On polynomial Pierre-Vincent Koseleff, Daniel Pecker

To cite this version:

Pierre-Vincent Koseleff, Daniel Pecker. On polynomial Torus Knots. Journal ofKnot Theory and Its Ramifications, World Scientific Publishing, 2008, 17 (12), pp.1525-1537. ￿10.1142/S0218216508006713￿. ￿hal-00108510v2￿

HAL Id: hal-00108510 https://hal.archives-ouvertes.fr/hal-00108510v2 Submitted on 18 Oct 2007

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2 The principal result

If n is odd, the Kn of type (2,n) is the boundary of a Moebius band twisted n times (see [Re, Ka, St]). The recently proved Tait’s conjectures allow us to characterize plane projections

Figure 1: Kn, n = 3, 5, 7. of Kn with the minimal number of crossings. Lemma 1 Let be a plane curve with n crossings parametrized by (t) = (x(t),y(t)). If is the C C C projection of a knot Kn then there exist real numbers s1 < < sn < t1 < < tn, such that · · · · · · (si)= (ti). C C Proof. Let be a plane projection of a knot of type Kn with the minimal number n of crossings. C Using the Murasugi’s theorem B ([Mu]) which says that a minimal projection of a prime alter- nating knot is alternating, we see that is alternating. C Then the Tait’s flyping conjecture, proved by Menasco and Thistlethwaite ([MT, Pr]), asserts that is related to the standard diagram of Kn by a sequence of flypes. Let us recall that a flype C is a transformation most clearly described by the following picture.

Figure 2: A flype

The standard diagram 0 of Kn has the property (cf [Re]) that there exist real numbers s1 < < S · · · sn

sa

s s′ A B C A C ′ t t B

Figure 3: on the part B

After the flype on B, we have new parameters corresponding to the crossing points satisfying

′ ′ ′ ′ ′ ′ ′ ′ sa

′ The transformed diagram has the same property: there exist real parameters s1 < < sn < ′ S ′ · · · t1 <

In this paper we shall consider polynomial knots, that is to say, polynomial embeddings IR 3 3 −→ IR , t (x(t),y(t), z(t)). Polynomial knots are non-compact subsets of IR . The closure of a 7→ 3 3 polynomial knot in the one point compactification S of the space IR is an ordinary knot (see [Va, Sh, RS] and figures at the end).

Lemma 2 Let be a plane polynomial curve with n crossings parametrized by (t) = (x(t),y(t)). C C Suppose that is the projection of Kn and deg x(t) deg y(t). Then we have deg x(t) 3. If C ≤ ≥ deg x(t) = 3, then deg y(t) n + 1. ≥ Proof. x(t) must be non-monotonic, so deg x(t) 2. Suppose that x(t) is of degree 2. Then ≥ x(ti) = x(si) implies that ti + si is constant, and so the parameter values corresponding to the crossing points are ordered as

s1 <

3 Proof of the main result

Our proof makes use of Chebyshev (monic) polynomials. 4 P. -V. Koseleff, D. Pecker, On polynomial Torus Knots 3.1 Chebyshev Polynomials sin((n + 1)θ) Definition 1 If t = 2 cos θ, let Tn(t) = 2 cos(nθ) and Vn(t)= . sin θ

Remark 1 Tn and Vn are both monic and have degree n. We have

V0 = 1, V1 = t, Vn+1 = t Vn Vn−1. (1) − We have also T0 = 2, T1 = t, Tn+1 = t Tn Tn−1. − n n−2 n−4 For n 2, let Vn = t + ant + bnt + . Using recurrence formula 1, we get ≥ · · ·

an+1 = an 1, bn+1 = bn an−1 − − so by induction, n n−2 1 n−4 Vn = t (n 1)t + (n 2)(n 3)t + . (2) − − 2 − − · · · We shall also need the following lemmas which will be proved in the next paragraph.

Lemma A. Let s = t be real numbers such that T3(s)= T3(t). For any integer k, we have 6 Tk(t) Tk(s) 2 kπ − = sin Vk−1(s + t). t s √3 3 −

Lemma B. Let n 3 be an integer. ≥ Let s1

3.2 Proof of the theorem Proof. We shall prove this result by reducing it to a contradiction. Suppose the plane curve C parametrized by x = P (t), y = Q(t) where deg P = 3, deg Q = n + 1 is a plane projection of Kn. By translation on t, one can suppose that P (t)= t3 αt+β. If the polynomial P was monotonic, − would have no crossings, which is absurd. Therefore α > 0. Dividing t by ρ = √3/√α, one C 3 3 has P (t) = ρ (t 3t)+ µ. By translating the origin and scaling x, one can now suppose that 3 − P (t)= t 3t = T3(t). − By translating the origin and scaling y, we can also suppose that Q(t) is monic and write

P (t)= T3(t), Q(t)= Tn+1(t)+ anTn(t)+ + a1T1(t). · · · By B´ezout theorem, the curve has at most (3 1)(n + 1 1)/2= n double points. As it has at C − − least n crossings, we see that it has exactly n crossings and therefore is a minimal crossing diagram of Kn. According to the lemma 1, there exist real numbers s1 < < sn, t1 < < tn, si < ti · · · · · · such that P (si)= P (ti), Q(si)= Q(ti). Let ui = ti + si, 1 i n. We have ≤ ≤ n Q(ti) Q(si) Tn+1(ti) Tn+1(si) Tk(ti) Tk(si) − = − + ak − . ti si ti si =1 ti si − − Xk − Accepted for publication in J. of Knot Th. and Ram. 5 so by lemma A, u1,...,un are the distinct roots of the polynomial n R(u)= εn+1 Vn(u)+ akεk Vk−1(u), (3) =1 Xk where 2 kπ εk = sin . √3 3

Remark 2 Note that εk = Vk−1(1) is the 6-period sequence ε0 = 0, ε1 = 1, ε2 = 1, 0, 1, 1,.... − − We have to consider several cases.

£ Case n 2 mod 3. εn+1 = 0 and R(u) has degree at most n 1. This is a contradiction. ≡ − £ Case n 1 mod 3. In this case, n 1 mod6 and εn+1 = εn = 1, εn−1 = 0. Thus R(u) can ≡ ≡ be written as

R(u) = Vn(u)+ anVn−1(u) an−2Vn−3(u) + a2V1(u)+ a1 n n−1 − n−2 −··· = u + anu (n 1)u + . − − · · · using equation 2. Therefore we get

ui = an, uiuj = (n 1), − − − 1≤i≤n 1≤i 6. i=1 i ≤ £ Case n 0 mod 3. In this last case we have n =3 mod 6,so εn+1 = 1, εn = 0 and εn−1 = 1, ≡ P − so

R(u) = Vn(u) an−1Vn−2(u) an−2Vn−3(u)+ a2V1(u) a1. − − − ···− − Let σi be the coefficients of n n k n−k R(u)= u + ( 1) σku . − =1 − Xk From the equation 2, we see that (n 2)(n 3) σ1 = 0, σ2 = (an−1 + n 1), σ4 = (n 3) an−1 + − − . − − − 2 n k Let Sk be the Newton sums i=1 ui of the roots of the polynomial R. Using the classical Newton formulas ([FS]), we obtain P 2 2 S1 = σ1 = 0, S2 = σ 2σ2 = 2σ2, S4 = 2σ 4σ4, 1 − − 2 − and then 2 S4 = 2(an−1 + 2) + 6n 18 6n 18. − ≥ − By the lemma B, we deduce that 22 + n 6n 18, i.e. n 8 so n = 3. 2 ≥ − ≤ 6 P. -V. Koseleff, D. Pecker, On polynomial Torus Knots 3.3 Proof of lemmas A and B We shall use the following lemma Lemma 3 (Lissajous ellipse) Let s = t be complex numbers such that 6 T3(t)= T3(s).

There exists a complex number α such that

s = 2 cos(α + π/3), t = 2 cos(α π/3). − Furthermore, α is real if and only if s and t are both real, and then t>s if and only if sin α> 0. Proof. We have T3(t) T3(s) − = t2 + s2 + st 3. (4) t s − − Then, if T3(t)= T3(s), t = s, we get 6 3 1 (t + s)2 + (t s)2 = 2(t2 + s2 + st) = 6. 2 2 −

That means (s,t) 2 2 2 t + s t s + − = 1. 2 √    2 3  1 Then there exists a complex number α such that t + s t s cos α = , sin α = − , (s, 2 cos α) 2 2√3 ±2 ±1 1 2 that is t = 2 cos(α π/3), s = 2 cos(α + π/3). ±1 − α is real if and only if cos α and sin α are both real that is to say, iff s and t are real. In this case: t>s sin α> 0. 2 ⇔ ±2 In order to prove lemma B, we shall use the following lemma which describes the geometrical configuration. Let us denote s(α) = 2 cos(α + π/3) and t(α) = 2 cos(α π/3). − Lemma 4 Let α, α′ [0,π] be such that s(α) < s(α′), and t(α) < t(α′). Then α > α′ and 2π α + α′ π ∈ > > . 3 2 3 Proof. We have 2 cos α = s(α)+ t(α) α′. α + α′ π α α′ t(α′) t(α) = 4sin( ) sin( − ) > 0, − 2 − 3 · 2 α + α′ π α α′ s(α′) s(α) = 4sin( + ) sin( − ) > 0. − 2 3 · 2 α + α′ π α + α′ π From α > α′ we get 0 < and + <π, that is to say 2 − 3 2 3 π α + α′ 2π < < . 3 2 3 Accepted for publication in J. of Knot Th. and Ram. 7 2

Proof of lemma B. Let s1 < > αn ]0,π[ such that si = s(αi), ti = t(αi) and we have · · · ∈ 2π α1 + α2 αn−1 + αn π > > α2 > > αn−1 > > . 3 2 · · · 2 3

At least two of the αi’s lie in the intervals ]0, π/2] or [π/2,π[. We have only two cases to consider: π π π > α1 > α2 , or αn−1 > αn > 0. ≥ 2 2 ≥ On the other hand, we get the equality

cos2 x + cos2 y = 1 cos2(x + y) + 2 cos x cos y cos(x + y). (5) −

π £ Case 1. αn−1 > αn > 0. 2 ≥ 1 We get cos αn 0, cos αn−1 0 and cos(αn−1 + αn) < so eq. 5 becomes ≥ ≥ −2

2 2 2 3 cos αn−1 + cos αn 1 cos (αn−1 + αn) ≤ − ≤ 4 and

n n−2 2 2 2 2 2 cos αi = cos α1 + cos αi + (cos αn−1 + cos αn) i=1 i=2 X X 1 3 1 1 + (n 3) + = (n + 4), ≤ − · 4 4 4 that is n n 2 2 S2 = u = (2 cos αi) n + 4. i ≤ i=1 i=1 X X π £ Case 2. π > α1 > α2 . ≥ 2 1 We get cos α1 0, cos α2 0 and cos(α1 + α2) < so eq. 5 becomes ≤ ≤ −2

2 2 2 3 cos α1 + cos α2 1 cos (α1 + α2) ≤ − ≤ 4 and similarly, we get n 2 S2 = (2 cos αi) n + 4. ≤ i=1 X Analogously, we get cos4 x + cos4 y (cos2 x + cos2 y)2, and we deduce: ≤ n 4 S4 = (2 cos αi) n + 22. ≤ i=1 X 8 P. -V. Koseleff, D. Pecker, On polynomial Torus Knots 2

Proof of lemma A. Let s

t = 2 cos(α π/3), s = 2 cos(α + π/3). − π We have s + t = 2 cos α and t s = 4sin sin α, so − 3 kπ sin kα sin Tk(t) Tk(s) 2 cos k(α π/3) cos k(α + π/3) · 3 − = − π− = π t s 4sin sin α  sin α sin − 3 · 3 2 kπ = sin Vk−1(2 cos α). √3 3 2

4 Parametrized models of K3,K5,K7 and K9

We get parametrizations of Kn: = (x(t),y(t), z(t)), with n crossings obtained for parameter C values satisfying the hypothesis of lemma 1. According to lemma A, we choose n distinct points 1 u1 < < un 1. We look for Q1 and Q2 of minimal degrees, such that − ≤ · · · ≤ Q1(t) Q1(s) Q2(t) Q2(s) R1(t + s)= − , R2(t + s)= − t s t s − − satisfy, for i = 1,...,n, i R1(ui) = 0, R2(ui) = ( 1) . − We then choose y(t) = Q1(t) and z(t) = Q2(t). We also add some linear combinations of T6i efficiently. We then obtain a knot whose projection is alternating, when R1 has no more roots in [ 2, 2]. As we have chosen symmetric ui’s, all of our curves are symmetric with respect to the − y-axis.

4.1 Parametrization of K3

We can parametrize K3 by x = T3(t),y = T4(t), z = T5(t). It is a Lissajous space curve (compare [Sh]). The plane curve (T3(t), T4(t)) has 3 crossing points. The plane curve (T3(t), T5(t)) has 4 crossing points corresponding to parameters (si,ti) with

s1

bottom view of K3 face view of K3

Figure 4: The K3

4.2 Parametrization of K5 Let us consider the curve of degree (3, 7, 8):

x = T3(t),

y = T8(t) 2 T6(t) + 2.189 T4(t) 2.170 T2(t), − − z = T7(t) 0.56 T5(t) 0.01348 T1(t). − − The curve (x(t),y(t)) has exactly 5 double points when the projection (x(t), z(t)) has exactly 6. Note here that deg z(t) < deg y(t). In conclusion we have found a curve of degree (3, 7, 8). Using

bottom view face view

Figure 5: Knot K5 our theorem, we see that this curve has minimal degree. A. Ranjan and R. Mishra showed the existence of such an example ([RS, Mi]).

4.3 Parametrization of K7 We choose

x = T3 (t) ,

y = T10 (t) 2.360 T8 (t) + 4.108 T6 (t) 6.037 T4 (t) + 7.397 T2 (t) , − − z = T11 (t) + 3.580 T7 (t) 3.739 T5 (t) T1 (t) . − − 10 P. -V. Koseleff, D. Pecker, On polynomial Torus Knots

Bottom view Zoom on the bottom view

Figure 6: Knot K7

The values of the parameters corresponding to the double 2 points are obtained as intersection points between the ellipse

t2 + s2 + st 3 = 0 1.5 −

t and the curve of degree 9: 1 (y(t) y(s))/(t s) = 0 − − 0.5 The curve (x(t),y(t)) has exactly 7 double points correspond- ing to cos(α)= 1/2, 3/10, 2/10, 0 . {± ± ± } In conclusion we have found a curve of degree (3, 10, 11). ±2 ±1.5 ±1 ±0.5 0 s Using our theorem, this curve has minimal degree.

4.4 Parametrization of K9 We choose polynomials of degree (3, 13, 14).

x = T3 (t) ,

y = T14 (t) 4.516 T12 (t) + 12.16 T10 (t) 24.46 T8 (t) + 39.92 T6 (t) − − 55.30 T4 (t) + 66.60 T2 (t) , − z = T13 (t) 2.389 T11 (t) 5.161 T7 (t) + 5.161 T5 (t) + 1.397 T1 (t) . − − The curve (x(t),y(t)) has exactly 9 double points corresponding to cos(α)= 1/2, 3/10, 2/10, 1/10, 0 . {± ± ± ± } One can prove that it is minimal under the assumption that the projection (x(t),y(t)) has exactly 9 double points.

Conclusion

We have found minimal degree polynomial curves for torus knots Kn, n = 3, 5, 7. For degree 9, one can prove that it is minimal under the assumption that the projection (x(t),y(t)) has exactly 9 double points. We have similar constructions for higher degrees.

Acknowledgments

We would like to thank Julien March´efor fruitful discussions on . Accepted for publication in J. of Knot Th. and Ram. 11

Bottom view Zoom on the bottom view

Figure 7: Knot K9

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