I've been studying basic concepts of inner product vector space, normed vector space and metric space. And all the inner products, norms and metrics are defined to be real-valued functions in my textbook. I doubted about it: why do I have to restrict the measurement standard to real number? Why not any ordered field? Why not… anything else?

I do know that there can be a lot of different ways to define 'distance': I'm studying topology, too. I'm just wondering if there is any more general definition to the inner product space, normed space, or metric space(my definition is just the standard one being taught at undergraduate level: only real-valued ones as I've mentioned).

For example, let $V$ be an $F$-vector space and $F_s$ be an ordered field embedded into $F$. Then I defined like this:

  1. $\langle v, v\rangle=0$ if and only if $v$ is the $0$ vector
  2. $\langle v, v\rangle>0$ (so the value is in $F_s$)
  3. $\langle av, u\rangle=a\langle v, u\rangle$ where $a\in F$
  4. $\langle v+u, z\rangle=\langle v, z\rangle+\langle u, z\rangle$
  5. $\langle v, u\rangle=\langle u, v\rangle$

We can define the normed vector space in this fashion as well. I think this is a more general definition: I couldn't find anything wrong about it. I mean, if we define $\|v\|=\langle v, v\rangle^{1/2}$ then indeed $(V, \|\,\|)$ becomes a normed vector space according to my definition, as it should in the ordinary definition (I think the Cauchy-Schwarz inequality which connects the two spaces also hold in this general definition: I've checked the proof in my textbook and it does not use any property of the real number).

Is my definition not right, or not useful? If so, then why? I know that the real number is the only (up to isomorphism) ordered field with the least upper bound property and the property that every increasing bounded sequence converges. But is that enough to justify that every metric spaces use real number? I want to be more convinced, then.

A little bit disorganized, but I just wanted to hear some other people's opinion about this. Thanks as always.

P.S. I also know about the complex inner product vector space. The inner product there is complex-valued, so I think it was the start of my questioning.

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    $\begingroup$ Your definition of the length of a vector as the square root of an inner product takes you outside the original scalar field unless the positive elements of the field all have square roots in the field. Most ordered fields do not have that property! Can you think of such an ordered field besides the real numbers? $\endgroup$ – KCd Dec 13 '14 at 12:57
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    $\begingroup$ If you replace your first two conditions by a condition called nondegeneracy ($\langle u,v\rangle = 0$ for all $v$ implies $u = 0$) then you have the concept of a nondegenerate symmetric bilinear form, which does not need the scalar field to be ordered at all. $\endgroup$ – KCd Dec 13 '14 at 13:00
  • $\begingroup$ Hm.. yes, come to think of it... the definition is very clumsy... $\endgroup$ – Taxxi Dec 13 '14 at 13:19
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    $\begingroup$ You'd have a hard time proving the existence of a bilinear form satisfying your conditions in the case $F=\mathbb{C}$ and $F_s=\mathbb{R}$. $\endgroup$ – egreg Dec 13 '14 at 15:27
  • $\begingroup$ Yes indeed. I think I cannot even try to prove it since my knowledge about fields is very shallow. Your example was the very question I had just after I came up with the definition above: if it were possible, then why do we use the complex inner product space? So I actually tried to see if it is possible but just gave it up since I don't have much spare time now. Also thanks for your edit! $\endgroup$ – Taxxi Dec 13 '14 at 15:33

why do I have to restrict the measurement standard to real number? Why not any ordered field?

You can use any ordered field, and the axioms will still make sense. The thing is, though, that most of our geometric intuition is built on the Archimedian property. But if we include this in our field, then it becomes a subfield of $\Bbb R$. In this case, we are really not restricting ourselves to $\Bbb R$ at all: it's the largest Archimdean ordered field!

Thinking about ordered fields that aren't Archimedian gets a little more weird. Are you prepared, for example, to have vectors $v,w$ such that $v$ and $w$ point in the same direction, and yet $vn$ is shorter than $w$ for every natural number $n$?

I mean, if we define $\|v\|=\langle v, v\rangle^{1/2}$ then indeed $(V, \|\, \|)$ becomes a normed vector space

No, not quite! Normally we want the norms to be in the base field, but the vector $(1,1)\in \Bbb Q^2$ would have a norm outside of $\Bbb Q$, with that definition. To fix things, you'd need something called a Pythagorean field, and that's enough to guarantee this definition of norm works.

If you don't ask for the norm to be in the base field, then you may not be able to carry out normalization, because dividing by the norm won't give you a vector in the field.

I know that the real number is the only (up to isomorphism) ordered field with the least upper bound property and the property that every increasing bounded sequence converges. But is that enough to justify that every metric spaces use real number? I want to be more convinced, then.

Yes, your intuition is along the right lines. The fact that $\Bbb R$ is "maximal" among Archimedian ordered fields makes it special. It's a smooth connected piece with no holes. In geometry this is important since it ensures that lines and circles cross where you expect them to. For example in $\Bbb Q^2$, you can find an example of a line and a circle that would intersect in $\Bbb R^2$, but they pass through each other in $\Bbb Q^2$ without touching.

Why not... anything else!!

Actually, geometers study generalizations of the norm in the form of bilinear forms over any field. The idea is that rather than focus on the norm, you instead focus on a (generalized) inner product. They can be quite different from what you're used to with run-of-the-mill normed real spaces. They can have, for example, nonzero vectors with length $0$ or even negative length.

Even more than that, "length" loses meaning totally when you're working in a field that isn't ordered: there's no such thing as positive or negative, there. Still, there is a huge theory for these types of spaces with (generalized) inner products.

  • $\begingroup$ Thanks. I indeed have proved that if a field is not Archimedian, it should have an element that is smaller than any integer fraction or an element greater than any integer. I tried to find a concrete example of such fields but Google only gave me examples of such fields that I cannot understand fully yet. I haven't proved that $R$ is maximal among the Archimedean fields, though. If the proof is at around my level (not that difficult), I might try it or find a reference of it: quite interesting :) $\endgroup$ – Taxxi Dec 13 '14 at 13:26
  • $\begingroup$ Also indeed Pythagorean field is the exact property that should be required in my definition! That's also interesting. So the conclusion seems to be, yes we can consider more general cases but still $R$ is so unique and has so many good properties that structures based on the real number deserves much attention, so that's why I'm seeing every structures being related to the real number at my level. $\endgroup$ – Taxxi Dec 13 '14 at 13:31
  • $\begingroup$ @TaxxiDriver Yes, it's just "the nicest" field to have when working with vectors using geometric intuition. $\endgroup$ – rschwieb Dec 13 '14 at 13:34
  • $\begingroup$ Another way to generalise normed vector spaces is when the underlying field is a normed field. This allows us to define a $k$-norm as a function into reals satisfying $\lVert av\rVert=\lvert a\rvert\cdot\lVert v\rVert$ (in addition to obvious axioms). If $k$'s norm is onto ${\bf R}_+$, you can then do normalisation (although it is non-unique, and most of the time this condition implies that $k$ is wildly inseparable, unless it is actually a subfield of the complex numbers with standard norm). Of course, this does not get us very far from reals in some ways. But then there are valued fields... $\endgroup$ – tomasz Dec 14 '14 at 6:49
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    $\begingroup$ The problem with Archimedeanness of real-closed fields is actually not so simple. Non-standard analysis approach, for example, redefines not only reals but ℕ too, making the answer dependent on whether we base Archimedeanness on standard natural numbers or “topical” ones. $\endgroup$ – Incnis Mrsi Dec 14 '14 at 12:24

You can define inner products for vector spaces and normed vector spaces over any ordered field.

You can define metrics over any ordered field: they are called generalised metrics.

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  • $\begingroup$ Thanks. I think I have to study more about fields themselves or bilinear... stuff or something like that. I get the idea that indeed any ordered field (possibly with some more conditions) can be used, though. $\endgroup$ – Taxxi Dec 13 '14 at 13:09
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    $\begingroup$ Why does it have to be a field for generalized metrics? As far as I can see, the metric axioms only use addition, not multiplication, so why additionally require a multiplication? $\endgroup$ – celtschk Dec 13 '14 at 18:57
  • $\begingroup$ @celtschk One can do bilinear forms for arbitrary commutative rings, but the theorems you can prove dwindle. Kaplansky talks about this in his book Linear algebra and geometry: a second course. He proves a few interesting theorems for bilinear forms over commutative local rings. But fields are the nicest since they allow $R^n$ to have all the transitivity properties one would hope for in a geometry. $\endgroup$ – rschwieb Dec 13 '14 at 20:48
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    $\begingroup$ @celtschk: You're right, there exist some papers which consider "metrics" with values in a complete lattice with a binary operation ("addition"). $\endgroup$ – mbork Dec 13 '14 at 21:17

Not any ordered field possesses the square root operation defined wherever $x ≥ 0$; rational numbers form an obvious counter-example known from ancient times. Yes, it is essentially the same problem as famous one faced by Pythagoreans, but in modern formulation: how to define number field to make things going well with geometry?

There can be numerous ways to tweak definition of an inner-product space, at expense of some nice properties (metric completeness, separability, symmetry of the inner product… ). We are even not obliged to compute the norm in the same field as the ground field of our vector space; the concept of valued field provides a possibility to do otherwise.

Real numbers are so popular not because they form “very correct” field, but because their arise from requirements deemed very natural in analysis: order, metric completeness, separability. Wherever you can sacrifice any of it, you are welcome to work over alternative fields.


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