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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?

Saurav Maheshkar
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T0mstone
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  • Right, in the algebraic setting, to get a non-trivial definition it should say 'such that the map from $V$ to $V^$ is not injective'. The Wikipedia article gives you an example where this can happen (so this isn't a vacuous definition). In the normed setting, one indeed considers the continuous dual, and there one introduces the notions of weakly non-degenerate* (i.e the induced map is injective) and and strongly non-degenerate (i.e the induced map is an isomorphism), and of course one requires the maps in question to be continuous. (authors may choose to emphasize/omit the weak/strong). – peek-a-boo Dec 13 '22 at 11:41
  • And in the normed setting, there of course exist strongly non-degenerate bilinear maps (the prototype, by Riesz's theorem, being inner products on a Hilbert space). – peek-a-boo Dec 13 '22 at 11:43

1 Answers1

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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.

Captain Lama
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  • Since there is a difference between the algebraic and the topological dual of a vector space, wouldn't regular bilinear forms be possible if they were referring to an isomorphism between V and its topological dual? (as some authors like Marsden actually talk about regular bilinear forms on infinite dimensional vector spaces, which however have to be reflexive) – whatever Jun 08 '23 at 17:31