0
$\begingroup$

I was wondering if there was a widely accepted generalization of inner product spaces where the inner product look something like $\langle\bullet , \bullet\rangle:V\times V \to \mathbb{F}$, where $\mathbb{F}$ does not have to equal to $\mathbb{R}$ or $\mathbb{C}$. Could you define a meaningful inner product space over $\mathbb{Q}$ or a finite field?

$\endgroup$
3
$\begingroup$

Yes it is called a bilinear form. A form because it takes value in the base field. You can guess what bilinear is.

https://en.wikipedia.org/wiki/Bilinear_form

$\endgroup$
  • $\begingroup$ I know of bilinear forms but is there anything a bit more resembling inner product than bilinearity? $\endgroup$ – Noam Jun 7 '16 at 20:37
  • $\begingroup$ You can have definiteness, it this case the form is called nondegenerate. But for the positivity of the inner product you don't have anything. Because for any arbitrary field there is no canonical ordering. For exemple in a finite field, an order preserved by addition and multiplication is trivial. The order of $\mathbb{Q}$ is better seen trought $\mathbb{R}$ $\endgroup$ – InfiniteLooper Jun 7 '16 at 20:44
1
$\begingroup$

Yes, you can do this. Choose a linearly independent basis for an as yet unconstructed vector space $V$. Define $V$ as the set of linear combinations of basis elements with coefficients drawn from any field $\mathbb{F}$ you like. (You'll need to make sure your basis elements can be multiplied by any of the fields' elements.) Now define your inner product by combining "the basis elements are orthonormal" with "this product is sesquilienar". (For self-conjugate fields such as $\mathbb{R}$, sesquilinearity is equivalent to bilinearity; see also Bleuderk's answer.) This uniquely defines an inner product that does what you've asked.

To consider an example, there's a well-known inner product with respect to which Hermite polynomials are orthonormal. The linear combinations with strictly rational coefficients comprise a vector space of functions over $\mathbb{F}=\mathbb{Q}$. The inner product will be over $\mathbb{Q}$ too, since$a_i,\,b_i\in\mathbb{F}\to \sum_ia_ib_i\in\mathbb{F}$. (We need a bit of care with inner products that could be infinite due to convergence conditions failing, but that's a familiar complication for infinite-dimensional vector spaces over $\mathbb{R}$ or $\mathbb{C}$ too.)

$\endgroup$
  • $\begingroup$ Can you define the phrase "self-conjugate field"? Googling this phrase didn't produce anything of worth... $\endgroup$ – goblin Oct 8 '16 at 1:58
  • $\begingroup$ @goblin Many fields have an involution $\ast$ satisfying $\left( ab\right)^\ast=b^\ast a^\ast$. Such an involution appears in the definition of sesquilinearity. For example, if an inner product is antilinear in its first argument we have $\left\langle\alpha u+\beta v | w\right\rangle = \alpha^\ast \left\langle u | w\right\rangle + \beta^\ast \left\langle v | w\right\rangle$. For $\mathbb{F}=\mathbb{C}$, $\ast$ is complex conjugation, so in general $z^\ast\ne z$. By contrast, for self-conjugate fields such as $\mathbb{F}=\mathbb{R}$ we have $x^\ast = x$ for all field elements $x$. $\endgroup$ – J.G. Oct 8 '16 at 6:47
0
$\begingroup$

This generalizes to a sesquilinear form on any module over a ring that admits an antiautomorphism. See Sesquilinear form on Wikipedia. The positive-definiteness of an inner product gets replaced by a constraint to a particular subset of the ring.

$\endgroup$

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.