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A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.

For finite-dimensional vector spaces $V,W$ over a field $F$, a bilinear pairing $B:V\times W\to F$ is perfect when the induced linear maps $V\to W^{\ast }$ and $W\to V^{\ast }$ are isomorphisms. Equivalently, both dimensions agree and the matrix of the pairing in any two bases is invertible. The evaluation pairing between a finite-dimensional vector space and its dual space is perfect. This notion allows two different spaces; a nondegenerate bilinear form is the case $V=W$.

For a closed oriented $d$-dimensional manifold over a field $F$, the cup product pairingis a perfect pairing between complementary degrees. The cap product with the fundamental class is an isomorphism by Poincare duality, and the universal coefficient theorem for cohomology over $F$ identifies the complementary cohomology with the full dual space of the resulting homology. Evaluation is therefore a perfect pairing. Taking the top-degree part of the same formula defines a nondegenerate bilinear form on the full graded cohomology. A boundary requires relative groups and Poincare-Lefschetz duality instead.

A linear subspace $W\subseteq X^{\ast }$ is separating if every nonzero $x\in X$ has $T(x)\ne 0$ for some $T\in W$. For finite-dimensional $X$, an evaluation map $X\to F^{\dim W}$ is injective, so the rank-nullity theorem forces $\dim W\ge \dim X=\dim X^{\ast }$ and $W=X^{\ast }$. For the infinite-dimensional polynomial space, the span of its coefficient functionals is a proper separating linear subspace of the dual space, which contains all sequences of coefficients, not just finitely supported ones.

The continuous dual space of $H^s({\mathbb{R}}^d)$ is $H^{-s}({\mathbb{R}}^d)$ under the extension of the L2 inner product. The Fourier transform definition $\Vert f{\Vert }_{H^s}^2=\int (1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d\xi$ proves boundedness of the pairing by the Cauchy-Schwarz inequality, and the Riesz representation theorem proves that every continuous functional has this form. Consequently a bounded linear operator $K:H^r\to H^{r+a}$ has an adjoint operator, relative to this pairing, from $H^{-r-a}$ to $H^{-r}$. For an operator that smooths by two orders for every $r$, this yields $K^{\ast }:H^2\to H^4$ by taking $r=-4$.