[PPT] - On adequacy and the crossing number of satellite knots Adri an Jim PowerPoint Presentation

SLIDE 1

On adequacy and the crossing number of satellite knots

Adri´ an Jim´ enez Pascual

The University of Tokyo

Tokyo Woman’s Christian University

23th December, 2017

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 1 / 19

SLIDE 2

Agenda

Preliminaries ↓ Link adequacy ↓ Link parallels ↓ Cable knots ↓ Main result ↓ Summary

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 2 / 19

SLIDE 3

Preliminaries

Definition (Satellite knot)

P: knot in ST. (Pattern) C: knot in S3 with framing 0. (Companion) e : ST ֒ → N(C): faithful embedding. Then eP is called a satellite knot (of C). From here on eP =: Sat(P, C).

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 3 / 19

SLIDE 4

Preliminaries

Definition (Satellite knot)

P: knot in ST. (Pattern) C: knot in S3 with framing 0. (Companion) e : ST ֒ → N(C): faithful embedding. Then eP is called a satellite knot (of C). From here on eP =: Sat(P, C).

P C

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 3 / 19

SLIDE 5

Preliminaries

Definition (Satellite knot)

P: knot in ST. (Pattern) C: knot in S3 with framing 0. (Companion) e : ST ֒ → N(C): faithful embedding. Then eP is called a satellite knot (of C). From here on eP =: Sat(P, C).

P C Sat(P, C)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 3 / 19

SLIDE 6

Problem

What is the minimal number of crossings with which Sat(P, C) can be drawn?

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 4 / 19

SLIDE 7

Problem

What is the minimal number of crossings with which Sat(P, C) can be drawn? Known facts: cr(Sat(P, C)) ≥ cr(C)/1013. (Lackenby, 2011)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 4 / 19

SLIDE 8

Problem

What is the minimal number of crossings with which Sat(P, C) can be drawn? Known facts: cr(Sat(P, C)) ≥ cr(C)/1013. (Lackenby, 2011)

Problem

Is the crossing number of a satellite knot bigger than that of its companion?

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 4 / 19

SLIDE 9

Problem

What is the minimal number of crossings with which Sat(P, C) can be drawn? Known facts: cr(Sat(P, C)) ≥ cr(C)/1013. (Lackenby, 2011)

Problem 1.67 (Kirby, 1995)

Is the crossing number of a satellite knot bigger than that of its companion?

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 4 / 19

SLIDE 10

Problem

What is the minimal number of crossings with which Sat(P, C) can be drawn? Known facts: cr(Sat(P, C)) ≥ cr(C)/1013. (Lackenby, 2011)

Problem 1.67 (Kirby, 1995)

Is the crossing number of a satellite knot bigger than that of its companion? Remarks: “Surely the answer is yes, so the problem indicates the difficulties of proving statements about the crossing number.”

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 4 / 19

SLIDE 11

Problem

What is the minimal number of crossings with which Sat(P, C) can be drawn? Known facts: cr(Sat(P, C)) ≥ cr(C)/1013. (Lackenby, 2011)

Problem 1.67 (Kirby, 1995)

Is the crossing number of a satellite knot bigger than that of its companion? Remarks: “Surely the answer is yes, so the problem indicates the difficulties of proving statements about the crossing number.” ↑ GOAL

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 4 / 19

SLIDE 12

Problem 1.65 (Kirby, 1995)

Is the crossing number cr(K) of a knot K additive with respect to connected sum, that is, is the equality cr(K1#K2) = cr(K1) + cr(K2) true?

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 5 / 19

SLIDE 13

Problem 1.65 (Kirby, 1995)

Is the crossing number cr(K) of a knot K additive with respect to connected sum, that is, is the equality cr(K1#K2) = cr(K1) + cr(K2) true? Known facts: Murasugi proved it is true for alternating knots. (Also Kauffman and Thistlethwaite)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 5 / 19

SLIDE 14

Problem 1.65 (Kirby, 1995)

Is the crossing number cr(K) of a knot K additive with respect to connected sum, that is, is the equality cr(K1#K2) = cr(K1) + cr(K2) true? Known facts: Murasugi proved it is true for adequate knots. (Also Kauffman and Thistlethwaite)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 5 / 19

SLIDE 15

Problem 1.65 (Kirby, 1995)

Is the crossing number cr(K) of a knot K additive with respect to connected sum, that is, is the equality cr(K1#K2) = cr(K1) + cr(K2) true? Known facts: Murasugi proved it is true for adequate knots. (Also Kauffman and Thistlethwaite) cr(K1#...#Kn) ≥ cr(K1)+...+cr(Kn)

152

. (Lackenby, 2011)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 5 / 19

SLIDE 16

Link adequacy

Definition

A state of a link is a function s : {c1, c2, ..., cn} → {−1, 1}.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 6 / 19

SLIDE 17

Link adequacy

Definition

A state of a link is a function s : {c1, c2, ..., cn} → {−1, 1}.

s(i) = −1 s(i) = +1 Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 6 / 19

SLIDE 18

Link adequacy

Definition

A state of a link is a function s : {c1, c2, ..., cn} → {−1, 1}.

s(i) = −1 s(i) = +1

The Kauffman bracket of a link with diagram D can be written as: D =

s
A

Pn

i=1 s(i)(−A−2 − A2)|sD|−1

.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 6 / 19

SLIDE 19

s+ is the state for which n

i=1 s+(i) = n

s− is the state for which n

i=1 s−(i) = −n

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 7 / 19

SLIDE 20

s+ is the state for which n

i=1 s+(i) = n

s− is the state for which n

i=1 s−(i) = −n

Definition

D is plus-adequate if |s+D| > |sD| for all s with n

i=1 s(i) = n − 2.

D is minus-adequate if |s−D| > |sD| for all s with n

i=1 s(i) = −n + 2.

D is adequate if plus-adequate and minus-adequate.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 7 / 19

SLIDE 21

Lemma 1 (Lickorish)

Let D be a link diagram with n crossings.

1 MD ≤ n + 2|s+D| − 2, with equality if D is plus-adequate, 2 mD ≥ −n − 2|s−D| + 2, with equality if D is minus-adequate. Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 8 / 19

SLIDE 22

Lemma 1 (Lickorish)

Let D be a link diagram with n crossings.

1 MD ≤ n + 2|s+D| − 2, with equality if D is plus-adequate, 2 mD ≥ −n − 2|s−D| + 2, with equality if D is minus-adequate.

Corollary 1 (Lickorish)

If D is adequate: B(D) = MD − mD = 2n + 2|s+D| + 2|s−D| − 4.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 8 / 19

SLIDE 23

Lemma 1 (Lickorish)

Let D be a link diagram with n crossings.

1 MD ≤ n + 2|s+D| − 2, with equality if D is plus-adequate, 2 mD ≥ −n − 2|s−D| + 2, with equality if D is minus-adequate.

Corollary 1 (Lickorish)

If D is adequate: B(D) = MD − mD = 2n + 2|s+D| + 2|s−D| − 4.

Lemma 2 (Lickorish)

Let D be a connected link diagram with n crossings. |s+D| + |s−D| ≤ n + 2, with equality if D alternating.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 8 / 19

SLIDE 24

Lemma 3

Let D be a diagram of an oriented link L. B(J(L)) = B(D) 4 .

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 9 / 19

SLIDE 25

Lemma 3

Let D be a diagram of an oriented link L. B(J(L)) = B(D) 4 . Proof. J(L) = (−A−3)wr(D)D

A2=t−1/2.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 9 / 19

SLIDE 26

Lemma 3

Let D be a diagram of an oriented link L. B(J(L)) = B(D) 4 . Proof. J(L) = (−A−3)wr(D)D

A2=t−1/2.

Theorem 1 (Lickorish)

Let D be a connected, n-crossing diagram of an oriented link L.

1 B(J(L)) ≤ n, 2 if D is alternating and reduced, B(J(L)) = n. Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 9 / 19

SLIDE 27

Link parallels

Definition

Let D be a diagram of an oriented link L. The r-parallel of D is the same diagram where each link component has been replaced by r parallel copies

f it, all preserving their “over” and “under” strands as in the original

diagram.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 10 / 19

SLIDE 28

Link parallels

Definition

Let D be a diagram of an oriented link L. The r-parallel of D is the same diagram where each link component has been replaced by r parallel copies

f it, all preserving their “over” and “under” strands as in the original

diagram.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 10 / 19

SLIDE 29

Lemma 4 (Lickorish)

Let D be a link diagram. If D is plus-adequate, Dr is also plus-adequate. If D is minus-adequate, Dr is also minus-adequate.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 11 / 19

SLIDE 30

Lemma 4 (Lickorish)

Let D be a link diagram. If D is plus-adequate, Dr is also plus-adequate. If D is minus-adequate, Dr is also minus-adequate. D adequate = ⇒ Dr adequate.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 11 / 19

SLIDE 31

Lemma 4 (Lickorish)

Let D be a link diagram. If D is plus-adequate, Dr is also plus-adequate. If D is minus-adequate, Dr is also minus-adequate. D adequate = ⇒ Dr adequate.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 11 / 19

SLIDE 32

Lemma 4 (Lickorish)

Let D be a link diagram. If D is plus-adequate, Dr is also plus-adequate. If D is minus-adequate, Dr is also minus-adequate. D adequate = ⇒ Dr adequate.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 11 / 19

SLIDE 33

We construct the parallel versions of the given results.

Lemma 1-bis (JP)

Let D be a link diagram with n crossings.

1 MDr ≤ nr2 + 2r|s+D| − 2, with equality if D is plus-adequate, 2 mDr ≥ −nr2 − 2r|s−D| + 2, with equality if D is minus-adequate. Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 12 / 19

SLIDE 34

We construct the parallel versions of the given results.

Lemma 1-bis (JP)

Let D be a link diagram with n crossings.

1 MDr ≤ nr2 + 2r|s+D| − 2, with equality if D is plus-adequate, 2 mDr ≥ −nr2 − 2r|s−D| + 2, with equality if D is minus-adequate.

Corollary 1-bis

If D is adequate: B(Dr) = MDr − mDr = 2nr2 + 2r|s+D| + 2r|s−D| − 4.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 12 / 19

SLIDE 35

Theorem 2 (JP)

Let L be an adequate oriented link. cr(Lr) ≥ r2 2 cr(L) + 2r − 1.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 13 / 19

SLIDE 36

Theorem 2 (JP)

Let L be an adequate oriented link. cr(Lr) ≥ r2 2 cr(L) + 2r − 1. Proof. cr(Lr) ≥ B(J(Lr)) = B(Dr) 4

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 13 / 19

SLIDE 37

Theorem 2 (JP)

Let L be an adequate oriented link. cr(Lr) ≥ r2 2 cr(L) + 2r − 1. Proof. cr(Lr) ≥ B(J(Lr)) = B(Dr) 4 ≥ r2 2 n + 2r − 1

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 13 / 19

SLIDE 38

Theorem 2 (JP)

Let L be an adequate oriented link. cr(Lr) ≥ r2 2 cr(L) + 2r − 1. Proof. cr(Lr) ≥ B(J(Lr)) = B(Dr) 4 ≥ r2 2 n + 2r − 1 ≥ r2 2 cr(L) + 2r − 1.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 13 / 19

SLIDE 39

Theorem 2 (JP)

Let L be an adequate oriented link. cr(Lr) ≥ r2 2 cr(L) + 2r − 1. Proof. cr(Lr) ≥ B(J(Lr)) = B(Dr) 4 ≥ r2 2 n + 2r − 1 ≥ r2 2 cr(L) + 2r − 1.

Corollary 2

If L is alternating: cr(Lr) ≥ r(r + 1) 2 cr(L) + r − 1. Proof uses Theorem 2 and Lemma 2.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 13 / 19

SLIDE 40

Theorem 2 (JP)

Let L be an adequate oriented link. cr(Lr) ≥ r2 2 cr(L) + 2r − 1. Proof. cr(Lr) ≥ B(J(Lr)) = B(Dr) 4 ≥ r2 2 n + 2r − 1 ≥ r2 2 cr(L) + 2r − 1.

Corollary 2

If L is alternating: cr(Lr) ≥ r(r + 1) 2 cr(L) + r − 1. Proof uses Theorem 2 and Lemma 2. ( |s+D| + |s−D| ≤ n + 2 )

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 13 / 19

SLIDE 41

Cable knots

Definition

D: diagram of C ⊂ S3. wr(D): writhe of D. We call (D; r) (the r-cable of D) to the r-parallel of D with −wr(D) full twists.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 14 / 19

SLIDE 42

Cable knots

Definition

D: diagram of C ⊂ S3. wr(D): writhe of D. We call (D; r) (the r-cable of D) to the r-parallel of D with −wr(D) full twists.

0 − framing 2 − parallel 2 − parallel composition T(6, 2)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 14 / 19

SLIDE 43

Lemma 5 (JP)

Let K ⊂ S3 be an oriented knot. B(J(K; r)) ≥ B(J(K r)).

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 15 / 19

SLIDE 44

Lemma 5 (JP)

Let K ⊂ S3 be an oriented knot. B(J(K; r)) ≥ B(J(K r)). Proof. Assume T(−wr(D)r, r)ST = r

i=0 αizi

ST , αr = 0. Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 15 / 19

SLIDE 45

Lemma 5 (JP)

Let K ⊂ S3 be an oriented knot. B(J(K; r)) ≥ B(J(K r)). Proof. Assume T(−wr(D)r, r)ST = r

i=0 αizi

ST , αr = 0.

Then, D; r = T(−wr(D)r, r)ST

zi

ST =Di = r

i=0 αiDi.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 15 / 19

SLIDE 46

Lemma 5 (JP)

Let K ⊂ S3 be an oriented knot. B(J(K; r)) ≥ B(J(K r)). Proof. Assume T(−wr(D)r, r)ST = r

i=0 αizi

ST , αr = 0.

Then, D; r = T(−wr(D)r, r)ST

zi

ST =Di = r

i=0 αiDi.

B(D; r) ≥ B(αrDr)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 15 / 19

SLIDE 47

Lemma 5 (JP)

Let K ⊂ S3 be an oriented knot. B(J(K; r)) ≥ B(J(K r)). Proof. Assume T(−wr(D)r, r)ST = r

i=0 αizi

ST , αr = 0.

Then, D; r = T(−wr(D)r, r)ST

zi

ST =Di = r

i=0 αiDi.

B(D; r) ≥ B(αrDr) ≥ B(αr) + B(Dr)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 15 / 19

SLIDE 48

Lemma 5 (JP)

Let K ⊂ S3 be an oriented knot. B(J(K; r)) ≥ B(J(K r)). Proof. Assume T(−wr(D)r, r)ST = r

i=0 αizi

ST , αr = 0.

Then, D; r = T(−wr(D)r, r)ST

zi

ST =Di = r

i=0 αiDi.

B(D; r) ≥ B(αrDr) ≥ B(αr) + B(Dr) = B(Dr).

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 15 / 19

SLIDE 49

P C

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 16 / 19

SLIDE 50

P C

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 16 / 19

SLIDE 51

P C

Theorem 3

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, let C ⊂ S3,

and let Sat(P, C) be their satellite knot. J(Sat(P, C)) =

M

k=0

βkJ(C; k).

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 16 / 19

SLIDE 52

P C

Theorem 3

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, let C ⊂ S3,

and let Sat(P, C) be their satellite knot. J(Sat(P, C)) =

M

k=0

βkJ(C; k).

Corollary 3

In particular, B(J(Sat(P, C))) ≥ B(βMJ(C; M)).

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 16 / 19

SLIDE 53

Main result

Theorem (JP)

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, and C ⊂ S3

adequate. cr(Sat(P, C)) ≥ B(βM) + M2 2 cr(C) + 2M − 1

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 17 / 19

SLIDE 54

Main result

Theorem (JP)

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, and C ⊂ S3

adequate. cr(Sat(P, C)) ≥ B(βM) + M2 2 cr(C) + 2M − 1 Proof. cr(Sat(P, C)) ≥ B(J(Sat(P, C)))

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 17 / 19

SLIDE 55

Main result

Theorem (JP)

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, and C ⊂ S3

adequate. cr(Sat(P, C)) ≥ B(βM) + M2 2 cr(C) + 2M − 1 Proof. cr(Sat(P, C)) ≥ B(J(Sat(P, C))) ≥ B(βMJ(C; M))

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 17 / 19

SLIDE 56

Main result

Theorem (JP)

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, and C ⊂ S3

adequate. cr(Sat(P, C)) ≥ B(βM) + M2 2 cr(C) + 2M − 1 Proof. cr(Sat(P, C)) ≥ B(J(Sat(P, C))) ≥ B(βMJ(C; M)) ≥ B(βMJ(C M))

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 17 / 19

SLIDE 57

Main result

Theorem (JP)

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, and C ⊂ S3

adequate. cr(Sat(P, C)) ≥ B(βM) + M2 2 cr(C) + 2M − 1 Proof. cr(Sat(P, C)) ≥ B(J(Sat(P, C))) ≥ B(βMJ(C; M)) ≥ B(βMJ(C M)) ≥ B(βM) + M2 2 cr(C) + 2M − 1.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 17 / 19

SLIDE 58

Main result

Theorem (JP)

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, and C ⊂ S3

adequate. cr(Sat(P, C)) ≥ B(βM) + M2 2 cr(C) + 2M − 1 Proof. cr(Sat(P, C)) ≥ B(J(Sat(P, C))) ≥ B(βMJ(C; M)) ≥ B(βMJ(C M)) ≥ B(βM) + M2 2 cr(C) + 2M − 1. ↑ ↑

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 17 / 19

SLIDE 59

Main result

Theorem (JP)

Let P ⊂ ST such that JST (P) = M

k=0 βkzk

ST with βM = 0, and C ⊂ S3

adequate. cr(Sat(P, C)) ≥ B(βM) + M2 2 cr(C) + 2M − 1 ≥ cr(C). Proof. cr(Sat(P, C)) ≥ B(J(Sat(P, C))) ≥ B(βMJ(C; M)) ≥ B(βMJ(C M)) ≥ B(βM) + M2 2 cr(C) + 2M − 1. ↑ ↑

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 17 / 19

SLIDE 60

Summary

Problem 1.65 (Kirby, 1995)

Is the crossing number cr(K) of a knot K additive with respect to connected sum, that is, is the equality cr(K1#K2) = cr(K1) + cr(K2) true? Murasugi proved it is true for adequate knots. (Also Kauffman and Thistlethwaite)

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 18 / 19

SLIDE 61

Summary

Problem 1.65 (Kirby, 1995)

Is the crossing number cr(K) of a knot K additive with respect to connected sum, that is, is the equality cr(K1#K2) = cr(K1) + cr(K2) true? Murasugi proved it is true for adequate knots. (Also Kauffman and Thistlethwaite)

Problem 1.67 (Kirby, 1995)

Is the crossing number of a satellite knot bigger than that of its companion? Remarks: “Surely the answer is yes, so the problem indicates the difficulties of proving statements about the crossing number.”

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 18 / 19

SLIDE 62

Summary

Problem 1.65 (Kirby, 1995)

Is the crossing number cr(K) of a knot K additive with respect to connected sum, that is, is the equality cr(K1#K2) = cr(K1) + cr(K2) true? Murasugi proved it is true for adequate knots. (Also Kauffman and Thistlethwaite)

Problem 1.67 (Kirby, 1995)

Is the crossing number of a satellite knot bigger than that of its companion? Remarks: “Surely the answer is yes, so the problem indicates the difficulties of proving statements about the crossing number.” It is true for adequate knots.

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 18 / 19

SLIDE 63

*(Arxiv: 1712.05635)

On adequacy and the crossing number of satellite knots

Adri´ an Jim´ enez Pascual

The University of Tokyo

Tokyo Woman’s Christian University

23th December, 2017

Adri´ an Jim´ enez Pascual (Univ. Tokyo) Adequacy and satellites 23 December, 2017 19 / 19