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Let $V$ be a vector space over a field $K$ and suppose we wish to define an inner product on $V$. Most books define inner products on $\mathbb R\mkern-0.5mu$- or $\mathbb C\mkern-0.5mu$-vector spaces only, so we must generalize the standard definition a bit.

In particular, we need $K$ to have an involution $*:K\to K,z\mapsto z^*$ and we need the subset $P=\bigl\{\langle x\,|\, x\rangle \,\bigm|\, x\in V\bigr\}\subseteq K$ to be ordered, so that positive-definiteness to make sense.

Consider the following tentative restrictions:

  1. $K$ has an involution (automorphism of characteristic $2$), $*:K\to K,z\mapsto z^*$.
  2. $\mathbb R\subseteq K$.

Under these restrictions, the standard definition of an inner product makes sense (replacing the complex conjugate function with the $*$ function above, and interpreting $0\leqslant a\in K$ to mean $a\in\mathbb R^+\subseteq K$, as was already the case for $\mathbb C$). My question is, can these restrictions be relaxed even further while having the inner product axioms still make sense and induce a norm?

In particular, I feel restriction $(2)$ above might be too strong. Perhaps the set $P$ defined above can be given a natural ordering so that positive-definiteness makes sense? Perhaps $\mathbb R$ is unnecessarily large? But note that $\mathbb Q$ may be too small because then square roots might not always be defined, so the standard norm $\|x\|=\sqrt{\langle x|x\rangle}$ is ill defined.

Background:

While similar questions have been asked twice on Math.SE, they have unfortunately gone unanswered, or answered under the false assumption that $K$ must be an ordered field (obviously wrong because $\mathbb C$ works just fine). The unanswered one links to a similar Mathoverflow question in which the answers also make this false assumption. Those answer also focus relaxing the definition of inner products to apply to more general fields, rather than restricting the fields such that the standard definition still applies to them; thus I feel my question as asked here is sufficiently different. I would like to see most generally what fields other than $\mathbb R$ and $\mathbb C$ can easily fit the standard definition.

Bernard
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WillG
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1 Answers1

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If you're absolutely intent of defining a norm by taking a square root, you can ask that the fixed field $K_0\subset K$ of the involution is Euclidean. That way you can always take a square root of $\langle x,x\rangle$ if you require that your hermitian form is definite positive.

This being said, I would like to point out that while geometrically it is of course useful to define such a norm, in algebraic settings we usually don't mind getting quadratic, or even higher-order norms. I don't know what applications you have in mind, but I think you should try to consider the possibility of just not taking a square root at all, which allows more general settings.

Captain Lama
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  • My abstract algebra is a bit rusty: Is the fixed field of an automorphism of characteristic $2$ guaranteed to be ordered? – WillG Apr 14 '20 at 21:47
  • Absolutely not. But a Euclidean field is, so "ordered" is not an extra assumption. – Captain Lama Apr 14 '20 at 21:48
  • One of my goals is to generalize Hilbert spaces to more general fields. So I at least need the inner product to induce a norm/metric/topology. – WillG Apr 14 '20 at 21:48
  • Also is every Euclidean field isomorphic to a unique subfield of $\mathbb R$? Because then at least there would be a canonical map into $\mathbb R$ that could be used define the generalized norm. – WillG Apr 14 '20 at 21:52
  • Just for consideration: you can define the usual euclidean topology without taking any square root. Using the "squared" norm is perfectly fine. – Captain Lama Apr 14 '20 at 21:53
  • Interesting. Suppose I define Hilbert spaces as inner product spaces whose induced "squared" norm gives rise to a complete metric space. Does that break anything important, and more specifically, is the "squared-norm-induced-topology" always the same as the "traditional-norm-induced-topology"? – WillG Apr 14 '20 at 22:07