# Topology of maximally writhed real algebraic knots

Oleg Viro introduced an invariant of rigid isotopy for real algebraic knots in which can be viewed as a first order Vassiliev invariant. In this paper we look at real algebraic knots of degree with the maximal possible value of this invariant. We show that for a given all such knots are topologically isotopic and explicitly identify their knot type.

^{†}

^{†}support: Research is supported in part by the SNSF-grants 159240, 159581 and the NCCR SwissMAP project (G.M.) and by RSF grant, project 14-21-00053 dated 11.08.14. (S.O.)

## Introduction

A real algebraic curve in is a (complex) one-dimensional subvariety in invariant under the involution of complex conjugation , . The conj-invariance is equivalent to the fact that can be defined by a system of homogeneous polynomial equations with real coefficients. The degree of is defined as its homological degree, i. e. the number such that . A curve of degree intersects a generic complex plane in points.

We denote the set of real points of by . We say that a real curve is smooth if it is a smooth complex submanifold of . In this case, is a smooth real submanifold of and if it is non-empty, we call it a real algebraic link or, more specifically, a real algebraic knot in the case when is connected.

Two real algebraic links are called rigidly isotopic if they belong to the same connected component of the space of smooth real curves of the same degree. A rigid isotopy classification of real algebraic rational curves in is obtained in [1] up to degree 5 and in [2] up to degree 6. Also we gave in [2] a rigid isotopy classification for genus one knots and links up to degree 6 (here we speak of the genus of the complex curve rather than the minimal genus of a Seifert surface of ).

In all the above-mentioned cases, a rigid isotopy class is completely determined by the usual (topological) isotopy class, the complex orientation (for genus one links), and the invariant of rigid isotopy introduced by Viro [4] (called in [4] encomplexed writhe). This invariant is defined as the sum of signs of crossings of a generic projection but the crossings with non-real branches are also counted with appropriate signs; see details in [4] (the definition of is also reproduced in [2]).

Let , , be the -torus link in the 3-sphere . If , we define the projective torus link .

Let . By the genus formula, this is the maximal possible value of for irreducible curves of degree which can be attained on rational curves only. So, if a real algebraic curve in is smooth, irreducible, and where (and hence the genus of is zero), then we call it maximally writhed or -curve. The main result is the following.

###### Theorem 1

Let be an -curve of degree , and . Then is isotopic to .

###### Corollary 1

A plane projection of an -curve from any generic real point has or real double points with real local branches.

###### Demonstration Proof

Follows from Murasugi’s result [3; Proposition 7.5] which states that any projection of a torus link , , has at least crossings. ∎

In Proposition 2 (see the end of the paper) we give a precision and a self-contained (i.e., not using [3]) proof to Corollary 1.

Conjecture 1. In Theorem 1, is rigidly isotopic to .

Conjecture 2. If an algebraic knot of degree in is isotopic to , then .

In a forthcoming paper we are going to give a proof of Conjecture 2 as well as a generalization of Theorem 1 for links of arbitrary genus.

The following differential geometric property of maximally writhed algebraic knots was communicated to us by Oleg Viro.

###### Proposition 1

Let be as in Theorem 1. Then the torsion of is everywhere positive.

## 1. -curves have everywhere positive torsion (proof of Proposition 1)

Recall that the sign of the (differential geometric) torsion of a curve , , coincides with the sign of and it does not depend on the parametrization if . The sign of the torsion of a curve in does not depend on a choice of positively oriented affine chart.

###### Lemma 1

Let be a real algebraic knot in of genus which is not contained in any plane. If the torsion of vanishes at a point , then there exists an arbitrarily small deformation of (in the class of real algebraic knots) which has points with negative torsion.

###### Demonstration Proof

We can always choose affine coordinates centered at such that a parametrization of at satisfies the condition . If for or , then can be perturbed so that and the -th derivative has any sign we want. Indeed, let be homogeneous coordinates such that , . Then the parametrization can be chosen so that , , where are real polynomials of degree , and . Then the desired perturbation of is just where and has the prescribed sign. ∎

###### Demonstration Proof of Proposition 1

By Lemma 1, it is enough to show that does not have points with negative torsion. Suppose it does. Then, in an appropriate affine chart, admits a parametrization of the form . This means that in a sufficiently small neighbourhood of the origin, the curve is approximated by a negatively twisted rational cubic curve. Hence there is a projection with a negative crossing (see [4; Section 1.4]). Since is the sum of the signs of all real crossings and the number of them is at most , a single negative crossing makes impossible to attain the equality . ∎

## 2. Uniqueness of -curves up to isotopy (proof of Theorem 1)

Let be as in Theorem 1. So, is a smooth rational curve in of degree and .

Given a point , let be the projection from and let be the restriction of to . If , then we extend to by continuity, thus where is the tangent line to at .

If , then we set . Note that

Recall that an algebraic curve (maybe, singular) of degree in is called hyperbolic with respect to a point (which may or may not belong to ), if any real line through intersects at real points counting the multiplicities. We denote:

It is easy to check that is either empty or a convex closed set. It is possible that contains only one point. In this case, the point should be singular. For example, if is a cuspidal cubic, then consists of the cusp only.

Similarly, we say that is hyperbolic with respect to a real line if, for any real plane passing through , each intersection point of and is real.

The following two properties of hyperbolic curves are immediate from the definition:

###### Lemma 2

Let be a real plane curve, , and . Then each local branch of at is smooth, real, and transverse to the line . The projection from defines a covering where is the normalization of . ∎

###### Lemma 3

Let be a real line in and . Then is hyperbolic with respect to if and only if . ∎

###### Lemma 4

If , then is a smooth point of and .

###### Demonstration Proof

By Proposition 1 the torsion of does not vanish at , hence the image of the germ under the projection is a smooth local branch of at . Suppose that has another local branch at which is the projection of a germ . Let be a real point on the line , . Then has (at least) two branches at one of whom is cuspidal. In this case by perturbing we may obtain either a non-real crossings or a pair of real crossings of opposite signs which contradicts the maximality of (cp. the end of the proof of Proposition 1). ∎

###### Lemma 5

If , then is the closure of a component of and is a smooth point of its boundary.

###### Demonstration Proof

Let . It is a smooth point of by Lemma 4. For a real singular point of , let be its contribution to , i.e. the sum of the signs of crossings of a nodal perturbation of . Then, similarly to [2; Proposition 21], we have

where are points close to on different sides of , and for is one half of the image of under the isomorphism (see [2; §6] for the choice of the orientations). It is clear that and . The sum of these two upper bounds is , thus implies the equality sign in the both estimates. It remains to note that implies .

The fact that is the closure of a component of the complement of follows from the discussion after (1) because is smooth on . ∎

Recall that the tangent line to at is denoted by . Let us set

###### Lemma 6

Suppose that is hyperbolic with respect to a real line and let . Then .

###### Demonstration Proof

Combine Lemma 2 and Lemma 3. ∎

###### Lemma 7

Let and be two distinct points on . Then .

###### Demonstration Proof

Let . Then is hyperbolic with respect to by Lemma 5 combined with Lemma 3, and we have by Lemma 4. Hence the result follows from Lemma 6. ∎

Thus is a disjoint union of a continuous family of real projective lines (topologically, circles) parametrized by . We are going to show that the pair is isotopic in to a hyperboloid with a projective torus link sitting in it. Note that is not smooth. It has a cuspidal edge along .

###### Lemma 8

There exist two real lines and such that is hyperbolic with respect to each of them, , and crosses without tangency at a pair of complex conjugate points.

Figure 1 Two perturbations of a cusp in the proof of Lemma 8\endcaption

###### Demonstration Proof

Let us choose a point and let . Then by Lemma 5 whence is hyperbolic with respect to by Lemma 3. Let . Then, again by Lemma 3, we have . The curve has a cusp at because the torsion at is nonzero. Let and be points close to and chosen on different sides of the osculating plane of at . Then and are obtained from by a perturbation of the cusp as shown in Figure 1 where is a solitary node of (a point where two complex conjugate branches cross). Then we set , , where the points and are chosen as in Figure 1. The fact that implies , , whence the hyperbolicity of with respect to by Lemma 3. ∎

###### Demonstration Proof of Theorem 1

Let and be as in Lemma 8.

The line is disjoint from by Lemma 6. Let be a real plane through . Again by Lemma 6, crosses each line , , at a single point. Let us denote this point by . Then is a continuous mapping. It is injective by Lemma 7 and its image (which is ) is disjoint from . Hence is a Jordan curve in the affine real plane . Let be the disk bounded by this Jordan curve and let where runs through all the real planes through . Then is fibered by disks over a circle which parametrizes the pencil of planes through . Since is orientable, this fibration is trivial, thus is a solid torus and . Each transversally crosses at real points, thus sits in and it realizes the homology class where is a generator of .

The same arguments applied to the line show that bounds a solid torus such that realizes the homology class where is a generator of . We conclude that the lift of on is and the result follows. ∎

We see that cuts into two solid tori and such that and .

###### Proposition 2

(Compare with Corollary 1). Let be a generic point of . Then has only real double points. If , then all the double points have real local branches and the interior of is non-empty. If , then one double point is solitary (i.e. has complex conjugate local branches), all the other double points have real local branches, and .

###### Demonstration Proof

Let us consider a generic path which relate the given point with a point on . It defines a continuous deformation of the knot diagram which is a sequence of Reidemeister moves (R1) – (R3). However, (R2) is impossible because it involves a negative crossing and (R1) smay occur only when passes through . Thus the number and the nature of double points does not change during the deformation. The projection from a point of is cuspidal and it is hyperbolic with respect to the cusp, so the result follows from Lemma 2.

Non-emptiness of the interior of in case , follows from the fact that can disappear only by a move (R3). This is however impossible because all crossings are positive and the boundary orientation on agrees with an orientation of due to Lemma 2 (see Figure 2). ∎

\botcaption

Figure 2 Impossibility of a move (R3) which eliminates \endcaption

### Acknowledgement

We are grateful to Oleg Viro for very useful discussions.

## References

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