On homotopy $K3$ surfaces constructed by two knots and their applications

1.6 MB·English

by  Masatsuna Tsuchiya

ON HOMOTOPY K3 SURFACES CONSTRUCTED BY TWO KNOTS
AND THEIR APPLICATIONS
MASATSUNA TSUCHIYA
5
1
0
Abstract. LetLHT bealefthandedtrefoilknotandK beanyknot. WedefineMn(K)
2
to be the homology 3-sphere which is represented by a simple link of LHT and LHT♯K

n withframings0andnrespectively. Startingwiththislink,weconstructhomotopyK3and
a
J spin rational homology K3 surfaces containing Mn(K). Then we apply the adjunction
0 inequality to show that if n > 2gsn(K)− 2, Mn(K) does not bound any smooth spin
2 rational 4-ball, and that under the same assumption the negative n-twisted Whitehead
double of LHT♯K is not a slice knot, where gsn(K) is the n-shake genus of K.
]
T
G
h. 1. Introduction
t
a Let K be a knot in S3. We define X (K) to be the 4-dimensional handlebody which
m n
has a handle decomposition represented by Figure 1.1, and define M (K) to be ∂(X (K)).
[ n n
The boundary M (K) is a homology 3-sphere. Note that the right side knot of this link is
n
1
the connected sum of a left handed trefoil knot LHT and K.
v
2
2
7
4
0
. Figure 1.1. X (K)
1 n
0
5
1 Definition 1.1 (r-shake genus of K). Let N be a 4-dimensional handlebody which is
: K,r
v constructed by attaching a 2-handle to D4 along a knot K with r-framing. We define
i
X the r-shake genus of K to be the minimal genus of smoothly embedded closed oriented
r surfaces in N representing the generator of H (N ). We denote the r-shake genus of
a K,r 2 K,r
K by gr(K).
s
Remark 1.2. Let g (K) be the 4-ball genus of K. Then we have gr(K) ≤ g (K), for any
4 s 4
r ∈ Z.
In this paper, we show the following
Theorem 1.3. If n > 2gn(K)−2, M (K) does not bound any smooth spin rational 4-ball.
s n
If K is an unknot, Theorem 1.3 is closely related to M. Tange’s result (see Remark 1.7),
who uses the Heegaard Floer homology HF+(M (U)) and the correction term d(M (U)),
n n
while we apply the adjunction inequality.
2010 Mathematics Subject Classification. Primary 57R65; Secondary 57M25.
1
2 MASATSUNATSUCHIYA
Corollary 1.4. If n > 2gn(K)−2, the negative n-twisted Whitehead double of LHT♯K is
s
not a slice knot, where LHT♯K is the connected sum of LHT and K.
Remark 1.5. By Corollary 1.4, if LHT♯K is a slice knot, g0(K) is not equal to 0.
s
If τ(K) = gn(K), Corollary1.4 is aspecial case of Hedden’s result (see [6, Theorem 1.5]),
s
where τ(K) is the τ-invariant of K.
We will prove Theorem 1.3 and Corollary 1.4 in Section 2.
Theorem 1.6. Let (a ,a ,a ,a ) be any permutation of (−1,−1,−2k−1,−2m−1), where
1 2 3 4
m and k are integers and k ≥ 0. Then the knot represented by Figure 1.2 is not a slice
knot, where a is the number of full twists (i = 1,2,3,4).
i
Figure 1.2.
We will prove Theorem 1.6 in Section 3.
Remark 1.7. LetU beanunknot. Y. Matsumotoasked in[7, Problem4.28]whether M (U)
0
bounds a contractible 4-manifold or not. By Gordon’s result [5], if n is odd, M (U) does
n
not bound any contractible 4-manifold (cf. [10, §3.1]). If n is equal to −6, N. Maruyama
[9] proved that M (U) bounds a contractible 4-manifold. If n is equal to 0, S. Akbulut
−6
[2] proved that M (U) does not bound any contractible 4-manifold. If n > −2, M. Tange
0
[12] proved that M (U) does not bound any negative definite 4-manifold by computing the
n
Heegaard Floer homology HF+(M (U)) and the correction term d(M (U)).
n n
Acknowledgements. The author would like to thank Tetsuya Abe, Mikio Furuta, Yukio
Matsumoto, Nobuhiro Nakamura and Motoo Tange for their useful comments and encour-
agement.
2. Proof of Theorem 1.3
Notation.
(1) We represent a knot K by Figure 2.1. If K is a right handed trefoil knot RHT,
X (RHT) is represented by Figure 2.2.
n
Figure 2.1. K Figure 2.2. X (RHT)
n
ON HOMOTOPY K3 SURFACES 3
(2) LetD (K,n)(resp.D (K,n))bethenegative(resp.positive) n-twistedWhitehead
− +
doubleof K, andwe represent D (K,n) by Figure2.3. For example, D (LHT,−6)
− −
is represented by Figure 2.4. To simplify the diagram, we usually use Figure 2.5
instead of Figure 2.4.
Figure 2.3. D (K,n) Figure 2.5. D (LHT,−6)
− Figure 2.4. D (LHT,−6) −
−
Let V (K) be the 4-dimensional handlebody represented by Figure 2.7.
n
Proposition2.1. The4-dimensionalhandlebodiesX (K) and V (K)have the samebound-
n n
ary.
Figure 2.6. X (K)
n Figure 2.7. V (K)
n
Proof. We show Proposition 2.1 by the following handle calculus:
Figure 2.8. X
n Figure 2.9.
Figure 2.10. Figure 2.11.
4 MASATSUNATSUCHIYA
Figure 2.13.
Figure 2.12.
Figure 2.14. V (K)
n
(cid:3)
Remark 2.2. By Figure 2.7, we can compute the Casson invariant λ(M (K)) as follows
n
(cf. [13, §2(vii)]):
λ(M (K)) = −n.
n
The Casson invariant, when reduced modulo 2, is the Rohlin invariant:
λ(M (K)) ≡ µ(M (K)) mod 2.
n n
Therefore, if n is odd, M (K) does not bound any smooth spin rational 4-ball.
n
Remark 2.3. If D (LHT♯K,n) is a slice knot, we have a smooth S2 with self intersection
−
−1 in V (K) representing the generator of H (V (K)). We can blow down this S2 to get a
n 2 n
smooth contractible 4-manifold W (K). For example, Casson showed that D (LHT,−6)
n −
is a slice knot. Therefore V (U) can be blown down to a contractible 4-manifold W (U)
−6 −6
represented by Figure 2.16.
Figure 2.16. W (U)
Figure 2.15. V (U) −6
−6
Let Y (K) be the4-dimensional handlebody represented by Figure2.17 with intersection
n
from 2E ⊕2(cid:0)0 1(cid:1)⊕(cid:0)0 1(cid:1)⊕1.
8 1 0 1 n
ON HOMOTOPY K3 SURFACES 5
Figure 2.17. Y (K)
n
Because the +1-framed knot in Figure 2.17 is a slice knot (RHT♯LHT), we can blow
it down. Then we have a smooth simply connected 4-dimensional handlebody Z (K)
n
represented by Figure 2.18 with intersection form 2E ⊕2(cid:0)0 1(cid:1)⊕(cid:0)0 1(cid:1). Note that Y (K)
8 1 0 1 n n
and Z (K) have the same boundary.
n
Figure 2.18. Z (K)
n
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Proposition 2.4. The 4-dimensional handlebodies X (K) and Z (K) have the same
n n
boundary.
Proof. BecauseY (K)andZ (K)havethesameboundary, weshowthatX (K)andY (K)
n n n n
have the same boundary by “S. Akbulut’s blowing up down process”(see [2, Figures 1–4]
and [1, Figures 9–19]) as follows:
Figure 2.19. X
n
Figure 2.20.
Figure 2.21.
Figure 2.22.
Figure 2.23.
Figure 2.24.
Figure 2.25. Figure 2.26.
ON HOMOTOPY K3 SURFACES 7
Figure 2.27. Figure 2.28.
Figure 2.29. Figure 2.30.
Figure 2.31. Figure 2.32.
Figure 2.33.
Figure 2.34.
8 MASATSUNATSUCHIYA
Figure 2.36.
Figure 2.35.
Figure 2.37.
Figure 2.38.
Figure 2.39. Figure 2.40.
ON HOMOTOPY K3 SURFACES 9
Figure 2.41.
Putting k = m = −1 in Figure 2.41, we get Figure 2.17. (cid:3)
Remark 2.5. If M (K) bounds a smooth contractible 4-manifold W (K) and n is even, we
n n
have a smooth homotopy K3 surface Z (K)∪ (−W (K)).
n ∂ n
Morgan and Szab´o proved the following adjunction inequality.
Theorem 2.6 (see [11, Corollary 1.2]). Let X be a smooth closed homotopy K3 surface
and g(x) be the minimal genus of smoothly embedded closed oriented surfaces representing
x, where x ∈ H (X;Z). For every x ∈ H (X;Z),x 6= 0,x·x ≥ 0, we have
2 2
2g(x)−2 ≥ x·x.
The author is informed from Mikio Furuta [4] that the following stronger version can be
proved essentially in the same way as Theorem 2.6.
Theorem 2.6′ ([4]). Let X be a smooth closed spin rational homology K3 surface and
g(x) be the minimal genus of smoothly embedded closed oriented surfaces representing x,
where x ∈ H (X;Z). For every x ∈ H (X;Z),x 6= 0,x·x ≥ 0, we have
2 2
2g(x)−2 ≥ x·x.
We will prove Theorem 1.3.
Proof of Theorem 1.3. Let N be a 4-dimensional handlebody constructed by attaching
K,n
a 2-handle to D4 along the n-framed knot K of Figure 2.18. Let x be a generator of
H (N ;Z). This homology class x is represented by smooth closed oriented surface Σ
2 K,n
with genus gn(K) and x·x = n. The handlebody N is a submanifold of Z (K).
s K,n n
Suppose that n > 2gn(K) − 2, M (K) (= ∂(Z (K))) bounds a smooth spin rational
s n n
4-ball W (K) and n is even, then we have a smooth spin rational homology K3 surface
n
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Z (K) ∪ (−W (K)) in which x persists and is represented by Σ with genus gn(K) and
n ∂ n s
x·x = n. We apply the adjunction inequality to x in Z (K)∪ (−W (K)). Then we get
n ∂ n
2gn(K)−2 ≥ x·x = n.
s
Thiscontradicts theassumption n > 2gn(K)−2. Thereforeifn > 2gn(K)−2andniseven,
s s
M (K) does not bound any smooth spin rational 4-ball. By Remark 2.2, we know that if
n
n is odd, M (K) does not bound any smooth spin rational 4-ball. Therefore we conclude
n
that if n > 2gn(K)−2, M (K) does not bound any smooth spin rational 4-ball. (cid:3)
s n
We will show Corollary 1.4.
Proof of Corollary 1.4. By Remark 2.3, if D (LHT♯K,n) is a slice knot, M (K) bounds a
− n
contractible 4-manifold. By Theorem 1.3, if n > 2gn(K)−2, M (K) does not bound any
s n
smooth spin rational 4-ball. Therefore if n > 2gn(K) − 2, D (LHT♯K,n) is not a slice
s −
(cid:3)
knot.
3. Proof of Theorem 1.6
We show that if (a ,a ,a ,a ) is any permutation of (−1,−1,−2k − 1,−2m − 1), the
1 2 3 4
knot represented by Figure 1.2 is not a slice knot, where m ∈ Z and k ∈ Z .
≥0
Proof of Theorem 1.6. Let Y be the 4-dimensional handlebody represented by Figure 3.3
with intersection form 2E ⊕3(cid:0)0 1(cid:1)⊕1. By the proof of Proposition 2.4, X (U) and Y
8 1 0 −6
have the same boundary.
If k ≥ 0 and the +1-framed knot in Figure 3.3 is a slice knot, then we can blow it down
and get a smooth simply connected 4-dimensional handlebody Z represented by Figure 3.4
with intersection form 2E ⊕ 3(cid:0)0 1(cid:1). Because M (U) bounds a contractible 4-manifold
8 1 0 −6
W (U) (see Remark 2.3), we have a homotopy K3 surface Z∪ (−W (U)). Let x be the
−6 ∂ −6
element of H (Z ∪ (−W (U));Z) which is generated by attaching a 2-handle along the
2 ∂ −6
2k-framed unknot in Figure 3.4. The homology class x is represented by a smooth S2 and
x·x = 2k. By Theorem 2.6, we have the following inequality:
−2 ≥ x·x = 2k.
This contradicts the assumption k ≥ 0. Therefore if k ≥ 0, the +1-framed knot in
Figure 3.3 is not a slice knot. By the handle calculus from Figure 3.5 to Figure 3.6, we
can show that if k ≥ 0, the +1-framed knot in Figure 3.6 is not a slice knot essentially
in the same way as above. Similarly we can prove that if k ≥ 0 and (a ,a ,a ,a ) is any
1 2 3 4
permutation of (−1,−1,−2k − 1,−2m− 1), the knot represented by Figure 1.2 is not a
(cid:3)
slice knot.
Figure 3.1. W−6(U) Figure 3.2. X−6(U)