Yuanan Diao, Gabor Hetyei and Kenneth Hinson- Tutte Polynomials of Signed Graphs and Jones...

8/3/2019 Yuanan Diao, Gabor Hetyei and Kenneth Hinson- Tutte Polynomials of Signed Graphs and Jones Polynomials of Som

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Tutte Polynomials of Signed Graphs and Jones Polynomials ofSome Large Knots

Yuanan DIAO, Gabor HETYEI and Kenneth HINSON

ABSTRACT

It is well-known that the Jones polynomial of a knot is closely related to the Tutte polynomial of a special graph

obtained from a regular projection of the knot. In this paper, we study the Tutte polynomials for signed graphs.

We show that if a signed graph is constructed from a simpler graph via k-thickening or k-stretching, then its

Tutte polynomial can be expressed in terms of the Tutte polynomial of the original graph, thus enabling us to

compute the Jones polynomials for some (special) large non-alternating knots.

Mathematical Subject Classification 2000: 57M25

Keywords: knots, Jones polynomials, Tutte polynomials, Signed graphs.

1. Introduction

A central issue in knot theory is to distinguish different knots from each other. For this purpose,various knot invariants have been developed and widely studied, among them are several wellknown knot polynomials such as the Jones polynomials and the Homfly polynomials [19, 24].Although none of the known knot polynomials is a complete knot invariant (a knot invariant isnot complete if some distinct knots yield the same knot invariant), they still provide the mostpowerful tools in this task. Another difficult issue is to tabulate the knots. Traditionally, thisis done according to the minimum crossing numbers of knots. For each given crossing numbern, one has to go through all possible knot projections with n crossings. So far, knots and linksup to 16 crossings have been completely tabulated [21], and alternating knots and links up to19 crossings have been tabulated [29].

In recent years, knot theory has found many applications in fields such as biology, chemistryand physics. Many applications involve not only the topology, but also the geometry of knots

and links. For a sample of such studies, see [1, 2, 6, 8, 15, 16, 17, 18]. One specific questionis about the ropelength of a knot. Intuitively, this can be thought of as the minimum lengthrequired to tie a knot with a physical flexible rope of unit radius (or diameter). For a morerigorous definition, see for example [12, 27]. The importance of this problem is obvious sinceknots and links are realized by rope-like materials in the real world (from fishing net to circularDNA). For knots with a small crossing number, computer simulations may be used to findthe approximate values of the ropelengths. For large knots, this becomes impractical. Some


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theoretical asymptotic studies have been carried out in this case and the best known resultstates that the ropelength of a knot with crossing number n is bounded above by O(n3/2) [14].However, the knots with the longest known ropelengths (constructed in [13]) are realized withropelengths linear in terms of their crossing numbers. This means that the general ropelengthupper bound lies between O(n) and O(n3/2). This gap proves to be very hard to close, despite

smaller upper bounds have been found for many large knot classes [9, 10] and some preliminarynumerical studies indicate the O(n) may not be the universal upper bound [11]. In order toprove that the actual upper bound is less than O(n3/2), one needs to come up with new methodsother than the one used in [14]. On the other hand, in order to prove that the general upperbound is larger than O(n), one would have to construct classes of knots and prove that theyrequire ropelengths more than the linear order of their crossing numbers. A main difficultyin constructing large knots is that if the knots are not alternating, then there are no readilyusable tools to determine their crossing numbers in general. Although the breadth of the Jonespolynomial provides a lower bound on the crossing number of a knot, the computation of itbecomes increasingly difficult as the number of crossings in a knot diagram increases. Currently,for knot diagrams with thousands of crossings, it is simply not possible to compute their Jones

polynomials using the existing programs and computers. This motivates the study of thispaper. We would like to find a way to compute the Jones polynomials for some special largenon-alternating knots. In doing so, we can obtain a lower bound on the crossing numbers ofthese knots and hence may be able to use such knots in our ropelength study in the future.

It is well-known that the Jones polynomial of a knot can be obtained from the Tuttepolynomial of a special graph obtained from a regular projection of the knot [23, 25, 26]. In thispaper, we will study the Tutte polynomials for signed graphs. We show that if a signed graphis constructed from a simpler graph via k-thickening or k-stretching, then its Tutte polynomialcan be expressed in terms of the Tutte polynomial of the original graph, thus enabling us to

compute the Jones polynomials for some (special) large non-alternating knots. In the nextsection, we will give some preliminaries on Tutte polynomials for signed graphs. In Section3, we will derive a formula for the Tutte polynomial of a signed graph G when the graph isobtained from a simpler graph via stretching and thickening. In Section 4, we will use theconnection between the Tutte polynomials and the Jones polynomials and some examples toshow how the Jones polynomials of some non-alternating knots constructed from simpler knotsmay be computed.

2. Preliminaries

In this section, we introduce the concept of Tutte polynomial for signed graph following closely

to [25, 26], with some slight modification. This is mainly for the convenience of our reader. Amore general and thorough treatment of Tutte polynomials for colored graphs can be found in[3].

Definition 1 Let G be a graph with edges labelled 1, 2, . . . , n, and let T be a spanning tree ofG. An edge e of T is said to be internally active if the label of e is the smallest label of any edge


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in G that connects the two components of T \ {e}. e is said to be internally inactive otherwise.On the other hand, an edge f of G \ T is said to be externally active if the label of f is thesmallest label of any edge in the unique cycle contained in T {f}. Otherwise, f is said to beexternally inactive.

Example 2 Figure 1 shows a graph with six edges and four vertices where a labelling of theedges is given. The edges of a spanning tree T are highlighted in the graph. With respect tothe tree T, edges 1, 4 and 5 are internal edges and edges 2, 3 and 6 are external edges. Edge 1is internally active since it has the smallest label, edge 4 is internally inactive since 4 is largerthan 3 in the cycle 134. Edge 5 is also internally inactive since 5 is larger than 2 in the cycle245. Edge 2 is externally active since 2 is the smallest in the cycle 245. Edge 3 is externallyinactive since 3 is larger than 1 in the cycle 134 and edge 6 is apparently externally inactivesince 6 is the largest of all the labels.

2

3

56

1

4

Figure 1: An example of labelled graph with a marked spanning tree.

Let G be a signed (and connected) graph, that is, each edge in G is assigned a + or sign.Let T be a spanning tree of G, for each edge e in G we will then assign one of the followingvariables to it according to the activities of e (with respect to the tree T):

sign of e activity variable assignment

+ internally active x+ internally active x+ externally active y+ externally active y+ internally inactive A

+ internally inactive A+ externally inactive B+ externally inactive B

Table 1: The variable assignment of an edge with respect to a spanning tree T.


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Definition 3 LetG be a connected signed graph. For a spanning tree T r ofG, letC(T r) be theproduct of the variable contributions from each edge of G according to the variable assignmentabove, then the Tutte polynomial T(G) is defined as the sum of all the C(T r)s over all possiblespanning trees of G.

Example 4 The graph given in Example 2 is assigned the signs as shown in Figure 2 below.

3

6

+

+

+

+

24

5

1

Figure 2: An example of signed and labelled graph with a marked spanning tree.

For the spanning tree shown in the figure, the total contributions of all the edges is easilycalculated to be x+y+A+AB+B. The complete list of the spanning trees of G is given inFigure 3. We leave it for our reader to verify that the Tutte polynomial of G is

4x+y+A+AB+B + 2y2+A+A

2

B+ + 2x2+A+B+B

2

+ y2+A2+AB + x

2+AB

2+B + y

2+yA

2+A + x

2+xB

2+B

+ y+A+A2

B2+ + y+A3+B

2

+ x+A2

B3+ + x+A2+B+B

2

.

Notice that the Tutte polynomial defined this way is labelling dependent. To remedy thesituation, we want to factor the polynomial ring Z[] := Z[A+, A, B+, B, x+, x, y+, y] withan appropriate ideal I, such that the formula for T(G) in Z[]/I becomes labelling independent.An exact description of all such ideals (for a larger class of signed graphs) was given by Bollobasand Riordan [3, Theorem 2]. (Here we state the two-colored version.)

Theorem 5 (Bollobas-Riordan) Assume I is an ideal ofZ[]. Then the homomorphic im-age of T(G) inZ[]/I is independent of the labelling of the edges of G if and only if

det x+ B+

x B

det A+ y+

A y

I,

ydet

A+ y+A y

ydet

A+ B+A B

I for {+, }, and

xdet

A+ y+A y

xdet

A+ B+A B

I for {+, }.


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p p

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4 2

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Figure 3: The complete set of spanning trees of the graph given in Figure 1.

Bollobas and Riordan denote the ideal generated by the differences listed in Theorem 5 byI0. The homomorphic image of T(G) in Z[]/I0 is the most general signed Tutte polynomialwhose definition is independent of the labelling. In Section 4 we describe how to factor Z[]by an ideal properly containing I0 to get the Jones polynomial. To simplify our calculations,we want to replace I0 with a larger ideal I1 in such a way that the effect of certain operationsof signed graphs is still describable in terms of Tutte polynomials (as elements ofZ[]/I1). Wewill keep our calculations as simple as possible since we want to obtain a homomorphic imagein an integral domain, and send A+, A, B+, B, x+, x, y+, and y into nonzero entries atthe end. Inspired by [3, Corollary 3] we make the following definition.

Definition 6 We consider the signed Tutte polynomial to be an element of Z[]/I1 where I1is the ideal generated by

det

x+ B+x B

det

A+ B+A B

and det

A+ B+A B

det

A+ y+A y

.

Clearly I1 properly contains I0 so our Tutte polynomial is labelling independent.

Finally, since the Tutte polynomial is labelling independent under our conditions, it is easyto see that we have the following recursive formula

T(G) = BT(G\e) + AT(G/e), (1)

where e is any given edge ofG with sign , G\e is the graph obtained from G by deleting e andG/e is the graph obtained from G by contracting e.


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3. The Tutte Polynomials of Stretched and Thickened Signed Graphs

Definition 7 Let G be a signed graph. The positive (negative) k-stretch of G is the graphG+(k ) (G(k )) obtained by replacing each positive (negative) edge in G by a path of lengthk such that each new edge in the path has also positive (negative) sign. The positive (negative) k-

thickening ofG is the graphG+(k ) (G(k )) obtained by replacing each positive (negative)edge in G by k parallel edges (connecting the same vertices incident to the edge being replaced),such that each new edge in the path is also of positive (negative) sign.

For unsigned graphs (corresponding to alternating knots), formulas for the k-thickening andk-stretch may be found in [5][Lemma 6.3.24], [23][(7.2) and (7.3)], and [22][(3.8) and (3.10)].Our first result is close to this known case and the proof we present could be used with littlechange to prove the old result. We could not find a similar argument in the literature, whichis surprising since this proof is closest to Tuttes original reasoning about internal and externalactivities.

Theorem 8 Consider all signed Tutte polynomials as elements ofZ. LetG be a connectedand signed graph and let us fix a labelling of the edges of G. Refine this labelling to labellingsof G+(k ), G+(k ), G(k ) and G(k ) in such a way that each label associated to anedge e in G is replaced by k consecutive labels of the k edges replacing e in G+(k ), G+(k ),G(k ) and G(k ). Then

T(G+(k ); A, A+, B, B+, x, x+, y, y+) =

TG; A, Ak+, B, B+ xk

+Ak

+

x+A+, x, x

k+, y, A

k1+ y+ +

xk1+

Ak1+

x+A+B+x+ , (2)

T(G(k ); A, A+, B, B+, x, x+, y, y+) =

T

G; Ak

, A+, Bxk

Ak

xA

, B+, xk

, x+, Ak1

y +xk1

Ak1

xA

Bx, y+

,(3)

T(G+(k ); A, A+, B, B+, x, x+, y, y+) =

T

G; A, A+yk+Bk

+

y+B+, B, B

k+, x, B

k1+ x+ +

yk1+

Bk1+

y+B+A+y+, y, y

k+

,

(4)

T(G(k ); A, A+, B, B+, x, x+, y, y+) =

T

G; Ayk

Bk

yB

, A+, Bk

, B+, Bk1

x +yk1

Bk1

yB

Ay, x+, yk

, y+

.(5)

Note that all Tutte polynomials above depend on the labellings chosen, since we have notfactored by any relation among our letters. For the sake of convenience for later use, we listthe substitutions in the theorem in the following table.

Proof. Each positive edge e of G is replaced by a path of k consecutive edges e1, . . . , ek inG+(k ). Consider a fixed spanning tree of G+(k ). This must contain at least k 1 elements


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Positive k-Stretching Negative k-Stretching

x+ xk+ x x

k

y+ Ak1+ y+ +

xk1+

Ak1+

x+A+B+x+ y A

k1

y +xk1

Ak1

xA

BxA+ A

k+ A A

k

B+ B+xk+Ak

+

x+

A+B B

xk

Ak

x

APositive k-Thickening Negative k-Thickening

x+ Bk1+ x+ +

yk1+

Bk1+

y+B+A+y+ x B

k1

x +yk1

Bk1

yB

Ayy+ y

k+ y y

k

A+ A+yk+Bk

+

y+B+A A

yk

Bk

yB

B+ Bk+ B B

k

Table 2: The variable substitutions in the Tutte polynomial of a signed graph under the k-

stretching and k-thickening operations.

of the set {e1, . . . , ek}. Let us call a positive edge e of G selected if all elements {e1, . . . , ek}belong to the spanning tree of G+(k ), otherwise call it not selected. For negative edges of G(which are also edges of G+(k )) call an edge selected or not selected depending on whetherthey belong to our fixed spanning tree of G+(k ). It is easy to see that the selected edges ofG must form a spanning tree ofG. Conversely, each spanning tree ofG+(k ) may be uniquelydetermined by making the following sequence of decisions:

(i) Select a spanning tree of selected edges in G;

(ii) For each unselected positive edge e pick exactly one of {e1, . . . , ek} to be not a part ofthe corresponding spanning tree of G+(k ).

The signed Tutte polynomial G+(k ) is a sum of terms over all spanning trees ofG+(k ).We may group these trees by the selection made in step (i) and thus obtain a sum over thespanning trees of G. Let us fix one spanning tree ofG and calculate the contribution of eachedge e to this sum.

It is not hard to verify that negative edges have the same attributes (internal/external and

active/inactive) in both G and G+

(k ). Hence when we calculate T(G+

(k )) from T(G) wemay leave the negative letters unchanged. The interesting changes occur in the contribution ofthe positive edges.

Consider first an internally active positive edge in G (contributing a factor of x+ to a termin T(G)). This is replaced by k edges, which are all internally active in all correspondingspanning trees G+(k ). Hence each letter x+ in T(G) needs to be replaced with a term xk+ in


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T(G+(k )). Similarly, internally inactive positive edges are replaced with k edges which areall internally inactive in all corresponding spanning trees of G+(k ). Hence each letter A+ inT(G) needs to be replaced with a term Ak+ in T(G

+(k )).

Describing the change in the contribution of an external positive edge e is more complex.We have k choices to mark exactly one element of {e1, . . . , ek} as not part of the spanning treeof G+(k ) and, correspondingly, each letter B+ and y+ in T(G) will have to be replaced bysome sum of k terms to get T(G+(k )). Assume that e is externally inactive in G, the labelsof e1, . . . , ek increase in this order, and consider the contribution of those spanning trees ofG+(k ) in which ei is missing from the tree. It is easy to see that the edges e1, . . . , ei1 areinternally active, ei is externally inactive, the remaining ei+1, . . . , ek are internally inactive inG+(k ). Thus we obtain a contribution ofxi1+ B+A

ki+ . Summing over i yields that each letter

B+ needs to be replaced with a factor of B+xk+Ak

+

x+A+to get T(G+(k )).

Consider finally the change in the contribution of an externally active positive edge of G.This case is similar to the externally inactive one, the only difference being that ei is externally

active in T(G+(k )) if and only if i = 1. Summing over i yields that each letter y+ needs tobe replaced with a factor of Ak1+ y+ +

xk1+

Ak1+

x+A+B+x+ to get T(G

+(k )).

The proofs of equations (3), (4) and (5) are similar and are left to the reader.

Remark 9 It should be noted that equation (4) is the dual of (2) in the sense that replacingeach graph with its dual, while leaving the signs of the edges unchanged, induces replacingk-stretch with k-thickening, swapping the letter x with y and swapping the letter A withB where {, +}. Hence, for planar graphs, equation (4) is also a consequence of (2). Ingeneral, one would need to observe that our proof may be repeated without substantial change

for matroids.

Theorem 8 suggests to consider the following endomorphisms of the polynomial ring Z[].The operator +k : Z[] Z[] sends the negative letters into themselves, A+ into A

k+, B+ into

B+xk+Ak

+

x+A+, x+ into x

k+, and y+ into A

k1+ y+ +

xk1+

Ak1+

x+A+B+x+. The operator

+k : Z[] Z[]

also sends the negative letters into themselves, and it sends A+ into A+yk+Bk

+

y+B+, B+ into B

k+, x+

into Bk1+ x+ +yk1+

Bk1+

y+B+A+y+, and y+ into y

k+. IfT(G) Z[] is a signed Tutte polynomial for

any fixed labelling of the edges ofG then, by Theorem 8, +k (T(G)) is a signed Tutte polynomialof T(G+(k )) and +

k(T(G)) is a signed Tutte polynomial of T(G+(k )). Unfortunately, we

may be given a polynomial whose congruence class modulo I1 is T(G) without this polynomialarising as a signed Tutte polynomial for any fixed labelling. This will not lead to inconsistency,because of the following fact of algebra.

Theorem 10 LetI1 be ideal given in Definition 6. Then the endomorphisms +k and

+k take

I1 into itself.


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Proof. We show the statement for +k only, the proof for +k may be obtained by swapping

the letter x with y and swapping the letter A with B, where {, +}. It is sufficient toshow that +k takes the generators of I1 into I1. Thus we want to show that

det+k (x+)

+k (B+)

+k (x) +k (B) = det

+k (A+) +k (B+)

+k (A) +k (B) = det

+k (A+) +k (y+)

+k (A) +k (y) , i.e.,

det

xk+ B+

xk+Ak

+

x+A+

x B

= det

Ak+ B+

xk+Ak

+

x+A+

A B

= det

Ak+ A

k1+ y+ +

xk1+

Ak1+

x+A+B+x+

A y

holds in Z[]/I1. The difference of the first two determinants is

B(xk+ A

k+) B+

xk+ Ak+

x+ A+(x A) =

xk+ Ak+

x+ A+(B(x+ A+) B+(x A))

which is a multiple of

B(x+ A+) B+(x A) = det

x+ B+x B

det

A+ B+A B

I1.

The difference of the second and third determinants is

Ak+(B y) A

B+

xk+ Ak+

x+ A+

Ak1+ y+ +

xk1+ Ak1+

x+ A+B+x+

= Ak+(B y) AB+x

k+ B+A

k+ A

k1+ x+y+ + A

k+y+ B+x

k+ A

k1+ B+x+

x+ A+

= Ak+(B y) AAk1+

B+A+ x+y+ + A+y+ B+x+x+ A+

= Ak+(B y) AAk1+ (B+ y+)

which is the Ak1+ -multiple of

A+(B y) A(B+ y+) = det

A+ B+A B

det

A+ y+A y

.

Corollary 11 The endomorphisms

+

k and

+

k induce endomorphisms of the ringZ

[]/I1.

Example 12 For the two simple signed graphs G and H with the given labelling shown inFigure 4 below, we have T(G) = x+B + y+A and T(H) = y+y. It follows that

T(G+(k )) = xk+B + (Ak1+ y+ +

xk1+ Ak1+

x+ A+B+x+)A


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and

T(H+(k )) = (Ak1+ y+ +xk1+ A

k1+

x+ A+B+x+)y.

Thus for the signed graphs G and H shown in Figure 4 (under the given labelling), we have

T(G) = xk+Bx

m

A

m

x A+ (Ak1+ y+ + x

k1

+ A

k1

+x+ A+

B+x+)Am (6)

and

T(H) = (Ak1+ y+ +xk1+ A

k1+

x+ A+B+x+)(A

m1

y +xm1

Am1

x ABx). (7)

G

+

2

1

H

+

H

+

1

2

G

+

12

k

k+1k+2

k+m

1

2

kk+1

k+2k+m

.

. . . . .

. . . .

. .. .

.

.. . .

.

Figure 4: Two simple signed graphs and the signed stretches obtained from them.

The Tutte polynomials of more complicated graphs obtained from G

and H

via thickeningcan be similarly obtained in this manner.

4. Converting the Tutte Polynomials into Jones Polynomials

The first task is to obtain a signed graph from a knot projection. From a knot K, we firstshade the regions in its projection either white or dark in a checkerboard fashion, so thatno two dark regions are adjacent, and no two white regions are adjacent. We consider theinfinite region surrounding the knot projection to be white. Note that as we move diagonallyover a knot crossing, we go from a white region to a white region, or from a dark region to a

dark region. Next we let the dark regions in our projection correspond to vertices in a planargraph G, and we let the knot crossings correspond to edges in G. So if we can move diagonallyover a knot crossing from one dark region to another, then these two dark regions and thecrossing will be represented in G as two vertices connected by an edge. Note that we mayobtain parallel edges from some knot projections. Now we have our unsigned graph. To obtainthe signed version, we look at each crossing in the knot projection. If, after the upper strandpasses over the lower, the dark region is to the left of the upper strand, then we denote this


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as a positive crossing. If the dark region is to the right of the upper strand, we denote it as anegative crossing. Then our signed graph is obtained by marking each edge of G with the samesign as the crossing of K to which it corresponds. See Figure 5 for an example.

The following theorem is due to Kauffman [25, 26].

Theorem 13 LetG be the graph obtained from a knot diagram D ofK as described above, thenT(G) equals the Kauffman bracket polynomial K under the following variable substitutions:

x+ A3, x A

3, y+ A3, y A

3

A+ A, A A1, B+ A

1, B A.

Furthermore, the Jones polynomial VK(t) of K can be obtained from

VK(t) = (A3)w(K)K (8)

by setting A = t1

4 , where w(K) is the writhe of the projection D.

Using Theorems 8 and 13, we are able to compute the Jones polynomials for some largenon-alternating knots. We will demonstrate this by a few examples. We will first do this for asmall knot so we can compare our result with the direct computation result using an existingsoftware. We will then do this for a much larger knot beyond the capacity of the existingprograms.

Our first knot is the one shown in Figure 5. In this projection diagram, we see 19 crossings

(10 positive and 9 negative with respect to the shaded regions). The corresponding signedgraph G of this diagram (using the shaded regions in the figure as the vertices) is shown in thefigure as well.

+

++

4

5

61

32

78

9+++

18

1617

__

_

15

14

13

_

_

_10

__

_

1112

+++ 19 +

Figure 5: A 19 crossing knot diagram and its corresponding signed graph G.

Denote the edge labelled by 19 in the figure by e. Let G\e be the graph obtained from Gby deleting e and let G/e be the graph obtained from G by contracting e, that is, the vertices


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G\e

++

4

5

61

32

78

9+++

18

1617

__

_

1514

13

_

_

_10

__

_

1112

+++

+

46

1012 1618

1315

7913

_

+

G/e

Figure 6: The graphs obtained from the graph G in Figure 5 by deletion and contraction.

incident to e are identified as one single vertex after e is removed. See Figure 6. By the recursiveformula (1), we have

T(G) = B+T(G\e) + A+T(G/e).

Notice that G\e and G/e can be obtained by thickening the graphs G and H in Example 12(for the case of k = 3). This means we need to make the following substitutions in T(G) andT(H):

A A(y2

+ yB + B2

) = A7 A3 + A,

A+ A+(y2+ + y+B+ + B

2+) = A

7 A3 + A1,

B B3

= A3,

B+ B3+ = A

3,

x B2

x + (y + B)Ay = A7 A3 A5,

x+ B2+x+ + (y+ + B+)A+y+ = A

7

A3

A5

,y y

3

= A9,

y+ y3+ = A

9

according to Theorems 8 and 13. After substituting t1

4 for A in Equation (8) (with w(K) =1), we obtain

VK(t) = t10(1 4t + 12t2 26t3 + 49t4 74t5 + 96t6 112t7 + 110t8

97t9 + 77t10 47t11 + 23t12 8t13 2t14 + 3t15 t16 + t17).

This matches the computation result obtained by using Knotscape [20]. Let us now calculatethe Jones polynomials for some larger knots constructed in a manner similar to the last exampleby modifying the knot in Figure 5 into k groups ofk positive twists and k groups ofk negativetwists for some odd positive integer k. (The example we just did is the case k = 3.) Use thefacts that the writhe of the knot is 1,

K = T(G) = B+T(G\e) + A+T(G/e) = A1T(G\e) + AT(G/e)


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and that T(G\e) and T(G/e) can be computed from T(G) and T(H) by the following substi-tutions

A Ayk

Bk

y B=

(A3)k Ak

A2 + A2,

A+ A+yk+

Bk+

y+ B+ = (A3)k Ak

A2 + A2 ,

B Bk

= Ak,

B+ Bk+ = A

k,

x Bk1

x +yk1

Bk1

y BAy = A

k+2 +(A3)k1 Ak1

A + A5,

x+ Bk1+ x+ +

yk1+ Bk1+

y+ B+A+y+ = A

(k+2) +(A3)k1 A(k1)

A1 + A5,

y yk = A

3k

,y+ y

3+ = A

3k.

For k = 5, we have

VK(t) = t26(1 6t + 26t2 91t3 + 275t4 737t5 + 1796t6 4021t7 + 8366t8 16284t9

+ 29818t10 51606t11 + 84676t12 132106t13 + 196368t14 278544t15 + 377546t16

489336t17 + 606846t18 720177t19 + 817720t20 887911t21 + 920952t22 911068t23

+ 857489t24 765053t25 + 643579t26 505933t27 + 366267t28 237242t29 + 128459t30

45354t31 11121t32 + 43431t33 56574t34 + 56418t35 48576t36 + 37646t37

26696t38 + 17478t39 10594t40 + 5941t41 3081t42 + 1466t43 637t44

+ 250t45 86t46 + 26t47 6t48 + t49).

For k = 7, we have

VK(t) = t50(1 8t + 43t2 183t3 + 666t4 2157t5 + 6370t6 17425t7 + 44654t8 108067t9

+ 248536t10 545847t11 + 1149387t12 2328122t13 + 4548764t14 8593271t15

+ 15728483t16 27941544t17 + 48253003t18 81115378t19 + 132896097t20 212430488t21

+ 331612373t22 505966329t23 + 755122019t24 1103084529t25 + 1578177868t26

2212528476t27 + 3040964638t28 4099238067t29 + 5421525110t30 7037249318t31

+ 8967357925t32 11220302612t33 + 13788073932t34 16642729060t35 + 19733901580t36

22987768175t37 + 26307888783t38 29578176531t39 + 32668072879t40 35439739995t41

+ 37756834115t42 39494183464t43 + 40547500331t44 40842187270t45 + 40340281916t46

39044758086t47 + 37000613964t48 34292527561t49 + 31039223375t50 27385045791t51

+ 23489568932t52 19516242140t53 + 15621197678t54 11943255900t55 + 8596013866t56


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5662618702t57 + 3193491734t58 1206952613t59 307642091t60 + 1384824407t61

2076750500t62 + 2446499548t63 2561637408t64 + 2488610216t65 2288339306t66

+ 2013210200t67 1705443587t68 + 1396707072t69 1108711902t70 + 854497057t71

640099923t72 + 466343256t73 330533561t74 + 227925671t75 152882699t76

+ 99711533t77 63198913t78 + 38898252t79 23227709t80 + 13441772t81 7528452t82

+ 4074514t83 2126998t84 + 1068635t85 515390t86 + 237854t87 104637t88 + 43667t89

17180t90 + 6321t91 2150t92 + 666t93 183t94 + 43t95 8t96 + t97).

For k = 9, the polynomial is too large to list, so we only list a few terms below:

VK(t) = t82(1 10t + 64t2 319t3 + 1345t4 5008t5 +

+ 20193935024459t97 101497138129454t98 +

5008t156 + 1345t157 319t158 + 64t159 10t160 + t161).

5. Ending Remarks and AcknowledgementIn the examples of the last section, we computed the Jones polynomials for knots with up to163 crossings in the diagram. The computation for the 163 crossing example took about 15minutes on a PC using Maple.

Some quick observations from the polynomials of the examples in the last section: Theknots constructed this way are non-alternating since the Jones polynomials are not alternating.It follows that the crossing numbers of these knots (at least for the ones we have computedabove) are at least 2k2 since the breadth of the polynomials is 2k2 1 and the crossing numberof a non-alternating knot is strictly larger than the breadth of its Jones polynomial.

Notice that starting from a simple signed graph G, there are many different ways to obtaina more complicated graph by choosing different numbers and orders for the stretching andthickening operations. So knots with structures more complicated than the ones in last sectionmay be constructed. For example, we may replace each positive sequence of the three twists inFigure 5 by 2k1 + 1 consecutive twists and make the total number of such twists to be 2k2 + 1.Similarly, we may replace each negative sequence of the three twists in Figure 5 by 2k3 + 1consecutive twists and make the total number of such twists to be 2 k4 + 1. This way we obtaina four parameter knot family.

Finally, we point out that it is possible to generalize our result so that the positive (negative)edges of G can be replaced by an entire new graph (with both positive and negative edges)using the graph tensor products (see Brylawski [4] and Brylawski and Oxley [5]). This will bedone in a future work.

Acknowledgement. This work was supported in part by NSF Grant #DMS-0310562 to Y.Diao. G. Hetyei is on leave from the Renyi Institute of the Hungarian Academy of Sciences.


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Department of Mathematics and Statistics, UNC Charlotte, Charlotte, NC 28223, [email protected], [email protected], [email protected]

Yuanan Diao, Gabor Hetyei and Kenneth Hinson- Tutte Polynomials of Signed Graphs and Jones...

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