Math Newb Wants to know what a Tensor is

[mathwonk]

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 function 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]

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]

jcsd, you made my day.

best wishes

mathwonk

 

[fizixx]

Tensors

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]

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]

curvature tensor

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~

 

[mathwonk]

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]

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]

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]

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]

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]

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 that's 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]

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]

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]

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]

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 https://www.physicsforums.com/showpost.php?p=396933&postcount=19 in the thread on "GR metric meaning".

 

[saski]

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 anyone 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 anyone 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]

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 statement - "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 length 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 statements 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 statement'. 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 definately 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 .

 

[gvk]

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.

 

[The Model]

Tensors

Hello everyone,

Check the link below:

http://profile.myspace.com/index.cfm?fuseaction=user.viewprofile&friendid=11555164

these guys wrote a song called "Differential Geometry" about tensors. The lyrics are probably inaccurate, but who cares! Personally, i think a tensor is really just a common vector except it exists in multiple plausible modalities, but hey, i can't see in 4 diemensions either!

best wishes to all:)

 

[The Model]

Math Nerds Unite!