Intersection form

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

1 Definition

After Poincaré, a closed oriented $n$${{Stub}} == Definition == <wikitex>; After Poincaré, a closed oriented $n$-manifold $N$ (PL or smooth) has a bilinear ''intersection product'' $$I_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to \Zz$$ defined on its homology. Take a triangulation $T$ of $N$. Denote by $T^*$ the [[Wikipedia:Poincare_duality#Dual_cell_structures|dual cell subdivision]]. Represent classes $x\in H_k(N;\Zz_2)$ and $y\in H_{n-k}(N;\Zz_2)$ by cycles $\overline x$ and $\overline y$ viewed as unions of $k$-simplices of $T$ and $(n-k)$-simplices of $T^*$, respectively. Define the modulo 2 intersection product modulo 2 $$ I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2 $$ by the formula $$ x\cap_{N,2} y = |\overline x\cap\overline y|\mod2. $$ This is well-defined because the intersection of a cycle and a boundary consists of an even number of points (by definition of a cycle and a boundary). Analogously (i.e. counting intersections with signs) one defines the intersection form $$ I_N=\cap_N=\cdot_N: H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to \Zz. $$ See [[Intersection_number_of_immersions|yet another geometric interpretation]], \cite[Chapter II]{Kirby1989}, \cite[$\S]{Skopenkov2015b}. Using the notion of ''transversality'', one can give an equivalent (and `more general') definition as follows. Take a $k$-chain $x\in C_k(N;\Zz)$ and an $(n-k)$-chain $y\in C_{n-k}(N;\Zz)$ which is ''transverse'' to $x$. The signed count of the intersections between $x$ and $y$ gives an intersection number \begin{equation} \label{eq:intersection} I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z. \end{equation} Using the notion of ''cup product'', one can give a dual (and so an equivalent) definition as follows. Define the intersection product $$ I_N: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz $$ by the formula $$ I_N(x,y) = \langle x \smile y , [X] \rangle , $$ i.e. the cup product of $x$ and $y$ is evaluated on the fundamental cycle given by the manifold $N$. The definition of a cup product is `dual' (and so is analogous) to the above definition of the intersection form on homology of a manifold, but is more abstract. However, the definition of a cup product generalizes to complexes (and so to topological manifolds). This is an advantage for mathematicians who are interested in complexes and topological manifolds (not only in PL and smooth manifolds). If $n=2k$, the intersection product is the ''intersection form'' denoted by $q_N$. A related important notion is [[Linking_form|linking form]]. </wikitex> == Properties == <wikitex>; Clearly, the intersection product is bilinear. Hence it vanishes on torsion elements. Thus it descends to a bilinear pairing $$H_k(X;\Zz) / \text{Torsion}\times H_{n-k}(X;\Zz) / \text{Torsion}\to\Zz.$$ on the free modules. This pairing is unimodular (in particular non-degenerate) by [[Wikipedia:Poincare_duality|Poincaré duality]]. We have $$I_N(x,y) = (-)^{k(n-k)}I_N(y,x).$$ Hence for $n=2k$ * If $k$ is even the pairing $q_N$ is symmetric: $q_N(x, y) = q_N(y, x)$. * If $k$ is odd the pairing $q_N$ is skew-symmetric: $q_N(x, y) = - q_N(y, x)$. <wikitex>; == Uni-modular bilinear forms== <wikitex>; Let $q$ and $q'$ be unimodular symmetric bilinear forms on underlying free $\mathbb{Z}$-modules $V$ and $V'$ respectively. The two forms $q$ and $q'$ are said equivalent if there is an isomorphism $f:V \to V'$ such that $q = f^* q'$. A form $q$ is called ''definite'' if it is positive or negative definite, otherwise it is called ''indefinite''. The ''rank'' of $q$ is the rank of the underlying $\mathbb{Z}$-module $V$. </wikitex> == Skew-symmetric bilinear forms == <wikitex>; The skew-symmetric hyperbolic form of rank $, $H_-(\Zz)$, is defined by the following intersection matrix $$ \left( \begin{array}{cc} ~0 & ~1 \ -1 & ~0 \end{array} \right) .$$ {{beginthm|Proposition}} Every skey-symmetri uni-modular bilinear form over $\Zz$, $q$, isomorphic to some the sum of some number of hyperbolic forms: $$ q \cong \oplus_{i=1}^r H_-(\Zz).$$ In particular the rank of $q$, in this case r$, is even. {{endthm}} </wikitex> == Symmetric bilinear forms == <wikitex>; The classification of uni-modular definite symmetric bilinear forms is a deep and difficult problem. However the situation becomes much easier when the form is indefinite. We begin by stating some fundamental invariants. Since $q$ is symmetric it is diagonalisable over the real numbers. If $b^+$ denotes the dimension of a maximal subspace on which the form is positive definite, and if $b^-$ is the dimension of a maximal subspace on which the form is negative definite, then the ''signature'' of $q$ is defined to be $$ \text{sign}(q) = b^+ - b^-. $$ The form $q$ may have two different ''types''. It is of type ''even'' if $q(x,x)$ is an even number for any element $x$. Equivalently, if $q$ is written as a square matrix in a basis, it is even if the elements on the diagonal are all even. Otherwise, $q$ is said of type ''odd''. </wikitex> === Classification of indefinite forms === <wikitex>; There is a simple classification result of indefinite forms \cite{Serre1970},\cite{Milnor&Husemoller1973}: {{beginthm|Theorem|(Serre (?))}} Two indefinite unimodular symmetric bilinear forms $q, q'$ over $\mathbb{Z}$ are equivalent if and only if $q$ and $q'$ have the same rank, signature and type. {{endthm}} There is a further invariant of a unimodular symmetric bilinear form $q$ on $V$: An element $c \in V$ is called a ''characteristic vector'' of the form if one has $$ q(c,x) \equiv q(x,x) \ (\text{mod} \ 2) $$ for all elements $x \in V$. Characteristic vectors always exist. In fact, when reduced modulo 2, the map $x \mapsto q(x,x) \in \mathbb{Z}/2$ is linear. By unimodularity there therefore exists an element $c$ such that the map $q(c,-)$ equals this linear map. The form $q$ is even if and only if <script type="math/tex">n$-manifold $N$$N$ (PL or smooth) has a bilinear intersection product

$\displaystyle I_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to \Zz$ $$\displaystyle I_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to \Zz$$

defined on its homology.

Take a triangulation

Tex syntax error

$T$ of $N$$N$.

Denote by $T^*$$T^*$ the dual cell subdivision.

Represent classes $x\in H_k(N;\Zz_2)$$x\in H_k(N;\Zz_2)$ and $y\in H_{n-k}(N;\Zz_2)$$y\in H_{n-k}(N;\Zz_2)$ by cycles $\overline x$$\overline x$ and $\overline y$$\overline y$ viewed as unions of $k$$k$-simplices of

Tex syntax error

$T$ and $(n-k)$$(n-k)$-simplices of $T^*$$T^*$, respectively.

Define the modulo 2 intersection product modulo 2

$\displaystyle I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2$ $$\displaystyle I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2$$

by the formula

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

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

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

See yet another geometric interpretation, [Kirby1989, Chapter II], [Skopenkov2015b, $\S$$\S$6].

Using the notion of transversality, one can give an equivalent (and `more general') definition as follows.

Take a $k$$k$-chain $x\in C_k(N;\Zz)$$x\in C_k(N;\Zz)$ and an $(n-k)$$(n-k)$-chain $y\in C_{n-k}(N;\Zz)$$y\in C_{n-k}(N;\Zz)$ which is transverse to

Tex syntax error

$x$. The signed count of the intersections between

Tex syntax error

$x$ and

Tex syntax error

$y$ gives an intersection number

(1)$I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z.$$I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z.$

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

$\displaystyle I_N: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz$ $$\displaystyle I_N: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz$$

by the formula

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

i.e. the cup product of

Tex syntax error

$x$ and

Tex syntax error

$y$ is evaluated on the fundamental cycle given by the manifold $N$$N$.

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 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).

If $n=2k$$n=2k$, the intersection product is the intersection form denoted by $q_N$$q_N$.

A related important notion is linking form.

2 Properties

Clearly, the intersection product is bilinear. Hence it vanishes on torsion elements. Thus it descends to a bilinear pairing

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

on the free modules. This pairing is unimodular (in particular non-degenerate) by Poincaré duality.

We have

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

Hence for $n=2k$$n=2k$

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

Uni-modular bilinear forms

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

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

3 Skew-symmetric bilinear forms

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

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

Proposition 6.1. Every skey-symmetri uni-modular bilinear form over $\Zz$$\Zz$, $q$$q$, isomorphic to some the sum of some number of hyperbolic forms:

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

In particular the rank of $q$$q$, in this case $2r$$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$$q$ is symmetric it is diagonalisable over the real numbers. If $b^+$$b^+$ denotes the dimension of a maximal subspace on which the form is positive definite, and if $b^-$$b^-$ is the dimension of a maximal subspace on which the form is negative definite, then the signature of $q$$q$ is defined to be

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

The form $q$$q$ may have two different types. It is of type even if $q(x,x)$$q(x,x)$ is an even number for any element

Tex syntax error

$x$. Equivalently, if $q$$q$ is written as a square matrix in a basis, it is even if the elements on the diagonal are all even. Otherwise, $q$$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 7.1 (Serre (?)). Two indefinite unimodular symmetric bilinear forms $q, q'$$q, q'$ over $\mathbb{Z}$$\mathbb{Z}$ are equivalent if and only if $q$$q$ and $q'$$q'$ have the same rank, signature and type.

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

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

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

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

Proposition 7.2. For a characteristic vector $c$$c$ of the unimodular symmetric bilinear form $q$$q$ one has

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

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

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

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

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

However, the form $q'$$q'$ is indefinite, so the above classification theorem applies. In particular, $q'$$q'$ is odd and has the same signature as $q$$q$, so it is equivalent to the diagonal form with $b^+ + 1$$b^+ + 1$ summands of (+1) and $b^- + 1$$b^- + 1$ summands of $(-1)$$(-1)$. This diagonal form has a characteristic vector $c'$$c'$ that is simply a sum of basis elements in which the form is diagonal. Of course $q'(c',c') = b^+ - b^-$$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 7.3. 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,

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

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

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

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

On the other hand, the matrix

Tex syntax error

$H$ given by

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

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

$\displaystyle k E_8 \oplus l H$ $$\displaystyle k E_8 \oplus l H$$

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

5 References

• [Kirby1989] R.C. Kirby, The topology of 4-manifolds, Lecture Notes in Math. 1374, Springer-Verlag, 1989. MR1001966 (90j:57012)
• [Milnor&Husemoller1973] J. Milnor and D. Husemoller, Symmetric bilinear forms, Springer-Verlag, New York, 1973. MR0506372 (58 #22129) Zbl 0292.10016
• [Serre1970] J. Serre, Cours d'arithmétique, Presses Universitaires de France, Paris, 1970. MR0255476 (41 #138) Zbl 0432.10001
• [Skopenkov2015b] A. Skopenkov, Algebraic Topology From Geometric Viewpoint (in Russian), MCCME, Moscow, 2015, 2020. Accepted for English translation by `Moscow Lecture Notes' series of Springer. Preprint of a part

6 External links

• The Wikipedia page on Poincaré duality

$ 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-manifold $N$$N$ (PL or smooth) has a bilinear intersection product

$\displaystyle I_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to \Zz$ $$\displaystyle I_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to \Zz$$

defined on its homology.

Take a triangulation

Tex syntax error

$T$ of $N$$N$.

Denote by $T^*$$T^*$ the dual cell subdivision.

Represent classes $x\in H_k(N;\Zz_2)$$x\in H_k(N;\Zz_2)$ and $y\in H_{n-k}(N;\Zz_2)$$y\in H_{n-k}(N;\Zz_2)$ by cycles $\overline x$$\overline x$ and $\overline y$$\overline y$ viewed as unions of $k$$k$-simplices of

Tex syntax error

$T$ and $(n-k)$$(n-k)$-simplices of $T^*$$T^*$, respectively.

Define the modulo 2 intersection product modulo 2

$\displaystyle I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2$ $$\displaystyle I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2$$

by the formula

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

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

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

See yet another geometric interpretation, [Kirby1989, Chapter II], [Skopenkov2015b, $\S$$\S$6].

Using the notion of transversality, one can give an equivalent (and `more general') definition as follows.

Take a $k$$k$-chain $x\in C_k(N;\Zz)$$x\in C_k(N;\Zz)$ and an $(n-k)$$(n-k)$-chain $y\in C_{n-k}(N;\Zz)$$y\in C_{n-k}(N;\Zz)$ which is transverse to

Tex syntax error

$x$. The signed count of the intersections between

Tex syntax error

$x$ and

Tex syntax error

$y$ gives an intersection number

(1)$I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z.$$I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z.$

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

$\displaystyle I_N: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz$ $$\displaystyle I_N: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz$$

by the formula

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

i.e. the cup product of

Tex syntax error

$x$ and

Tex syntax error

$y$ is evaluated on the fundamental cycle given by the manifold $N$$N$.

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 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).

If $n=2k$$n=2k$, the intersection product is the intersection form denoted by $q_N$$q_N$.

A related important notion is linking form.

2 Properties

Clearly, the intersection product is bilinear. Hence it vanishes on torsion elements. Thus it descends to a bilinear pairing

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

on the free modules. This pairing is unimodular (in particular non-degenerate) by Poincaré duality.

We have

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

Hence for $n=2k$$n=2k$

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

Uni-modular bilinear forms

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

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

3 Skew-symmetric bilinear forms

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

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

Proposition 6.1. Every skey-symmetri uni-modular bilinear form over $\Zz$$\Zz$, $q$$q$, isomorphic to some the sum of some number of hyperbolic forms:

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

In particular the rank of $q$$q$, in this case $2r$$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$$q$ is symmetric it is diagonalisable over the real numbers. If $b^+$$b^+$ denotes the dimension of a maximal subspace on which the form is positive definite, and if $b^-$$b^-$ is the dimension of a maximal subspace on which the form is negative definite, then the signature of $q$$q$ is defined to be

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

The form $q$$q$ may have two different types. It is of type even if $q(x,x)$$q(x,x)$ is an even number for any element

Tex syntax error

$x$. Equivalently, if $q$$q$ is written as a square matrix in a basis, it is even if the elements on the diagonal are all even. Otherwise, $q$$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 7.1 (Serre (?)). Two indefinite unimodular symmetric bilinear forms $q, q'$$q, q'$ over $\mathbb{Z}$$\mathbb{Z}$ are equivalent if and only if $q$$q$ and $q'$$q'$ have the same rank, signature and type.

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

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

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

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

Proposition 7.2. For a characteristic vector $c$$c$ of the unimodular symmetric bilinear form $q$$q$ one has

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

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

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

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

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

However, the form $q'$$q'$ is indefinite, so the above classification theorem applies. In particular, $q'$$q'$ is odd and has the same signature as $q$$q$, so it is equivalent to the diagonal form with $b^+ + 1$$b^+ + 1$ summands of (+1) and $b^- + 1$$b^- + 1$ summands of $(-1)$$(-1)$. This diagonal form has a characteristic vector $c'$$c'$ that is simply a sum of basis elements in which the form is diagonal. Of course $q'(c',c') = b^+ - b^-$$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 7.3. 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,

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

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

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

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

On the other hand, the matrix

Tex syntax error

$H$ given by

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

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

$\displaystyle k E_8 \oplus l H$ $$\displaystyle k E_8 \oplus l H$$

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

5 References

• [Kirby1989] R.C. Kirby, The topology of 4-manifolds, Lecture Notes in Math. 1374, Springer-Verlag, 1989. MR1001966 (90j:57012)
• [Milnor&Husemoller1973] J. Milnor and D. Husemoller, Symmetric bilinear forms, Springer-Verlag, New York, 1973. MR0506372 (58 #22129) Zbl 0292.10016
• [Serre1970] J. Serre, Cours d'arithmétique, Presses Universitaires de France, Paris, 1970. MR0255476 (41 #138) Zbl 0432.10001
• [Skopenkov2015b] A. Skopenkov, Algebraic Topology From Geometric Viewpoint (in Russian), MCCME, Moscow, 2015, 2020. Accepted for English translation by `Moscow Lecture Notes' series of Springer. Preprint of a part

6 External links

• The Wikipedia page on Poincaré duality