2
$\begingroup$

The title says it all. I'm wondering if there is any infinite dimensional vector space over some infinite ordered field such that we cannot impose any inner product on it at all. I understand that this is possible if our field is finite, because in that case no subfield can be ordered. So I'd be much obliged if anyone could give me an example of such a vector space, or prove that no such field exists? Thanks in advance. Cheers!

$\endgroup$
  • 1
    $\begingroup$ Pick a basis, and take that basis to be an orthonormal basis. $\endgroup$ – Qiaochu Yuan Jan 5 '15 at 5:30
5
$\begingroup$

As long as you can choose a Hamel basis for your vector space (this is where axiom of choice has to intervene), then you can always define an inner product in the usual way $\sum_i x_i\bar{y_i}$. Keeping in mind that every vector can be represented as a ''finite'' linear combination of the (Hamel) basis vectors. Thus in that sense every infinite dimensional vector space will have an inner product structure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.