Purpose of Field and Rings

[conway]

Do they not exist not only as abstract toys for mathmaticans, but also as tools to describe our "space time" continuum and the realitys and interactions that exist therein. If the space time contiuumn is measured "relatively" speaking from a "perspective". Then should not operations that exist amongst elements and vectors in fields and rings also exist as relative to a perspective? If the idea of all field and rings and the study of infinity in these given fields and rings, is described as "successors and predecessors related by morphisms (functions)," then should not zero, existing as an "element" in fields and rings, not also posses a "successors and predecessors related by morphisms (functions)," so as to approach it on these "given" fields and rings?

[Xerxes]

Conway, first note that in mathematics the term "field"is context-dependent. In abstract algebra a ring is defined as a set together with a pair of closed binary operations (typically + and x), each with an identity (respectively here 0 and 1) and an inverse for the operation + but none for x.

 

A field in abstract algebra is defined as a ring with the additional property of an inverse for the operation x.

 

Rings and fields in this sense comprise numbers of various sorts (mathematicians call them scalars) which, in applications, are generally taken to be Real or Complex.

 

In linear algebra, on the other hand, a field assigns to each and every point in some chosen space a value, generally scalar-, vector- or tensor-valued. Here relativistic effects do NOT apply to scalar fields.

 

So to your question: In the sense of abstract algebra, relativistic considerations are out of context - they simply do not apply.

 

Moreover, the assertion that every set contains the empty set as a predecessor set can be used in the set-theoretic construction of the Ordinals and not, as far as I am aware, in the construction of a field of any description.

 

The convention that, in the construction of the Ordinals, the empty set be re-labelled as 0 and its successor set as 1 etc is carried over to the Real numbers (with 0 as a field additive identity) seems to depend only on the simple observation that for any set [math]S[/math] that [math]S \cup \emptyset = S[/math].

 

Confused? You should be! This is a HUGE area of mathematics, which some of us have spent (wasted?) years studying

Edited January 1, 2016 by Xerxes

[conway]

Xerxes

 

Excellent reply! Thank you.

 

 

 

"In linear algebra, on the other hand, a field assigns to each and every point in some chosen space a value, generally scalar-, vector- or tensor-valued. Here relativistic effects do NOT apply to scalar fields."

 

 

In this sentence you chose to use the words value, and that these values are assigned to a "point in chosen SPACE". So then if the space time continuum is "space", with the "value" being time, and it is relative. Then should not all "mathematical" fields represent this intrinsic property of "space" and "value" - time or otherwise" being relative. For example. If we allow that all numbers are defined as value and space, and that , that value or space is defined or undefined, infinite or finite, fractional or whole, positive or negative, then it is possible to show a relativistic relationship with all multiplication and division......that is your "functions" used in fields and rings. Naturally a redefining of numbers is necessary.

 

 

 

So then all numbers are defined as being composites of Space and Value.

Space being labeling of quantities of dimensions

Value being labeling of quantities of existence other than dimensions

 

If you will Xerxes you may take the following as an addition to all current field axioms.

 

For every A in S there exist a Z1(space) and a Z2(value), such that any A in operation of multiplication and division is only representing Z1 or Z2 in any given equation.

 

 

 

Naturally this axiom is tentative. I could pen it down further, but I will wait and see what happens here. Confused? You should be, Ive spent much time thinking about this idea of "relative mathematics" (maybe wasted).

Edited January 1, 2016 by conway

[studiot]

Conway, first note that in mathematics the term "field"is context-dependent. In abstract algebra a ring is defined as a set together with a pair of closed binary operations (typically + and x), each with an identity (respectively here 0 and 1) and an inverse for the operation + but none for x.

 

A field in abstract algebra is defined as a ring with the additional property of an inverse for the operation x.

 

Rings and fields in this sense comprise numbers of various sorts (mathematicians call them scalars) which, in applications, are generally taken to be Real or Complex.

 

In linear algebra, on the other hand, a field assigns to each and every point in some chosen space a value, generally scalar-, vector- or tensor-valued. Here relativistic effects do NOT apply to scalar fields.

 

So to your question: In the sense of abstract algebra, relativistic considerations are out of context - they simply do not apply.

 

Moreover, the assertion that every set contains the empty set as a predecessor set can be used in the set-theoretic construction of the Ordinals and not, as far as I am aware, in the construction of a field of any description.

 

The convention that, in the construction of the Ordinals, the empty set be re-labelled as 0 and its successor set as 1 etc is carried over to the Real numbers (with 0 as a field additive identity) seems to depend only on the simple observation that for any set [math]S[/math] that [math]S \cup \emptyset = S[/math].

 

Confused? You should be! This is a HUGE area of mathematics, which some of us have spent (wasted?) years studying

 

I think you might like to check your mathematics dictionary as I would say you have the underlined almost the other way round.

 

However I agree that the topic is hugely enormous and very confusingly full of special terminology.

 

For this reason I offered (and unfortunately was rejected) my rough guide to sets in pengkuan's recent two threads.

 

In fact there are several recent threads going about this very subject and its terminology by different authors so I was going to point this out in conway's last thread, untill it was closed a second time, though I do not know why.

[Xerxes]

I think you might like to check your mathematics dictionary as I would say you have the underlined almost the other way round.

So, you truly believe that I learned my mathematics from some "dictionary" ? I can assure you this is not the case.

 

I can also assure you that the definition I gave of rings vs, fields is correct.

 

If you want a lesson on integral domains, rings and fields just ask. Otherwise, do the other thing......

[studiot]

There is no call to be rude about it.

 

Just tell me what the multiplicative inverse of the zero element is in the field of real numbers please.

 

That is what is 1/0 ?

Edited January 1, 2016 by studiot

[andrewcellini]

There is no call to be rude about it.

 

Just tell me what the multiplicative inverse of the zero element is in the field of real numbers please.

 

That is what is 1/0 ?

quote from wiki:

 

Finally, a field is a commutative ring in which one can divide by any nonzero element: an example is the field of real numbers.

 

https://en.wikipedia.org/wiki/Ring_(mathematics)

[ajb]

Do they not exist not only as abstract toys for mathmaticans, but also as tools to describe our "space time" continuum and the realitys and interactions that exist therein.

Fields and rings are some of the basic algebraic structures that are used to build all sorts of further mathematical structures, some of which can be directly used in Einsteinian relativity.

 

 

If the space time contiuumn is measured "relatively" speaking from a "perspective". Then should not operations that exist amongst elements and vectors in fields and rings also exist as relative to a perspective?

No, the operations that define a given ring or field are fixed and that is it. It is how you use these structures in relativity that is important, but the basic structures themselves are not changed.

 

 

In the context of relativity, one is usually dealing with smooth real Riemannian manifolds and sometimes complex ones. The basic structure of a manifold, in particular its structure sheaf (basically the ring of functions) is given and fixed. The tensor algebra and so on is also defend from the smooth manifold structure. It is how you use and interpret these things that is important in physics but the basic structure are classically not 'flexible': the basic operations in algebra do not depend on anything in physics.

 

 

In fact there are several recent threads going about this very subject and its terminology by different authors so I was going to point this out in conway's last thread, untill it was closed a second time, though I do not know why.

Indeed, I think this thread is an attempt to discuss thing that we have already discussed and to some extent rejected.

[studiot]

 

andrewcellini

Finally, a field is a commutative ring in which one can divide by any nonzero element: an example is the field of real numbers.

 

Exactly.

Thank you for demonstrating my point to Xerxes.

 

ajb

Indeed, I think this thread is an attempt to discuss thing that we have already discussed and to some extent rejected.

 

 

 

Yes, but conway is currently suspended and cannot answer for himself, particularly in the now closed thread where his interpretation of the nature of a set was superior to that of Sato (and actually confirmed in the link that Sato provided)

 

It is regretable that acrimony creeps into discussion with conway, all too readily, but he is not always wrong.

 

I am hoping to avoid such in the discussion with Xerxes in this thread as I only posted in the hope of a gentle nudge to tighten up his wording to exclude countercases, once conway had weighed into the discussion. I had previously intended to do this by PM.

[Strange]

Exactly.

Thank you for demonstrating my point to Xerxes.

 

As far as I can see, that shows Xerxes was right and you were wrong.

 

Xerxes said:

A field in abstract algebra is defined as a ring with the additional property of an inverse for the operation x.

Which is what that definition says (division is the inverse operation).

 

Whereas you said he had the definition of field and ring the wrong way round. Which is clearly wrong.

 

(You seem to have now retreated to the irrelevant point that division is only defined for non-zero elements. No one ever said it was so that appears to be a strawman argument.)

 

 

 

[studiot]

 

As far as I can see, that shows Xerxes was right and you were wrong.

 

Xerxes said:

Which is what that definition says (division is the inverse operation).

 

Whereas you said he had the definition of field and ring the wrong way round. Which is clearly wrong.

 

(You seem to have now retreated to the irrelevant point that division is only defined for non-zero elements. No one ever said it was so that appears to be a strawman argument.)

 

 

 

One counterexample is enough to disprove any assertion.

 

All that was required was to improve the wording.

 

Both rings and fields have two binary operations defined.

 

There are rings that have no zero element, but as far as I know, there are no fields that possess this characteristic.

[Strange]

 

One counterexample is enough to disprove any assertion.

 

Saying that Xerxes had the definitions of ring and field swapper isn't a counterexample.

 

Neither is a strawman (i.e. a counter to a nonexistent example).

[Xerxes]

There are rings that have no zero element,

If you know of one (or more) rings with no additive identity, I implore you to share them with us.

[studiot]

Oh, what a minefield.

 

Generalisation v restriction is where I think so many discussions go astray.

 

If we start with sets we have operations from one set to another - morphisms.

These can be very general and apply to sets of non mathematical interest so we restrict the sets under discussion to certain types of set.

 

Groups.

These are sets of elements of mathematical interest which have one defined operation between them.

 

This operation is often called multiplication, but as we shall see this can lead to substantial confusion so I will call this the group star operation.

 

For this star operation group axioms guarantee that

 

1) For any two elements a , b the star operation will produce a*b = c

2) There is an element e such that a*e = e*a = a for every element in the group.

3) There is another element in the group a such that aa^-1 = e for each element in the group.

 

[aside]

Note there is no guarantee that a*b = b*a.

This possibilty is the underlying pure mathematical reason for the Heisenberg Uncertainty Principle, from which it can be derived.

[/aside]

 

Now for the tricky bit.

 

We are discussing numbers.

So we restrict our set further to fit with numbers and their properties.

 

For the real numbers the above axioms

 

Are satisfied by the arithmetic operation of addition

Are not satisfied by the arithmetic operation of multiplication (no inverse for the zero element)

But we add the property that a*b = b*a, making the group abelian.

 

So the group star operation is addition though some call it multiplication

 

Let us move on and restrict still further, adding a second operation and forming the group into a ring.

We already have arithmetical addition so we add arithmetical multiplication.

 

So this ring has an (arithmetical) additive inverse but no (arithmetical) inverse

 

 

as I said.

 

Now please let us stop squabbling and move on.

Edited January 2, 2016 by studiot

[ajb]

If you know of one (or more) rings with no additive identity, I implore you to share them with us.

The structure would not be a ring. A ring has to have a abelian group structure under 'addition'.

[Xerxes]

Of course, hence my somewhat sarcastic appeal.

 

But lookee here. I quite myself

In linear algebra, on the other hand, a field assigns to each and every point in some chosen space a value, generally scalar-, vector- or tensor-valued. Here relativistic effects do NOT apply to scalar fields.

Although this is not be best definition of a (say) a vector field, it tells us all we need to know.

 

We can still have fun with it, though.

 

Suppose I denote our "space" by [math]M[/math]. (Some of you you will know why I make this choice, but for now it is unimportant).

 

Suppose that I denote the collection of all vector fields on [math]M[/math] as [math]\mathcal{X}[/math]. By construction there may be any number of possible fields on [math]M[/math]i.e there may be any number of "choices" of the single vector [math]v[/math] at [math]p \in M[/math].

 

Now since, by the definition of a vector space restricted to [math]p \in M[/math]I will have that

 

1. [math]v = u+w[/math] in this vector space

 

2. [math]\alpha v = z [/math] in this space

 

it follows that, for the vector fields [math]X,\,\,Y \in \mathcal{X}[/math], then (say)[math]\alpha X + \beta Y[/math] for scalars [math] \alpha,\,,\,\beta[/math]is also a vector field.

 

BUT this is nothing more nor less than the definition of a vector space. Which leads to the curious (but true) conclusion that [math]\mathcal{X}[/math] is a vector space. And since by construction [math]X,\,\,Y \in \mathcal{X} [/math] and since elements in a vector space are vectorswe are left with the curious conclusion that a vector field is a vector!!

 

Same is true for scalar and tensor fields, so next time you hear a mathematician tutting (tsk, tsk)when a physicist talks about the metric tensor (which is both a field and a vector) or the curvature tensor, bear this in mind.

[studiot]

Xerxes.

 

Consider post#16 quoted

 

Yes a tensor conforms to the vector axioms for vectors.

 

It is also true that group elements may themselves be operations.

[ajb]

Suppose I denote our "space" by [math]M[/math]. (Some of you you will know why I make this choice, but for now it is unimportant).

But what is a vector field on a "space"?

 

If you are talking about a smooth manifold (of finite dimension) then we are all clear that all the definitions coincide. In more generality you need to specify what you are talking about.

 

Same is true for scalar and tensor fields, so next time you hear a mathematician tutting (tsk, tsk)when a physicist talks about the metric tensor (which is both a field and a vector) or the curvature tensor, bear this in mind.

You mean that we can consider the vector space of covariant or contravariant tensors (on some smooth manifold) of some given valency? If so then sure.

 

However, if on were to just say 'a vector on M', then most people would have in mind a vector field.

[uncool]

I think that Xerxes may be referring to the physics definition of "field", as in "quantum field theory" or "electromagnetic field" (which corresponds to a vector bundle for mathematicians, I think). The "space" would be the base space, usually a manifold (often spacetime), and the "vector space at each point" the stalk of the bundle.

Edited January 4, 2016 by uncool

[Xerxes]

As to quantum fields I know literally nothing, but otherwise yes.

 

As to my fudging by using the term "space", yes I was well aware I was doing something slightly worse than hand-waving.

 

In mitigation, I did not want to use the "m-word" without explaining what a manifold is, which in turn I didn't feel I could do justice to without first going through topological spaces. A big ask in a single post!!

 

As a matter of interest - suppose in my last post one assumes my "space" a manifold (which my notation clearly telegraphed to the initiated), would it not be sufficient that the manifold is a least [math]C^2[/math] i.e. not necessarily smooth?

[ajb]

As a matter of interest - suppose in my last post one assumes my "space" a manifold (which my notation clearly telegraphed to the initiated), would it not be sufficient that the manifold is a least [math]C^2[/math] i.e. not necessarily smooth?

In order to define vector fields?

 

Maybe, but that would not be enough to fully develop the differential calculus. I do not have any working knowledge of manifolds that are not smooth but just C^k.

 

I think that Xerxes may be referring to the physics definition of "field", as in "quantum field theory" or "electromagnetic field" (which corresponds to a vector bundle for mathematicians, I think). The "space" would be the base space, usually a manifold (often spacetime), and the "vector space at each point" the stalk of the bundle.

A field in physics is understood as a section of a fibre bundle (usually all in the smooth category). You might be able to weaken this to just sections of a fibred manifold, but I an not aware of people in physics really doing that.

Edited January 5, 2016 by ajb