View source

You do not have permission to edit this page, for the following reason:

You can view and copy the source of this page:

{{Stub}} == Introduction == <wikitex>; For a closed (topological or smooth) manifold $X$ of dimension $2n$ the intersection form $$ q_X: H^{n}(X;\mathbb{Z}) \times H^{n}(X;\mathbb{Z}) \to \mathbb{Z} $$ 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 $X$. 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]]. * 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: $q_X(x, y) = - q_X(y, x)$. </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 $2$, $H_-(\Zz)$, 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 $2r$, 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 $0$ 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: 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. </wikitex> === Examples, Realisations of indefinite forms === <wikitex>; 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 $-E_8$. </wikitex> == References == {{#RefList:}} [[Category:Theory]] [[Category:Surgery]]

Return to Intersection form.