Intersection form

Revision as of 13:01, 14 February 2021

1 Introduction

Let ${{Stub}} == Introduction == <wikitex>; 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''' $$I_N=\cap_N=\cdot_N=\lambda_N\colon H_k(N;\Zz) \times H_{n-k}(N;\Zz) \to \Zz$$ 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 [[Stiefel-Whitney_characteristic_classes|characteristic classes]] and [[Linking_form|linking form]]. These are important invariants used in the classification of [[Manifold|manifolds]]. In this page $T$ is a triangulation (or a cell subdivision) of $N$, and $T^*$ is the [[Wikipedia:Poincare_duality#Dual_cell_structures|dual cell subdivision]]. The exposition follows \cite[Chapter II]{Kirby1989}, \cite[$\S, $\S]{Skopenkov2015b}, \cite[$\S, $\S]{SkopenkovS}. </wikitex> == Definition of the intersection product== <wikitex>; In this subsection we mostly omit $\Zz_2$-coefficients. '''A short direct definition of a homology group with $\Zz_2$-coefficients.''' For an integer $s$ denote by $C_s(T)=C_s(T;\Zz_2)$ the set (the $\Zz_2$-space) of arrangements of zeroes and units on the $s$-dimensional cells of $T$ (so $C_s(T)=\{0\}$ if there are no $s$-dimensional cells in $T$). Denote by $\partial=\partial_s\colon C_s(T) \to C_{s - 1}(T)$ the extension over $C_s(T)$ of the map taking an $s$-dimensional cell $\sigma$ of $T$ to the boundary of $\sigma$. Denote $$Z_s(T)=Z_s(T;\Zz_2):=\ker\partial_s\quad\text{and}\quad B_s(T)=B_s(T;\Zz_2):=\mathrm{im}\partial_{s + 1}.$$ Then the $s$-th homology group of $T$ with $\Zz_2$-coefficients is defined as $H_s(T)=H_s(T;\Zz_2):=Z_s(T)/ B_s(T)$. This depends only on $N$, not on $T$. '''A short direct definition of the modulo 2 intersection product.''' For modulo 2 chains $x\in C_k(T)$ and $y\in C_{n-k}(T^*)$ define the ''modulo 2 intersection number'' by the formula $$ I_{T,2}(x,y)=x\cap_{T,2} y=\langle\, x \, , \, y\, \rangle := |x\cap y|\mod2\in\Zz_2. $$ 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 product'' $$ I_{N,2}=\cap_{N,2}: H_k(N;\Zz_2) \times H_{n-k}(N;\Zz_2) \to \Zz_2\quad\text{by}\quad I_{N,2}([x],[y]):=I_{N,2}(x,y).$$ 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). '''Sketch of a short direct definition of the intersection product.''' Analogously, counting intersections with signs, one defines the '''intersection number''' $$I_N(x,y)=x\cdot y=\langle\, x \, , \, y\, \rangle \in \Z$$ 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'' $I_N: H_k(N;\Zz)\times H_{n-k}(N;\Zz)\to\Zz$. {{beginthm|Remark}} (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|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: $$I_N(x,y) = \langle x^*\smile y^*,[N]\rangle \in \Z,$$ where $x^*\in H^{n-k}(N)$, $y^*\in H^n(N)$ are the [[Poincare isomorphism|Poincaré duals]] of $x$, $y$, and $[N]$ is the fundamental class of the manifold $N$. We can also define the ''cup (cohomology intersection) product'' $$ I_N^*: H^k(N;\Zz) \times H^{n-k}(N;\Zz) \to \Zz \quad\text{by}\quad I_N^*(p,q) = \langle p \smile q , [N] \rangle . $$ 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). See \cite[Remark 2.3]{Skopenkov2005}. {{endthm}} </wikitex> == Simple properties == <wikitex>; 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'' $$H_k(N;\Zz) / \text{Torsion}\times H_{n-k}(N;\Zz) / \text{Torsion}\to\Zz.$$ on the free modules. We have $$I_N(x,y) = (-1)^{k(n-k)}I_N(y,x).$$ 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: $q_N(x, y) = - q_N(y, x)$. </wikitex> == Poincaré duality == <wikitex>; {{beginthm|Theorem}}[Poincaré duality] (a) The modulo 2 intersection product is non-degenerate. (b) The integer intersection pairing is unimodular (in particular non-degenerate). {{endthm}} ''Proof of (a).'' (This proof is well-known to specialists but is, in this short and explicit form, absent from textbooks. This proof is based on a text written by E. Kogan in frame of the course `Algorithms for recognition of realizability of hypergraphs' taught by A. Skopenkov.) We use orthogonal complements with respect to the modulo 2 intersection product $C_s(T)\times C_{n-s}(T^*)\to\Zz_2$ <!--Let $s$ be an integer. For an $s$-cell $c$ of $T$ denote by $c^*$ the dual $(n-s)$-cell of $T^*$. For a chain $C = \sum_i c_i \in C_s(T)$, where $c_i$ are $s$-cells of $T$ denote $C^*:=\sum_{i} c_i^* \in C_{n - s}(T^*)$. For a bilinear form $\gamma\colon V_1 \times V_2$ denote $\ker_L\gamma := \phantom{}^\bot V_2$, $\ker_R\gamma := V_1^\bot$.--> It suffices to prove that $$\phantom{}^\bot Z_{n-s}(T^*)= B_s(T) \text{and}\quad Z_s(T)^\bot=B_{n-s}(T^*).$$ Let us prove the left-hand equality; the right-hand equality is proved analogously. Since $I_{N,2}$ is non-degenerate, we only need to check that $B_s(T)^\bot=Z_{n-s}(T^*)$. The inclusion $B_s(T)^\bot \supset Z_{n-s}(T^*)$ is obvious. The opposite inclusion follows because ''if $I_{N,2}(\partial c,d)=0$ for an $(s+1)$-cell $c$ of $T$ and a chain $d\in Z_{n-s}(T^*)$, then $\partial d$ does not involve the cell $c^*$ dual to $c$''. </wikitex> == Definition of signature == <wikitex>; 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 $$ \sigma(q)={\rm sign}(q) = b^+(q) - b^-(q). $$ If $n$ is divisible by 4, the '''signature''' $\sigma(N)={\rm sign}(N)$ is defined to be the signature of the intersection form of $N$. </wikitex> == Equivalence of bilinear forms== <wikitex>; 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 $V$. </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 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. 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''. </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$ 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.

In this page $T$ is a triangulation (or a cell subdivision) of $N$, and $T^*$ is the dual cell subdivision.

The exposition follows [Kirby1989, Chapter II], [Skopenkov2015b, $\S$6, $\S$10], [SkopenkovS, $\S$6, $\S$10].

2 Definition of the intersection product

In this subsection we mostly omit $\Zz_2$-coefficients.

A short direct definition of a homology group with $\Zz_2$-coefficients. For an integer $s$ denote by $C_s(T)=C_s(T;\Zz_2)$ the set (the $\Zz_2$-space) of arrangements of zeroes and units on the $s$-dimensional cells of $T$ (so $C_s(T)=\{0\}$ if there are no $s$-dimensional cells in $T$). Denote by $\partial=\partial_s\colon C_s(T) \to C_{s - 1}(T)$ the extension over $C_s(T)$ of the map taking an $s$-dimensional cell $\sigma$ of $T$ to the boundary of $\sigma$. Denote

$$\displaystyle Z_s(T)=Z_s(T;\Zz_2):=\ker\partial_s\quad\text{and}\quad B_s(T)=B_s(T;\Zz_2):=\mathrm{im}\partial_{s + 1}.$$

Then the $s$-th homology group of $T$ with $\Zz_2$-coefficients is defined as $H_s(T)=H_s(T;\Zz_2):=Z_s(T)/ B_s(T)$. This depends only on $N$, not on $T$.

A short direct definition of the modulo 2 intersection product. For modulo 2 chains $x\in C_k(T)$ and $y\in C_{n-k}(T^*)$ define the modulo 2 intersection number by the formula

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

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 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).

Sketch of a short direct definition of the intersection product. 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 $I_N: H_k(N;\Zz)\times H_{n-k}(N;\Zz)\to\Zz$.

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). See [Skopenkov2005, Remark 2.3].

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

Proof of (a). (This proof is well-known to specialists but is, in this short and explicit form, absent from textbooks. This proof is based on a text written by E. Kogan in frame of the course `Algorithms for recognition of realizability of hypergraphs' taught by A. Skopenkov.)

We use orthogonal complements with respect to the modulo 2 intersection product $I_{T,2}:C_s(T)\times C_{n-s}(T^*)\to\Zz_2$. It suffices to prove that

$$\displaystyle \phantom{}^\bot Z_{n-s}(T^*)= B_s(T) \text{and}\quad Z_s(T)^\bot=B_{n-s}(T^*).$$

Let us prove the left-hand equality; the right-hand equality is proved analogously. Since $I_{T,2}$ is non-degenerate, we only need to check that $B_s(T)^\bot=Z_{n-s}(T^*)$. The inclusion $B_s(T)^\bot \supset Z_{n-s}(T^*)$ is obvious. The opposite inclusion follows because if $I_{N,2}(\partial c,d)=0$ for an $(s+1)$-cell $c$ of $T$ and a chain $d\in Z_{n-s}(T^*)$, then $\partial d$ does not involve the cell $c^*$ dual to $c$.

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

Tex syntax error

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.

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

• [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
• [Skopenkov2005] A. Skopenkov, A classification of smooth embeddings of 4-manifolds in 7-space, Topol. Appl., 157 (2010) 2094-2110. Available at the arXiv:0512594.
• [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
• [SkopenkovS] Template:SkopenkovS

10 External links

• The Wikipedia page on Poincaré duality

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

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.

In this page $T$ is a triangulation (or a cell subdivision) of $N$, and $T^*$ is the dual cell subdivision.

The exposition follows [Kirby1989, Chapter II], [Skopenkov2015b, $\S$6, $\S$10], [SkopenkovS, $\S$6, $\S$10].

2 Definition of the intersection product

In this subsection we mostly omit $\Zz_2$-coefficients.

A short direct definition of a homology group with $\Zz_2$-coefficients. For an integer $s$ denote by $C_s(T)=C_s(T;\Zz_2)$ the set (the $\Zz_2$-space) of arrangements of zeroes and units on the $s$-dimensional cells of $T$ (so $C_s(T)=\{0\}$ if there are no $s$-dimensional cells in $T$). Denote by $\partial=\partial_s\colon C_s(T) \to C_{s - 1}(T)$ the extension over $C_s(T)$ of the map taking an $s$-dimensional cell $\sigma$ of $T$ to the boundary of $\sigma$. Denote

$$\displaystyle Z_s(T)=Z_s(T;\Zz_2):=\ker\partial_s\quad\text{and}\quad B_s(T)=B_s(T;\Zz_2):=\mathrm{im}\partial_{s + 1}.$$

Then the $s$-th homology group of $T$ with $\Zz_2$-coefficients is defined as $H_s(T)=H_s(T;\Zz_2):=Z_s(T)/ B_s(T)$. This depends only on $N$, not on $T$.

A short direct definition of the modulo 2 intersection product. For modulo 2 chains $x\in C_k(T)$ and $y\in C_{n-k}(T^*)$ define the modulo 2 intersection number by the formula

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

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 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).

Sketch of a short direct definition of the intersection product. 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 $I_N: H_k(N;\Zz)\times H_{n-k}(N;\Zz)\to\Zz$.

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). See [Skopenkov2005, Remark 2.3].

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

Proof of (a). (This proof is well-known to specialists but is, in this short and explicit form, absent from textbooks. This proof is based on a text written by E. Kogan in frame of the course `Algorithms for recognition of realizability of hypergraphs' taught by A. Skopenkov.)

We use orthogonal complements with respect to the modulo 2 intersection product $I_{T,2}:C_s(T)\times C_{n-s}(T^*)\to\Zz_2$. It suffices to prove that

$$\displaystyle \phantom{}^\bot Z_{n-s}(T^*)= B_s(T) \text{and}\quad Z_s(T)^\bot=B_{n-s}(T^*).$$

Let us prove the left-hand equality; the right-hand equality is proved analogously. Since $I_{T,2}$ is non-degenerate, we only need to check that $B_s(T)^\bot=Z_{n-s}(T^*)$. The inclusion $B_s(T)^\bot \supset Z_{n-s}(T^*)$ is obvious. The opposite inclusion follows because if $I_{N,2}(\partial c,d)=0$ for an $(s+1)$-cell $c$ of $T$ and a chain $d\in Z_{n-s}(T^*)$, then $\partial d$ does not involve the cell $c^*$ dual to $c$.

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

Tex syntax error

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.

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

• [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
• [Skopenkov2005] A. Skopenkov, A classification of smooth embeddings of 4-manifolds in 7-space, Topol. Appl., 157 (2010) 2094-2110. Available at the arXiv:0512594.
• [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
• [SkopenkovS] Template:SkopenkovS

10 External links

• The Wikipedia page on Poincaré duality