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I've often wondered why it seems inner products only ever use $\mathbb{R}$ or $\mathbb{C}$ for the base field of their vector space, or sub-fields like the algebraic or construct-able numbers. According to Wikipedia, these are the only fields we can use which have the right properties. Why? Why not use other, stranger, more complicated fields? What breaks down if we do so, and is there a meaningful generalization? What does it say about the inner product to have these restrictions?

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  • $\begingroup$ Can you document where Wikipedia says these are the only fields which we can use for inner products? Or do they simply omit consideration of other possibilities? $\endgroup$
    – Lee Mosher
    Jun 18, 2018 at 13:20
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    $\begingroup$ Anyway, inner products are certainly used for other fields, for example finite fields. $\endgroup$
    – Lee Mosher
    Jun 18, 2018 at 13:21
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    $\begingroup$ Here is where the Wikipedia article says that it only works for real and complex numbers: en.wikipedia.org/wiki/Inner_product_space >There are various technical reasons why it is necessary to restrict the basefield to R and C in the definition. Briefly, the basefield has to contain an ordered subfield in order for non-negativity to make sense,[5] and therefore has to have characteristic equal to 0 (since any ordered field has to have such characteristic). This immediately excludes finite fields. $\endgroup$ Jun 18, 2018 at 13:24
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    $\begingroup$ >The basefield has to have additional structure, such as a distinguished automorphism. More generally any quadratically closed subfield of R or C will suffice for this purpose, e.g., the algebraic numbers or the constructible numbers. However in these cases when it is a proper subfield (i.e., neither R nor C) even finite-dimensional inner product spaces will fail to be metrically complete. In contrast all finite-dimensional inner product spaces over R or C, such as those used in quantum computation, are automatically metrically complete and hence Hilbert spaces. $\endgroup$ Jun 18, 2018 at 13:24

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The comments point out is that your original assumption isn't really correct (although the reals and complexes are very common choices).

One thing that you often want is for the field to not have characteristic two.

If you do, the relationship between "squared distance" (by which I mean $\| v \|^2 = v \cdot v$) and inner product gets messed up: you can no longer recover the inner product from the distance via the polarization identity, which reads (in two forms): $$ \| x+y \|^2 = \| x\|^2 + 2 x \cdot y + \|y\|^2 \\ \frac{1}{2}\left( \| x+y \|^2 - \| x\|^2 - \|y\|^2 \right) = x \cdot y $$ Recovering the inner product from the distance requires dividing by $2$, which isn't possible in characteristic $2$. (This peculiar annoyance, for integer quadratic forms, leads down a whole interesting trail to something called the Arf invariant, which actually matters in the topology of 4-manifolds, and perhaps $4k$-manifolds...it's been so long that I've now forgotten.)

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  • $\begingroup$ I've edited my question to account for certain subfields. I guess the motivation for my question is how inner products of complex vector fields are sequilineear, and I wanted to know if there was a generalisation of sequilinearlity for arbitrary fields and what that meant conceptually, even if just to put the complex and real cases in context. But I found that the normal definition apparently implies the base space can only be real or complex (which you have denied). So how do we generalize? $\endgroup$ Jun 18, 2018 at 13:35
  • $\begingroup$ I defer to @LeeMosher, who's more current on such things than I am. It's a pity you changed your question, because it makes the answers you've already gotten look foolish, which isn't very polite. $\endgroup$ Jun 18, 2018 at 13:40
  • $\begingroup$ My apologies, I had thought it was the right thing to do given the site's focus on useful, targeted questions and answers. I thought that the oversight in my original question made it inane and not worthwhile. $\endgroup$ Jun 18, 2018 at 13:42
  • $\begingroup$ What you can do is to keep the original form of the question on top, but then add a new heading like "Additional comments" or "Edited question" below that. $\endgroup$
    – Lee Mosher
    Jun 18, 2018 at 13:56
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If the field is $\mathbb{R}$ or $\mathbb{C}$ you have the order structure of $\mathbb{R}$ at your disposal: directly if you get $\mathbb{R}$ or by requiring that the product is actually hermitian in the case of $\mathbb{C}$.

This allows you to prove Cauchy-Schwarz inequality (you have nothing similar in the context of a generic field), and then it follows that $||x||:={\langle x,x\rangle}^{1/2}$ is actually a norm so, if you define $d(x,y):=||x-y||$, you have a (homogeneous and translation invariant) metric at your disposal on your initial vector space that allows you to use the instruments of metric spaces.

This is not the whole story: using orthogonality (that you can state also in the case of a generic field) and the fact that you have an order structure (and now you need $\mathbb{R}$ or, thanks to the hermitian trick, $\mathbb{C}$) you can prove Bessel inequality, that allows you, for example, to prove Riemann-Lebesgue lemma. Also, if you require the completeness of the metric, what you get from Bessel inequality is that, given an arbitrary orthonormal set, for each element of the space, the corresponding Fourier series is norm convergent and, if the orthonormal set is actually complete, this series converges in norm to the element that originates it: you have got Parseval identity. Then you have the machinery of Fourier series at your disposal.

A final note: you may wonder why not using just an ordered field at this point. In fact, also in this case, you get both a generalized form of Cauchy-Schwarz and Bessel inequality. However, now a problem pops up: if your ordered field hasn't the least upper bound property, you can't get only from Bessel inequality the convergence of the Fourier series. So, because there exists (up to isomorphisms) just only one ordered field that has the the least upper bound property (i.e. $\mathbb{R}$), basically you have that the Fourier series machinery works only if your ordered field is $\mathbb{R}$.

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