The Foundations of Relativity

**George Jones** · Dec25-05, 11:11 PM

Originally Posted by robphy

These N vector spaces are isomorphic copies of a single vector space... for example, V_1 and V_2 must have the same dimensionality [which doesn't seem to be required according to what has been written].

Right - there is nothing in the general definition that says that the N vector spaces can't be N copies of the same vector space, but there also is nothing in the general definition that says that the N vector spaces have to be N copies of the same vector space.

Also, this defintion only works for finite-dimensional vector spaces, like in relativity. For tensor products of infinte-dimensional vectors spaces, which occur in quantum theory, a different definition is needed. (The 2 definitions agree for finite-dimensiona vector spaces.)

How is the book "A Course in Modern Mathematical Physics"? I'm fairly sure that I will soon order it.

Regards,
George

**robphy** · Dec25-05, 11:50 PM

Originally Posted by George Jones

Right - there is nothing in the general definition that says that the N vector spaces can't be N copies of the same vector space, but there also is nothing in the general definition that says that the N vector spaces have to be N copies of the same vector space.

Ah, yes... so when permitting distinct vector spaces, one would probably use different sets of indices... as is done with "soldering forms".

**Oxymoron** · Dec26-05, 12:35 AM

Posted by pervect:

Yes. The vectors v_i all live in the same vector space. In GR this is usually the tangent space of some manifold. So we start with a manifold (which we haven't defined in this thread, that's a whole topic in itself - but for an illustrative example, picture a general manifold to be a curved n-dimensional surface, and for a specific example imagine that we have the 2-d surface of some 3-d sphere).

Given the manifold, there is also some tangent space on this manifold (for the example above, just imagine a plane that's tangent to the sphere). This tangent space is the vector space V that all the vectors "live in". There is also a "dual space" V* with the same number of dimensions that the duals of the vectors live in.

A tensor is just the functional you described which maps a certain number of dual vectors and a certain other number of vectors to a scalar - but all the vectors v_i live in the same vector space V, and all the dual vectors v_j live in the same dual space V*.

This is really a very minor distinction, otherwise you seem to be on the right track. But since you seem to be a mathemetican (or at least approaching the topic in the same way that a mathemtician does), I thought I'd try to be extra precise.

Perfect, exactly what I wanted to hear! Well written. BTW, you are correct, I am a mathematician - well at least I have just graduated from a Bachelor of Maths. Anyway, this description really helps.

Posted by George Jones:

How is the book "A Course in Modern Mathematical Physics"? I'm fairly sure that I will soon order it.

I ordered it over the internet about 3 weeks ago. It bridges the gap between undergraduate and graduate mathematical physics really well. I found it very well structured and written. Its about 600 pages and starts off with group theory and vector spaces. Then it moves into inner product spaces and algebras. Then it moves on to exterior algebra which I found very interesting and then has a chapter on tensors - introducing them in two different ways (which we have been discussing here) and finishes with applications to special relativity. The second part of the book (the final 300 pages) starts with topology and measure theory and some work on distibution functions which sets the stage for applications to quantum theory. There is also a chapter on Hilbert spaces. Finally chapters on differential geometry, forms, and manifolds (which I haven't read yet) which finishes with Riemannian curvature, connections, homology and a final chapter on Lie groups (all of which I haven't read). To sum up, its my favourite book at the moment. Very well written.

**Oxymoron** · Dec26-05, 01:00 AM

Ok, so Im pretty sure I understand what covariant and contravariant tensors are. A covariant tensor or type (0,2) is a map

$LaTeX Code: T:V\\times V \\rightarrow \\mathbb{R}$

Now if you take, say a two dimensional vector space

which has two basis components $LaTeX Code: \\{e_1,e_2\\}$ then the dual vector space

, which (as robphy and pervect pointed out) has the same dimension as

, then has two basis components. Now am I correct in assuming that the basis components of the dual space is written in Greek? As in, $LaTeX Code: \\{\\epsilon_1,\\epsilon_2\\}$ ?

So a vector $LaTeX Code: \\omega \\in V^*$ may be written as a sum of it's basis components: $LaTeX Code: \\omega = \\alpha\\epsilon_1 + \\beta\\epsilon_2$ ?

Extending this idea to

-dimensional vector spaces we have that $LaTeX Code: e_1,e_2,\\dots,e_n$ is a basis for

and $LaTeX Code: \\epsilon_1,\\epsilon_2,\\dots,\\epsilon_n$ is a basis for

.

As we have already discussed I assume that when writing, say, the product of the two basis vectors $LaTeX Code: (\\epsilon)(e)$ and by including the indices we would write

$LaTeX Code: \\epsilon^i(e_j)$

So I would write the

index superscripted because the $LaTeX Code: \\epsilon$ basis vector came from the dual vector space, and the

index is subscripted because

came from the vector space. Is this the reason for superscripting and subscripting indices - to make a distinction about which space we are in? Because after all, they are by no means identical bases, even if the vector space and its dual are equal?

My last question for now is, why is the product

$LaTeX Code: \\langle \\epsilon^i,e_j \\rangle = \\delta_j^i$

equal to the Kronecker delta? The Kronecker delta equal 1 if the indices are the same, and zero if the indices are different. Lets say that the vector space,

has

dimensions and the dual space,

has

dimensions. Then,

$LaTeX Code: \\delta_j^i = 1 + 1 + \\dots + 1 + 1^* + 1^* + \\dots + 1^*$

where there are

1's in the first sum and

1*'s in the second sum. Therefore

$LaTeX Code: \\delta_j^i = n + m = \\dim(V) + \\dim(V^*)$

which should equal the product of the basis vectors, $LaTeX Code: \\epsilon^i$ and

. Could this be the reason?

**pervect** · Dec26-05, 04:50 AM

Originally Posted by Oxymoron

Ok, so Im pretty sure I understand what covariant and contravariant tensors are. A covariant tensor or type (0,2) is a map
$LaTeX Code: T:V\\times V \\rightarrow \\mathbb{R}$

Now if you take, say a two dimensional vector space which has two basis components $LaTeX Code: \\{e_1,e_2\\}$ then the dual vector space , which (as robphy and pervect pointed out) has the same dimension as , then has two basis components. Now am I correct in assuming that the basis components of the dual space is written in Greek? As in, $LaTeX Code: \\{\\epsilon_1,\\epsilon_2\\}$ ?

Usually basis one-forms are written as $LaTeX Code: \\{\\omega^1, \\omega^2 \\}$ , a different greek letter choice, and more importantly superscripted rather than subscripted.

So a vector $LaTeX Code: \\omega \\in V^*$ may be written as a sum of it's basis components: $LaTeX Code: \\omega = \\alpha\\epsilon_1 + \\beta\\epsilon_2$ ?

If you write out a vector as a linear sum of multiples of the basis vectors as you do above, it's traditional to write simply

$LaTeX Code: x^i \\, e_i$ . Repeating the index i implies a summation, i.e.

$LaTeX Code: \\sum_{i=1}^{n} x^i e_i$

Extending this idea to -dimensional vector spaces we have that $LaTeX Code: e_1,e_2,\\dots,e_n$ is a basis for and $LaTeX Code: \\epsilon_1,\\epsilon_2,\\dots,\\epsilon_n$ is a basis for .

If you write out a one-form in terms of the basis one-forms, it's
$LaTeX Code: x_i \\omega^i$

Is this the reason for superscripting and subscripting indices - to make a distinction about which space we are in? Because after all, they are by no means identical bases, even if the vector space and its dual are equal?

Yes. It also leads to fairly intuitive tensor manipulation rules when you get used to the notation.

My last question for now is, why is the product
$LaTeX Code: \\langle \\epsilon^i,e_j \\rangle = \\delta_j^i$
equal to the Kronecker delta? The Kronecker delta equal 1 if the indices are the same, and zero if the indices are different. Lets say that the vector space, has dimensions and the dual space, has dimensions. Then,
$LaTeX Code: \\delta_j^i = 1 + 1 + \\dots + 1 + 1^* + 1^* + \\dots + 1^*$
where there are 1's in the first sum and 1*'s in the second sum. Therefore
$LaTeX Code: \\delta_j^i = n + m = \\dim(V) + \\dim(V^*)$
which should equal the product of the basis vectors, $LaTeX Code: \\epsilon^i$ and . Could this be the reason?

In an orthonormal basis, $LaTeX Code: e_i \\cdot e_j = \\delta_j^i$ . This is not true in a general basis, only in an orthonormal basis.

$LaTeX Code: \\omega^1 e_1$ is just different notation for $LaTeX Code: e_1 \\cdot e_1$ , so it will be unity only if the basis is normalized. Similarly only when the basis vectors are orthogonal will $LaTeX Code: \\omega^i e_j$ be zero when i is not equal to j.

**Oxymoron** · Dec26-05, 05:15 AM

Posted by Pervect.

Usually basis one-forms are written as...

What is a one-form?

~~pmb_phy~~ · Dec26-05, 07:40 AM

Originally Posted by Oxymoron

What is a one-form?

A 1-form is a mapping (i.e. function) which maps vectors to scalars. If "a" is a 1-form and "B" a vector and "s" the scalar then the typical notation is

s = <a, B>

Pete

**George Jones** · Dec26-05, 07:53 AM

Originally Posted by Oxymoron

Because after all, they are by no means identical bases, even if the vector space and its dual are equal?

A vector space and its dual are not equal, but they have equal dimension. Any 2 vector spaces that have equal dimension are isomorphic, but without extra structure (like a metric), there is no natural basis independent isomorphism.

The Kronecker delta equal 1 if the indices are the same, and zero if the indices are different.

Yes.

Lets say that the vector space, has dimensions and the dual space, has dimensions. Then,
$LaTeX Code: \\delta_j^i = 1 + 1 + \\dots + 1 + 1^* + 1^* + \\dots + 1^*$
where there are 1's in the first sum and 1*'s in the second sum. Therefore
$LaTeX Code: \\delta_j^i = n + m = \\dim(V) + \\dim(V^*)$

Careful - this isn't true.

My last question for now is, why is the product
$LaTeX Code: \\langle \\epsilon^i,e_j \\rangle = \\delta_j^i$
equal to the Kronecker delta?

Given an n-dimensional vector space

, the dual space

is defined as

$LaTeX Code: V* = \\left\\{f: V \\rightarrow \\mathbb{R} | f \\mathrm{is linear} \\right\\}.$

The action of any given linear between vector spaces is pinned down by finding/defining its action on a basis for the vector space that is the domain of the mapping. A dual vector is a linear mapping between the vector spaces

and

, so this is true in this case.

Let $LaTeX Code: \\left\\{ e_{1}, \\dots , e_{n} \\right\\}$ be a basis for

, and define $LaTeX Code: \\omega^{i}$ by: 1) $LaTeX Code: \\omega^{i} : V \\rightarrow \\mathbb{R}$ is linear; 2) $LaTeX Code: \\omega^{i} \\left( e_{j} \\right) = \\delta^{i}_{j}$ . Now let $LaTeX Code: v = v^{j} e_{j}$ (summation convention) be an arbitrary element of

. Then

$LaTeX Code: \\omega^{i} \\left( v \\right) = \\omega^{i} \\left( v^{j} e_{j} \\right) = v^{j} \\omega^{i} \\left( e_{j} \\right) = v^{j} \\delta^{i}_{j} = v^{i}.$

Consequently, $LaTeX Code: \\omega^{i}$ is clearly an element of

, and $LaTeX Code: \\omega^{i} \\left( e_{j} \\right) = \\delta^{i}_{j}$ by definition!

Exercise: prove that $LaTeX Code: \\left\\{\\omega_{1}, \\dots , \\omega_{n} \\right\\}$ is a basis for the vector space

.

What is a one-form?

I like to make a distinction between a tensor and a tensor field. A tensor field (of a given type) on a diifferentiable manifold

is the smooth assignment of a tensor at each $LaTeX Code: p \\in M$ .

A one-form is a dual vecor field. Note, however, that some references call a dual vector a one-form. See the thread "One-forms" in the Linear & Abstract Algebra forum. I tried to sum up the situation in posts #11 and #23.

Regards,
George

**Oxymoron** · Dec26-05, 09:14 AM

Posted by George Jones:

Careful - this isn't true.

I read what I wrote again and it sounds wrong indeed. For one, how can the dimension of

be different from

as I have implied by m and n. However, I would like some clarification on the incorrectness it.

Exercise: prove that $LaTeX Code: \\{\\omega_1,\\dots,\\omega_n\\}$ is a basis for the vector space .

Well, each $LaTeX Code: \\omega^i$ is linearly independant of the dual basis vectors, $LaTeX Code: \\epsilon^i$ from what you wrote, that

$LaTeX Code: \\omega^i\\epsilon^j = \\delta^_j$

Im not sure how to show that they span though, I mean, I could probably do it, but Im not sure which vectors and bases and stuff to use.

Posted by George Jones:

A one-form is a dual vecor field. Note, however, that some references call a dual vector a one-form. See the thread "One-forms" in the Linear & Abstract Algebra forum. I tried to sum up the situation in posts #11 and #23.

I read that, and it makes sense, especially as I was almost up to that point.

If you have a tensor of type

then it can be written as

$LaTeX Code: T:V\\times V \\rightarrow \\mathbb{R}$

which is covariant. If $LaTeX Code: \\omega$ and $LaTeX Code: \\rho$ are elements of

(which means that they are simply linear functionals over

right?) then we can define their 'tensor' product as

$LaTeX Code: \\omega \\otimes \\rho (u,v) = \\omega(u) \\rho(v)$

At first this was hard to get my head around. My first thought was, $LaTeX Code: \\omega \\otimes \\rho$ was multiplied by

and so what was this

thing? But then I though this is just like

$LaTeX Code: \\phi(u,v) = \\phi(u)\\phi(v)$

in group theory - its just a mapping! Where $LaTeX Code: \\omega \\otimes \\rho$ was the 'symbol' (or $LaTeX Code: \\phi$ in this case) representing the tensor product acting on two variables from $LaTeX Code: V \\times V$ .

My next question at this stage is "how does one define a basis on $LaTeX Code: V\\times V$ (can I assume that the notation $LaTeX Code: V^{(0,2)}$ means $LaTeX Code: V \\times V$ ?

Well, if

has dimension

, then $LaTeX Code: V^{(0,2)}$ has dimension

. So let $LaTeX Code: \\epsilon^i$ be a basis of

, then

$LaTeX Code: \\epsilon^i \\otimes \\epsilon^j$

forms a basis for $LaTeX Code: V^{(0,2)}$ , yes?

Now, is $LaTeX Code: \\epsilon^i \\otimes \\epsilon^j$ a tensor?

My main issue with dealing with these basis vectors is I want to define the metric tensor next, and Im thinking that a sound understanding of how to define bases for these vector spaces and tensors is a logical stepping stone.

**pervect** · Dec26-05, 03:03 PM

Originally Posted by Oxymoron

.
My next question at this stage is "how does one define a basis on $LaTeX Code: V\\times V$ (can I assume that the notation $LaTeX Code: V^{(0,2)}$ means $LaTeX Code: V \\times V$ ?

I've never seen that notation used, at least in physics.

Well, if has dimension , then $LaTeX Code: V^{(0,2)}$ has dimension . So let $LaTeX Code: \\epsilon^i$ be a basis of , then
$LaTeX Code: \\epsilon^i \\otimes \\epsilon^j$
forms a basis for $LaTeX Code: V^{(0,2)}$ , yes?

In tensor noation we use subscripts for vectors, so we'd usually write that

is a basis of V (we would write $LaTeX Code: \\omega^i$ as a basis of V*)

Now, is $LaTeX Code: \\epsilon^i \\otimes \\epsilon^j$ a tensor?

$LaTeX Code: e_i \\otimes e_j$ is an element of $LaTeX Code: V \\otimes V$ , not a map from $LaTeX Code: V \\otimes V$ to a scalar.

**George Jones** · Dec26-05, 04:33 PM

Originally Posted by Oxymoron

However, I would like some clarification on the incorrectness it.

As you said,

$LaTeX Code: \\delta^{i}_{j} = \\left\\{\\begin{array}{cc}0,&\\mbox{ if } i \\neq j \\\\1, & \\mbox{ if } i = j \\end{array}\\right,$

but $LaTeX Code: \\delta^{i}_{j}$ is not expressed as a sum. $LaTeX Code: \\delta^{i}_{j}$ can be used in sums, e.g.,

$LaTeX Code: \\sum_{i = 1}^{n} \\delta^{i}_{j} = 1,$

and

$LaTeX Code: \\sum_{i = 1}^{n} \\delta^{i}_{i} = n.$

Im not sure how to show that they span though, I mean, I could probably do it, but Im not sure which vectors and bases and stuff to use.

First, let me fill in the linear independence argument. $LaTeX Code: \\left\\{\\omega_{1}, \\dots , \\omega_{n} \\right\\}$ is linearly independent if $LaTeX Code: 0 = c_{i} \\omega^{i}$ implies that the

for each

. The zero on the left is the zero function, i.e.,

for $LaTeX Code: v \\in V$ . Now let the equation take $LaTeX Code: e_{j}$ as an argument:

$LaTeX Code: 0 = c_{i} \\omega^i \\left( e_{j} \\right) = c_{i} \\delta^{i}_{j} = c_{j}.$

Since this is true for each

, $LaTeX Code: \\left\\{\\omega_{1}, \\dots , \\omega_{n} \\right\\}$ is a linearly independent set of covectors.

Now show spanning. Let $LaTeX Code: f : V \\rightarrow \\mathbb{R}$ be linear. Define scalars $LaTeX Code: f_{i}$ by $LaTeX Code: f_{i} = f \\left( e_{i} \\right)$ . Then

$LaTeX Code: f \\left( v \\right) = f \\left( v^{i} e_{i} \\right) = v^{i} f \\left( e_{i} \\right) = v^{i} f_{i}$ .

Now show that $LaTeX Code: f_{i} \\omega^{i} = f$ :

$LaTeX Code: f_{i} \\omega^{i} \\left( v \\right) = f_{i} \\omega^{i} \\left( v^{j} e_{j} \\right) = f_{i} v^{j} \\omega^{i} \\left( e_{j} \\right) = f_{i} v^{j} \\delta^{i}_{j} = f_{i} v^{i} = f \\left( v \\right).$

Since this is true for every

, $LaTeX Code: f_{i} \\omega^{i} = f$ .

its just a mapping! Where $LaTeX Code: \\omega \\otimes \\rho$ was the 'symbol' (or $LaTeX Code: \\phi$ in this case) representing the tensor product acting on two variables from $LaTeX Code: V \\times V$ .

Exactly!

Becoming used to the abstractness of this approach takes a bit of time and effort.

My next question at this stage is "how does one define a basis on $LaTeX Code: V\\times V$ (can I assume that the notation $LaTeX Code: V^{(0,2)}$ means $LaTeX Code: V \\times V$ ?

I think you mean $LaTeX Code: V* \\otimes V*$ . $LaTeX Code: V \\times V$ is the set of ordered pairs where each element of the ordered pairs comes from

. As a vector space, this is the external direct sum of

with itself. What you want is the space of all (0,2) tensors, i.e.,

$LaTeX Code: V* \\otimes V* = \\left\\{ T: V \\times V \\rightarrow \\mathbb{R} | T \\mathrm{is linear} \\right\\}.$

Well, if has dimension , then $LaTeX Code: V^{(0,2)}$ has dimension . So let $LaTeX Code: \\epsilon^i$ be a basis of , then
$LaTeX Code: \\epsilon^i \\otimes \\epsilon^j$
forms a basis for $LaTeX Code: V^{(0,2)}$ , yes?

Yes.

[QUOTE}Now, is $LaTeX Code: \\epsilon^i \\otimes \\epsilon^j$ a tensor?[/quote]

Yes, $LaTeX Code: \\epsilon^i \\otimes \\epsilon^j$ is, for

and

, an element of $LaTeX Code: V* \\otimes V*$ . $LaTeX Code: \\epsilon^i$ and $LaTeX Code: \\epsilon^j$ are specific examples of your $LaTeX Code: \\omega$ and $LaTeX Code: \\rho$ .

Regards,
George

**Oxymoron** · Dec27-05, 09:08 PM

Thanks George and Pervect. Your answers helped me a lot. So much in fact I have no further queries on that. Which is good.

But now I want to move along and talk about the metric tensor. Is a metric tensor similar to metric functions in say, topology or analysis? Do they practically do the same thing? That is, define a notion of distance between two objects? Or are they something completely abstract?

I had a quick look over the metric tensor and there seems to be several ways of writing it. The first method was to introduce an inner product space. Then to define a functional as

$LaTeX Code: g\\,:\\,V\\times V \\rightarrow \\mathbb{R}$

defined by

$LaTeX Code: g(u,v) = u\\cdot v$

As we have already discussed this is a bilinear covariant tensor of degree two. Now this is not the general metric tensor I have read about, instead this is the metric tensor of the inner product. Are the two different? Or is the metric tensor intertwined with some sort of inner product always?

I understand that to introduce the idea of a metric we need some sort of mathematical tool which represents distance. In this case the inner product usually represents 'length' of an element. Is this the reason for introducing the metric tensor like this? Could you go further and instead of an inner product, simply define the metric tensor as an arc length or something general like this?

**George Jones** · Dec27-05, 11:37 PM

What is an inner product? I ask this because I want to compare and contrast whatever answer you give with a "metric" tensor.

Regards,
George

**Oxymoron** · Dec27-05, 11:43 PM

In my understanding to have an inner product you need a vector space

over the field $LaTeX Code: \\mathbb{R}$ . Then the inner product over the vector space is a bilinear mapping:

$LaTeX Code: g\\,:\\,V\\times V \\rightarrow \\mathbb{R}$

which is symmetric, distributive, and definite.

**George Jones** · Dec28-05, 12:34 AM

So, isn't an inner product on a vector space

a (0,2) tensor, i.e., an element of $LaTeX Code: V* \\otimes V*$ ?

Too exhausted to say more - took my niece and nephew skating (their first time; 3 and 4 years old), and pulling them around the ice completely drained me.

Regards,
George

**pervect** · Dec28-05, 01:50 AM

Well, I think of distances as being quadratic forms. Quadratic forms are in a one-one correspondence with symmetric bilinear forms.

http://mathworld.wolfram.com/SymmetricBilinearForm.html

the defintion of which leads you directly to your defintion $LaTeX Code: M : V \\otimes V -> \\mathbb{R}$ , except for the requirement of symmetry.

There's probably something deep to say about symmetry, but I'm not quite sure what it is. In GR we can think of the metric tensor as always being symmetric, so if you accept the symmetry as being a requirement, you can go directly from quadratic forms to symmetric bilinear forms.

Of course you have to start with the assumption that distances are quadratic forms, I'm not sure how to justify something this fundamental offhand.

[add]
I just read there may be a very small difficulty with the above argument, see for instance

http://www.answers.com/topic/quadratic-form