I'm a bit confused about what it means for a bilinear form to be degenerate in the context of infinite-dimensional vector spaces.
According to Wikipedia,
[A] degenerate bilinear form $f(x, y)$ on a vector space $V$ is a bilinear form such that the map from $V$ to $V^∗$ (the dual space of $V$) given by $v \mapsto (x \mapsto f(x, v))$ is not an isomorphism. An equivalent definition when V is finite-dimensional is that it has a non-trivial kernel: there exist some non-zero $x$ in $V$ such that
$$
f(x, y) = 0 \;\text{ for all }\; y \in V
$$
But since an infinite-dimensional vector space is never isomorphic to its dual, then a bilinear form on an infinite dimensional vector space is always degenerate.
That doesn't seem right, so what is going on here?
I don't think it's talking about the continuous dual space, since $V$ is introduced as any vector space, i.e. no topology is given.
Of course I'm aware you can define "degenerate" to mean whatever you want,
so my question is specifically about which definition is commonly used
for a bilinear form to be "degenerate",
specifically when not only talking about finite-dimensional vector spaces.
Is there even such a thing, or is "degenerate" a notion that only really applies to
bilinear forms on finite-dimensional vector spaces?
Best Answer
Of course this depends on authors, but usually (especially when working over rings and not just fields), one says that a bilinear form is nondegenerate if it induces an injection from $V$ to its dual, and that it is regular when it induces an isomorphism.
For finite-dimensional vector spaces this is the same thing of course, but usually there is a difference, and in particular for infinite-dimensional vector spaces the notion of a regular bilinear form becomes void, as you noticed yourself, but you can still ask whether a form is nondegenerate.