QUADRUPLE PRODUCTS OF 2-DIMENSIONAL ... - surgery.matrix.jpsurgery.matrix.jp/math/4classes/4classes.pdf · QUADRUPLE PRODUCTS OF 2-DIMENSIONAL HOMOLOGY CLASSES OF SIMPLY-CONNECTED

QUADRUPLE PRODUCTS OF 2-DIMENSIONAL HOMOLOGY CLASSES OF

SIMPLY-CONNECTED 4-MANIFOLDS

MASAYUKI YAMASAKI

Abstract. This is a translation of Part C of the author’s master’s thesis [5], with some correc-

tions. Let W be a simply-connected 4-manifold, and let x1, x2, x3, x4 be 2-dimensional homology

classes of W . If xi · x = 0 for every x ∈ H2(W,Z) and i = 1, 2, 3, 4, and all the Matsumoto triple

〈xi, xj , xk〉 vanish for all distinct i, j, k’s, we define a quadruple product 〈x1, x2;x3, x4〉. It is de-

fined in Z/I(x1, x2;x3, x4) where I(x1, x2;x3, x4) is an ideal of Z generated by the sets 〈x1, x3, x〉,〈x1, x4, x〉, 〈x2, x3, x〉, 〈x2, x4, x〉 ( x ∈ H2(W,Z) ). If x1, x2, x3, x4 can be represented by disjoint

immersions, then this quadruple product vanishes. In the appendix, a relation with the second order

non-repeating intersection invariant of Schneider-Teichner [4] is discussed. This version was created

on November 15, 2013.

1. Introduction and the definition of a quadruple product

Let W be a simply-connected oriented smooth 4-manifold. We consider a “separation problem”

of 2-dimensional homology classes of W which asks when a given set of homology classes can be

represented by disjoint immersions of 2-spheres. Such a problem can be extended to a separation

problem of 2-dimensional homotopy classes of arbitrary 4-manifolds, but we restrict ourselves to the

simply-connected case. The following facts were known:

• Two 2-dimensional homology classes x1, x2 can be separated if and only if their intersection

number x1 · x2 is 0. {This is true if we use the non simply-connected intersection number of

Wall.}• Three 2-dimensional homology classes x1, x2, x3 can be separated if and only if their Mat-

sumoto triple 〈x1, x2, x3〉 vanishes modulo an ideal I(x1, x2, x3) of Z ([2][6]). Here, I(x1, x2, x3)

is defined by:

I(x1, x2, x3) = { x1 · y1 + x2 · y2 + x3 · y3 | y1, y2, y3 ∈ H2(W,Z) } .

{This was later extended to the non simply-connected case by R. Schneiderman and P. Te-

ichner [3].}

Now suppose we have four homology classes x1, x2, x3, x4 ∈ H2(W,Z) which satisfy the following

two conditions:

xi · x = 0 for all x ∈ H2(W,Z) and for all i = 1, 2, 3, 4.(1.1)

〈xi, xj , xk〉 = 0 for all i, j, k = 1, 2, 3, 4 which are mutually distinct.(1.2)

Remark 1.1. Here 〈xi, xj , xk〉 is defined as an integer, under the condition (1.1).

Definition 1.2. (1.1) I(x1, x2;x3, x4) is the ideal of Z generated by 〈x1, x3, x〉, 〈x1, x4, x〉, 〈x2, x3, x〉,〈x2, x4, x〉 ( x ∈ H2(W,Z) ). Note that 〈xi, xj , x〉’s are not integers; they are well-defined modulo

1

2 MASAYUKI YAMASAKI

{x · y|y ∈ H2(W,Z)}. So we regard them as ‘sets’, not integers.

(2) I(x1, x2, x3, x4) is the ideal of Z generated by the sets 〈xi, xj , x〉 (i 6= j, x ∈ H2(W,Z)). It includes

the intersection numbers x · y for all x, y ∈ H2(W,Z).

To each quadruple x1, x2, x3, x4 we associate an integer 〈x1, x2;x3, x4〉. We will prove its well-

definedness modulo I(x1, x2;x3, x4) in §2. Once the well-definedness is proved, the following proposi-

tion is obvious from the definition.

Proposition 1.3. If x1, x2, x3, x4 can be represented by disjoint immersed 2-spheres, then

〈xσ(1), xσ(2);xσ(3), xσ(4)〉 = 0

for any permutation σ of the set {1, 2, 3, 4}.

We will also prove the following formulas in §3:

〈x1, x2;x3, x4〉 = 〈x2, x1;x3, x4〉 = 〈x1, x2;x4, x3〉 = 〈x3, x4;x1, x2〉 mod I(x1, x2;x3, x4).(1.3)

〈x1, x2;x3, x4〉+ 〈x1, x3;x2, x4〉+ 〈x1, x4;x2, x3〉 = 0 mod I(x1, x2, x3, x4).(1.4)

Let us start the defintion of 〈x1, x2;x3, x4〉. When we consider 〈x1, x2;x3, x4〉, we do not treat the

four numbers 1, 2, 3, 4 equally. We say that 1 and 2 are companions of each other, and that 3 and 4

are companions of each other. For i = 1, 2, 3, 4, the companion of i is denoted by ı; i.e. 1 = 2, 2 = 1,

3 = 4, 4 = 3.

First we represent x1, x2, x3, x4 by smoothly immersed 2-spheres S1, S2, S3, S4 which are generic

in the sense that all the self- and mutual-intersections of Si’s are transeverse. These transverse

intersection points are called order 0 intersection points. In pictures, order 0 intersection points will

be represented by small circles ◦.Suppose i 6= j. Since the intersection number Si · Sj is 0, the intersection Si ∩ Sj consists of the

same number of positive intersection points {p{i,j}λ }λ and negative intersection points {q{i,j}λ }λ. Draw

smooth embedded arcs γ{i,j}λ,i , γ

{i,j}λ,j connecting p

{i,j}λ and q

{i,j}λ on Si, Sj , respectively. Since W is

simply-connected, the union γ{i,j}λ,i ∪ γ

{i,j}λ,j bounds a smoothly immersed 2-disk with corners in W ,

say ∆{i,j}λ which is transverse to Si and Sj along γ

{i,j}λ,i and γ

{i,j}λ,j . When we consider orientation, we

use notations like γ(i,j)λ,i , γ

(i,j)λ,j , ∆

(i,j)λ , etc. We use the following orientation convention for these:

• γ(i,j)λ,i : the direction from p

{i,j}λ to q

{i,j}λ is positive,

• γ(i,j)λ,j : the direction from q

{i,j}λ to p

{i,j}λ is positive,

• [∆(i,j)λ ] = [the outward normal vector]×[∂∆

(i,j)λ ], where ∂∆

(i,j)λ = γ

(i,j)λ,i ∪γ

(i,j)λ,j and [ ]’s denote

the orientations.

Thus γ(j,i)λ,i , γ

(j,i)λ,j , ∆

(j,i)λ have the opposite orientations of γ

(i,j)λ,i , γ

(i,j)λ,j , ∆

(i,j)λ , respectively. The unions⋃

λ γ{i,j}λ,i ,

⋃λ ∆{i,j}λ ,

⋃λ γ

(i,j)λ,i ,

⋃λ ∆

(i,j)λ will be denoted γ

{i,j}i , ∆{i,j}, γ(i,j)

i , ∆(i,j), respectively. We

assume that all interections are generic; the following intersections are called order 1 intersection

points and will be represented by small black disks • in pictures:

• the intersection points of ∆(∗,∗) and S∗, and

• the intersection points of ∂∆(∗,∗)’s.

QUADRUPLE PRODUCTS OF 2-DIMENSIONAL HOMOLOGY CLASSES 3

Definition 1.4. Suppose 1 ≤ i, j ≤ 4 be distinct integers, then X(j)i denotes the subset Si ∪∆{i,k} ∪

∆{i,l}, where {k, l} = {1, 2, 3, 4} − {i, j}.

Now set (i, j, k, l) to be one of (4, 3, 2, 1), (3, 4, 1, 2), (2, 1, 4, 3), (1, 2, 3, 4) until we finish the defi-

nition of 〈x1, x2;x3, x4〉 in 1.6. The integers i and j are companions, and k and l are companions. By

the assumption 〈xi, xj , xk〉 = 0, we have an equality of integers 1:∑λ

∆(i,j)λ · Sk +

∑µ

∆(j,k)µ · Si +

∑ν

∆(k,i)ν · Sj

+∑ ∂∆

(k,i)ν · ∂∆

(i,j)λ

Si+∑ ∂∆

(i,j)λ · ∂∆

(j,k)µ

Sj+

∑ ∂∆(j,k)µ · ∂∆

(k,i)ν

Sk= 0 ,

whereA ·BC

denotes the intersection number of A and B with respect to the orientation of C.

Let {p(l)α }α be the positive order 1 intersection points and {q(l)

α }α be the negative ones. We can

connect p(l)α and q

(l)α by piecewise smooth arcs γ

(l)α,i on X

(l)i = Si ∪ ∆{i,j} ∪ ∆{k,i}. This is always

possible because any of the intersection points is a point of Si, or it can be connected to a point in Si

by a curve in X(l)i : let a be either p

(l)α or q

(l)α . There are six cases: (i) a ∈ ∆

(i,j)λ ∩Sk, (ii) a ∈ ∆

(j,k)µ ∩Si,

(iii) a ∈ ∆(k,i)ν ∩ Sj , (iv) a ∈ ∂∆

(k,i)ν ∩ ∂∆

(i,j)λ , (v) a ∈ ∂∆

(i,j)λ ∩ ∂∆

(j,k)µ , (vi) a ∈ ∂∆

(j,k)µ ∩ ∂∆

(k,i)ν .

The following are typical local pictures of the arcs γ(l)α,i around a in these six cases:

∆(i,j)λ

Si

Ska

γ(l)α,i

(i)∆

(j,k)µ

Sia

γ(l)α,i

(ii)∆

(k,i)ν

Si

Sja

γ(l)α,i

(iii)

∆(k,i)ν

∆(i,j)λ

Si

aγ(l)α,i

(iv)∆

(i,j)λ

∆(j,k)µ

Sj

a

γ(l)α,i

(v)∆

(j,k)µ

∆(k,i)ν

Sk

a

γ(l)α,i

(vi)

We require that these arcs γ(l)α,i satisfy the following conditions:

• the arcs are in general position with respect to the curves γ{∗,∗}i ,

• in particular, they do not pass through the points p(l)β ’s and q

(l)β ’s except possibly at the end

points,

• when they pass through a self-intersection point of X(l)i , they do not transfer to the other

branch,

• they do not meet Sl, and

• in the case (iv) above, the arc stays in Si near a.

We can also connect p(l)α and q

(l)α by piecewise smooth arcs γ

(l)α,j on X

(l)j = Sj ∪∆{i,j}∪∆{j,k} and γ

(l)α,k

on X(l)k = Sk ∪∆{j,k} ∪∆{k,i} which satisfy similar conditions as above. To define for 〈x1, x2;x3, x4〉,

we use only the arcs γ(l)α,i’s and γ

(l)α,j ’s. (Recall that we are assuming that i and j are companions.)

1This is one of the two places where we essentially use the very strong condition (1.1). The other place is (2.4.2).

4 MASAYUKI YAMASAKI

The orientation convention for these arcs is as follows:

• γ(l)α,i : the direction from p

(l)α to q

(l)α is positive, and

• γ(l)α,j : the direction from q

(l)α to p

(l)α is positive.

The union γ(l)α,i ∪ γ

(l)α,j bounds a disk in W , say ∆

(l)α , with the orientation given by

[∆(l)α ] = [the outward normal vector]× [∂∆(l)

α ] .

Again we assume that all intersections are generic except in certain cases described in Remark 1.7 be-

low. The following intersection points will be called order 2 intersection points and will be represented

by × in pictures:

• the intersection points of ∆(∗) and S∗,

• the intersection points of ∂∆(∗,∗) and ∂∆(∗),

• the intersection points of ∂∆(∗)’s, and

• the intersection points of ∆(∗,∗)’s.

Key ingredients used in the consruction of the ∆(l)’s are summerized in Table 1 below.

∆(1) ∆(2) ∆(3) ∆(4)

∆(4,3) ∆(3,4) ∆(2,1) ∆(1,2)

relevant disks ∆(3,2) ∆(4,1) ∆(1,4) ∆(2,3)

∆(2,4) ∆(1,3) ∆(4,2) ∆(3,1)

the orientation ofthe boundary:

+→ − on X(1)4 X

(2)3 X

(3)2 X

(4)1

− → + on X(1)3 X

(2)4 X

(3)1 X

(4)2

Table 1

Remark 1.5. If l = 1, 2, then ∂∆(l) is in X(l)3 ∪X

(l)4 , and if l = 3, 4, then ∂∆(l) is in X

(l)1 ∪X

(l)2 .

Definition 1.6. The quadruple product 〈x1, x2;x3, x4〉 is defined to be the following sum:

(1.5)

4∑l=1

∆(l) · Sl +

4∑i=1

∑l 6=i,ı

∂∆(l,i) · ∂∆(l)

Si+γ

(4)1 · γ(2)

3

∆(1,3)+γ

(2)4 · γ(3)

1

∆(4,1)+γ

(4)2 · γ(1)

3

∆(2,3)+γ

(1)4 · γ(3)

2

∆(4,2)

+γ

(3)2 · γ(4)

1

∆(1,2)+γ

(3)1 · γ(4)

2

∆(2,1)+γ

(1)4 · γ(2)

3

∆(3,4)+γ

(1)3 · γ(2)

4

∆(4,3)+ ∆(1,3) ·∆(4,2) + ∆(4,1) ·∆(2,3),

where γ(2)3 =

⋃α γ

(2)α,3, and so on.

Thus, to define 〈x1, x2;x3, x4〉, we used ∆(∗,∗)-disks and also ∆(∗)-disks. Their boundary curves

may intersect each other and the configuration is rather complicated. To better understand these

arcs on Si, set j = ı and {k, l} = {1, 2, 3, 4} − {i, j}; so i and j are companions, and k and l are

companions. On Si, there are γ{i,j}i -arcs, γ

{i,k}i -arcs, and γ

{i,l}i -arcs. There are also γ

(k)i -arcs and

γ(l)i -arcs, but there are no γ

(j)i -arcs on Si. Note that the endpoints of a component arc of γ(k) ∩ Si

are in γ{i,j} ∪ γ{i,l} ∪ (∆{j,l} ∩ Si), and that the endpoints of a component arc of γ(l) ∩ Si are in

γ{i,j} ∪ γ{i,k} ∪ (∆{j,k} ∩ Si).


S

i

S

k

S

l

S

j

S

j

S

k

S

l

fi;lg

i

fi;jg

i

fi;kg

i

(k)

i

(l)

i

(k)

i

(l)

i

(k)

i

(l)

i

(l)

i

(k)

i

�

fj;kg

�

fk;lg

�

fj;lg

The following is a typical picture of ∆{i,j}, when i and j are companions:

(k)

i

(l)

i

(k)

j

(l)

j

fi;jg

i

fi;jg

j

�

fi;jg

fi;lg

i

fi;kg

i

fj;lg

j

fj;kg

j

×�

f�;�g

S

i

S

j

S

k

S

l

p q

and the following is a typical picture of ∆{i,j}, when i and l are companions:

γ(k)i

γ(l)i

γ(k)j

γ(l)j

γ{j,l}j γ

{j,k}j

γ{i,l}i γ

{i,k}i

γ{i,j}i

γ{i,j}j

∆{i,j}

×

∆{∗,∗}Si Sj Sk Slp q

Remark 1.7. Assume that i and j are companions. If both p(l)α and q

(l)α are intersection points of

∆(i,j)λ and Sk, then we may assume that the arcs γ

(l)α,i and γ

(l)α,j are the same and that the ‘disk’ ∆

(l)α

degenerates to an arc. Such a disk is said to be degenerate, and only its boundary curves contribute

to (1.6).

2. Well-definedness

We show the well-definedness of 〈x1, x2;x3, x4〉 modulo I(x1, x2;x3, x4). The following is a list of

the choices involved in the definition, listed in the backward order.

(1) the choice of ∆(l)α fixing the boundary

(2) the choice of the boundary ∂∆(l)α

(3) the choice of pairing p(l)α and q

(l)α

(4) the choice of ∆(i,j)λ

(5) the choice of γ(i,j)λ,i

6 MASAYUKI YAMASAKI

(6) the choice of pairing p(i,j)λ and q

(i,j)λ

(7) the choice of Si

2.1. The choice of (1). Let us fix (2)–(7) and choose two disks ∆(l)α and ∆

(l)α . By condition (1.1),

we have (∆(l)α ∪ −∆

(l)α ) · Sl = 0. Therefore ∆

(l)α · Sl = ∆

(l)α · Sl. So this does not change (1.6).

2.2. The choice of (2). Let us fix (3)–(7) and change the boundary ∂∆(l)α . Let us assume that l = 4.

Then the boundary ∂∆(4)α is made up of an arc γ

(4)α,1 on X

(4)1 and an arc γ

(4)α,2 on X

(4)2 .

2.2.1. We first look at the effects of changing the arc γ(4)α,1. Note that this change of arc can be achieved

by a ‘piecewise smooth’ regular homotopy on X(4)1 = S1 ∪∆(1,2) ∪∆(3,1). We break up the homotopy

into pieces with support (a) in S1, (b) in the interior of ∆(1,2), (c) in the interior of ∆(3,1), and (d) in

a small neighborhood of γ(1,2) ∪ γ(3,1), and consider the following typical cases:

2.2.1.a. The homotopy hits an intersection point x of S4 and S1 as in the local picture of S1 below.

(4)

�;1

(4)

�;1

x

S

4

(4;1)

�;1

"

"

0

=)

S

1

We modify ∆(4)α by attaching the shadowed region to it and then making it transverse to S1 by pushing

the interior of the shadowed region off S1. This only affects the terms ∆(4)α ·S4 and (∂∆

(4,1)λ ·∂∆

(4)α )/S1;

we obtain two new intersection points with sign ε and ε′:

(2.1) [∆(4)α ]× [S4] = ε[W ], [γ

(4,1)λ,1 ]× [γ

(4)α,1] = ε′[S1] .

Let δ = ±1 be the sign of the intersection point x. Recall that the orientation of the arc γ(4,1)λ,1 is

defined to be the direction from the negative intersection point to the positive one. Therefore the

orientation of ∆(4)α near x is given by [∆

(4)α ] = −δ[γ(4,1)

λ,1 ]× [γ(4)α,1]; therefore, by (2.1), we obtain

(2.2) −δ[γ(4,1)λ,1 ]× [γ

(4)α,1]× [S4] = ε[W ] .

On the other hand, at the point x, we have [S1]× [S4] = δ[W ]; therefore, by (2.1), we have

(2.3) δ[γ(4,1)λ,4 ]× [γ

(4)α,4]× [S4] = ε′[W ] .

By comparing (2.2) and (2.3), we obtain ε+ε′ = 0. This implies that the following sum is unchanged:

∆(4) · S4 +∂∆(4,1) · ∂∆(4)

S1.

Therefore (1.6) is unchanged.

2.2.1.b. The homotopy hits an intersection point x of S4 and ∆(1,2)λ as in the local picture of ∆

(1,2)λ

below.


(4)

�;1

(4)

�;1

x

S

4

(3)

�;2

(3)

�;1

�

(1;2)

�

"

"

0

=)

We modify ∆(4)α by attaching the shadowed region to it, and this only affects the terms ∆

(4)α · S4 and

(γ(3)β,2 · γ

(4)α,1)/∆

(1,2)λ ; we obtain new intersection points with sign ε and ε′:

(2.4) [∆(4)α ]× [S4] = ε[W ], [γ

(3)β,2]× [γ

(4)α,1] = ε′[∆(1,2)

λ ].

Let δ = ±1 be the sign of the intersection of ∆(2,1)λ and S4 at x; i.e. [∆

(2,1)λ ] × [S4] = δ[W ]. Recall

that the orientation of γ(3)β,2 is defined to be the orientation from the positive intersection point of

∆(2,1)λ and S4. Then the orientation of ∆

(4)α is given by [∆

(4)α ] = δ[γ

(3)β,2] × [γ

(4)α,1]; therefore, by (2.4),

we obtain

δ[γ(3)β,2]× [γ

(4)α,1]× [S4] = ε[W ] , [γ

(3)β,2]× [γ

(4)α,1]× [S4] = −ε′δ[W ] ,

which implies that ε+ ε′ = 0. Therefore (1.6) is unchanged.

2.2.1.c. The homotopy hits an intersection point x of S4 and ∆(1,3)λ as in the local picture of ∆

(1,3)λ

below.

(4)

�;1

(4)

�;1

x

S

4

(2)

�;3

"

"

0

=)

�

(1;3)

�

This only affects the terms ∆(4)α · S4 and (γ

(4)α,1 · γ

(2)β,3)/∆

(1,3)λ ; we obtain new intersection points with

sign ε and ε′:

[∆(4)α ]× [S4] = ε[W ], [γ

(4)α,1]× [γ

(2)β,3] = ε′[∆(1,3)

λ ] .


(1,3)λ ]× [S4] = δ[W ]. Then the

orientation of ∆(4)α is given by [∆

(4)α ] = δ[γ

(2)β,3]× [γ

(4)α,1]; therefore, we obtain

δ[γ(2)β,3]× [γ

(4)α,1]× [S4] = ε[W ] , δ[γ

(2)β,3]× [γ

(4)α,1]× [S4] = −ε′[W ] ,


2.2.1.d-1. The homotopy moves a corner point of γ(4)α,1 along a subarc of γ

(1,2)1 ∪ γ(3,1)

1 . First consider

a move along a subarc of γ(1,2)λ,1 as in the following picture.

8 MASAYUKI YAMASAKI

�

(1;2)

�

�

(4;1)

�

S

1

�

(4)

�

(1;2)

�;1

(4;1)

�;1

(3)

�;2

(3)

�;1

(4)

�;1

x

=)

�

(1;2)

�

�

(4;1)

�

S

1

�

(4)

�

(4;1)

�;1

(3)

�;2

(4)

�;1

y

z

The terms of (1.6) changes only if the subarc intersects some γ(4,1)µ,1 . Suppose x is such an intersection

point. It is an endpoint of some γ(3)β,2. Therefore, after the move, we have a new intersection point

y of γ(4,1)µ,1 and γ

(4)α,1, and a new intersection point z of γ

(3)β,2 and γ

(4)α,1. Recall that ∆(3)-disks were

constructed by setting (i, j, k, l) = (2, 1, 4, 3) in the previous section. So x is considered to be an

intersection point of sign δ = ([γ(2,1)λ,1 ] × [γ

(1,4)µ,1 ])/[S1]. Let w denote a vector at x pointing to the

inward direction of ∆(1,2)λ . Since the direction of γ

(3)β,2 is + → −, we may assume [γ

(3)β,2] = δ[w] near

x, by making γ(3)β,2 normal to γ

(1,2)λ,1 if necessary. On the other hand, we have [∆

(1,2)λ ] = [−w]× [γ

(1,2)λ,1 ]

near x. Suppose [γ(4)α,1] = ε[γ

(2,1)λ,1 ] at y; then, we have [γ

(4)α,1] = −ε[γ(2,1)

λ,1 ] at z. Now we can compute

the signs of intersection at y and z as follows:

[γ(4,1)µ,1 ]× [γ

(4)α,1] = ε[γ

(2,1)λ,1 ]× [γ

(1,4)µ,1 ] = εδ[S1] at y,

[γ(3)β,2]× [γ

(4)α,1] = −εδ[−w]× [γ

(1,2)λ,1 ] = −εδ[∆(1,2)

λ ] at z.

So this move does not change any of the terms of (1.6). The case of a move along a subarc of γ(3,1)1

can be handled in a similar way as above.

2.2.1.d-2. The homotopy moves a part of γ(4)α,1 from S1 to ∆(1,2) ∪∆(3,1), or from ∆(1,2) ∪∆(3,1) to S1.

For example, let us consider the following move of γ(4)α,1, where x is an intersection of γ

(4,1)µ,1 and

γ(1,2)λ,1 :

�

(1;2)

�

�

(4;1)

�

S

1

(1;2)

�;1

(4;1)

�;1

(3)

�;2

(4)

�;1

x

y

=)

�

(1;2)

�

�

(4;1)

�

S

1

(4;1)

�;1

(3)

�;2

(4)

�;1

z

In the d-1 case, new intersections y and z are created, but this replaces an intersection y of γ(4,1)µ,1

and γ(4)α,1 by an intersection z of γ

(4)α,1 and γ

(3)β,2. Also if we set [γ

(4)α,1] = ε[γ

(2,1)λ,1 ] at y; then, we have

[γ(4)α,1] = ε[γ

(2,1)λ,1 ] at z. These are the only differences between d-1 and d-2 cases. Following almost

similar computations, the relevant signs at y and z can be verified to be the same in this case.

Therefore, the value of (1.6) is unchanged.

2.2.2. We next look at the effects of changing the arc γ(4)α,2. This change of arc can be achieved by a

piecewise smooth regular homotopy on X(4)2 = S2 ∪∆(1,2) ∪∆(2,3). We consider the following typical

cases:

2.2.2.a. The homotopy hits an intersection point x of S4 and S2 as in the local picture of S2 below.


(4)

�;2

(4)

�;2

x

S

4

(4;2)

�;2

"

"

0

=)

S

2

This only affects the terms ∆(4)α · S4 and (∂∆

(4,2)λ · ∂∆

(4)α )/S2; we obtain two new intersection points

with sign ε and ε′:

[∆(4)α ]× [S4] = ε[W ], [γ

(4,2)λ,2 ]× [γ

(4)α,1] = ε′[S2] .

Let δ = ±1 be the sign of the intersection point x; i.e. [S2] × [S4] = δ[W ]. Then the orientation of

∆(4)α is given by [∆

(4)α ] = −δ[γ(4,2)

λ,2 ]× [γ(4)α,2]; therefore, we obtain

−δ[γ(4,2)λ,2 ]× [γ

(4)α,2]× [S4] = ε[W ] , δ[γ

(4,2)λ,2 ]× [γ

(4)α,2]× [S4] = ε′[W ] ,

and we obtain ε+ ε′ = 0.

2.2.2.b. The homotopy hits an intersection point x of S4 and ∆(2,1)λ as in the local picture of ∆

(2,1)λ

below.

(4)

�;2

(4)

�;2

x

S

4

(3)

�;1

"

"

0

=)

(3)

�;2

�

(2;1)

�


(3)β,1 · γ

(4)α,2)/∆


sign ε and ε′:

[∆(4)α ]× [S4] = ε[W ], [γ

(3)β,1]× [γ

(4)α,2] = ε′[∆(2,1)

λ ] .


(2,1)λ ]× [S4] = δ[W ]. Then the


(4)α ] = −δ[γ(3)

β,1]× [γ(4)α,2]; therefore, we obtain

δ[γ(3)β,1]× [γ

(4)α,2]× [S4] = −ε[W ] , δ[γ

(3)β,1]× [γ

(4)α,2]× [S4] = ε′[W ] ,


2.2.2.c. The homotopy hits an intersection point x of S4 and ∆(2,3)λ as in the local picture of ∆

(2,3)λ

below.

10 MASAYUKI YAMASAKI

(4)

�;2

(4)

�;2

x

S

4

(1)

�;3

"

"

0

=)

�

(2;3)

�


(4)α,2 · γ

(1)β,3)/∆


sign ε and ε′:

[∆(4)α ]× [S4] = ε[W ], [γ

(4)α,2]× [γ

(1)β,3] = ε′[∆(2,3)

λ ] = −ε′[∆(3,2)λ ] .


(3,2)λ ]× [S4] = δ[W ]. Then the


(4)α ] = −δ[γ(1)

β,3]× [γ(4)α,2]; therefore, we obtain

δ[γ(1)β,3]× [γ

(4)α,2]× [S4] = −ε[W ] , δ[γ

(1)β,3]× [γ

(4)α,2]× [S4] = ε′[W ] ,

which implies that ε+ ε′ = 0. Therefore (1.6) is unchanged,

2.2.2.d. The homotopy changes γ(4)α,2 in a small neighborhood of γ

(1,2)2 ∪ γ(2,3)

2 . The argument is

similar to 2.2.1.d; so we consider only the case of a homotopy which moves a corner point of γ(4)α,2

along a subarc of γ(2,1)λ,2 as in the following picture.

�

(2;1)

�

�

(4;2)

�

S

2

�

(4)

�

(2;1)

�;2

(4;2)

�;2

(3)

�;1

(3)

�;2

(4)

�;2

x

=)

�

(2;1)

�

�

(4;2)

�

S

2

�

(4)

�

(4;2)

�;2

(3)

�;1

(4)

�;2

y

z

Here x is an intersection point of the subarc and γ(4,2)ν,2 of sign δ = ([γ

(4,2)ν,2 ]× [γ

(2,1)λ,2 ])/[S2]. After the

move, we have a new intersection point y of γ(4,2)ν,2 and γ

(4)α,2, and a new intersection point z of γ

(3)β,1

and γ(4)α,2. Let w denote a vector at x pointing to the inward direction of ∆

(2,1)λ . Since the direction

of γ(3)β,1 is − → +, we may assume [γ

(3)β,1] = δ[−w] near x, by making γ

(3)β,1 normal to γ

(2,1)λ,2 if necessary.

On the other hand, we have [∆(2,1)λ ] = [−w]× [γ

(2,1)λ,2 ] near x. Suppose [γ

(4)α,2] = ε[γ

(2,1)λ,2 ] at y; then, we

have [γ(4)α,2] = −ε[γ(2,1)

λ,2 ] at z. Now we can compute the signs of the intersection at y and z as follows:

[γ(4,2)ν,2 ]× [γ

(4)α,2] = [γ

(4,2)ν,2 ]× ε[γ(2,1)

λ,2 ] = δε[S2] at y,

[γ(3)β,1]× [γ

(4)α,2] = −εδ[−w]× [γ

(2,1)λ,2 ] = −εδ[∆(2,1)

λ ] at z.

So this move does not change the value of (1.6).

Now the case l = 4 is done. The cases l = 1, 2, 3 can be treated in the same way as above.2

2In fact there are symmetries (1.3) in the definition of the quadruple product, as will be discussed in §3. Therefore

there is no need to check the l = 1, 2, 3 cases.


2.3. The choice of (3). Let us fix (4)–(7) and change the choice of pairing p(l)∗ and q

(l)∗ . It suffices

to consider the case of changing the pairs {p(l)α , q

(l)α }, {p(l)

β , q(l)β } into the pairs {p(l)

α , q(l)β }, {p

(l)β , q

(l)α }.

We connect the two arcs γ(l)α,i and γ

(l)β,i by a thin band Bi on X

(l)i , and the two arcs γ

(l)α,j and γ

(l)β,j by

a thin band Bj on X(l)j , to obtain arcs connecting the new pairs. Choose a disk ∆

(l)β for the pair

{p(l)β , q

(l)α }, as shown in the picture below, and define a disk ∆

(l)α for the pair {p(l)

α , q(l)β } by attaching

the two bands and the original disks ∆(l)α , ∆

(l)β to ∆

(l)β . This may affect the terms concerning ∆(l),

∂∆(l), γ(l)i , and γ

(l)j . But the newly introduced intersections occur in pairs of opposite signs, so they

cancel each other; therefore (1.6) is unchaged.

q

(l)

�

p

(l)

�

p

(l)

�

(l)

�;i

(l)

�;j

�

(l)

�

q

(l)

�

(l)

�;i

(l)

�;j

�

(l)

�

b

�

(l)

�

B

j

B

i

2.4. The choice of (4). Let us fix (5)–(7) and assume we have two choices ∆{i,j}λ and ∆

{i,j}λ . By

spinning the neighborhood of the boundary of ∆{i,j}λ , we may assume that these two disks are identical

near the boundary [1, p.87]. This modification introduces only intersections of the disk and the spheres

Si, Sj ; so it does not change the sum (1.6). Now there exists an immersed sphere S0 such that ∆{i,j}λ

is regularly homotopic to the connected sum ∆{i,j}λ #S0 fixing a neighbourhood of the boundary. We

will study the effects of such a regular homotopy and of taking connected sum with S0.

2.4.1. First let us consider a modification of ∆{i,j}λ by a regular homotopy fixing a neighbourhood of

the bounday.

As long as the regular homotopy does not hit the boundary of the other surface, their intersection

number does not change. Of course there may be births or deaths of pairs of intersection points

with opposite signs, and there may be births and deaths of ∆(l)-disks (l 6= i, j) as in the picture

below, where {i, j, k, l} = {1, 2, 3, 4}. We may assume that there are no γ(∗,∗)-arcs or γ(∗) in the

neighborhood.

S

k

�

fi;jg

�

=)

S

k

�

fi;jg

�

y

x

If {i, j} = {1, 2} or {3, 4}, then the ∆(l)-disk arising from the intersection points x and y can be

chosen to be degenerate in the sense of Remark 1.6, and it does not contribute to any of the terms

of (1.6). If {i, j} = {1, 3}, {1, 4}, {2, 3} or {2, 4}, then we can use the little Whitney disk as the

∆(l)-disk, and it does not contribute to (1.6), either.


Next let us consider a regular homotopy which hits the boundary. The picture below shows a

typical regular homotopy which pushes a neighborhood of an intersection point x in ∆{i,j}λ off ∆

{k,l}µ .

New intersection points y and z of ∆{i,j}λ and Sk are created.

S

l

S

k

�

fi;jg

�

�

fk;lg

�

=)

S

l

S

k

�

fi;jg

�

�

fk;lg

�

x

y

z

If {i, j, k, l} 6= {1, 2, 3, 4}, then the intersections which appear in the picture do not contribute to

(1.6) and so (1.6) is unchanged. So assume that {i, j, k, l} = {1, 2, 3, 4}.If {i, j} = {1, 2} or {3, 4}, then the intersection point x is not counted in (1.6); we can choose a

degenerate ∆(l)-disk contained in ∆{i,j}λ for the intersection points y and z, and this new disk does

not contribute to (1.6). So (1.6) is unchanged.

Now let us assume that {i, j} = {1, 3}, {1, 4}, {2, 3} or {2, 4}. Then {i, j} = {k, l}; here, we are

using the notation 1 = 2, 2 = 1, 3 = 4, 4 = 3 introduced in §1. In this case, the intersection x in

the picture before modification contributes to (1.6). And, in the picture after modification, we need

to take an arc γ(l)k on Sk and an arc on ∆

{i,j}λ to construct a ∆(l)-disk for y and z. So we have an

intersection point w of ∂∆(l,k)µ and ∂∆(l) as in the picture below. The region painted gray is the

∆(l)-disk for y and z.

S

l

S

k

�

fi;jg

�

�

(l;k)

�

y

z w

v

n

We will show that the intersection points x and w have the same sign in (1.6). Let us consider the case

l = 4 (and hence k = 1 or k = 2). We will check the contribution ε of x in the term ∆(l,k) ·∆(k,l) before

the modification and the contribution ε′ of w in the term (∂∆(l,k) · ∂∆(l))/Sk after the modification:

[∆(l,k)µ ]× [∆(k,l)] = ε[W ] , [∂∆(l,k)

µ ]× [∂∆(l)] = ε′[Sk] ,

and show that ε = ε′. Let v be the vector pointing to the right and n be the vector pointing

downward, as in the picture above, and let u be the vector pointing to the hidden fourth dimension

such that the orientation [∆(k,l)λ ] coincides with [v]×[u] near x before the modification. By convention,

[∆(l,k)µ ] = [n]× [γ

(l,k)µ,k ]. Therefore we have

[n]× [γ(l,k)µ,k ]× [v]× [u] = ε[W ] .


After the modification, on the other hand, we have [∆(k,l)λ ] = −[n]× [u] near the point y. Suppose

the sign of the intersection of ∆(k,l)λ and Sk at y is δ = ±1: [∆

(k,l)λ ] × [Sk] = δ[W ]. When we

determine the orientation of ∆(l), we look at the intersections of ∆(k,l) and Sk if k = 1, and look at

the intersections of ∆(l,k) and Sk if k = 2. This implies that, according to Table 1, the orientation

[γ(l)k ] is equal to −δ[v] near w for both k = 1, 2. Therefore we have

[γ(l,k)µ,k ]× (−δ[v]) = ε′[Sk] .

This, combined with the identity −[n]× [u]× [Sk] = δ[W ], implies the identity

[n]× [γ(l,k)µ,k ]× [v]× [u] = ε′[W ] .

Therefore, we conclude that ε = ε′ and that (1.6) is unchanged. The other cases l = 1, 2, 3 can be

handled similarly.

2.4.2. Now suppose that a disk ∆{i,j}λ is changed to ∆

{i,j}λ = ∆

{i,j}λ #S0, where S0 is an immersed

2-shpere.

We define the orientation of S0 in the following way. If {i, j} = {1, 2}, {3, 4}, then either of the

two orientations of S0 will do. If {i, j} = {1, 3}, {1, 4}, {2, 3} or {2, 4}, then we use the orientation

that comes from the orientaion of ∆(i′,j′)λ , where (i′, j′) = (1, 3), (4, 1), (2, 3), (4, 2), respectively.

These are the disks which appear in the last two terms of (1.6).

Let {k, l} = {1, 2, 3, 4} − {i, j}. Recall that we assumed S0 · Sk = S0 · Sl = 0 3; therefore, the

intersection numbers of ∆(i,j)λ and Sk, Sl do not change and we have new ∆(k)-disks, deonted ∆

(k)new,

and new ∆(l)-disks, denoted ∆(l)new, for the new intersection points.

If {i, j} = {1, 2}, {3, 4}, then note that these disks can be chosen to be degenerate and that the

intersections of ∆{i,j} and ∆{k,l} are not counted in (1.6); therefore, (1.6) is unchanged.

S

0

S

k

S

l

�

fi;jg

�

�

fk;lg

�

(k)

new

�

(l)

new

+

�

+

�

If {i, j} = {1, 3}, {1, 4}, {2, 3} or {2, 4}, let (i′, j′) be as above and (k, l) be (j′, i′). Then the

following are the terms that may change.

• ∆(k) · Sk and ∆(l) · Sl increase by ∆(k)new · Sk and ∆

(l)new · Sl, respectively,

• ∆(k,l) ·∆(i′,j′) increases by ∆(k,l) · S0.

• ∂∆(l,k) · ∂∆(l)

Skand

∂∆(k,l) · ∂∆(k)

Slincrease by

∂∆(l)new · ∂∆(k,l)

Skand

∂∆(k,l) · ∂∆(k)new

Sl, respec-

tively

•γ

(k)i′ · γ

(l)j′

∆(i′,j′)increase by

∂∆(k)new · ∂∆

(l)new

S0.

3This is the second place where we essentially use the very strong condition (1.1).


S

0

S

k

S

l

�

(i

0

;j

0

)

�

�

(k;l)

�

(k)

new

�

(l)

new

+

�

+

�

+

�

We claim that the sum of these changes is equal to (a representative of) the Matsumoto triple

〈xk, xl, x0〉, where x0 ∈ H2(E,Z) is the homology class represented by S0. Actually, we claim that

∆(k)new works as ∆(l,0), and that ∆

(l)new works as ∆(0,k). Let us consider the case {i, j} = {1, 3} (i.e.

{k, l} = {2, 4}). Then (i′, j′) = (1, 3) and (k, l) = (4, 2). To determine the orientation of ∆(4)new, we

use the disk ∆(3,1) = −∆(1,3); therefore the sign of an intersection point of S0 and S2 is the opposite

of the sign of the same intersection of ∆(3,1) and S2. Thus the orientation of ∂∆(4)new is from + to −

on S2 and is from − to + on S0, which coincides with that of ∆(2,0). To determine the orientation of

∆(2)new, we use the disk ∆(1,3); therefore the sign of an intersection point of S0 and S4 is the same as

the sign of the same intersection of ∆(1,3) and S4. Thus the orientation of ∂∆(2)new is from + to − on

S0 and is from − to + on S4, which coincides with that of ∆(0,4). Therefore, the sum above of the

changes in this case can be written as:

∆(4,2) · S0 + ∆(2,0) · S4 + ∆(0,4) · S2 +∂∆(0,4) · ∂∆(4,2)

S4+∂∆(4,2) · ∂∆(2,0)

S2+∂∆(2,0) · ∂∆(0,4)

S0,

which is equal to the definition of the Matsumoto triple 〈x4, x2, x0〉. Therefore, (1.6) is unchanged

modulo I(x1, x2;x3, x4). The other three cases can be handled similarly.

2.5. The choice of (5). Let us fix (6), (7) and change the arc γ{i,j}λ,i on Si.

. . . this part needs a complete rewriting . . .

So we only need to consider three types of isotopies of the arc γ{i,j}λ,i described in the following

pictures:

fi;jg

�;1

fi;jg

�;1

(�)

�;i

(�)

�;i

=)

Type 0.

S

i

fi;jg

�;1

fi;jg

�;1

�

fk;lg

�

\ S

i

=)

Type 1.

S

i


fi;jg

�;1

fi;jg

�;1

fi;�g

�;i

=)

Type 2.

S

i

The disk ∆{i,j}λ is modified first by adding the shadowed region and then making it transverse to Si.

First observe that an isotopy of Type 0 does not change any of the terms in (1.6). Next observe that,

if j = ı, then the isotopies of Type 1 and Type 2 do not change any of the terms in (1.6), either. So

we assume that j 6= ı and consider the isotopies of Type 1 and Type 2. We further assume that i = 1.

The cases i = 2, 3, 4 can be handled similarly.

2.5.1. We consider an isotopy of Type 1. We omit the subscripts λ and µ which identify the individual

arc/disk from the notation.

(1)

f1;jg

1

�

f2;jg

\ S

1

�

f1;jg

(4)

1

S

1

(2)

f1;jg

1

�

f3;4g

\ S

1

�

f1;jg

S

1

(3)

f1;jg

1

�

f2;�|g

\ S

1

�

f1;jg

(j)

1

S

1

v

If (1) {k, l} = {2, j} or (2) {k, l} = {3, 4}, then no terms in (1.6) are changed. So let us consider the

case (3) {k, l} = {2, }. There is a new intersection point x of ∆{1,j} and ∆{2,}, and there is a new

intersection point y of γ{1,j}1 and γ

(j)1 . We check the contributions of these two points.

First let us consider the j = 3 case. Set ε, ε′, and δ as follows:

[∆(1,3)]× [∆(4,2)] = ε[W ] at x, [S1] = ε′[γ(3,1)1 ]× [γ

(3)1 ] at y, [S1]× [∆(4,2)] = δ[W ] at x.

Table 1 says that the direction of γ(3)1 is from the negative intersection of ∆(4,2) and S1 to the positive

one; therefore, we have [γ(3)1 ] = −δ[v], where v is the vector pointing to the right as in the picture

above. So we have

[S1] = ε′[γ(3,1)1 ]× (−δ[v]) = ε′δ[γ(1,3)

1 ]× [v],

and hence

ε′δ[γ(1,3)1 ]× [v]× [∆(4,2)] = δ[W ].

Since [∆(1,3)] = [v]× [γ(1,3)1 ] at y (and also at x), we obtain an identity

−ε′[∆(1,3)]× [∆(4,2)] = [W ] at x,

which implies that ε+ ε′ = 0; therefore, (1.6) is unchanged.

Next, let us consider the j = 4 case. Set ε, ε′, and δ as follows:

[∆(4,1)]× [∆(2,3)] = ε[W ] at x, [S1] = ε′[γ(4,1)1 ]× [γ

(4)1 ] at y, [S1]× [∆(2,3)] = δ[W ] at x.

Table 1, in this case, implies that [γ(4)1 ] = δ[v]. Arguing in a similar manner to the j = 3 case, we can

conclude that ε+ ε′ = 0 and that (1.6) is unchanged.


2.5.2. Now we consider an isotopy of Type 2. First note that the existence of arcs of type γ(∗)∗,1

on S1 does not affect the value of (1.6), because some intersections are just disregarded and those

intersections which are counted appear in pairs of opposite signs. So we omit these from the pictures

in the following discussion, including those arcs which meet γ{1,∗}∗,1 . We consider the following typical

cases:

(1)

f1;jg

1S

2

(�|)

1

(j)

1

S

1

f1;2g

1

(2)

f1;jg

1

S

j

(�|)

1

S

1

f1;jg

1

(3)

f1;jg

1

S

�|

(j)

1

S

1

f1;�|g

1

Here, the intersections of γ{i,j} and γ(k) which are counted in (1.6) are marked by red circles. In

cases (1) and (3), we have new disks ∆() and ∆(2), respectively. In the case (2), no new ∆(l)-disks

are created and (1.6) is unchanged.

Now let us take a closer look at the case (1). We consider the j = 3 case. Since the arc γ(1,2)1 is

embedded in S1, its tubular neighborhood in W is trivial; in fact a framing of the normal bundle of

γ(1,2)1 in W can be constructed as follows: the first vector field along γ

(1,2)1 is the unit normal vector

field of γ(1,2)1 in S1, the second vector field is the unit inward normal vector field of γ

(1,2)1 in ∆(1,2),

and the third vector field is any unit vector field which is normal to the previous two vector fields and

also to γ(1,2)1 . Therefore, we may assume that a neighborhood of γ

(1,2)1 in S1 is a part of a horizontal

plane in R3 and a neighborhood of γ(1,2)1 in ∆(1,2) is also in R3 and is normal to S1 as in the following

picture:

S

1

(3;1)

1

(1;2)

1

S

2

(3)

1

(3)

1

S

2

�

(3;1)

�

(1;2)

�

(4)

new

(4)

new;2

u

u

n

v

S2 is drawn as two line segments; it actually extends to the “time” direction. ∆(3,1) was constructed

by first attaching the “shadowed region”, denoted D, to the original ∆(3,1) and then pushing D in the

downward direction to avoid intersection with ∆(1,2). This new disk ∆(3,1) meets S2 in two points of

the opposite signs, say p and q, which are immediately below the two intersection points of D and S2.

Connect p to a point p′ on γ(1,2)2 which is positioned above p using a straight line segment in S2 ∩R3.

Push the interior of this line segment into the “past” fixing the endpoints p and p′ to obtain an arc

in S2. We obtain a disk ∆(4)new by moving this arc along γ

(1,2)1 and then taking the trace of the move.

This disk is painted in blue in the picture above. The only parts of this disk contained in R3 are the

two horizontal parallel edges, and the rest is in the past. For each intersection of a γ(3)1 -arc and γ

(1,2)1 ,


we obtain an intersection point of γ(3,1)1 and γ

(3)1 , say a, and an intersection point of γ

(3)1 and γ

(4)new,2,

say b, as marked by red circles. We will calculate their contributions to the terms (γ(3,1)1 · γ(3)

1 )/S1

and (γ(3)1 · γ(4)

2 )/∆(2,1) of (1.6)

Let v be a tangent vector of S1 pointing to the right, n be a downward vector, and u be tangent

vector of S1 which points away from γ(1,2)1 . Set δ and δ′ as follows:

[γ(1,2)1 ] = δ[v], [D] = δ′[S1],

where D is given the orientation compatible with that of the original ∆(3,1). Then [∆(3,1)] = δ′[S1]

near the intersection points p and q. Let us recall that theorientation for γ(1,2)1 is from the positive

intersection point of S1 and S2 to the negative one, and that the orientation for γ(4)new,2 is from the

negative intersection point of ∆(3,1) and S2; therefore, we have

[γ(4)new,2] = −δ′[γ(1,2)

1 ] = −δ′δ[v] .

Also note the following:

• There is a common α = ±1 such that [γ(3)1 ] = α[u] (near a), and [γ

(3)1 ] = α[n] (near b).

• [S1] = δ′[D] = δ′[u]× [γ(3,1)1 ] near a.

• [∆(2,1)] = [n]× [γ(2,1)1 ] = [n]× (−[γ

(1,2)1 ]) = −δ[n]× [v] near b.

Now the contribution of the intersection point a is given by:

[γ(3,1)1 ]× α[u]

δ′[u]× [γ(3,1)1 ]

= −α[u]× [γ(3,1)1 ]

δ′[u]× [γ(3,1)1 ]

= −αδ′ ,

and the contribution of the intersection point b is given by:

α[n]× (−δ′δ[v])

−δ[n]× [v]= αδ′ .

Therefore (1.6) is unchanged, when j = 3.

When j = 4, the picture looks like this:

S

1

(1;4)

1

(1;2)

1

S

2

(4)

1

(4)

1

S

2

�

(1;4)

�

(1;2)

�

(3)

new

(3)

new;2

u

u

n

v

To check that (1.6) is unchanged by this move, print out the proof in the j = 3 case, and make

necessary changes using a red pen.

The case of (3) can be handled in the same way. I only show the pictures in the cases j = 3 and

j = 4 below.


S

1

(1;3)

1

(1;4)

1

S

4

(3)

1

(3)

1

S

4

�

(1;3)

�

(1;4)

�

(2)

new

(2)

new;4

u

u

n

v

S

1

(4;1)

1

(1;3)

1

S

3

(4)

1

(4)

1

S

3

�

(4;1)

�

(1;3)

�

(2)

new

(2)

new;3

u

u

n

v

2.6. The choice of (6). Let us fix (7) and change the pairings p{i,j}∗ , q

{i,j}∗ . It suffices to consider

the case of changing the pairs {p(i,j)λ , q

(i,j)λ }, {p(i,j)

µ , q(i,j)µ } into the pairs {p(i,j)

λ , q(i,j)µ }, {p(i,j)

µ , q(i,j)λ }.

We connect the two arcs γ(i,j)λ,i and γ

(i,j)µ,i by a thin band Bi on Si, and the two arcs γ

(i,j)λ,j and γ

(i,j)µ,j

by a thin band Bj on Sj to obtain arcs connecting the new pairs. Here we require that these bands

meet the γ∗,∗ arcs only along the ends so that the new arcs are disjoint embeddings in Si and Sj .

Choose a disk ∆(i,j)µ for the pair {p(i,j)

µ , q(i,j)λ }, as shown in the picture below, and define a disk ∆

(i,j)λ

for the pair {p(i,j)λ , q

(i,j)µ } by attaching the two bands and the original disks ∆

(i,j)λ , ∆

(i,j)µ to ∆

(i,j)µ .

Then slightly modify ∆(i,j)λ so that it is in general position with the disks ∆

(i,j)µ , Si, and Sj .

q

(i;j)

�

p

(i;j)

�

p

(i;j)

�

(i;j)

�;i

(i;j)

�;j

�

(i;j)

�

q

(i;j)

�

(i;j)

�;i

(i;j)

�;j

�

(i;j)

�

b

�

(i;j)

�

B

j

B

i

b

�

(i;j)

�

b

�

(i;j)

�

B

i

B

j

S

i

S

j

Let {k, l} = {1, 2, 3, 4} − {i, j}. Let us check the possible changes in (1.6).

(1) An intersection point of ∆(i,j)µ and Sk is accompanied by an intersection point of ∆

(i,j)λ and Sk

with the opposite sign; and hence we need to consider a new disk ∆(l)new. These two intersection

points can be connected by three arcs as in the picture below; a green arc on Xi, a blue arc


on Xj , and a red arc on Xk. The disk ∆(l)new for these points is bounded by two of the three

arcs, which are selected according to Table 1. Since any two of the three arcs bounds a vary

narrow (and possibly long) strip, we may assume that ∆(l)new misses Sl.

b

�

(i;j)

�

b

�

(i;j)

�

B

i

B

j

S

i

S

j

S

k

(2) When there are intersection points of ∆(i,j)µ and Sl, the same argument as above works.

(3) An intersection point of ∆(i,j)µ and ∆(k,l) is accompanied by an intersection point of ∆

(i,j)λ

and ∆(k,l) with the opposite sign. So either they are not counted (j = ı ) or the contributions

of these points cancel each other (j 6= ı ); so (1.6) is unchanged, anyway.

b

�

(i;j)

�

b

�

(i;j)

�

B

i

B

j

S

i

S

j

�

(k;l)

(4) If Bi meets ∂∆(j) (transversely), then this affects the terms (∂∆(j,i)λ · ∂∆(j))/Si and (∂∆

(j,i)µ ·

∂∆(j))/Si of (1.6), but the changes cancel each other. The intersection of Bj and ∂∆(i) can

be handled in the same way.

Thus (1.6) is unchaged under this operation.

2.7. The choice of (7). Let us change Si’s. Two immersions from S2 to W which represent the

same element of H2(W ) ∼= π2(W ) are regularly homotopic after locally introducing a finite number

of Whitney’s self-intersection points to one of them. Such introduction does not change (1.6); so we

only need to consider changes caused by regular homotopies.

2.7.1. We consider the changes of the configuration of Si and ∆(i). There are three typical cases. The

first case is an introduction of two intersection points of Si and the interior of ∆(i) which obviously

does not change (1.6):

�

(i)

S

i

=)

�

(i)

S

i

y

x

The second case is a finger move of Si across the boundary of ∆(i) so that an intersection point x

is replaced by two new intersection points y and z of Si and Sj (j 6= i, ı):


S

j

or �

(j;k)

S

i

�

(i)

=)

n

v

S

i

�

(i)

x

y

z w

Corresponding to these two new intersection points, we make a very small ∆(i,j)-disk whose interior

has no intersection with Sk or Sl ({k, l} = {1, 2, 3, 4} − {i, j}). We claim that the changes of the

terms ∆(i) · Si and (∆(i,j) ·∆(i))/Sj cancel each other and (1.6) is unchanged. Let ε be the sign of

the intersection point x of ∆(i) and Si and let ε′ be the sign of the intersection point w of the line

segment zy = γ(i,j)j :

[∆(i)]× [Si] = ε[W ] at x, [γ(i,j)j ]× [γ

(i)j ] = ε′[Sj ] at w.

Let n be the downward vector and v be the horizontal vector pointing to the left as in the picture

above. Then [∆(i)] = [n]× [γ(i)j ]. Suppose [v]× [γ

(i)j ] = δ[Sj ]; then the sign of the intersection at z is

given by εδ. Therefore we have [γ(i,j)j ] = εδ[v]. Now

ε′[Sj ] = [γ(i,j)j ]× [γ

(i)j ] = εδ[v]× [γ

(i)j ] = ε[Sj ],

and we obtain ε = ε′.

The third case is a finger move of Si across the boundary of ∆(i) so that an intersection point x of

sign ε is replaced by two new intersection points y and z of Si and ∆(j,k), where j 6= i, ı and k 6= i, j.

See the picture above. In this case we have a new ∆(l)-disk, which is degenerate and contained in

∆(j,k) if k and j are companions, and is a small Whitney disk if k and j are not companions. Anyway

its interior does not meet Sl; so we only have a contribution ε′ coming from the line segment yz = γ(l)j .

As in the second case, we can show that ε = ε′; so (1.6) is unchanged.

2.7.2. We consider the changes of the configuration of Si and ∆(j,k). There are two typical cases. The

first case is an introduction of two intersection points of Si and the interior of ∆(i,j) which obviously

does not change (1.6):

�

(j;k)

S

i

=)

�

(j;k)

S

i

y

x

The second case is a finger move of Si across the boundary of ∆(j,k)λ so that an intersection point

x of sign ε is replaced by two new intersection points y and z of Si and Sj :

S

k

S

j

S

i

�

(j;k)

�

=)

S

k

S

j

S

i

�

(j;k)

�

x

x

0


We need to connect these two intersection points by a curve γ(i,j)µ,j on Sj and by a curve γ

(i,j)µ,i on Sj .

The curve γ(i,j)λ,j should not meet γ

(i,j)µ , so it needs to go around one of the endpoints of γ

(i,j)µ as in the

picture above. The new ∆(i,j)-disk is specified using dashed lines in the picture. It has an intersection

point x′ with Sk. Originally, the intersection point x was paired with another intersection point y

of different sign. They were connected by two arcs and they together bounded a ∆(l)-disk whose

boundary consists of two of the three arcs colored green, blue, and red. Now, after the modification,

we need to connect x′ and y and make a new ∆(l)-disk. This is done by choosing the corresponding two

arcs from the three arcs painted gree, blue, and red on the right, and add the obvious disk bounded

by the two arcs to the original disk. The produces a singular disk; so modify slightly near x to get

an immersion. Since the added part is in a small neighborhood of γ(j,k)λ , its interior does not meet

Sl. Its boundary may have new intersection points with γ(∗)-arcs but they occur in pairs of opposite

signs so they do not contribute to (1.6). Thus (1.6) is unchanged.

This completes the proof of the well-definedness of 〈x1, x2;x3, x4〉.

3. Some Properties

Proposition 3.1. The quadruple product 〈x1, x2;x3, x4〉 has the following symmetry.

(1) 〈x1, x2;x3, x4〉 = 〈x2, x1;x3, x4〉 = 〈x1, x2;x4, x3〉 = 〈x3, x4;x1, x2〉 mod I(x1, x2;x3, x4).

(2) 〈x1, x2;x3, x4〉+ 〈x1, x3;x2, x4〉+ 〈x1, x4;x2, x3〉 = 0 mod I(x1, x2, x3, x4) .

Proof. (1) We prove 〈x1, x2;x3, x4〉 = 〈x2, x1;x3, x4〉. The recipe for computing both sides are sum-

merized in the next table. The left half is Table 1; the right half is obtained from the left half by

switching 1 and 2 in Table 1.

〈x1, x2;x3, x4〉〈x2, x1;x3, x4〉∆(1) ∆(2) ∆(3) ∆(4) ∆(1) ∆(2) ∆(3) ∆(4)

∆(4,3) ∆(3,4) ∆(2,1) ∆(1,2) ∆(3,4) ∆(4,3) ∆(1,2) ∆(2,1)

relevant disks ∆(3,2) ∆(4,1) ∆(1,4) ∆(2,3) ∆(2,3) ∆(1,4) ∆(4,1) ∆(3,2)

∆(2,4) ∆(1,3) ∆(4,2) ∆(3,1) ∆(4,2) ∆(3,1) ∆(2,4) ∆(1,3)

the orientation ofthe boundary:

+→ − on X(1)4 X

(2)3 X

(3)2 X

(4)1 X

(1)3 X

(2)4 X

(3)1 X

(4)2

− → + on X(1)3 X

(2)4 X

(3)1 X

(4)2 X

(1)4 X

(2)3 X

(3)2 X

(4)1

Table 2

Let us use the same ∆{i,j}-disks to compute both sides. The pair of a positive intersection point

p(l)α and a negative intersection point q

(l)α for 〈x1, x2;x3, x4〉 becomes a pair of a negative intersection

point q(l)α and a positive intersection point p

(l)α for 〈x2, x1;x3, x4〉; so we can use the same arcs to join

the pairs and can use the same disk ∆(l)α = ∆

(l)α . Since the recipe for the orientation of the boundary


of ∆(l)α is the opposite, we have identities ∆

(l)α = ∆

(l)α as oriented disks. Therefore, we have

〈x2, x1;x3, x4〉

=

4∑l=1

∆(l) · Sl +

4∑i=1

∑l 6=i,ı

∂∆(l,i) · ∂∆(l)

Si+γ

(4)2 · γ(1)

3

∆(2,3)+γ

(1)4 · γ(3)

2

∆(4,2)+γ

(4)1 · γ(2)

3

∆(1,3)+γ

(2)4 · γ(3)

1

∆(4,1)

+γ

(3)1 · γ(4)

2

∆(2,1)+γ

(3)2 · γ(4)

1

∆(1,2)+γ

(2)4 · γ(1)

3

∆(3,4)+γ

(2)3 · γ(1)

4

∆(4,3)+ ∆(2,3) ·∆(4,1) + ∆(4,2) ·∆(1,3)

=

4∑l=1

∆(l) · Sl +

4∑i=1

∑l 6=i,ı

∂∆(l,i) · ∂∆(l)

Si+γ

(4)1 · γ(2)

3

∆(1,3)+γ

(2)4 · γ(3)

1

∆(4,1)+γ

(4)2 · γ(1)

3

∆(2,3)+γ

(1)4 · γ(3)

2

∆(4,2)

+γ

(3)2 · γ(4)

1

∆(1,2)+γ

(3)1 · γ(4)

2

∆(2,1)+γ

(1)4 · γ(2)

3

∆(3,4)+γ

(1)3 · γ(2)

4

∆(4,3)+ ∆(1,3) ·∆(4,2) + ∆(4,1) ·∆(2,3)

= 〈x1, x2;x3, x4〉 .

The other identities can be checked in the same way.

(2) Let us use the same ∆{∗,∗}-disks to compute the three terms 〈x1, x2;x3, x4〉, 〈x1, x3;x2, x4〉, and

〈x1, x4;x2, x3〉. Table 3 gives the recipe for the computation.

〈x1, x2;x3, x4〉〈x1, x3;x2, x4〉〈x1, x4;x2, x3〉

∆(1) ∆(2) ∆(3) ∆(4) ∆(1) ∆(2) ∆(3) ∆(4) ˜∆(1) ˜

∆(2) ˜∆(3) ˜

∆(4)

∆(4,3) ∆(3,4) ∆(2,1) ∆(1,2) ∆(3,4) ∆(4,3) ∆(1,2) ∆(2,1) ∆(4,3) ∆(3,4) ∆(2,1) ∆(1,2)

∆(3,2) ∆(4,1) ∆(1,4) ∆(2,3) ∆(2,3) ∆(1,4) ∆(4,1) ∆(3,2) ∆(3,2) ∆(4,1) ∆(1,4) ∆(2,3)

∆(2,4) ∆(1,3) ∆(4,2) ∆(3,1) ∆(4,2) ∆(3,1) ∆(2,4) ∆(1,3) ∆(2,4) ∆(1,3) ∆(4,2) ∆(3,1)

X(1)4 X

(2)3 X

(3)2 X

(4)1 X

(1)4 X

(2)3 X

(3)2 X

(4)1 X

(1)3 X

(2)4 X

(3)1 X

(4)2

X(1)3 X

(2)4 X

(3)1 X

(4)2 X

(1)2 X

(2)1 X

(3)4 X

(4)3 X

(1)2 X

(2)1 X

(3)4 X

(4)3

Table 3

Note that the orientation convention for ∆{∗,∗}-disks for 〈x1, x3;x2, x4〉 is the opposite of the

conventions for the other two terms. So a positive intersection points p = p(l) of ∆{∗,∗} and S∗

for 〈x1, x2;x3, x4〉 is also a positive intersection point ˜p = ˜p(l) for 〈x1, x4;x2, x3〉, but is a negative

intersection point q = q(l) for 〈x1, x3;x2, x4〉. Similarly, we have q = p = ˜q.

Let {i, j, k, l} = {1, 2, 3, 4}. For each pair p(l) and q(l), prepare three arcs on X(l)i , X

(l)j , X

(l)k . To

compute the three quadruple products of x1, x2, x3, x4, we choose two arcs from the three arcs and

construct disks bounded by them. The disks used to compute 〈x1, x2;x3, x4〉, 〈x1, x3;x2, x4〉, and

〈x1, x4;x2, x3〉 are denoted ∆(l), ∆(l), and˜∆(l), respectively. Their boundary arcs are denoted by

γ(l)∗ , γ

(l)∗ , and ˜γ

(l)∗ , respectively.

We give orientations to the arcs and the disks according to Table 3. The three disks ∆(l), ∆(l),

and˜∆(l) can be patched together to form an oriented 2-dimensional sphere S2 :


�

(1)

e

�

(1)

e

e

�

(1)

(1)

4

(1)

3

~

~

(1)

3

~

(1)

4

~

(1)

2

~

~

(1)

2

~

~p = ~q = p q = ~p =

~

~q

�

(2)

e

�

(2)

e

e

�

(2)

(2)

3

(2)

4

~

~

(2)

4

~

(2)

3

~

(2)

1

~

~

(2)

1

~

~p = ~q = p q = ~p =

~

~q

�

(3)

e

�

(3)

e

e

�

(3)

(3)

2

(3)

1

~

~

(3)

1

~

(3)

2

~

(3)

4

~

~

(3)

4

~

~p = ~q = p q = ~p =

~

~q

�

(4)

~

�

(4)

~

~

�

(4)

(4)

1

(4)

2

~

~

(4)

2

~

(4)

1

~

(4)

3

~

~

(4)

3

~

~p = ~q = p q = ~p =

~

~q

This means that the sum ∆(l) · Sl + ∆(l) · Sl +˜∆(l) · Sl = S2 · Sl is 0 by the assumption (1.1).

Next let us consider the sum of the intersections on S1:

γ(3,1)1 · γ(3)

1 + γ(4,1)1 · γ(4)

1 + γ(2,1)1 · γ(2)

1 + γ(4,1)1 · γ(4)

1 + γ(2,1)1 · ˜γ(2)

1 + γ(3,1)1 · ˜γ(3)

1 .

This is 0, because we have γ(4)1 = −γ(4)

1 , ˜γ(3)1 = −γ(3)

1 , and ˜γ(2)1 = −γ(2)

1 . The sums of the intersections

on S2, S3, S4 are also 0.

Next let us consider the 24 terms of intersections on ∆{∗,∗}’s. We pic up the 4 terms of intersections

on ∆{2,3}:

γ(4)2 · γ(1)

3

∆(2,3)+γ

(4)3 · γ

(1)2

∆(3,2)+

˜γ(1)3 · ˜γ

(4)2

∆(2,3)+

˜γ(1)2 · ˜γ

(4)3

∆(3,2).

This is 0, because we have ˜γ(4)2 = −γ(4)

2 and ˜γ(1)3 = −γ(1)

3 . The sum of the remaining 20 terms is also

0.

Finally the sum of the following remaining terms is obviously 0:

∆(1,3) ·∆(4,2) + ∆(4,1) ·∆(2,3) + ∆(1,2) ·∆(4,3) + ∆(4,1) ·∆(3,2) + ∆(1,2) ·∆(3,4) + ∆(3,1) ·∆(4,2) ,

and this completes the proof. �

Example 3.2. Consider the following almost trivial link L of 4-components in S3, and let W be the

4-manifold obtained by attaching 0-framed 2-handles to the 4-disk D4 along L. If x1, x2, x3, x4 are

the elements of H2(W,Z) corresponding to the cores of the 2-handles, then we have xi · xj = 0 for all

i, j and 〈xi, xj , xk〉 = 0 for all distinct i, j, k. Some of their quadruple products are non-trivial:

〈x1, x2;x3, x4〉 = 0, 〈x1, x3;x2, x4〉 = 1, 〈x1, x4;x2, x3〉 = −1,

and hence these four elements cannot be represented by disjoint immersions.


1

2

3

4

Appendix A. Higher Order Intersection Invariants of Schneiderman-Teichner

In this appendix, a relation with the second order non-repeating intersection invariant of Schneider-

Teichner [4] is discussed.

Let A = A1, A2, A3, A4 be immersed 2-spheres in a simply-connected 4-manifoldW which represent

homology classes x1, x2, x3, x4, and assume that these homology classes satisfy (1.1) and (1.2). Then

the order 0 non-repeating intersection invariant λ0(A) and the order 1 non-repeating intersection

invariant λ1(A) vanish; and A has a second order non-repeating Whitney towerWA. The second order

non-repeating intersection invariant λ2(WA) is defined to be the sum of signed trees corresponding

to the non-repeating second order intersection points of WA. If we introduce the AS (antisymmetry)

relation, λ2(WA) can be written as a sum:

λ2(WA) = α1t1 + α2t2 + α3t3,

where ti’s are the trees described in the following picture:

t

1

: t

2

: t

3

:

2 1 3 2 4 2

3 4 1 4 3 1

We can calculate the contributions to 〈x1, x2;x3, x4〉, 〈x1, x3;x2, x4〉, 〈x1, x4;x2, x3〉 of an intersection

point corresponding to εti (ε = ±1, i = 1, 2, 3) as in Table 4.

εt1 εt2 εt3

〈x1, x2;x3, x4〉 0 −ε +ε

〈x1, x3;x2, x4〉 −ε +ε 0

〈x1, x4;x2, x3〉 +ε 0 −εTable 4

We pick up three typical examples of an intersection point which contributes εt1 to λ2(A).

The first example is an intersection point p of sign ε of W((1,2),3) and A4. Consult the following

picture:


+�

+

�

A

1

A

2

A

3

A

4

p

W

(1;2)

W

((1;2);3)

We compute the contribution of p to the three quadruple products. Note that W(1,2) = ∆(1,2). Please

consult Table 3.

• The disk ∆(4) for 〈x1, x2;x3, x4〉 is degenerate, because its boundary lies in ∆(1,2), so the

contribution of p is 0.

• The orientation on A3 of the boundary curve of ∆(4) for 〈x1, x3;x2, x4〉 is from the positive

intersection point of A3 and ∆(2,1) (i.e. the negative intersection point of A3 and W(1,2)) to

the other intersection point; therefore ∆(4) = −W((1,2),3), and p contributes −ε.

• The orientation on A3 of the boundary curve of˜∆(4) for 〈x1, x4;x2, x3〉 is from the negative

intersection point of A3 and ∆(1,2) (i.e. the negative intersection point of A3 and W(1,2)) to

the other intersection point; therefore˜∆(4) = W((1,2),3), and p contributes ε.

The second example is an intersection point p of sign −ε of W((4,3),1) and A2. Consult the following

picture:

+�

+

�

A

4

A

3

A

1

A

2

p

W

(4;3)

W

((4;3);1)

We compute the contribution of p to the three quadruple products. Note that W(4,3) = ∆(4,3).

• The disk ∆(2) for 〈x1, x2;x3, x4〉 is degenerate, because its boundary lies in ∆(3,4), so the

contribution of p is 0.

• The orientation on A1 of the boundary curve of ∆(2) for 〈x1, x3;x2, x4〉 is from the negative

intersection point of A1 and ∆(4,3) (i.e. the positive intersection point of A1 and W(4,3)) to

the other intersection point; therefore ∆(2) = W((4,3),1), and p contributes −ε.

• The orientation on A1 of the boundary curve of˜∆(2) for 〈x1, x4;x2, x3〉 is from the negative

intersection point of A1 and ∆(3,4) (i.e. the positive intersection point of A1 and W(4,3)) to

the other intersection point; therefore˜∆(2) = −W((4,3),2), and p contributes ε.


The third example is an intersection point p of W(1,2) = ∆(1,2) and W(3,4) = ∆(3,4) of sign ε. We

can verify Table 4 in this case by just looking at the terms of (1.6) displayed at the end of the proof

of (1.4) in the previous section.

We can continue checking the validity of Table 4 in this manner.

Now, we can calculate the three quadruple products modulo I(x1, x2, x3, x4) as follows:

〈x1, x2;x3, x4〉 ≡ α3 − α2, 〈x1, x3;x2, x4〉 ≡ α2 − α1, 〈x1, x4;x2, x3〉 ≡ α1 − α3.

Next suppose that B = B1, B2, B3, B4 is another set of immersed 2-spheres in W , representing

the same classes x1, x2, x3, x4. Then B has a second order non-repeating Whitney tower WB , and

its second order non-repeating intersection invariant can be written as follows:

λ2(WB) = β1t1 + β2t2 + β3t3 .

Then we should have the following identities:

α3 − α2 ≡ β3 − β2, α2 − α1 ≡ β2 − β1, α1 − α3 ≡ β1 − β3 mod I(x1, x2, x3, x4) .

This means that, if we further introduce the IHX (Jacobi) realtion (t1 + t2 + t3 = 0) and the mod

I(x1, x2, x3, x4) relation in addition to the AS relation introduced earlier, then λ(WA) and λ(WB)

are the same. So the second order non-repeating intersection invariant λ2 of Schneiderman-Teichner

is an invariant of the homology classes modulo the relations above, under the assumptions (1.1) and

(1.2). This is weaker than the conjecture given in §8 of [4].

References

[1] M. Freedman and R. Kirby, A geometric proof of Rochlin’s theorem, Proc. Sympos. Pure Math., vol. 32, Part 2,

Amer. Math. Soc., Providence, R.I., 1978, 85–98.

[2] Y. Matsumoto, Secondary intersectional properties of 4-manifolds and Whitney’s trick, Proc. Sympos. Pure Math.,

vol. 32, Part 2, Amer. Math. Soc., Providence, R.I., 1978, 99–107.

[3] R. Schneiderman and P. Teichner, Higher order intersection numbers of 2-spheres in 4-manifolds, Alg. and

Geom. Topology, 1 (2001), 1–29.

[4] R. Schneiderman and P. Teichner, Pulling apart 2-spheres in 4-manifolds, preprint (2012) arXiv:1210.5534v1

[math.GT]

[5] M. Yamasaki, On quadruple product of simply-connected oriented 4-manifolds, Part C of the author’s master’s

thesis submitted to Dept. of Math., University of Tokyo (March 1978); a pdf file (in Japanese) is available at

http://surgery.matrix.jp/math/4classes/.

[6] M. Yamasaki, Whitney’s trick for three 2-dimensional homology classes of 4-manifolds, Proc. Amer. Math. Soc.,

75 (1979), 365–371.

Department of Applied Science, Okayama University of Science, Okayama, Okayama 700-0005, Japan,

E-mail: [email protected]

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QUADRUPLE PRODUCTS OF 2-DIMENSIONAL ... - surgery.matrix.jpsurgery.matrix.jp/math/4classes/4classes.pdf · QUADRUPLE PRODUCTS OF 2-DIMENSIONAL HOMOLOGY CLASSES OF SIMPLY-CONNECTED