Some invariants of pretzel links
SOME INVARIANTS OF PRETZEL LINKS
DONGSEOK KIM AND JAEUN LEE
Abstract. We show that nontrivial classical pretzel knots L(p, q, r) are hyperbolic
with eight exceptions which are torus knots. We find Conway polynomials of n-
pretzel links using a new computation tree. As applications, we compute the genera
of n-pretzel links using these polynomials and find the basket number of pretzel
links by showing that the genus and the canonical genus of a pretzel link are the
same.
1. Introduction
Let L(p1, p2, . . . , pn) be an n-pretzel link in S
3 where pi ∈ Z represents the number of
half twists as depicted in Figure 1. In particular, if n = 3, it is called a classical pretzel
link, denoted by L(p, q, r). If n is odd, then an n-pretzel link L(p1, p2, . . . , pn) is a knot
if and only if none of two pi’s are even, a pretzel knot is denoted by K(p1, p2, . . . , pn).
If n is even, then L(p1, p2, . . . , pn) is a knot if and only if only one of the pi’s is even.
Generally the number of even pi’s is the number of components, unless pi’s are all
odd. Since pretzel links have nice structures, they have been studied extensively. For
example, several polynomials of pretzel links have been calculated [13, 15, 22, 28]. Y.
Shinohara calculated the signature of pretzel links [34]. Two and three fold covering
spaces branched along pretzel knots have been described [4, 19]. For notations and
definitions, we refer to [2].
A link L is almost alternating if it is not alternating and there is a diagram DL of
L such that one crossing change makes the diagram alternating; we call DL an almost
alternating diagram. One of the classifications of links is that they are classified by
hyperbolic, torus or satellite links [2]. First we show that classical pretzel links are
prime and either alternating or almost alternating. Menasco has shown that prime
alternating knots are either hyperbolic or torus knots [24]. It has been generalized by
Adams that prime almost alternating knots are either hyperbolic or torus knots [1].
It is known that no satellite knot is an almost alternating knot [17]. Thus, we can
classify classical pretzel knots completely by hyperbolic or torus knots.
Let L be a link in S3. A compact orientable surface F is a Seifert surface of L if
the boundary of F is L. The existence of such a surface was first proven by Seifert
using an algorithm on a diagram of L, named after him as Seifert’s algorithm [33].
The genus of a link L can be defined by the minimal genus among all Seifert surfaces
2000 Mathematics Subject Classification. 57M25, 57M27.
Key words and phrases. pretzel links, Conway polynomial, Seifert surfaces, genus, basket number.
The first author was supported by Korea Research Foundation Grant funded by Korea Government
(MOEHRD, Basic Research Promotion Fund) (KRF-2006-351-C00005). The second author was
supported in part by Com2MaC-KOSEF(R11-1999-054).
http://arxiv.org/abs/0704.1432v1
2 DONGSEOK KIM AND JAEUN LEE
p1 p2 p3 . . . pn−1 pn
Figure 1. An n-pretzel link L(p1, p2, . . . , pn)
of L, denoted by g(L). A Seifert surface F of L with the minimal genus g(L) is
called a minimal genus Seifert surface of L. A Seifert surface of L is canonical if it
is obtained from a diagram of L by applying Seifert’s algorithm. Then the minimal
genus among all canonical Seifert surfaces of L is called the canonical genus of L,
denoted by gc(L). A Seifert surface F of L is said to be free if the fundamental group
of the complement of F , namely, π1(S
3 −F) is a free group. Then the minimal genus
among all free Seifert surfaces of L is called the free genus for L, denoted by gf (L).
Since any canonical Seifert surface is free, we have the following inequalities,
g(L) ≤ gf(L) ≤ gc(L).
There are many interesting results about the above inequalities [5, 8, 21, 26, 29, 32].
Gabai has geometrically shown that the minimal genus Seifert surface of n-pretzel
links can be found as a Murasugi sum using Thurston norms and proved that the
Seifert surfaces obtained by applying Seifert’s algorithm to the standard diagram of
L(2k1 + 1, 2k2 + 1, . . . , 2kn + 1) and L(2k1, 2k2, . . . , 2kn) are minimal genus Seifert
surfaces [12]. There is a classical inequality regarding the Alexander polynomial and
the genus g(L) of a link L: G. Torres showed the following inequality,
2g(L) ≥ degree∆L − µ+ 1(1)
where ∆L is the Alexander polynomial of L and µ is the number of components
of L [36]. R. Crowell showed that the equality in inequality (1) holds for alternat-
ing links [8]. Cimasoni has found a similar inequality from multi-variable Alexander
polynomials [6]. In fact, we can find the genera of oriented n-pretzel links from the in-
equality (1) and the Conway polynomial found in section 3, i.e., we will show that the
equality in inequality (1) holds for all n-pretzel links with at least one even crossing.
For pretzel links L(2k1, 2k2, . . . , 2kn) with all possible orientations, Nakagawa showed
that a genus and a canonical genus are the same [28]. The idea of Nakagawa [28] can
be extended to arbitrary n-pretzel links, i.e., we can show that these three genera
g(L), gf(L) and gc(L) are the same.
Some of Seifert surfaces of links feature extra structures. Seifert surfaces obtained
by plumbings annuli have been studied extensively for the fibreness of links and sur-
faces [10–12, 14, 25, 29, 31, 35]. Rudolph has introduced several plumbed Seifert sur-
faces [30]. Let An ⊂ S
3 denote an n-twisted unknotted annulus. A Seifert surface F is
a basket surface if F = D2 or if F = F0 ∗α An which can be constructed by plumbing
SOME INVARIANTS OF PRETZEL LINKS 3
Figure 2. An alternating diagram of L(q,−1, r).
An to a basket surface F0 along a proper arc α ⊂ D2 ⊂ F0 [30]. A basket number of
a link L, denoted by bk(L), is the minimal number of annuli used to obtain a basket
surface F such that ∂F = L [3, 16]. As a consequence of the results in section 4 and
a result [3, Corollary 3.3], we find the basket number of n-pretzel links.
The outline of this paper is as follows. In section 2, we mainly focus on the classical
pretzel links L(p, q, r). We find Conway polynomial of n-pretzel links in section 3. In
section 4, we study the genera of n-pretzel links. In section 5, we compute the basket
number of n-pretzel links.
2. Classical pretzel links L(p, q, r)
2.1. Almost alternating. One can see that L(p, q, r) is alternating if p, q, r have the
same signs. Since every alternating link (including any unlink) has an almost alter-
nating diagram, we are going to show that every nontrivial pretzel link has an almost
alternating diagram. Since the notation depends on the choice of +,− crossings, it is
sufficient to show that L(−p, q, r) has an almost alternating diagram where p, q, r are
positive. In particular, one might expect that L(−1, q, r) is almost alternating, but
surprisingly it is also alternating.
Theorem 2.1. For positive integers p, q and r, L(−1, q, r) is an alternating link and
L(−p, q, r) has an almost alternating diagram.
Proof. One can see that L(q,−1, r) is isotopic to L(q−2, 1, r−2) as shown in Figure 2.
For the second part, see Figure 3. �
Theorem 2.2. All nontrivial pretzel knots K(p, q, r) are either torus knots or hyper-
bolic knots.
Proof. The key ingredient of theorem is that prime alternating (almost alternating)
knots are either hyperbolic or torus knots [24, Corollary 2] ( [1, Corollary 2.4], respec-
tively). Since every pretzel knot has an almost alternating diagram by Theorem 2.1,
we need to show that all nontrivial classical pretzel knots are prime. Since no two of
p, q, r are even, there are two cases : i) all of them are odd, ii) exactly one is even.
4 DONGSEOK KIM AND JAEUN LEE
Figure 3. An almost alternating diagram of L(p,−q, r).
i) p ≡ q ≡ r ≡ 1 (mod 2). For this case, we can use the genus of K = K(p, q, r).
Suppose K = K1#K2. Since a Seifert surface of K is the punctured torus, it has
genus 1 as described in the left top of Figure 2. But 1 = g(K) = g(K1)+g(K2). Thus
one of g(K1) or g(K2) has to be 0, i.e., one of Ki is trivial. Therefore K cannot be
decomposed as a connected sum of two nontrivial knots.
ii) Suppose that p is even i.e., p = 2l, and q, r are odd. Then it is easy to see that
the left two twisting parts form a prime tangle (except when |p| = 2l and |q| = 1).
The right part is an untangle, but since r is odd, we can use a result of Lickorish [23,
Theorem 3] to conclude that K(2l, q, r) is prime. For the above exceptional cases,
we can assume that |r| = 1 because we can choose |q| ≥ |r|. So all possible cases
are K(2l,±1,∓1), K(2l, 1, 1) and K(2l,−1,−1). But the first one is the unknot and
the other two can be deformed to K(p, q, r) of all odd crossings, i.e, K(2l,−1,−1) =
K(2l − 1, 1, 1) and K(2l, 1, 1) = K(2l + 1,−1,−1). This completes the proof. �
2.2. Prime torus pretzel knots. The primary goal of this section is to decide
which classical pretzel knots are torus knots. For convenience, the (m,n) torus link is
denoted by T(m,n). One can see that all 2-string torus links are alternating. C. Adams
has conjectured that only (3, 4) and (3, 5) torus knots are almost alternating [1]. One
can see that K(−2, 3, 3) is the (3, 4) torus knot and K(−2, 3, 5) is the (3, 5) torus
knot. Since the branched double cover of a torus link is a Seifert fibred space with the
base surface S2 and at most three exceptional fibers, and the branched double cover
of a nontrivial n-pretzel link is a Seifert fibred space with n exceptional fibers, there
will be no torus knot of the form K(p1, p2, · · · , pn) for n ≥ 4 and |pi| ≥ 2.
To find all torus knots, we use the Jones polynomials of K(2l, q, r) because the
genera of pretzel knots tell us that no K(p, q, r), with p, q, r all odd, is a torus knot
except the unknot and trefoil, and it is known that K(p,−1,1) is the unknot and
K(±1,±1,±1) are trefoils, which are the only torus knots of genus 1. Remark that
the genus of an (m,n) torus knot is (m − 1)(n − 1)/2. The Jones polynomial of an
(m,n) torus link (m ≤ n) is given by equation (2) if m is odd, by equation (3) if
4 ≤ m is even, and by equation (4) if m = 2 and n is even. This is due to the original
work by Jones [18] but still there is no combinatorial proof for these formulae.
SOME INVARIANTS OF PRETZEL LINKS 5
−t(m−1)(n−1)/2[tm+n−2 + tm+n−4 + · · ·+ tn+1 − tm−1 − · · · − t2 − 1],(2)
−t(m−1)(n−1)/2[tm+n−2 + tm+n−4 + · · ·+ tn − tn−1 − · · · − t2 − 1],(3)
−t(n−1)/2[tn − tn−1 + tn−2 − · · · − t3 + t2 + 1].(4)
Using a formula for the Jones polynomials of n-pretzel knots in [22], we find the
following lemma. Since the Jones polynomial of the mirror image L of L can be found
by VL(t) = VL(t
−1), we may assume q, r are positive integers.
Lemma 2.3. Let l, q, r be positive integers. Let k = 2l + q + r.
VK(2,1,r) = t
(r+1)/2−(2+1)(tr+2+1 − 2tr+2 + 2tr+1 − · · ·+ 2t3 − t2 + t− 1),
VK(2l,q,r) = t
(q+r)/2−(2l+1)(tk − 2tk−1 + 3tk−2 − 4tk−3 + · · · − 3t2 + t− 1), if l ≥ 1,
VK(2l,−q,r) = −t
(−4l−3q+r)/2(tq+r − tq+r−1 + · · · − t+ 1) if q > 1,
VK(−2,1,r) = −t
(r+1)/2(tr+2 − tr+1 + tr − · · ·+ t3 − t2 − 1),
VK(−2,3,3) = −t
3(t5 − t2 − 1),
VK(−2,3,5) = −t
4(t6 − t2 − 1),
VK(−2,3,r) = −t
(3+r)/2(t3+r−2 − tr + · · · − t2 − 1) if r ≥ 7,
VK(−2,q,r) = −t
(q+r)/2(−tq+r−1 + 2tq+r−2 − · · · − t2 − 1) if q, r ≥ 5,
VK(−2l,q,r) = −t
(q+r)/2(at∗ + · · · ± t∓ 1) if l, q, r > 1.
By comparing Jones polynomials of pretzel knots in Lemma 2.3 and Jones polyno-
mials of torus knots in equation (2), (3) and (4), we find the following theorem.
Theorem 2.4. The following are the only nontrivial pretzel knots which are torus
knots.
1) K(p,±1,∓1) are unknots for all p.
2) K(±1,±1,±1) are (2,±3) torus knots.
3) K(±2,∓1,±r) are (2,±r ∓ 2) torus knots.
4) K(∓2,±3,±3), K(∓2,±3,±5) are (3,±4), (3,±5) torus knots, respectively.
Proof. We only need to consider K(2l, q, r). We can see that K(2,−1,r) can be de-
formed to K(0,1,r − 2) by a move shown in Figure 2. The coefficient of t1 and the
second leading coefficient of the Jones polynomial of a torus knot are zero, but by
Lemma 2.3 these are possible only for K(−2, 3, 3), K(−2, 3, 5) and their mirror im-
ages. But the number of terms in the Jones polynomials of these knots is 3, and only
(3, n) torus knots have this property. By comparing the terms of the highest degree, we
conclude that K(∓2,±3,±3) and K(∓2,±3,±5) are the remaining non-alternating
torus knots. �
6 DONGSEOK KIM AND JAEUN LEE
2.3. Minimal genus Seifert surfaces. When one applies Seifert’s algorithm to a
diagram of a link L, in general one may not get a minimal genus Seifert surface. In
fact, Moriah found infinitely many knots which have no diagram on which Seifert’s
algorithm produces a minimal genus surface [26]. But it is known that a minimal
genus Seifert surface can be obtained from an alternating diagram by applying Seifert’s
algorithm [27] and more generally, alternative links [20]. We prove that the Seifert
surface obtained by applying Seifert’s algorithm to the diagram in Figure 4 of a pretzel
knot K(p, q, r) is a minimal genus Seifert surface. Since K(2l, q, r) and its mirror
image are alternating, without loss of a generality, we only need to find Alexander
polynomials of K(−2l, q, r) and K(−2l, q,−r).
Lemma 2.5. Let l, q, r be positive integers.
∆K(−2l,q,r)(t) = t
−(q+r)/2(ltq+r − (2l − 1)tq+r−1 + · · · − (2l − 1)t+ l),
∆K(−2l,q,−r)(t) = t
−(q+r−2)/2(tq+r−2 − 2tq+r−3 + · · · − 2t+ 1).
Proof. One can prove inductively the lemma by the following recurrence formulae
which come from the skein relations, and the formulae for the Alexander polynomial
of the (2, p) torus links.
∆T(2,±p)(t) = t
(1−p)/2(tp−1 − tp−2 + · · · − t+ 1) if p is odd,
∆T(2,±p)(t) = t
(1−p)/2(−tp−1 + tp−2 + · · · − t+ 1) if p is even,
∆K(−2,q,±r)(t) = ∆T(2,q)(t)∆T(2,r)(t) + (t
−1/2 − t1/2)∆T(2,q±r)(t),
∆K(−2l,q,±r)(t) = ∆K(−2(l−1),q,±r)(t) + (t
−1/2 − t1/2)∆T(2,q±r)(t).
Theorem 2.6. The surface obtained by applying Seifert’s algorithm to the pretzel knot
K(p, q, r) as in Figure 4 is a minimal genus Seifert surface, if 1/|p|+1/|q|+1/|r| ≤ 1.
Proof. We consider two cases : i) all of p, q, r are odd, ii) exactly one of p, q, r is even.
For the first case, the first Seifert surface in Figure 4 is clearly a minimal genus since
its genus is 1 unless K(p, q, r) is the unknot. But it can not be the unknot by the
hypothesis. For the second case, we can consider K(−2l, q,±r), K(−2l, q,±r) or their
mirror images, where l, q, r are positive. Their canonical Seifert surfaces are given in
Figure 4. To prove these surfaces are minimal genus Seifert surfaces, first we find
2g(K(−2l, q,±r)) ≥ q+r−1±1 using the Alexander polynomials of K(−2l, q, r) and
K(−2l, q,−r) given in Lemma 2.5 and inequality (1). But the genus of the second
Seifert surface in Figure 4 is (q+r)/2, and the third surface in Figure 4 is (q+r−2)/2.
It completes the proof. �
By combining Theorem 2.4 and Theorem 2.6, we find the following corollary.
Corollary 2.7. The genus of K(p, q, r) is as follows.
1) K(p,±1,∓1), K(±2,∓1,±3) have genus 0 for all p.
SOME INVARIANTS OF PRETZEL LINKS 7
, ±p ±q 2l
±p 2l ∓r
Figure 4. Minimal genus Seifert surfaces of the pretzel knots K(p, q, r).
2) K(p, q, r) has genus 1 if p ≡ q ≡ r ≡ 1 (mod 2) and we are not in case 1).
3) K(±2,∓1,±r) has genus (|r − 2| − 1)/2.
4) K(∓2l, q, r) has genus (|q|+ |r|)/2 if q, r have the same sign and we are not in any
of the previous cases.
5) K(∓2l, q, r) has genus (|q| + |r| − 2)/2 if q, r have different signs and we are not
in cases 1), 2) or 3).
For classical pretzel links, one can see that L(2l1, 2l2, 2l3) has genus 0. For L(2l1, 2l2,
r), we are going to see more interesting results for the genus because there is a freedom
to choose orientations of the components. But, Lemma 2.5 remains true for arbitrary
integers q, r, so we can find the following corollary.
Corollary 2.8. The genus of the link L(2l1, 2l2, r), where |l1| ≥ |l2| , l1, l2 > 0(unless
we indicate differently) and r ≥ 0, is as follows.
1) L(2l1, 2l2,±r) has genus 0 if r ≡ 0 (mod 2) and l1, l2 are nonzero integers.
2) L(±2,±2l2,∓1) has genus (|2l2 − 2| − 2)/2.
3) K(∓2l1,∓2l2,∓r) has genus (|l2| + |r| − 1)/2 if we are not in one of the previous
cases.
4) K(∓2l1,∓2l2,±r) has genus (|l2| + |r| − 3)/2 if we are not in any of the previous
cases.
5) K(∓2l1,±2l2,∓r) has genus (|l2|+ |r| − 3)/2 if we are not in case 1).
8 DONGSEOK KIM AND JAEUN LEE
o1 o2 o3 . . . o2k o2k+1
e1 o2 o3 . . . o2k o2k+1
e1 o2 o3 . . . o2k−1 o2k
(8, 9)
o1 o2 o3 . . . o2k−1 o2k
Figure 5. all oriented n-pretzel knots L(p1, p2, . . . , pn)
6) K(∓2l1,±2l2,±r) has genus (|l2|+ |r|−1)/2 if we are not in case 1) and |l1| > |l2|,
or has genus (|l2|+ |r| − 3)/2 if we are not in case 1) and |l1| = |l2|.
Proof. We follow the proof of Theorem 2.4 and Theorem 2.6 carefully ; if r = ±1, the
link will have two representatives by the move we used in the proof of Theorem 2.1,
we get the result, with a note that we have a freedom to choose an orientation of the
component which goes through two even crossing boxes. �
3. Conway polynomials of n-pretzel links
To find the polynomial invariants of n-pretzel links, we will use a computation tree
: a computation tree of a link polynomial PL is an edge weighted, rooted binary tree
whose vertices are links, the root of the tree is L, two vertices L1, L2 are children of
a vertex Lp if
PLp = w(Lp(1))PL1 + w(Lp(2))PL2 ,
and w(Lp(i)) is the weight on the edge between Lp and Li. One can see that the link
polynomial PL can be computed as follows,
Lp∈P(Lv)
w(Lp(i))PLv ,
where L is the set of all vertices of valence 1 and P(Lv) is the set of all vertices of the
path from the root to the vertex Lv. In general, it is easy to find PL if we repeatedly
use the skein relations until each vertex Lv becomes an unlink. Moreover, one can
replace links by other for a convenience of the computation. For instance, J. Franks
and R. F. Williams used braids to find a beautiful result on Jones polynomial [9].
SOME INVARIANTS OF PRETZEL LINKS 9
To compute Conway Polynomials of n-pretzel links, we will use a new notation for
n-pretzel links which will be used for vertices of a computation tree. We called a
rectangle in Figure 5 a box and the link moves in the same direction in a box if it
has the orientation as in the second box from the left of the diagram (6) of Figure 5,
in the opposite directions if it has the orientation as in the first box from the left
of the diagram (6) in Figure 5. If we have a box for which two strings move in
the opposite directions and we use the skein relation at this box, then the resulting
links have either less number of the boxes or less number of crossings. One can
see that an opposite direction can be happened only for a box with even number of
crossings (but this is not sufficient) except in the case that n is even and all the pi’s
are odd (we will handle this case separately). Suppose we have at least one even
crossing box. We may assume that it is p1 = 2l1. Let us remark that the Conway
polynomial vanishes for split links. The following is our new notation for n-pretzel
links. From a given n-pretzel link L with an orientation O, we can represent L by a
vector in (Z×Z2)
n such as (pǫ11 , p
2 , . . . , p
n ), where ǫi = 1(−1) if the link moves in the
same(opposite, respectively) direction in the box corresponding to pi with respect to
the given orientation O. Write p1i = pi. First we find the following recursive formula,
∇L(pǫ11 ,p
2 ,...,p
,...,p
. . . ˆ∇T
. . .∇T
− liz∇
2 ,...,
,...,p
where the term under ˆ is deleted.
By repeatedly using above formulae, we can make a computation tree that there is
no negative ǫi for the representative at each vertex of valence 1. Then, we can expand
(. . . , pi, . . .) into (. . . , pi ± 1(= p
i), . . .) and (. . . , pi ± 2, . . .) with suitable weights on
edges, 1 or ±z where |pi| > |p
i|. We can keep on expanding at the crossings until
all the entries in the vectors of vertices of valence 1 are either 0 or ±1. At this
stage, if it has more than two 0’s then we stop the expansion and change the vertex
to zero because it is a split link. If it has only one zero, it is a composite link of
T(2,pi)’s. Otherwise, we change the vector to an integral value m, the sum of the signs
of entries in the vector. In fact, it is the closed braid of two strings represented by σm1 .
Therefore, we can compute the Conway polynomial of a link L using this computation
tree and the Conway polynomial of closed 2-braids.
3.1. Conway polynomial of n-pretzel knots. The general figures of n-pretzel
knots are given in Figure 5 (the right-top one is a two components link) where e1 =
2l, oi = 2ki + 1. We can see that there is at most one box in which the knot moves in
opposite directions. But for a two component link, all boxes might move in opposite
directions for the orientation which is not in Figure 5. Counterclockwise from the
top-right, we get representatives, (o−11 , o
2 , . . . , o
2k ), (o1, o2, . . . , o2k), (o
1 , o
2 , . . . ,
o−12k+1), (e
1 , o2, o3, . . . , o2k+1) and (e1, o2, o3, . . . , o2k). By using a computation tree
for these representatives, we find Theorem 3.1. For convenience, we abbreviate ∇T(2,n)
by ∇n throughout the section.
10 DONGSEOK KIM AND JAEUN LEE
Theorem 3.1. Let e′1 = sign(e1)(|e1| − 1), o
i = sign(oi)(|oi| − 1), α =
i=2 sign(oi)
and β = sign(e1). The Conway polynomials of n-pretzel knots in Figure 5 are
∇L(o1,o2,o3,...,on) =
(n−1)/2
2i,(5)
∇L(e1,o2,o3,...,on) = ∇o2∇o3 . . .∇on [1− lz[−
]],(6)
∇L(e1,o2,o3,...,on) = ∇o2∇o3 . . .∇on [∇e′1 +∇e1 [−
β + α
]],(7)
∇L(o1,o2,o3,...,on) =
(n+1)/2
2i−1,(8)
∇L(o1,o2,o3,...,on) = ∇o1∇o2 . . .∇on [∇
i=1 sign(oi)
],(9)
where for L(o1, o2, o3, . . . , on) we have two possible orientations because it is a two
components link, so we get (8) for (o−11 , o
2 , . . . , o
2k ) and (9) for (o1, o2, . . . , o2k).
Proof. We will only prove (6) but one can prove the other by a similar argument. In
the computation tree, we use skein relation at crossings until vertices of valence 1 in
the computation tree up to this point will be (c1, c2, . . . , cn) where ci is either 0 or
±1. Since the Conway polynomials of split links vanish, we may assume there are
no than one 0’s. The first term in the parenthesis comes from the case where all |ci|
are 1 because it is again the (2, α) torus link horizontally. It is a two component link
with linking number −α/2, so its Conway polynomial is −(α/2)z. For the case where
only one ci = 0, the values on edges to the vertex will contribute exactly ∇o′
and the
vertex is the composite link of (2, oj) torus knots j = 2, . . . , n except i. �
3.2. Conway polynomials of n-pretzel links. Since we have already handled links
of all odd crossings, we assume that n-pretzel links have at least one even crossing
box. Let L(p1, p2, . . . , pn) be an n-pretzel link and let s be the number of even pi’s.
Then it is a link of s components. The Conway polynomial ∇L depends on the choice
of the orientation of L. There are 2s−1 possible orientations of L. But one can easily
see that the link always moves in the same direction in all boxes of odd crossings for
arbitrary orientation. For further purpose, we will calculate the Conway polynomial of
the pretzel link with the following orientations. For the existence of such orientations,
we will prove it in Lemma 4.2 : if n − s is even, then there exists an orientation O
of L such that the link L moves in the opposite directions in all boxes of even pi.
If n − s is odd, then there exists an orientation O of L such that the link L moves
in the opposite directions in all boxes of even pi except one pt but without loss of a
generality we assume that p1 = pt.
Theorem 3.2. Let L(p1, p2, . . . , pn) be a pretzel link with the above orientation O.
Let pei = 2li be all even and poj = 2kj +1 be all odd. Let s be the number of even pi’s
SOME INVARIANTS OF PRETZEL LINKS 11
and let α =
i=1 sign(poi) and β = sign(p1). Let p
i = sign(pi)(|pi| − 1). If n− s is
even, then the Conway polynomial of L(p1, p2, . . . , pn) is
(−li)]z
∇poi )[−
∇p′oi
] + [
j=1,j 6=i
(−lj)]z
If n− s is odd, then the Conway polynomial of L(p1, p2, . . . , pn) is
(−li)]z
∇poi)∇p1 [−
α + β
∇p′oi
] + [
j=2,j 6=i
(−lj)]z
Proof. It is clear by choosing (p−1e1 , p
, . . ., p−1es , po1 , . . ., pon−s) and (pe1, p
, . . ., p−1es ,
po1 , . . ., pon−s), respectively. �
More generally, we get the following results by taking (p−1e1 , p
, . . ., p−1et , pet+1, . . .,
pes , po1, . . ., pon−s) for a representative of L(p1, p2, . . . , pn) induced by an orientation
Theorem 3.3. Let pei = 2li be all even and poj = 2kj + 1 be all odd. Let s be
the number of even pi. Let t be the number of even pi in the corresponding boxes in
which the link moves in the opposite direction, say pei where i = 1, 2, . . ., t. and let
j=1 sign(poj) and β =
i=t+1 sign(pei). Let p
i = sign(pi)(|pi| − 1). Then the
Conway polynomial of L(p1, p2, . . . , pn) with the orientation O is
(−li)]z
∇poi)(
∇pej )[−
α + β
i=t+1
∇p′ei
∇p′oj
] + [
j=1,j 6=i
(−lj)]z
4. Genera of n-pretzel links
We will consider the genus of an n-pretzel link with at least one even crossing box.
Let FL be a Seifert surface of an n-pretzel link L. For the rest of the section, let
χ(FL) be the Euler characteristic of FL, V be the number of Seifert circles, E be the
number of crossings and F be the number of the components of L.
4.1. Genera of n-pretzel knots with one even pi. We divide into two cases : i)
n is odd, ii) n is even. For the first case: n is odd, we can see that the degree of
∇K(e1,o1,o2,...,on) is
degree(∇oi) = 2 +
(|oi| − 1),
and the coefficient of this leading term is −lα/2 from Theorem 3.1.
12 DONGSEOK KIM AND JAEUN LEE
Suppose α is nonzero. Then the Seifert surface F obtained by applying Seifert’s
algorithm to the diagram in Figure 5 is a minimal genus surface. The genus of the
Seifert surface FK is
g(FK) =
[2− χ(FK)] =
(2− V + E − F )
[2− (|e1|+ n− 2) + (|e1|+
|oi|)− 1] =
(|o1| − 1)]
degree ∇K(e1,o1,o2,...,on).
Suppose α = 0. This means that we have the same number of positive and negative
twists on odd twists. If we look at the Conway polynomial in equation 6, we drop
exactly one in degree with new leading coefficient 1. It is sufficient to show that the
degree of the following term is negative. Remark that ∇oi = ∇−oi .
−lz[−
] = −l[0 +
] = −l[
sign(oi)(∇|oi| −∇|oi|−2)
∇|oi|
= −l[
(sign(oi) +
∇|oi|−2
∇|oi|
)] = −l[
∇|oi|−2
∇|oi|
We hope to find a minimal surface of this genus. For the first case, the sign of
an n-pretzel is (±,±, . . . ,±, even,∓,∓, . . . ,∓). The rule is to use the move from
the outmost pair. Then the moves in Figure 6 will increase V by two but will not
change E, F (= 1); thus we get a surface with one less genus. If we represent the
move by the Conway notation for algebraic links [7], it is either (. . . ,−a, . . . , b, . . .) ⇒
(. . . , (−1)(−a+1), . . . , (b−1)(1), . . .) or (. . . , a, . . . ,−b, . . .)⇒ (. . . , (1)(a−1), . . .,(−1)
(−b+ 1),. . .) where the sign sum of the oi’s between a, b has to be zero.
For the general case, if we only look at the signs of the odd twists from o1, we can
find a pair oi, oj such that we can apply the move we described above. The resulting
diagram satisfies the same hypothesis with strictly smaller twisted bands. Inductively
we get a well-defined sequence of moves which makes the desired diagram on which
Seifert’s algorithm will produce a minimal genus surface. Figure 6 shows the effect
on V,E. This completes the case i).
For the second case, n is even, we can see that the degree of ∇K(e1,o1,o2,...,on) is
1 + degree(∇e1) +
degree(∇oi) = |e1|+
(|oi| − 1),
and the coefficient of the leading term is −sign(e1)(α + β)/2 from Theorem 3.3.
Suppose α+β is nonzero. Then the Seifert surface F obtained by applying Seifert’s
algorithm to the diagram in Figure 5 is a minimal genus surface. The genus of the
Seifert surface FK is
SOME INVARIANTS OF PRETZEL LINKS 13
o3on−1 ∼=
Figure 6. How to modify a diagram in Figure 5 to find a minimal
genus diagram of L(p1, p2, . . . , pn).
g(FK) =
[2− χ(FK)] =
(2− V + E − F )
[2− (n) + [|e1|+
(|oi|)]− 1] =
[|e1|+
(|o1| − 1)]
degree∇K(e1,o1,o2,...,on).
Suppose α + β = 0. This means that we have the same number of positive and
negative twists. As we did before we drop exactly one in the degree of the Conway
polynomial in equation 7 with new leading coefficient 1. All arguments are the same
if we change the term in parentheses in the equation as follows.
[∇e′1 +∇e1(−
β + α
)] = ∇e1 [−
β + α
We can find a minimal surface of this genus by the same method as shown in
Figure 6 if we handle the even crossing box together. This gives us the following
theorem.
Theorem 4.1. Let K(p1, o2, o3, . . . , on) be an n-pretzel knot with one even p1. Let α
i=2 sign(oi) and β= sign(p1). Suppose |p1|, |oi| ≥ 2. Let
(|oi| − 1).
14 DONGSEOK KIM AND JAEUN LEE
Figure 7. Boundary orientation of L(p1, p2, . . . , pn).
Then the genus g(K) of K is
g(K) =
(δ + 2) if n is odd and α 6= 0,
δ if n is even and α = 0,
(|p1|+ δ) if n is even and α + β 6= 0,
(|p1|+ δ)− 1 if n is even and α + β = 0.
4.2. Genera of n-Pretzel links. Intuitively, if we have more even pi’s with opposite
directions, then we will have a surface of smaller genus. So we want to choose an
orientation which forces all the even pi’s to move in the opposite directions, but this
may not be possible for all cases.
Lemma 4.2. Let L(p1, p2, . . . , pn) be an n-pretzel link and let s be the number of even
pi’s. If n− s is even, then there exists an orientation of L such that the link L moves
in opposite directions in all boxes of even pi. If n − s is odd and a given pt is even,
then there exists an orientation of L such that the link L moves in opposite directions
in all boxes of even pi’s except the one corresponding to pt.
Proof. If all pj between two even pi and pk are odd, the number of these pj ’s odd
(mod 2) will decide the boundary orientation as depicted in Figure 7.
If the number of odd crossing boxes is even, we can orient the link such that the
link moves oppositely in all boxes of even crossings. Otherwise there is just one box
for which the link moves in the same direction. So starting from pt will complete the
proof. �
Let us denote the orientation we choose in Lemma 4.2 by O′. From Theorem 3.2,
we can do almost the same comparison by using equation (1). But we have to be
careful to use (1) for links. Since it was defined for oriented links, we can interpolate
it as follows.
g(L) = minO{min{genus of F(L,O) | FL,O is a Seifert surface
with the orientation O}}.
where the first O runs over all possible orientations of L. So (1) gives us an inequality
on the second minimum of the fixed orientation O and ∇(L,O).
SOME INVARIANTS OF PRETZEL LINKS 15
We divide into two cases : i) n− s is even, ii) n− s is odd. For the first case, n− s
even, we can see that the degree of ∇L(p1,p2,...,pn) is
degree(∇pmi ) + 1 = s+
(|pmi | − 1) + 1,
and the coefficient of this leading term is −α/2 from Theorem 3.2.
Suppose α is nonzero. Then the Seifert surface F obtained by applying Seifert’s
algorithm with the fixed orientation O′ is a minimal genus surface of (L,O′). Let us
find the genus of the Seifert surface F(L,O′).
2g(FL) = 2− χ(FL) = 2− (V − E + F )
= 2− (n− s) + (
(|pmj | − 1)) + [
|pij |+
(|pmi|)]− s
= 2 +
(|pki| − 1) = degree(∇L)− s + 1.
For the rest of the cases of the arguments are parallel to the argument for knots.
Next, we explain how pt will be chosen for the rest of the article.
Remark 4.3. First, we look at the minimum of the absolute value of pei over all
even crossings. If the minimum is taken by the unique pei or by pei’s of the same
sign, we choose it for pt. If there are more than two pei’s with different signs and the
same absolute value, then we look at the value α, the sign sum of odd crossings. If
it is neither 1 nor −1, then we pick the positive one for pt. If α = 1(−1), pick the
negative(positive) one for pt.
For the second case, n− s odd, we find pt as described the above. For the last two
cases, we will drop the genus by 1. Denote the orientation we chose here by O1.
Lemma 4.4. For an arbitrary orientation O, degree∇(L,O) ≥ degree∇(L,O1).
Proof. If we count tO, the number of even crossings in which the link moves in the
opposite directions with respect to O, we can see that tO ≤ tO1. If we look at the
Conway polynomial in Theorem 3.3, we have that i) we can ignore the second term,
ii) increasing t by 1 will change the degree of the second term by −(|pi| − 2), and by
hypothesis, |pi| ≥ 2. �
Theorem 4.5. Let L(p1, o2, . . . , os, es+1, . . . , en) be an n-pretzel link with at least one
even pi. Let α =
i=2 sign(poi) and β = sign(pt). Suppose |oi|, |ej| ≥ 2. Let pt be
the integer described in Remark 4.3. Let l be the number of even pi’s. Let
(|oi| − 1).
Then the genus g(L) of L is
16 DONGSEOK KIM AND JAEUN LEE
g(L) =
δ + 1 if n− s is even and α 6= 0,
δ if n− s is even and α = 0,
(|pt|+ δ) if n− s is odd and α + β 6= 0,
(|pt|+ δ)− 1 if n− s is odd and α + β = 0.
Proof. It follows from Theorem 3.2, 3.3 and Lemma 4.4. �
5. The basket numbers of pretzel links
First let us recall a definition of the basket number. Let An ⊂ S
3 denote an n-
twisted unknotted annulus. A Seifert surface F is a basket surface if F = D2 or if
F = F0 ∗αAn which can be constructed by plumbing An to a basket surface F0 along
a proper arc α ⊂ D2 ⊂ F0. A basket number of a link L, denoted by bk(L), is the
minimal number of annuli used to obtain a basket surface F such that ∂F = L. For
standard definitions and notations, we refer to [30]. Throughout the section, we will
assume all links are not splitable, i.e., Seifert surfaces are connected. Otherwise, we
can handle each connected component separately.
For the basket number and the genus of a link, there is a useful theorem.
Theorem 5.1 ( [3]). Let L be a link of l components. Then the basket number of L
is bounded by the genus and the canonical genus of L as,
2g(L) + l − 1 ≤ bk(L) ≤ 2gc(L) + l − 1.
Since we have found that a minimal genus surface of a pretzel link L of genus g(L)
can be obtained by applying Seifert algorithm on a diagram of L, i.e., g(L) = gc(L), we
find that the basket number of a pretzel link L is 2g(L)+l−1, i.e., bk(L) = 2g(L)+l−1.
Theorem 5.2. Let K(p1, o2, o3, . . . , on) be an n-pretzel knot with one even p1. Let α
i=2 sign(oi) and β= sign(p1). Suppose |p1|, |oi| ≥ 2. Let
(|oi| − 1).
Then the basket number bk(K) of K,
bk(K) =
δ + 2 if n is odd and α 6= 0,
δ if n is even and α = 0,
|p1|+ δ if n is even and α+ β 6= 0,
|p1|+ δ − 2 if n is even and α+ β = 0.
Theorem 5.3. Let L(p1, o2, . . . , os, es+1, . . . , en) be an n-pretzel link with at least one
even pi. Let α =
i=2 sign(poi) and β = sign(pt). Suppose |oi|, |ej| ≥ 2. Let pt be
the integer described in Remark 4.3. Let l be the number of even pi’s. Let
(|oi| − 1).
SOME INVARIANTS OF PRETZEL LINKS 17
Then the basket number bk(L) of L,
bk(L) =
δ + l + 1 if n− s is even and α 6= 0,
δ + l − 1 if n− s is even and α = 0,
|pt|+ δ + l − 1 if n− s is odd and α+ β 6= 0,
|pt|+ δ + l − 3 if n− s is odd and α+ β = 0.
Acknowledgments
The author would like to thank Cameron Gordon for helpful discussion, valuable
comments on this work. The TEX macro package PSTricks [37] was essential for
typesetting the equations and figures.
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Department of Mathematics, Yeungnam University, Kyongsan, 712-749, Korea
E-mail address : dongseok@yumail.ac.kr
Department of Mathematics, Yeungnam University, Kyongsan, 712-749, Korea
E-mail address : julee@yu.ac.kr
ftp://ftp
	1. Introduction
	2. Classical pretzel links L(p,q,r)
	2.1. Almost alternating
	2.2. Prime torus pretzel knots
	2.3. Minimal genus Seifert surfaces
	3. Conway polynomials of n-pretzel links
	3.1. Conway polynomial of n-pretzel knots
	3.2. Conway polynomials of n-pretzel links
	4. Genera of n-pretzel links
	4.1. Genera of n-pretzel knots with one even pi
	4.2. Genera of n-Pretzel links
	5. The basket numbers of pretzel links
	Acknowledgments
	References
