Meridional rank of whitehead doubles

Abstract

We prove that the meridional rank and the bridge number of the Whitehead double of a prime algebraically tame knot coincide. Algebraically tame knots are a broad generalization of torus knots and iterated cable knots.

Full text

MERIDIONAL RANK OF WHITEHEAD DOUBLES
EDERSON R. F. DUTRA
Abstract. We prove that the meridional rank and the bridge number of the
Whitehead double of a prime algebraically tame knot coincide. Algebraically
tame knots are a broad generalization of torus knots and iterated cable knots.
1. Introduction
Let k⊆S3be a knot. The bridge number of k, denoted by b(k), is defined as
the minimal number of bridges in a bridge presentation of k. A meridian of kis
an element of π1(S3\k)which is represented by a curve freely homotopic to the
boundary path of a disk in S3that intersects kin exactly one point. The meridional
rank of k, denoted by w(k), is defined as the minimal number of meridians needed
to generate π1(S3\k). It is known that any bridge presentation of kcontaining b
bridges yields a generating set of π1(S3\k)consisting of bmeridians. Therefore we
always have the inequality w(k)≤b(k).
It was asked by S. Cappell and J. Shaneson [16, Problem 1.11] as well as by
K. Murasugi whether the equality b(k) = w(k)always holds. To this day no coun-
terexamples are known but the equality has been verified for many classes of knots
(using a variety of techniques): knots whose group is generated by two merid-
ians [10], (generalized) Montesinos links [9, 17], torus knots [18], iterated cable
knots [12, 13], knots whose exterior is a graph manifold [7], links of meridional rank
3whose double branched covers are graph manifolds [8], and recently for twisted
and arborescent links [3, 4].
An interesting result proved recently by R. Blair, A. Kjuchukova, R. Velazquez,
and P. Villanueva [6] states that the bridge number of any knot equals its Wirtinger
number, an invariant closely related to the meridional rank. This result therefore
establishes a weak version of Cappell and Shaneson’s question, and also points to
new approaches to this problem.
In this article we show that the conjuncture holds for Whitehead doubles of prime
algebraically tame knots and we also show that the class of prime algebraically tame
knots is closed under braid satellites.
Theorem 1.1. Let kbe a Whitehead double of a prime algebraically tame knot.
Then it holds w(k) = b(k).
Theorem 1.2. The class of prime algebraically tame knots is closed under taking
satellites with braid pattern.
Research supported by FAPESP, São Paulo Research Foundation, grants 2018/08187-6 and
2021/12276-7.
1
arXiv:2212.13081v1 [math.GT] 26 Dec 2022

2 EDERSON R. F. DUTRA
Acknowledgments. I would like to thank the faculty of mathematics of the Uni-
versity of Regensburg for the hospitality during my one year visit as a pos-doctoral
guest in Regensburg, in special to prof. Stefan Friedl and Filip Misev for helpful
conversations during the preparation of this article and for Claudius Zibrowius for
a nice office atmosphere.
2. Basic definitions
In this section we fix the notation needed to prove Theorem 1.1 and Theorem 1.2.
We closely follow the notation from [11]. Let k⊆S3be a knot. The knot exterior
of kis defined as
E(k) := S3\Int V (k)
where V(k)is a regular neighborhood of k. Let x0∈∂E(k) = ∂V (k). The group of
kis defined as G(k) := π1(E(k), x0)and the peripheral subgroup of kis defined as
P(k) = π1(∂E(k), x0).
A meridian of kwas previously defined as an element of G(k)represented by a
curve freely homotopic to the boundary path of a disk which intersects kin exactly
one point. For our purposes, however, we need a fixed meridian.
Let µ⊂∂V (k)be the boundary of a meridional disk of V(k)such that x0∈µ.
Define m:= [µ]∈P(k). Thus mis a meridian of kas defined previously and any
other meridian of kis conjugate to m.
Let also λ⊆∂E(k)be a longitudinal curve of k, i.e. a simple closed curve that
bounds an orientable surface in E(k), such that x0=λ∩µ. Define l:= [λ]∈P(k).
Observe that {m, l}is a basis for P(k).
2.1. Satellite knots. We recall the construction of satellite knots. Let k1⊆S3
be a non-trivial knot. Let V⊆S3be a standardly embedded unknotted solid
torus and let k0a knot contained in the interior of Vsuch that k0intersects each
meridional disk of Vin at least one point. For any homeomorphism hfrom Vonto
a regular neighborhood V(k1)of k1we call the knot kh:= h(k0)asatellite knot with
companion k1and pattern (V, k0), see Figure 1. We will be specially concerned with
(V, k0)
h
V(k1)
kh
Figure 1. Whitehead double of the figure-eight knot.
the following classes of satellite knots:

MERIDIONAL RANK OF WHITEHEAD DOUBLES 3
•If (V, k0)is the pattern described in Figure 1, then khis called a Whitehead
double of k1.
•If k0lies in a torus T⊆Int(V)which is parallel to ∂V , then khis called a
cable knot on k1.
•If k0is a closed braid standardly embedded in Vas shown in Figure 2, i.e.
any meridional disk p×D2⊆Vintersects k0in exactly npoints, then kh
is called a satellite knot with braid pattern.
Figure 2. k0is the closed braid ˆ
βwhere β=σ2σ1−1σ2
2.
The exterior of a satellite knot k=h(k0)clearly decomposes as
E(k) = h(V\Int V (k0)) ∪E(k1)and h(V\Int V (k0)) ∩E(k1) = h(∂V ) = ∂E (k1)
where V(k0)⊆Int(V)is a regular neighborhood of k0in V. The Theorem of Seifert
and van-Kampen implies that
G(k) = π1(V\Int V (k0)) ∗(α,C,ω)G(k1)
where
•C:= hme, lei∼
=Z⊕Z.
•α:C→π1(V\Int V (k0)) is given by
α(me) = mVand ω(le) = lV
where mVis a meridian of Vand lVis a meridian of the solid torus S3\Int V
(and therefore a longitude of V).
•ω:C→G(k1)is defined by
ω(me) = h∗(mV)and ω(le) = h∗(lV)
where h∗is the induced isomorphism π1(∂V )→P(k1).
Observe that we can always take h∗(mV)as the fixed meridian of k1and therefore
denote h∗(mV)simply by m1.
The main tool that we are going to use in this paper is the theory of folding
in graph of groups developed in [22] which is a slightly variation of the theory
developed in [15]. For this reason, we will always consider the amalgamated free
product
G(k) = π1(V\Int V (k0)) ∗(α,C,ω)G(k1)
as a graph of groups Ahaving a pair of vertices v0and v1such that
Av0=π1(V\Int V (k0)) and Av1=G(k1)
and a single edge pair {e, e−1}with α(e) = v0and ω(e) = v1such that
Ae=Ae−1:= C=hme, lei.
The boundary monorphisms αe:Ae→Av0and ωe:Ae→Av1coincide with the
monorphisms defined before, that is, αe:= αand ωe:= ω, see Figure 3.

4 EDERSON R. F. DUTRA
Av0=π1(V\Int V (k0)) αe
←− Ae=⟨me, le⟩ωe
−→ Av1=G(k1)
v0
e
v1
Figure 3. The graph of groups A.
2.2. The group of the braid space and cable space. Assume that k0⊂Vis a
closed braid, i.e. k0=ˆ
βwhere βis an n-braid whose associate permutation τ∈Sn
is cycle of length n. The braid space of βis defined as
BS(β) := V\Int V (ˆ
β)
where V(ˆ
β)is a regular neighborhood of k0=ˆ
β. We can construct BS(β)as a
mapping torus in the following way. Let D2={z∈C| kzk ≤ 1}and
Σ = D2\Int(δ1∪. . . ∪δn)
where δ1is a small disk in the interior of D2,δi=ρi−1
0(δ1)for 2≤i≤nand where
ρ0:D2→D2is a rotation by an angle of 2π/n about the origin x0:= 0. It is
known that there is a homeomorphism ρ: (Σ, x0)→(Σ, x0)such that ρ(δi) = δτ(i)
for all 1≤i≤nand
BS(β)∼
=Σ×[0,1]/(x, 0) ∼(ρ(x),1)
In the case of a cable pattern the homeomorphism ρis a rotation about x0through
an angle of 2π(m/n)for some integer msuch that gdc(m, n) = 1. In this case we
denote BS(β)by CS (n, m).
Denote the free generators of π1(Σ, x0)corresponding to the boundary paths of
the removed disks δ1, . . . , δnby x1, . . . , xnrespectively. Let t∈π1(BS(β), x0)be
the element represented by the loop x0×I⊂BS(β). Therefore,
π1(BS(β), x0) = π1(Σ, x0)o Z
where the action of Z=htion π1(Σ, x0)is given by
txit−1=aixτ(i)a−1
ifor 1≤i≤n
for some words a1, . . . , an∈π1(Σ, x0). Observe that in the case of a cable space
ai= 1 for all 1≤i≤n. Any element of π1(BS(β), x0)is therefore uniquely
written in the form wtzwith w∈π1(Σ, x0)∼
=F(x1, . . . , xn)and z∈Z. Note also
that π1(Σ, x0) = hhx1ii (normal closure) as any two elements of {x1, . . . , xn}are
conjugate in π1(BS(β), x0).
Let Adenote π1(BS(β), x0). We say that a subgroup U≤Ais meridional if
Uis generated by finitely many conjugates of x1. For example, π1(Σ, x0)≤Ais
meridional. We will need the following result from [7] which explains the behavior
of the meridional subgroups with respect to the peripheral subgroups
PV:= π1(∂V (k0), x0)and CV:= π1(∂ V, x0).
Observe that CVis generated by mV, lVand PVis generated by x1, tn.
Lemma 2.1. Let U=hg1x1g−1
1, . . . , gkx1g−1
kiwith k≥0and g1, . . . , gk∈A.
Then either U=π1(Σ, x0)(and in this case k≥n) or Uis freely generated by
h1x1h−1
1, . . . , hmx1h−1
mwith m≤kand h1, . . . , hm∈Asuch that:
(1) for any g∈Aone of the following holds:
(a) gPVg−1∩U={1}.

MERIDIONAL RANK OF WHITEHEAD DOUBLES 5
(b) gPVg−1∩U=ghx1ig−1and gx1g−1is in Uconjugate to hlx1h−1
lfor
some l∈ {1, . . . , m}.
(2) for any g∈Athe subgroups gCVg−1and Uintersect trivially.
Remark 2.2.The previous lemma tells us that the minimal number of conjugates
of x1needed to generate a given meridional subgroup U≤Acoincides with the
rank of U, i.e. the minimal number of elements needed to generate U.
2.3. The centralizer of the meridian. This subsection is devoted to show a well-
known fact, see [2, Theorem 2.5.1] for example, that says that almost no element
in the knot group commutes with the meridian.
Lemma 2.3. Let kbe a non-trivial prime knot and g∈G(k). Then
gmg−1=miff g∈P(k).
Proof. It is proved in Lemma 3.1 of [21] that the peripheral subgroup of kis mal-
normal in G(k), i.e. gP (k)g−1∩P(k) = 1 for all g∈G(k)\P(k), unless kis a torus
knot or a cable knot or a composite knot. The last case does not occur since kis
assumed to be prime. In the case of a torus knot it is shown in Lemma 3.2 of [21]
that gmg−1∩P(k) = 1 for all g∈G(k)\P(k).
We therefore need to consider the case where kis a cable knot. Thus G(k)splits
as π1(A, v0)where Ais the graph of groups described in the previous subsection.
Observe that the meridian of kis represented by the reduced A-path x1∈Av0=
π1(CS (n, m), x0), that is, m= [x1]∈π1(A, v0).
Let g∈G(k) = π1(A, v0)such that gmg−1=m. We can write g= [p]where
p=a0, e, a1, . . . , a2l−1, e−1, a2l
is a reduced A-path of length 2l≥0. Then gmg−1=mimplies that the A-paths
a0, e, a1, . . . , e−1, a2l·x1·a−1
2l, e, . . . , a−1
1, e−1, a−1
0and x1
are equivalent, see [22, p.612] or [15, Definition 2.3]. If l≥1then we can apply a
reduction to px1p−1. Since pis reduced we conclude that a2lx1a−1
2lis conjugate in
the free group Fn=hx1, . . . , xni ≤ Av0to an element of αe(Ae) = CV. A simple
homology argument shows that this cannot occur. Thus l= 0 and so the equality
gmg−1=mreduces to a0x1a−1
0=x1. Write a0=utzwhere u∈Fnand z∈Z.
Then
a0x1a−1
0=uxτz(i)u−1.
Hence a0x1a−1
0=x1iff τz(1) = 1. The second equality holds only when z=nk
for some k. Since ux1u−1=x1in Fnwe conclude that u=xw
1for some w∈Z.
Therefore, g=xw
1(tn)kwhich shows that g∈PV=P(k).
2.4. Algebraically tame knots. Let k⊆S3be a knot. We call a subgroup
U≤G(k)meridional if Uis generated by finitely many meridians of k. The minimal
number of meridians needed to generated a meridional subgroup U, denoted by
w(U), is called the meridional rank of U. Observe that the knot group G(k)is
meridional and w(G(k)) equals the meridional rank w(k)of k.
Definition 2.4. A meridional subgroup U=hg1mg1, . . . , grmgri ≤ G(k)of merid-
ional rank r:= w(U)is called tame if the following hold:
(1) the meridians gimgiand gjmgj(1≤i6=j≤r) are not conjugate in U.

6 EDERSON R. F. DUTRA
(2) for any g∈G(k)either U∩g P (k)g−1= 1 or
U∩gP (k)g−1=ghmig−1
and gmg−1is in Uconjugate to gimgifor some 1≤i≤r.
Definition 2.5. A non-trivial knot kis called algebraically tame if any meridional
subgroup of G(k)that is generated by less than b(k)meridians is tame.
Remark 2.6.Observe that for any algebraically tame knot kthe equality w(k) = b(k)
holds since otherwise G(k)would be a tame subgroup.
Remark 2.7.A (possibly) larger class of knots (called meridionally tame knots)
is defined in [7]. The author does not know any example of a prime knot that is
algebraically tame but not meridionally tame.
Example 2.8. Whitehead doubles are never algebraically tame since their exterior
contains a properly immersed π1-injective pair of pants in which two boundary
components are mapped to meridional curves of k, see [1].
Proposition 2.9. Two bridge knots, torus knots and prime three bridge knots are
algebraically tame.
Proof. First assume that kis a 2-bridge knot. Thus we need to show that any
meridional subgroup generated by a single meridian is tame. Let U=hama−1i ≤
G(k). We can assume that a= 1. Since kis neither a cable knot nor a composite
knot, Lemma 2.3 implies that gP (k)g−1∩U6= 1 iff g∈P(k). Therefore
gP (k)g−1∩U6= 1 iff gP (k)g−1∩U=hmi.
Clearly gmg−1is conjugate in Uto m.
Assume now that kis a torus knot. It is shown in Theorem 1.2 of [18] that
any meridional subgroup of kof meridional rank < b(k)is freely generated by
meridians. This shows condition (1) of tameness. Condition (2) is implicit in [18]
and an explicit argument is given in [7, Lemma 6.1].
The algebraic tameness of prime 3-bridge knots follows from the proof of Propo-
sition 7.1 of [7]. 
3. The pattern space of a Whitehead double
In this section we study the fundamental group of the pattern space E:=
V\Int V (k0)where (V, k0)is the Whitehead pattern and V(k0)is a regular neigh-
borhood of k0in the interior of V. Let Σ⊆Ebe the properly embedded pair
Σ
Figure 4. The pattern space E=V\Int V (k0).

MERIDIONAL RANK OF WHITEHEAD DOUBLES 7
of pants shown in Figure 4 and let E0be the space obtained from Eby cutting
along Σ, see Figure 5. The group π1(E0, e0)(resp. π1(Σ, σ0)) is freely generated
by the elements x1and x2(resp. y1and y2) whose representatives are described in
Figure 5. It follows from the theorem of Seifert and van-Kampen that π1(E, e0)
splits as an HNN extension:
π1(E, e0) = π1(E0, e0)∗(α,π1(Σ,σ0),ω )
where:
•α:π1(Σ, σ0)→π1(E0, e0)is given by
α(y1) = x2x−1
1x−1
2and α(y2) = x1.
•ω:π1(Σ, σ0)→π1(E0, e0)is given by
ω(y1) = x2x−1
1x−1
2x1x−1
2and ω(y2) = x2.
x2
x1
x2
x2x−1
1x−1
2x1x−1
2
x1
x2x−1
1x−1
2
y1
y2
Σ
E0
ω:π1(Σ, σ0)→π1(E0, e0)
α:π1(Σ, σ0)→π1(E0, e0)
σ0
e0
lω
lα
Figure 5. The fundamental groups of E0and Σ.
A presentation for π1(E, e0)is therefore given by
π1(E, e0) = hx1, x2, lV|lVx2x−1
1x−1
2x1x−1
2l−1
V=x2x−1
1x−1
2and lVx2l−1
V=x1i
where lVis represented by the path lαl−1
ω(which is a longitude of V). The meridian
of V(which coincides with the outer boundary of Σ) represents the element
mV:= x2x−1
1x−1
2x1=α(y1y2) = ω(y1y2) = x2x−1
1x−1
2x1x−1
2x2.
Define CV:= hmV, lVi∼
=π1(∂V , e0)≤π1(∂E, e0).
For the next lemmas we denote the subgroups α(π1(Σ, σ0)) and ω(π1(Σ, σ0)) of
π1(E0, e0)by Uαand Uωrespectively.
Lemma 3.1. Uαand Uωare conjugacy separable in π1(E0, e0)meaning that the
intersection Uα∩gUωg−1is trivial for all g∈π1(E0, e0).
Proof. The lemma follows by a simple homology argument. 
Lemma 3.2. Let a∈π1(E0, e0). Then the following hold:
(1) ahmVia−1∩Uα6= 1 iff a∈Uα. Similarly, ahmVia−1∩Uω6= 1 iff a∈Uω
(2) ahx1ia−1∩Uα6= 1 iff a=uxε
2xk
1with u∈Uα,ε∈ {0,1}and k∈Z.
Similarly, ahx2ia−1∩Uω6= 1 iff a=u(x2x−1
1)εxk
2with u∈Uω,ε∈ {0,1}
and k∈Z.

8 EDERSON R. F. DUTRA
(3) aUαa−1∩Uα6= 1 iff a=u1xε
2u2with u1, u2∈Uαand ε∈ {−1,0,1}.
Similarly, aUωa−1∩Uω6= 1 iff a=u1(x2x−1
1)εu2with u1, u2∈Uωand
ε∈ {−1,0,1}
Proof. We prove the above claims for Uα. The argument for Uωis similar. We
identify π1(E0, e0)with the fundamental group of the rose R2with two petals.
Let p: (Γ, u0)→(R2, v)be the covering corresponding to Uα=hx1, x2x−1
1x−1
2i,
see Figure 6.
(1) The result follows from the fact a lifting emof mk
V= (x2x−1
1x−1
2x1)kis closed
iff emstars (and therefore also ends) at the vertex u0.
u0
x1
x2
x1
x2x1x
−1
2
p
v
u1
Figure 6. The covering corresponding to Uα=hx2x1x−1
2, x1i.
(2) The argument is similar to the argument given in (1). A lifting of ˜
lof xz
1is
closed iff ˜
lstarts at u0or at u1. Therefore any lifting of astarting at u0either
terminates at u0(which implies that alies in Uα) or terminates at u1(which implies
that a=ux2xk
1with u∈Uαand k∈Z).
(3) First observe that reduced words (in the basis {x1, x2}of π1(E0, e0)) that rep-
resent elements of Uαare of the form
(*) xn0
1·x2xn1
1x−1
2·xn2
1·. . . ·xnd−1
1·x2xnd
1x−1
2
with n0, nd∈Zand n1, . . . , nd−1∈Z\ {0}.
After multiplying aon the right and on the left by elements of Uαwe can assume
that whenever a=u1a0u2with u1, u2∈Uαand |a0|≤|a|then u1=u2= 1, where
|a|denotes the word length of awith respect to the basis {x1, x2}. We will refer to
this assumption as the length assumption on a.
We will show that a=xε
2for some ε∈ {−1,0,1}. First observe that if a=xs
2
with s∈Zthen (*) implies that aUαa−1∩Uα6= 1 iff s∈ {−1,0,1}.
Assume that ais not a power of x2. The length assumption on aimplies that a
can be written as xr
2·b·xs
2with r, s ∈Z\ {0}and the first and last letters in bare
x±1
1. If s6=−1then for any u∈Uthe reduced word that represents aua−1contains
the initial sub-word xr
2b. From (*) we conclude that that r= 1. But b=xl
1cfor
some l6= 0 and some c(which might be trivial). Thus
a=x2xl
1cxs
2=x2xl
1x−1
2·x2cxs
2.

MERIDIONAL RANK OF WHITEHEAD DOUBLES 9
with x2xl
1x−1
2∈Uand |x2cxs
2|<|a|, a contradiction to the length assumption.
Assume now that s=−1. The assumption on bimplies that b=cxl
1for some
l6= 0. Thus
a=xr
2bx−1
2=xr
2cxl
1x−1
2=xr
2cx−1
2·x2xl
1x−1
2
with x2xl
1x−1
2∈Uand |xr
2cx−1
2|<|a|which contradicts the length assumption. 
Remark 3.3.It follows from item (1) that ahmVia−1∩Uαis either trivial or equal
to ahmVia−1. Item (2) implies that ahx1ia−1∩Uαis either trivial or equal to
ahx1ia−1. The same claims also holds for Uω.
From Lemma 3.2(3) we immediately obtain the following corollary.
Corollary 3.4. Nπ1(E0,e0)(Uα) = Uαand Nπ1(E0,e0)(Uω) = Uωwhere Nπ1(E0,e0)(U)
denotes the normalizer of U≤π1(E0, e0).
We will consider subgroups of π1(E, e0)that are generated by conjugates of x1
and x2. We call such subgroups meridional. We define the meridional rank ¯w(U)
of a meridional subgroup U≤π1(E, e0)as the minimal number of conjugates of x1
and x2needed to generate U.
Definition 3.5. A meridional subgroup U=hg1xi1g−1
1, . . . , glxilg−1
li ≤ π1(E, e0)
with l= ¯w(U)is good if there is a partitioned J1, . . . , Jdof {1, . . . , l}such that:
(1) U=U1∗. . . ∗Udwhere Us(1≤s≤l) is generated by {gkxikg−1
k|k∈Js}.
(2) for any h∈π1(E , e0)one of the following holds:
(a) U∩hCVh−1= 1.
(b) U∩hCVh−1=hhmVh−1iand there are 1≤s≤dwith |Js| ≥ 2and
u∈Usuch that hmVh−1lies in uUsu−1.
The main result of this section says that meridional subgroups are almost good
in the sense that they are contained in a larger good subgroup whose rank does not
increase.
Proposition 3.6. Any meridional subgroup U≤π1(E, e0)is contained in a good
meridional subgroup ¯
U≤π1(E, e0)with ¯w(¯
U)≤¯w(U).
3.1. Proof of Proposition 3.6. We consider the HNN extension
π1(E, e0) = π1(E0, e0)∗(α,π1(Σ,σ0),ω )
as a graph of groups Ahaving a single vertex vwith group Av=π1(E0, e0)and
a single edge pair e, e−1with α(e) = ω(e) = vand group Ae=Ae−1=π1(Σ, σ0).
The boundary monorphisms αe, ωe:Ae→Avare given by αe=αand ωe=ω.
e
Ae=π1(Σ, σ0) = F(y1, y2)
ωe:y17→ x2x−1
1x−1
2x1x−1
2
y27→ x2
αe:y17→ x2x−1
1x−1
2
y27→ x1
v
Av=π1(E0, e0) = F(x1, x2)
Figure 7. The graph of groups A.

10 EDERSON R. F. DUTRA
The idea of the proof is to look at A-graphs that possibly represent good merid-
ional subgroups of π1(E, e0)∼
=π1(A, v). We star by defining A-graphs of cyclic and
non-cyclic type which will serve as the building blocks of the A-graph we are going
to consider.
We say that a folded A-graph Bis of cyclic type if the following hold:
(1) the underlying graph of Bis an interval f1, . . . , fdof length d≥0.
f1
fd
fd−1
f2
ud
ud−1
u1
u0
u2
Figure 8. The underlying graph of an A-graph of cyclic type.
(2) for each u∈V B ={u0, . . . , ud}the corresponding vertex group Bu≤Av
is generated by a single conjugate of x1or x2.
(3) for each f∈EB ={f±1
1, . . . , f ±1
d}the corresponding edge group Bf≤Ae
is generated by a single conjugate of y1or y2.
We say that a folded A-graph Bis of non-cyclic type if the following hold:
(1) the underlying graph Bof Bcontains an interval f1, . . . , fdof length d≥2
with u0=α(f1)and ui=ω(fi)for i= 1, . . . , d such that
B\ {f±1
1, . . . , f ±1
d}=l1∪l2∪l3∪l4
where lk(1≤k≤4) is an interval of length dk≥0with u0=α(l1) = α(l2)
and ud=α(l3) = α(l4), see Figure 9.
ud−1
u1
u0
u2
l1
l2
l3
l4
ud
f1
f2
fd−1
fd
Figure 9. The underlying graph of an A-graph of non-cyclic type.
(2) for each i∈ {1,2,3,4}the sub-A-graph Biof Bthat is carried by the
interval liis an A-graph of cyclic type.
(3) for each 1≤i≤d,Bfi=Aei=Ae=hy1, y2iwhere ei:= [fi]∈EA.
(4) for each 1≤i≤d−1,Bui=Av=hx1, x2i.
(5) the vertex groups at u0=α(f1)and ud=ω(fd)are given by:
Bu0=of1αe1(Ae1)o−1
f1=of1hαe1(y1), αe1(y2)io−1
f1≤Av0
and
Bud=t−1
fdωed(Aed)tfd=t−1
fdhωed(y1), ωed(y2)itfd≤Av0.

MERIDIONAL RANK OF WHITEHEAD DOUBLES 11
We say that a connected sub-A-graph B0of Bis non-degenerate if B0contains
at least one vertex with Bu=Av, or in other words, if B0contains at least one of
the vertices u1, . . . , ud−1.
Let Bbe an A-graph of cyclic or non-cyclic type. It follows from the definition
combined with [15, Proposition 2.4] that the subgroup U:= U(B, u0)≤π1(A, v)
represented by Bis meridional. If Bis of cyclic type, then we further see that
U(B, u0)is generated by a single conjugate of x1or x2; hence w(U) = 1. If Bis of
non-cyclic type, then is not hard to see that the sub-A-graph B0carried by f1, . . . , fd
carries the fundamental group of B, i.e. the natural inclusion of B0into Binduces
an isomorphism π1(B0, u0)→π1(B, u0); hence U=U(B0, u0). Moreover, to B0
we associate a manifold XB0whose fundamental group is canonically isomorphic
to π1(B0, u0)as shown in Figure 10. Since Band therefore also B0are folded we
u0
f1
u1
f2
ud
fd
ud−1
fd−1
Figure 10. The manifold XB0.
conclude from [22, Proposition 6] that U=U(B0, u0)is isomorphic to π1(B, u0)and
hence isomorphic to π1(B0, u0)∼
=π1(XB0). Thus Uis generated by dconjugates
of x1and x2. Since the first homology group of XB0is isomorphic to Zdwe can
conclude that w(U) = rank(U) = rank(π1(XB0)) = d.
Let Bbe an arbitrary A-graph. We define Ess(B)as the A-graph that is obtained
from Bby removing all vertices and edges with non-trivial group. Any component
of Ess(B)is called an essential piece of B.
Definition 3.7. We say that an A-graph Bis good if the following hold:
(G1) the underlying graph of Bis a finite tree.
(G2) if B0is an essential piece of B, then B0is either an A-graph of cyclic type
(in this case we say that B0is a cyclic essential piece of B) or B0is a non-
degenerate sub-A-graph of some A-graph of non-cyclic type (in this case we
say that B0is a non-cyclic essential piece of B).
The set of vertices and edges with non-trivial group in a good A-graph can be
decomposed as follows. Let Bbe a good A-graph. We define
VB:= {u∈V B |Bu6= 1}and EB:= {f∈EB |Bf6= 1}.
The set VBis partitioned into three types of vertices, namely, the set of:
(1) cyclic vertices which consists of all vertices u∈V B such that Buis gener-
ated by a single conjugate of x1or x2.

12 EDERSON R. F. DUTRA
(2) almost full vertices which consists of all vertices u∈V B for which there is
an edge f∈StarB(u)such that Bf=hy1, y2iand
Bu=hofαe0(y1)o−1
f, ofαe0(y2)o−1
fi
where e0= [f]∈EA. Observe that Bω(f)=Av.
(3) full vertices which consists of all vertices usuch that Bu=Av=hx1, x2i.
Similarly, EBis partitioned into two types of edges, namely, the set of:
(1) cyclic edges which consists of all edges f∈EB such that Bfis generated
by a single conjugate of y1or y2.
(2) full edges which consists of all edges f∈EB such that Bf=hy1, y2i.
Lemma 3.8. Let Bbe a good A-graph and let u∈V B and f∈StarB(u). Assume
that e0= [f]∈EA ={e, e−1}. Then the following hold:
(1) If Buis generated by a single conjugate of x1(in particular uis cyclic) and
ofαe0(Ae0)o−1
f∩Bu6= 1
then e0=eand up to auxiliary moves we can assume that Bu=hx1iand
of=x−ε
2with ε∈ {0,1}.
Similarly, if Buis generated by a single conjugate of x2and
ofαe0(Ae0)o−1
f∩Bu6= 1
then e0=e−1and up to auxiliary moves we can assume that Bu=hx2i
and of= (x2x−1
1)−εwith ε∈ {0,1}.
(2) If Bu6=Avis generated by two conjugates of x1(in particular uis almost
full) and
Bu∩ofαe0(Ae0)o−1
f6= 1
then e0=eand up to auxiliary moves of=xε
2with ε∈ {−1,0,1}.
Similarly, if Bu6=Avis generated by two conjugates of x2(in particular u
is almost full) and
Bu∩ofαe0(Ae0)o−1
f6= 1
then e0=e−1and up to auxiliary moves of= (x2x−1
1)ηwith η∈ {−1,0,1}.
Proof. Assume that Bu=hgx1g−1i ≤ Avwith g∈Av=π1(E0, e0). Lemma 3.1
implies that e0=e. Lemma 3.2(2) implies that
o−1
fg=αe(c)xε
2xk
1
with c∈Ae,ε∈ {0,1}and k∈Z. Therefore
of=g−1xk
1x−ε
2αe(c).
This implies that after applying auxiliary moves to Bbased on uand on fwe can
assume that Bu=hx1iand of=x−ε
2.
(2) After applying an auxiliary move of type A0 based on uwe can assume that
Bu=αe(Ae) = hαe(y1), αe(y2)i=hx2x−1
1x−1
2, x1i ≤ Av.
By hypothesis ofαe0(Ae0)o−1
f∩Bu6= 1. Thus, by Lemma 3.1, it holds e0=eand
by Lemma 3.2(3) it holds
of=uxε
2αe(c)

MERIDIONAL RANK OF WHITEHEAD DOUBLES 13
with b∈Buand c∈Ae. Therefore, after applying auxiliary moves based on uand
of f, we can assume that of=xε
2.
Collapsing essential pieces. In addition to auxiliary moves as defined in [22,
subsection 1.5] we will also need another prepossessing move which is the inverse
of a type IIA fold.
Let Bbe a good A-graph. Let B0be an essential piece of Bsuch that B0contains
at least one edge (and hence Bcontains at least one edge with non-trivial group)
and let u∈V B0. Choose u0∈V B0\ {u}such that:
(i) valB0(u0) = 1 (in a finite tree there are at least two vertices with this
property).
(ii) if u00 ∈V B0\ {u}with valB0(u00 ) = 1 then, up to conjugacy, Bu0is a
subgroup of Bu00 .
Let f0∈EB0be the edge starting at u0. Denote u00 := ω(f)∈V B0and e0= [f0]∈
EA. We define a new A-graph ¯
Bas follows:
1. if u0is not full, then from the definition of A-graphs of cyclic and non-cyclic
type we see that Bu0=of0αe0(Bf0)o−1
f0. In this case we replace the vertex
group Bu0by ¯
Bu0= 1 and the edge group Bf0≤Ae0by ¯
Bf0= 1, see
Figure 11.
f′
u′
u′′
f′
u′
u′′
¯
Bu′= 1
Figure 11. Case 1: u0is not full.
2. if u0is full, then there are two cases:
(a) u00 is full or u00 is almost full and equal to u. Observe that if the second
option occurs, then the choice of u0guarantees that B0contains only
the edge f0. As u0is full we have
Bu=Av=hx1, x2i
=of0hx1, x2io−1
f0
=hof0αe0(y2)o−1
f0, of0xio−1
f0i
where i∈ {1,2}depends on e0:= [f0]. In this case we do the following:
(i) replace the edge group Bf0=Ae0=hy1, y2iby ¯
Bf0= 1.
(ii) replace the vertex group Bu0=Avby ¯
Bu0=hof0xio−1
f0i.
(iii) If u00 is full then we do not change the vertex group at u00,
i.e. ¯
Bu00 =Bu00 , and if u00 is almost full then we replace
Bu00 =ht−1
f0ωe0(y1)tf0, t−1
f0ωe0(y2)tf0i
by ¯
Bu00 =ht−1
f0ωe0(y2)tf0i, see Figure 12.

14 EDERSON R. F. DUTRA
f′
u′
u′′
f′
u′
u′′
¯
Bu′=⟨of′xio−1
f′⟩
¯
Bu′′ =Bu′′ =Av
f′
u′
u′′ =u
f′
u′
u′′ =u
¯
Bu′=⟨of′xio−1
f′⟩
¯
Bu′′ =⟨t−1
f′ωe′(y2)tf′⟩
Figure 12. Case (2.a): u00 is full or u00 is almost full and u00 =u.
(b) u00 is not full and distinct from u. The choice of u0implies that uis
cyclic and valB0(u00 ) = 2. Let f00 be the edge in StarB0(u00)distinct
from (f0)−1and e00 := [f00]. The previous lemma implies that there is
h∈Bu0=Avand i, j ∈ {1,2}such that
Bu0=Av=hof0αe0(yj)o−1
f0, hxih−1i
and
of00 αe00 (Bf00 )o−1
f00 =ht−1
f0ωe0(yj)tf0i,
see Figure 13. In this case we replace the edge group Bf0=Ae=
hy1, y2iby ¯
Bf0= 1 and the vertex group Bu0by ¯
Bu0=hhxih−1i.
f′
f′′
u′′′
u′′
u′
t−1
f′ωe′(yj)tf′
hxih−1
of′αe′(yj)o−1
f′
Figure 13. Case (2.b): u00 is not full and u00 6=u
It is not hard to see that ¯
Bis indeed good. We say that ¯
Bis obtained from Bby
unfolding along the edge f. After repeating this procedure finitely many times we
obtain a good A-graph e
Bin which all edge groups in B0are trivial and e
Bu=Bu

MERIDIONAL RANK OF WHITEHEAD DOUBLES 15
if uis cyclic or full (in B) and e
Buis generated by a conjugate of x1or x2if uis
almost full (in B). We say that e
Bis obtained from Bby collapsing B0into u. It is
clear from the definition that the labels of edges are not affected in these moves.
Let Bbe a good A-graph. The rank of B, denoted rank(B), is defined as the
rank of π1(B, u0). The number of full vertices (resp. full edges) in Bis denoted by
vfull(B)(resp. ef ull (B)).
Lemma 3.9. Let Bbe a good A-graph with base vertex u0∈V B. Let B1,...,Br
(resp. Br+1,...,Bl) be the cyclic (resp non-cyclic) essential pieces of B. Then
rank(B) = r+
l
X
i=r+1
(vfull(Bi) + 1)
and the subgroup U(B, u0)≤π1(A, v) = π1(E, e0)represented by Bis generated by
rank(B)conjugates of x1and x2.
If Bis folded, then U(B, u0)is good and w(U(B, u0)) = rank(B).
Proof. For each i= 1, . . . , l let uibe an arbitrary vertex of Bi. Let further pi
be a B-path from u0to uisuch that its underlying path is reduced. Since the
group of any edge and the group of any vertex that is not in Ess(B)is trivial, the
fundamental group of Bsplits as
π1(B, u0) = p1π1(B1, u1)p−1
1∗. . . ∗plπ1(Bl, ul)p−1
l,
where piπ1(Bi, ui)p−1
iconsists of all elements of π1(B, u0)represented by paths of
the form pipp−1
iwhere pis a closed path in Bibased at ui.
Grushko’s Theorem [14] implies that rank(B) = Prank(Bj). As observed be-
fore the rank of any cyclic piece is 1and the rank of any non-cyclic piece Br+iis
vfull(Bi)+1. Therefore we have the right formula for rank(B).
Assume now that Bis folded. It follows from [22, Proposition 6] that U(B, u0)
is isomorphic to π1(B, u0). From this fact we readily conclude that w(U(B, u0)) =
rank(B)and that U(B, u0) = U1∗. . . ∗Ulwhere
Ui:= φ(piπ1(Bi, ui)p−1
i)∼
=π1(Bi, ui)for i= 1, . . . , l.
This shows that condition (1) of the definition of good subgroups of π1(E, e0)is
satisfied.
It remains to show that (2) also holds. Let h∈π1(E , e0) = π1(A, v). Since the
underlying graph of Bis a tree and lVis represented by the closed A-path 1, e, 1,
we conclude that
U(B, u0)∩hCVh−1≤hhmVih−1.
Assume that hmz
Vh−1∈U(B, u0)for some z6= 0. The Normal Form Theorem [15,
Proposition 2.4] combined with the foldedness of Bimplies that there must be
a vertex u∈V B such that Bucontains amz
Va−1for some a∈Av. Since no
cyclic subgroup of Avcontains mz
Vit follows that uis either almost full or full
(in particular ulies in a non-cyclic essential piece of B). Thus Bualso contains
amVa−1. This shows that for any h∈π1(E, e0)one of conditions (2.a) or (2.b)
from the definition of good subgroups holds. 
proof of Proposition 3.6. Let U=hg1xi1g−1
1, . . . , gkxikg−1
kibe a meridional sub-
group of π1(E, e0)of meridional rank k:= w(U). Since π1(E , e0)splits as π1(A, v),
where Ais the graph of groups described above, each gs(1≤s≤k) can be written
as gs= [ps]where psis a (non-necessarily reduced) closed A-path of positive length.

16 EDERSON R. F. DUTRA
We define B0as follows (see figure 28). Start with a vertex u0and for each
1≤s≤kwe glue an interval lssubdivided into length(ps)segments to u0. The
label of each lsis defined so that µ(ls) = ps. The vertex group at ω(ls)is hxisiand
the remaining vertex and edge groups are trivial.
Note that B0is good since all components of Ess(B0)are (degenerate) A-graphs
of cyclic type. Note further that π1(B0, u0)is freely generated by
[l1xi1l−1
1],...,[lkxikl−1
k]
and, by definition, [lsxisl−1
s]is mapped onto gsxisg−1
s. Therefore U(B0, u0) = U
and k=w(U) = rank(B0).
B0
u0
= 1
B0
ω(l1)=⟨xi1⟩
B0
ω(ls)=⟨xis⟩
B0
ω(lk)=⟨xik⟩
l1
ls
lk
Figure 14. The A-graph B0.
Definition 3.10. We define the complexity of a good A-graph Bas the tuple
c(B)=(rank(B),|EB|,|EB| − ef ull (B),|EB|−|EB|)∈N4
0.
Recall that efull(B)denotes the number of full edges in Band EBdenotes the set
of edges in Bwith non-trivial group. Throughout the proof we assume that N4
0is
equipped with the lexicographic order.
A straightforward inspection of the various cases reveals that all auxiliary moves
preserve both goodness and the complexity, i.e. if B0is obtained from a good A-
graph by an auxiliary move then B0is also good and c(B0) = c(B).
Now choose a good A-graph Bsuch that the following hold:
(1) U≤U(B, u0)for some vertex u0of B.
(2) the complexity of Bis minimal among all good A-graphs satisfying condi-
tion (1).
Since B0is good and U(B0, u0) = Uwe conclude that rank(B)≤rank(B0) =
k=w(U). Thus w(U(B, u0)) ≤w(U). It remains to show that U(B, u0)is good.
According to Lemma 3.9 it suffices to show that Bis folded. We will argue by
contradiction. Thus assume that Bis not folded. Since the underlying graph of B
is a tree, only folds of type IA and folds of type IIA can be applied to B.
(IA) Assume that a fold of type IA can be applied to B. Thus there are edges f1
and f2in Bwith u:= α(f1) = α(f2)and e0= [f1] = [f2]∈EA such that
of2=bof1αe0(c)
for some b∈Buand some c∈Ae0. After applying auxiliary moves to B(see [22,
section 1.5]) we can assume that the fold is elementary, that is, of2=of1.

MERIDIONAL RANK OF WHITEHEAD DOUBLES 17
Let B0be any A-graph that is obtained from Bby collapsing the essential pieces
containing ω(f1)and ω(f2); hence, there are no edges with non-trivial group in
B0starting at w(f1)or ω(f2). Since the label of f1and f2is not affected, the
elementary fold that identifies f1and f2can also be applied to B0.
Let B00 be the A-graph obtained from B0by this fold and let ybe the image of
ω(f1)(hence also of ω(f2)) under the fold. According [22, Proposition 7] we have
U(B00, u00
0) = U(B, u0). Since the vertex groups B0
ω(f1)and B0
ω(f2)are replaced by
B00
y=hB0
ω(f1), B0
ω(f2)i ≤ A[y]
we see that B00
yis generated by at most four conjugates of x1and x2. If B00
yis not
cyclic, then we replace B00
yby B000
y=Av=hx1, x2i. In this way we obtain a good
A-graph B000 such that
U≤U(B00, u0)≤U(B000 , u000
0).
The complexity decreases since rank(B000 )≤rank(B)and |EB000|=|EB| − 2.
(IIA) Assume that no fold of type IA can be applied to B. Since Bis not folded
a fold of type IIA can be applied to B. Thus there is f∈EB with u:= α(f),
u0:= ω(f)and e0:= [f]∈EA ={e, e−1}such that
(3.1) ofαe0(Bf)o−1
f6=ofαe0(Ae0)o−1
f∩Bu.
Since all edges with non-trivial group lie in some essential piece and these A-graphs
are folded by definition, we conclude that Bf= 1. Thus (3.1) means that
(3.2) ofαe0(Ae0)o−1
f∩Bu6= 1.
To simplify the argument we assume that e0=e. The case where fis of type e−1
is entirely analogous. We need to consider various cases depending on the vertex
groups Buand Bu0.
Case 1: uis full, that is, Bu=Av=hx1, x2i.The assumption that no fold of type
IA can be applied to Bimplies that there is at most one edge in StarB(u)\ {f}
which is necessarily of type e−1. We need to consider all possibilities for Bu0.
(a) Bu0is trivial. We define B0as the A-graph that is obtained from Bby replacing
the edge group Bf= 1 by B0
f=Ae=hy1, y2iand the vertex group Bu0= 1 by
B0
u0=t−1
fωe(Ae)tf=t−1
fhωe(y1), ωe(y2)itf≤Av,
see Figure 15. Then B0is good since this move simply “enlarges” the essential piece
u
u′
f
Bu′= 1
u
u′
f
Figure 15. Case (1.a): B0
u0=t−1
fhωe(y1), ωe(y2)itf

18 EDERSON R. F. DUTRA
that contains u. Moreover, c(B0)< c(B)since rank(B0) = rank(B),|EB0|=|EB|
and efull(B0) = ef ull (B)+2.
(b) u0is cyclic. We first consider the case StarB(u0)∩EBis empty. We define
B0as the A-graph that is obtained from Bby replacing the edge group Bf= 1
by B0
f=Ae=hy1, y2iand the vertex group Bu0by B0
u0=Av=hx1, x2i, see
Figure 16. It is not hard to see that B0is good. The complexity decreases since
rank(B0) = rank(B),EB0=EB and ef ull (B0) = ef ull(B)+2.
u
f
u
u′
f
u′
Bu′=⟨gxjg−1⟩
Figure 16. Case (1.b).
Assume now that StarB(u0)∩EB6=∅. It follows immediately from the definition
of A-graphs of (non-)cyclic type that StarB(u0)∩EBcontains exactly one edge, say
h, such that
Bh=hcyic−1iand Bu0=ohhαe00 (cyic−1)io−1
h
for some i∈ {1,2}and some c∈Aewhere e00 := [h]∈EA. Let B0be the A-graph
obtained from Bby doing the following:
(a) replace the edge group Bh=hcyic−1iby B0
h= 1.
(b) replace the edge group Bf= 1 by B0
f=Ae=hy1, y2i.
(c) replace the vertex group Bu0=ohhαe00 (cyic−1)io−1
hby
Bu0=t−1
fωe(Ae)tf=t−1
fhωe(y1), ωe(y2)itf≤Av.
f
h
u
u′′
u′
f
h
u′′
u′
u
Figure 17. Case (1.b).
It is not hard to see that B0is good. Moreover, c(B0)< c(B)since rank(B0) =
rank(B),|EB0|=|EB|and ef ull(B0) = efull (B)+2.

MERIDIONAL RANK OF WHITEHEAD DOUBLES 19
(c) u0is almost full. By definition there is an unique edge h∈EB such that
Bh=Ae00 =hy1, y2i,Bu00 =Av=hx1, x2iand
Bu0=ohαe00 (Ae00 )o−1
h=ohhαe00 (y1), αe00 (y2)io−1
h
where e00 := [h]∈EA and u00 =ω(h)∈V B. After unfolding along edges with cyclic
group in the essential piece that contains u0we can assume that StarB(u0)∩EB=
{h}.
If e00 =ethen we define B0as the A-graph that is obtained from Bby replac-
ing the edge group Bf= 1 by B0
f=Aeand the vertex group Bu0by B0
u0=Av=
hx1, x2i. Thus B0is good and u0becomes a full vertex, see Figure 18. The complex-
f
h
u
f
h
u′′
u′
u
u′′
u′
Figure 18. Case (1.c).
ity decreases since rank(B0) = rank(B),EB0=EB and ef ull(B0) = ef ull (B)+2.
If e00 =e−1then the previous move does not yield a good marked A-graph since
f−1and hcan then be folded and so cannot lie in an A-graph of non-cyclic type.
In this case we define a new good A-graph in which a fold of type IA can be applied
to.
First let B0be the A-graph that is obtained from Bby collapsing the essential
piece that contains uinto u. Thus B0
u=Bu=hx1, x2isince uis full. After applying
an auxiliary move of type A2 based on the edge fto B0we can assume that of= 1.
Now let B00 be the A-graph that is obtained from B0by pushing the element
x1along f, that is, we replace B0
u=hx1, x2iby B00
u=hx2iand B0
u0=Bu0by
B00
u0=Av, see Figure 19. The assumption that StarB(u0)contains only hensures
f
h
u
f
h
u′′
u′
u
u′′
u′
B′′
u
=⟨x2⟩
Figure 19. Case (1.c).
that B00 is good. Moreover, rank(B0) = rank(B).

20 EDERSON R. F. DUTRA
Since f−1and hare of same type and B00
u0=Avit follows that a fold of type
IA that identifies f−1with hcan be applied to B00. After folding these edges we
obtain a good A-graph B000 . To see that the rank decreases observe that the vertex
groups B00
u=hx2iand B00
u00 =Av=hx1, x2iare replaced by Av=hx1, x2i. Thus
c(B000)< c(B).
(d) u0is full, that is, Bu0=Av=hx1, x2i.In this case we simply replace Bf= 1
by B0
f=Ae=hy1, y2i, see Figure 20. The assumption that no fold of type IA
guarantees that B0is good. Moreover, rank(B0)≤rank(B)−1which implies that
c(B0)< c(B).
u
u′
f
u
u′
f
Figure 20. Case (1.d).
Case 2: uis almost full. By definition, there is a unique edge h∈S tarB(u)such
that Bh=Ae00 =hy1, y2i,Bu00 =Av=hx1, x2iand
Bu=ohαe00 (Ae00 )o−1
h=ohhαe00 (y1), αe00 (y2)io−1
h,
where e00 = [h]∈EA and u00 =ω(h)∈V B. Since
ohαe00 (Ae00 )o−1
h∩ofαe(Ae)o−1
f=Bu∩ofαe(Ae)o−1
f6= 1
it follows from Lemma 3.8(2) that e00 =e. Lemma 3.8(2) further implies that
Bu∩ofαe(Ae)o−1
f=ofhαe(yi)io−1
f
for some i∈ {1,2}. We can assume that u0is not full since otherwise we apply
the argument given in Case 1 to f−1. We distinguish two cases depending on the
vertex group Bu0.
(a) Bu0= 1.We define B0as the A-graph obtained from Bby doing the following:
(1) replace Bf= 1 by B0
f=hyii.
(2) replace Bu0= 1 by B0
u0=ht−1
fωe(yi)tfi.
The construction of B0is described in Figure 21. The assumption that no fold of
type IA can be applied to Bensure that B0is good. Moreover, the complexity
decreases as rank(B0) = rank(B),|EB0|=|EB|,efull(B0) = ef ull (B)and |EB0|=
|EB|+ 2.
(b) Bu06= 1.Let B(u)(resp. B(u0)) denote the essential piece that contains u
(resp. u0). We first construct B0by collapsing B(u0)into u0(hence B0
u06= 1) and
by collapsing cyclic edges in B(u)so that StarB0(u)∩EB0={h}.
Let now B00 be the A-graph that is obtained from B0by replacing the vertex
group B0
u=Buby B00
u=Av, see Figure 22. Note that the rank increases by one,

MERIDIONAL RANK OF WHITEHEAD DOUBLES 21
h
f
u′
u
u′′
Bu′= 1
h
f
u′
u
u′′
Figure 21. Case (2.a).
hence c(B00)> c(B). The edges fand hcan be folded since uis full in B00 and f
and hare of same type. The A-graph that is obtained by this fold is good and its
rank is at most rank(B)since the vertices u0and u00 are identified. The complexity
decreases because the number of edges decreases.
h
f
u′
u
u′′
h
f
u′
u
u′′
Figure 22. Case (2.b). The construction of B00 .
Case 3: uis cyclic. As in the previous case we can assume that u0is not full. As
uis cyclic we have Bu=hgxig−1ifor some g∈Avand some i∈ {1,2}. Since
the intersection Bu∩ofαe(Ae)o−1
fis non-trivial, Lemma 3.1 implies that i= 1.
Lemma 3.8(1) tells us that after auxiliary moves we have Bu=hx1iand of=x−ε
2
for some ε∈ {0,1}. We distinguish two cases depending on the vertex group Bu0.
(a) Bu0= 1.In this case we simply replace the edge group Bf= 1 by B0
f=hyii
and the vertex group Bu0= 1 by
B0
u0=ht−1
fωe(yi)tfi ≤ Av.
Observe that i= 1 if ε= 1 and i= 2 is ε= 0. The new A-graph is clearly
good. The complexity decreases because rank(B0) = rank(B),|EB0|=|EB|,
efull(B0) = ef ull (B)and |EB0|=|EB|+ 2.
(b) Bu06= 1.Let B(u)(resp. B(u0)) denote the essential piece of Bthat contains
the vertex u(resp. the vertex u0). We need to consider various cases depending on
B(u)and B(u0).
B(u)and B(u0)are cyclic essential pieces. We first collapse B(u)and B(u0)into u
and u0respectively. Then we replace the vertex group Bu0by B0
u0=Av=hx1, x2i,
the vertex group Buby αe(Ae) = hx2x−1
1x−1
2, x1i ≤ Avand the edge group Bf= 1

22 EDERSON R. F. DUTRA
by B0
f=Ae=hy1, y2i, see Figure 23. This A-graph is good and its complexity
decreases since the number of full edges increase by two. Note that this move turns
u(resp. u0) into an almost full vertex (resp. full vertex).
f
u′
u
f
u′
u
Figure 23. Case (3).
B(u)is a non-cyclic essential piece. After collapsing B(u0)into u0we can assume
that the vertex group Bu0is cyclic and StarB(u0)∩EB=∅. Let u00 be the closest
full vertex in B(u)to uand let f1, . . . , fkbe the shortest path from uto u00. After
collapsing B(u)into uwe can assume that Bu=hx1iand Bu00 =hgxig−1ifor some
g∈Avand some i∈ {1,2}. It follows from Lemma 3.8 that [fi] = eif i∈ {1, . . . , k}
is odd and [fi] = e−1if i∈ {1, . . . , k}is even.
Let B0be the A-graph obtained from Bby replacing the vertex groups Buand
Bu0by B0
u=Av=hx1, x2iand B0
u0=Av=hx1, x2iand replacing the edge group
Bf= 1 by B0
f=Ae=hx1, x2i. Thus uand u0(resp. f) are full vertices (resp. full
edge) in B0. Observe further that B0is good and represents a meridional subgroup
that contains U(B, u0)because we are replacing vertex groups and edge groups by
larger ones. The only issue we need to solve is the rank which increase by one in
going from Bto B0. We remedy this as follows. Since uand u0are full vertices and
[fi] = ewe conclude that the vertex u00 =ω(fk)folds onto uor u0depending on
the parity of d, see Figure 24. Thus the rank of the resulting A-graph clearly drops
back to rank(B). Since the number of edges decreases we see that c(B0)< c(B).
B(u)is cyclic and B(u0)is non-cyclic. We modify Bto fall in the previous case.
After unfolding along some edges in B(u0)we can assume that {h}=StarB(u0)∩
EB. Observe that u0either cyclic or almost full.
Let B0be the A-graph obtained from Bby doing the following:
(1) replace all vertex and edge groups in B(u)by trivial groups.
(2) replace Bu0by B0
u0=ht−1
fωe(y1)tfi.
(3) replace Bhby Bh= 1.
Thus B0is good and the rank is not affected. Now a fold along the edge h−1can
be applied to B0. To see that we fall in the previous case observe that the essential
piece of B0that contains u0is cyclic (as it contains only the vertex u0) and the
essential piece of B0that contains u00 := ω(h)is non-cyclic since u0is not full in
B.
4. Meridional rank of Whitehead doubles
In this section we prove Theorem 1.1. We follow the notation introduced in
Section 2. We will first compute the bridge number of the Whitehead double of

MERIDIONAL RANK OF WHITEHEAD DOUBLES 23
f
u′
u
fk−1
fk
Bu′′ =⟨gxig−1⟩
f1
k
folds of
type IA
Figure 24. Case (2.b). u00 folds onto uor u0.
an arbitrary knot. Thus let k1be an arbitrary non-trivial knot and let kbe a
Whitehead double of k1. We compute b(k)using the notion of plat presentations.
We say that a knot k0is a plat on 2n-strings (or simply a 2n-plat) if k0is the
union of a braid with 2nstrings and 2nunlinked and unknotted arcs which connect
pairs of consecutive strings of the braid at the top and at the bottom endpoints,
see Figure 25. Any knot admits a plat presentation, that is, it is ambient isotopic
to a plat. According to [5, Theorem 5.2] the minimal nsuch that k0admits a plat
presentation with 2n-strings coincides with the bridge number of k0.
Figure 25. A 6-plat.
We show that b(k)=2b(k1). A plat presentation with 2(2b(k1))-strings for kis
described in Figure 26. Hence b(k)≤2b(k1). On the other hand, as the index of the
Whitehead pattern (V, k0)is 2, it follows from [19] (see also [20]) that b(k)≥2b(k1).
We now show that if k1is prime and algebraically tame then w(k) = b(k). Let A
be the graph of groups described in subsection 2.2. Recall that
Av0=π1(E, e0) = hx1, x2, lV|lV·x2x−1
1x−1
2x1x−1
2·l−1
V=x2x−1
1x−1
2, lV·x2·l−1
V=x1i.
By choosing the base point appropriately we can assume that m= [x1]∈π1(A, v0)
is the fixed meridian of k.

24 EDERSON R. F. DUTRA
Figure 26. Whitehead double of k1.
We say that an A-graph Bis tame if the following hold:
(T0) Bis π1-surjective meaning that the induced homomorphism
φ:π1(B, u0)→π1(A, v0)∼
=G(k)
is surjective for some (and therefore) any vertex u0of type v0.
(T1) the underlying graph Bis a finite tree.
(T3) edge groups are either trivial or equal to hmei.
(T3) for each vertex u∈V B of type v0the vertex group Bu≤Av0is good.
(T4) for each vertex u∈V B of type v1the following hold:
(a) Bu=hg1m1g−1
1, . . . , grm1g−1
riwhere r=w(Bu)< b(k1).
(b) for each i= 1, . . . , r there is an edge fi∈StarB(u)∩EBsuch that
gim1g−1
iis (in Bu) conjugate to ofim1o−1
fi=ofiαe(me)o−1
fi.
The complexity of a tame A-graph is defined as the triple
c(B) := (c1(B),|EB|,|EB|−|EB|)∈N3
0
where
c1(B) := X
u∈V B
[u]=v0
¯w(Bu)
Recall that ¯w(Bu)denotes the minimal number of conjugates (in Av0) of x1and x2
needed to generate Bu. The set N3
0is equipped with the lexicographic order.
A straightforward inspection of the various cases reveals that all auxiliary moves
preserve tameness and complexity, i.e. if B0is obtained from a tame A-graph Bby
an auxiliary move then B0is also tame and c(B0) = c(B).
proof of Theorem 1.1. We want to show that w(k) = b(k). The proof will be con-
tradiction. Thus assume that
G(k) = hg1mg−1
1, . . . , glmg−1
li
with l < b(k). Each gican be written as gi= [pi]where piis a non-necessarily
reduced A-path from v0to v0of positive length.
We define an A-graph B0as follows. Start with a vertex u0of type v0and for
each 1≤s≤lwe glue an interval lssubdivided into length(ps)segments to u0.
The label of lsis defined so that µ(ls) = ps. The vertex group of ω(ls)is hmiand
the remaining vertex and edge groups are trivial. Therefore
φ([lsml−1
s]) = [µ(ls)mµ(ls)−1]=[psmp−1
s] = gsxisg−1
s,

MERIDIONAL RANK OF WHITEHEAD DOUBLES 25
and B0is π1-surjective. Observe that B0is tame since the only non-trivial vertex
groups are cyclically generated by the meridian m= [x1]∈Av0, and so are good
subgroups of Av0.
The claim that w(k)< b(k)therefore implies the existence of a tame A-graph.
Now choose a tame A-graph Bsuch that c(B)is minimal. Since B0is tame it
follows that c1(B)≤c1(B0) = l < b(k).
Since all edge groups in Bare proper subgroups of Ae=hme, leiwe conclude
that Bis not folded. As the underlying graph of Bis a tree we conclude that a fold
of type IA or a fold of type IIA can be applied to B. We consider these two cases
separately.
Fold of type IA. In this case a pair of distinct edges f1and f2starting at a
common vertex u:= α(f1) = α(f2)∈V B can be folded. After auxiliary moves we
can assume that the fold is elementary, that is, f1and f2have same label:
(a, e0, b) := (of1,[f1], tf1)=(of2,[f2], tf2)∈A[u]×EA ×A[ω(f1)].
Put y1:= ω(f1)and y2:= ω(f2).
Let B0be the A-graph that is obtained from Bby folding f1and f2. By definition,
the edge groups Bf1and Bf2are replaced by
B0
f=hBf1, Bf2i ≤ Ae
and the vertex groups By1and By2are replaced by
B0
y=hBy1, By2i ≤ A[y],
see Figure 27. Observe that B0
fis either trivial or generated by meand that
B0
y≤A[y1]=A[y2]is meridional, that is, generated by conjugates of x1and x2if
[y1] = v0and generated by conjugates of m1if [y1] = v1. We consider two cases
Bu
By1
By2
B′
u=Bu
B′
y=⟨By1, By2⟩
f1
f2
(a, e′, b)
(a, e′, b)
Bf
IA
(a, e′, b)
Figure 27. An elementary fold of type IA.
depending on the type e0of f1and f2.
Case 1: e0=e. In this case uis of type v0and y1and y2are of type v1. The
only tameness condition that is not trivially satisfied is condition (T4.a). Condition
(T4.b) says that there are w(Byi)edges in
Si:= StarB(yi)∩EB={f∈EB |Bf6= 1 and α(f) = yi}.
Let f∈S1∪S2. Since Bf=hmeiit follows that t−1
fmVtf∈Bω(f). Condition (2) in
the definition of good subgroups of Av0=π1(E , e0)then implies that ¯w(Bω(f))≥2.
Therefore, if ω(f)6=ω(f0)for all f∈S1and all f0∈S2(or equivalently, if f−1
1∪f−1
2

26 EDERSON R. F. DUTRA
is not contained in S1∪S2) then
2w(B0
y)≤2w(By1)+2w(By2)
≤2|S1|+ 2|S2|
≤X
f∈S1∪S2
¯w(Bω(f))
≤c1(B)
and so 2w(B0
y)≤c1(B)≤l < b(k) = 2b(k1)which implies that w(B0
y)< b(k1). If
f−1
1∈S1and f−1
2∈S2then, since Byi≤Av1is tame and the meridian
t−1
fim1tf1=b−1m1b
lies in Byi, we can assume that b−1m1bis part of a minimal meridional generating
set of Byi. Thus
w(B0
y)≤w(By1) + w(By2)−1.
The same computation as above with the summation taken over the set S1∪S2\
{f−1
1}shows that w(B0
y)< b(k1). Therefore Bis tame. Since no vertex of type v0
is affected we see that c1(B0)≤c1(B)and since the number of edges drops by two
we conclude that c(B0)< c(B).
Case 2. e0=e−1. Thus uis of type v1and y1and y2are of type v0. It follows
from Proposition 3.6 that B0
y=hBy1, By2i ≤ Av0is contained in a good subgroup
B00
y≤Av0such that ¯w(B00
y)≤¯w(B0
w).
Let B00 be the A-graph that is obtained from B0by replacing the vertex group
B0
yby B00
y≤Av0. Thus B00 is tame and
c1(B00)−c1(B) = ¯w(B00
y)−¯w(By1)−¯w(By2)≤0.
Since two edges are identified we conclude that c(B00)< c(B)which contradicts our
choice of B.
Fold of type IIA. We can assume that no fold of type IA is applicable to B. By
definition of type IIA folds, there is an edge f∈EB such that
Bf6=α−1
e0(o−1
fBxof)
where e0:= [f]∈EA and x:= α(f)∈V B. Let ydenote the vertex ω(f)∈V B.
We consider two cases depending on the type e0of f.
Case 1. e0=e−1. Thus xis of type v1and Bxis a tame subgroup of Av1=G(k1).
Assume that
Bx=hh1m1h−1
1, . . . , hrm1h−1
ri
where r=w(Bx). From
α−1
e−1(o−1
fBxof)6= 1
it follows that
ofαe−1(Ae)o−1
f∩Bx6= 1.
The tameness of the meridional subgroup Bx≤Av1implies that
ofαe−1(Ae)o−1
f∩Bx=ofhm1io−1
f
and ofm1o−1
fis in Bxconjugate to him1h−1
ifor some 1≤i≤r. Condition (T4.b)
says that there is g∈StarB(x)∩EBsuch that ogm1o−1
gis in Buconjugate to
him1h−1
i. Hence, ogm1o−1
gand ofm1o−1
fare conjugate in Bx. Lemma 2.3 then

MERIDIONAL RANK OF WHITEHEAD DOUBLES 27
implies that of=aogcfor some a∈Bxand some c∈P(k1). This equality implies
that a fold of type IA can be applied to Bsince P(k1) = αe−1(Ae−1) = ωe(Ae),
which contradicts our assumption on B.
Case 2. e0=e.Hence xis of type u0and so Bxis a good subgroup of Av0. Thus
ofαe(Ae)o−1
f∩Bu=ofCVo−1
f∩Bu=ofhmVio−1
f
since ofαe(Ae)o−1
f∩Buis non-trivial.
Let B0be the A-graph that is obtained from Bby replacing Byby
B0
y=hBy, t−1
fm1tfi
and Bf= 1 by B0
f=hmei. The only tameness condition that is not trivially
satisfied is condition (T4.a), i.e. w(B0
y)< b(k1). To see this we apply the same
computation as in the case of a fold of type IA based on an edge of type e. The
complexity clearly decreases since no vertex of type v0is affected, |EB0|=|EB|
and |EB|=|EB|+ 2.
(of, e, tf)
By
Bf= 1
B′
f=⟨me⟩
IIA
Bx
(of, e, tf)
B′
y=⟨By, t−1
fm1tf⟩
Bx
Figure 28. An elementary fold of type IIA when [f] = e.
5. Algebraically tame knots and braid satellites
In this section we prove Theorem 1.2. We follow the notation from Section 2.
Let k1be a prime algebraically tame knot and let kbe a braid satellite of k1with
braid pattern β. We want to show that kis algebraically tame. To this end we
must show that any meridional subgroup of G(k)that is generated by less than b(k)
meridians is tame where nis the number of strand of β. Let Abe the graph of
groups described in Section 2. Recall that
Av0=F(x1, . . . , xn)ohti
where the action of hti∼
=Zin F(x1,...xn)is given by
txit−1=aixτ(i)a−1
ifor 1≤i≤n
where a1, . . . , an∈F(x1, . . . , xn)and τ∈Snis the permutation associated to β.
By choosing the base point appropriately we can assume that m= [x1]∈π1(A, v0)
is the fixed meridian of k.
Thus let U≤G(k)be a meridional subgroup with r:= w(U)< b(k)(it follows
from [19] that b(k) = nb(k1)). In order to show that Uis tame we study A-graphs
that represent U. However, we do not consider arbitrary ones but only those that
are “nice” in the following sense. We say that an A-graph Bis benign if:
(B0) Brepresents U, that is, U(B, u0) = Ufor some vertex u0∈V B of type
v0∈V A.
(B1) the underlying graph of Bis a finite tree.
(B2) the edge groups are either trivial or equal to hmei.
(B3) for each vertex u∈V B of type v0the corresponding vertex group Buis
meridional (generated by conjugates in Av0of x1).

28 EDERSON R. F. DUTRA
(B4) for each vertex u∈V B of type v1the following hold:
(a) Bu=hg1m1g−1
1, . . . , grm1g−1
risuch that r=w(Bu)< b(k1).
(b) for each i= 1, . . . , r there is an edge fi∈StarB(u)∩EBsuch that
gim1g−1
iis (in Bu) conjugate to ofim1o−1
fi.
The complexity of a benign A-graph Bis defined as the triple
c(B) := (c1(B),|EB|,|EB|−|EB|)
where
c1(B) = X
u∈V B
[u]=v0
rank(Bu).
The same construction as in the previous section shows that there is a benign
A-graph such that c1(B0) = r=w(U). Now choose a benign A-graph Bsuch
that c(B)is minimal. An argument completely analogous to the one given in
the previous section, with the obvious adjustments and using Lemma 2.1 instead of
Proposition 3.6 and the notion of good subgroups, shows that if Bis not folded then
there is a benign A-graph that represents Uand has smaller complexity. Therefore
Bmust be folded.
We now use the foldedness of Bto show that Uis tame. According to Lemma 2.1,
for each u∈V B with [u] = v0, the corresponding vertex group Buis freely gener-
ated by
gu,1x1g−1
u,1, . . . , gu,rux1g−1
u,ru
where ru:= rank(Bu)and gu,i ∈Av0. For each such ulet γube a reduced B-path
from u0to u. Then
S:= {mu,i := [µ(γu·gu,ix1g−1
u,i ·γ−1
u)] |u∈V B s.t. [u] = v0and 1≤i≤ru}
is a meridional generating set of Uof minimal size. First observe that, as Bis
folded, mu,i is conjugate to mu0,i0iff u=u0and gu,ix1g−1
u,i is in Bucojugate to
gu,i0x1g−1
u,i0. Lemma 2.1 therefore implies that i=i0and so the meridians mu,i are
pairwise non conjugate. Next we show that for any g∈G(k)either gP (k)g−1∩U= 1
or gP (k)g−1∩U=ghmig−1and gmg−1is in Uconjugate to some of the meridians
in S. Assume that gpg−1∈Ufor some non-trivial peripheral element p∈P(k).
The foldedness of Bimplies that there is u∈V B of type v0such that Bucontains
a conjugate (in Av0) of p. Lemma 2.1 implies that:
(1) p∈ hx1iand cx1c−1∈Bufor some c∈Av0.
(2) cx1c−1is in Buconjugate to gu,ix1g−1
u,i for some 1≤i≤ru.
Therefore gP (k)g−1=ghmig−1and gmg−1is in Uconjugate mu,i for some uand
some ias before. This completes the proof of the theorem.
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Universidade Federal de São Carlos, São Carlos, Brazil
Email address:edersondutra@dm.ufscar.br