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Let $Q:V\to {\mathbb{F}}_2$ be a quadratic form whose polar bilinear form of a quadratic form $B$ is nondegenerate, so $\dim V=2m$. For a symplectic basis $e_1,f_1,\dots ,e_m,f_m$, defineThis does not depend on the symplectic basis. The hyperbolic plane $Q(x,y)=xy$ has Arf invariant zero; $Q(x,y)=x^2+xy+y^2$ has Arf invariant one and vanishes only at zero. Every such quadratic form is an orthogonal direct sum of hyperbolic planes, with one copy of the latter plane replacing a hyperbolic plane when the Arf invariant is one.^[ref]

A useful basis-independent test isOn one two-dimensional summand with symplectic basis $e,f$, write $Q(xe+yf)=ax+by+xy$; the four terms sum to $2(-1{)}^{ab}$. Multiplication over the orthogonal direct sum proves the formula and the invariance of its sign. Consequently the number of zeros is $2^{2m-1}+(-1{)}^{\mathrm{Arf}(Q)}{2}^{m-1}$. For a fixed $B$ and $m\ge 1$, the numbers of binary quadratic refinements with Arf invariant zero and one are respectively $2^{m-1}(2^m+1)$ and $2^{m-1}(2^m-1)$.

Bill Casselman writes research and expository mathematics connecting algebraic structure with concrete calculations. His author-hosted Introduction to quadratic forms by Bill Casselman explains quadratic forms over general fields, including the distinctions required in characteristic two. It develops decomposition results using totally isotropic subspaces and the hyperbolic plane.^[ref]

For example, over $\mathbb{R}$ the quadratic form $Q(x,y,z)=xy+z^2$ separates into a hyperbolic plane on the first two coordinates and an anisotropic quadratic form on the last. The $x$-axis consists of isotropic vectors, but it is not invisible to the whole form: its associated bilinear form pairs $(1,0,0)$ nontrivially with $(0,1,0)$. Pairing such complementary directions produces the hyperbolic part of an orthogonal direct sum. The example helps distinguish a vector on which a quadratic form vanishes from a vector in the radical of its associated bilinear form.^[ref]

The cup product on the second integral cohomology of a complex K3 surface is an even unimodular lattice of signature $(3,19)$. Here $U$ is the hyperbolic plane and $E_8(-1)$ is the negative-definite form of the E8 lattice. In the M-theory K3 duality, this lattice matches the heterotic electric charge lattice. The self-dual harmonic two-forms of a Ricci flat K3 metric span its positive three-plane.

For a complex K3 surface, $L=H^2(X,\mathbb{Z})$ is torsion-free and has its integral cup product pairingThe K3 intersection lattice isthe even unimodular integral quadratic lattice of signature $(3,19)$. Here $U$ is the hyperbolic plane with Gram matrixand $E_8(-1)$ is the negative of the positive-definite E8 lattice. Thus the rank is $3\cdot 2+2\cdot 8=22$ and the pairing has determinant of absolute value one.

These invariants can be checked from the topology and Hodge theory. The trivial canonical bundle and $H^1(X,{\mathcal{O}}_X)=0$ give $\chi ({\mathcal{O}}_X)=2$. The Noether formula gives $c_2(X)=24$, which is the topological Euler characteristic. Since $X$ is simply connected, $b_1=b_3=0$, so $24=2+b_2$ and $b_2=22$. Integral Poincare duality gives unimodularity; the universal coefficient sequence and $H_1(X,\mathbb{Z})=0$ give torsion-freeness of $H^2(X,\mathbb{Z})$.

Its second Stiefel–Whitney class is the reduction modulo two of $c_1(X)$, hence zero. The characteristic-class identitymakes the lattice even. Finally the real and imaginary parts of a holomorphic two-form give two positive directions, a Kähler class gives a third, and the primitive real $(1,1)$ classes are negative definite by the Hodge–Riemann bilinear relations. Therefore the signature is $(3,19)$. The classification of indefinite even unimodular lattices then identifies the displayed lattice uniquely. In particular the Picard group is only the integral $(1,1)$ part; it should not be confused with this rank-twenty-two topological lattice.

The symmetric bilinear form associated with a quadratic form is obtained by polarization:To diagonalize it, argue by induction on $\dim V$. If $\varphi (v,v)=0$ for every $v$, polarization gives $\varphi =0$, so every basis works. Otherwise choose $e$ with $\varphi (e,e)\ne 0$. Thenand diagonalize the restriction to $e^{\perp }$ inductively. Thus some basis gives a diagonal matrix with positive, negative, and zero entries. By Sylvester's law of inertia, their counts $n_{+},n_{-},n_0$ are basis-independent. In the convention relevant here, the signature of a quadratic form is

Suppose $R$ lies in the radical of a bilinear form, so $\varphi (r,v)=0$ for all $r\in R,v\in V$. Then $q(r)=0$ andHence $q^{\prime }(v+R)=q(v)$ is a well-defined quadratic form on the quotient vector space $V/R$. In an adapted diagonal basis, quotienting by $R$ merely removes zero diagonal directions, so $n_{+}$ and $n_{-}$, and therefore the signature, are unchanged.

Now let $W=\mathrm{span}(e,f)$. Its Gram matrix iswhich has determinant $-1$. Thus $W$ is nondegenerate and has one positive and one negative square: it is a hyperbolic plane (quadratic form) and has signature zero. Nondegeneracy gives the orthogonal direct sumSignature is additive under orthogonal direct sums, so