Math "Newb" Wants to know what a Tensor is [Archive]

chroot

Jul22-04, 03:50 PM

Even more generally, a tensor is a sort of mathematical machine that takes one or more vectors and produces another vector or number.

A tensor of rank (0,2), often just called rank 2, is a machine that chews up two vectors and spits out a number.

A tensor of rank (0,3) takes three vectors and produces a number.

A tensor of rank (1,2) takes two vectors and produces another vector.

Hopefully you see the pattern.

You actually already know what a (1,1) tensor is -- it's nothing more than a good ol' matrix. It accepts one vector and produces another vector.

If you're working in three dimensions, a (1,1) tensor can be represented by its nine components. Here's a simple (1,1) tensor.

T = \left(
\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}
\right)

You already know what this guy does -- it takes a vector and gives you back the exact same vector. It's the identity matrix, of course. You would use it as such:

\vec v = T \vec v

If the world were full of nothing but (1,1) tensors, it'd be pretty easy to remember what T means. However, there are many different kinds of tensors, so we need a notation that will help us remember what kind of tensor T is. We normally use something "abstract index notation," which sounds more difficult than it is. Here's our (1,1) tensor, our identity matrix, laid out in all its regalia:

T^a_b

The a and b are referred to as indices. The one on the bottom indicates the tensor takes one vector as "input." The one of the top indicates it produces one vector as "output."

Tensors don't have to accept just vectors or produce just vectors -- vectors are themselves just a type of tensor. Vectors are tensors of type (0,1). In full generality, tensors can accept other tensors, and produce new tensors. Here are some complicated tensors:

R^a{}_{bcd}\ \ \ \ G_{ab}

The second one, G_{ab} is a neat one to understand. You should already understand from its indices that it is a type (0,2) tensor, which means it accepts two vectors as input and produces a number as output. It's called the metric tensor, and represents an operation you already know very well -- the dot product of two vectors!

In normal Euclidean 3-space, G_{ab} is just the identity matrix. You can easily demonstrate the following statement is true by doing the matrix multiplication by hand:

\vec u \cdot \vec v = G_{ab} \vec u \vec v

The metric tensor is more complicated in different spaces. For example, in curved space, it's certainly not the identity matrix anymore -- which means the vector dot product is no longer what you're used to either when you're near a black hole. Tensors are used extensively in a subject called differential geometry, which deals with, among other topics, curved spaces. General relativity, Einstein's very successful theory which explains gravity as the curvature of space, is cast in the language of differential geometry.

So there you have it: tensors are the generalization of vectors and matrices and even scalars. (Scalars, by the way, are considered to be type (0,0) tensors.)

I should mention that there not all mathematical objects with indices are tensors -- a tensor is a specific sort of object that has the transformation properties described by others in this thread. To be called a tensor, an object much transform like a tensor. Don't worry though, you're not going to run into such objects very often.

- Warren

mathwonk

Jul23-04, 06:53 PM

wow! this is the simplest clearest explanation of tensors i have seen. thank you.

pmb_phy

Jul24-04, 03:43 AM

See - http://www.geocities.com/physics_world/ma/intro_tensor.htm

Pete

pmb_phy

Jul24-04, 03:58 AM

Even more generally, a tensor is a sort of mathematical machine that takes one or more vectors and produces another vector or number.

That's the definition of a general tensor. You neglected to add that the machine must be linear. There are also general tensors which accepts one-forms as an arguement.

An (M,N) tensor is a linear function of M one-forms and N-vectors into the real numbers.

An affine tensor is another kind of tensor. Examples of affine tensors are Cartesian tensors and Lorentz tensors. These are defined according tothow their components transform under a restricted transformation in a given space. E.g. the transfrormation relavent to a Cartesian tensor is the orthogonal transformation. See -- http://www.geocities.com/physics_world/ma/orthog_trans.htm

The transformation relavent to a Lorentz tensor is the Lorentz transformation.

For details see

Tensors, Differential Forms, and Variational Principles, Lovelock & Rund, Dover Pub., (1989)

Pete

chroot

Jul24-04, 04:08 AM

Of course I left out a bit pmb, I wasn't trying to get into the concepts of one-forms versus vector fields just yet. :smile: Thanks for the additions, though.

- Warren

pmb_phy

Jul24-04, 09:45 AM

Of course I left out a bit pmb, I wasn't trying to get into the concepts of one-forms versus vector fields just yet. :smile: Thanks for the additions, though.

- Warren
You're welcome chroot. I had a feeling you were being sparse intentionaly but wanted to give the StonedPanda more of the details just in case he was overly stoned and needed that much more. :biggrin:

The part about affine tensors I find to be an important addition because not knowing the use of the different kinds of tensors can be confusing. For instance: In D'Inverno's text Introducing Einstein's Relativity even the author goofed up on this subtle point. I.e. D'Inverno defines tensors in a manner which is equivalent to how you have defined them, i.e. as general tensors (by "equivalent" I refer to the definition of tensor as that whose components transform in a particular way). D'Inverno then goes on to define the angular momentum tensor (an important tensor in SR) on page 118

L^{\mu\nu} = x^{\mu}p^{\nu} - x^{\nu}p^{\mu}

This is certainly not a general tensor since xu isn't a general tensor. However it is a Lorentz tensor and therefore Luv is a Lorentz tensor and not a general tensor.

Pete

chroot

Jul26-04, 03:07 PM

I have split off Epsilon Pi's responses to a new thread in Theory Development,

http://www.physicsforums.com/showthread.php?t=36791

because they are not relevant to this discussion.

- Warren

StonedPanda

Aug13-04, 01:59 AM

You're welcome chroot. I had a feeling you were being sparse intentionaly but wanted to give the StonedPanda more of the details just in case he was overly stoned and needed that much more.
I definitely picked the wrong name for this forum!

But thanks!

At what level or mathematics does one encounter tensors?

MoonUnit

Aug19-04, 10:16 PM

L^{\mu\nu} = x^{\mu}p^{\nu} - x^{\nu}p^{\mu}

This is certainly not a general tensor since xu isn't a general tensor. However it is a Lorentz tensor and therefore Luv is a Lorentz tensor and not a general tensor.

Is there any chance you could enlighten me here - I follow what was said earlier regarding sub/superscript representing the vectors that are input and output. But, in the equation quoted there appears to be no input vector, but an output - what is going on there?

I'm basically asking for a clarification of the meaning of the notation when you just have superscripts, not for any particularly detailed mathematical explanation, as some of these maths threads are way over my head.

The other thing is I'm a bit unclear on the exact meanings of covariant and contravariant. Does anyone know a website, or can give an explanation where these terms are explained geometrically rather than in mathspeak. (To put it in context here - I love physics, but I'm not particularly good at maths when it is explained to me in terms of maths.... I guess I just need to see what is happening physically when these terms are defined.)

chroot

Aug19-04, 11:38 PM

A tensor with just a single superscript is nothing more than an ordinary vector.

x^\mu = \left(
\begin{array}{c}
x^0\\
x^1\\
x^2\\
x^3
\end{array}
\right)

where x^0 and so on are just numbers (scalars).

The contravariant vector is the normal vector you're used to working with. Covariant vectors are dual to contravariant vectors. Contravariant vectors are column vectors, while variant vectors are row vectors. If you multiply a covariant vector by a contravariant vector, you get a number out:

x^\mu x_\mu = a

- Warren

pmb_phy

Aug20-04, 04:48 AM

chroot

Aug20-04, 01:47 PM

I'm actually interested in MoonUnit's last paragraph:
The other thing is I'm a bit unclear on the exact meanings of covariant and contravariant. Does anyone know a website, or can give an explanation where these terms are explained geometrically rather than in mathspeak.
I'm not too clear on the geometrical meaning myself, though I can go through the motions and manipulate tensor expressions just fine.

A contravariant vector exists in the tangent space of a specific point in the manifold being considered. In other words, if you have a basketball (the surface of which is a 2D manifold), and you glue little toothpicks tangent to it, each toothpick is a contravariant vector defined at the point it is glued to the basketball. This much makes intuitive sense to me -- I can just as easily think "tangent" whenever someone says "contravariant." If you take any point on the basketball, the set of all tangent vectors you could glue to it there forms a (2-dimensional) tangent space at that point. Contravariant vectors at that point belong to that tangent space.

Now, I know covariant vectors live in cotangent spaces, but I'm not really clear on how to visualize a cotangent space. I have read most of John Baez' book "Gauge Fields, Knots, and Quantum Gravity," in which he makes an earnest attempt to help the reader visualize a covariant vector -- but it falls flat on me. I just can't understand what he's trying to say.

Does anyone have a clear explanation of how to "visualize" a covariant vector? Is it really even possible to visualize it?

- Warren

robphy

Aug20-04, 02:39 PM

Does anyone have a clear explanation of how to "visualize" a covariant vector? Is it really even possible to visualize it?

A covector can be visualized as a stack of parallel-planes. (Generally, a pair of parallel planes is sufficient. Visualize "twin blades".)
The spacing between planes is inversely-proportional to the size of the covector.
Think: local approximations of "equipotential surfaces".

The contraction of a vector and a covector can be interpreted in terms of the number of piercings (MTW's "bongs of a bell") of the stack by the vector.

chroot

Aug20-04, 02:52 PM

robphy

Aug20-04, 03:10 PM

robphy,

That's basically how Baez tried to explain it, and I basically still don't get it. Where are these parallel (hyper)planes supposed to be? Equipotential of what?

Are the hyperplanes of the covariant vector perpendicular to the contravariant vector?

- Warren

Take the electrostatic field as a physical application.

The electric potential \phi is a scalar field.
The object -\nabla_a \phi is covector field (a field of one-forms).
Visually, the covectors are these twin-blades tangent to the equipotentials of \phi .

Consider a displacement vector d^a.
Suppose that d^a is parallel to an equipotential. The contraction -\nabla_a \phi d^a (that is \vec E\cdot \vec d) is zero. Thus, the potential difference from tail to tip is zero.

Suppose that d^a is perpendicular to an equipotential (i.e. parallel to the gradient vector). The contraction -\nabla_a \phi d^a (that is \vec E\cdot \vec d) is nonzero because d^a pierces some planes in the stack. Thus, the potential difference from tail to tip is nonzero.

With d^a fixed, suppose that electric potential \phi is doubled in strength. We expect the potential difference from tail to tip to double. To obtain twice the number of piercings for the same d^a, the spacing between the twin-blades must be halved.

Here is a nice article by Jancewicz:
http://arxiv.org/abs/gr-qc/9807044

Other tensors (e.g., differential forms) can be visualized similarly.

chroot

Aug20-04, 03:15 PM

Thanks robphy,

I'll try to chew on this a little more. I feel odd that I have a good grasp of the mathematical difference between covariant and contravariant vectors, but am not be able to describe them in any other way... :redface:

- Warren

pmb_phy

Aug21-04, 04:11 AM

Does anyone have a clear explanation of how to "visualize" a covariant vector? Is it really even possible to visualize it?

robphy's description is a specific one and, of course, is correct as that description goes. However such a description tends to let one confuse the tangent space with the dual space. It can give the false impression that covariant and contravariant vectors are the same kind of animals. e.g. in Euclidean geometry the covariant basis vectors are typically defined as

\bold e^i = \nabla x^i

where the xi are coordinates. The contravariant vectors are typically defined as

\bold e_i = \frac{\partial \bold r}{\partial x^i}

When you visualize these two objects it gives you the impression that you can represent one as a sum of the others since each are vectors in the same space. But that is not a true statement in general. In general, covariant and contravariant vectors lie in different spaces altogether (e.g. such an addition would be like trying to represent a bra as a linear sum of kets). It requires a metric just to relate the covariant vectors to the contravariant vectors. With no metric no such relationship exists.

I wish there were a general way to visualize 1-forms but I haven't seen one yet.

Pete

MoonUnit

Aug21-04, 04:12 AM

A tensor with just a single superscript is nothing more than an ordinary vector.

Thank you Chroot!

I've been baffled by that for a while now - and I'm no longer baffled.

Yet again the curse of learning yet another notation.... I really wish we could get some linguists in to give mathematical language (symbolic and descriptive) an overhaul to redefine the way things are written so as to make a more precise language for maths, which makes intuitive sense without requiring quite as much raw indoctrination as we have to endure now :)

And thanks to everyone for the discussion on the geometrical meaning - it does make it somewhat clearer, although I'm still working my way through robphy's post.

I do have another query though, and that's just a clarification of the expressions of the type "(stuff) exists in the (other stuff) space" - like "contravariant vector exists in the tangent space" as chroot said. What IS the tangent space - is it a (2D) surface which basically follows the surface of the basketball? How do I generalise it so that when someone I come across a new one, I might be able to picture it :) What steps do you go through in your head for visualising (or just translating) "in a (whatever) space". What advantage does thinking of tangent space have over just thinking of all the tangential vectors on the surface? There must be an application of it, or some subtlety I'm missing that explains the need for tangent space and so on. In QM we touched on momentum space, but all that seemed to be was just a graph in which psi(x) vs. x became phi(p) vs. p, and momentum was the x-axis, yet it really didn't explain what "momentum space" was. I almost get it but not quite....

Finally definitions again - contravariant and covariant.... they are contra/covariant with respect to what?

Thanks for your input - it is helpful! Sorry if I'm asking really dense questions, but most of my friends don't really care to discuss this kind of thing....

chroot

Aug21-04, 04:36 PM

If you take my basketball again, and consider some specific point p on it, you may glue an infinite number of different toothpicks to that point, all pointing in different directions. That's a 2D vector space -- a plane.

The term "tangent space" is just used to mean the 2D vector space tangent to the basketball at some point p.

The reason we refer to vectors living in different spaces is simple. If you imagine my basketball again, there's one tangent space at a specific point p, e.g. on the equator, and a different tangent space at e.g. a point q like the north pole. A vector at p and a vector at q are quite literally in totally different vector spaces. You can't add them, or otherwise relate them, because they literally exist in different spaces. Every point on the basketball's surface has its very own tangent space.

You can relate the tangent spaces at two distinct points with a special mechanism called "parallel transport." In parallel transport, you effectively "push" the vector along the surface of the manifold from one point p to another point q. The vector is naturally changed by this process; a vector tangent to the basketball's surface in one place remains tangent to its surface at all points when you push it around the surface, and may point in a totally different direction after being pushed around. Parallel transport provides a rigorous definition of a relationship between vectors in the tangent space at p and vectors in the tangent space at q.

- Warren

robphy

Aug21-04, 05:44 PM

In general, covariant and contravariant vectors lie in different spaces altogether (e.g. such an addition would be like trying to represent a bra as a linear sum of kets). It requires a metric just to relate the covariant vectors to the contravariant vectors. With no metric no such relationship exists.

I wish there were a general way to visualize 1-forms but I haven't seen one yet.

Pete

Yes, it is true that covectors and vectors lie in different spaces, and, as you say, cannot be added together. (That is why these two directional quantities have different representations.) However, they can be contracted since a covector is a linear map on the space of vectors to the reals. No metric is required for this operation.

Although my example with the electrostatic field was a little loose (where I snuck in the Euclidean metric in a few places), I can make it more precise if requested.

As you suggest, given a metric, one can map a vector to a covector. If that metric is invertible, one can map a covector to a vector. (In index-speak, this is lowering and raising of an index.) In fact, there is a geometric construction that can take a vector (represented by an arrow) and a Euclidean metric (represented by an ellipse [generally, an ellipsoid]) to produce the covector (represented by the twin-blades) obtained by lowering the index of the vector. That same construction can be done in reverse to raise the index of the covector. Thus, this construction shows that these representations are duals of each other.

In addition, for covectors (represented as twin-blades), there is an analogue of the parallelogram rule for adding vectors.

Some references for this visualization of a 1-form is, in addition to MTW 2.5,
http://www.ucolick.org/~burke/forms/draw.ps (need ghostview)
http://www.ucolick.org/~burke/forms/tdf.ps
http://content.aip.org/JMAPAQ/v24/i1/65_1.html
http://www.ee.byu.edu/ee/forms/forms-home.html
which I believe all derive from Schouten's Tensor Analysis for Physicists.

mathwonk

Aug22-04, 01:47 PM

Ohmigosh, I worked for over an hour on how to picture a covector and a differential form, and the web connection just shut down and lost it all.

Well, that spares you a long post.

Here is just the last sentence from it: Professor Bott (a famous engineer - turned - topologist] used to say a cocyle, i.e. differential form, is something that "hovers over the space, looking for a cycle [i.e. a path], and when it sees one, it pounces on it gobbles it up, and spits out a number".

What is more visual than that!? Sort of like a hungry bird of prey.

Most of my post was devoted to recalling how the familiar coordinate functions x and y on the euclidean plane are covectors, and the grid of parallel lines on a graph paper are the corresponding parallel families of "level sets" for those covectors, i.e. sets where the coordinate takes the same value.

This lets us picture at least the level sets of covectors within the vector space it self.

Beer-monster

Aug27-04, 05:29 PM

Okay I've read this post and the other refering to tensors, and followed the links, and I'm still lost.

Could someone please tell me (without technical words and maths jargon :biggrin: ) what a tensor is, what a one form is and how they relate to each other and a vector. Also what is dual space?

Sorry to say this (and quite embarassed really) but somehow I've managed to get through 2 years of a physics degree and the maths involved (though linear algebra was never my strong suit) without understanding the meaning of words like homogenous, covector etc. Not quite sure how that happened. I can somehow understand what happening until someone uses technical mathematical terms.

So is it possible to get a qualitive conceptual picture of these terms (i.e not refering to it as a function that does such and such just yet, as that never seems to work for me). Perhaps I'm just an idiot :yuck: , or have a weird brain. Maybe someone could explain why we need these terms and what they describe in relation to GR and other physical situations. That seems to have worked with other things (though I could go through the motions, I never really understood vector calculus, line intergrals etc until I studied Maxwell's laws and EM field theory). One of my texts says that one-folds and dual space is like bra-kets in QM, so is a one fold like a complex vector (it exists in another form of space as a complex wavefunction exists in imaginary space)

Sorry if it seems like I'm asking a lot, but I would really like a deeper insight into GR and cosmology and seem to be getting no where. Anyway I'll leave it there coz I'm feeling rather stupid, right now. :wink:

robphy

Aug28-04, 08:02 AM

These aren't definitions... just ways one can think about things. (What I am saying is nothing new... you'll probably find these in the various threads on this topic.)

Hopefully, you remember matrix multiplication. We'll stick to three dimensions.

Think of a vector \overline v as a "column vector" \left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right) with components v_1, v_2, v_3 . Of course, you can add such column vectors together and do scalar-multiplication on them.

Think of a dual-vector (or covector) \underline u as a "row vector" \left( u_1\quad u_2\quad u_3 \right) with components u_1, u_2, u_3 . Of course, you can add such row vectors together and do scalar-multiplication on them.

What makes the "space of dual-vectors" the dual to the "space of vectors" is that there is a rule that: given a vector \overline v,

there is a dual-vector \underline u that can produce a scalar, denoted as \underline u \overline v, which you calculate by matrix multiplication: \underline u \overline v=\left( u_1\quad u_2\quad u_3 \right)\left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right)=s
such that a linear combination of dual-vectors a \underline t+ b\underline u applied to \overline v results in ( a \underline t+ b\underline u)\overline v=(a \underline t)\overline v+ (b\underline u)\overline v=a(\underline t \overline v) + b(\underline u \overline v)
that is, a linear combination of the scalars that you get from each dual-vector applied to \overline v separately.

In QM, the bra <\phi| is a dual-vector and the ket |\psi> is a vector.

Hopefully, this helps get you started. If so, maybe the other posts in these various threads are more digestible.

poophead

Sep11-04, 06:46 PM

I'm a bit confused now about the commutivity (if such a thing exists) of tensors. I think part of my confusion stems from my lack of familiarity with this summation notation.

For matrices, I know that AB != BA but what about for tensors?

More specifically, I'm curious about when indices don't necessarily repeat throughout the whole expression.

Does \Lambda^{\mu}_{\ \rho} H^{\rho\sigma}a_{\nu}\Lambda^{\nu}_{\sigma} give the same thing as \Lambda^{\mu}_{\ \rho}\Lambda^{\nu}_{\sigma} H^{\rho\sigma}a_{\nu} ?

da_willem

Sep12-04, 04:00 AM

Yes! Because the summation convention requires you multiply elements with the same (dummy-)index (one sub- and one super-scripted) and sum over all allowed values of the dummy index. So it does not matter in what order the tensors appear; simply because you multiply normal numbers for wich ab=ba !

Example: A_\nu B^\nu = A_0 B^0 + A_1 B^1 +A_2 B^2 +A_3 B^3 = B^\nu A_\nu

Where the last equality comes from the fact that because e.g. A_0 B^0 = B^0 A_0 . For they are just numbers.

poophead

Sep12-04, 05:07 PM

Ohhh.... excellent. Thanks for the explanation, it's all starting to make sense now. :smile:

EDIT:

Oops, I had just one more question.

What the heck is T^{\mu\nu} ? How exactly is that different from T^\mu_{\ \nu}? Are they basically the same thing or are we talking about differences in terms of contravariant and covariant?

da_willem

Sep13-04, 06:37 AM

These are different, and indeed the difference is that the latter has a co- and a contra-variant component while the firts has two contravariant components. You can relate the two by the metric: T^a_b=g_{bc} T^{ac} (This is called 'lowering an index'). In Euclidean space these are the same because the metric is the identity matrix, but in a general Riemannian space the components can be different.

mathwonk

Sep13-04, 07:35 PM

chroot, are you actually a cigar?

gvk

Sep16-04, 02:41 PM

I'm not too clear on the geometrical meaning myself, though I can go through the motions and manipulate tensor expressions just fine.

A contravariant vector exists in the tangent space of a specific point in the manifold being considered. In other words, if you have a basketball (the surface of which is a 2D manifold), and you glue little toothpicks tangent to it, each toothpick is a contravariant vector defined at the point it is glued to the basketball. This much makes intuitive sense to me -- I can just as easily think "tangent" whenever someone says "contravariant." If you take any point on the basketball, the set of all tangent vectors you could glue to it there forms a (2-dimensional) tangent space at that point. Contravariant vectors at that point belong to that tangent space.

Now, I know covariant vectors live in cotangent spaces, but I'm not really clear on how to visualize a cotangent space. I have read most of John Baez' book "Gauge Fields, Knots, and Quantum Gravity," in which he makes an earnest attempt to help the reader visualize a covariant vector -- but it falls flat on me. I just can't understand what he's trying to say.

Does anyone have a clear explanation of how to "visualize" a covariant vector? Is it really even possible to visualize it?

- Warren
First of all, covariant and contravariant vectors are not different vectors. They represent ONE VECTOR (an arrow :-) in two different coordinate systems (dual, or reciprocal, or skew, or...coordinates). The reciprocal system is equally satisfactory for representing vectors, but 'contravariant' vector looks exactly the same as 'covariant'. So "visualize" them as ONE tangent arrow (toothpick) if you wish. Two paralell blades, probably, mean direct and reciprocal coordinate planes, which may have complement scale or orientation, but, of course, should be parallel (no less no more).
Secondly, any quantity that we wish to define, be it scalar, vector, or tensor, must be independent of the special coordinate system. We shell adopt this as fundamental principal. However, its representation will depend on the particular system.

chroot

Sep16-04, 02:53 PM

jcsd

Sep16-04, 04:12 PM

First of all, covariant and contravariant vectors are not different vectors. They represent ONE VECTOR (an arrow :-) in two different coordinate systems (dual, or reciprocal, or skew, or...coordinates). The reciprocal system is equally satisfactory for representing vectors, but 'contravariant' vector looks exactly the same as 'covariant'. So "visualize" them as ONE tangent arrow (toothpick) if you wish. Two paralell blades, probably, mean direct and reciprocal coordinate planes, which may have complement scale or orientation, but, of course, should be parallel (no less no more).
Secondly, any quantity that we wish to define, be it scalar, vector, or tensor, must be independent of the special coordinate system. We shell adopt this as fundamental principal. However, its representation will depend on the particular system.

I think what may of confused you is that for a Euclidean vector space the contravariant and covariant vectors are the 'same' (i.e. in any given frame a pair of dual vectors have the same compoents) as the compoents of the metric are simply the compoents of an identity matrix.

But in general a vector and it's one-form belong to different spaces, infact the dual space of some linear vector space S is the set of all linear functions that map a vector in S to some (in genral complex) number and it also constitutes a linear vector space in it's own right. Further the components of a pair of dual vectors Aν and Aν in the same frame are not in general the same (also belonging to differnet spaces they have different bases).

mathwonk

Sep16-04, 04:25 PM

a "vector" is the derivative of a curve, i.e. of a map from R^1 to R^n, whereas a covector (dual vector) is the opposite, the gradient of a function from R^n to R^1. (at one point)

That's why usually a vector is a column, and a covector is a row. So one way to remember them is by the sound "covector" equals "rowvector".

Geometrically a vector is a tangent vector to a curve, while a covector (dual vector, gradient vector) is a normal vector to a level surface.

(Unfortunately covectors are only covariant in differential geometry, they are contravariant in category theory, algebraic topology, and the rest of mathematics.)

gvk

Sep17-04, 10:49 AM

The difference between a vector and a dual vector is not just a change of coordinate system, gvk. Vectors and dual vectors live in entirely different spaces, and are certainly not the same vector.

I know what you're trying to say: a vector is related to its dual via a one-to-one mapping.

- Warren
No, not only this I'm trying to say, Warren.
You completely forgot who asked "what a tensor is?" and what is the level of this person. He/she is a student of High School! If you will teach them in such manner, they never understand anything about vector and tensor calculus at all, and will never ask you again.
It is the same as you start to teach them QM before they learn anything about classical mechanics. But educational sequence in math plays even more important role than in natural sciences. Here everything should be sequential and well understood in elementary manner.
So, they first need understand that "a vector is an arrow or column", and "a tensor is a matrix" and what are the properties of these notions in Euclidean space. This is most important, because they may use the tensors in the future just to caclulate stress and deformations in materials. :smile:

mathwonk

Sep17-04, 12:13 PM

chroot

Sep17-04, 01:28 PM

You completely forgot who asked "what a tensor is?" and what is the level of this person.
You were replying to a question of mine, not the original poster's.
He/she is a student of High School! If you will teach them in such manner, they never understand anything about vector and tensor calculus at all, and will never ask you again.
Purposefully simplifying material for pedagogical purposes does not give you license to say things that are patently false. You could say "vectors and dual vectors are closely related, and can be thought of as different "representations" of the same fundamental thing -- that would be fine. Saying that vectors and dual vectors are related through a coordinate transformation, on the other hand, is quite wrong, and would probably do more harm to the student's understanding than good.

- Warren

gvk

Sep20-04, 03:41 PM

You were replying to a question of mine, not the original poster's.
Purposefully simplifying material for pedagogical purposes does not give you license to say things that are patently false. You could say "vectors and dual vectors are closely related, and can be thought of as different "representations" of the same fundamental thing -- that would be fine. Saying that vectors and dual vectors are related through a coordinate transformation, on the other hand, is quite wrong, and would probably do more harm to the student's understanding than good.
- Warren

Do you think the student should start first to learn non-euclidean geometry,
modern notion 'dual space', and then come back to Euclidean case?
I don't think this is a right way. I deeply convince and say again: math education should be in sequential order, any new stuff should overlap the current level of knowledge without gaps, any new material should be accompanied with lots of simple examples, any new notions should be explained in connection with old ones. Student should learn first the classical stuff with contravarient, covarient notations for vectors and tensors (using direct and reciprocal systems), and then turn to more complex.

And I did not tell that "vectors and dual vectors are related through a coordinate transformation". This is your interpretation.
Read carefully: "covariant and contravariant vectors are not different vectors. They represent ONE VECTOR (an arrow :-) in two different coordinate systems". I meant, of course, euclidean case.
Is that 'patently false'?

gvk

Sep20-04, 04:29 PM

i cannot agree with you less, gvk. this high school, student is also a student of multivariable calculus at UCLA. As such I think he is capable of understanding what things mean, not just the symbols used for them.
What are you talking about?
I have a son in High School of MN, and have 30 yr. experience of teaching (in other country), and can tell you that US school's education in math is about 1.5-2,5 years behind the Europe's. Moreover, it's 1yr. behind what I learned at the same age as son many many years ago.
At the age 14 we knew all elementary and trigonometric functions, algebra (linear and quadratic equations), geometry on the plane and much much more. Now, in 10th grade, they start to learn so called 'integrated math' which has only 1/4 what i mentioned.
How do you expect they will to know the term "rank" of matrix or "homogeneous" transformation at the end of HS?
I guess, it is impossible with this level of math education, even a guy attended multivariable calculus at UCLA, he still missed a lots of math.

chroot

Sep20-04, 04:45 PM

gvk,

A vector and its dual do not even exist in the same vector space, even when the spaces are Euclidean. It is indeed patently false to assert they are in any way related through a coordinate transform, no matter what kind of spaces you're dealing with. I feel sorry for your students if you are willing to commit such grave errors for nothing more than simplifying your pedagogy.

- Warren

mathwonk

Sep21-04, 12:05 AM

You do not seem to realize, gvk, that not all high schools in the US are the same low level. You are of course correct that the average high school and even most high schools, are at a low level, but there are some high schools which are different. And even at average high schools there are students who are different.

I have for example taught at a high school where at least some of my 10th graders learned completeness of real numbers, countability and uncountability, archimedean axioms, and limits, with complete proofs. Other high schoolers there studied Galois theory with me. In another class I taught linear algebra, matrices, vector geometry of n dimensions, and calculus of several variables including integration of differential forms, There are other high schools in US where students are at even higher levels.

I have had one high school student who led my class in graduate PhD preparation level algebra at a major state university. He graduatd from university and high school simultaneously.

Your comments may be correct for a generic case but do not seem to apply to the student we are discussing.

gvk

Oct18-04, 11:58 AM

Could you please give me a little bit more detail about those high schools (city, state). I never heard about such unique cases here.
Sorry for digressions in posts. When I read a reply from StonedPanda and it sounded to me as a discouragment:
"I definitely picked the wrong name for this forum!"

MiGUi

Oct18-04, 12:13 PM

So then, I begin to realize that the low level is a worldwide problem not only in Spain happens. Well, this is a very poor consolation, but... still is

gvk

Oct18-04, 12:27 PM

A vector and its dual do not even exist in the same vector space, even when the spaces are Euclidean.
- Warren

Euclidean vector space with metric is coincided with its dual, so it is the same, R = R*.
The most students learn from elementary examples and only later move to the general definitions. It is not "my pedagogy", it's a natural way of learning. (I am sure you were on the same way too. By the way, in what age and where did you learn about dual space and covarience? Highschool or college?). However, in more general cases, you are right.

chroot

Oct18-04, 12:59 PM

Euclidean vector space with metric is coincided with its dual, so it is the same, R = R*.
The most students learn from elementary examples and only later move to the general definitions. It is not "my pedagogy", it's a natural way of learning. (I am sure you were on the same way too. By the way, in what age and where did you learn about dual space and covarience? Highschool or college?). However, in more general cases, you are right.
I continue to disagree. A vector space and its dual space are not the same, even when the two spaces are both Euclidean. Just because two objects have the same components, for example, does not mean they are really the same. You still need to use the metric to convert between them -- it just happens that the metric is the identity matrix.

You seem to missing the point, anyway. If you're going to teach students anything about dual vector spaces, you should teach them correct things. I'm not saying you should lump all the complexity of dual spaces on them at once, but saying "vectors and dual vectors are related through a coordinate transform" is just patently false. Thankfully, you're not a teacher.

- Warren

mathwonk

Oct19-04, 08:08 PM

gvk, try to understand what chroot is saying. there is big difference between saying there is an isomorphism between two spaces and saying they are the same.

i.e. a "metric" or dot product on euclidean space allows us to make a nice one to one identification of the space and its dual by matching up the vector v with the operator

v.( ), but that does not mean that a vector and dotting with that vector are the same.

This is the common problem with epople struggling elsewhere on this site with understanding "tensors". If you just lok at the notational representation of an object and not at what it means, or how it behaves, you lose most of the understanding, and many things look the same.

The key property of any identification is how it behaves under mappings, and a dualk space behaves exactlky the opposite from the roiginal space in this regard.

I.e. if f:V-->W is a linear mapping, then there is a natural mapping in the other direction on duals f*:W*-->V* taking the functional n:W-->R, acting on vectors in W, to the functional f*(n) = nof: V-->W-->R.

If you use the dot product to identify R^n with its dual, then a linear map of R^n given by a matrix T, should correspond under this identification to the linear map given by the transpose matrix, not to the same matrix.

If the spaces had become "equal" one might think one could use the same matrix.

best wishes,

PS: Some good high schools in US include the Paideia School in Atlanta, Andover and Exeter in New Hampshire I believe, Brooklyn technical high school and Bronx High shool of science in New York. Other students at average high schools attend university while still in hogh school. My older son e.g. easily exceled in calculus classes at Georgia Tech before being accepted to both Stanford and Harvard. I believe the high school student who was the top performer in my university level algebra class was from Gainesville high in Georgia. Westminster in Atlanta is also very strong, and there are many others. New Trier high school in Winnetka? Illinois is very famous.

mathwonk

Oct19-04, 08:35 PM

here is an example of when two different vector spaces can be regarded as almost the same: let V be any vector space over the real scalars, and V* = Hom(V,R) = its dual space, the space of all linear functions from V to R. Then let V** = (V*)*

= Hom(V*,R) be the dual of V*, the space of all linear mappings from V* to R.

Then it is possible to identify V with V**, when V is finite dimensional, so that they are essentially the same. Just let a vector v in V be the map fron V* to R, given by "evaluation at v". I.e. if n:V-->R is a linear operator in V*, define v(n) = n(v).

This, when V is finite dimensional, is an isomorphism with a nice property:

when ever there is given a linear map T:V-->W, the associated map T**:V**-->W**

taking an operator s:V*-->R, in V**, to the operator T**(s) = (soT*):W*-->R, in W**. i.e. such that, if t:W-->R is an element of W*, then (T**(s))(t) = (soT*)(t)

= s(T*(t)) = s(toT).

confusing isn't it?

But anyway, one can "easily show" that for any two maps (SoT):V-->W-->U,

that (SoT)** = S** o T**:V**-->W**-->U**.

The point is that not only can one make the spaces V and V** correspond, oine can also make maps between them correspond in a natural way.

In aprticular, under the isomorphisms above of V-->V** and W-->W**, for any map T:V-->W, the compositions V-->V**-->W**,

and V-->W-->W**, are equal.

Challenge: There is no way in the world to do this for dual spaces! I.e. there is no way whatever to choose an isomorphism V-->V* from each space to its dual space, and a compatible correspondence between maps taking T to T*, with the nice properties above.

this is called category theory, and these well behaving guys, like the correspondence V-->V**, are called "natural transformations". Basically it means you should be aware of how maps between your spaces behave, as much as or more than, how points in your spaces look.

You might enjoy the original article laying out these ideas, by I think Eilenberg and MacLane. now about 50 or 60 years old.

Peterdevis

Oct22-04, 03:37 AM

May be this text can clearify the meanings of covariance and contravariance for the less mathematical people

http://www.mathpages.com/home/kmath398.htm

jcsd

Oct22-04, 06:49 AM

If a quantity is covariant it means it transforms like the basis vectors, contrvriant means it traNsforms oppositely to basis vectors.

mathwonk

Oct22-04, 10:37 PM

look, if f:X-->Y is a differentiable map, and v is a tangent vector to X, and df is the derivative of f, then df(v) is a tangent vector to Y. this means v transforms "covariantly" in category language, i.e. in the same direwction as the map f, i.e.; from X to Y.

If on the other hand, q is a covector on Y, i.e. a function that sends a tangent vector on Y to a number, then df*(q) is a covector on X, as follows: if v is a tangent vector on X, then df*(q)(v) = q(df(v)). thus df* takes covectors on Y to covectors on X, i.e. covectors go in the oposite direction from the map f.

thats all there is to it.

of course the confusion is that in differential geometry, which i suppose means also in physics, the terminology is backwards.

quetzalcoatl9

Oct27-04, 04:21 PM

If on the other hand, q is a covector on Y, i.e. a function that sends a tangent vector on Y to a number, then df*(q) is a covector on X, as follows: if v is a tangent vector on X, then df*(q)(v) = q(df(v)). thus df* takes covectors on Y to covectors on X, i.e. covectors go in the oposite direction from the map f.

mathwonk,

i couldnt agree with you more. this well-defined behavior of differential forms with respect to pullback is one of the most intruiging and useful properties of forms. a more subtle point is also that these are still well defined even if the inverse mapping is not! tensors in general, do not enjoy this property.

i have nothing against the physicists way of doing math, but personally i find the terminology and way that they go about doing tensor analysis extremely confusing. when i was first learning this stuff (and am still learning it) i started by reading books that taught tensors the classical way: "something is a tensor if it transforms in blah blah way under a coordinate change blah blah". i did this for some time until i realized that this is an extremely confusing way of going about it...i then picked up frankel's and bishop's books on the subject and everything made perfect sense.

i think that one of the main hurdles is that in classical tensor analysis, ppl dealt with the components of a tensor rather than the tensor itself! this is still a major point of confusion in the subject. furthermore, why bother saying that a tensor is something that changes in a certain way with coordinate change, when it is nearly always impractical to work in terms of coordinates anyway??

Peterdevis

Nov8-04, 08:04 AM

jcsd

Nov8-04, 08:30 AM

This is correct.

from http://www.mathpages.com/rr/s5-02/5-02.htm

No it isn't; as has been pointed out before so-called contravariant and covariant vectors belong to two different vector spaces (infact a covariant vector is infact a linear function of a cotravariant vector and vice versa).

IMHO this confusion arises because people are used to delaing with Cartesian tensors where the distinction is isn't obvious.

All your link is really saying is that there exists a certain one-to-one correpsondnace between a vector space and it's dual vector space that maps a vector to it's dual vector, this one-to-one corresponadance is called the metric and is a tensor of rank (0,2). Clearly identifying a pair of dual vectors as a single vector is incorrect.

Peterdevis

Nov8-04, 09:31 AM

All your link is really saying is that there exists a certain one-to-one correpsondnace between a vector space and it's dual vector space that maps a vector to it's dual vector, this one-to-one corresponadance is called the metric and is a tensor of rank (0,2)

No, it isn't! The problem is that most math people only see one to one correspondances and not what object tensors really are. Tensors are primally
object who are independant of the coördinatesystem.
When you talking about covariant or contravariant vectors, you talk about the components of the tensor, expressed in a certain basis. Both components and basis are coördinate dependent!
When you take "tangent vectors" as basic you call the vectorcomponents covariant because when you change the coördinate system the components transform by the covariant transformationrule (and the basic vectors bij the contravarianttransformation rule).
By not seeing that covariant and contravariant components describing the same object, you don't understand the power of the concept of tensors.

jcsd

Nov8-04, 09:53 AM

What is this object they are describing? As I said earlier the only object I can see that they are describing is a pair of dual vectors, but from it's very name you can see that we think of this object as two differet objects!

These days we don't tend to define tensors by their tranformation rules, we instea dprefer to define them multilienar functions of vectors and covectors (vector = contravriant vector, covector = covaraint vector), so already in our definition of a tensor we've described vectors and covectors as different objects! Also by seeing them as the same object you may well miss out on the fact that covectors are linear functions which map a single vector to a number or that a vector is a linear function that maps a covector to a number.

jcsd

Nov8-04, 10:01 AM

Note that when we raise or lower the index of a vector or a covector what we are really doing is multiplying metric by the vector/covector.

pete66

Nov8-04, 12:05 PM

Wow I read this entire thread and I fully don't understand tensors. I don't understand a lot of the notation (is that the word? :-) or even the application.

I'm 22 and come from more of a design background, I haven't really done any maths since I was about 15 but tonight I was hoping I'd be able to get the hang of "general relativity". (I was never bad at maths, btw, just impossibly slack.)

I have a solid grasp of 3d modelling/animation software as far as x,y,z verticies can animate along a timeline while being joined (via "vectors"?) to unlimited numbers of other verticies. Will this help me comprehend tensors?

Specific question, what do the little right pointing arrows signify in maths?

Cheers
You guys are amazing.
Pete

PS John Napier (1550 - 1617) is a direct great^x grandad of mine! Lol, and I've never studied logarithms until tonight.

da_willem

Nov9-04, 07:58 AM

Specific question, what do the little right pointing arrows signify in maths?

On top of a letter it signifies that this is a vector, but I assume that is not your question.

V-->W signifies a 'mapping' from V to W. This is another name for a transformation. It is usually a function that takes objects from a space V to W. In linear algebra for example the mapping is really a matrix multiplication.It brings one vector from a certain vector space (V) to another (W).

Astronuc

Nov10-04, 09:04 PM

quetzalcoatl9

Nov11-04, 12:43 AM

"An nth-rank tensor in m-dimensional space is a mathematical object that has n indices and components and obeys certain transformation rules. Each index of a tensor ranges over the number of dimensions of space. However, the dimension of the space is largely irrelevant in most tensor equations (with the notable exception of the contracted Kronecker delta). Tensors are generalizations of scalars (that have no indices), vectors (that have exactly one index), and matrices (that have exactly two indices) to an arbitrary number of indices.
:

i'm going to go out on a limb here and say that this definition is the whole problem with the way tensors are looked at...it doesn't really tell you anything meaningful. just saying that something follows certain transformation rules does not help you...

a tensor is a multilinear mapping, as follows:

T: V^* \times V^* \times V^* \times ... \times V \times V \times V ... \rightarrow R

where the number of V^*'s is some k, and the number of V's is some l, then we have a tensor of rank (k,l).

so it maps some combination of vectors and covectors into reals, and it does so in a multilinear way. so a (2,2) tensor, for instance, would be:

T(\alpha, \beta, \vec{v}, \vec{w}) = a_i b_i v_j w_j T(dx^{\alpha i}, dx^{\beta i}, \partial_{v j}, \partial_{w j})

and let's look at the metric tensor:

G = g_{ij} dx_i \otimes dx_j

let's feed it two vectors:

G(\vec{v}, \vec{w}) = g_{ij} dx^i(\vec{v}) dx^j(\vec{w}) = g_{ij} dx^i(v^i \partial_i) dx^j(w^j \partial_j) = g_{ij} v^i w^j dx^i(\partial_i) dx^j(\partial_j) = g_{ij} v^i w^j

most importantly, and the whole point that is lost with the wolfram definition, is that g_{ij} is NOT the metric tensor, it is the components of the metric tensor. The metric tensor is the whole expression for G.

I can prove it to you: G is invariant under a coordinate change, that is it will always map you to the same values no matter what coordinates you are in. g_{ij} on the other hand, will be different in various coordinate systems. look at the tensor G and you will see that this _must_ be the case.

this is no different than confusing the components of a vector, with a vector itself! This is the problem with the so-called "classical definition": it defines a tensor in terms of what it's components do which is unnatural and quite frankly not very useful.

Peterdevis

Nov11-04, 05:48 AM

The question is:

G = g_{ij} dx_i \otimes dx_j =g^{ij} \partial_{i} \otimes \partial_{j} ?

quetzalcoatl9

Nov11-04, 05:56 PM

The question is:

G = g_{ij} dx_i \otimes dx_j =g^{ij} \partial_{i} \otimes \partial_{j} ?

let's see:

G^{-1}(\alpha, \beta) = g^{ij} \partial_i(\alpha) \partial_j(\beta) = g^{ij} \alpha(\partial_i) \beta(\partial_j) = g^{ij} v_i w_j = g^{ij} g_{ij} v^j w_j = v^j w_j = g_{ij} v^j w^i

by symmetry of the inner product (by definition) however, g_{ij} = g_{ji} so we can write:

g_{ij} v^j w^i = g_{ij} v^i w^j = G(\vec{v}, \vec{w})

so G^{-1}(\alpha, \beta) = G(\vec{v}, \vec{w})

which defines our relationship between vectors and covectors using this so-called metric tensor.

Peterdevis

Nov22-04, 10:39 AM

What is this object they are describing? As I said earlier the only object I can see that they are describing is a pair of dual vectors, but from it's very name you can see that we think of this object as two differet objects!

I don't think you have read the link http://www.mathpages.com/rr/s5-02/5-02.htm. When you look at the drawing you see the object P and his covariant en contravariant expression. So one object two expressions!

quetzalcoatl9

Nov22-04, 08:13 PM

I don't think you have read the link http://www.mathpages.com/rr/s5-02/5-02.htm. When you look at the drawing you see the object P and his covariant en contravariant expression. So one object two expressions!

I believe that both yourself and jcsd are correct.

jcsd is saying that tensors are multilinear mappings that are coordinate independent, and that we can define everything in terms of linear functionals mapping to reals. this is true. infact we can even go so far as to say that \alpha is the covector such that G^{-1}( . , \alpha) = G( . , \vec{v}) is satisfied for some vector \vec{v} (where the first arguments are fixed by bilinearity) and use the metric tensor as a linear transformation to the dual space.

Peterdevis is saying that the components and basis are very much coordinate dependent (by themselves), and that this is important to understand the transformation laws. this is also true.

Am I not mistaken here?

I will add one thing though: our basis vectors and the components of our covectors, even when used tensorially, are NOT the same even though they follow the same transformation laws. they live in different spaces altogether. example: we can pushforward the basis of our tangent vectors (as defined by our differential map or the Jacobian) but we cannot push forward the components of our covectors, we must use the pullback.

jcsd

Nov23-04, 08:33 AM

That's what I am trying to say we're talking vectors that live in different vector spaces, so we really shouldn't see them as the same object even if there exists an importnat bijection between them (the metric tensor). Infact it's worth noting thta the scalr product isn't always defined, in these cases are we then discussing incomplete objects?

Peter Deis isn'tthat far off the mark he's just viewing tensors by the old definition. I have an old maths textbook thta tlaks aboiut the contravariant and covariant componets of a vector., but I can't say I liket that approach at all (especially as it talks about vector spaces and linear operators in a basis indepednt way, but when it comes to tensors it suddenly starts to define tem purely in terms of how the compoents change between basis, even though it notes that (some) linear operators are tensors).

mathwonk

Nov23-04, 04:57 PM

In my opinion, anybody who thinks that the properties of covariance or contravariance are not intrinsic properties of an object, is quite innocent of what is going on in differential geometry and manifold theory.

Perhaps they are confused by the phenomenon of a Riemannian metric, i.e. a smoothly varying dot product on the tangent bundle, since if one has a dot product, one can artificially represent a vector as a covector, by dotting with it. But then it is a different object. I.e. the operation of dotting with a vector is not the same object as the vector it self.

As to convincing such people:

They said it couldn't be done.
They laughed when I said I would do it.
They said that it couldn't be done.
I rolled up my sleeves and went to it.

I struggled, I strove, I strained.
I fought at it day and night.
They said that it couldn't be done.
They were right.

quetzalcoatl9

Nov23-04, 09:46 PM

I have an old maths textbook thta tlaks aboiut the contravariant and covariant componets of a vector., but I can't say I

jcsd,

i also have many older books that explain tensors this way, and have also been puzzled as to why they would want to approach the subject in this manner. as mathwonk pointed out, the real power of differential geometry will remain a mystery in that context.

it is also interesting to note that differential forms will be well-defined with respect to pullback, even if the inverse mapping isn't. tensors in general (and certaintly not vectors) do not enjoy this property, they are entirely dependent upon the mapping to be bijective. so certain topological mappings (such as irreversible deformations) can still be modelled using forms - this is awesome, and something that i am trying to learn more about now.

mathwonk

Nov24-04, 05:03 PM

the point quetzalcoatl9 is making is extremely important, and leads to the powerful invariant called de rham cohomology. i.e. suppose we have a vector assigned to each point of X, i.e. a vector field on X, and a smooth mapf from X to Y. if we want to assign a corresponding vector field on Y via the map f, we are out of luck. i.e. take a point y of Y. we want to push over some vector associated to a point of X. but what point do we choose? there is no way to say. we have the same problem with a covector field on X.

but if we have a "covector field" s on Y the story is quite different. i.e. to each point of Y we have assigned a linear function on tangent vectors. now to assign such a covector field on X is easy. if x is a point of X and v is a tangent vector at x, then to evaluate our covector field on v, just apply the derivative (or differential) of f to v and push it over to a tangent vector to Y at f(x). this number is defined to be the value of the pullback covector field, f*(s) at v.

Then we can show also that exterior derivative commutes with pullback. Now define the de rham cohomology group of dergee 1 to be the space of differentials s such that ds = 0, modulo thiose which are of form dg for some fucntion g. then the remarks above show that any smooth map f:X-->Y induces a linear map from the de rahm group of Y to that of X.

thus we have a functor on manifolds. the existence of this functor depends on the fact that covector fields are "contravariant" in category language, or "covariant" in differential geometry language, in either case the point is that they go in the opposite direction to the original map.

it should be obvious to anyone reading this discussion (the choir?) that covariance and contravariance are clearly intrinsic and absolutely essential properties of the objects being discussed.

the existence of this functor has powerful consequences.

for example, by stokes theorem, the de rham cohomology of a disk is zero, and since the form dtheta is non zero on a circle, the de rham cohomology of a circle is not zero. this implies the brauer non retraction theotrem as follows:

if there were a smooth map from the disc onto its boundary, leaving the boundary fuxed, it would induce the identity map of de rham cohomology group of the circle, which also factored through the azero group of the disc, an absurdity.

this then implires the brauer fixed point theorem for the disc.

a similar argument with the solid angle form implies that the antipodal map of the sphere cannot be deformed smoothly nito that identity map of the sphere and this implies that there are no smooth vector fields on the sphere having no zero vectors anywhere.

the subject goes on and on and on..... to say whole books have been written about it, is a huge understatement, such as bott - tu, on differential forms, recapturing large parts of algebraic topology via de rham cohomology.

when I was young and energetic, I taught the stuff on vector fields on spheres and brauer theorems, in my sophomore several variables calculus class at central washignton state college.

jcsd

Nov27-04, 10:07 PM

I don't really have much to add, except to say that I've found the discussions I've had with mathwonk and his posts on this subject extremely enlightening vis a vis tensors.

mathwonk

Nov28-04, 01:06 AM

jcsd, you made my day.

best wishes

mathwonk

fizixx

Nov28-04, 08:06 PM

chroot (Warren)....

That was a wonderful explanation of Tensors. I appreciated the time you took to write it. I did not have the benefit of learning about tensors in school. It's just not something they taught when I was learning about physics. Since I've seen the notation and even knew a few tidbits of info, but that's about it. At some point I decided to just teach myself. Something scientists are good at doing, most of the time, in their careers to help fill in the gaps left by formal education. This lacking always made me feel like I'm missing something. I'm hoping to parse off some time so that I can pursue this segue of learning, even tho I do not use it in my work.

Your explanation is clear and concise. Makes me wonder how far ahead I would have been had I had more teachers like you in school!

Thanks again......many accolades to you!

:)

Even more generally, a tensor is a sort of mathematical machine that takes one or more vectors and produces another vector or number.

A tensor of rank (0,2), often just called rank 2, is a machine that chews up two vectors and spits out a number.

A tensor of rank (0,3) takes three vectors and produces a number.

A tensor of rank (1,2) takes two vectors and produces another vector.

Hopefully you see the pattern.

You actually already know what a (1,1) tensor is -- it's nothing more than a good ol' matrix. It accepts one vector and produces another vector.

If you're working in three dimensions, a (1,1) tensor can be represented by its nine components. Here's a simple (1,1) tensor.

T = \left(
\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}
\right)

You already know what this guy does -- it takes a vector and gives you back the exact same vector. It's the identity matrix, of course. You would use it as such:

\vec v = T \vec v

If the world were full of nothing but (1,1) tensors, it'd be pretty easy to remember what T means. However, there are many different kinds of tensors, so we need a notation that will help us remember what kind of tensor T is. We normally use something "abstract index notation," which sounds more difficult than it is. Here's our (1,1) tensor, our identity matrix, laid out in all its regalia:

T^a_b

The a and b are referred to as indices. The one on the bottom indicates the tensor takes one vector as "input." The one of the top indicates it produces one vector as "output."

Tensors don't have to accept just vectors or produce just vectors -- vectors are themselves just a type of tensor. Vectors are tensors of type (0,1). In full generality, tensors can accept other tensors, and produce new tensors. Here are some complicated tensors:

R^a{}_{bcd}\ \ \ \ G_{ab}

The second one, G_{ab} is a neat one to understand. You should already understand from its indices that it is a type (0,2) tensor, which means it accepts two vectors as input and produces a number as output. It's called the metric tensor, and represents an operation you already know very well -- the dot product of two vectors!

In normal Euclidean 3-space, G_{ab} is just the identity matrix. You can easily demonstrate the following statement is true by doing the matrix multiplication by hand:

\vec u \cdot \vec v = G_{ab} \vec u \vec v

The metric tensor is more complicated in different spaces. For example, in curved space, it's certainly not the identity matrix anymore -- which means the vector dot product is no longer what you're used to either when you're near a black hole. Tensors are used extensively in a subject called differential geometry, which deals with, among other topics, curved spaces. General relativity, Einstein's very successful theory which explains gravity as the curvature of space, is cast in the language of differential geometry.

So there you have it: tensors are the generalization of vectors and matrices and even scalars. (Scalars, by the way, are considered to be type (0,0) tensors.)

I should mention that there not all mathematical objects with indices are tensors -- a tensor is a specific sort of object that has the transformation properties described by others in this thread. To be called a tensor, an object much transform like a tensor. Don't worry though, you're not going to run into such objects very often.

- Warren

mathwonk

Nov29-04, 09:23 PM

I agree that chroot's explanation is wonderful. it is the best clearest one i have ever seen. I love it and learned something from it immediately.

The other shoe: one reason for that is that it describes only such simple tensors that one does not need the concept of tensors to understand them.

if you really want to understand tensors, i sugest you try to understand the curvature tensor. this concept was apparently the reason for the invention of tensors and the first tensor invented by riemann.

(another entry in the thread that can never die.)

fizixx

Nov30-04, 06:50 PM

mathwonk.....

That is a good suggestion, but perhaps since you suggested this, can you offer a particular bit of info or website to get those of us, myself included, a place to sink our teeth into please?

fiz~ :smile:

mathwonk

Nov30-04, 07:27 PM

i do not feel i myself understand curvature well. my attempts to make sense of the definitions i have seen, were in my posts to the thread "proof that grad V is a (1,1) tensor" have you seen those?

I also continue to recommend as the best possible text on differential geometry the second volume of michael spivak's opus on the topic.

there is also a nice short book by manfredo do carmo. other books i have consulted and recommend include milnor's books on morse theory and characteristic classes, and the little undergraduare book on differential topology by guillemin and pollack, which grew out of milnor's great text, topology from the differentiable viewpoint.

Peterdevis

Dec1-04, 02:15 AM

if you really want to understand tensors, i sugest you try to understand the curvature tensor. this concept was apparently the reason for the invention of tensors and the first tensor invented by riemann.

I think the curvature tensor let us see that covariant en contravariant components refers to the same object.
You can describe the curvature by a (1,3)tensor but the (0,4) tensor contains the same information about the curvation of the manifold

mathwonk

Dec1-04, 08:54 PM

peterdevis: we seem not to be on the same page somehow. let me ask what you think of categories and functors? say contravariant versus covariant functors? i.e. hom(X, ) versus hom( ,X)?

are they some of your favorite objects? if not, and you want to make friends with them, try my humble post on the "what are catergories?" thread.

(indeed in two posts on also page 4 of the present thread I have tried to make an extremely precise statement via category theoretic ideas, to the effect that there is no way in the world to identify all vector spaces with their duals in a natural way. what do you think of that discussion?)

FSC729

Dec5-04, 05:48 PM

Hello Stoned Panda, (I'm amazed panda's can learn and type english) anyway, the best way for anyone to truly learn what a certain mathematical concept is, is to use it.

Practice using the definition of a tensor, apply it to a simple problem, work with it, understand it from your perspective. Look for as many examples of a tensor as possible and rederive the examples without looking at the solution.

http://fsc729.ifreepages.com

saski

Dec6-04, 01:49 AM

Hiya, Peterdevis and mathwonk

In the spirit of chroot's beautifully economical characterization of tensors, the Riemann curvature tensor is a machine which takes a bivector area and returns a rotation matrix. In other words, it's a (1,1)-tensor-valued 2-form.

The motivation is "parallel transport around a closed curve". What is being transported here is an arbitrary tangent vector or, equivalently, a tangent basis. The curve it is transported around is considered the boundary of a surface, like a soap-film picked up on a loop. Since the curvature is a 2-form, Stokes' theorem equates the integral of curvature over the area of the film to net rotation of the basis as it traverses the loop, regardless of the shape of the film.

To take the best-known illustration, let's consider the game of trying to hold constant the orientation of a spear, as you traverse a closed curve. On a flat surface you can do this, on a curved surface you can't.

On flat ground, you can proceed as follows. You are handed the spear; you turn yourself in its direction; you walk 100 paces. Then you turn 90 degrees right, making sure that you are keeping the spear stationary; and you walk 100 paces. After you have walked the four sides of a square, you are back where you started, and the spear has never once changed direction.

But now try it while sailing from the north pole to the equator, along the equator and back to the pole, making an equilateral, right-angled spherical triangle. Start your journey in the direction the spear is pointing. What you'll find is that, while the spear is at all times pointing south, by the time you get back to the pole it has been rotated 90 degrees from its = original orientation.

You can play this game at:

Spherical Geometry Demo
John M Sullivan, University of Illinois
http://torus.math.uiuc.edu/jms/java/dragsphere/

The Gaussian curvature in this case is the angle turned (pi/2) divided by the area circumscribed (1/8 * 4 pi a^2), so equals (1/a^2). It's easy to do that on a sphere since its curvature is constant.

For general Riemannian spaces, of dimension >= 2, the equivalent is the Riemann curvature tensor. It has an input slot for tangent area and it yields a rotation matrix. You get the components by contracting it with the unit tangent bivectors, and contracting that with the unit tangent vectors.

The calculations can be done in terms of metric and connection, as below, but it's rather messy. It is more neatly expressible in terms of forms, but I'm still getting to grips with that.

Parallel Transport, Covariant Differentiation, and Curvature Tensors
David Boozer
http://www.its.caltech.edu/~boozer/physics/geometry3/geometry3.html

Tensors and Relativity: Chapter 6
Peter Dunsby
http://vishnu.mth.uct.ac.za/omei/gr/chap6/chap6.html

Peterdevis

Dec6-04, 05:12 AM

peterdevis: we seem not to be on the same page somehow

Maybe, I must reformulate my point of view.
As physisist i'm interested in a description of reality independent of a reference system. So i organise "events" (x,y,z,t) as a manifold and now i put one every event a container (I called this a tensor, but this confuse everybody so i call it now a container) who contains some kind of information that only depends of the event. But to calculate reality , I have too translate that information into a reference system (because thats the only way we can measure things) This I can do in a covariant or a contravariant way.
(By the way , I also don't like to define them with the transformation rules)
So If (in a matimatical way) a vector and a oneform lives in two different vectorspaces I will accept that.
But for a good phuysical understanding I think it's necessary to see that they refer both to the "container".

are they some of your favorite objects? if not, and you want to make friends with them, try my humble post on the "what are catergories?" thread.

Now , I don't. But thanks for the reference. I will studiing them with joy.

PS: Is my English legible

mathwonk

Dec15-04, 12:11 PM

Dear Saski,

Thank you very much for the beautiful anmd insightful discussion of the curvature tensor. I love the spear metaphor. At last we are getting somewhere at understanding curvature. Unfortunately it seems that what you said is not entirely true. At least not in the technical sense of being correct, although it does apparently contain much valuable insight. Of course sometimes it is more helpful to say something false which has some truth in it, than to say something correct which is unenlightening.

I refer to your statement: "In the spirit of chroot's beautifully economical characterization of tensors, the Riemann curvature tensor is a machine which takes a bivector area and returns a rotation matrix."

As you probably know, the curvature matrix is actually not a rotation matrix, but perhaps you meant "infinitesimal rotation". I presume you fully understand what the truth is and were merely simplifying it for us.

Your answer puzzled me for some time, and here is what I have come up with as of now. Since the curvature tensor is a (1,1) tensor valued 2 form, it assigns an endomorphism of the tangent space to an oriented pair of orthonormal unit vectors. I made the mistake of assuming in the case of a surface in three space, that this endomorpohism was the differential of the gauss map, especially since that endomorphism has as its eigenvalues the two principal sectional curvatures, and thus has as determinant the gaussian curvature of the surface at the point.

On the other hand you claimed that the endomorphism was a rotation, which is incompatible with my assumption since a rotation is not a diagonalizable endomorphism, and has usually no real eigenvalues. In other words, in an oriented orthonormal basis I was claiming the endomorphism assigned by the curvature had as matrix a diagonal matrix with diagonal entries a,b where the product ab was the Gaussian curvature. A rotation matrix through angle t on the other hand is a matrix with diagonal entries cos(t) and off diagonal entries sin(t) and -sin(t). So which is correct?

Well since the gaussian curvature is a scalar, and can be any scalar, it cannot be represented by a rotation matrix since a rotation matrix is essentially a point on the circle, while the gaussian curvature is essentially a point on the real line. So a rotation matrix does not have enough information to capture the curvature. On the other hand it also should not be the differential of the gauss map, since that mtrix has in it not only the curvature ab, but the individual eigenvalues a and b. So this is too much information, indeed gauss discovery was that the metric rpeserves precisely the product of the Euler's sectional curvatures, not the sectional curvatures themselves.

In fact it seems the endomorphism is one with a skew symmetric matrix! So how is this related to the two other candidates above? Well the entries in the skew symmetric amtrix are of cousre zeroes on the diagonal, and the off diagonal entries are apparently just the curvature ab, and -ab. And how is this connected up to the beautiful idea of rotating the spear? Well algebraists know that the vector space of skew symmetric matrices is just the lie algebra, i.e. the tangent space at the identity, of the linear group of rotation matrices.

So there is still the real possibility, and of course strong likelihood, that the curvature matrix is a velocity vector of a curve of rotation matrices, defined in some way locally by parallel transport, as Saski suggested.

How does this sound Saski?

robphy

Dec15-04, 01:52 PM

Saski and Mathwonk,

Since you are both thinking about the interpretation of the Riemann tensor, maybe I can get you to think of an analogous interpretation of the Weyl (conformal) Curvature tensor.

mathwonk

Dec15-04, 02:45 PM

how is it usually described or defined??

i googled this whicha int much:

The Weyl tensor has the special property that it is invariant under conformal changes to the metric. That is, if g' = f g for some positive scalar function then W' = W. For this reason the Weyl tensor is also called the conformal tensor. It follows that a necessary condition for a Riemannian manifold to be conformally flat is that the Weyl tensor vanish. It turns out that in dimensions >= 4 this condition is sufficient as well

robphy

Dec18-04, 02:59 PM

Mathwonk,
Yes, the Weyl Curvature tensor is that part of the Riemann tensor which is conformally invariant and is also traceless. You may be interested in the post of Rob Woodside http://www.physicsforums.com/showpost.php?p=396933&postcount=19 in the thread on "GR metric meaning".

saski

Dec21-04, 04:21 AM

Hiya mathwonk

I was away for a few days, came back and settled down to sort out the relations
between curvature and the several kinds of derivatives on amanifold; and
I'm still going.

Anyway, my apology for imprecision above. I think the following is correct,
though I can't prove it explicitly:

(1) To every closed curve on a manifold, integration of the Riemann tensor
(over the area of any surface bounded by the curve) assigns an isometric
transformation of tangent bases (at the points of the curve).

(2) At any one point, to any tangent bivector the Riemann tensor assigns
a generator of the isometry group: an "infinitesimal transformation", if you like.

So, although it is correct to say the Riemann tensor is a (1,1)-tensor-valued
2-form, it may be more revealing to say that it's a 2-form which maps the
tangent bivectors to the Lie algebra of the isometry group.

Now the Gauss Map is a construction I haven't seen for years, if at all. It
must display some aspect of the Riemann curvature in general - but I'm not
sure whether the parallel transport business leads to a simple way of doing
it.

For a 2D surface, the Gauss map is the unit sphere in 3D. For what it's worth,
I've found in

Exact Solutions of Einstein's Field Equations
Ed. E. Schmutzer
Cambridge University Press, 1980

the result

R_a_b_c_d = K (g_a_c g_b_d - g_a_d g_b_c)

where K is the Gaussian curvature, the Riemann tensor having only one
independent component. That doesn't get us terribly far.

The real key is the concept of the connection, which does have a direct
relation to the Gauss map, although I'm having a hard time making it
explicit. I think the connection is essentially the differential of
the Gauss map, but do treat the following with caution.

Explanations of the Gauss map that I've seen, especially:

http://www.mathwright.com/librarya/ccurves/ccurves4.htm

present it from the viewpoint of embedding a curve in R^2 or surface in
R^3.That makes a normal to the n-manifold in question a vector in
R^{n+1}, a sensible way to look at it. However from a
completely intrinsic point of view, I would characterise the "normal"
as the manifold "surface element": i.e. for any tangent basis
e_i, i=1,...,n , it is the wedge product of all the
unit tangents

e_1 \wedge e_2 \wedge ... \wedge e_n

Remove any one term from this and apply Hodge duality to what remains. We get
the same term again, maybe up to a factor of -1.

Take the exterior derivative of the complete product, and we get a series of
terms

e_1 \wedge ... \wedge de_k \wedge ... e_n

each expressible via the Hodge duality as

<e_k, de_k>

That series is what the differential of the Gauss map would be.

de_k is the complete differential of e_k; the derivative
of e_k in the direction of e_l is

<e_l, de_k>

Despite appearances, this is a vector; and its component in the
de_m direction is

<e_m, <e_l, de_k>>

The set {de_k} *is what's called the connection, and the above
for all k,l,m are its components. In relativity work, using index notation,
the connection is written as the Christoffel symbol

\Gamma^k_{lm}

Now the point of all this is that formally we have taken the tangent basis
{e_i} and differentiated it. That's something that it makes
sense to do, only *if we have a function assigning one tangent basis to each
point of the manifold, and this function is smooth, i.e. actually has a
differential. That is actually the case, for example, when we have a
parametric curve in 2D andcan define its Serret-Frenet basis from its
velocity and acceleration vectors; or for another example, the Darboux
basis for a surface. This concept of a "frame field" and its differential
is generalized to the Cartan formalism of "the moving frame".

Simply, de_k is an element of the Lie algebra of the Lie group
which preserves the metric: the connected isometry group SO(n). In other
words de_k tells us what happens to e_k *as we move
across the manifold, by giving an infinitesimal transformation matrix
<e_m, <e_l, de_k>> * for each direction e_l .

This is the mechanism which "rotates the spear". I had better try and make
this precise. To do so, I'm going to set out the whole geometric intuition as
best I can; so bear with me on what may seem a digression.

In many developments of differential geometry, we stay within a completely
intrinsic picture of a given manifold; we do not treat it by giving it an
embedding in a higher R^n. That will be the case here.

Consider a 2D surface, smooth but irregularly shaped; say a bowling pin. We
are concerned with what happens to a tangent vector ("the spear") as we move
it around the surface.

Obviously the spear _could_ wobble all over the place; it could be the tangent
vector to any curve we care to draw. However there is a natural way to move
it, so that it has no local change of direction: this is called "parallel
transport".

Let's look at a spear on a flat surface with a cartesian coordinate system.
Move it in a straight line. We will know its direction is unchanged, that is *
it is parallel to its former orientation, if its components are unchanged.
That defines parallel transport on a flat surface. But can we define the
equivalent on a bowling pin?

Yes we can: by transporting it along a geodesic curve - which is a locally
flat piece of the bowling pin's surface.

Take a flat sheet of paper and roll it into a cylinder. Despite having two
edges identified, it is still intrinsically flat, and the component criterion
for parallel transport still holds.

OK? Then take a long strip of paper, and roll it up like a roll of sticky
tape. If we printed cartesian coordinates with one axis down the centre, the
axis would overlay itself, right through the roll.

Lay the sticky tape down around a cylinder, perpendicular to the axis: it
overlays itself. Lay it down at an angle and it comes back parallel to
itself. This shows that the cylinder is flat.

Lay it on a sphere, and it will follow a great circle and come back to overlay
itself. It will crinkle a little at the edges, but we can make the crinkling
as small as we like, by using a narrower tape. Try to make it follow a line
of constant latitude at 30 degrees though, and it can't be done without
constantly bending and crinkling it to fit. Our sticky tape is a portable
standard of local parallelism: it is locally flat. It follows a geodesic
curve; and if you want to find the geodesic curve on a bowling pin, in a
given direction and through a given point, just pull out the sticky tape and
see where it goes.

But here's the crucial thing. On a curved surface, you cannot generally get
the sticky tape to stay parallel to a set of coordinate lines, no matter what
coordinate system you draw on it. If you start laying tape on the sphere
along the 30-degree-north line, the tape will diverge as it follows its own
great circle. The tangent bases along the tape and the tangent bases defined
by the spherical coordinates will relatively rotate. And the rate of
rotation, expressed in degrees per inch, or as a skew-symmetric
rotation-group generator matrix, is given by the connection (contracted with
the direction of the tape).

Just to align this with standard notation, the connection is often given the
symbol

\nabla

The connection along a given tangent, defined by a vector u is

\nabla_u

The infinitesimal change that makes to a vector v is

\nabla_u v

The component of that change measured by the 1-form w is

<w, \nabla_u v>

and the component of the change, when the vectors are the unit tangents and
1-forms defined by a coordinate system, is the Christoffel component

\Gamma^c_{ab} = <w^c, \nabla_(e_b) e_a>

Now out of all this, we want a definition of the curvature of the manifold at
a point. We're interested in the geometric invariants. For this purpose we
don't care what happens to a particular vector, let alone the components of
its variation. What counts is the connection in a given direction:

\nabla_u

This measures the rate at which the coordinate basis vectors are rotating with
respect to parallel transport. The connection in general,
\nabla , gives the rate of rotation corresponding to any rate of
movement across the manifold.

Now what we really want to know is the amount of rotation that happens around
a closed path, in the small limit. We choose as a path the circumference of
the area defined by two tangent vectors u and v. (Or strictly speaking, the
limit of the area defined by orbits tangent to u and v at their
intersections, as the lengths of the orbits approach zero.)

I'm going to quote the result, rather than provide the proof I read in Misner,
Thorne & Wheeler's _Gravitation_, which has some subtle features. Maybe there
is a neater proof around.

Assuming we have zero for the commutator \left u,v \right, which is
the case for coordinate line tangents, the net rotation per unit area is

\nabla_v \nabla_u - \nabla_u \nabla_v

That is the definition of the Riemann tensor.

And it's also as far as I've got. For an n-manifold, these rotations are in
SO(n). I'm certain there must be a map, based on these, which takes the
complete volume elements e_1 \wedge e_2 \wedge ... \wedge e_n to
the surface of the unit sphere in *R^{n+1}... i.e. the Gauss
map ... but I haven't seen it yet.

Thanks for the prompt. More anon, I hope.

gvk

Dec27-04, 09:52 AM

I was absent for long time and when return found the conclusion:
mathwonk
"In my opinion, anybody who thinks that the properties of covariance or contravariance are not intrinsic properties of an object, is quite innocent of what is going on in differential geometry and manifold theory. Perhaps they are confused by the phenomenon of a Riemannian metric, i.e. a smoothly varying dot product on the tangent bundle, since if one has a dot product, one can artificially represent a vector as a covector, by dotting with it. But then it is a different object. I.e. the operation of dotting with a vector is not the same object as the vector it self."

There is a joke about great mathematician who started a course of logic for the sophomores.
"Logic is the laws of thinking - he said. "Now I will explain you what is the law, and what is the thinking.
I will not explain you what is 'of'".

Here you have to explain what do you mean saying 'intrinsic properties', because what is 'covariance or contravariance' and what is the 'object' are well known.
Usually, under 'intrinsic properties' of the 'object' physicists (or people who using differential geometry for applications) means what were described by Peterdevis:
"Tensors are primally object who are independant of the coördinatesystem....,
....
As physisist i'm interested in a description of reality independent of a reference system....,
....
So If (in a mathematical way) a vector and a oneform lives in two different vectorspaces I will accept that.
But for a good phuysical understanding I think it's necessary to see that they refer both to the "container"."

In other words, physisists means the object can be measured somehow and this measurements do not depend on coordinate system or other way we measure it or what we think about it this is why we called them "invariant properties".
Invariant properties of physical objects is of primary importance and it has been the leading idea in physics and mathematics since 19 century.

Now I try to explain that the statment - "the operation of dotting of metric with a vector is not the same object as the vector itself" is simply incorrect.
It is very clear from its physical meaning and formal proof that it is invariant object can be found in hundreds (may be thousands!) old and new textbook on tensor analysis and physics! And there is nothing confusing for the people who understand the physical reality. Using the math language which you, mathwonk, prefere it can be shown that the space with metric has the mapping g : V —> V*, where g is an element of GL(n,K). And this mapping is linear and canonical isomorphism with respect to g. (The canonical isomorphism V ->V** always exists, canonical isomorphism V —> V* appears for space with a metric). And this isomorphism does not allow to distinct the covariant and contravariant vectors and associate the invariant value V^i V_i with intrinsic property of the 'object' (one object, not two different objects!), viz. squared lenght of the vector.
The presentation of the vector \vec V = V^i e_i = V_i e^i , which one can find in the classical books on absolute differential calculus including the books by Ricci, Levi-Chivita, and E.Cartan, simply means exactly the same. In such context, this identification is perfectly correct.
In respond to the critic about patently false statments in old books. I don't think the fathers of tensor analysis were confused by the phenomenon of Riemannian metric. On the contrary, they thought more deep and explained more clear than many modern 'interpretators'. They considered Riemannian and Euclidean spaces as a good starting point of our imagination and intuition. Euclidean case is very important and not 'just happen'. The modern geometry and our geometrical sense would not exist if we did not have them.
In mathematics, enough to remind that the definition of topological manifolds hinges on Euclidean space for local coordinate maps.
However, the situation in quantum mechanics became different, and a discovery that the space on atomic level does not have a metric changed our understanding on space. Other properties, in particular, linearity of the space was still hold and the linear functional f: V->K which mapped a vector to the scalar was some sort of 'invariant substitution' of metric. In those cases, the contravariant and covariant vectors are different. And nobody (including myself and, I guess, Peterdevis) argue with that.
But those vectors do not describe the physical 'object' itself as something which can be measured. The physical interpretation has only the product of contravariant and covariant vectors (braket) which is invariant. Of course, from mathematical point of view any mathematical structure can be called 'the object', but its real interpretation won't be possible.
Vector space, as defined in algebra, does not have any tool to measure anything. To be consisted, we have to say that vector in such space does not have the lenght, and the school definition of vector as a line segment having a direction and magnitude, is also 'patently false statment'. Is't it? Should we eliminate this and similar notions from our education?
In my opinion, it would be wrong. And not only because there is still a room for the spaces with metric, but mostly because, on this way the most people could not understand anything about any space, and definitly could not image it. Indeed, when we draw the vector or covector, we live in Euclidean physical space. This means that not only are addition of vectors and multiplication by scalars defined, but the lengths of the vectors, angles between them, and the area of all figures. Our diagrams carry compelling information about these metric properties and we perceive them automatically, though they are in no way reflected in the general mathematical axiomatics of direct or dual spaces. It is impossible to draw the vector which does not have a length. The length means for us some kind of invariant not changing with a change the coordinate system. If we could live in vector space (as it defined in algebra) no one can also tell that one vector is longer than another or that a pair of vectors forms a right angle unless the space is equipped with a special inner product.
So, from this point of view, visualization of vectors and covectors (tangent and cotangent spaces without metric) is, strictly speeking, impossible. This visualization can be done only with help of scalar product. But if the scalar product is introduced one immedeatly can see that cotangent vector can be received from tangent by g_{i,j}, and those vectors are one object having invariant scalar module V^i V_i.

Mathwonk, it seems you don't share the point of view that the driving engine of the differential geometry and other parts of mathematics is the physical reality, but not the abstract definitions. Or I'm wrong.

P.S. Thank you very much for HS information. I have requested more detail about the math programs of these schools.
It's interesting, especially in the contrast between what the people ask here, and what our students are learning in normal public schools :smile:.

gvk

Dec27-04, 05:26 PM

Peterdevis wrote above the curvature as a good example of invariant properties of real object. I want to add some simple example.
Everybody knows that the notion of the curvature means some sort of invariant properties of nonlinear manifolds or difference between linear and nonlinear spaces.

The curvature and torsion are the only two curvature invariants known for 1-D manifolds (smooth curves in 3-D space).
The curvature and torsion describe so called 'external curvature invariants' of curves and were well known before Gauss and Riemann.
There is no any 'internal curvature' for the smooth curves in n-D space, but there is only one 'internal metric invariant' or infinitesimal segment of the curve:

\mid V \mid ^2=V^i V_i=g_{i,j} V^i V^j =g^{i,j} V_i V_j

(The free usage of covarient and contravarient components of vector \vec V emphasizes that they belong to one object. Result does not depend on type of used components.) It equals to 1 no matter how the curve is curved (p.s. vector \vec V is calculated with respect to the natural parameter).

It is obvious from here that the curves do not have 'internal geometry' and if somebody stay inside the 1-D manifold he can not tell: does the curve has a curvature or not. On the contrary, 2-D (and higher D) manifolds have 'internal geometry' and Gauss and Riemann were first who realized this feature and introduced the notions of 'internal curvature'.
By the way, two works of Gauss and Riemann were reproduced with nice comments of Spivak in his geometry books (vol. 3 or 4, sorry don't
remember exactly, may be mothwonk can correct).

It is interesting to note that the search of 'internal curvature' was started with the same notion -'internal metric invariant' or small length of curve on the 2-D manifolds.
The 2-D surface in the small scale is Euclidean space, and 2-dimentions allows to make a small closed circles, which is impossible for 1-D. Two dimentions also give us the second 'internal invariant' - viz. angle between two tangent (or cotangent) vectors. If we take other vector \vec W from the same point on the surface with respect to the natural parameter, it is also satisfied to the eq.

\mid W \mid ^2=W^i W_i=g_{i,j} W^i W^j = g^{i,j} W_i W_j= 1.

The scalar product of these two vectors is

\cos(A)=W^i V_i=g_{i,j} W^i V^j=W_i V^i= g^{i,j} W_i V_j

(here the same usage of covarient and contravarient components of vector is possible, because they describe the one object and resulting angle does not depend on such usage). This is the second 'internal angle invariant'. Now we can calculate any geometrical property on this surface.
In fact, the length of any line whatever is found can be caclulated by integration from the length of its infinitesimal 'internal metric invariants', the
area of a figure can be calculated by breaking it up into elementary parallelograms with known 'internal angle invariants', and so on.
And the wonderfull thing was discovered by Gauss when he calculated the sum of angles for the small triagle: the difference (A+B+C - Pi) between the sum of three angles and Pi, is the area of triagle with the coefficient of proportianality called Gauss curvature.
Gauss curvature for 2-D is directly connected to the Riemann curvature tensor, viz.

K = R_{1,2,1,2} / det(g) .

The another name of this procedure is 'parallel transport', which was described above by Saski in detail.