A More Abstract Definition of an Inner Product Space?

In summary: The restriction to real and complex numbers is merely done for convenience and because in practice that's what you would be working with, but that it's actually possible to define the concept of inner product space in more abstract terms. I want to know how such a definition would look like.
  • #1
Laton
6
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An inner product space is often simply described as a vector space with the addition of an inner product, but when it comes to the formal definition, the basefield seems to always be restricted to the fields of real and complex numbers. The Wikipedia article on inner product spaces remarks that this is done for "various technical reasons". I understand that the concept of non-negativity makes difficulties; I'm also not sure how conjugate symmetry could be expressed in more abstract terms. What I would like to know is: how would a more abstract formal definition of an inner product space look like? Is it possible at all?
 
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  • #2
Certainly you could define an inner product space over a rational vector space immediately immediately. Let's see what happens

The reason why inner product spaces are so nice is because you can get a metric by defining [tex] |v|=\sqrt{\left<v,v \right>}[/tex] This requires your inner product to take real numbers. Furthermore, you want to be able to normalize vectors to get unit vectors: This requires your field to contain all the real numbers. For example, we could take the set [tex]\mathbb{Q}^2[/tex] with the normal inner product but the vector (1/2,1/2) can't be normalized because that would be [tex] (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})[/tex] which isn't actually an element of our vector space

The only nice fields containing the real numbers are the real numbers and the complex numbers, so those are the two fields you take vector spaces over
 
  • #3
Thank you for your reply.

Office_Shredder said:
The reason why inner product spaces are so nice is because you can get a metric by defining [tex] |v|=\sqrt{\left<v,v \right>}[/tex] This requires your inner product to take real numbers.
I don't think I understand. How is that a metric. This looks like the naturally induced norm of an inner product space. A metric would take two parameters, wouldn't it. I also don't understand why it requires the inner product to take real numbers. The inner product takes vectors, not scalars.

Office_Shredder said:
Furthermore, you want to be able to normalize vectors to get unit vectors: This requires your field to contain all the real numbers. For example, we could take the set [tex]\mathbb{Q}^2[/tex] with the normal inner product but the vector (1/2,1/2) can't be normalized because that would be [tex] (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})[/tex] which isn't actually an element of our vector space
I don't recall any definition of inner product space to mention a norm. If you define a norm on it, I guess it's a normed inner product space. How does that pertain in any way to my question?
And besides, doesn't to normalize a vector simply mean to divide a vector by its norm? So I might be able to choose a norm such that it is possible to normalize all the vectors in the vector space.

Office_Shredder said:
The only nice fields containing the real numbers are the real numbers and the complex numbers, so those are the two fields you take vector spaces over
I understand why you work with real and complex numbers in practice, but that's not what my question is about. I was under the impression that the restriction to real and complex numbers in the definition of inner product space was merely done for convenience and because in practice that's what you would be working with, but that it's actually possible to define the concept of inner product space in more abstract terms. I want to know how such a definition would look like.
 
  • #4
Laton said:
... I want to know how such a definition would look like.

Try the section on Hermitian forms in Serge Lang's Algebra.
 
  • #5
As a concrete example, it is interesting to see how the inner product is generalised to work over quaternions.

First define conjugation by a+bi+cj+dk -> a-bi-cj-dk then the definition of inner product works much the same though you have to take a bit of care to deal with non-commutativity.
 
  • #6
bpet said:
Try the section on Hermitian forms in Serge Lang's Algebra.
There, too, the conjugate symmetry is stated as [tex]\left<x,y\right> = \overline{\left<y,x\right>}[/tex], but I don't really understand what the conjugate means when we're not talking about complex numbers.

BruceG said:
As a concrete example, it is interesting to see how the inner product is generalised to work over quaternions.

First define conjugation by a+bi+cj+dk -> a-bi-cj-dk then the definition of inner product works much the same though you have to take a bit of care to deal with non-commutativity.
That's interesting. So the conjugate can exist for non-complex numbers. Do I have to define it myself or is there a definition that works for all fields?

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I'm also curious to know how many people here think that the restriction to [tex]\mathbb{R}[/tex] and [tex]\mathbb{C}[/tex] is just part of the definition of inner product space and how many people think the concept itself is more abstract and above such restrictions. Most Wikipedia articles I've seen put that restriction in the definition.
 
  • #7
Bowen & Wang (Introduction to Vectors and Tensors, Vol 1) define an inner product only for a vector space over the complex field, or one of its subfields. I haven't yet seen any alternative definition for an inner product space over a general field.
 
  • #8
Laton said:
An inner product space is often simply described as a vector space with the addition of an inner product, but when it comes to the formal definition, the basefield seems to always be restricted to the fields of real and complex numbers. The Wikipedia article on inner product spaces remarks that this is done for "various technical reasons". I understand that the concept of non-negativity makes difficulties; I'm also not sure how conjugate symmetry could be expressed in more abstract terms. What I would like to know is: how would a more abstract formal definition of an inner product space look like? Is it possible at all?

One generalization is a bilinear form. In this case one does not need a field but only a commutative ring. The reals and complexes and their subfields have the notion of a norm, a measure of vector length, that arises from the inner product. With a general bilinear form you may not have a norm.
 
  • #9
Laton said:
That's interesting. So the conjugate can exist for non-complex numbers. Do I have to define it myself or is there a definition that works for all fields?

I'm also curious to know how many people here think that the restriction to [tex]\mathbb{R}[/tex] and [tex]\mathbb{C}[/tex] is just part of the definition of inner product space and how many people think the concept itself is more abstract and above such restrictions. Most Wikipedia articles I've seen put that restriction in the definition.

In the many applications of inner product spaces e.g. functional analysis, quantum mechanics the underlying field just act as the scalars of the theory. Complex numbers are used as the default scalars because they are the unique complete algebraically-closed field so using C guarantees everything will work nicely - there is no really loss of generality in using C as the default field.

One way to generalise the notion of scalar numbers is normed division algebras: there is a procedure called the Cayley-Dickson construction which generates a normed division algebra from ordered pairs of elements of an existing normed division algebra. This gives us the series R->C->H->O. Conjugation at one level is defined in terms of the previous level by (a,b)* = (a*,-b).

Note though that beyond C, we are no longer dealing with fields since they are not commutative.
 
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  • #10
lavinia said:
One generalization is a bilinear form. In this case one does not need a field but only a commutative ring. The reals and complexes and their subfields have the notion of a norm, a measure of vector length, that arises from the inner product. With a general bilinear form you may not have a norm.

My current tentative definitions of form: 1.a. a scalar-valued function of some number of vectors; 1.b. a vector field whose values are such functions; 2.a. an alternating, continuous, linear scalar-valued function of some number of tangent vectors to a manifold; 2.b. a vector field whose values are such functions; 3. as in non-technical English.

From this, it seems to be also used, in a generalisation of 1.a., for a function [itex]f:M \times ... \times M \rightarrow R[/itex] where [itex]M[/itex] is the underlying set of a module over a commutative ring, and [itex]R[/itex] that of its base ring.
 
  • #11
A bilinear form is an R-bilinear map
[tex]V\times W\to R,[/tex]
where V is a left R-module and W a right R-module.
 
  • #12
Thanks for all the replies.

Rasalhague said:
Bowen & Wang (Introduction to Vectors and Tensors, Vol 1) define an inner product only for a vector space over the complex field, or one of its subfields. I haven't yet seen any alternative definition for an inner product space over a general field.
I see. That makes sense. I see now why it is difficult to define it in more abstract terms.

lavinia said:
One generalization is a bilinear form. In this case one does not need a field but only a commutative ring. The reals and complexes and their subfields have the notion of a norm, a measure of vector length, that arises from the inner product. With a general bilinear form you may not have a norm.
But an inner product is not bilinear, it's sesquilinear. Am I missing something here? A bilinear form may be more general, but I wasn't asking for more general, I was asking for more abstract.

BruceG said:
In the many applications of inner product spaces e.g. functional analysis, quantum mechanics the underlying field just act as the scalars of the theory. Complex numbers are used as the default scalars because they are the unique complete algebraically-closed field so using C guarantees everything will work nicely - there is no really loss of generality in using C as the default field.

One way to generalise the notion of scalar numbers is normed division algebras: there is a procedure called the Cayley-Dickson construction which generates a normed division algebra from ordered pairs of elements of an existing normed division algebra. This gives us the series R->C->H->O. Conjugation at one level is defined in terms of the previous level by (a,b)* = (a*,-b).

Note though that beyond C, we are no longer dealing with fields since they are not commutative.
But if we no longer have a field, we don't have a vector space. But the inner product space without the inner product should still be a vector space.

---

Another question I have:
An inner product space is a vector space together with an inner product. How is the "together with" formalized? Is an inner product space an ordered pair? Or maybe a 4-tuple (vector set, vector addition, scalar multiplication, inner product)?
 
  • #13
Laton said:
I wasn't asking for more general, I was asking for more abstract.
You asked for something like an inner product space where the associated field isn't necessarily the real numbers or the complex numbers. I would describe that as "more general", not "more abstract". I don't know what "more abstract" would mean here.

If V is a vector space over a field F, and t is an automorphism of F, you could define your "inner product" as a map [itex](x,y)\mapsto\langle x,y\rangle[/itex] such that

[tex]\langle x+y,z\rangle=\langle x,z\rangle+\langle y,z\rangle[/tex]

[tex]\langle x,y+z\rangle=\langle x,y\rangle+\langle x,z\rangle[/tex]

[tex]\langle ax,by\rangle=t(a)b\langle x,y\rangle[/tex]

for all x,y,z in V and all a,b in F, and also

[tex]\forall x\langle x,y\rangle=0\Rightarrow y=0[/tex]

[tex]\forall y\langle x,y\rangle=0\Rightarrow x=0[/tex]

I stumbled across something similar today, in a book on quantum logic. It's on the middle of page 23. I haven't seen anything like this anywhere else, so I don't know how useful it is. Note that the author is talking about a "vector space" over a division ring, not a field. I think most people would call that a module, not a vector space. A division ring is like a field, except that multiplication isn't necessarily commutative.

Edit: I suppose we also need something that corresponds to [itex]\langle x,y\rangle^*=\langle y,x\rangle[/itex]. The first idea that occurs to me is [itex]t(\langle x,y\rangle)=\langle y,x\rangle[/itex].

Laton said:
An inner product space is a vector space together with an inner product. How is the "together with" formalized? Is an inner product space an ordered pair? Or maybe a 4-tuple (vector set, vector addition, scalar multiplication, inner product)?
I usually define a vector space as a 4-tuple (V,F,a,s), where V is the set, F is the field, a is the addition operation and s is the scalar multiplication function (i.e. multiplication by a scalar). That makes it convenient to define an inner product space as a pair (X,i) where X is such a 4-tuple and i is the inner product. But it really doesn't matter if you do it this way or any of the other ways which seem reasonable to you, like defining the inner product space to be the 5-tuple (V,F,a,s,i). Instead of including the field F in the list, you could mention its underlying set along with the addition and multiplication operations on it, and if you want to, also the multiplicative inverse operation and additive inverse operation. You have lots of options, and it doesn't matter much which one you choose. The only reason to prefer some of the options over the others is that some of them might make it obvious which functions between two structures of the type we're defining we intend to call "homomorphisms", while others don't.
 
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  • #14
BruceG said:
Note though that beyond C, we are no longer dealing with fields since they are not commutative.

Umm, how about the field of rational functions?

Laton said:
There, too, the conjugate symmetry is stated as [tex]\left<x,y\right> = \overline{\left<y,x\right>}[/tex], but I don't really understand what the conjugate means when we're not talking about complex numbers.

I guess, every conjugate would be of the form [tex]\overline{x}=y-z[/tex] for some y,z with x=y+z (at least, in all rings for which [tex]1+1[/tex] is invertible). But then if we defined, say, [tex]\overline{\sqrt{2}}=-\sqrt{2}[/tex] this would imply that the hermitian form is trivially zero (so it fails the +ve definite axiom for inner products). I wonder, what field elements other than [tex]\sqrt{-1}[/tex] can satisfy [tex]\bar{z}=-z[/tex] and still allow an inner product to be defined?
 
  • #15
Thanks for the replies.

Fredrik said:
You asked for something like an inner product space where the associated field isn't necessarily the real numbers or the complex numbers. I would describe that as "more general", not "more abstract". I don't know what "more abstract" would mean here.
When I wrote "more abstract" I imagined – with hindsight, this may have been a silly idea – "inner product space" being defined for any arbitrary field so that when applied to the complex field, everything would automatically fall into place in such a way that the resulting definition would be equivalent to the common definition of the concept, the one where the field is restricted to the complex numbers in the first place. I hope that I'm making sense.
 
  • #16
Laton said:
but I don't really understand what the conjugate means when we're not talking about complex numbers.
Field automorphisms! An invertible function s from your field to itself with the property that
  • s(x+y) = s(x) + s(y)
  • s(xy) = s(x) s(y)

Some examples:
  • Complex conjugation
  • In the field of numbers of the form [itex]a + b \sqrt{2}[/itex], the transformation that sends [itex]\sqrt{2}[/itex] to [itex]-\sqrt{2}[/itex]
  • In the field of rational functions in x over any field where [itex]2 \neq 0[/itex], the transformation sending x to 2x+3.
  • In a finite field of characteristic p, every automorphism is either the map sending a to ap, or one of its iterates.

Often, you are interested in fixing some field. The complex numbers have gazillions of automorphisms. But only two of them leave the real numbers unchanged. (the identity map and complex conjugation) The real numbers are rather weird -- the only automorphism of the reals is the identity map.
 
  • #17
Thanks. That sort of makes sense. I guess I have to read up on field automorphisms.
 
  • #18
Laton said:
Thanks. That sort of makes sense. I guess I have to read up on field automorphisms.

Specifically, automorphisms of order <=2. The identity [tex]2x = (x+\bar{x})+(x-\bar{x})[/tex] will come in handy, as will the properties of ordered fields (for positiveness to be well defined).

HTH
 

Related to A More Abstract Definition of an Inner Product Space?

1. What is an inner product space?

An inner product space is a mathematical concept that consists of a vector space and an inner product defined on that space. The inner product is a mathematical operation that takes two vectors as inputs and produces a scalar value as output. It is a generalization of the dot product in three-dimensional Euclidean space.

2. How is an inner product space different from a vector space?

An inner product space differs from a vector space in that it has an additional mathematical operation, the inner product, defined on it. This operation allows for the measurement of angles, lengths, and orthogonality between vectors in the space.

3. What are the properties of an inner product in an inner product space?

The inner product in an inner product space must satisfy several properties, including linearity in the first argument, conjugate symmetry, and positive definiteness. These properties ensure that the inner product is a meaningful and useful operation in the space.

4. Why is an abstract definition of an inner product space useful?

An abstract definition of an inner product space allows for the generalization of the concept to spaces other than three-dimensional Euclidean space. This allows for the use of inner products in a wide range of mathematical fields, including linear algebra, functional analysis, and quantum mechanics.

5. Can an inner product space be finite-dimensional?

Yes, an inner product space can be finite-dimensional. In fact, most of the commonly used inner product spaces, such as Euclidean space and complex vector spaces, are finite-dimensional. However, the concept of an inner product space is not limited to finite dimensions and can also be applied to infinite-dimensional spaces.

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