Vector or not vector?

by BobbyFluffyPric
Tags: vector
HW Helper
PF Gold
P: 4,119
 Quote by pmb_phy Consider a flat space which consists only of a square. I.e. it is a finite manifold. Let the metric be the Euclidean one. I can still define position vectors on this space, however now there is no closure. I can't arbitrarily multiply any vector by an arbitrary number and expect the result to be within the space. Pete
You can certainly still label the "positions"... assigning pairs of numbers in some coordinate system. But what gives you the right to call them "vectors" if their sum or multiplication by scalars is not well defined... following (say) Schutz's GMoMP definition?

If you wish to assign to each point a vector, somehow representing its position, then that vector is in the vector space R2 you have attached to each point... which can be added to other vectors in the vector-space-based-at-that-point.
HW Helper
PF Gold
P: 4,119
 Quote by pmb_phy You haven't really said anything here my friend towards an example/proof in your favor. All you've done is shown that Schutz defined the term "vector space". He did not say that all types of geometric vectors are in any vector space. He even defined a geometric tensor (as a tangent vector) without mentioning the space that its in. I'm 99.99% sure that I'm correct here. Just to make myself get more towards 100% sure I spoke with a friend of mine last night. He was a physics prof since 1960 until recently (1995?). You may have heard of him: Dr. Robert W. Brehme. His book is recommended reading by Schutz in his gr text. He's written tons of stuff on relativity, since that's his area. He tried to explain it to me in terms I could relay to you but I was certain such a relay wouldn't work. I.e. I was 100% sure that someone would come back with the response that hurkyl provided. I'll try to get a better example. Best wishes Pete
From Schutz's "A First Course in General Relativity" p. 39, (eq. 2.8) [i.e. from the rest of that section you first quoted]
 Vectors in spacetime obey the usual rules: if A and В are vectors and $\mu$ is a number, then A + В and $\mu$А are also vectors, with components. $$\vec A+\vec B \rightarrow_{\cal O} (A^0+B^0,A^1+B^1,A^2+B^2,A^3+B^3)$$ $$\mu \vec A \rightarrow_{\cal O} (\mu A^0,\mu A^1 ,\mu A^2 ,\mu A^3)$$ Thus, vectors add by the usual parallelogram rule.
I can also try to find the definition of a vector from one of my relativity professors.
P: 2,954
 Quote by robphy From Schutz's "A First Course in General Relativity" p. 39, (eq. 2.8) [i.e. from the rest of that section you first quoted] ...
)Odd, since it doesn't seem to apply to the very simple example I gave.

I can also try to find the definition of a vector from one of my relativity professors.[/QUOTE]Why? I wasn't looking toward my friend for supporting me. I was looking to him to give me an example which will clarify this issue. The rest I mentioned because the source of an example is sometimes important to (very few) people, but I put it there anyway. Bad move on my part!

Let's put this on hold until I determined if I made an error or I can come back with a strong determining arguement (which I seem to lacking right now).

Cheers Rob

Pete
P: 2,954
 Quote by robphy You can certainly still label the "positions"... assigning pairs of numbers in some coordinate system. But what gives you the right to call them "vectors" if their sum or multiplication by scalars is not well defined... following (say) Schutz's GMoMP definition?
The position vector is well known for being the prototype of a geometric vector. I guess I should have made it a circle. Change "square" to "circle". This will allow the vectors to fullfill the defining transformation equation.
 If you wish to assign to each point a vector, somehow representing its position, then that vector is in the vector space R2 you have attached to each point... which can be added to other vectors in the vector-space-based-at-that-point.
In general the resulting vectors won't be in the manifold on which they are defined.

I'm curious. How is it you came to believe that all geometric vectors are the same kind of vectors which are elements of a vector space?

Pete
HW Helper
PF Gold
P: 4,119
 Quote by pmb_phy The position vector is well known for being the prototype of a geometric vector.
The displacement vector [from one point to another] is the prototype of a geometric vector. Look at p. 36 in Schutz's "First Course", the first page in Chapter 2. The distinction between positions and displacements goes back to earlier discussions in this thread, as well as the distinction between an affine space and a vector space [from a much earlier thread].

Linearity is the key feature of a vector.
P: 2,954
 Quote by robphy The displacement vector [from one point to another] is the prototype of a geometric vector.
That's what I said. The difference between these two vectors is that the position vector is just the displacement vector with a point designated as origin.

The position vector is defined in places such as Ohanian's GR text.

I'm curious as to your position. Any two elements in a vector space must be able to be added to yield another vector. How do you see this in working on a curved manifold when only vectorss defined at the same point can be added? I assume you're response will be related to parallel transporting vectors defined at other points. But this is seems to me to be an over qualification, i.e. you will need something besides a vector and a scalar to create the resulting vector.

Pete
 Sci Advisor HW Helper PF Gold P: 4,119 Maybe I need a clarification of your question. Are you asking whether or not a "geometric vector" (for now, let's instead call it a gvector) [which you defined as a quantity that transforms like a displacement vector], must be an element of a vector-space [using Schutz's GMoMP definition]? If so, what is your answer? To me, a vector is an element of a vector space... and I believe it can be shown to be compatible with the above "transformation"-viewpoint [probably due to F.Klein]. In a vector space [like Euclidean space or Minkowski spacetime], I can define a displacement vector between two points... or, if I choose an origin, I can define a position vector of a distant point. In a general manifold, I can't as easily make such definitions that join two points on that manifold with a vector. Can one define 2D position-vectors on a sphere? The best I can do is to talk about vectors in the tangent vector-spaces attached to the different points... and, as you say, addition can only take place in the same vector space... possibly using a parallel-transported vector from another point's tangent vector space. When the manifold is a vector-space, the distinction between the tangent vector-spaces and the vector-space manifold can be blurred... parallel transport isn't much of an issue.
P: 2,954
 Quote by robphy Maybe I need a clarification of your question. Are you asking whether or not a "geometric vector" (for now, let's instead call it a gvector) [which you defined as a quantity that transforms like a displacement vector],
Hi Rob

Let me take some time and think this over. But I'm curious as to why you call it my definition? Had it been me that originated such a definition and it found its way into all the physics texts then I could see your point. But I'm going by what I gather from the literature, e.g. from all the books I've read on the subject and what comes up in conversation with other relativists I know. If it was me that created ut ya think I'd get some royalties huh?

Best wishes and I'll get back to this later.

Pete
P: 2,954
Hi Rob

Okay. I found the passage that I read which is one of the resources I came acrossed. This is from Schutz's "Geometrican Methods of Mathematical Physics[/i], page 15 towards bottom of page
 The norm n(x) on Rn is called the Euclidean norm. When we regard Rn as a vector space with this norm then we denote it by En and call it Euclidean space.
That's pretty much what I said at the start as I recall (or something pretty darn close to it).

Pete
HW Helper
PF Gold
P: 4,119
Quote by pmb_phy
Hi Rob

Okay. I found the passage that I read which is one of the resources I came acrossed. This is from Schutz's "Geometrican Methods of Mathematical Physics[/i], page 15 towards bottom of page

 The norm n(x) on Rn is called the Euclidean norm. When we regard Rn as a vector space with this norm then we denote it by En and call it Euclidean space.
That's pretty much what I said at the start as I recall (or something pretty darn close to it).

Pete
It may be what you said at the start... but I have no issue with this comment.

My issue is with your statement in #10 "Some types of vectors may belong to a vector space but they don't all need to be."

As I said in #3 and #25 , "a vector is an element of a vector space..."
P: 2,954
 Quote by robphy It may be what you said at the start... but I have no issue with this comment. My issue is with your statement in #10 "Some types of vectors may belong to a vector space but they don't all need to be." As I said in #3 and #25 , "a vector is an element of a vector space..."
Howdy Rob

Then let me ask you this. A great deal of textual sources define vectors according to their transportation properties. That is neccesary condition to be a vector according to those sources. Now consider an element of a vector space. This property of transformation is not a condition to be a vector. Therefore an element of a vector space does not imply that the element is also a vector whose components transforms as a vector.

Do you or do you not agree with this. If not then please elaborate. Thanks.

That is what I've been trying to get across. I guess I failed.

Pete
Mentor
P: 6,226
 Quote by pmb_phy Then let me ask you this. A great deal of textual sources define vectors according to their transportation properties. That is neccesary condition to be a vector according to those sources. Now consider an element of a vector space. This property of transformation is not a condition to be a vector. Therefore an element of a vector space does not imply that the element is also a vector whose components transforms as a vector.
The transformstion properties are derived from vector space properties, so vector space is more fundamental.
Emeritus
PF Gold
P: 16,099
 Quote by pmb_phy Do you or do you not agree with this. If not then please elaborate. Thanks.
It depends on the context.

In a setting that adopts the convention that the word "vector" is used exclusively to refer to tangent vectors of manifolds, and furthermore that tangent vectors are only to be thought of as coordinate-chart-dependent-tuples-of-'components', I would completely agree with what you have written in that post.

But the tangent vectors at a point P are elements of a vector space: the tangent space at P. So this isn't an example of vectors that are not elements of a vector space.
P: 2,954
 Quote by George Jones The transformstion properties are derived from vector space properties, so vector space is more fundamental.
How would you derive the transformation properties for vector spaces? There are vector spaces in which no such requirement need be defined or cannot be defined.

Pete
Mentor
P: 6,226
 Quote by pmb_phy How would you derive the transformation properties for vector spaces? There are vector spaces in which no such requirement need be defined or cannot be defined. Pete
Let $V$ be an n-dimensional real vector space, and let $\left\{ e_{1}, \dots, e_{n} \right\}$ be a basis for $V.$ If $L: V \rightarrow V$ is an invertible linear transformation on $V,$ then $\left\{ e'_{1}, \dots, e'_{n} \right\}$ with $e'_{i} = Le_{i}$ is also a basis for $V.$ Each member of the primed basis can be written as linear combintion of elements of the unprimed basis:

$e'_{i} = L^{j} {}_{i} e_{j}.[/tex] Any [itex]v$ in $V$ can be expanded with respect to both bases:

$$v = v^j e_j = v'^i e'_i = v'^i L^{j} {}_{i} e_{j}.$$

Thus, for example,

$$v^j = v'^i L^{j} {}_{i}.$$

Some books blur the distinction between (tangent) vectors (at a point) and vector fields. The set of vector fields is not a vector space, but is a generalization (a module over the ring of scalar fierlds) of a vector space. Again, transformation properties are derived properties.

Consider an n-dimensional real differentiable manifold, and suppose further that the bases above are coordinate tangent vector fields that arise from 2 overlapping charts:

$$e_{i} =\frac{\partial}{\partial x^{i}}$$

and

$$e'_{i} =\frac{\partial}{\partial x'^{i}}.$$

Then, by the chain rule, the change of basis relation is

$$e'_{i} = L^{j} {}_{i} e_{j} = \frac{\partial x^{j}}{\partial x'^{i}} e_{j},$$

and

$$v^j = \frac{\partial x^{j}}{\partial x'^{i}} v'^i.$$
P: 2,954
 Quote by George Jones Let $V$ be an n-dimensional real vector space, and let $\left\{ e_{1}, \dots, e_{n} \right\}$ be a basis for $V.$ If $L: V \rightarrow V$ is an invertible linear transformation on $V,$ then $\left\{ e'_{1}, \dots, e'_{n} \right\}$ with $e'_{i} = Le_{i}$ is also a basis for $V.$ Each member of the primed basis can be written as linear combintion of elements of the unprimed basis: [itex]e'_{i} = L^{j} {}_{i} e_{j}.[/tex]
What gives you the idea that there are more than one coordinate systems for a particular vector space or that a vector is something that can be represented as an object with a single index? Rember that a vector in a vector space can be a lot of things like a second rank tensor or a 10x10 matrix. And why do you think vector transformations exists for all kinds of vectors?

I'll get back after I do some more searching on this topic. The idea is to get a good example.
Mentor
P: 6,226
 Quote by pmb_phy What gives you the idea that there are more than one coordinate systems for a particular vector space
It's a basic fact of elementary linear algebra that there are infinitely many distinct bases for any vector space.

 or that a vector is something that can be represented as an object with a single index? Rember that a vector in a vector space can be a lot of things like a second rank tensor or a 10x10 matrix.
All vector spaces be they spaces of 10x10 matrices or anything else, can be transformed in the way I outlined.

 And why do you think vector transformations exists for all kinds of vectors?
Because of the linear algebra fact I stated above.

Now, for the space of 10x10 matrices there are (at least) two types of transformations that can induced by a change of basis, depending on whether the change of basis is for the space of matrices (this treats the matrices as single-index objects), or for the space on which the matrices act (this treats the matrices as two-index objects).

This discussion is getting somewhat surreal, so I'll think I'll bow out now.
 P: 2,954 I can't, for the life of me, understand why you think that any vector can be transformed from one coordinate system to another when such a transformation may be entirely meaningless for a particuar kind of vector. E.g. I meant for the matrices to be ten dimensional, not 10x10 (In case you were thinking about 4-D spacetime). As I said, I'm looking for such an example right now and will post it when I find it. But such a criteria never exist for a vector space where such a critera always exists for geometrical vectors. A good example which comes to mind now is the vector space whose elements belong to Fourier series, i.e. a sequance of sines and cosines in increasing frequency which can approximate any function of a given interval. How would you transform these elements to another coordinate system when, even when you can write down something that lokks like a transformation (God only knows what that transformation looks like or means) such a transformation is meaningless. Pete

 Related Discussions General Math 11 Calculus & Beyond Homework 3 Introductory Physics Homework 3 General Math 3 Differential Geometry 1