MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Math people:

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: someone alerted me to a Wikipedia page that addresses this question ( 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$\ldots$"

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 $\mathbb{C}$.

Stefan (STack Exchange FAN)

share|cite|improve this question
Doesn't the wiki page go a long way towards clearing this up? – Mark Grant May 2 '13 at 12:31
One could easily define an inner product for a vector space over a formally real field ... Your condition would then be $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$ for nonzero vectors $\mathbf x$. – Gerald Edgar May 2 '13 at 12:55
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.) – Henry Cohn May 2 '13 at 14:30
It's also worth keeping in mind that people often look at analogues of symmetric and even Hermitian inner products in cases without positive definiteness. (For example, to define unitary groups over finite fields.) This is certainly somewhat different from the real/complex/quaternionic case, but it's not like there's a sharp dividing line conceptually. Instead, you just give up more properties. – Henry Cohn May 2 '13 at 14:36
@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 $x$. – Stefan May 2 '13 at 15:24

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.

share|cite|improve this answer
@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? – Stefan May 2 '13 at 20:40
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. – Ben McKay May 2 '13 at 21:05
@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). – Stefan May 2 '13 at 21:21
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. – Mlazhinka Shung Gronzalez LeWy May 2 '13 at 22:00
@Franklin : good observation. – Stefan May 4 '13 at 0:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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