal knots and, more generally, satellite knots of a regular and non-trivial simple minimal knot. We show that Theorem 6 The knot 8 17 cannot be the knot of a simple minimal knot or a minimal knot which is the satellite knot of a regular and non-trivial simple minimal knot. otice that the first candidate for a counterexample in the Rolfsen classification is the knot 8 17 which is the first negative amphicheiral knot. Proof. If the knot is simple, then Theorem 4 shows that it cannot be either chiral or negative amphicheiral. Suppose now that there is a satellite knot K of a simple minimal knot isotopic to Then K is contained in a tubular neighborhood T of its companion knot which is a simple minimal knot K(, p, q, φ). The linking number κ of 30

31 the satellite knot with a meridian of the tube T is larger than two. Indeed, as the parameter t [0, 1[ of the knot increases, the projection onto the circle S ɛ of the corresponding point on the knot parametrized by t, turns in the same direction But, we know from [3] that the Alexander polynomial of K is a multiple of P (x κ ) where P is the Alexander polynomial of the companion knot K(, p, q, φ) which we have assumed to be nontrivial. This is impossible since P (z) = z 6 4z 5 + 8z 4 11z 3 + 8z 2 4z ew Examples of Simple Minimal Knots Most minimal knots given here can be described with the help of the Rolfsen table i.e. they are knots that are the connected sum of prime knots whose (minimal) crossing number is at most ten. In some cases however, we use the Hoste-Thistlewaite table provided in the program KnotTheory and give examples of minimal knots with crossing number 15. In each case we compute the Alexander and Jones polynomials and represent the minimal braid. Knots that are in the Rolfsen table are drawn using KnotPlot. We need to set some conventions. First we use Rolfsen s notation for the Alexander polynomial : [a 0 + a 1 + a a n ] := x n. n i=0 a i ( x + 1 x) i. Secondly, we will number the strands of the minimal braid top down according to the order of the strands on the left side of the braid ( i.e. y k < y l if y k (0) < y l (0)). 8.1 Additional Properties of K(, p, q, φ) Let us first show that the same knots appear for periodic values of p and q in K(, p, q, φ). As simple minimal knots are phase invariant up to mirror symmetries, we will drop the phase term and denote them by K(, p, q). Lemma 15 If { p p mod p p mod 2q 31 (10)

32 then K(, p, q) is isotopic to K(, p, q). Indeed, the braid diagrams are the same, and by the crossing number formula in Lemma 5, the signs are also identical. In particular K(, p + 2q, q) = K(, p, q) (equal means isotopic). Lemma 16 For all, p, q, p >, q >, K(, p, q) = K(, q, p). Proof. We obtain K(, q, p) from K(, p, q) by adding the phase π/2 and by interchanging last two coordinates of R 4. The invariance of the knot with respect to the phase yields the result. It is easy to show that Lemma 17 For any m, B(, mq, q) is represented by a word in the braid group of the form ( σ α 1 1 σ α ) 1 q 1 with α i = ±1 ( If α i = +1, i = 1,..., 1 we recover the torus knot, ). Remark 1 It is known that Torus knots are not Lissajous knots ( cf. [1]) ; this is not so for simple minimal knots. All torus knots are minimal knots: T (a, b) can be realized as K(a, b, b). Moreover if the minimal braid has two strands, K(2, p, q) is either trivial if p q ( cf. Proposition 7) or the torus knot T (q, 2). Remark 2 There is an infinite number of minimal representations of the trivial knot : for any and q, K(, + q, q) is isotopic to the trivial knot. Let us now describe some non-trivial examples with > 2; we begin with knots K(, p, q) where is odd. 8.2 Minimal Knots with 3 or 5 strands Let us first consider knots whose minimal braid has three strands. The braid of K(3, p, q) is of the form b(, p, q) = q i=1 σ α i 1 σ β i 2, 32

33 α i, β i = ±1. If the first two crossings of the braid corresponds to the crossings of the strands (1, 2) and (1, 3) then the k-th pair of crossings corresponds to the intersection of the couple of strands (τ k (1), τ k (2)) and (τ k (1), τ k (3)) where τ is the cyclic permutation (1, 2, 3) K(3,., 4) The number of crossings of the braid diagram is at most q( 1) = 8; hence, if these knots are prime, they must be in the Rolfsen table. If they are not prime then they are connected sums of knots that are in the Rolfsen table. From lemma 15 we know that knots appear with a period of 24 with respect to p : in fact we find thata smaller period occurs and that five different knots appear. 1. The torus knot T (3, 4) = K(3, 4, 4p) = K(3, 20, 4) This case generalizes to all K(, p, p) or K(, ap, p) for suitable a. These knots are clearly reversible. 2. The square knot 3 1 # 3 1 = K(3, 5, 4) = K(3, 4, 5). This is the first example in the literature of a minimal knot that is not toric (cf. [11]). This is the smallest knot with respect to the lexicographic ordering on, p, q. It is fully amphicheiral and P = [ 1 + 1] The trivial knot K(3, 7, 4) = 1 ( from Remark 2). otice that K(3, k, 4) = 1 for k = 5, 11, 13, The connected sum of the figure eight knot and the square knot : 4 1 #(3 1 # 3 1 ) = K(3, p, 4). ( p = 8, ). This is a knot of type K(, aq, q) and from Lemma 17 it is either the torus knot T (, p) or a knot whose braid representatiove is ( ) 4. σ 1 σ2 1 This knot is fully amphicheiral. 5. The first non toric and non trivial prime knot : the eight knot 4 1 = K(3, 10, 4) = K(3, 22, 4)...; it is fully amphicheiral and P = [ 3 + 1] K(3,., 5) The number of crossings of the minimal braid is at most 10 which still allows us to check in the Rolfsen table. Altogether only four knots are found. 33

34 Figure 4: Minimal braid and knot of K(3,5,4) Figure 5: One of many minimal representation of a trivial knot: K(3,7,4) 34

35 Figure 6: The eight knot realized as K(3,10,4) 1. The torus knot T (3, 5) realized by K(3, 5, 5). 2. The prime knot = K(3, 7, 5); it is reversible. As in the case of Lissajous knot, the appearance of this knot for values as small as (3, 7, 5) suggests that it is difficult to find a topological characterization of minimal knots (see also K(4, 11, 5) described below). 3. The trivial knot K(3, 8, 5) = The connected sum 6 2 # 6 2 = K(3, 10, 5). This is a knot of type K(, aq, q) ; (= K(3, 20, 5), but K(3, 25, 5) = K(3, 35, 5) is the torus knot T (3, 5)) K(3,., 7) The number of crossings of the minimal braid is at most 14; we need the Hoste-Thistlewaite table. Altogether only six different knots appear : 1. K(3, 7, 7) = T (3, 7). 2. K(3, 8, 7) = 5 1 # 5 1 and P = [ ] K(3, 10, 7) = 1. 35