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Usually X is a space of functions. We define a polynomial function from the unit circle S1 in R2 to R3 to be a function which is given by polynomials of some degree d in the stan- dard coordinates in R2. The set of all such functions forms a finite-dimensional vector space Pd, and we consider all generic properties relative to Pd for some d > 0 with the usual Cartesian topology and measure. A function K : S1 → R3 is a knot if it is injective, and since this is a generic property, we are justified in referring to polynomial functions in general position as polynomial knots. More generally, we may define k(S1) to be the disjoint union of k unit circles, consider the vector space Pd,k of k- tuples of polynomial functions, and define a link to be an in- jective function from k(S1) → R3 for some k. Figure 1 The concept of a polynomial link is not an essential ingre- dient in this paper, but the following lemmas, whose proofs are easy, make it a useful one: π V. L(·,b) has all of the previous properties at all but one Lemma 2.2. Given an arbitrary smooth function f : k(S1) → point of S2 and has three of the previous properties at R3, there is a sequence of polynomial links (of varying degree) the remaining point p. In this case, as a varies from one π whose values and first derivatives converge uniformly to those side of b to the other, the structure of L(·,a) near p is of f. We may choose the sequence to be in general position. characterized by one of the corresponding diagrams in Figure 1. Lemma 2.3. A property P of members of a finite-dimensional Proof. We define the algebraic dimension of a subset S of a vector space is a polynomial property or an algebraically vector space V to be the Krulldimension of the ring of polyno- generic property if there exists some non-trivial polynomial p mial functions restricted to S. (The Krull dimension of a com- on the vector space such that P is true at all points for which mutative ring is the maximum length of an ascending chain of p is non-zero. All polynomial properties are generic. prime ideals [3].) We will need two basic facts about algebraic dimension: The algebraic dimension image of a set S under a If L is a polynomial link with k components, we define a projection (or more generally a polynomial map) is less than projection function π : k(S1) × k(S1) − ∆ → S2, where ∆ is L or equal to the algebraic dimension of S, and the complement the diagonal, by: of a set of algebraic codimension 1 or more is a polynomial property. In the following discussion we will also use codi- π (a,b) = (L(a) − L(b))/|L(a) − L(b)|. L mension to mean the difference of the dimension of a pair of π π sets. We view L as a family of maps L(·,b) parameterized by the second variable. For simplicity, we consider only the case of knots. Observe R3 The main result of this section is the following lemma. Nei- that in the vector space of ordered quadruples of points in , ther the lemma nor the proof have more mathematical content the subset for which the four points are collinear has algebraic than equivalentlemmas in [5] and [6], and the key idea is orig- codimension 4. Given four points a, b, c, and d on the unit inally due to Reidemeister [7], so the proof here is sketched to circle, the space of knots K of degree d (for d ≥ 2) projects R3 some extent. onto the space of quadruplesof points in . Therefore the set of knots K of degree d for which K(a), K(b), K(c), and K(d) π Lemma 2.4. With L and L defined as above, it is a polyno- are collinear has codimension 4 as well, as does the analogous mial property for L to be a smooth embedding, i.e. its deriva- set in the space of quintuples (K,a,b,c,d), where a, b, c, and tive does not vanish anywhere. It is also a polynomialproperty d are four distinct pointson the circle. By projection,the setof of L for there to exist a finite set of points of k(S1), called the pairs (K,a) for which there exists b, c, and d such that K(a), set of special points, whose complement is the set of generic K(b), K(c), and K(d) are collinear has codimension at least points, such that for a generic point a and a special point b: 1. Except for an algebraic subset of the set of knots, the set of a for a knot K for which b, c, and d can be found with this π I. L(·,a) is a smooth immersion of a 1-manifold with ends, property is polynomial, i.e. finite. Such a b, c, and d would where the ends correspond to the tangent directions of π have to exist in order for L(·,a) to be three-to-one. Thus, part L ata. III of the lemma is proved for knots. The rest of the lemma can proved in the same fashion, II. π (·,a) does not pass through the two tangent directions. L namely by keeping track of the codimension of certain sets. π Informally, a set of algebraic codimension n can be called an III. L(·,a) is everywhere one-to-one or two-to-one. n-fold coincidence. Parts I and II of the lemma hold because, π 2 IV. If L(·,a) is two-to-one at a point of S , it is self- given points a and b on a link L, it would take a 2-fold coin- transverse at that point. cidence for the tangent to L at b to contain a, and allowing a 3 to vary, it would take a 1-fold coincidence in the choice of a, or allowing a to vary, a 1-fold coincidence for the choice of b. Part IV of the lemma holds because, given a, b, and c on a link L, it would take a 3-fold coincidence for a, b, and c to be collinear and for the tangent lines at b and c to be coplanar. For part V, the case when condition IV of the lemma fails typifies the method of proof. Informally, at a special point π a for which L(·,a) is somewhere three-to-1, three arms of the projection of the link meet at a point and it would take a coincidence for there to be a fourth arm at the point or for (a) two of the arms to have the same slope. Near a the front two arms cross at a point and it would take a coincidence for that crossing to travel parallel to the third arm instead of passing through it. Geometrically, it would take a 6-fold coincidence for five given points on a link L to be collinear, and it would take a 5- fold coincidence for four given points on L to be collinear and for two of the tangent lines to be coplanar. So in either case it would take a 1-fold coincidence in the choice of L for such a set of points to exist. Finally, consider collinear four points a, b, c, and d on L and let la, lb, lc, and ld be the tangent lines at these points. The set of lines that intersect la, l , and lc sweeps b (b) out a surface, and it is a 1-fold coincidence in the choice of ld for it to be tangent to that surface. If it is not tangent, then π L(·,a) will look as it does in case IV of Figure 1. Figure 2

3. KNOTS IN GENERAL POSITION points (a,b) with the property that at least one point of K lies The arguments in this section follow that of [6] and [5]. between K(a) and K(b). Lemma 2.4 has implications about The only new feature is the notion of topological non-trivial the local structure of O. For fixed a, the set La of all (a,b) in quadrisecants, which we will need to generalize the main the- M is a line segment which wraps around M as in Figure 2(a). orem to arbitrary knots. The intersection O ∩ Lb is a finite set. If b is a generic point, π We begin with a simple lemma and a definition: the topology of K(·,a), and therefore the topology of O ∩ La, cannot change as we vary a slightly. But if a is a special point, Rn Lemma 3.1. Let C be a compact set in . Then not every the topology of O ∩ La changes as illustrated in Figure 2(b). point of C lies between two other points ofC. For example, if a is a special point at which condition IV of Lemma 2.4 for π (·,a) fails, then there exist three points b, c, Proof. If p is any point in Rn, then a point q ∈ C which is K and d so that a, b, c, and d, in that order, make a quadrisecant farthest from p has this property, because if q lay between of K. The trisecants a, c, d and a, b, d represent the same two other points, one of them would be still farther away. point of O, and if condition V of Lemma 2.4 holds, they rep- resent arms of O that cross. Meanwhile the trisecant a, b, c Definition 3.2. A secant of a link L with no extra interior in- represents a point of O that lies elsewhere along La. tersections with L is topologically trivial if its endpoints lie on It follows that O is the image of a self-transverse smooth the same component of L, and if it, together with one of the immersion of a 1-manifold, and a self-crossing corresponds two arcs of this component, bounds a disk whose interior does to a quadrisecant. If C is a curve of points in O which does not intersect L. The disk may intersect itself and the secant. not “make turns” at the self-crossings, then C is a continuous A quadrisecant ad with middle points b and c is topologically curve of trisecants. trivial if any of the secants ab, bc, and cd are. Similarly for a The significance of O is that it is an obstruction to the fol- trisecant. lowing construction: Recall that on a M¨obius strip, there are two kinds of properly embedded arcs, non-separating arcs and Lemma 3.3. A knot in general position has a topologically separating arcs. Suppose that A is a non-separating arc of M non-trivial quadrisecant. which avoids O. Then A corresponds to a family of secants whose interiors do not intersect K. This family of secants in- Proof. Let K be a polynomial knot in general position. Let M duces a map D from the unit disk to R3 whose boundary is K be the set of unordered pairs of points of S1, or equivalently and whose interior does not intersect K. By Dehn’s lemma, K the set of secants of K. M is topologically a M¨obius strip. is trivial. We define O to be the subset M consisting of those pairs of Suppose that K has no quadrisecants. Then O is an em- 4 bedded 1-manifold. By elementary homology theory, if O ob- apply the proof of the previous lemma. But the middle points structs all non-separating arcs, there is a circular component may lie on some other component H of L. In this case, the C of O which winds around M either one or two times. The secants of C induce a map f from a surface E to R3, where E curve C is a continuous family of trisecants. We consider the is either an annulus or a M¨obius strip, depending on whether 1 corresponding families of points {a,b}t and mt , with t ∈ S , C winds once or twice around K. We may choose f so that the such that K(mt ) lies between K(at ) and K(bt ). If C winds median of E maps to the middle points of the trisecants of C. once around M, the endpoints travel half way around S1 and The set of lines l perpendicular to H at a given point p is then switch places, and since mt is trapped between them, it homeomorphic to a circle, and the corresponding set T of all must jump discontinuously, a contradiction. If C winds twice ordered pairs (l, p) is homeomorphic to a torus. We may or- around M, the endpointseach wind once around S1, and there- thogonally project each trisecant t ∈ C to a line perpendicular fore so does mt . Thus, every point of K lies between two other to H, i.e. a memberof T , thereby obtaining a map f from C to points, which contradicts Lemma 3.1. T. Since C is a circle, this map has an ordered pair of winding ω ω Topological non-triviality is achieved by a modification of numbers ( 1, 2) which are well-defined up to an orientation this construction. Let O′ be the subset of O consisting of of C. There are three cases to consider, depending on the val- topologically non-trivial trisecants and quadrisecants which ues of the winding numbers. are non-trivial at the middle secant. Observe that O′ is also ω ω Suppose that 1 = 2 = 0. We construct a disk whose the image of a smooth immersion: If O′ contains a self- boundary is K and whose interior avoids L. The map f inter- intersection point of O but does not contain all four arms of sects H at the median, but it may also intersect K at some other the self-intersection, then it must contain exactly two arms, points, because C may include some quadrisecants which are and they must be opposite rather than adjacent. In this case topologically trivial on one side. In this case we can modify the self-intersection point is a quadrisecant which is topolog- f according the prescription in Figure 3(a) to obtain a map f ′ ically trivial on one side. Therefore if O′ has a self-crossing, which avoids K in the interior and which agrees with f in a it corresponds to a topologically non-trivial quadrisecant. If neighborhoodof the median. Since both winding numbers are there are no such quadrisecants, O′ must also have a circular zero, we may now homotop f ′ in a neighborhood of H to ob- component C with all of the properties mentioned above, pro- tain a map f ′′ which avoids H and is constant on the median vided that O′ is also an obstruction to all non-separating arcs of E. Finally, we identify the median of E to a point to obtain A. a space E′ and a map f ′′′. If E is a M¨obius strip, E′ is a disk, Let A be a non-separating arc which avoids O′. We may but if E is an annulus, E′ is two disks identified at a point. Ei- choose A to be transverse to O. As before, we construct the ther way, we obtain the desired spanning disk, which we may disk DA from the secants of A, but this time DA does not avoid convert to an embedded disk by Dehn’s Lemma. K. Consider a point where A intersects a topologically triv- ω ω Suppose instead that 2 = 0 but 1 6= 0. Then we extend E ial trisecant T . By hypothesis there exists a disk DT which to a line bundle E′ and extend f linearly to a map f ′ : E′ → R3. bounds a secant of T and an arc of K. Using DT and a tubular ′ ′ We can homotop f in a neighborhood of H without changing neighborhoodof K, we mayalter DA to obtain a disk DA which its values in E′\E to a map f ′′ which has constant value p on avoids K in the vicinity of T , according to Figure 3(a). We the zero section of E′, but we cannot make f ′′ avoid H. Let may similarly modify DA in the vicinity of a quadrisecant Q p ∈ H. As before, we identify the zero section of E′ to a point which is topologically trivial in the middle, as in Figure 3(b). and obtain a space E′′, and correspondingly alter f ′′ to obtain In this fashion we obtain a disk D whose interior avoids K as a map f ′′′ : E′′ → R3. This time the intersection number be- before, and Dehn’s lemma applies. ′′′ ω ′′′ tween H and f at p is 1. But since f is a closed map from the pseudo-manifold E′′ to R3, it induces a well-defined homology class in the infinite homology of R3. H induces 4. LINKS IN GENERAL POSITION another such homology class, and by elementary homology theory, the total intersection number between f ′′′ and H must The result of this section is a completion of analogous re- be zero. The map f ′′′ must intersect H at another point, and sults in [5] and [6]. The arguments there roughly correspond therefore f ′′ does also. Suppose that f ′′(x) is this point, with to the omega1 6= 0 case of the proof, although the argument in x ∈ E′. The point x cannot be in E, therefore f ′(x)= f ′′(x). [6] is somewhat more general than this special case. Since f ′ is linear on the fibers of E′, the image under f ′ of the Lemma 4.1. Every non-trivial link L in general position has fiber containing x yields a quadrisecant Q. The quadrisecant a topologically non-trivial quadrisecant. Q is necessarily topologically non-trivial, because if the inter- section points of Q are labeled in order as a,b,c, and x, then Proof. We may assume without loss of generality that no com- b,x ∈ H and a,c ∈ K. ω ponent of L bounds a disk whose interior avoids L. The only remaining possibility is that 2 6= 0. In this case, Let K be a component of L. Let MK be the M¨obius strip of every point of H lies between two points of K. We may repeat ′ secants of K, and let OK be the corresponding set of topolog- the whole argumentwith each componentof L playing the role ically non-trivial trisecants and quadrisecants which are non- of K, thereby obtaining a function f from components of L to ′ trivial in the middle. As before, OK must be an obstruction to components of L such that every point of f (K) lies between non-separating arcs A, and we obtain a circle C which winds two points of K. The map f must have at least one circular around M. If the middle points of C also lie on K, we may orbit, and we may set C to be the set in R3 which is the union 5

(a)

(b)

Figure 3

′ R3 R3 of all components of L in this orbit. Evidently, C is a compact T1 to obtain a diffeomorphism h : − K → − h(K). Be- set and every point of C lies between two other points of C,a cause of the distance condition, we can continuously extend h′ contradiction by Lemma 3.1. to K by setting it equal to h on K. This continuous extension is the desired map. The proof in the case of links is similar. 5. ARBITRARY TAME KNOTS AND LINKS We are now in a position to prove Theorem 1.3. In fact, we Definition 5.1. A link L in R3 is tame if there exists a home- can prove something slightly stronger: omorphism h of R3 which carries L to a polynomial link, or equivalently a piecewise linear or smooth link. Theorem 5.3. If L is a non-trivial tame link in R3, then L has a quadrisecant, none of whose component secants are subsets of L. Lemma 5.2. If L is a tame link, there exists a homeomor- R3 phism h of which maps L to a smooth link with h smooth Proof. Let L be a non-trivial, tame link and let h be a homeo- R3 on − L. morphism given by Lemma 5.2. Let N be a tubular neighbor- ′ Proof. Let K be a tame knot and let h be an arbitrary homeo- hood of h(L), let N be the normal bundle of h(L), and choose ′ morphism such that h(K) is smooth. Using a tubular neighbor- a diffeomorphism n : N → N . Consider a sequence of links Li such that h(L ) is disjoint from L and n(h(L )) is a smooth sec- hood of h(K), we can choose Ti, with i ≥ 1, to be a sequence i i ′ −1 tion. Choose the sequence so that h(Li) converges smoothly of nested, parallel tori converging to h(K). Let Ti = h (Ti). By the theory of triangulationsand smoothingsof 3-manifolds to h(L), i.e. n(h(Li)) converges smoothly to the zero section. ′′ Since h is a diffeomorphismoutside of L, we may choose each (see [4, p. 217]), there exists a sequence of smooth tori Ti , ′′ ′ ′ Li to be a polynomial link in general position. with each Ti lying between Ti and Ti+1, and a sequence of ′ ′′ ′−1 −1 By hypothesis, each Li has the same isotopy type as L, and diffeomorphisms hi : Ti → Ti such that h i and h |T are iso- i in particular each Li is non-trivial. Therefore each Li has a R3 topic as maps from Ti to −K. Furthermore, we can arrange topologically non-trivial quadrisecant Qi. By compactness, ′−1 −1 ∞ that the distance between h i and h goes to zero as i → . {Qi} has a convergent subsequence in the space of line seg- ′ R3 By the isotopy condition, the hi’s may be extended smoothly ments in ; we may suppose without loss of generality that ′ ′ to the each region between Ti and Ti+1 and the region outside the original sequence converges. The resulting limit is a se- 6

At least one of these loops must be non-contractible,therefore they cannot both be shorter. Therefore each component secant of Q must have one point which lies outside the Ti’s. Finally, suppose that P(x,y,z) is a non-trivial polynomial ∂ whose zero set contains Ti for all i. Then the restriction of P to the line containing Q must be non-trivial and must have at least 8 real roots, counting multiplicity. Therefore P has degree at least 8. The authoronce believedthat the loop in a solid toruswhich is the shortest non-zero multiple of the core is necessarily ho- motopic to the core. However, this is false by an example of Doug Jungreis. We can consider the region S in R3 which Figure 4 consists of the set of points (x,y,z) such that: 2 2 ε |x − sin(L1y)/L1 − sin(L2z)/L2| < , where L is very large, L is much larger still, and ε is much cant of L. We must show that the endpoints and middle points 1 2 smaller than 1/L . The region S could be described as a cor- of the quadrisecants do not converge together. 2 rugated sheet, and it has the property that if a,b ∈ S and the For each i, let Si be a topologically non-trivial secant of Li straight-line distance from a to b is greater than 1, then this and suppose that the Si’s converge to a point p on L. Let B ′ distance is much less than the length of the shortest path in S be a round, open ball in N centered at n(h(p)). Then there from a to b. If M is a smooth M¨obius strip in R3 whose tan- exists an i such that Si and an arc A of Li with the same end- gent plane varies slowly, we can approximate M with a solid −1 −1 points as Si are both contained in h (n (B)). For each point torus T which is topologically a tubular neighborhood of M s ∈ n(h(Si)), we consider the line segment from s to t, where but which is geometrically quite different: T is the union of ′ t is the point in n(h(Li)) which lies in the same fiber of N a thick tube centered around the boundary of M and a corru- as s, as illustrated in Figure 4. Since n(h(Li)) is a section, gated sheet which approximates the interior of M, as shown in t is unique. The union of these line segments is the image Figure 5. Clearly the shortest non-trivial loop in T stays close of a spanning disk of n(h(A ∪ Si)) which does not intersect to the boundary of M and is therefore homotopically twice the n(h(Si)). Therefore Si is topologically trivial, a contradiction. core. The proof that the limit of the Si’s is not a subset of L is It is easy to show that the bound in Corollary 1.4 is the best similar. possible: If we choose two numbers r1 > r2, then the surface given by: Corollary 1.4 follows from this theorem: 2 2 2 2 2 2 2 2 2 (x + y + z − r1 − r2) − 4(x + y )r1 = 0 Proof. Let {Ti} be a non-trivially linked collection of solid is a torus. If r > 2r , we can multiply two such surfaces tori. For each i and each n > 0, let li,n be the shortest non- 1 2 together to obtain two linked tori. contractible loop in Ti which is homotopically n times the core of Ti. Let li be a shortest member of the set {li,n}. Ifwe let D and D′ be two disjoint, non-separating disks in T for some i, i 6. QUESTIONS OPEN TO THE AUTHOR then we see that the length of li,n is bounded below by n times ′ the distance between D and D . Therefore li exists, although it may not be unique. The most serious shortcoming of Corollary 1.4 is the fact R3 Suppose that for some a b S1, l a l b . Then we that it only applies to closed surfaces in , while the usual , ∈ i( )= i( ) R 3 can divide l into two loops from l a to itself. At least one context for studying degrees of real algebraic surfaces is P . i i( ) R3 R 3 of these loops must be non-contractible and both loops are We may view a subset of as a subset of P which is dis- R 2 shorter, which is a contradiction. Thus, each l is an embed- joint from the “plane at infinity”, which is a copy of P . We i R 3 ding. If we let L be the union of the images of the l ’s, then L may define a flat plane in P to be the image of the plane at i infinity under a projective transformation of RP3, anda topo- is a satellite link of the Ti’s. By a theorem in knot theory [8, p. 113], L must be a non-trivial link if the T ’s are. logical plane to be the image of the plane at infinity under i R 3 Since a geodesic in a smooth manifold with smooth bound- a homeomorphism of P . This brings us to the following ary must be C1 (see [1]; a proof was also suggested to the generalization of the results of this paper: author by Tom Ilmanen), L must be a tame link. By the pre- Conjecture 6.1. If a non-trivial link in RP3 is disjoint from ceding theorem, L must have a quadrisecant Q such that no some topological plane, then it has four collinear points. component secant of Q is contained in L. Suppose that a com- R 3 ponent secant S of Q were contained entirely inside some Ti. Conjecture 6.2. If an algebraic surface in P is disjoint Let p be a path which goes from one endpoint of S to the from some topological plane and bounds a collection of non- other. Then we can divide li into two paths q1 and q2 to make trivially linked solid tori, then the surface has degree at least two loops q1 p and q2 p whose composition is homotopic to li. 8. 7

Figure 5

The following questions have also eluded the author: quadrisecants. Conjecture 6.3. If an algebraic surface in R3 is a smooth torus which is knotted on the outside, then it has degree at Question 6.5. What is the lowest possible degree of a poly- least eight. nomial surface in R3 which is the boundary of the tubular neighborhood of a trefoil knot? Conjecture 6.4. Every wild arc in R3 has infinitely many

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