The title is the question: What fields can be used for an inner product space?

This question has been discussed in Math Stack Exchange with no definitive resolution. A similar question appeared here, and an answer was accepted, but someone pointed out a serious problem with the answer.

I am using the standard definition of inner product, which includes $\langle \mathbf{x}, \mathbf{x} \rangle > 0$ for all non zero vectors $\mathbf{x}$.

It seems to me that any field of prime characteristic does not make sense, because it does not have an order relation that respects addition. It also seems to me that the field $\mathbb{F}$ can be any ordered field or any subfield of $\mathbb{C}$ that is stable under complex conjugation (for any non-algebraists like me, the word "stable" seems to be standard here. Anyone else would use the word "closed"). I do not know if any other fields are possible. Of course, an ordered field may or may not be a subfield of $\mathbb{R}$.

It seems to be rare for people, even mathematicians, to use any field other than $\mathbb{R}$ or $\mathbb{C}$ for an inner product space.

Can anyone clear this up?

EDIT: Mark Grant alerted me to a Wikipedia page that addresses this question (http://en.wikipedia.org/wiki/Inner_product_space#Definition). Let me quote the relevant part : "There are various technical reasons why it is necessary to restrict the basefield to $\mathbb{R}$ and $\mathbb{C}$ in the definition. Briefly, the basefield has to contain an ordered subfield[citation needed] (in order for non-negativity to make sense) and therefore has to have characteristic equal to 0 (since any ordered field has to have such characteristic). This immediately excludes finite fields. The basefield has to have additional structure, such as a distinguished automorphism. More generally any quadratically closed subfield of $\mathbb{R}$ or $\mathbb{C}$ will suffice for this purpose, e.g., the algebraic numbers…."

The Wikipedia article fails to explain why the basefield has to have additional structure. They do not define a "distinguished automorphism" or provide a link to a definition. I am not an algebraist. I Googled the term and I could not find a definition of "distinguished automorphism". I did find links to papers and books that probably do contain a definition. The article states it is "necessary" to restrict the basefield to $\mathbb{R}$ and $\mathbb{C}$ in the definition but then contradicts itself by at least suggesting that the basefield can be any quadratically closed subfield of $\mathbb{R}$ or $\mathbb{C}$.

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    $\begingroup$ Doesn't the wiki page en.wikipedia.org/wiki/Inner_product_space#Definition go a long way towards clearing this up? $\endgroup$
    – Mark Grant
    May 2, 2013 at 12:31
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    $\begingroup$ One could easily define an inner product for a vector space over a formally real field en.wikipedia.org/wiki/Formally_real_field ... Your condition would then be $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$ for nonzero vectors $\mathbf x$. $\endgroup$ May 2, 2013 at 12:55
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    $\begingroup$ In the Hermitian case "field" is overly restrictive (the skew field of quaternions is important too). Even in the symmetric case, it can be convenient to look at rings: for example, integral lattices are often thought of as free abelian groups with integer-valued, positive-definite, symmetric bilinear forms. (Of course integers are real numbers, so this can be thought of as a special case of real inner products, but the restriction to a subring changes how things feel.) $\endgroup$
    – Henry Cohn
    May 2, 2013 at 14:30
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    $\begingroup$ @darij : Thanks. Apparently "ordered field" refers to a field together with the order relation while a "formally real" field is a field that can be endowed with an order relation with the required properties. I don't really understand your first comment. I don't see an obvious order relation for formal real-valued rational functions in real variables $x$ and $y$ similar to the standard order relation for rational functions in just $x$. $\endgroup$
    – Stefan
    May 2, 2013 at 15:24
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    $\begingroup$ @Stefan: Because the order is in general not unique. In order for the condition to make sense, you have to fix a particular order, but then you can as well come out and say that the structure you are working with is an ordered field, rather than just a formally real field. By the way, the order on a formally real field is unique if and only if every element or its negation can be written as a sum of squares, but I don’t know if such fields have a special name. $\endgroup$ May 2, 2013 at 20:26

2 Answers 2


Of course if you insist on condition $\langle \mathbf{x},\mathbf{x}\rangle > 0$, and not merely $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$, then you must have an order.

Let $F$ be a formally real field. Then $$ \langle \mathbf{x}, \mathbf{y}\rangle = \sum_{j=1}^n x_j y_j $$ can be a reasonable inner product on $F^n$. According to an ordering for $F$ (indeed, any ordering, since there may be more than one) we have $\langle \mathbf{x}, \mathbf{x}\rangle > 0$ if $\mathbf x \ne \mathbf 0$.

Another part that you quote is what would be required for metric completeness. Do you want that? If $F$ is a proper subfield of $\mathbb R$, then even the one-dimensional space is not complete.

Something weaker than completeness will be enough to carry out the Gram-Schmidt process. It requires only that square-roots of $\langle \mathbf{x}, \mathbf{x}\rangle$ exist.

  • $\begingroup$ @Gerald : Thanks. I don't really care about metric completeness for this question. Either I forgot to omit that part from the Wikipedia quote, or someone put it back in. What about the mysterious "additional structure, such as a distinguished automorphism", that the Wikipedia article refers to? Do we need to impose that, is it there already, or is the article wrong? $\endgroup$
    – Stefan
    May 2, 2013 at 20:40
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    $\begingroup$ I think that the additional structure is supposed to be complex conjugation. Recall that a Hermitian inner product changes by conjugation when you swap its arguments. So you might allow yourself the freedom to pick some map from the field to itself to play the role of complex conjugation. $\endgroup$
    – Ben McKay
    May 2, 2013 at 21:05
  • $\begingroup$ @Gerald: I don't even care for this question whether square roots of $\langle \mathbf{x},\mathbf{x}\rangle$ exist, or whether you can carry out the Gram-Schmidt process, just the basic definition of inner product (with positive-definiteness). $\endgroup$
    – Stefan
    May 2, 2013 at 21:21
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    $\begingroup$ Orthogonal bases are useful enough and perhaps you are happy with them not being of normalized vectors. You may run Gram-Schmidt without normalization and I think you don't need square roots of positive elements for that. $\endgroup$
    – O.R.
    May 2, 2013 at 22:00
  • $\begingroup$ @Franklin : good observation. $\endgroup$
    – Stefan
    May 4, 2013 at 0:48

The positive definiteness axiom can be replaced with an axiom asserting the existence of an "orthonormal basis".

Let $R$ be a $*$-ring. We consider a left $*$-module over $R$ which we call $M$. We ask what it means for a Hermitian form $\langle -, - \rangle$ to be an inner product over $M$.

In finite dimensions, you would say that $\langle -, - \rangle$ is an inner product if there exists a finite basis $B=\{b_1,b_2,\dotsc,b_n\}$ of $M$ such that $\langle b_i, b_j \rangle = \delta_{ij}$ where we have used the Kronecker delta symbol.

In possibly infinite dimensions, you would ask that your "orthonormal basis" only be "densely spanning". What I mean by this is that there should exist a subset $B \subset M$ for which $\langle b, b' \rangle=\begin{cases}1, & b = b' \\ 0, & b \neq b' \end{cases}$, and for which every $0 \neq v \in M$ should admit some $b \in B$ for which $\langle b, v \rangle \neq 0$.

For the usual $*$-rings of $\mathbb R$ and $\mathbb C$ (with their usual involutions) this recovers the usual notion of an inner product space. The proof uses Zorn's Lemma.

In foundations not admitting the Axiom of Choice, this definition may be superior to the usual one. It's certainly easier to do maths over an inner product space once you have an orthonormal basis.

In answer to your question, the appropriate setting is $*$-fields, and not fields. And all $*$-fields can admit inner product spaces.

  • $\begingroup$ It seems that, at least in finite dimensions, the datum of "vector space with specified orthonormal basis" is little more than "vector space with basis": if I just have a basis, then it is orthonormal for an appropriate inner product (at least if we agree that the ground $*$-field supports inner products). So what do I gain by calling the basis orthonormal rather than just a basis? Or, perhaps stated differently, what do I gain by restricting my attention to these sorts of ground $*$-fields? $\endgroup$
    – LSpice
    Feb 4, 2023 at 14:50
  • $\begingroup$ @LSpice It's not a "vector space with specified basis", it's actually a vector space and a Hermitian form which admits some basis which is orthonormal with respect to it. The point being, that given an inner product, there may be lots of bases which are orthonormal with respect to it, and we don't care which one it is. $\endgroup$
    – wlad
    Feb 4, 2023 at 15:00
  • $\begingroup$ I don't know what you mean by "these sorts of ground $*$-fields". I'm suggesting that you can use any and all $*$-fields whatsoever. There are still unique advantages though to $\mathbb R$, $\mathbb C$ and $\mathbb H$ (equipped with their usual involutions), as expressed by Solèr's theorem. $\endgroup$
    – wlad
    Feb 4, 2023 at 15:04
  • $\begingroup$ @LSpice I've read your comment and I still don't understand it. I think you've misunderstood something in what I said. Can you clarify? $\endgroup$
    – wlad
    Feb 4, 2023 at 19:02
  • $\begingroup$ You seemed to propose "VS with Hermitian form that admits an orthonormal basis". I am saying that, even if we strengthen that to "VS with a Hermitian form and a preferred orthonormal basis", then it seems that we are doing nothing more than specifying an abstract vector space with a preferred basis, because you can recover the Hermitian form from the basis and "sequilinearity". So it seems at first that anything one could prove about general vector spaces with Hermitian forms using orthonormal bases could be proven about bare vector spaces using bases, and I am asking why that's not so. $\endgroup$
    – LSpice
    Feb 4, 2023 at 19:16

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