Intersection form

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Contents

1 Introduction

Let N be a closed oriented n-manifold (PL or smooth). After Poincaré one studies the intersection number of transverse submanifolds or chains in N. The intersection number gives a bilinear intersection product

\displaystyle I_N=\cap_N=\cdot_N=\lambda_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to  \Zz

defined on the homology of N. For n=2k this is the intersection form of N denoted by q_N. For n=4k the signature of this form is the signature \sigma(N) of N. The intersection product is closely related to the notions of characteristic classes and linking form. These are important invariants used in the classification of manifolds.

2 Definition

Take a triangulation T of N. Denote by T^* the dual cell subdivision. Represent classes [x]\in H_k(N;\Zz_2) and [y]\in H_{n-k}(N;\Zz_2) by cycles x and y viewed as unions of k-simplices of T and (n-k)-simplices of T^*, respectively. Define the modulo 2 intersection number by the formula

\displaystyle  I_{N,2}(x,y)=x\cap_{N,2} y=\langle\, x \, , \, y\, \rangle := |x\cap y|\mod2\in\Zz_2.

Define the modulo 2 intersection product

\displaystyle  I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2\quad\text{by}\quad  I_{N,2}([x],[y]):=I_{N,2}(x,y).

This product is well-defined because the intersection of a cycle and a boundary consists of an even number of points (by definition of a cycle and a boundary).

Analogously, counting intersections with signs, one defines the intersection number

\displaystyle I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z

of integer chains x\in C_k(T;\Zz) and y\in C_{n-k}(T^*,\Zz). Clearly, the product of a cycle and a boundary is zero. Hence this defines the above intersection product I_N: H_k(N;\Zz)\times H_{n-k}(N;\Zz)\to\Zz.

See [Kirby1989, Chapter II], [Skopenkov2005, Remark 2.3], [Skopenkov2015b, \S6].

Remark 2.1. (a) Using the notion of transversality, one can give an equivalent (and `more general') definition as follows. Take a k-chain x\in C_k(N;\Zz) and an (n-k)-chain y\in C_{n-k}(N;\Zz) which are transverse to each other. The signed count of the intersections between x and y gives the intersection number I_N(x,y). A particular case is intersection number of immersions. Then define the intersection product I_N by I_N([x],[y]):=I_N(x,y).

(b) Using the notion of cup product, one can give a dual (and so an equivalent) definition:

\displaystyle I_N(x,y) = \langle x^*\smile y^*,[N]\rangle \in \Z,

where x^*\in H^{n-k}(N), y^*\in H^n(N) are the Poincaré duals of x, y, and [N] is the fundamental class of the manifold N. We can also define the cup (cohomology intersection) product

\displaystyle  I_N^*: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz \quad\text{by}\quad  I_N^*(p,q) = \langle p \smile q , [N] \rangle .

The definition of a cup product is `dual' (and so is analogous) to the above definition of the intersection product on homology, but is more abstract. However, the definition of a cup product generalizes to complexes (and so to topological manifolds). This is an advantage for mathematicians who are interested in complexes and topological manifolds (not only in PL and smooth manifolds).

3 Simple properties

The following properties are easy to check using the simple direct definition; they also follow from simple properties of the cup product.

The intersection product is bilinear. Hence it vanishes on torsion elements. Thus it descends to a bilinear (integer) intersection pairing

\displaystyle H_k(N;\Zz) / \text{Torsion}\times H_{n-k}(N;\Zz) / \text{Torsion}\to\Zz.

on the free modules.

We have

\displaystyle I_N(x,y) = (-1)^{k(n-k)}I_N(y,x).

Hence for n=2k

  • If k is even the form q_N is symmetric: q_N(x, y) = q_N(y, x).
  • If k is odd the form q_N is skew-symmetric: q_N(x, y) = - q_N(y, x).


4 Poincaré duality

Theorem 4.1.[Poincaré duality] (a) The modulo 2 intersection product is non-degenerate.

(b) The integer intersection pairing is unimodular (in particular non-degenerate)




5 Definition of signature

Let q be a symmetric bilinear form on a free \mathbb{Z}-module. Denote by b^+(q) (b^-(q)) the number of positive (negative) eigenvalues.

Note that since q is symmetric, it is diagonalisable over the real numbers, so b^+(q) (b^-(q)) is the dimension of a maximal subspace on which the form is positive (negative) definite.

Then the signature of q is defined to be

Tex syntax error
If n is divisible by 4, the signature
Tex syntax error
is defined to be the signature of the intersection form of N.

6 Equivalence of bilinear forms

Let q and q' be unimodular bilinear forms on underlying free \mathbb{Z}-modules V and V' respectively. The forms q and q' are called equivalent or isomorphic if there is an isomorphism f:V \to V' such that q = f^* q'.

The rank of q is the rank of the underlying \mathbb{Z}-module V.

7 Skew-symmetric bilinear forms

The skew-symmetric hyperbolic form of rank 2, H_-(\Zz), is defined by the following intersection matrix

\displaystyle  \left( \begin{array}{cc} ~0 & ~1 \\ -1 & ~0 \end{array} \right) .

Proposition 7.1. Every skey-symmetri uni-modular bilinear form over \Zz, q, isomorphic to the sum of some number of hyperbolic forms:

\displaystyle  q \cong \oplus_{i=1}^r H_-(\Zz).

In particular the rank of q, in this case 2r, is even.

8 Symmetric bilinear forms

The classification of uni-modular definite symmetric bilinear forms is a deep and difficult problem. However the situation becomes much easier when the form is indefinite. Fundamental invariants are rank, signature and the following two.

A form q is called definite if it is positive or negative definite, otherwise it is called indefinite.

A form q may have two different types. It is of type even if q(x,x) is an even number for any element x. Equivalently, if q is written as a square matrix in a basis, it is even if the elements on the diagonal are all even. Otherwise, q is said of type odd.

8.1 Classification of indefinite forms

There is a simple classification result of indefinite forms [Serre1970],[Milnor&Husemoller1973]:

Theorem 8.1 (Serre (?)). Two indefinite unimodular symmetric bilinear forms q, q' over \mathbb{Z} are equivalent if and only if q and q' have the same rank, signature and type.

There is a further invariant of a unimodular symmetric bilinear form q on V: An element c \in V is called a characteristic vector of the form if one has

\displaystyle  q(c,x) \equiv q(x,x) \ (\text{mod} \ 2)

for all elements x \in V. Characteristic vectors always exist. In fact, when reduced modulo 2, the map x \mapsto q(x,x) \in \mathbb{Z}/2 is linear. By unimodularity there therefore exists an element c such that the map q(c,-) equals this linear map.

The form q is even if and only if 0 is a characteristic vector. If c and c' are characteristic vectors for q, then there is an element h with c' = c + 2h. This follows from unimodularity. As a consequence, the number q(c,c) is independent of the chosen characteristic vector c modulo 8. One can be more specific:

Proposition 8.2. For a characteristic vector c of the unimodular symmetric bilinear form q one has

\displaystyle  q(c,c) \equiv \text{sign}(q) \ (\text{mod} \ 8)

Proof: Suppose c is a characteristic vector of q. Then c + e_+ + e_- is a characteristic vector of the form

\displaystyle  q' = q \oplus \begin{pmatrix} 1  & \ 0 \\ 0 & -1 \end{pmatrix},

where e_+, e_- form basis elements of the additional \mathbb{Z}^2 summand with square \pm 1. We notice that

\displaystyle  q(c,c) = q'(c+e_+ + e_-, c+e_+ + e_-) .

However, the form q' is indefinite, so the above classification theorem applies. In particular, q' is odd and has the same signature as q, so it is equivalent to the diagonal form with b^+ + 1 summands of (+1) and b^- + 1 summands of (-1). This diagonal form has a characteristic vector c' that is simply a sum of basis elements in which the form is diagonal. Of course q'(c',c') = b^+ - b^-. The claim now follows from the fact that the square of a characteristic vector is independent of the chosen characteristic vector modulo 8.

Corollary 8.3. The signature of an even (definite or indefinite) form is divisible by 8.

8.2 Examples, Realisations of indefinite forms

We shall show that any indefinite form permitted by the above theorem and corollary can be realised.

All possible values of rank and signature of odd forms are realised by direct sums of the forms of rank 1,

\displaystyle  b^+ (+1) \oplus b^- (-1) .

An even positive definite form of rank 8 is given by the E_8 matrix

\displaystyle  E_8 = \left( \begin{array}{c c c c c c c c} \ 2 \ & \ 1\  & \ 0\  & \ 0\ & \ 0\ & \ 0\ & \ 0\ & \ 0\ \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 2 \\ \end{array}  \right) .

Likewise, the matrix -E_8 represents a negative definite even form of rank 8.

On the other hand, the matrix H given by

\displaystyle  H = \begin{pmatrix} \ 0 \ & \ 1 \ \\ 1 & 0  \end{pmatrix}

determines an indefinite even form of rank 2 and signature 0. It is easy to see that the direct sums

\displaystyle  k E_8 \oplus l H

with k \in \mathbb{Z}, l \in \mathbb{N}=\{1,2,\dots\} realise all unimodular symmetric indefinite even forms that are allowed by the above classification result. Here we use the convention that k E_8 is the k-fold direct sum of E_8 for positive k and (-k) (-E_8) is the |k|-fold direct sum of the negative definite form -E_8.


9 References

10 External links

$ is a characteristic vector. If $c$ and $c'$ are characteristic vectors for $q$, then there is an element $h$ with $c' = c + 2h$. This follows from unimodularity. As a consequence, the number $q(c,c)$ is independent of the chosen characteristic vector $c$ modulo 8. One can be more specific: {{beginthm|Proposition|}} For a characteristic vector $c$ of the unimodular symmetric bilinear form $q$ one has $$ q(c,c) \equiv \text{sign}(q) \ (\text{mod} \ 8) $$ {{endthm}} Proof: Suppose $c$ is a characteristic vector of $q$. Then $c + e_+ + e_-$ is a characteristic vector of the form $$ q' = q \oplus \begin{pmatrix} 1 & \ 0 \ 0 & -1 \end{pmatrix}, $$ where $e_+, e_-$ form basis elements of the additional $\mathbb{Z}^2$ summand with square $\pm 1$. We notice that $$ q(c,c) = q'(c+e_+ + e_-, c+e_+ + e_-) . $$ However, the form $q'$ is indefinite, so the above classification theorem applies. In particular, $q'$ is odd and has the same signature as $q$, so it is equivalent to the diagonal form with $b^+ + 1$ summands of (+1) and $b^- + 1$ summands of $(-1)$. This diagonal form has a characteristic vector $c'$ that is simply a sum of basis elements in which the form is diagonal. Of course $q'(c',c') = b^+ - b^-$. The claim now follows from the fact that the square of a characteristic vector is independent of the chosen characteristic vector modulo 8. {{beginthm|Corollary|}} The signature of an even (definite or indefinite) form is divisible by 8. {{endthm}} === Examples, Realisations of indefinite forms === ; We shall show that any indefinite form permitted by the above theorem and corollary can be realised. All possible values of rank and signature of ''odd'' forms are realised by direct sums of the forms of rank 1, $$ b^+ (+1) \oplus b^- (-1) . $$ An even positive definite form of rank 8 is given by the $E_8$ matrix $$ E_8 = \left( \begin{array}{c c c c c c c c} \ 2 \ & \ 1\ & \ 0\ & \ 0\ & \ 0\ & \ 0\ & \ 0\ & \ 0\ \ 1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 2 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 2 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 2 & 1 & 0 & 1 \ 0 & 0 & 0 & 0 & 1 & 2 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 2 \ \end{array} \right) . $$ Likewise, the matrix $-E_8$ represents a negative definite even form of rank 8. On the other hand, the matrix $H$ given by $$ H = \begin{pmatrix} \ 0 \ & \ 1 \ \ 1 & 0 \end{pmatrix} $$ determines an indefinite even form of rank 2 and signature 0. It is easy to see that the direct sums $$ k E_8 \oplus l H $$ with $k \in \mathbb{Z}, l \in \mathbb{N}=\{1,2,\dots\}$ realise all unimodular symmetric indefinite even forms that are allowed by the above classification result. Here we use the convention that $k E_8$ is the $k$-fold direct sum of $E_8$ for positive $k$ and $(-k) (-E_8)$ is the $|k|$-fold direct sum of the negative definite form $-E_8$. == References == {{#RefList:}} == External links == * The Wikipedia page on [[Wikipedia:Poincare duality#Bilinear_pairings_formulation|Poincaré duality]] [[Category:Theory]] [[Category:Definitions]] [[Category:Forgotten in Textbooks]] [[Category:Surgery]]N be a closed oriented n-manifold (PL or smooth). After Poincaré one studies the intersection number of transverse submanifolds or chains in N. The intersection number gives a bilinear intersection product

\displaystyle I_N=\cap_N=\cdot_N=\lambda_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to  \Zz

defined on the homology of N. For n=2k this is the intersection form of N denoted by q_N. For n=4k the signature of this form is the signature \sigma(N) of N. The intersection product is closely related to the notions of characteristic classes and linking form. These are important invariants used in the classification of manifolds.

2 Definition

Take a triangulation T of N. Denote by T^* the dual cell subdivision. Represent classes [x]\in H_k(N;\Zz_2) and [y]\in H_{n-k}(N;\Zz_2) by cycles x and y viewed as unions of k-simplices of T and (n-k)-simplices of T^*, respectively. Define the modulo 2 intersection number by the formula

\displaystyle  I_{N,2}(x,y)=x\cap_{N,2} y=\langle\, x \, , \, y\, \rangle := |x\cap y|\mod2\in\Zz_2.

Define the modulo 2 intersection product

\displaystyle  I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2\quad\text{by}\quad  I_{N,2}([x],[y]):=I_{N,2}(x,y).

This product is well-defined because the intersection of a cycle and a boundary consists of an even number of points (by definition of a cycle and a boundary).

Analogously, counting intersections with signs, one defines the intersection number

\displaystyle I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z

of integer chains x\in C_k(T;\Zz) and y\in C_{n-k}(T^*,\Zz). Clearly, the product of a cycle and a boundary is zero. Hence this defines the above intersection product I_N: H_k(N;\Zz)\times H_{n-k}(N;\Zz)\to\Zz.

See [Kirby1989, Chapter II], [Skopenkov2005, Remark 2.3], [Skopenkov2015b, \S6].

Remark 2.1. (a) Using the notion of transversality, one can give an equivalent (and `more general') definition as follows. Take a k-chain x\in C_k(N;\Zz) and an (n-k)-chain y\in C_{n-k}(N;\Zz) which are transverse to each other. The signed count of the intersections between x and y gives the intersection number I_N(x,y). A particular case is intersection number of immersions. Then define the intersection product I_N by I_N([x],[y]):=I_N(x,y).

(b) Using the notion of cup product, one can give a dual (and so an equivalent) definition:

\displaystyle I_N(x,y) = \langle x^*\smile y^*,[N]\rangle \in \Z,

where x^*\in H^{n-k}(N), y^*\in H^n(N) are the Poincaré duals of x, y, and [N] is the fundamental class of the manifold N. We can also define the cup (cohomology intersection) product

\displaystyle  I_N^*: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz \quad\text{by}\quad  I_N^*(p,q) = \langle p \smile q , [N] \rangle .

The definition of a cup product is `dual' (and so is analogous) to the above definition of the intersection product on homology, but is more abstract. However, the definition of a cup product generalizes to complexes (and so to topological manifolds). This is an advantage for mathematicians who are interested in complexes and topological manifolds (not only in PL and smooth manifolds).

3 Simple properties

The following properties are easy to check using the simple direct definition; they also follow from simple properties of the cup product.

The intersection product is bilinear. Hence it vanishes on torsion elements. Thus it descends to a bilinear (integer) intersection pairing

\displaystyle H_k(N;\Zz) / \text{Torsion}\times H_{n-k}(N;\Zz) / \text{Torsion}\to\Zz.

on the free modules.

We have

\displaystyle I_N(x,y) = (-1)^{k(n-k)}I_N(y,x).

Hence for n=2k

  • If k is even the form q_N is symmetric: q_N(x, y) = q_N(y, x).
  • If k is odd the form q_N is skew-symmetric: q_N(x, y) = - q_N(y, x).


4 Poincaré duality

Theorem 4.1.[Poincaré duality] (a) The modulo 2 intersection product is non-degenerate.

(b) The integer intersection pairing is unimodular (in particular non-degenerate)




5 Definition of signature

Let q be a symmetric bilinear form on a free \mathbb{Z}-module. Denote by b^+(q) (b^-(q)) the number of positive (negative) eigenvalues.

Note that since q is symmetric, it is diagonalisable over the real numbers, so b^+(q) (b^-(q)) is the dimension of a maximal subspace on which the form is positive (negative) definite.

Then the signature of q is defined to be

Tex syntax error
If n is divisible by 4, the signature
Tex syntax error
is defined to be the signature of the intersection form of N.

6 Equivalence of bilinear forms

Let q and q' be unimodular bilinear forms on underlying free \mathbb{Z}-modules V and V' respectively. The forms q and q' are called equivalent or isomorphic if there is an isomorphism f:V \to V' such that q = f^* q'.

The rank of q is the rank of the underlying \mathbb{Z}-module V.

7 Skew-symmetric bilinear forms

The skew-symmetric hyperbolic form of rank 2, H_-(\Zz), is defined by the following intersection matrix

\displaystyle  \left( \begin{array}{cc} ~0 & ~1 \\ -1 & ~0 \end{array} \right) .

Proposition 7.1. Every skey-symmetri uni-modular bilinear form over \Zz, q, isomorphic to the sum of some number of hyperbolic forms:

\displaystyle  q \cong \oplus_{i=1}^r H_-(\Zz).

In particular the rank of q, in this case 2r, is even.

8 Symmetric bilinear forms

The classification of uni-modular definite symmetric bilinear forms is a deep and difficult problem. However the situation becomes much easier when the form is indefinite. Fundamental invariants are rank, signature and the following two.

A form q is called definite if it is positive or negative definite, otherwise it is called indefinite.

A form q may have two different types. It is of type even if q(x,x) is an even number for any element x. Equivalently, if q is written as a square matrix in a basis, it is even if the elements on the diagonal are all even. Otherwise, q is said of type odd.

8.1 Classification of indefinite forms

There is a simple classification result of indefinite forms [Serre1970],[Milnor&Husemoller1973]:

Theorem 8.1 (Serre (?)). Two indefinite unimodular symmetric bilinear forms q, q' over \mathbb{Z} are equivalent if and only if q and q' have the same rank, signature and type.

There is a further invariant of a unimodular symmetric bilinear form q on V: An element c \in V is called a characteristic vector of the form if one has

\displaystyle  q(c,x) \equiv q(x,x) \ (\text{mod} \ 2)

for all elements x \in V. Characteristic vectors always exist. In fact, when reduced modulo 2, the map x \mapsto q(x,x) \in \mathbb{Z}/2 is linear. By unimodularity there therefore exists an element c such that the map q(c,-) equals this linear map.

The form q is even if and only if 0 is a characteristic vector. If c and c' are characteristic vectors for q, then there is an element h with c' = c + 2h. This follows from unimodularity. As a consequence, the number q(c,c) is independent of the chosen characteristic vector c modulo 8. One can be more specific:

Proposition 8.2. For a characteristic vector c of the unimodular symmetric bilinear form q one has

\displaystyle  q(c,c) \equiv \text{sign}(q) \ (\text{mod} \ 8)

Proof: Suppose c is a characteristic vector of q. Then c + e_+ + e_- is a characteristic vector of the form

\displaystyle  q' = q \oplus \begin{pmatrix} 1  & \ 0 \\ 0 & -1 \end{pmatrix},

where e_+, e_- form basis elements of the additional \mathbb{Z}^2 summand with square \pm 1. We notice that

\displaystyle  q(c,c) = q'(c+e_+ + e_-, c+e_+ + e_-) .

However, the form q' is indefinite, so the above classification theorem applies. In particular, q' is odd and has the same signature as q, so it is equivalent to the diagonal form with b^+ + 1 summands of (+1) and b^- + 1 summands of (-1). This diagonal form has a characteristic vector c' that is simply a sum of basis elements in which the form is diagonal. Of course q'(c',c') = b^+ - b^-. The claim now follows from the fact that the square of a characteristic vector is independent of the chosen characteristic vector modulo 8.

Corollary 8.3. The signature of an even (definite or indefinite) form is divisible by 8.

8.2 Examples, Realisations of indefinite forms

We shall show that any indefinite form permitted by the above theorem and corollary can be realised.

All possible values of rank and signature of odd forms are realised by direct sums of the forms of rank 1,

\displaystyle  b^+ (+1) \oplus b^- (-1) .

An even positive definite form of rank 8 is given by the E_8 matrix

\displaystyle  E_8 = \left( \begin{array}{c c c c c c c c} \ 2 \ & \ 1\  & \ 0\  & \ 0\ & \ 0\ & \ 0\ & \ 0\ & \ 0\ \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 2 \\ \end{array}  \right) .

Likewise, the matrix -E_8 represents a negative definite even form of rank 8.

On the other hand, the matrix H given by

\displaystyle  H = \begin{pmatrix} \ 0 \ & \ 1 \ \\ 1 & 0  \end{pmatrix}

determines an indefinite even form of rank 2 and signature 0. It is easy to see that the direct sums

\displaystyle  k E_8 \oplus l H

with k \in \mathbb{Z}, l \in \mathbb{N}=\{1,2,\dots\} realise all unimodular symmetric indefinite even forms that are allowed by the above classification result. Here we use the convention that k E_8 is the k-fold direct sum of E_8 for positive k and (-k) (-E_8) is the |k|-fold direct sum of the negative definite form -E_8.


9 References

10 External links

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