Knots with Hopf crossing number at most one
Maciej Mroczkowski
math.GT 57M25, 57M27
                   



Institute of Mathematics

Faculty of Mathematics, Physics and Informatics

University of Gdansk, 80-308 Gdansk, Poland

e-mail: mmroczko@mat.ug.edu.pl


We consider diagrams of links in S^2 obtained by projection from S^3 with the Hopf map
and the minimal crossing number for such diagrams. Knots admitting diagrams with at most one crossing are classified.
Some properties of these knots are exhibited. In particular, we establish which of these knots are algebraic and, for such knots, give an answer
to a problem posed by Fiedler in <cit.>.

Knots with Hopf crossing number at most one
    Maciej Mroczkowski
    December 30, 2023
===========================================



Mathematics Subject Classification 2010: 57M25, 57M27 



§ INTRODUCTION

Let p:S^3→ S^2 be the Hopf map in the Hopf fibration of S^3.
For a link L in S^3, we study the minimal number of crossings of p(L) among generic projections in S^2.
Let us call this number the Hopf crossing number of L and denote it by h(L). The classical crossing number is denoted by c(L). 

The Hopf crossing number was considered by Fiedler <cit.> in the context of algebraic links coming from singularities of complex curves.
For an algebraic link L realized as the intersection of a complex plane algebraic curve X with a small sphere S^3 with a singularity of X
at its center, he considered the canonical Hopf fibration from S^3 to P^1≈ S^2, i.e. the fibration induced by the intersection of S^3 
with the complex lines through the center of S^3. He defined C_alg(L) as the minimal number of crossings under this Hopf fibration 
for all generic realizations of L as X∩ S^3.
He computed C_alg for torus knots T(p,q), 1<p<q, q<2p and found lower bounds for C_alg of other algebraic knots and links.

For an algebraic link L one has h(L)≤ C_alg(L) as the link does not have to be realized as the intersection of a complex curve with a sphere.
In fact Fiedler poses a problem: is this inequality always an equality for algebraic links?

In this paper we classify knots with h(K)≤ 1 (see Theorem <ref>) and show that, for such knots, 
the answer to Fiedler's problem is positive (see Theorem <ref>).
In order to achieve it, we use arrow diagrams introduced in <cit.>, which are equivalent to 
gleams introduced in <cit.>.

In section <ref> we present the arrow diagrams of links in S^3 for which the minimal crossing number
is the Hopf crossing number.

In section <ref> we consider several examples of arrow diagrams of knots.

In section <ref> we compute the Jones polynomial of knots K with h(K)≤ 1.

In section <ref> we classify knots K with h(K)≤ 1 and study some of their properties.
In particular, we find all knots K with at most 10 crossings (in the classical sense) satisfying h(K)≤ 1.

Finally, in section <ref> we establish which knots K, with h(K)≤ 1, are algebraic and show that, for such knots, h(K)=C_alg(K).



§ ARROW DIAGRAMS OF LINKS

Let T=S^1× D be the solid torus. An arrow diagram of a link L in T is obtained by cuting
T along 1× D×, 1∈ S^1, and projecting L from [0,1]× D thus obtained onto D. One keeps the information
about over- and undercrossings in this projection. Also, the points in L∩ 1× D are projected
onto arrows pointing to the part of L that is close to 0 in [0,1]× D.
It was shown in <cit.> that there are five Reidemeister moves connecting any two diagrams which represent
the same link in T: three classical (_1, _2 and _3) 
and two extra ones (_4 and _5), presented in Figure <ref>.



Given the solid torus T, S^3 is obtained from T by gluing to it a second solid torus T', see Figure <ref>. 
In this figure, T is shown. As a basis for H_1(∂ T) we choose a standard longitude-meridian basis
of a torus embedded in a standard way in S^3.
In the decomposition T=S^1× D, used for the arrow diagrams as above, we take as S^1 the curve passing through
A and A' (i.e. a fiber in the Hopf fibration of S^3). Such S^1, when oriented, represents (1,1) in the chosen basis of H_1(∂ T).
The solid torus T' is glued to T in such a way that its meridional disk D', shown vertically in the middle
of the figure, is glued to T along a (1,0) curve in ∂ T. Here, we choose the orientations of S^1 and ∂ D
indicated on the figure by arrows.



Any link L in S^3 can be pushed inside T. Its projection onto D gives an arrow diagram of L,
as decribed above. If two links L and L', lying in T, are ambiently isotopic in S^3, one can pass from L to L' by a series
of ambient isotopies in T and gliding of arcs along the disc D'. Such a gliding gives rise to one extra Reidemeister move, 
called the _∞ move, shown in Figure <ref>. Indeed, consider the small arc c above D in Figure <ref>.
It projects onto an arc in D as on the left of Figure <ref>. When c is glided over D' and slightly beyond, its projection in D
will be such as the one depicted on the right of Figure <ref>. Indeed, going around ∂ D in counterclockwise direction, one meets in succesion:
the unchanged endpoints of c, the arrow where the glided c crosses D next to D∩ D' (the arrow is oriented in clockwise direction as we go opposite
to the orientation of the fibers S^1), a point next to A' projecting to a point next to A and, finally, a point next to B' projecting to a point next to B.



Let S^2 be obtained from D and D' by identifying the points of ∂ D and ∂ D' belonging to the same fiber 
(such as A and A' or B and B' in Figure <ref>). Then the Hopf map, p:S^3→ S^2, maps T onto D⊂ S^2 and T' onto D'⊂ S^2. 
As any link L in S^3 can be pushed into T without changing the number of crossings of p(L), the Hopf crossing number h(L) is the minimum of crossings
in all arrow diagrams of L.

For any link L its classical diagrams form a subset of its arrows diagrams (consisting of diagrams
with no arrows). Thus, obviously, h(L)≤ c(L).

As in our later computations we use the Kauffman bracket, it is necessary to consider framed links. Such a link is equipped with a framing i.e.
a smooth family of unit vectors orthogonal to the link.
An arrow diagram represents a framed link where the framing is orthogonal to the fibers S^1 of the Hopf fibration.
All Reidemeister moves except _1 preserve this framing. For _∞, notice that when we glide an arc, such as c in Figure <ref>, through
D', the framing stays orthogonal to the fibers during the gliding.

Let D be an arrow diagram. Denote by w(D) the writhe of the diagram obtained from D by ignoring all arrows.
Denote by w(D) the writhe of the framed link represented by D. If D' has no arrows and is obtained from D by Reidemeister moves
including a moves _1 increasing the writhe and b moves _1 decreasing the writhe, then w(D)=w(D')+b-a.

For an oriented diagram an arrow is positive if it points in the same direction as the orientation.
Otherwise it is negative. An arrow is removable if it can be eliminated with an _∞
move followed by an _4 move. Thus, an arrow is removable if it is not separated from the boundary of
the diagram by any arc and is clockwise (so it will be opposite to the arrow created by _∞).


Suppose D is an oriented arrow diagram of a knot with all arrows being removable. Suppose that there are a>0
positive and b≥ 0 negative arrows. Then w(D)=w(D)+a(a-1)+b(b-1)-2ab.

Denote D by D_a,b. Eliminate a positive arrow by _∞ and _4, then push with _2 and _5
the other arrows through the new arc so that they are again removable. Denote the resulting diagram by
D_a-1,b. This is shown in Figure <ref> and one sees that the remaining a-1 positive
arrows contribute +2 each to w(D_a-1,b) whereas the b negative arrows contribute -2 each.
Thus w(D_a-1,b)=w(D_a,b)+2(a-1)-2b.
We apply the same moves to D_a-1,b getting
w(D_a-2,b)=w(D_a-1,b)+2(a-2)-2b.
Continuing until a=0 we get w(D_0,b)=w(D_a,b)+2(a-1)-2b+2(a-2)-2b+...+2-2b+0-2b=w(D_a,b)+a(a-1)-2ab.

Now reverse the orientation of D_0,b to get D_b,0 (w is unchanged).
By the first part w(D_0,0)=w(D_b,0)+b(b-1).
As D_0,0 has no arrows w(D_0,0)=w(D_0,0).

We have not used _1 moves, so w(D)=w(D_a,b)=w(D_0,0).
Thus w(D)=w(D)+a(a-1)-2ab+b(b-1).






We will later need to estimate, from an arrow diagram of a knot K, the classical crossing number c(K). The following lemma will be useful:

 Suppose that a knot K has an arrow diagram D with k crossings, a+b arrows of which a are removable and b are not removable
but are not separated from the boundary of the diagram.

Then c(K)≤ k+a(a-1) if b=0 and c(K)≤ k+b-1+(a+b)(a+b-1) if b>0.

If b=0, remove all a arrows just as in the proof of the preceding lemma.
This creates 2(a-1)+2(a-2)+...+2=a(a-1) crossings and all arrows have been eliminated.

If b>0, create b-1 kinks with _1 moves next to each arrow that is not removable except for one (so do nothing if b=1).
Then push such arrows into the kinks with _5. This increases the number of crossings by b-1. Now a+b-1 arrows are removable. 
Remove these arrows as in the proof of the preceding lemma. This creates 2(a+b-2)+2(a+b-3)+...+2 crossings. 
Push the last non removable arrow across the a+b-1 arcs with _2 and _5 creating 2(a+b-1) crossings.
At this stage there are k+b-1+2(a+b-1)+2(a+b-2)+..+2=k+b-1+(a+b)(a+b-1) crossings.
Now create a kink, push the last arrow into it with _5 and eliminate the arrow with _∞. As there are no more arrows, the arc created
by _∞ going next to the boundary of the diagram can be shrinked above the rest of the diagram and the crossing coming from the last kink
created can be eliminated with _1. So we get the required number of crossings.






§ EXAMPLES: ARROW DIAGRAMS OF TORUS KNOTS AND SOME OTHER KNOTS

In this section we consider some arrow diagrams of torus knots and a couple of other knots that will be needed later.

Consider a torus knot T(n,m), n<m, gcd(n,m)=1, lying on the boundary of the solid torus T (see Figure <ref>), 
going in a uniform way n times in the longitudinal and m times in the meridional direction. Thus, its homology class in H_1(∂ T) is (n,m),
in the standard longitude-meridian basis of H_1(∂ T).

The Hopf map projects T onto D and T(n,m) onto the circle ∂ D.
Push the knot into T and perturb it slightly, to get a generic projection with all crossings near each other (see Figure <ref>). 
We may choose the cut in T (which creates the arrows) and the perturbation of the knot in such a way that there is an arrow just before all crossings and pointing
to the undercrossings. All other arrows are uniformly spaced on the resulting curve.
Now each arrow corresponds to going once along a Hopf fiber i.e. going around a curve (1,1). Going around the disk corresponds to 
a meridional curve (0,1). Thus, going a times around the disk, with b arrows, will correspond to a torus knot of type a(0,1)+b(1,1)=(b+a,a)
(the addition and multiplication in H_1(∂ T)). In Figure <ref> the first knot is of type (1,1)+(1,1)+(1,1)+(0,1)=(3,4) i.e. the knot T(3,4).
The second knot is of type 3(1,1)+2(0,1)=(3,5), the third is of type 5(1,1)+2(0,1)=(5,7) and the fourth is of type 4(1,1)+3(0,1)=(4,7).



It follows that an oval with n counterclockwise arrows on it is the torus knot T(n,n+1). Thus h(T(n,n+1))=0.
More generally we see that T(n,n+k) has a diagram with k-1 crossings so h(T(n,n+k))≤ k-1.
Indeed, consider a diagram D going around the disk k times with k-1 crossings, generalizing the diagrams of Figure <ref>
(for example k=3 for the last diagram in this figure).
Suppose that there are n arrows on D spaced uniformly, where the first arrow is just before the crossings. Then D is a diagram of the
torus knot of type k(0,1)+n(1,1)=(n,n+k).
Fiedler has shown in <cit.>, that C_alg(T(n,n+k))=k-1 if k<n. It does not imply that
h(T(n,n+k))=k-1 if k<n. We will get the last equality for k≤ 3, see Proposition <ref>.

If k>n, there are less arrows contributing (1,1) than meridional curves contributing (0,1) in the arrow diagrams such as in Figure <ref>.
Consider the same diagram D with k-1 crossings as in the preceding paragraph: going around the disk k times with n
arrows spaced uniformly, the first arrow just before the crossings. As n<k, the second arrow will not be on the most nested kink, so this
kink can be removed with _1, yielding a diagram with k-2 crossings. Thus, h(T(n,n+k))≤ k-2 if k>n. 
It is possible that more nested kinks may be removed. The most extreme case is T(2,2+k) with a diagram containing only two arrows, 
so that all the most nested kinks up to the second arrow can be removed. One checks easily that h(T(2,2+k))≤k-1/2.


Three arrow diagrams of the torus knot T(2,5) are shown in Figure <ref>.







Consider the knot with a diagram shown in Figure <ref>. In this figure Reidemeister moves are used to show
that it is the knot 9_42. In the last row of this figure arcs that are moved in the diagrams are thickened.
The third diagram from the end is transformed by a rotation by π of a solid torus containing the knot indicated with
dashed lines.






Denote  the first arrow diagram in Figure <ref> by K_2,1. Denote a similiar diagram with 3 counterclockwise arrows on the right loop instead of 2
by K_3,1 (see Figure <ref> below). Later, we will see that K_2,1 and K_3,1 share the same Jones polynomial.
By eliminating all arrows the HOMFLY polynomial of these two knots can be computed with KnotPlot<cit.>. One gets for K_2,1 and K_3,1 respectively:

    P_K_2,1=-2l^-2-3-2l^2+m^2l^-2+4m^2+m^2l^2-m^4


    P_K_3,1=5+10l^2+8l^4+2l^6-10m^2-25m^2l^2-14m^2l^4-m^2l^6
    
    +6m^4+22m^4l^2+7m^4l^4-m^6-8m^6l^2-m^6l^4+m^8l^2

Thus these two knots are distinct (including mirror images) and because of the assymetry of the l terms K_3,1 is chiral.
One can check that the reductions of the two HOMFLY polynomials to the Jones polynomial give indeed the same polynomial.





§ JONES POLYNOMIAL OF KNOTS K WITH H(K)≤ 1

In this section we compute the Jones polynomial of knots with Hopf crossing number at most 1.
This will allow us to classify such knots in the next section.

For n≥ 0, let T_n be an oval with n counterclockwise arrows on it. It is the right handed torus knot T(n,n+1) (see the previous section).
The Jones polynomial of a torus knot T(m,n) is given by <cit.>:

    V_T(m,n)=t^(m-1)(n-1)/21-t^m+1-t^n+1+t^m+n/1-t^2VT1

Thus, for right handed torus knots T(n,n+1) the Jones polynomial is: 

    V_T_n=t^n(n-1)/21-t^n+1-t^n+2+t^2n+1/1-t^2VT2

We extend to n∈ the definition of T_n: if n<0, T_n is an oval with |n| clockwise arrows on it.
If n<0, T_n can be transformed with _∞ and _4 into T_-n-1.
In the same way, for n∈, let T'_n be an oval with n clockwise arrows, if n≥ 0, and |n| counterclockwise arrows, if n<0.
One can transform T'_n into T_n-1 with _∞ and _4, so these diagrams represent the same knot.
One checks easily that in the equation (<ref>) V_T_n and V_T_-n-1 are equal for all n.
Thus, this equation is also valid for negative n.

An arrow diagram of a knot with no crossings is just an oval with some arrows on it. Opposite arrows can be removed with _4.
Thus, from the preceding paragraph we get immediately (see also <cit.>):


A knot K satisfies h(K)=0 if and only if K is a torus knot T(n,n+1).


We use the Kauffman bracket in order to compute the Jones polynomial (see <cit.>).
Substituting t=A^-4, for a framed knot, <K>=(-A)^3w(K)V(K),
where <>, V and w stand for the Kauffman bracket, Jones polynomial and writhe of the knot, respectively.
Until the end of this section we will use both A and t, sometimes in the same formulas, with the understanding that t=A^-4.

Notice that the Kauffman bracket and Kauffman relations (smoothing a crossing and removing a trivial component)
are defined for the orthogonal projection of framed links from ^3 to ^2. However, the Kauffman is also defined for arrow diagrams, which, 
as classical diagrams, represent framed links in S^3. Also, Kauffman relations hold for arrow diagrams. Indeed, to apply a Kauffman relation to an arrow
diagram, we may first remove all arrows with appropriate Reidemeister moves, apply the relation to a classical diagram, then recreate the removed
arrows with Reidemeister moves. 


The writhe of T_n is n(n+1), for n∈.

If n=0, then it is trivial. If n>0, then apply _∞ to T_n to get T'_n+1. This does not change the writhe. 
Orient the diagram so that all the arrows of T'_n+1 are positive and removable as in the definition before lemma <ref>. 
From this lemma, w(T'_n+1)=w(T'_n+1)+(n+1)n=n(n+1).

If n<0, orient the diagram so that there are |n| postive removable arrows. From the same lemma,
w(T_n)=w(T'_|n|)=|n|(|n|-1)=(-n)(-n-1)=n(n+1).



Using equation (<ref>) and the last lemma, one gets: 

    <T_n>=A^3n(n+1)t^n(n-1)/21-t^n+2-t^n+1(1-t^n)/1-t^2


Let us write <T_n>=U_n+V_n where U_n contains 1-t^n+2 and V_n
contains -t^n+1(1-t^n) of the numerator of the bracket.

Let us also write <T_n>=U'_n+V'_n where U'_n contains 1-t^n+1 and V'_n
contains -t^n+2(1-t^n-1) of the numerator of the bracket.


Let n∈. Then V_n+A^2U_n-2=0 and V'_n+A^-2U'_n-2=0.


    V_n=-t^n+1 A^3n(n+1) t^n(n-1)/21-t^n/1-t^2=-A^v1-t^n/1-t^2


    A^2U_n-2=A^2 A^3(n-2)(n-1)t^(n-2)(n-3)/21-t^n/1-t^2=A^u1-t^n/1-t^2

One has to check that v=u so these two terms cancel each other.

    v=-4(n+1)+3n(n+1)-2n(n-1)=n^2+n-4


    u=2+3(n-2)(n-1)-2(n-2)(n-3)=n^2+n-4

The proof for V'_n and U'_n-2 is similar. In fact, using the same v and u as above one has:

    V'_n=-tA^v1-t^n-1/1-t^2


    A^-2U'_n-2=tA^2U'_n-2=tA^u1-t^n-1/1-t^2

As v=u, the equality follows.




For n,m∈, m>0 let:

    S_n,m   =    <T_n>+A^2<T_n-2>+...+A^2m<T_n-2m>
    
    S'_n,m   =   <T_n>+A^-2<T_n-2>+...+A^-2m<T_n-2m>

Then:

    S_n,m   =   A^3n(n+1)t^n(n-1)/2[1-t^n+2-t^(m+1)(n-m+1)(1-t^n-2m)/1-t^2]
    
    S'_n,m   =   A^3n(n+1)t^n(n-1)/2[1-t^n+1-t^(m+1)(n-m+2)(1-t^n-2m-1)/1-t^2]


From Lemma <ref>, S_n,m=U_n+A^2mV_n-2m, as the other terms cancel each other. Thus:

    S_n,m   =   A^3n(n+1)t^n(n-1)/21-t^n+2/1-t^2
          +A^2mA^3(n-2m)(n-2m+1)t^(n-2m)(n-2m-1)/2(-t^n-2m+11-t^n-2m/1-t^2)

For the second term:

    A^2m A^3(n-2m)(n-2m+1)=A^3n(n+1)A^2m A^-6m(n-2m+1)-6mn
       =   A^3n(n+1)A^-12mn+12m^2-4m=A^3n(n+1)t^3mn-3m^2+m

and

    t^(n-2m)(n-2m-1)/2t^n-2m+1=t^n(n-1)/2t^-m(n-2m-1)-mn+n-2m+1
       =   t^n(n-1)/2t^-2mn+2m^2-m+n+1

Thus, factoring out A^3n(n+1)t^n(n-1)/2 in the second term one gets:

    -t^-2mn+2m^2-m+n+1+3mn-3m^2+m=-t^mn-m^2+n+1=-t^(m+1)(n-m+1)

For the second part, again from Lemma <ref>, S'_n,m=U'_n+A^-2mV'_n-2m. Thus:

    S'_n,m   =   A^3n(n+1)t^n(n-1)/21-t^n+1/1-t^2
          +A^-2mA^3(n-2m)(n-2m+1)t^(n-2m)(n-2m-1)/2(-t^n-2m+21-t^n-2m-1/1-t^2)

The difference between the coefficient of the second term in S'_n,m compared to S_n,m is the multplication by A^-2m instead of A^2m
(which corresponds to multiplying by A^-4m=t^m) and -t^n-2m+2 instead of -t^n-2m+1. Overall, this corresponds
to multiplying by t^m+1.
Thus, factoring out A^3n(n+1)t^n(n-1)/2 in the second term one gets:

    -t^(m+1)(n-m+1)t^m+1=-t^(m+1)(n-m+2)





From the definition of the Kauffman bracket (see <cit.>) one gets easily the two following relations:


    -7pt
        < g r a p h i c s >
    =A^2 -7pt
        < g r a p h i c s >
    
    +(A^-1-A^3) -7pt
        < g r a p h i c s >



    -7pt
        < g r a p h i c s >
    =A^-2-7pt
        < g r a p h i c s >
    
    +(A-A^-3) -7pt
        < g r a p h i c s >



For a≥ 0, b≥ 0, let K_a,b, K'_a,b and K”_a,b be the framed knots shown on the left of Figure <ref>, where a next
to an arrow stands for a arrows (and the same for b).
In this figure the Kauffman bracket of these knots is calculated using the previous two relations (for simplicity, the bracket is omitted in the figure). 
For example, applying _5 and _4 moves, in the first case one gets:

<K_a,b>=A^2<K_a-1,b-1>+(A^-1-A^3)<T_a+b> 

Iterating the first relation until b=0 and removing the positive kink with -A^3 in the Kauffman bracket one gets:


    1<K_a,b>=(A^-1-A^3)[<T_a+b>+A^2 <T_a+b-2>+…
    …+A^2b-2<T_a-b+2>]-A^2b+3<T_a-b>



Symmetrically, exchanging A and A^-1 and T_k with T'_k=T_k-1, one gets for K':


    2<K'_a,b>=(A-A^-3)[<T_a+b-1>+A^-2 <T_a+b-3>+…
    …+A^-2b+2<T_a-b+1>]-A^-2b-3<T_a-b-1>


For K” one gets:


    3<K”_a,b>=(A^-1-A^3)[<T_a-b-1>+A^2 <T_a-b+1>+…
    …+A^2b-2<T_a+b-3>]-A^2b+3<T_a+b-1>





Let n=a+b, then: 

    <K_a,b>   =   A^3n(n+1)t^n(n-1)/2A^-1[1-t^n+2-t^b(a+2)(1-t^a-b+2)/1-t^2.
          .-t^-11-t^n+2-t^(b+1)(a+1)(1-t^a-b)/1-t^2]
    
    <K'_a,b>   =   A^3n(n-1)t^(n-1)(n-2)/2A[1-t^n-t^b(a+2)(1-t^a-b)/1-t^2.
          .-t1-t^n-t^(b+1)(a+1)(1-t^a-b-2)/1-t^2]
    
    <K”_a,b>   =   A^3n(n-1)A^2b+3t^(n-1)(n-2)/2[t^n+21-t^n-2-t^ba(1-t^a-b-2)/1-t^2.
          .-1-t^n-t^(b+1)(a+1)(1-t^a-b-2)/1-t^2]



Using S_n,m and S'_n,m as in Lemma <ref>, equation (<ref>) can be rewritten as:

    K_a,b=A^-1S_n,b-1-A^3S_n,b=A^-1(S_n,b-1-t^-1S_n,b)

Similarily, equation (<ref>) can be rewritten as:

    K'_a,b=AS'_n-1,b-1-A^-3S'_n-1,b=A(S'_n-1,b-1-tS'_n-1,b)

Equation (<ref>) can be rewritten as:

    K”_a,b=A^-1A^2b-2S'_n-3,b-1-A^3A^2bS'_n-1,b=A^2b-3S'_n-3,b-1-A^2b+3S'_n-1,b

The result follows from the formulas for S and S' from Lemma <ref>. For the first term of <K”_a,b> one uses
the identity that is easily checked:

    A^2b-3A^3(n-3)(n-2)t^(n-3)(n-4)/2=A^3n(n-1)A^2b+3t^(n-1)(n-2)/2t^n+2




To get the Jones polynomial from the Kauffman bracket it suffices to compute the writhes of the knots.


w(K_a,b)=1+a(a+1)+b(b+1)-2ab

w(K'_a,b)=-1+a(a-1)+b(b-1)-2ab

w(K”_a,b)=1+a(a-1)+b(b+1)+2ab

In order to use lemma <ref> we want all arrows to be removable. This is the case for K'_a,b.
The single crossing of K'_a,b is negative. Also we can orient K'_a,b in such a way that the a arrows on the
left loop are positive and the b arrows on the right loop are negative.
From lemma <ref> we get:

w(K'_a,b)=w(K'_a,b)+a(a-1)+b(b-1)-2ab=-1+a(a-1)+b(b-1)-2ab

In K_a,b none of the arrows are removable. Let D_a,b be the diagram obtained from K_a,b by putting with _1 a negative kink
next to each arrow then pushing each arrow with _5 into each such kink which become positive.
The _1 moves change the writhe so that w(K_a,b)=w(D_a,b)+a+b. Because of the positive kinks w(D_a,b)=w(K_a,b)+a+b=1+a+b,
as the single crossing of K_a,b is positive.
Now in D_a,b all arrows are removable and it can be oriented so that there are a postive and b negative arrows.
Applying again lemma <ref> we get:

w(K_a,b)=w(D_a,b)+a+b=w(D_a,b)+a(a-1)+b(b-1)-2ab+a+b=a(a-1)+b(b-1)-2ab+a+b+a+b+1=1+a(a+1)+b(b+1)-2ab

In K”_a,b the a arrows on the left kink are removable and the b arrows on the right kink are not. As in the previous case, we make these b
arrows removable by _1 and _5 moves getting a diagram D”_a,b with w(K”_a,b)=w(D”_a,b)+b and 
w(D”_a,b)=w(K”_a,b)+b=1+b, as the single crossing of K”_a,b is positive.
D”_a,b can be oriented so that it has a+b positive removable arrows and from lemma <ref> we get:

w(K”_a,b)=w(D”_a,b)+b=w(D”_a,b)+(a+b)(a+b-1)+b=(a+b)(a+b-1)+2b+1=
1+a(a-1)+ab+b(a+b-1)+2b+1=1+a(a-1)+2ab+b(b+1)





Let n=a+b, then:

    V_K_a,b   =   t^a^2+b^2-4ab-a-b/2[1-t^n+2-t^(b+1)(a+1)(1-t^a-b)/1-t^2.
          .-t1-t^n+2-t^b(a+2)(1-t^a-b+2)/1-t^2]
    
    V_K'_a,b   =   t^a^2+b^2-4ab-3a-3b/2[t1-t^n-t^(b+1)(a+1)(1-t^a-b-2)/1-t^2.
          .-1-t^n-t^b(a+2)(1-t^a-b)/1-t^2]
    
    V_K”_a,b   =   t^b t^(n-1)(n-2)/2[-t^n+21-t^n-2-t^ba(1-t^a-b-2)/1-t^2.
          .+1-t^n-t^(b+1)(a+1)(1-t^a-b-2)/1-t^2]


As V_K=(-A)^-3w(K)<K> let us compute the contributions of the writhe from Lemma <ref> 
times the coefficients in front of the parentheses for
the Kauffman bracket of K_a,b, K'_a,b and K”_a,b from Proposition <ref>.

For V_K_a,b it is:

    (-A)^-3(1+a(a+1)+b(b+1)-2ab)A^3n(n+1)t^n(n-1)/2A^-1=


    -A^-3-3a^2-3a-3b^2-3b+6ab+3a^2+3b^2+6ab+3a+3b-2a^2-2b^2-4ab+2a+2b-1=


    -A^-4-2a^2-2b^2+8ab+2a+2b=-tt^a^2+b^2-4ab-a-b/2

Multpliying the terms in the parenthesis of <K_a,b> by -t we get the required Jones polynomial.

For V_K'_a,b one has:

    (-A)^-3(-1+a(a-1)+b(b-1)-2ab)A^3n(n-1)t^(n-1)(n-2)/2A=


    -A^3-3a^2+3a-3b^2+3b+6ab+3a^2+3b^2+6ab-3a-3b-2a^2-2b^2-4ab+6a+6b-4+1=


    -A^-2a^2-2b^2+8ab+6a+6b=-t^a^2+b^2-4ab-3a-3b/2


For V_K”_a,b one has:

    (-A)^-3(1+a(a-1)+b(b+1)+2ab)A^3n(n-1)A^2b+3t^(n-1)(n-2)/2=


    -A^-3-3a^2+3a-3b^2-3b-6ab+3a^2+3b^2+6ab-3a-3b+2b+3-2a^2-2b^2-4ab+6a+6b-4=


    -A^-4-2a^2-2b^2-4ab+6a+2b=-A^-4b-2(a+b-1)(a+b-2)=-t^bt^(n-1)(n-2)/2







§ CLASSIFICATION OF KNOTS K WITH H(K)≤ 1



Let K be a knot with h(K)=1. Then K is one of K_a,b, K'_a,b with a≥ b≥ 1 or
K”_a,b with a>1 and b≥ 1.

By assumption K has an arrow diagram with one crossing i.e. with two loops. If one loop is nested in the other, perform an _∞
move to get two unnested loops (so the diagram now looks like ∞ with a positive or negative crossing). Let us call the loops
of the diagram of K a left loop and a right loop.
On each loop, we eliminate opposite arrows with _4. Now, if there are no arrows on one of the loops it can be eliminated with _1 and
h(K)=0, a contradiction. Thus, there are some arrows on each loop.

Suppose first that the arrows on each loop are counterclockwise (as in K_a,b). If the unique crossing is different from K_a,b,
then we push one arrow from the left loop to the right with _5 changing the crossing. Then we eliminate opposite arrows with _4.
There should be arrows on each loop otherwise, as above, h(K)=0. If there are more arrows on the left loop than on the right loop, then
rotate the diagram of K by π. The crossing is unchanged, the arrows are counterclockwise and we get K_a,b with a≥ b≥ 1.

Suppose now that the arrows on each loop are clockwise (as in K'_a,b). By the same method as above we can change the crossing if needed
and put more arrows on the left by rotating the diagram. Thus we get K'_a,b with a≥ b≥ 1.

Finally suppose that the arrows are counterclockwise on one loop and clockwise on the other (as in K”_a,b). First, modify as above the crossing
if needed. If the counterclockwise arrows are on the left loop, then rotate the diagram by π. In this way we get K”_a,b. If there is only
one arrow on the left loop it can be pushed to the right loop and h(K)=0. Thus we may assume that a>1. Notice that in this case it
is not necessary that a≥ b.



We will use the Jones polynomials from Theorem <ref> to distinguish the knots K_a,b, K'_a,b and K”_a,b.
First, one can consider their Jones polynomials multiplied by 1-t^2 and by a power of t so that 
the lowest coefficient has degree zero. Denote by J_K such polynomial obtained from the Jones polynomial V_K of a
knot K. For K_a,b, a≥ b≥ 1 one has:

    J_K_a,b=1-t-t^a+b+2+t^a+b+3+t^ab+2b+1-t^ab+a+b+1-t^ab+a+b+3+t^ab+2a+1

In this formula, we want to have increasing powers of t and check for possible cancellations. One has to consider some
special cases.

First, suppose that b=1. Then:

    J_K_a,1=1-t+t^a+4-t^2a+2-t^2a+4+t^3a+1JK1

In the last forumla the powers are increasing except for
a=1 giving J_K_1,1=1-t+t^5-t^6 and a=2 or a=3 
giving J_K_2,1=J_K_3,1=1-t+t^7-t^8.

Suppose now that b>1.
If a=b, a=b+1 or a=b+2, one has, respectively,

    J_K_b,b=1-t-t^2b+2+t^2b+3+t^b^2+2b+1-t^b^2+2b+3JK2


    J_K_b+1,b=1-t-t^2b+3+t^2b+4+t^b^2+3b+1-t^b^2+3b+2+t^b^2+3b+3-t^b^2+3b+4JK3


    J_K_b+2,b=1-t-t^2b+4+t^2b+5+t^b^2+4b+1-t^b^2+4b+3JK4

Finally, if a>b+2,

    J_K_a,b=1-t-t^a+b+2+t^a+b+3+t^ab+2b+1-t^ab+a+b+1-t^ab+a+b+3+t^ab+2a+1JK5


Similarily, for K'_a,b, a≥ b≥ 1 one has:

    J_K'_a,b=-1+t+t^a+b-t^a+b+1+t^ab+2b-t^ab+a+b-t^ab+a+b+2+t^ab+2a


Again, we consider special cases.
Suppose first that b=1. Then one has:

    J_K'_a,1=-1+t+t^a+1-t^2a+1-t^2a+3+t^3aJK'1

In the last formula the powers are increasing except for
a=1 giving J_K'_1,1=-1+t+t^2-t^5, a=2 
giving J_K'_2,1=-1+t+t^3-t^5+t^6-t^7
and a=3 giving J_K'_3,1=-1+t+t^4-t^7.

Suppose now that b>1.
If a=b, a=b+1 or a=b+2, one has, respectively,

    J_K'_b,b=-1+t+t^2b-t^2b+1+t^b^2+2b-t^b^2+2b+2JK'2


    J_K'_b+1,b=-1+t+t^2b+1-t^2b+2+t^b^2+3b-t^b^2+3b+1+t^b^2+3b+2-t^b^2+3b+3JK'3


    J_K'_b+2,b=-1+t+t^2b+2-t^2b+3+t^b^2+4b-t^b^2+4b+2JK'4

Finally, if a>b+2,

    J_K'_a,b=-1+t+t^a+b-t^a+b+1+t^ab+2b-t^ab+a+b-t^ab+a+b+2+t^ab+2aJK'5


Consider the third family K”_a,b now. We suppose that a>1, b≥ 1 but it is not
necessary that a≥ b. One has:

    J_K”_a,b=1-t^a+b-t^a+b+2+t^2a+2b-t^ab+a+b+1+t^ab+a+b+2+t^ab+2a-1-t^ab+2a

Suppose first that b=1. One has:

    J_K”_a,1=1-t^a+1-t^a+3+t^2a+3+t^3a-1-t^3aJK”1

In the last formula the powers are increasing except for
a=2 giving J_K”_2,1=1-t^3-t^6+t^7,
a=3 giving J_K”_3,1=1-t^4-t^6+t^8
and a=4 giving J_K”_4,1=1-t^5-t^7+2t^11-t^12.

Suppose now that b>1. As a special case suppose that a=2, then,

    J_K”_2,b=1-t^b+2-t^b+4+t^2b+3-t^3b+3+t^3b+4JK”2

Now suppose that a>2 (and b>1).
There may be some cancellations in the four last terms of the general formula. One has:
(ab+2a-1)-(ab+a+b+1)=a-b-2. So there are some terms with the same degree if a-b is 1, 2 or 3.

If a≤ b,

    J_K”_a,b=1-t^a+b-t^a+b+2+t^2a+2b+t^ab+2a-1-t^ab+2a-t^ab+a+b+1+t^ab+a+b+2JK”3

If a=b+1, a=b+2 or a=b+3, one has, respectively,

    J_K”_b+1,b=1-t^2b+1-t^2b+3+t^4b+2+t^b^2+3b+1-2t^b^2+3b+2+t^b^2+3b+3JK”4
 

    J_K”_b+2,b=1-t^2b+2-t^2b+4+t^4b+4JK”5


    J_K”_b+3,b=1-t^2b+3-t^2b+5+t^4b+6-t^b^2+5b+4+2t^b^2+5b+5-t^b^2+5b+6JK”6

Finally, if a>b+3,

    J_K”_a,b=1-t^a+b-t^a+b+2+t^2a+2b-t^ab+a+b+1+t^ab+a+b+2+t^ab+2a-1-t^ab+2aJK”7



The knots K_a,b, K'_a,b, a≥ b≥ 1 and K”_a,b, a>1, b≥ 1 are distinct.
Any knot K with h(K)=1 is such a knot.


If two such knots are the same then their polynomials J must be the same, so it is sufficient to check
that all these polynomials are different (except for K_2,1 and K_3,1 which are distinguished by the HOMFLY polynomial,
see example <ref> in section <ref>).

First we notice that J_K'_a,b starts with -1 whereas J_K_a,b and J_K”_a,b start with 1.
Also, the second term in K_a,b is -t whereas in K”_a,b it is -t^a+b and a+b≥ 3.
Thus these three families of knots are distinguished by J.

∙ Consider the family K_a,b. In the five formulas for J above there are 6 or 8 terms except for the
special cases K_1,1, K_2,1 and K_3,1 for which there are 4 terms. Thus, these special cases are
distinct from the rest (and K_1,1 is distinct from K_1,2 and K_1,3).

Consider the formulas with six terms i.e. equations (<ref>), (<ref>) and (<ref>).
In equation (<ref>) the last term has coefficient 1, which distinguishes it from the other two equations
with last coefficient -1.
For these two equations, suppose that J_K_b,b=J_K_c+2,c for some integers b>1 and c>1.
By comparing the different terms in these polynomials we must have:

2b+2=2c+4 and b^2+2b+1=c^2+4c+1. From the first equation b=c+1 so b^2+2b+1=c^2+4c+4≠ c^2+4c+1.

Also, the knots K_a,1 are distinguished by a, K_b,b are distinguished by b and K_b+2,b are
also distinguished by b.

Now consider the formulas with eight terms i.e. equations (<ref>) and (<ref>).
These two subfamilies are distinguished by the coefficient of the last term (-1 and 1).
The knots K_b+1,b are distinguished by b.
For the general J_K_a,b of equation (<ref>), notice that the difference between the powers of the second
and third term determines a+b, whereas the difference between the powers of the last two terms determines a-b.
Together, this determines the couple (a,b) so all such knots are distinct.

∙ Consider the family K'_a,b. In the five formulas for J above there are 6 or 8 terms except for the
specials cases K'_1,1, K'_3,1 for which there are 4 terms. Thus, these special cases are
distinct from the rest (and K'_1,1 is distinct from K'_3,1). Also, K'_2,1 for which there are 6 terms
is distinct from the other knots with a>3 in equation (<ref>) and from the knots in equations
(<ref>) and (<ref>) because the third term of K'_2,1 is t^3 whereas in these two equations
the power of the third term is even.

Consider the formulas with six terms i.e. equations (<ref>), (<ref>) and (<ref>).
In equation (<ref>) the last term has coefficient 1, which distinguishes it from the other two equations
with last coefficient -1.
For these two equations, suppose that J_K'_b,b=J_K'_c+2,c for some integers b>1 and c>1.
By comparing the different terms in these polynomials we must have:

2b=2c+2 and b^2+2b=c^2+4c. From the first equation b=c+1 so b^2+2b=c^2+4c+3≠ c^2+4c.

Also, the knots K'_a,1 are distinguished by a, K'_b,b are distinguished by b and K'_b+2,b are
also distinguished by b.

For the remaining subfamilies with eight terms the proof that they all have distinct polynomials J is exactly like for K_a,b
in the previous case.

∙ Consider the family K”_a,b. For J one has 4, 5, 6, 7 or 8 terms.
The knots with 4 terms are K”_2,1 (with second term -t^3), K”_3,1 (with second term -t^4)
and the knots K”_b+2,2 of equation (<ref>) (with second term -t^2b+2 and b>1 so 2b+2≥ 6).
Thus, all these knots are distinct.

The only knot with 5 terms is K”_4,1 so it is distinct from all other knots.

The knots with 6 terms occur in equations (<ref>) and (<ref>). These two subfamilies are distinguished
by the coefficient of the last term (-1 and 1). Also, K”_a,1 are distinguished by a and K”_2,b are distinguished by b.

The knots with 7 terms occur in equations (<ref>) and (<ref>). Again, the two subfamilies are distinguished 
by the coefficient of the last term (1 and -1) and inside each family the knots are distinguished by b.

Finally, the knots with 8 terms occur in equations (<ref>) and (<ref>). Once again, the two subfamilies are distinguished
by the coefficient of the last term (1 and -1). Inside each family the second term determines a+b and the difference between the powers of the
seventh and the sixth terms determine b-a or a-b. This together determines the couple (a,b)

For the second part of the theorem, it follows from Lemma <ref> 
that if h(K)=1 then K is a knot in one of the three listed families. One has to check that these knots do not have the Hopf crossing
number equal to 0, i.e. that they are not right handed torus knots T(n,n+1) (see Lemma <ref>). 
From equation (<ref>) the J polynomials of T(n,n+1) are equal to  1-t^n+1-t^n+2+t^2n+1. 
Consider all knots in the three families having J polynomials with four terms. 
One checks easily that all these J polynomials are different from the J polynomials of the knots T(n,n+1).



Now we identify the knots K with h(K)=1 and the classical crossing number c(K)≤ 10 in the Rolfsen knot table<cit.>.


There are eight knots K up to 10 crossings with h(K)=1, namely K_1,1=4_1,
K_2,1=9_42, K'_1,1=3_1,
K'_2,1=5_2, K'_3,1=10_132, K'_2,2=10_145,
K”_2,1=5_1=T(2,5) and K”_3,1=10_124=T(3,5).

The span of the Jones polynomial is a lower bound on the number of crossings.
It is equal to the span of the polynomial J minus 2, as one divides by
1-t^2. Thus it is sufficient to consider the knots with span of J not greater than 12. 
Apart from the listed knots there are also the knots
K_3,1, K_2,2, K'_4,1, K”_4,1, K”_2,2 and K”_4,2.
Except for K_3,1, checking their Jones polynomial shows that these knots have more than 10
crossings. The knot K_3,1 was considered in example <ref> in section <ref>.
As it is distinct from K_2,1=9_42 and 9_42 and there are no knots up to 10 crossings sharing the Jones
polynomial with 9_42, the knot K_3,1 has more than 10 crossings.

Using Lemma <ref> one checks that K_1,1, K_2,1, K'_1,1, K'_2,1 and K”_2,1
 have diagrams with at most 9 crossings. Their Jones polynomials allow to identify these knots.
For K'_3,1 and K'_2,2  the lemma gives an upper bound of 13 crossings.
Drawing their diagrams one can easily reduce the number of crossings to 11. Using their DT codes and KnotFinder<cit.>
one identifies K'_3,1 as 10_132 and K'_2,2 as 10_145.

Finally K”_3,1 is the torus knot T(3,5)=10_124 (apply _∞ to the diagram of T(3,5) in Figure <ref> 
to get K”_3,1).



Let us study now the effect of taking the mirror image of a knot K with h(K)≤ 1. 
We show that, except for the knots 3_1 and 4_1, the Hopf crossing number of such a mirror image is greater than 1.
The right handed trefoil knot satisfies h(3_1)=h(T_2,3)=0 and h(3_1)=h(K'_1,1)=1.
The figure eight knot 4_1=K_1,1 is amphicheiral.


Let K be a knot with h(K)≤ 1, K≠ T(2,3), K≠ K_1,1 and K≠ K'_1,1. Let K be the mirror of K.
Then h(K)>1. In particular all such knots K are chiral and none is a mirror image of another one. 

Notice that as V_K(t)=V_K(t^-1), J_K(t)=-t^s J_K(t^-1) where s is the degree of J_K(t).
Thus,  J_K is obtained from J_K by symmetrically reversing the coefficients and multiplying by -1.
In particular the number of terms in J_K and J_K is the same. Let us use in this proof the notation g(K)
for the ordered tuple of successive gaps between the powers of J_K. For instance, as the J polynomials of T(n,n+1) are equal to 
1-t^n+1-t^n+2+t^2n+1, we have g(T(n,n+1))=(n+1,1,n-1). Obviously, g(K) is obtained from g(K)
by reversing the order of the gaps.

Suppose that K satisfies the hypothesis of the proposition. We need to show that J_K is not equal to any J of
a knot with h(K)≤ 1. We consider case by case J_K with 4,5,...,8 terms.

∙ The knots K with J_K having 4 terms (excluding K_1,1 and K'_1,1) are T(n,n+1) (with Hopf crossing number 0),
K_2,1, K_3,1, K'_3,1, K”_2,1 and K”_b+2,b, b≥ 1 (the exceptional case K”_3,1 can be included in this family). 
The gaps for J of these knots are g(K_2,1)=g(K_3,1)=(1,6,1), g(K'_3,1)=(1,3,3), g(K”_2,1)=(3,3,1) 
and g(K”_b+2,b)=(2b+2,2,2b). 

Reversing any of these gaps one never gets (n+1,1,n-1) so these are not mirror images of 
torus knots T(n,n+1). This is also true for the knots T(n,n+1) as the equation (n+1,1,n-1)=(n'-1,1,n'+1) has no solutions.
Thus h(K)>0 (notice that this is obviously also true for knots K having J_K with more than 4 terms).
Furthermore, this implies that h(T(n,n+1))>1 for n>2.

From examples <ref> and <ref> in section <ref> it follows that K_2,1, K_2,1, 
K_3,1 and K_3,1 are pairwise distinct (K_2,1=9_42 is chiral). 
Thus, in particular, h(K_2,1)>1 and h(K_3,1)>1.
K'_3,1 is not the mirror image of K”_2,1 because the first knot is 10_132 and the second is 5_1.
Finally, for K”_b+2,b the last even 2b cannot be equal to 1 or 3 and the equation
(2b+2,2,2b)=(2b',2,2b'+2) has no solutions. Thus h(K)>1 for all knots with J_K having 4 terms.

∙ The knot K”_4,1 is the only knot with J_K”_4,1 having five terms and this polynomial is changed after reversing
the coefficients and multiplying by minus one. So K”_4,1 is chiral and h(K”_4,1)>1.

∙ Consider the gaps of knots K with J_K having 6 terms. These are:

    g(K'_2,1)=(1,2,2,1,1)


    g(K_a,1)=(1,a+3,a-2,2,a-3),  a≥ 4


    g(K_b,b)=(1,2b+1,1,b^2-2,2),  g(K_b+2,b)=(1,2b+3,1,b^2+2b-4,2),  b>1


    g(K'_a,1)=(1,a,a,2,a-3),  a≥ 4


    g(K'_b,b)=(1,2b-1,1,b^2-1,2),  g(K'_b+2,b)=(1,2b+1,1,b^2+2b-3,2), b>1
 

    g(K”_a,1)=(a+1,2,a,a-4,1),  a≥ 5


    g(K”_2,b)=(b+2,2,b-1,b,1), b>1

None of the gaps starts with 1,1 so h(K'_2,1)>1.
We get the same inequality for K_b,b, K_b+2,b, K'_b,b and K'_b+2,b because the last gap for these knots is 2
and none of the gaps starts with 2.
The reversed gap of K”_a,1 is (1,a-4,a,2,a+1) and the reversed gap of K”_2,b is (1,b,b-1,2,b+2).
None of these gaps can be equal to (1,a'+3,a'-2,2,a'-3) or (1,a',a',2,a'-3).
The last possibility of gaps symmetry to exclude is when a=4 so g(K_a,1)=(1,7,2,2,1) and g(K'_a,1)=(1,4,4,2,1) but
these gaps are not symmetric and reversing the first one does not give the second one.

∙ The two families of knots K with J_K having 7 terms have coefficient 2 or -2 in the sixth term.
Thus J_K has 2 or -2 in the second term so h(K)>1.

∙ Consider the gaps of knots K with J_K having 8 terms. These are:

    g(K_b+1,b)=(1,2b+2,1,b^2+b-3,1,1,1), b>1


    g(K_a,b)=(1,a+b+1,1,ab+b-a-2,a-b,2,a-b-2), b>1,a>b+2


    g(K'_b+1,b)=(1,2b,1,b^2+b-2,1,1,1), b>1


    g(K'_a,b)=(1,a+b-1,1,ab+b-a-1,a-b,2,a-b-2), b>1,a>b+2


    g(K”_a,b)=(a+b,2,a+b-2,ab-2b-1,1,b-a+1,1), a≤ b,a>2


    g(K”_a,b)=(a+b,2,a+b-2,ab-a-b+1,1,a-b-3,1), b>1,a>b+3

None of the gaps starts with 1,1,1 so h(K_b+1,b)>1 and h(K'_b+1,b)>1.
The reversed gaps of K”_a,b end with (...,a+b-2,2,a+b) (for both cases of K”_a,b) so
the last gap a+b is larger than the gap a+b-2.
The gaps of K_a,b and K'_a,b end with (...,a-b,2,a-b-2) so the last gap a-b-2 is smaller than
a-b. Thus h(K”_a,b)>1.
Finally the next to last gap of K_a,b and K'_a,b is 2 which cannot be the second gap for these knots.



The Jones polynomials of knots K with h(K)=1 have relatively simple form (the J polynomials have at most 8 terms).
We show now that the coefficients of these Jones polynomials are bounded in absolute value by 2.


If K is a knot with h(K)=1 then its Jones polynomial can have as coefficients only ± 1 and ± 2.

Consider the knots K_a,b.
If a+b is even we can write the J polynomials grouping terms in couples:

    J_K_a,b=(1-t^a+b+2)-t(1-t^a+b+2)+t^ab+2b+1(1-t^a-b)-t^ab+a+b+3(1-t^a-b-2)

The Jones polynomial is obtained from J by dividing by 1-t^2 and multiplying by t to some power
(the last operation only shifts the coefficients so we can ignore it).
Dividing the first couple 1-t^a+b+2 by 1-t^2 contributes 1 to some coefficients of the Jones polynomial.
The same is true for the third couple, whereas for the second and the fourth dividing by 1-t^2 contributes -1.
Adding these contributions we see that the coefficients are bounded in absolute value by 2.

If a+b is odd we use a different grouping:

    J_K_a,b=1-t^ab+a+b+1-t(1-t^ab+2b)-t^a+b+2(1-t^ab+a-b-1)+t^a+b+3(1-t^ab)

Again, we get 1 from two couples and -1 from the other two couples, which gives the required bound
on the Jones polynomial coefficients.

For the two other families of knots the reasoning is similar so let us just group in couples the terms of J polynomials.
For K'_a,b, suppose that a+b is even:

    J_K'_a,b=-1(1-t^a+b)+t(1-t^a+b)+t^ab+2b(1-t^a-b)-t^ab+a+b+2(1-t^a-b-2)

Suppose now that a+b is odd:

    J_K'_a,b=-1(1-t^ab+2b)+t(1-t^ab+a+b-1)+t^a+b(1-t^ab+2)-t^a+b+1(1-t^ab+a-b-1)

For K”_a,b, suppose that a+b is even:

    J_K”_a,b=1-t^a+b-t^a+b+2(1-t^a+b-2)-t^ab+a+b+1(1-t^a-b-2)+t^ab+a+b+2(1-t^a-b-2)

Suppose now that a+b is odd:

    J_K”_a,b=1-t^ab+a+b+1-t^a+b(1-t^ab+2)-t^a+b+2(1-t^ab+a-b-3)+t^2a+2b(1-t^ab-2b)





Let k>0, then h(T(n,n+k))=k-1 for k≤ 3 except for h(T(2,5))=1.
Also, h(T(n,n+k))≥ 2 for k>3.

For k=1 or 2 the equality follows from Lemma <ref> and section <ref>.
From equation (<ref>) the J polynomial of T(n,n+k) is 1-t^n+1-t^n+k+1+t^2n+k.
If k>2 the J polynomials are different from the J polynomials of T(n,n+1) and T(n,n+2).
One checks case by case that they are also different from the J polynomials with four terms of all other knots K
with h(K)=1 except for K”_2,1 which is 5_1=T(2,5).
Thus, excluding T(2,5), if k>2 then h(T(n,n+k))≥ 2 and, as in section <ref> we saw that
h(T(n,n+3))≤ 2, it follows that h(T(n,n+3))=2.
 




§ ALGEBRAIC KNOTS K WITH H(K)≤ 1


An algebraic knot is an iterated torus knot of type {(p_1,q_1),(p_2,q_2),…,(p_s,q_s)}.
Denote such a knot K_s.
It is constructed inductively: K_1 is the torus knot of type (p_1,q_1), and, for 2≤ i≤ s, K_i
is a (p_i,q_i) cable of K_i-1.
Also, for every i∈{1,…,s}, gcd(p_i,q_i)=1, 1<p_i<q_i, and, for every i∈{1,…,n-1}, p_ip_i+1q_i<q_i+1 (see <cit.>).

In order to detect which knots K, with h(K)≤ 1, are algebraic knots, we use a formula for the colored Jones polynomial of the (p,q) cabling
of a knot K, denoted K(p,q), from <cit.> (see also <cit.>):


    V_K(p,q)(n)=A^-pq(n^2-1)_j=-n-1/2^n-1/2A^4qj(jp+1)V_K(2pj+1)Kpq



Note, that the conventions in <cit.> are: J for our V, t for our A, s for p and r for q.
The colored Jones V_K(n) is normalized so that V_U(n)=[n] for the unknot U, where:

    [n]=A^2n-A^-2n/A^2-A^-2

The usual Jones polynomial of K, denoted V_K, equals V_K(2)/[2] and V_K(1)=1 for every K.
Also, by convention, V_K(-n)=-V_K(n).

We have,

    [2](A^2-A^-2)=A^4-A^-4=t^-1-t=t^-1(1-t^2)

Let J_K(n)=(A^2-A^-2)V_K(n). For n=2, it is the polynomial J_K of section <ref>, up to some power of t.
Indeed, by definition of J_K, we have, for some a∈:

    J_K=t^a(1-t^2)V_K=t^a t [2](A^2-A^-2)V_K=t^a+1J_K(2)


Let K_s be an algebraic knot of type {(p_1,q_1),…,(p_s,q_s)}.
We want to compute the highest power of A in J_K_s and a few gaps between the following decreasing powers of A.

Let l_1=-2(q_1-1)(p_1-1)+2 and, for 1<i≤ s, l_i=-2q_ip_i+2q_i+p_il_i-1.


For any i≥ 1, -2q_ip_i<l_i<0

Clearly -2q_1p_1<l_1<0. Assume the lemma is true up to i-1.

    l_i=-2q_ip_i+2q_i+p_il_i-1>-2q_ip_i+2q_i-2p_iq_i-1p_i-1>-2q_ip_i

The last inequality comes from q_i>p_ip_i-1q_i-1.
Also, l_i is clearly negative by induction.




For s≥ 2, 4q_s-4(p_1+q_1)p_2… p_s+2l_s-1>0

First we check it for s=2:

    4q_2-4(p_1+q_1)p_2-4(q_1-1)(p_1-1)+4>4q_1p_2p_1-4(p_1+q_1)p_2-4q_1p_1+4p_1+4q_1=


    =4(p_2-1)q_1p_1-4(p_1+q_1)(p_2-1)=4(p_2-1)(q_1p_1-q_1-p_1)>0

Assume s>2. By the previous lemma and using q_s>p_sp_s-1q_s-1, it is sufficient to prove

    4p_sp_s-1q_s-1-4(p_1+q_1)p_2… p_s-4p_s-1q_s-1>0

or

    (p_s-1)p_s-1q_s-1-(p_1+q_1)p_2… p_s>0

As p_s-1/p_s≥1/2, it is sufficient to show

    (p_1+q_1)p_2… p_s-1/p_s-1q_s-1<1/2

One checks easily by induction on s>2 that

    (p_1+q_1)p_2… p_s-1/p_s-1q_s-1≤p_1+q_1/q_2 <2q_1/p_1p_2q_1≤1/2





Let n≥ 1. The highest power of A in J_K_s(n) is (n-1)l_s+2.
For n≥ 2, the three consecutive gaps between the four highest powers of A are:

4p_1… p_s(n-1)+4, 4(q_1-p_1)p_2… p_s(n-1), 4p_1… p_s(n-1)-4.

If n=1, one has for any knot K, J_K(1)=(A^2-A^-2)V_K(1)=A^2-A^-2 and the highest power of A is indeed 2.
For the rest of the proof we assume that n≥ 2.

Assume that s=1 so K_1 is a torus knot, K_1=T(p,q), p<q (to simplify the notation we use p and q instead of p_1 and q_1 in the case s=1).
Using [n](A^2-A^-2)=A^2n-A^-2n in formula (<ref>), with K the unknot, we get:

    J_T(p,q)(n)=A^-pq(n^2-1)_j=-n-1/2^n-1/2A^4qj(jp+1)(A^4pj+2-A^-4pj-2)

For a given j>0, consider the four terms in the sum coming from ± j:

    A^4qpj^2(A^4qj+4pj+2-A^4qj-4pj-2-A^-4qj+4pj-2+A^-4qj-4pj+2)

If j≥ 1, we check that the term with the lowest degree coming from ± j has higher degree than any of the terms coming from ±(j-1):

    4qpj^2-4qj-4pj+2>4qp(j-1)^2+4q(j-1)+4p(j-1)+2

or

    4(qj-1)(pj-1)-2>4(qj-q+1)(pj-p+1)-2

which is clearly true if j≥ 1.

Thus, the highest power of A comes from j=n-1/2 and equals:

    -pq(n^2-1)+pq(n-1)^2+2q(n-1)+2p(n-1)+2=(n-1)(-2pq+2q+2p)+2=


    =(n-1)(-2(p-1)(q-1)+2)+2=(n-1)l_1+2

The first three gaps come from j=±n-1/2 and are:
4p(n-1)+4, 4(q-p)(n-1) and 4p(n-1)-4.

Suppose now that s>1 and the proposition is true for all K_i, i<s.
Formula (<ref>) holds when we replace V(n) with J(n), since the second polynomial is A^2-A^-2 times the first.
Using induction on s and this modified formula, the term for j=n-1/2 gives a power of A:

    -p_sq_s(n^2-1)+p_sq_s(n-1)^2+2q_s(n-1)+((p_s(n-1)+1)-1)l_s-1+2=


    =(n-1)(-2p_sq_s+2q_s)+p_s(n-1)l_s-1+2=(n-1)l_s+2

Also, for j=n-1/2, the highest gap in J_K_s-1(2p_s j+1)=J_K_s-1(p_s(n-1)+1) is, by induction,

    4p_1...p_s-1((p_s(n-1)+1)-1)+4=4p_1...p_s-1p_s(n-1)

Similarly, the two following gaps are: 4(q_1-p_1)p_2...p_s(n-1) and 4p_1...p_s(n-1)-4.

We will prove now that the other terms in the sum in the formula for J_K_s 
(i.e. for j<n-1/2) have sufficently low powers so they do not perturb the four highest powers coming from the term for j=n-1/2.
Denote by d(j) the highest power of A for the j term in the sum (we ignore the common factor in front of the sum).

Let j≥1/2. We compare d(j) and d(-j).
By induction the highest power of A in J_K_s-1(2p_sj+1) is 2p_sjl_s-1+2. As
J_K_s-1(2p_s(-j)+1)=-J_K_s-1(2p_sj-1), its highest power of A is (2p_sj-2)l_s-1+2. The difference between these powers is -2l_s-1.
The coefficients in front of these two terms are A^4q_sj(jp_s+1) and A^4q_s(-j)((-j)p_s+1) giving a difference of -8q_sj.
Thus, the overall difference is

    d(-j)-d(j)=-8q_sj-2l_s-1

This means that the drop in power of A between j and -j is

    d(j)-d(-j)=8q_sj+2l_s-1≥ 4q_s+2l_s-1>0
 
the last inequality from Lemma <ref>. For j=n-1/2 the drop is 4q_s(n-1)+2l_s-1.
We check that even after substracting the three first highest gaps for j=n-1/2, the drop is still positive. 
The sum of these three gaps is 4(p_1+q_1)p_2… p_s(n-1). Using Lemma <ref> twice (once for l_s-1<0 together with n-1≥ 1),

    4q_s(n-1)-4(p_1+q_1)p_2… p_s(n-1)+2l_s-1≥


    (n-1)(4q_s-4(p_1+q_1)p_2… p_s+2l_s-1)>0

This shows that the term for j=-n-1/2 does not affect the four highest powers of A coming from the term j=n-1/2.

Now let j≥ 1 be an index in the sum. Since, for n=2, j=±1/2, this means that n≥ 3. 
We compare d(j) and d(j-1). For J_K_s-1 the difference between the term for j-1 and j
is, by induction, 

    ((2p_s(j-1)+1-1)l_s-1+2)-((2p_sj+1-1)l_s-1+2)=-2l_s-1p_s

For the coefficients the difference is

    4q_s(j-1)((j-1)p_s+1)-4q_sj(jp_s+1)=-4q_s(1+p_s(2j-1))

Thus the drop in power of A is

    d(j)-d(j-1)=4q_s(1+p_s(2j-1))+2l_s-1p_s>p_s(4q_s+2l_s-1)>0

The last inequality comes from Lemma <ref>.

For j=n-1/2 the drop is 4q_s(1+p_s(n-2))+2l_s-1p_s.
We check again that after substracting the three first gaps, the drop, say D, is still positive i.e.

    D=4q_s(1+p_s(n-2))-4(p_1+q_1)p_2… p_s(n-1)+2l_s-1p_s>0

We have:

    D/p_s(n-1)=4q_s1+p_s(n-2)/(n-1)p_s-4(p_1+q_1)p_2… p_s-1+2l_s-1/n-1


As n≥ 3, n-2/n-1≥1/2, so 1+p_s(n-2)/(n-1)p_s≥1/2. Also p_s≥ 2, 2/n-1≤ 1 and l_s-1<0, so:

    D/p_s(n-1)≥1/2(4q_s-4(p_1+q_1)p_2… p_s-1p_s+2l_s-1)>0

the last inequality from Lemma <ref> again. Thus D>0.

In conclusion, we have shown that: the term for j_0=n-1/2 gives the wanted 4 highest powers of A; the terms for j_0-1 and -j_0 do
not affect these 4 powers; all other j terms have lower powers than the term for j_0-1 (since d(j)-d(j-1)>0 if j≥ 1 and 
d(j)-d(-j)>0 if j>0). Thus, we get the wanted 4 highest powers of A for J_K_s(n). 




If K_s is an algebraic knot with h(K_s)≤ 1, K not a torus knot (i.e. s≥ 2), then K_s is doubly iterated of type
{(p_1,p_1+1),(2,q_2)} and it belongs to the family K”_a,b.

Recall that t=A^-4. Thus, the gaps between the highest powers of A become gaps between the lowest powers of t divided by 4 for the J polynomials.
As J_K_s equals J_K_s(2), up to a power of t, the first gap of J_K_s equals the first gap of J_K_s(2) which is p_1… p_s+1.
We check the gaps for K_a,b, K'_a,b and K”_a,b in section <ref>.   
As the first gaps of J of K_a,b and K'_a,b are equal to 1, these knots are not algebraic. 
The second gap of the J_K_s is (q_1-p_1)p_2… p_s. For K”_a,b, the second
gap equals 2 (except for the torus knot K”_2,1=T(2,5)). 
Thus, if K”_a,b is an algebraic knot, which is not a torus knot, then s=2, q_1-p_1=1 and p_2=2. 



We want to find necessary conditions on q_2 in the preceding corollary in order to have h(K_s)≤ 1.
Let K be of type {(p,p+1),(2,q)} and let K' be the torus knot T(p,p+1).
Using formula (<ref>),

    V_K(2)=A^-6q(A^4qV_K'(3)+V_K'(-1))=A^-6q(A^4qV_K'(3)-1)

Thus,

    J_K(2)=(A^2-A^-2)V_K(2)=A^-6q(A^4qJ_K'(3)-A^2+A^-2)

In the proof of Proposition <ref> is a formula for J_T(p,q)(n). Specializing to K' and n=3,

    J_K'(3)=A^-8p(p+1)(A^4p(p+1)(A^8p+6-A^2-A^-6+A^-8p-2)+A^2-A^-2)


    =A^-4p^2+4p+6(1-A^-8p-4-A^-8p-12+A^-16p-8+A^-4p^2-12p-4-A^-4p^2-12p-8)

Thus,

    J_K(2)=A^-6q+4q-4p^2+4p+6(1-A^-8p-4-A^-8p-12+A^-16p-8+


    +A^-4p^2-12p-4-A^-4p^2-12p-8-A^-4q+4p^2-4p-4+A^-4q+4p^2-4p-8)

Switching to t and ignoring the factor in front, so that the lowest term is 1, we get:

    J_K=1-t^2p+1-t^2p+3+t^4p+2+t^p^2+3p+1-t^p^2+3p+2-t^q-p^2+p+1-t^q-p^2+p+2JK

We get now a necessary condition on q in order to have h(K)≤ 1:


Let K be algebraic of type {(p,p+1),(2,q)}. If h(K)≤ 1 then q=2p(p+1)+1 and K=K”_p+1,p.

Let q_0=2p(p+1)+1.
As K is algebraic, q≥ q_0. We have,

    q_0-p^2+p+1=p^2+3p+2

Thus, for q_0, there are 7 increasing terms in equation (<ref>) with the sixth term having coefficient -2.
If q>q_0, there are 8 increasing terms in this equation.
We use the J polynomials of K”_a,b in section  <ref> to determine which K”_a,b may be equal to K.
Because of the five first terms of J_K, the only candidates are K”_b+1,b, b≥ 2 with 7 terms in J and
K”_a,b, 2<a≤ b, with 8 terms. We want to reject the second possibility in order to reject q>q_0.

Notice that for q_0 the last term in equation (<ref>) is t^p^2+3p+3 and if q>q_0 the power of the last
term is greater than p^2+3p+3.
If K”_a,b=K, then, comparing the second terms of J in equations (<ref>) and (<ref>), we get a+b=2p+1.
As a≤ b, one checks easily that ab≤ p(p+1). The last term of J for K”_a,b is t^ab+a+b+2 and we have,

    ab+a+b+2≤ p(p+1)+2p+1+2=p^2+3p+3

So K”_a,b cannot be equal to K if q>q_0.

Finally, for J_K, the first gap in equation (<ref>) is 2p+1, whereas for J_K”_b+1,b it is 2b+1.
Thus, if K=K”_b+1,b, then p=b, so K=K”_p+1,p.



Notice: we cannot conclude from the last proposition, that the knots K”_p+1,p are algebraic. It will be shown to be true in Theorem <ref>.

In the remaining part of this section we will show that, if K is algebraic and h(K)≤ 1, then C_alg(K)=h(K). This
gives a positive answer, for knots with h(K)≤ 1, to the problem of Fiedler (see the introduction). Because of Corollary <ref>
and Proposition <ref>, we have to show that if K is a doubly iterated knot of type {(p,p+1),(2,2p(p+1)+1)}, then C_alg(K)≤ 1.
We also need to check that for torus knots K in Proposition <ref> satisfying h(K)≤ 1, we have C_alg(K)≤ 1.

Let  be a small sphere in ^2, centered at the origin. Let (x,y)∈, with x=r_xe^iθ_x, y=r_ye^iθ_y and
r_x^2+r_y^2=ϵ^2. Intersecting  with complex lines through the origin gives the Hopf map p:→ P^1≈ S^2, 
p(x,y)=[x,y]. 

We can visualize p with the help of Figure <ref>.  is decomposed into an infinite family of tori and two S^1's. 
For fixed r_x and r_y with r_xr_y>0, (x,y) lies on a torus such as ∂ T (see Figure <ref>). Let us choose the solid torus T 
in such a way that (x,y)∈∂ T if r_x=r_y (T is 'half' of S^3, the other half being T').

If θ_x varies, θ_x∈[0,2π], we get the longitudes of the tori. Similarly, for θ_y∈[0,2π], we get the meridians of the tori. 
We choose the parametrization in such a way that, if (x,y)∈ D then θ_x=0.
Increasing r_x (and keeping r_x^2+r_y^2=ϵ^2), we get increasingly thinner tori inside T.
The two S^1's correspond to (r_x,r_y)=(ϵ,0) (the fiber going through the center of D) and 
(r_x,r_y)=(0,ϵ) (the fiber going through the center of D'). 

Now, p(x,y)=[r_xe^iθ_x, r_ye^iθ_y]=[r_x,r_ye^i(θ_y-θ_x)].
If r_x>0 the map: f: (x,y)→ (r_x,r_ye^i(θ_y-θ_x)) is well defined. Let us restrict f to {(x,y)∈,r_x≥ r_y}.
Then f maps the solid torus T onto D and, by definition, it represents the Hopf map p, restricted to T.
Thus, if L is a link in T and f(L) generic in D, then an arrow diagram of L is obtained from f(L) by adding some arrows and 
information of under/over at the crossings. In particular, the number of crossings of this arrow diagram equals the number of crossings of f(L).
Using f:T→ D we can now estimate C_alg(K) for algebraic knots K with h(K)≤ 1.

Let X be a complex plane algebraic curve with a singularity at the origin of ^2. Let ϵ<<1 be small and let K=X∩.
Assume that, for some p∈, the Puiseux expansion of X in 0 (see <cit.>) is:

    y=x^p+1/p(1+x^1/2p)

We use the notations of <cit.>, where y=x^q_1/p_1(a_1+x^q_2/p_1p_2), so q_1=p+1,p_1=p,q_2=1 and p_2=2.
It is shown there that K is an iterated torus knot of type {(p_1,α_1),(p_2,α_2)}
with α_1=q_1 and α_2=q_2+p_1p_2α_1 (Proposition 1A.1 in <cit.>).

As, in our case, α_2=1+2p(p+1), K is of type {(p,p+1),(2,1+2p(p+1))}.
Notice, that p=1 is not excluded, in which case K is a (2,5) cable of the trivial torus knot of type (1,2), i.e. K is
the torus knot T(2,5).

We can choose a parametrization of X with t=re^iθ:

x=t^2p

y=t^2p+2(1+t)

For (x,y)∈ K=X∩, we have,

    r_x^2+r_y^2=r^4p+r^4p+4|1+re^iθ|^2=r^4p+r^4p+4(1+r^2+2rcosθ)=ϵ^2


We assumed that ϵ is small. It can be checked easily that, for each θ∈ [0,2π), there is a unique r satisfying the preceding
equality. Denote it by r_θ. It is also easily checked, that r_θ is increasing for θ∈[0,π] and decreasing for θ∈[π,2π].

Now x=r_θ^2pe^2ipθ and y=r_θ^2p+2e^i(2p+2)θ(1+r_θ e^iθ). As ϵ<<1 is arbitarily small, we can also assume that 
r_θ <<1. Then it is clear that r_x>r_y, so (x,y)∈ T and:

    f(x,y)=(r_θ^2p,r_θ^2p+2e^2iθ(1+r_θ e^iθ))


We see that for θ∈[0,2π], r_θ^2p will be increasing from 0 to π, then decreasing from π to 2π.
Also, for θ∈[0,2π], the argument of the second coordinate of f(x,y) goes from 0 to 4π, because of e^2iθ, and it does
so monotonically. Indeed, d/dθarg(e^2iθ)=2, whereas it is easily checked that |d/dθarg(1+r_θ e^iθ)|<1 for
r_θ<<1. Thus, for θ∈[0,2π], f(x,y) is a curve going twice around the center of D, first closer to it (as r_x increases), then away from it
(as r_x decreases). It follows that f sends K onto a curve with 1 crossing. This means that C_alg(K)≤ 1. 
From Lemma <ref>, h(K)≠ 0. As h(K)≤ C_alg(K), we get C_alg(K)=1.

Finally, we consider the torus knots in Proposition <ref> satisfying h(K)≤ 1. These are the torus knots T(n,n+1), T(n,n+2) and
T(2,5). The knot T(2,5) has been considered above as a (2,5) cable of the trivial knot, so C_alg(T(2,5))=1. 
For T(n,n+k), k=1 or 2, Fiedler states in <cit.> (remark on page 260), that C_alg(T(n,n+k))=k-1, with T(n,n+1) being realized
by the curve X:z^n=w^n+1 and T(n,n+2) by a slight perturbation of the curve with equation z^n=w^n+2. The second case can be checked
using the following paramatrization for the perturbation of equation z^n=w^n+2: z=t^n and w=t^n+2(1+t).
With the same proof as for the doubly iterated torus knots above, one can show that this perturbation gives the knot T(n,n+2) 
with a projection on the disk D with 1 crossing.

We resume the results of this section in the following theorem:


The family of algebraic knots K with h(K)≤ 1 consists of: torus knots T(n,n+1), if h(K)=0; torus knots T(n,n+2), T(2,5) or K”_p+1,p
if h(K)=1. The knots K”_p+1,p are doubly iterated torus knots of type {(p,p+1),(2,2p(p+1)+1)}. Furthermore, for all such knots C_alg(K)=h(K).

The only remaining thing to check is that the knots K”_p+1,p are indeed algebraic. Let K be an algebraic knot of type {(p,p+1),(2,2p(p+1)+1)}.
We have shown above that C_alg(K)=1. But h(K)≤ C_alg(K). By Proposition <ref>, K must be equal to K”_p+1,p.



We conclude with two remarks.


It can be shown directly, using Reidemeister moves, that K”_p+1,p is doubly iterated of type {(p,p+1),(2,2p(p+1)+1)}.
For example, we can transform K”_3,2 with _∞ and _4 into a diagram shown in Figure <ref>.
Then one could remove the two couples of arrows in this figure with appropriate Reidemeister moves to get a cable of the trefoil knot T(2,3) and
check that it is the (2,13) cable. This generalizes to any number p of couples of arrows (for p=1 we get the middle diagram of Figure <ref>).
However, this proves only that such doubly iterated torus knots K satisfy h(K)=1 and we have shown above that C_alg(K)=1, which is stronger.

It can also be checked that changing the crossing in Figure <ref> gives a (2,11) cable of the trefoil, which is also the knot K”_4,1.
More generally, we would get the family K”_p+2,p-1, p≥ 2, of doubly iterated knots of type {(p,p+1),(2,2p(p+1)-1)}. Notice,
that these are not algebraic knots.




 Fiedler has shown in <cit.>, Theorem 1, the following inequality for K algebraic of type {(p_1,q_1),…,(p_n,q_n)}, q_1<2p_1:

    C_alg(K)≥(q_1-p_1)p_2… p_n-1

The right hand side of this inequality equals one if: q_1-p_1=2 and n=1 (torus knots T(p_1,p_1+2)) or q_1-p_1=1, p_2=2 and n=2 
(type {(p_1,p_1+1),(2,q_2)}).
In the second case, it follows from Proposition <ref> that C_alg(K)≥ h(K)>1, if q_2>2p_1(p_1+1)+1. Thus, for such q_2, Fiedler's
inequality is not sharp. This can be compared to Theorem 2 in <cit.>, which gives a sharper inequality then Theorem 1 for doubly iterated torus knots
with p_1 and q_1 both odd.




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