Are Scalar Quantities in Physics really 1-D Affine Spaces in disguise?

[Rising Eagle]

We could assign the vectors to be linear functionals, tensors, polynomial or Fourier (sin, cos, e) functions, rows or columns of a matrix or some other discrete functions, normally distributed random variables, gradient and other linear operators; anything that obeys the linearity axioms. We may even assign the vectors to be the Real numbers. Name other examples if you can think of them.

I believe we can add to the list Linear Time Invariant and Linear Shift Invariant systems as studied in engineering, random processes, matrices, many different types of functionals, and, interestingly homogeneous linear and differential equations too. I never thought of it before, but equations themselves can be added and scaled and so qualify. Maybe they don't even have to be linear or homogeneous either; just any general equation. Not sure about inequalities or greater than/less than relationships, though.

I vote for Linear Elements as the term for elements of a Linear Space. And I would like to call measurements of Scalar Quantities (scalar values) Affinitors or something similar if Numbers turns out not to be the best model for such measurements.

Another possibility is to call the elements of a Linear Space a Linear Form as a take off on 1-form or differential form. Any linear element in general would be a Linear Form.

 

[Studiot]

If you don't exclude zero you allow the following paradox.

It is axiomatic that for any a, c in F there exists a 'b' in f such that

a*b = b*a = c

1 is in F
0 is in F

set a = 0, c =1

therefore there exists a 'b' such that

0*b = b*0 = 1

leading to b is the inverse or reciprocal of zero.

 

[jgens]

If you don't exclude zero you allow the following paradox.

Do you understand the difference between saying that the multiplication on $F$ is commutative and saying that the multiplication on $F$ makes $F$ into an abelian group? The first statement is true and the second statement is false; they are not analogous.

There is no paradox in saying that the multiplication on $F$ is commutative. If you think there is, then find the flaw in this argument: Clearly multiplication is commutative on $F \setminus \{0\}$ by hypothesis and for all $x \in F$ the equality $0x = 0 = x0$.

 

[Studiot]

This area of pure maths (in fact any area) is not my speciality and we are off to the seaside for the day so perhaps Frederik or Hurkyl will explain.

I think the reasoning is as follows:

Yes composition by zero is commutative.

However you need to specifiy this separately not inherit it from group properties.

I will post a more complete explanation if no one else has done so.

 

[jgens]

This area of pure maths (in fact any area) is not my speciality and we are off to the seaside for the day so perhaps Frederik or Hurkyl will explain.

Luckily this happens to be the area of math that I am best at.

However you need to specifiy this separately not inherit it from group properties.

Actually no. If $(F,+,\cdot)$ is a field, then the multiplication on $F$ is a map $\cdot:F \times F \rightarrow F$. This internal law of composition is clearly commutative and is defined on all of $F$ and not just on $F \setminus \{0\}$. So multiplication on $F$ is commutative even if we include $0$.

If we want to talk about group structures, then this is a different story. The pair $(F,\cdot)$ is not a group since $0$ has no inverse while the pair $(F \setminus \{0\},\cdot)$ is a group by the definition of a field. So you would be correct if you were claiming that we need to exclude zero in order to get a group under multiplication; however, we were not talking about getting a group under multiplication, and instead were talking about whether or not the multiplication operation is commutative over all of $F$.

 

[micromass]

If you want a neat math term for $(F,\cdot)$, then you could always say that it's a commutative monoid.

But jgens is right: saying that $(F,\cdot)$ is commutative is perfectly alright. However, this fact does not immediately follow from $(F\setminus \{0\},\cdot)$ being an abelian group!

 

[Studiot]

and instead were talking about whether or not the multiplication operation is commutative over all of F.

You might have been, I never have been.

I stated my position clearly in post#30.

You added a further case in which multiplication is commutative in post #31

I offered an explanation as to why this case was normally excluded in post#32

From post#33 onwards you seem to want to argue that I am saying multiplication including the zero element is not commutative.

I never have said this. In fact in post#39 I said the opposite.

All I have been doing is seeking an explanation as to why respected mathematical dictionaries and most algebra, group theory and galois theory textbooks that I have access to exclude multiplicative zero in the definition of a field.

I don't know if my explanation is correct or if there is another one.

Can you provide an explanation please?