Understanding the Structure and Transformation of Tensors in Spacetime

[robphy]

Are you sure? In the book I am reading, the formulation of a tensor as a multilinear map requires several vector spaces V_1,\dots,V_N. Maybe I am wrong and I don't understand their version of 'vector space'.

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].

 

[pervect]

Are you sure? In the book I am reading, the formulation of a tensor as a multilinear map requires several vector spaces V_1,\dots,V_N. Maybe I am wrong and I don't understand their version of 'vector space'.

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.

I have to hope that I have not violated Born's dictum here, which is to never write more precisely than you can think - I usually take a more "physical" approach than the rather abstract approach I am taking here.

 

[George Jones]

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]

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]

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]

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

T:V\times V \rightarrow \mathbb{R}

Now if you take, say a two dimensional vector space V which has two basis components \{e_1,e_2\} then the dual vector space V^*, which (as robphy and pervect pointed out) has the same dimension as V, 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, \{\epsilon_1,\epsilon_2\}?

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

Extending this idea to n-dimensional vector spaces we have that e_1,e_2,\dots,e_n is a basis for V and \epsilon_1,\epsilon_2,\dots,\epsilon_n is a basis for V^*.

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

\epsilon^i(e_j)

So I would write the i index superscripted because the \epsilon basis vector came from the dual vector space, and the j index is subscripted because e_j 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

\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. Let's say that the vector space, V has n dimensions and the dual space, V^* has m dimensions. Then,

\delta_j^i = 1 + 1 + \dots + 1 + 1^* + 1^* + \dots + 1^*

where there are n 1's in the first sum and m 1*'s in the second sum. Therefore

\delta_j^i = n + m = \dim(V) + \dim(V^*)

which should equal the product of the basis vectors, \epsilon^i and e_j. Could this be the reason?

 

[pervect]

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

Now if you take, say a two dimensional vector space V which has two basis components \{e_1,e_2\} then the dual vector space V^*, which (as robphy and pervect pointed out) has the same dimension as V, 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, \{\epsilon_1,\epsilon_2\}?

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

So a vector \omega \in V^* may be written as a sum of it's basis components: \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

x^i \, e_i. Repeating the index i implies a summation, i.e.

\sum_{i=1}^{n} x^i e_i

Extending this idea to n-dimensional vector spaces we have that e_1,e_2,\dots,e_n is a basis for V and \epsilon_1,\epsilon_2,\dots,\epsilon_n is a basis for V^*.

If you write out a one-form in terms of the basis one-forms, it's
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
\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. Let's say that the vector space, V has n dimensions and the dual space, V^* has m dimensions. Then,
\delta_j^i = 1 + 1 + \dots + 1 + 1^* + 1^* + \dots + 1^*
where there are n 1's in the first sum and m 1*'s in the second sum. Therefore
\delta_j^i = n + m = \dim(V) + \dim(V^*)
which should equal the product of the basis vectors, \epsilon^i and e_j. Could this be the reason?

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

\omega^1 e_1 is just different notation for 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 \omega^i e_j be zero when i is not equal to j.

 

[Oxymoron]

Posted by Pervect.

Usually basis one-forms are written as...

What is a one-form?

 

[pmb_phy]

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]

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

Careful - this isn't true.

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

Given an n-dimensional vector space V, the dual space V* is defined as

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 V and V*, so this is true in this case.

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

\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, \omega^{i} is clearly an element of V*, and \omega^{i} \left( e_{j} \right) = \delta^{i}_{j} by definition!

Exercise: prove that \left\{\omega_{1}, \dots , \omega_{n} \right\} is a basis for the vector space V*.

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 M is the smooth assignment of a tensor at each 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 "https://www.physicsforums.com/showthread.php?t=96073"" in the Linear & Abstract Algebra forum. I tried to sum up the situation in posts #11 and #23.

Regards,
George

 

[Oxymoron]

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 V be different from V^* as I have implied by m and n. However, I would like some clarification on the incorrectness it.

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

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

\omega^i\epsilon^j = \delta^_j

Im not sure how to show that they span though, I mean, I could probably do it, but I am 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 (0,2) then it can be written as

T:V\times V \rightarrow \mathbb{R}

which is covariant. If \omega and \rho are elements of V^* (which means that they are simply linear functionals over V right?) then we can define their 'tensor' product as

\omega \otimes \rho (u,v) = \omega(u) \rho(v)

At first this was hard to get my head around. My first thought was, \omega \otimes \rho was multiplied by (u,v) and so what was this (u,v) thing? But then I though this is just like

\phi(u,v) = \phi(u)\phi(v)

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

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

Well, if V has dimension n, then V^{(0,2)} has dimension n^2. So let \epsilon^i be a basis of V, then

\epsilon^i \otimes \epsilon^j

forms a basis for V^{(0,2)}, yes?

Now, is \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 I am thinking that a sound understanding of how to define bases for these vector spaces and tensors is a logical stepping stone.

 

[pervect]

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

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

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

In tensor noation we use subscripts for vectors, so we'd usually write that e_i is a basis of V (we would write \omega^i as a basis of V*)

Now, is \epsilon^i \otimes \epsilon^j a tensor?

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

 

[George Jones]

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

As you said,
\delta^{i}_{j} = \left\{\begin{array}{cc}0,&amp;\mbox{ if } i \neq j \\1, &amp; \mbox{ if } i = j \end{array}\right,
but \delta^{i}_{j} is not expressed as a sum. \delta^{i}_{j} can be used in sums, e.g.,
\sum_{i = 1}^{n} \delta^{i}_{j} = 1,
and
\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 I'm not sure which vectors and bases and stuff to use.

First, let me fill in the linear independence argument. \left\{\omega_{1}, \dots , \omega_{n} \right\} is linearly independent if 0 = c_{i} \omega^{i} implies that the c_i = 0 for each i. The zero on the left is the zero function, i.e., 0(v) = 0 for v \in V. Now let the equation take e_{j} as an argument:
0 = c_{i} \omega^i \left( e_{j} \right) = c_{i} \delta^{i}_{j} = c_{j}.
Since this is true for each j, \left\{\omega_{1}, \dots , \omega_{n} \right\} is a linearly independent set of covectors.

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

its just a mapping! Where \omega \otimes \rho was the 'symbol' (or \phi in this case) representing the tensor product acting on two variables from 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 V\times V (can I assume that the notation V^{(0,2)} means V \times V?

I think you mean V* \otimes V*. V \times V is the set of ordered pairs where each element of the ordered pairs comes from V. As a vector space, this is the external direct sum of V with itself. What you want is the space of all (0,2) tensors, i.e.,
V* \otimes V* = \left\{ T: V \times V \rightarrow \mathbb{R} | T \mathrm{is linear} \right\}.

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

Yes.

Now, is \epsilon^i \otimes \epsilon^j a tensor?

Yes, \epsilon^i \otimes \epsilon^j is, for i and j, an element of V* \otimes V*. \epsilon^i and \epsilon^j are specific examples of your \omega and \rho.

Regards,
George

 

[Oxymoron]

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

g\,:\,V\times V \rightarrow \mathbb{R}

defined by

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]

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]

In my understanding to have an inner product you need a vector space V over the field \mathbb{R}. Then the inner product over the vector space is a bilinear mapping:

g\,:\,V\times V \rightarrow \mathbb{R}

which is symmetric, distributive, and definite.

 

[George Jones]

So, isn't an inner product on a vector space V a (0,2) tensor, i.e., an element of 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]

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 M : V \otimes V -&gt; \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

 

[masudr]

The metric tensor helps us "lower" or "raise" indices, thus allowing us to make scalars out of tensors. For example, say we want a scalar out of two rank 1 tensors A^\mu, B^\nu. We can go for

g_{\mu\nu}A^\mu B^\nu.

This is usually the inner product between A and B.

EDIT: The metric has other important functions too.

 

[Careful]

Hi,

It seems nobody answered yet a particular part of your original question which is wether the geometry of spacetime can also be given a distance formulation. It can (in the achronal case), and here are the axioms:
(a) d(x,y) >= 0 and d(x,x) = 0
(b) d(x,y) > 0 implies that d(y,x) = 0.
(c) d(x,z) >= d(x,y) + d(y,z) if d(x,y)d(y,z) > 0

Notice that d can also take the value infinity. d gives a partial order defined by x < y if and only if d(x,y) > 0 as you can easily verify. There is an old approach based upon (a suitably differentiable and causally stable) d to general relativity which is the world function formulation by Synge.

Cheers,

Careful

 

[Oxymoron]

Posted by Careful:
It seems nobody answered yet a particular part of your original question which is wether the geometry of spacetime can also be given a distance formulation. It can (in the achronal case), and here are the axioms:

Thanks Careful. I was hoping there was some sort of metric structure on spacetime.

Posted by George Jones:

George Jones So, isn't an inner product on a vector space V a (0,2) tensor, i.e., an element of V^* \otimes V^*?

Well, after all that we have discussed, I suppose it is.

Posted by Masudr:

The metric tensor helps us "lower" or "raise" indices, thus allowing us to make scalars out of tensors.

So what you are saying is, I can turn a covariant tensor into a contravariant tensor (or a vector into a dual vector) by multiplying it by metric tensor.

 

[George Jones]

TI can turn a covariant tensor into a contravariant tensor (or a vector into a dual vector) by multiplying it by metric tensor.

Let v be a 4-vector. Use the metric g to define the covector \tilde{v} associated with v: for every 4-vector w
\tilde{v} \left( w \right) := g \left( v , w \right).
This is the abstract, index-free version of index lowering. To see this, let \left\{ e_{1}, \dots , e_{n} \right\} be a basis for V, and let \left\{ e^{1}, \dots , e^{n} \right\} be the associated basis for V*. Write g_{ij} = g \left( e_{i} , e_{j} \right).

Write v in terms of the basis for V,
v = v^{i} e_{i},
and \tilde{v} in terms of the basis for V*,
\tilde{v} = v_{i} e^{i},.
Then,
\tilde{v} \left( e_{j} \right) = v_{i} e^{i} \left( e_{j} \right) = v_{i} \delta_{j}^{i} = v_{j}.
But, by definition,
\tilde{v} \left( e_{j} \right) = g \left( v , e_{j} \right) = g \left( v^{i} e_{i} , e_{j} \right) = v^{i} g \left( e_{i} , e_{j} \right) = v^{i} g_{ij}.
Combining these results gives
v_{j} = v^{i} g_{ij}.
Much more to follow.

Regards,
George

 

[Oxymoron]

SUMMARY SO FAR...

A tensor is a multilinear map. In fact, it is just a generalization of a linear functional that was in linear algebra. Therefore, a tensor, in it barest form, is

T\,:\,V_1 \times V_2 \times \dots \times V_N \times V^*_1 \times V^*_2 \times \dots \times V^*_M \rightarrow \mathbb{R}

where V_1,V_2,\dots,V_N are vector spaces over a field \mathbb{R} and V^*_1,V^*_2,\dots,V^*_M are the corresponding dual vector spaces over the same field.

The tensor written above is a tensor of type (M,N). As a result, a tensor of type (0,1) is simply a linear functional

T\,:\, V \rightarrow \mathbb{R}

For every vector space V there exists a dual vector space V^* consisting of all linear functionals on V. From now on we will refer to linear functionals on V as covectors, or 1-forms.

Let V be a finite dimensional vector space. Its dimension, n, is the number of basis vectors on V which are needed to uniquely define a linear functional. Thus,

\dim(V) = n

and there are n basis vectors, \{e_1,e_2,\dots,e_n\}. Likewise for the corresponding dual vector space V, whose dimension is \dim(V^*) = n. The basis of the dual vector space consists of n basis vectors \{\epsilon^1,\epsilon^2,\dots,\epsilon^n\} satisfying

\epsilon_i(e_j) = \delta^i_j

where d^i_j is the Kronecker delta.

Now let's add more structure to our vector space by defining an inner product on it. As a result, V becomes an inner product space with

g\,:\, V \times V \rightarrow \mathbb{R}

defined as

g(u,v) = u\cdot v

This bilinear functional is actually a covariant tensor of degree 2 or simply, the metric tensor. Covariant tensors are of the form:

T\,:\,V \times V \times \dots \times V \rightarrow \mathbb{R}

and contravariant tensors are of the form

T\,:\,V^*\times V^* \times \dots \times V^* \rightarrow \mathbb{R}.

A tensor can also be symmetric as in the case of the metric tensor. Such tensors have the following property

g(u,v) = g(v,u)

that is, if we reverse which way we operate on the elements the sign of the tensor does not change.

One of the fundamental concepts of tensors is the ability to raise and lower the indices. If we have a contravariant tensor, T^i and g_{ij}, as we have seen, is a covariant tensor of degree 2. Then

g_{ij}T^i = T_j

and we say that taking the inner product with the metric tensor has lowered a contravariant index to a covariant index. It can be shown that g is invertible, so

g^{ij}T_j = T^i

where g^{ij} = (g_{ij})^{-1}.

 

[pervect]

SUMMARY SO FAR...
A tensor is a multilinear map. In fact, it is just a generalization of a linear functional that was in linear algebra. Therefore, a tensor, in it barest form, is
T\,:\,V_1 \times V_2 \times \dots \times V_N \times V^*_1 \times V^*_2 \times \dots \times V^*_M \rightarrow \mathbb{R}
where V_1,V_2,\dots,V_N are vector spaces over a field \mathbb{R} and V^*_1,V^*_2,\dots,V^*_M are the corresponding dual vector spaces over the same field.
The tensor written above is a tensor of type (M,N). As a result, a tensor of type (0,1) is simply a linear functional
T\,:\, V \rightarrow \mathbb{R}
For every vector space V there exists a dual vector space V^* consisting of all linear functionals on V. From now on we will refer to linear functionals on V as covectors, or 1-forms.
Let V be a finite dimensional vector space. Its dimension, n, is the number of basis vectors on V which are needed to uniquely define a linear functional. Thus,
\dim(V) = n
and there are n basis vectors, \{e_1,e_2,\dots,e_n\}. Likewise for the corresponding dual vector space V, whose dimension is \dim(V^*) = n. The basis of the dual vector space consists of n basis vectors \{\epsilon^1,\epsilon^2,\dots,\epsilon^n\} satisfying
\epsilon_i(e_j) = \delta^i_j
where d^i_j is the Kronecker delta.

This is very good, and extremely well written, except for a minor but potentially important detail

A basis is , in my understanding, is defined as a set of vectors that are linearly independent and span the space - a basis does not have to be orthonormal, e.g.

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

Thus it is not necessarily true that e_i \cdot e_j = \delta^i_j.

IF e_i \cdot e_j = \delta^i_j THEN \epsilon^i(e_j) = \delta^i_j

So the above statement is not necessarily true, all we need to say at this point is that the \epsilon^i span the dual space and are linearly independent, any other claim is imposing more structure on them than is necessarily true.

Now let's add more structure to our vector space by defining an inner product on it.

After we've done this, we can talk about the e_i being orthonormal, and we can also make the remark that this is equivalent to \epsilon^i (e_j) = \delta^i_j

 

[Oxymoron]

After we've done this, we can talk about the e_i being orthonormal, and we can also make the remark that this is equivalent to \epsilon^i (e_j) = \delta^i_j

You're right. I should have written this first. Thankyou for the compliments though.

Before I go on, I think I left something out. We can form a tensor space by collecting all tensors of a fixed type (r,s). This space is actually a vector space, and the tensors can be added and subtracted together by real numbers. The problem for our tensor space is in defining a basis for it. This is where we need a new operation called the tensor product, denoted as \otimes.

If T is a (k,l) tensor and S is a (m,n) tensor we define T\otimes S as a (k+m,l+n) tensor, by

T\otimes S (\omega^1,\dots,\omega^k,\dots,\omega^{k+m},V^1,\dots,V^l,\dots,V^{l+n})
= T(\omega^1,\dots,\omega^k,V^1,\dots,V^l)S(\omega^{k+1},\dots,\omega^{k+m},V^{l+1},\dots,V^{l+n})

where \omega^i and V^i are distinct dual vectors and vectors. That is, we define the tensor product of T and S by first acting T on the appropriate set of dual vectors and vectors, and then act S on the remainder, and multiply everything together. But this operation is not commutative.

We can now construct a basis for the space of all fixed type tensors, by taking the tensor product of basis vectors and dual vectors. The resulting basis will consist of all the tensors of the following form:

e_{i_1}\otimes \dots \otimes e_{i_k} \otimes \epsilon^{j_1} \otimes \dots \otimes \epsilon^{j_l}

Therefore, every tensor T of the fixed type (k,l) has a unique expansion:

T = T^{i_1\dots i_k}_{j_1\dots j_l} e_{i_1} \otimes \dots \otimes e_{i_k} \otimes \epsilon^{j_1} \otimes \dots \otimes \epsilon^{j_l}

where T^{i_1\dots i_k}_{j_1\dots j_l} = T(\epsilon^{i_1},\dots,\epsilon^{i_r}, e_{j_1}, \dots, e_{j_s}), which are simply the components of the tensor with respect to the basis of V.

But expressing a tensor T as T^{i_1\dots i_k}_{j_1\dots j_l} e_{i_1} is just like expressing a vector by its components - merely a shortcut.

 

[masudr]

Yes that looks good and fairly complete, so far. Some of my understanding might also help.

A nice thing to remember is that components of a vector transform like a vector, but the basis vectors transform like components of a covector; since the vector is actually the components multiplied by the basis vector, our vector is invariant! Similar considerations apply to covector components and basis covectors (although, the opposite, of course). And that's why, after all, tensors are objects which do not change under co-ordinate transformations (don't get confused: the components transform, but the basis transforms in the opposite way).

Another thing to note is that often the simplest basis for a vector space at some point is the co-ordinate basis, namely the vectors pointing along the co-ordinate lines at any point. And a covector basis can be defined as those linear functionals, when acting on the corresponding co-ordinate basis, gives us the Kronecker delta.

PS. I should formalise this in LaTeX, but it's pretty late here.

 

[George Jones]

You're right. I should have written this first.

Actually, in the spite of the presence of a Kronecker delta, the relation \epsilon^i (e_j) = \delta^i_j has nothing to do either with orthogonality, or with a metric tensor. Given a basis for a vector space V, this relation can always be used to define a basis for the vector space V*, the algebraic dual of V. No metric tensor is needed. Also, the relation does not define a metric tensor.

Orthogonality is a condition between vectors in the same vector space, and the e_j and the \epsilon^i live in different vector spaces. The initial basis for V need not be an orthonormal one in order to use the above relation to define a basis for V*.

The construction that I outlined towards the end of post #40 made no use of a metric tensor.

Regards,
George

 

[pervect]

Actually, in the spite of the presence of a Kronecker delta, the relation \epsilon^i (e_j) = \delta^i_j has nothing to do either with orthogonality, or with a metric tensor. Given a basis for a vector space V, this relation can always be used to define a basis for the vector space V*, the algebraic dual of V. No metric tensor is needed. Also, the relation does not define a metric tensor.

Why do you say that?

It seems to me that the relation does define a dot product, and hence the metric tensor, in a very natural way.

If \epsilon^j(e_i) = \delta^i_j, you have defined a particular mapping from basis vectors to basis co-vectors. e_i is associated with \epsilon^i, the co-vector with the same index.

Now it is possible that you do not want to make use of this relationship, but if you don't want to use it, why specify it? I.e. \epsilon^j(e_i) = \delta^i_j has the purpose of singling out a unique co-vector that is associated with a vector with the same index. If there is some other purpose for writing this relationship down, please enlighten me, because I'm missing the point :-(.

Given that we actually make use of this association, we now have a map from the basis vectors to the basis co-vectors - for every e_i, we have singled out a unique \epsilon^i with the same index, thus we have assigned a unique co-vector to every basis vector.

Because an arbitrary vector can be defined by a weighted linear sum of basis vectors, we have also defined a map from every vector u to a unique dual vector, found by substituting all the e_i with the \epsilon^i and keeping the same linear weights.

Given a mapping between vectors and duals, we have defined a dot product.
Given two vectors u and v, we use the above mapping to find u*. u* is a map from a vector to a scalar. Applying this map u* to the vector v, we then have a scalar. Thus we have a mapping - a linear mapping, though I've skipped over proving this - from two vectors (u,v) to a scalar. This linear mapping between two vectors and a scalar defines the dot product, and a metric tensor.

[add]
Something I should add - in this dot product we have defined above, we can now ask - what is e_i \cdot e_j? A little work shows the answer is \delta^i_j. Thus our basis vectors are orthonormal.

The way I view the end result is that if we have a vector space, from any set of basis vectors we can form a consistent dot product in which those basis vectors are orthonormal. However, when dealing with physics, we have a physical notion of distance which imposes a particular dot product, one that arises from the physics (our ability to measure distances). Thus we restrict the mathematical notions of possible dot product to those that are physically meaningful.

 

[George Jones]

Why do you say that?

Let me start with an example. Let (V,g) be Minkowski vector space, and let {e_0, e_1, e_2, e_3} be an orthonormal(i.e, 1 = g(e_0 , e_0) = -g(e_1 , e_1) = -g(e_2 , e_2) = -g(e_3 , e_3)) basis. There is nothing that prohibits taking u = v in an inner product. Let u = v = e_0 + e_1. Using the bilinearity and symmetry of g gives

g(u,u) = g( e_0 + e_1 , e_0 + e_1) = g(e_0 , e_0) +2g(e_0 , e_0) + g(e_1 , e_1) = 1 + 0 + (-1) = 0.

u is a lightlike vector as expected.

Now calculate the inner product using your construction. This gives

(e^0 + e^1) (e_0 + e_1) = e^0(e_0 + e_1) + e^1(e_0 + e_1) = 1 + 0 + 0 + 1 = 2

Regards,
George

 

[George Jones]

I had a similar http://groups.google.ca/group/sci.physics.research/msg/5405b4c4304b8304?dmode=source&hl=en" sci.phyics.research as well.

Regards
George