BOUNDARY-SUM IRREDUCIBLE FINITE ORDER CORKS
MOTOO TANGE

Abstract. We prove for any positive integer n there exist boundarysum irreducible Zn -corks with Stein structure. Here ‘boundary-sum irreducible’ means the manifold is indecomposable with respect to boundarysum. We also verify that some of the finite order corks admit hyperbolic
boundary by HIKMOT.

1. Introduction
1.1. G-corks. Cork (C, g) is a pair of a compact contractible (Stein1) 4manifold C and a diﬀeomorphism g on the boundary ∂C that g cannot
extend to the inside C as a smooth diﬀeomorphism. Cork twist means the4-dimensional surgery by the following cut-and-paste
X ′ = (X − C) ∪g C.
The manifold presented by the diagram as in Figure 1 becomes a cork.
The map g is the 180◦ rotation about the horizontal line in the picture. In
particular, C(1) is the first cork which was used by Akbulut. Here a box
with the integer x stands for the x-fold right handed full twist.

Figure 1. The handle diagram of C(m).
In the definition of the original cork the condition g 2 = id∂C is included.
Recently, in some papers the order of gluing map g is generalized to finite
order ([11], [2]), infinite order [7] or generally any group G in [2]. In terms ofDate: October 25, 2017.
1991 Mathematics Subject Classification. 57R55, 57R65.
Key words and phrases. Finite order cork, 4-manifold, boundary-sum irreducible, verification of hyperbolic structure.
The author was partially supported by JSPS KAKENHI Grant Number 17K14180.
1This condition is included in some original papers by Akbulut et.al., for example [1]12

MOTOO TANGE

the view by Auckly, Kim, Melvin, and Ruberman [2], if a group G smoothlyand eﬀectively acts on the boundary of a contractible 4-manifold C and any
non-trivial diﬀeomorphism g ∈ G cannot smoothly extend to the inside C,then the pair (C, G) is called a G-cork.
As examples of finite order corks, the author [11] gave pairs of (Xn,m , τn,m )for X = C, D, E and or generally, X = X(x) for {∗, 0}-sequence x ̸=(0, · · · , 0) or (∗, · · · , ∗), where we call such a sequence x non-trivial. The
diﬀeomorphism τn,m is the 2π/n-rotation with respect to the diagram. InX . We
the paper [11], we put the index X on τn,m , like the notation τn,mremove the indexes if it is understood from the context. We describe Cn,min Figure 2. Dn,m is obtained by exchanging all the dots and 0-framings

Figure 2. The handle decomposition of Cn,m .
in Cn,m . En,m is obtained by modifying Cn,m as in Figure 3. The concrete
diagrams for these examples are described in [11].

Figure 3. The modification.

BOUNDARY-SUM IRREDUCIBLE FINITE ORDER CORKS

Theorem 1.1 ([11]). For X = C, D, E or X(x), for any non-trivial {∗, 0}sequence x, (Xn,m , τn,m ) is a finite order cork. Furthermore, Cn,m is a
Zn -cork with Stein structure.
Auckly, Kim, Melvin, and Ruberman [2] gave the examples of G-corks forany finite subgroup G of SO(4).
Theorem 1.2 ([2]). Let G be any finite subgroup in SO(4). Then there
exists a G-cork.
Let Y1 , Y2 be two n-manifolds with boundary. We call the surgery ofattaching an n-dimensional 1-handle along two neighborhoods of pi ∈ ∂Yiboundary-sum and the resulting manifold as X1 ♮X2 . Their Stein corks in [2]were constructed by the boundary-sum of several copies of C(1). They alsoannounce the existence of finite order cork with hyperbolic boundary in [2].
We say that an n-manifold X with boundary is boundary-sum irreducible
if X = X1 ♮X2 , then X1 or X2 is homeomorphic to an n-disk. If X is notboundary-sum irreducible, then we call X boundary-sum reducible. Here a4-manifold X is called irreducible if for any connected-sum decomposition
X = X1 #X2 , X1 or X2 is a homotopy n-sphere. We call a 3-manifold Yprime if for any connected-sum decomposition Y = Y1 #Y2 , Y1 or Y2 is a3-sphere. The following holds.
Lemma 1.3. Let X be a 4-manifold. If X is irreducible and ∂X is prime,then X is boundary-sum irreducible.
The problem of whether the examples Xn,m in Theorem 1.1 are boundarysum irreducible corks or not has remained. Our main theorem answers thisquestion for the case of X(x)n,m .
Theorem 1.4. For any integer m and positive integer n. There exist
boundary-sum irreducible Zn -corks (Cn,m , τn,m ) with Stein structure. For
any non-trivial {∗, 0}-sequence x, (X(x)n,m , τn,m ) are boundary-sum irreducible finite order corks.
Another variation (En,1 , τn,1 ) is boundary-sum irreducible Zn -corks.
Indeed, X(x)n,m is irreducible and ∂X(x)n,m is prime. This result meansthat (X(x)n,m , τn,m ) is a diﬀerent finite order cork from the one used in
Theorem A in [2]. We do not know whether our examples are diﬀerent fromtheir finite order corks with hyperbolic boundary.
We can show the following result which follows immediately from theproof of Theorem 1.4. We set Yn,m := ∂Cn,m . Clearly this 3-manifold is′:=diﬀeomorphic to any ∂Xn,m (x), for any {∗, 0}-sequence x. We set Yn,m
∂En,m .
′ are prime
Theorem 1.5. Let n, m be integers as above. Then Yn,m and Yn,1
homology spheres.

Furthermore we can prove the following hyperbolicity. In [2] they suggested any Yn,m would be a hyperbolic 3-manifold.

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Theorem 1.6. Let n, m be integers with 0 ≤ m ≤ 2 and 1 ≤ n ≤ 4. Yn,m′and Yn,mare hyperbolic 3-manifolds.
These are direct results by the computer software HIKMOT [5]. It is′proven that Y1,m = Y1,m= ∂C(m) are hyperbolic 3-manifolds in [8], by using
the fact that these are Dehn surgeries of the pretzel knot P r(−3, 3, −3). We
put a question here.
Question 1.7. Let X be X(x) for non-trivial {∗, 0}-sequence or E.
• Is (Xn,m , τn,m ) finite order cork with Stein structure?
• Is (Xn,m , τn,m ) finite order cork with hyperbolic boundary?
Notice that it is not known at all what reflection for the exotic structures
does a cork twist by a cork with hyperbolic boundary give. This theme isleft up to a future study of exotic 4-manifolds.
Acknowledgements
I thank for Akitoshi Takayasu and Hidetoshi Masai for helps of the installation of HIKMOT and some useful comments and some strategies to findhyperbolic solution. I also thank for Kouichi Yasui, Kouki Sato, TakahiroOba, and Robert Gompf for giving me useful advice, comments and suggestion.
′2. Primeness of Kn,m and Kn,m.
′Yn,m and Yn,mare n-fold cyclic branched covers of Y (m) := ∂C(m) with′the branch locus Kn,m and Kn,mrespectively. See Figure 4 for Kn,m . The′picture of Kn,m is obtained by modifying the diagram of Kn,m ⊂ Y (m) inFigure 4 according to Figure 3. In this picture, the slice disks of Kn,m′and Kn,mintersect with the 0-framed 2-handle at 2n points. Let d(K) bethe top degree of the symmetrized Alexander polynomial ∆K (t).

Lemma 2.1. For any integer m and positive integer n, the Alexander poly.
′′
nomials of Kn,m and Kn,mare ∆Kn,m = 2tn − 5 + 2t−n and ∆Kn,m=n−n′6t − 13 + 6t . Furthermore, the genera of Kn,m and Kn,m are n.
Note that the computation in the case of K1,m was done in [8].
Proof. Kn,m has the genus n Seifert surface Σn,m as in Figure 4. We
compute the Seifert matrix for Σn,m . We take the generators {λi , µi |i =1, · · · , n} in H1 (Σn,m ) as in Figure 5.
+We define λ+i and µi to be the parallel transforms in the one side of theneighborhood of Σn,m . Consider the order of the generators as
λ 1 , λ2 , · · · , λn , µ 1 , µ2 , · · · , µn .
The (r, s)-entry of the Seifert matrix Sn,m is the linking number lk(x+r , xs ),where xi is the i-th generator above. Here we have the following:{2 i≤j+++lk(λi , λj ) = 0, lk(µi , µj ) = 0, lk(λi , µj ) =1 i>j

BOUNDARY-SUM IRREDUCIBLE FINITE ORDER CORKS{andlk(µ+i , λj ) =1 i≤j
2 i > j.

These calculations are done
the linking indicated in Figure 6.
( by considering)
On An
The Seifert matrix Sn is, where On is the n × n zero matrix, An
Bn Onand Bn are n × n matrices satisfying the following:{{2 j≥i
1 j≥i
An = (aij ), aij =and Bn = (bij ), bij =1 j < i.
2 j < i.
Then we have∆Kn,m(= det(tSn − SnT ) = det

OntBn − ATn

tAn − BnTOn)= (−1)n det(tAn − BnT ) det(tBn − ATn )= det(tAn − BnT ) det(An − tBnT ).

2t − 2 j > i
We set (αij ) = tAn − BnT , where αij = 2t − 1 i = j We define det(tAn −t − 1 j < i.
TBn ) to be αn . By expanding αn and deforming it, we have
1 2t − 2···· · · 2t − 2
−t 2t − 1 2t − 2 · · · 2t − 2.. ..
.
.  = αn−1 + tβn−1 ,αn = det  0 t − 1 2t − 1
 .
..
..
..
 ..
.
. 2t − 2
.
0 t−1···
t − 1 2t − 1
where βn−1 is the (n − 1) × (n − 1) matrix satisfying the following:
2t − 2 2t − 2···
· · · 2t − 2
 t − 1 2t − 1 2t − 2 · · · 2t − 2 .
.. ..
.
t − 1 2t − 1. βn−1 = det  ..
 .
..
..
..
 ..
.
. 2t − 2
.
t−1 t−1···t − 1 2t − 1
0 2t − 2···
· · · 2t − 2
−t 2t − 1 2t − 2 · · · 2t − 2.. ..
.
.  = tβn−2= det  0 t − 1 2t − 1
 .

..
..
..
 ..
.
. 2t − 2
.
0 t−1···
t − 1 2t − 1()2t − 2 2t − 2β2 = det= 2t(t − 1).
t − 1 2t − 1∑
Thus βn−1 = 2tn−2 (t − 1), therefore, we have αn = 2t2 − 1 + nk=3 2tk−1 (t −1) = 2tn − 1.

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By using the following equality
det(An − tBnT ) = (−t)n det(1/tAn − BnT )we have det(An − tBnT ) = (−t)n (2t−n − 1) = (−1)n (2 − tn ). Therefore, wehave.
∆Kn,m (t) = (2tn − 1)(−1)n (2 − tn ) = 2tn − 5 + 2t−n .
Hence, since d(Kn,m ) coincides with the genus of Σn,m , we can see that thesurface is the minimal Seifert surface. Thus, we have g(Kn,m ) = n.
′In the case of Kn,m, we can do the similar computation to above by taking
the corresponding generators
surface.
( in the ′Seifert)
OAnn , where O is the n × n zero matrix,The Seifert matrix Sn′ isnBn′ OnA′n and Bn′ are n × n matrices satisfying the following:{{−2 j ≥ i
−3 j ≥ i′′′An = (aij ), aij =and Bn = (bij ), bij =−3 j < i.
−2 j < i.
Then we have′∆Kn,m(=T
det(tSn′ − S ′ n ) = det

OntBn′ − A′ TntA′n − B ′ TnOn).
= 6tn − 13 + 6t−n .
2Figure 4. Kn,m in Y (m)Lemma 2.2. Let K1 and K2 be two knots in two homology spheres Y1 , Y2
respectively. Theng(K1 #K2 ) = g(K1 ) + g(K2 )holds.

BOUNDARY-SUM IRREDUCIBLE FINITE ORDER CORKS

Figure 5. Generators of H1 (Σn,m ).

+Figure 6. lk(λ+i , µj ) and lk(λi , µj )This is a classical result, however, we prove it here again.
Proof. Let S ⊂ Y1 #Y2 be the embedded separating sphere for Y1 andY2 . We suppose that S is separating K1 #K2 i.e., (K1 #K2 ) ∩ S are twopoints. Let Σ be the minimal genus Seifert surface of K1 #K2 . The setof the intersection Σ ∩ S consists of finite circles and single arc connecting
the two points in the general position. We take the inner most circle C
not including the arc in the interior. C bounds a disk in Σ because Σ isthe minimal genus surface. Then by cutting the disk and capping two newdisks, we can decrease the number of the intersection circles. We call the new7

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embedded surface Σ again. This cut-and-past process preserves the genusof Σ. The isotopy class of Σ may be changed. By iterating this process wevanish all the intersection circles. Then Σ = Σ1 ♮Σ2 is obtained andg(K1 #K2 ) = g(Σ) = g(Σ1 ) + g(Σ2 ) ≥ g(K1 ) + g(K2 )holds. Conversely, since g(K1 )+g(K2 ) ≥ g(K1 #K2 ), we obtain g(K1 #K2 ) =g(K1 ) + g(K2 ).
2Let K be a knot in a homology sphere. If ∆K (t) cannot be decomposed
as in ∆K (t) = f1 (t)f2 (t) and fi (t) agrees with the Alexander polynomial ofa knot in a homology sphere, then we call ∆K (t) A-irreducible.
Lemma 2.3. Let K be a knot in a homology sphere Y . If ∆K (t) is Airreducible and g(K) = d(K), then K is prime.
Proof. Suppose that K is not prime. Then K is isotopic to a composite
knot K1 #K2 . Then ∆K (t) = ∆K1 (t)∆K2 (t). Hence, d(K) = d(K1 ) + d(K2 )holds. Since g(K) = d(K), we have d(K) = g(K) = g(K1 ) + g(K2 ) ≥d(K1 ) + d(K2 ). Therefore, g(K1 ) + g(K2 ) = d(K1 ) + d(K2 ) holds. From theinequalities g(Ki ) ≥ d(Ki ), g(Ki ) = d(Ki ) holds for i = 1, 2.
On the other hand, since K is A-irreducible, ∆K1 (t) = 1 or ∆K2 (t) = 1.
Thus g(K1 ) = 0 or g(K2 ) = 0 holds. This means that K is prime.
2We prove Kn,m is a prime knot.
′′
Lemma 2.4. Kn,m and Kn,m
are prime knots in Yn,m , Yn,mrespectively.
′Proof. The Alexander polynomials of Kn,m and Kn,m
are 2tn − 5 + 2t−nn−nand 6t − 13 + 2t . These polynomials are A-irreducible. Because, the.
polynomials are completely decomposed as ∆Kn,m = 2tn − 5 + 2t−n = (2tn −.
.
′1)(tn − 2) and ∆Kn,m= 6tn − 13 + 6t−n = (2tn − 3)(3tn − 2) as a polynomial
over Z. Any factor of these decompositions is an irreducible polynomial by
Eisenstein’s criterion and is not an Alexander polynomial of a knot in a′homology sphere. Since the genus of Kn,m and Kn,m
is n, from Lemma 2.3,′Kn,m and Kn,m are prime.
23. Boundary-sum irreducibility of X(x)n,m .
Before proving Theorem 1.4, we prove Lemma 1.3.
Proof. Suppose that X 4 is boundary-sum reducible. Then there exists
a decomposition X = X1 ♮X2 and Xi is not homeomorphic to a 4-disk.
Suppose that either ∂X1 or ∂X2 is diﬀeomorphic to S 3 . We may assume∂X1 ∼= S 3 . Then X is connected-sum X̂1 #X2 , where X̂1 is X1 capped oﬀby a 4-disk D4 and X̂1 is not homeomorphic to S 4 . Then X is irreducible.
Therefore we get the desired assertion.
Here we prove Theorem 1.4.
Proof. For any {∗, 0}-sequence we set X = X(x). The irreducible decomposition of Xn,m is already done and unique. Use Freedman’s classification[3] for the double of X. Thus Xn,m is irreducible.
We prove that Yn,m is a prime 3-manifold. Yn,m is the n-fold cyclic
branched cover of Y (m) along Kn,m . Namely, Yn,m /⟨τ ⟩ = Y (m), whereBOUNDARY-SUM IRREDUCIBLE FINITE ORDER CORKSτ = τn,m . If S ⊂ Yn,m is an embedded 2-sphere, we assume that up toisotopy, S satisfies either of the following conditions for any g ∈ ⟨τ ⟩ due to[9] and [6]:• g(S) ∩ S = ∅• g(S) = S.
Suppose that the first condition is satisfied. S does not intersect withthe branch locus, namely, S is projected to a sphere in Y (m). Since Y (m)is a prime 3-manifold due to [8], then the sphere bounds a 3-ball in Y (m).
Hence, lifting the ball to Yn,m , we can find a 3-ball in Yn,m with the boundaryS.
Suppose that the second condition is satisfied. Then the action restricts
on S. The action is orientation-preserving, because if the action on S isorientation-reversing, then the quotient space has a connected-sum component of L(2, 1). Then in the general position, S transversely intersects withthe branch locus at finite points. By this argument, we can rule out the casewhere the branch locus is included in S.
This means that ⟨τ ⟩ acts on S with the fixed points discrete. The finite
action of the 2-sphere is conjugate to the rotation in SO(3) up to homotopy
due to [10]. In particular, the fixed points are two points. Let S ′ be animage of S into Y (m) and S ′ ∩ Kn,m are two points. Since Y (m) is prime,
S ′ bounds a 3-disk D in Y (m). D ∩ Kn,m is the trivial arc, because Kn,mis prime knot in Y (m). Since the branched cover along the trivial arc is a3-disk, S bounds a 3-disk in Yn,m .
In each case, any embedded sphere in Yn,m bounds a 3-disk. This meansYn,m is prime and it follows that Xn,m is boundary-sum irreducible.
′ is the n-fold branched cover over K′Yn,1n,1 in Y (1) = C(1). This argu′ . This means Y ′ is prime. Thus, for any n, (Ement works for Yn,1n,1 , τn,1 )n,1is boundary-sum irreducible Zn -order cork.
2Here, we put a quick proof of Theorem 1.5.
Proof of Theorem 1.5. It immediately follows from the latter of the proof.
2
For any integer m with m ̸= 1, we do not know whether En,m is boundarysum irreducible or not. We need to prove the primeness of Y ′ (m). The Dehn
surgery diagram of Y ′ (m) is drawn in Figure 7. This manifold is a Dehn
surgery of S13 (P r(−3, 3, −3)).

4. Proof of hyperbolicity.
Finally, we prove Theorem 1.6.
Proof.
The output “True” for the program HIKMOT means that the3-manifold admits hyperbolic structure [4]. To get True-output, we needapply Algorithm 2 in [4]. The data after using the algorithm are updated
in the site [12]. We can get “True” for these four examples by running thedata by HIKMOT.
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isotopy

blow upslidingsliding & canceling

Figure 7. The Dehn surgery diagram of Y ′ (m).
References[1] S. Akbulut and K. Yasui Cork twisting exotic Stein 4-manifolds, J. Diﬀerential Geom.
93 (2013), no. 1, 1–36.
[2] D. Auckly, H. J. Kim, P. Melvin, and D. Ruberman, Equivariant corks, Algebr. Geom.
Topol. 17 (2017), no. 3, 1771–1783.
[3] M. Freedman, The topology of four-dimensional manifolds, J. Diﬀ. Geom. 17 (1982),
357–453[4] N. Hoﬀman, K. Ichihara, M. Kashiwagi, H. Masai, S. Oishi, and A. Takayasu, Verified
computations for hyperbolic 3-manifolds, Exp. Math. 25 (2016), no. 1, 66–78.
[5] HIKMOT, http://www.oishi.info.waseda.ac.jp/~takayasu/hikmot/[6] M. J. Dunwoody, An equivariant sphere theorem, Bull. London Math. Soc. 17 (1985),no. 5, 437–448.
[7] R. Gompf, Infinite order corks, Geom. Topol. 21 (2017), no. 4, 2475–2484.
[8] C. Karakurt, T. Oba, and T. Ukida, Planar Lefschetz fibrations and Stein structureswith distinct Ozsváth-Szabó invariants on corks, Topology Appl. 221 (2017), 630–637.
[9] W. Meeks, L. Simon, and S. T. Yau, Embedded minimal surfaces, exotic spheres, andmanifolds with positive Ricci curvature, Ann. of Math. (2) 116 (1982), no. 3, 621–659.

BOUNDARY-SUM IRREDUCIBLE FINITE ORDER CORKS[10] S. Smale, Diﬀeomorphisms of the 2-Sphere, Proceedings of the American Mathematical Society Vol. 10, No. 4 (Aug., 1959), pp. 621–626[11] M. Tange, Finite order corks. Internat. J. Math. 28 (2017), no. 6, 1750034, 26. pp.
[12] Supplemental data for ”Boundary-sum irreducible finite order cork”. http://www.
math.tsukuba.ac.jp/~tange/program.html
Institute of Mathematics, University of Tsukuba, 1-1-1 Tennodai, Tsukuba,
Ibaraki 305-8571, Japan
E-mail address: tange@math.tsukuba.ac.jp

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