Intersection form

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(Introduction)
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== Introduction ==
== Introduction ==
<wikitex>;
<wikitex>;
For a closed oriented manifold (topological or smooth) $X$ of dimension $2n$ the intersection form
+
Let $X$ be a closed oriented manifold (PL or smooth) of dimension $2n$.
+
+
Take a triangulation $T$ of $X$.
+
Denote by $T^*$ the [[Wikipedia:Poincare_duality#Dual_cell_structures|dual cell subdivision]].
+
Represent classes $x,y\in H_n(X;\Zz_2)$ by cycles $\overline x$ and $\overline y$ viewed as unions of $n$-simplices of $T$ and of $T^*$, respectively.
+
Define the intersection form modulo 2
+
$$
+
\cap_{X,2}: H_n(X;\mathbb{Z}_2) \times H_n(X;\mathbb{Z}_2) \to \mathbb{Z}_2
+
$$
+
by the formula
+
$$
+
x\cap_{X,2} y = |\overline x\cap\overline y|\mod2.
+
$$
+
This 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 (i.e. counting intersections with signs) one defines the intersection form
+
$$
+
q_X=\cap_X=\cdot_X: H_n(X;\mathbb{Z}) \times H_n(X;\mathbb{Z}) \to \mathbb{Z}.
+
$$
+
Clearly, this form is bilinear.
+
+
Hence $q_X$ vanishes on torsion elements.
+
Thus $q_X$ descends to a bilinear pairing on the free module $H_n(X;\mathbb{Z}) / \text{Torsion}$.
+
Denote the latter pairing also by $q_X$.
+
This pairing is uni-modular (in particular non-degenerate) by [[Wikipedia:Poincare_duality|Poincaré duality]].
+
* If $n$ is even the pairing $q_X$ is symmetric: $q_X(x, y) = q_X(y, x)$.
+
* If $n$ is odd the pairing $q_X$ is skew-symmetric: $q_X(x, y) = - q_X(y, x)$.
+
+
Using the notion of ''cup product'', one can give a dual (and so an equivalent) definition as follows.
+
Define the intersection form
$$
$$
q_X: H^{n}(X;\mathbb{Z}) \times H^{n}(X;\mathbb{Z}) \to \mathbb{Z}
q_X: H^{n}(X;\mathbb{Z}) \times H^{n}(X;\mathbb{Z}) \to \mathbb{Z}
$$
$$
is obtained by the formula
+
by the formula
$$
$$
q_X(x,y) = \langle x \smile y , [X] \rangle ,
q_X(x,y) = \langle x \smile y , [X] \rangle ,
Line 12: Line 41:
i.e. the cup product of $x$ and $y$ is evaluated on the fundamental cycle given by the manifold $X$.
i.e. the cup product of $x$ and $y$ is evaluated on the fundamental cycle given by the manifold $X$.
By bilinearity $q_X$ vanishes on torsion elements, hence the map $q_X$ descends to a bilinear pairing on the free module $H^{n}(X;\mathbb{Z}) / \text{Torsion}$ which we also denote $q_X$. This pairing is uni-modular (in particular non-degenerate) by [[Wikipedia:Poincare_duality|Poincaré duality]].
+
The definition of a cup product is `dual' (and so is analogous) to the above definition of the intersection form on homology of a manifold, but is more abstract.
* If $n$ is even the pairing $q_X$ is symmetric: $q_X(x, y) = q_X(y, x)$.
+
However, the definition of a cup product generalizes to complexes and topological manifolds.
* If $n$ is odd the pairing $q_X$ is skew-symmetric: $q_X(x, y) = - q_X(y, x)$.
+
This is an advantage for mathematicians who are interested in complexes and topological manifolds (not only in PL and smooth manifolds).
</wikitex>
</wikitex>

Revision as of 13:40, 7 March 2019

This page has not been refereed. The information given here might be incomplete or provisional.

Contents

1 Introduction

Let X be a closed oriented manifold (PL or smooth) of dimension 2n.

Take a triangulation T of X. Denote by T^* the dual cell subdivision. Represent classes x,y\in H_n(X;\Zz_2) by cycles \overline x and \overline y viewed as unions of n-simplices of T and of T^*, respectively. Define the intersection form modulo 2

\displaystyle  \cap_{X,2}: H_n(X;\mathbb{Z}_2) \times H_n(X;\mathbb{Z}_2) \to \mathbb{Z}_2

by the formula

\displaystyle  x\cap_{X,2} y = |\overline x\cap\overline y|\mod2.

This 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 (i.e. counting intersections with signs) one defines the intersection form

\displaystyle  q_X=\cap_X=\cdot_X: H_n(X;\mathbb{Z}) \times H_n(X;\mathbb{Z}) \to \mathbb{Z}.

Clearly, this form is bilinear.

Hence q_X vanishes on torsion elements. Thus q_X descends to a bilinear pairing on the free module H_n(X;\mathbb{Z}) / \text{Torsion}. Denote the latter pairing also by q_X. This pairing is uni-modular (in particular non-degenerate) by Poincaré duality.

  • If n is even the pairing q_X is symmetric: q_X(x, y) = q_X(y, x).
  • If n is odd the pairing q_X is skew-symmetric: q_X(x, y) = - q_X(y, x).

Using the notion of cup product, one can give a dual (and so an equivalent) definition as follows. Define the intersection form

\displaystyle  q_X: H^{n}(X;\mathbb{Z}) \times H^{n}(X;\mathbb{Z}) \to \mathbb{Z}

by the formula

\displaystyle  q_X(x,y) = \langle x \smile y , [X] \rangle ,

i.e. the cup product of x and y is evaluated on the fundamental cycle given by the manifold X.

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

2 Uni-modular bilinear forms

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

A form q is called definite if it is positive or negative definite, otherwise it is called indefinite. The rank of q is the rank of the underlying \mathbb{Z}-module V.

3 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 3.1. Every skey-symmetri uni-modular bilinear form over \Zz, q, isomorphic to some 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.

4 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. We begin by stating some fundamental invariants.

Since q is symmetric it is diagonalisable over the real numbers. If b^+ denotes the dimension of a maximal subspace on which the form is positive definite, and if b^- is the dimension of a maximal subspace on which the form is negative definite, then the signature of q is defined to be

\displaystyle  \text{sign}(q) = b^+ - b^-.

The 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.

4.1 Classification of indefinite forms

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

Theorem 4.1 (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 4.2. 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 4.3. The signature of an even (definite or indefinite) form is divisible by 8.

4.2 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.


5 References

6 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:Surgery]]X be a closed oriented manifold (PL or smooth) of dimension 2n.

Take a triangulation T of X. Denote by T^* the dual cell subdivision. Represent classes x,y\in H_n(X;\Zz_2) by cycles \overline x and \overline y viewed as unions of n-simplices of T and of T^*, respectively. Define the intersection form modulo 2

\displaystyle  \cap_{X,2}: H_n(X;\mathbb{Z}_2) \times H_n(X;\mathbb{Z}_2) \to \mathbb{Z}_2

by the formula

\displaystyle  x\cap_{X,2} y = |\overline x\cap\overline y|\mod2.

This 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 (i.e. counting intersections with signs) one defines the intersection form

\displaystyle  q_X=\cap_X=\cdot_X: H_n(X;\mathbb{Z}) \times H_n(X;\mathbb{Z}) \to \mathbb{Z}.

Clearly, this form is bilinear.

Hence q_X vanishes on torsion elements. Thus q_X descends to a bilinear pairing on the free module H_n(X;\mathbb{Z}) / \text{Torsion}. Denote the latter pairing also by q_X. This pairing is uni-modular (in particular non-degenerate) by Poincaré duality.

  • If n is even the pairing q_X is symmetric: q_X(x, y) = q_X(y, x).
  • If n is odd the pairing q_X is skew-symmetric: q_X(x, y) = - q_X(y, x).

Using the notion of cup product, one can give a dual (and so an equivalent) definition as follows. Define the intersection form

\displaystyle  q_X: H^{n}(X;\mathbb{Z}) \times H^{n}(X;\mathbb{Z}) \to \mathbb{Z}

by the formula

\displaystyle  q_X(x,y) = \langle x \smile y , [X] \rangle ,

i.e. the cup product of x and y is evaluated on the fundamental cycle given by the manifold X.

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

2 Uni-modular bilinear forms

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

A form q is called definite if it is positive or negative definite, otherwise it is called indefinite. The rank of q is the rank of the underlying \mathbb{Z}-module V.

3 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 3.1. Every skey-symmetri uni-modular bilinear form over \Zz, q, isomorphic to some 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.

4 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. We begin by stating some fundamental invariants.

Since q is symmetric it is diagonalisable over the real numbers. If b^+ denotes the dimension of a maximal subspace on which the form is positive definite, and if b^- is the dimension of a maximal subspace on which the form is negative definite, then the signature of q is defined to be

\displaystyle  \text{sign}(q) = b^+ - b^-.

The 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.

4.1 Classification of indefinite forms

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

Theorem 4.1 (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 4.2. 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 4.3. The signature of an even (definite or indefinite) form is divisible by 8.

4.2 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.


5 References

6 External links

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