Intersection form

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	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''.		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''.
−
−
−	{{beginthm|Proposition|}} If the unimodular symmetric bilinear form $q$ is even then its signature is divisible by 8.
−	{{endthm}}
−	Proof:
	</wikitex>		</wikitex>
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	{{beginthm|Theorem|(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. {{endthm}}		{{beginthm|Theorem|(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. {{endthm}}
−	</wikitex>
−	== Examples ==	+
		+	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
		+	$$
		+	q(c,x) \equiv q(x,x) \ (\text{mod} \ 2)
		+	$$
		+	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:
		+
		+	{{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:
		+
		+	</wikitex>
		+	== Examples, Realisations of indefinite forms ==
	<wikitex>;		<wikitex>;

---

Revision as of 16:15, 7 June 2010

Contents 1 Introduction 2 Algebraic invariants 3 Classification of indefinite forms 4 Examples, Realisations of indefinite forms 5 References

1 Introduction

For a closed (topological or smooth) manifold $<!-- COMMENT: To achieve a unified layout, along with using the template below, please OBSERVE the following: besides, $...$ and $$...$$, you should use two environments: - For statements like Theorem, Lemma, Definition etc., use e.g. {{beginthm|Theorem 1|(Milnor)}} ... ... ... {{endthm}}. - For references, use e.g. {{cite|Milnor1958b}}. END OF COMMENT --> == Introduction == <wikitex>; For a closed (topological or smooth) manifold $X$ of dimension n$ the intersection form $$ q_X: H^{2n}(X;\mathbb{Z}) \times H^{2n}(X;\mathbb{Z}) \to \mathbb{Z} $$ is obtained by the formula $$ q_X(x,y) = \langle x \smile y , [X] \rangle , $$ i.e. the cup product of $x$ and $y$ is evaluated on the fundamental cycle given by the manifold $X$. $q$ descends to a bilinear pairing on the free module $H^{2n}(X;\mathbb{Z}) / \text{Torsion}$. It is a symmetric and unimodular (in particular non-degenerate) pairing, the latter follows from Poincaré duality. </wikitex> == Algebraic invariants == <wikitex>; 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. As $q$ 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 $$ \text{sign}(q) = b^+ - b^-. $$ 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''. </wikitex> == Classification of indefinite forms == <wikitex>; There is a simple classification result of indefinite forms: {{beginthm|Theorem|(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. {{endthm}} 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 $$ q(c,x) \equiv q(x,x) \ (\text{mod} \ 2) $$ 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 <script type="math/tex">X$ of dimension $4n$ the intersection form

is obtained by the formula

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

$q$ descends to a bilinear pairing on the free module $H^{2n}(X;\mathbb{Z}) / \text{Torsion}$. It is a symmetric and unimodular (in particular non-degenerate) pairing, the latter follows from Poincaré duality.

2 Algebraic invariants

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.

As $q$ 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

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.

3 Classification of indefinite forms

There is a simple classification result of indefinite forms:

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

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:

Proof:

4 Examples, Realisations of indefinite forms

5 References

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

$ 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: == Examples, Realisations of indefinite forms == ; == References == {{#RefList:}} [[Category:Theory]] {{Stub}}X of dimension $4n$ the intersection form

is obtained by the formula

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

$q$ descends to a bilinear pairing on the free module $H^{2n}(X;\mathbb{Z}) / \text{Torsion}$. It is a symmetric and unimodular (in particular non-degenerate) pairing, the latter follows from Poincaré duality.

2 Algebraic invariants

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.

As $q$ 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

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.

3 Classification of indefinite forms

There is a simple classification result of indefinite forms:

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

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:

Proof:

4 Examples, Realisations of indefinite forms

5 References

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