8 SELMA ALTINOK AND MOHAN BHUPAL

that ki = ki − ki = ki for i = 1, . . . , l. Thus, by Lemma 3.1, the page-genus of

OB(Z(T )) is given by

g(Z(T )) = 1 +

l

i=1

(wi − 2)zi + (zi − 1)ki

2

.

Substituting vi = wi + ti into (3.3), where ti is the number of non-Tjurina

components intersecting Ei, we obtain

g(Y + Z(T )) − g(Y ) =

l

i=1

(wi + ti + ki − 2)zi − ki

2

−

n

i=l+1

ki − ki

2

= g(Z(T )) − 1 +

l

i=1

tizi

2

−

n

i=l+1

ki − ki

2

= g(Z(T )) − 1 +

l

i=1

tizi

2

+

n

i=l+1

zφ(i)

2

= g(Z(T )) − 1 +

l

i=1

tizi,

where the penultimate equality follows from the fact that ki − ki = zφ(i) for i =

l + 1, . . . , n, which in turn follows from (3.2). Now, appealing to Theorem 3.10 in

[12], we see that zi = 1 whenever ti = 0. This completes the proof.

Lemma 3.3. Let Y =

∑n

i=1

miEi be an element of E + and Ei be a non-Tjurina

component for Y that does not intersect any Tjurina component for Y . Then

genus(OB(Y + Ei)) = genus(OB(Y )) + vi − Y · Ei − 1.

Proof. For convenience we will assume that E1 is our non-Tjurina component.

Deﬁne ki, for i = 1, . . . , n, by (3.1) and deﬁne ki by replacing (m1, . . . , mn) by

(m1 + 1, m2, . . . , mn) in (3.1). Arguing as in the proof of Lemma 3.2, we now

obtain that

g(Y + E1) − g(Y ) =

(v1 + k1 + k1 − 2) − (k1 − k1)

2

−

n

i=2

ki − ki

2

=

v1 + 2k1 − 2

2

+

v1

2

= v1 + k1 − 1,

where the second equality follows from the fact that ki − ki = 1 for the indices

of the v1 curves intersecting E1, and ki − ki = 0 for all the other indices. This

completes the proof.

Proof of Theorem 1.1. Use Theorem 2.6, Lemma 3.2, Lemma 3.3 and Re-

mark 2.8.

Proof of Theorem 1.2. First note that the number of binding components

of an open book OB(Y ) for Y ∈ E

+

is given by

bc(OB(Y )) = −

n

i=1

Y · Ei.

8

that ki = ki − ki = ki for i = 1, . . . , l. Thus, by Lemma 3.1, the page-genus of

OB(Z(T )) is given by

g(Z(T )) = 1 +

l

i=1

(wi − 2)zi + (zi − 1)ki

2

.

Substituting vi = wi + ti into (3.3), where ti is the number of non-Tjurina

components intersecting Ei, we obtain

g(Y + Z(T )) − g(Y ) =

l

i=1

(wi + ti + ki − 2)zi − ki

2

−

n

i=l+1

ki − ki

2

= g(Z(T )) − 1 +

l

i=1

tizi

2

−

n

i=l+1

ki − ki

2

= g(Z(T )) − 1 +

l

i=1

tizi

2

+

n

i=l+1

zφ(i)

2

= g(Z(T )) − 1 +

l

i=1

tizi,

where the penultimate equality follows from the fact that ki − ki = zφ(i) for i =

l + 1, . . . , n, which in turn follows from (3.2). Now, appealing to Theorem 3.10 in

[12], we see that zi = 1 whenever ti = 0. This completes the proof.

Lemma 3.3. Let Y =

∑n

i=1

miEi be an element of E + and Ei be a non-Tjurina

component for Y that does not intersect any Tjurina component for Y . Then

genus(OB(Y + Ei)) = genus(OB(Y )) + vi − Y · Ei − 1.

Proof. For convenience we will assume that E1 is our non-Tjurina component.

Deﬁne ki, for i = 1, . . . , n, by (3.1) and deﬁne ki by replacing (m1, . . . , mn) by

(m1 + 1, m2, . . . , mn) in (3.1). Arguing as in the proof of Lemma 3.2, we now

obtain that

g(Y + E1) − g(Y ) =

(v1 + k1 + k1 − 2) − (k1 − k1)

2

−

n

i=2

ki − ki

2

=

v1 + 2k1 − 2

2

+

v1

2

= v1 + k1 − 1,

where the second equality follows from the fact that ki − ki = 1 for the indices

of the v1 curves intersecting E1, and ki − ki = 0 for all the other indices. This

completes the proof.

Proof of Theorem 1.1. Use Theorem 2.6, Lemma 3.2, Lemma 3.3 and Re-

mark 2.8.

Proof of Theorem 1.2. First note that the number of binding components

of an open book OB(Y ) for Y ∈ E

+

is given by

bc(OB(Y )) = −

n

i=1

Y · Ei.

8