what is a tensor [Archive]

mathwonk

Aug10-04, 01:41 PM

pmb phy, With all respect, I think you may be misunderstanding what you have seen.

For instance on the first page of the site you just referred us to, equation (1) displays the "metric tensor" in exactly the form I gave for a second order (or rank) tensor, namely it has an expression as a homogeneous polynomial of degree 2.

Later on, this same site describes a zero rank tensor as a scalar valued function that keeps the same value when coordinates are changed, and a first order tensor as one which transforms by the usual first order chain rule, when coordinates are changed, and a second order tensor as one which transforms by the chain rule for second derivatives when coordinates are changed, etc...

Perhaps the confusion is that I was referring to the appearance of a (symmetric) tensor in a given coordinate system, and your sources emphasize the way these representations change, under change of coordinates.

Unfortunately many sources emphasize the appearance or representation of tensors rather then their conceptual meaning. The essential content of a tensor (at a point) is its multilinearity.

Globalizing them, leads to a family or "field" of such objects at different points, and then to the necessity of changing coordinates, at which point the way in which they change appearance under coordinate change becomes of interest.

It seems odd to me at least to define them this way however. On the other hand, you are right that some features of tensors, or even vectors, are invisible except when one changes coordinates.

E.g. a vector and a covector at a point both look like an n tuple of numbers, but when you change coordinates one changes by the transpose of the matrix changing the other.

Of course conceptually they differ even at a point, as one is a tangent vector and one is a linear form acting on tangent vectors.

Do you buy any of this?

pmb_phy

Aug10-04, 04:09 PM

For instance on the first page of the site you just referred us to, equation (1) displays the "metric tensor" in exactly the form I gave for a second order (or rank) tensor, namely it has an expression as a homogeneous polynomial of degree 2.

Equation 1 contains the components is the metric tensor. The components in that particular case were all zero except for g11 and g2 which equal 1.

Perhaps the confusion is that I was referring to the appearance of a (symmetric) tensor in a given coordinate system, and your sources emphasize the way these representations change, under change of coordinates.

I think you're confusing the tensor with the expression the components of the tensor appears in. A general tensor is a geometric object which is linear function of its variables which maps into scalars. For example: Let g be the metric tensor. Its a function of two vectors. The boldface notation represents the tensor itself and not components in a particular coordinate system. An example of this would be the magnitude of a vector.

A^2 = \boldf g \left \boldf A, \boldf B \right

When you represent the vector in a basis and using the linearlity of the tensor then you get the usual expression in terms of components.

Unfortunately many sources emphasize the appearance or representation of tensors rather then their conceptual meaning. The essential content of a tensor (at a point) is its multilinearity.

There are two ways of looking at tensors. I've been meaning to make a new web page to emphasize the geometric meaning but am unable to do so at this time. Plus I'm still thinking of the best way to do that.

E.g. a vector and a covector at a point both look like an n tuple of numbers, but when you change coordinates one changes by the transpose of the matrix changing the other.

The terms "covariant" and "contravariant" can have different meanings in the same context depending on their usage. For example: A little mentioned notion is that a single vector can have covariant and contravariant components. For details please see

http://www.geocities.com/physics_world/co_vs_contra.htm

Of course conceptually they differ even at a point, as one is a tangent vector and one is a linear form acting on tangent vectors.

Do you buy any of this?

Some of it.

Pete

mathwonk

Aug10-04, 06:30 PM

Pete, let me try again. we may be getting closer together here.

An equation like summation gjk dxj dxk, as on the site you referenced, is a covariant tensor of second rank, because it is a second degree homogeneous polynomial in the expressions dxj, dxk, which are themselves covariant tensors of first rank. i.e. it is of rank 2, because there are two of them multiplied together. this results in a bilinear operator, which is linear in each variable separately.

now WHICH second rank tensor it is, is determined by what the coefficients are, or the "components" if you like, namely the gjk.

the usual metric tensor on the euclidean plane is dx1dx1 + dx2dx2, so the only non zero components are g11 = 1, g22 = 1.

But there are many other riemannian metrics given by other choices of the gjk.

I have tried to explain the conceptual idea of tensors at greater length in some other threads in this forum also with the word tensor in them. let me know what you think of them.

Brutally, if T is the tangent bundle and T^ the dual bundle of a manifold, then sections of T are contravariant tensors of rank one, and sections of T^ are covariant tensors of rank 1.

A section of (T^ tensor T^), where this is the bundle of tensor products of the dual spaces, is a covariant tensor of second rank, such as a metric.

if we consider only one tangent space isomorphic to R^n, its dual has basis dx1,...dxn, and the second tensor product of the duals has basis dxjdxk, for all j,k, (where the order matters).

Any second rank tensor can be expressed in terms of this basis and the coordinates or coefficients or componets are called gjk.

When we change coordinates, we get a new standard basis for the second tensor product and hence the coefficients of our basis expansion change. I/.e/; the gjk change into some other matrix valued function g'jk, in the way specified.

To define a tensor by saying how the components transform instead of what bundle it is a section of, is like defining a duck by the way it walks. Of course we all know the old saying: if it transforms like a tensor, then it is a tensor.

peace,

roy

mathwonk

Aug10-04, 07:27 PM

Dear competing tensor advocates.

I just checked out the site recommended above by Tom McCurdy in post 3.

http://en.wikipedia.org/wiki/Tensor

It tries to discuss both viewpoints on tensors and even relate them. I recommend it also. It seems from that discussion that my viewpoint is the so called modern one.

"The modern (component-free) approach views tensors initially as abstract objects, expressing some definite type of multi-linear concept. Their well-known properties can be derived from their definitions, as linear maps or more generally; and the rules for manipulations of tensors arise as an extension of linear algebra to multilinear algebra."

I still like my duckwalk joke, though.

pmb_phy

Aug11-04, 10:09 AM

An equation like summation gjk dxj dxk, as on the site you referenced, is a covariant tensor of second rank, because it is a second degree homogeneous polynomial in the expressions dxj, dxk, which are themselves covariant tensors of first rank.

Why do you call the summation a second rank tensor? It is not. gjk is a tensor of rank two. The differentials dxk are tensors of rank one. The summation is a contraction of a tensor of rank two with two tensors of rank one giving a tensor of rank zero.

i.e. it is of rank 2, because there are two of them multiplied together.

No. They are not multiplied together. They are summed. That is a huge difference.

if we consider only one tangent space isomorphic to R^n, its dual has basis dx1,...dxn, and the second tensor product of the duals has basis dxjdxk, for all j,k, (where the order matters).

I recommend learning how to use subscripts and superscripts on this forum. That way I can tell if you're using them or not. I don't see why you're refering to the differentials as a basis. A basis is a vector and not a component like dxj.

Pete

mathwonk

Aug11-04, 12:43 PM

Pete,

Let me try once again. I could be wrong of course, but I am not convinced by what you are saying. I am confident however that the miscommunication is due either to my ignorance, which hopefully is curable, or to a difference in language about tensors.

Now first of all you say that gjk is a tensor (subscript). To me this is like saying an n tuple of numbers is a vector. Just as an n tuple of numbers represents a vector, in terms of some basis, so also a matrix like (gjk) (subscripts) represents a tensor in terms of a basis.

Now if I denote by {...,ej,....} (sub) a basis of tangent vectors, then an n tuple of numbers like (,,,aj ...) represents

the vector: summation ajej.

Here it does not matter whether the indices are up or down, because we know what the objects are, namely, the ej are tangent vectors, so we know how they transform, whether I indicate it by sub or super scripts. Also the aj are numbers, and I have written a summation. Einstein's convention as I recall it, is that one can save on writing summation notation, if one uses oppositely placed scripts to signal summation automatically. I am not doing this.

Another convention is that subscripts are used like for tangent vectors ej, to denote that they are classical contravariant vectors (that is covariant in category theory). whereas superscripts are used like your elegantly written dx^j above to denote classically covariant vectors, i.e. covectors.

In modern terminology I believe this is achieved by saying that the ej are basic sections of the tangent bundle, whose variance is known to transform by the jacobian matrix, and saying the dx^j are (basic) sections of the cotangent bundle, or dual tangent bundle, whose sections are known to transform by the transpose of the jacobian matrix.

(These opposite transformation laws are used on the second site you referred me to, to define contravariant and covariant vectors.)

As to the word "basis", it is used by me in the sense of vector space theory. I.e. any space of objects closed under additon and scalar multiplication, is called a vector space. E.g. a tensor space is also an example of a vector space but I have refrained from using that term in that way on this post since it is not used that way by physicists it seems.

A basis for an abstract vector space is any collection of elements of the vector space, (e.g. if it is a basis of a tensor space, they will be tensors), such that every element of the space has a unique expression as a sum of scalar multiples of the given basis.

E.g. in R^n, if we denote the standard basis vectors (0,...0,1,0,....0) with a 1 in the jth place, by the symbol ej (sub), then the collection e1,...,en is a basis for the tangent space to R^n, simply because every tangent vector (a1,...,an) can be written uniquely as

summation: ajej.

(Here I have committed the apparent contradiction of referring to an n tuple of numbers (a1,...,an) as a tangent vector. But that is because R^n is the one vector space in the whole world, whose vectors really are n tuples of numbers. The concept of a basis is a way to represent elements of other vector spaces as elements of R^n, i.e. as n tuples of numbers.)

If I consider on the other hand the dual space (R^n)^ = linear maps from R^n to R, then this space is isomorphic to R^n, but the elements transform differently. One way to signal this would be to denote their coefficients by super scripts, but this is unnecessary, if we simply choose different symbols for them, such as dx^j.

These symbols are well chosen, because the basic elements of the dual space are the differentials of functions, and the simplest (coordinate) functions on R^n are the functions x^j.

(Thank you for your patience in bearing with what is no doubt extremely familiar to you. We may get somewhere yet however.)

Thus a covector like dx^j acts on a vector like ek by dot product in terms of their coordinate representations, or more intrinsically, by noting that dx^j(ek) = kronecker delta ?jk. Now I agree this pairing or contraction is signaled by the fact that the scripts of the dx's are up and those of the e's are down.

Now here is the source of the confusion for me in the equation (1) we were discussing.
But I will postpone it a little longer to clarify further my use of notation, and "basis".

In addition to the dual space T^ = (R^n)^, of R^n whose elements are apparently rank 1 covariant (old terminology) tensors, there is another space formed by taking the tensor product of this space with itself,

called T^ tensor T^,

whose elements are rank 2 covariant tensors. By definition, this space may be defined as the space of all bilinear maps from TxT to R. As such it is an abstract vector space, although it would be a sin to call its elements "vectors" in a physics forum, because to a physicist that word is reserved for rank 1 tensors.

Nonetheless T^ tensor T^ is a linear space, (that is a better word), and it has a basis, i.e. a set of elements such that all other elements can be written in terms of these.

E.g. such a basis is given by the tensor product dx^jdx^k of the two rank 1 tensors dx^j and dx^k. In gneral, if f,g are rank 1 covariant tensors, and if v,w is a pair of contravariant vectors, then the value of (f tensor g) on (v,w) is f(v)g(w).

Thus there is a tensor multiplication taking pairs of elements of T^ to one element of T^ tensor T^. Then it is a theorem, easily checked, that the set of products (dx^j)tensor(dx^k), for all pairs j,k, is a basis for the space of second rank covariant tensors.

In particular every such tensor can be written in terms of these. Thus a general second rank covariant tensor would be written as

summation gjk(sub if you like) dx^j dx^k, where I have omitted the tensor sign.

(Thus you are right there is a summation here of the tensors gjk dx^j dx^k,

but there is also a tensor product, of dx^j times dx^k.)

In particular, the standard scalar product on euclidean space would be written as

summation (kronecker delta) dx^j dx^k = dx^1dx^1 +.....+ dx^ndx^n.

Such an object acts on pairs of tangent vectors and spits out a number. E.g. on the pair (a1,...,an),(b1,...,bn) = (summation ajej, summation bkek),

it spits out of course summation ajbj.

Thus I am interpreting the object in equation (1) at the referenced site as representing the second rank covariant tensor:

summation gjk dx^j dx^k

whose value on the pair of vectors (summation ajej, summation bkek),

is the number (matrix product):

(,,,aj,....) (gjk) (...,bk,...)T = summation (over j,k), gjk aj bk.

where here the T means transpose.

Now we can also consider the tensor product of the tangent space with itself and get the space (T tensor T) of second rank contravariant vectors, for which a basis is given by the products {ej tensor ek}.

Then we can consider that a covariant rank 2 vector like:

summ gjk dx^j dx^k

acts not on the pair (summation ajej, summation bkek),

but rather on the contravariant vector, their product:

summation ajbk (ej tensor ek).

Then the value in coordinates is given by:

summation(over j,k) ajbk gjk, a number.

Now I am trying to see how to possibly interpret equation (1) as you have done.

E.g. if I represent a contravariant rank 2 vector:

summation gjk ej tensor ek,

simply by the matrix gjk,

and represent the covariant tensor:

summation dx^j tensor dx^j,

by the same expression:

summation dx^j tensor dx^j,

then I suppose I could believe that the equation (1) represents the number

obtained by evaluating the covariant 2 tensor:

summation dx^j tensor dx^j,

on the contravariant 2 tensor:

summation gjk ej tensor ek.

This would violate two principles I hold sacred: first the symbols gjk are never used for contravariant tensors, but always for covariant tensors.

second and more important: the object being represented there is not a number as you say, but a tensor, the metric tensor. They say so right on the site. (let me check that and get back to you.)

Anyway I appreciate your sincere and patient attempt to communicate with me.

We are struggling since it seems you apparently speak primarily "indices" and i speak only "index free", so if someone with dual langauage capabilities would jump in, it might help, but maybe we are doing as well as can be expected.

OK I have been back your site, and to my mind it confirms what I have been saying.

E.g. it says there that the expression G (bold) = summation gjk dx^j dx^k is a tensor, whose components are the numbers gjk. (The matrix gjk is therefore not itself a tensor.)

From the basis point of view, this means that this tensor G is written as a linear combination of the basic tensors dx^j dx^k, using the components or coefficients gjk.

He does not say it there, but those basic tensors themselves are products of the rank 1 tensors dx^j and dx^k.

This is fun, and I hope we are helping each other. I know I appreciate your patience with my ignorance of common longstanding notation and practice in physics.

best regards,

roy

pmb_phy

Aug11-04, 02:08 PM

I'm sorry, but due to major back problems it is impossibe for me to sit down and read such a long post. I'll have to spend some time reading and absorbing what you wrote bit by bit due to the short amounts of time I can sit in front of the computer.

In the meantime can you please find and post a reference to where you've seen/learned the definition(s) which you hold to be true.
Thanks

Pete

mathwonk

Aug11-04, 03:08 PM

These are the definitions I believe to have been standard in mathematical treatments of differential geometry since the 1960's, for example in Michael Spivak's little book Calculus on Manifolds, or his large treatise Differential Geometry. I will look for an internet source, but i suspect the one on wikipedia would suffice.

http://en.wikipedia.org/wiki/Tensor

I will check it out more carefully.

I sympathize with the back problems as I also have them. Mine are helped by sitting only in an old captain's chair I inherited from my grandfather, but there must be others out there.

mathwonk

Aug11-04, 03:31 PM

Pete,

Here is a quote from wikipedia:

"There are equivalent approaches to visualizing and working with tensors; that the content is actually the same may only become apparent with some familiarity with the material.

* The classical approach
The classical approach views tensors as multidimensional arrays that are n-dimensional generalizations of scalars, 1-dimensional vectors and 2-dimensional matrices. The "components" of the tensor are the indices of the array. ......

* The modern approach
The modern (component-free) approach views tensors initially as abstract objects, expressing some definite type of multi-linear concept. Their well-known properties can be derived from their definitions, as linear maps or more generally; and the rules for manipulations of tensors arise as an extension of linear algebra to multilinear algebra.

This treatment has largely replaced the component-based treatment for advanced study, in the way that the more modern component-free treatment of vectors replaces the traditional component-based treatment after the component-based treatment has been used to provide an elementary motivation for the concept of a vector. You could say that the slogan is 'tensors are elements of some tensor space'.

* The intermediate treatment of tensors article attempts to bridge the two extremes, and to show their relationships."

Here is the link for the intermediate article, but it is pretty sketchy.

http://en.wikipedia.org/wiki/Intermediate_treatment_of_tensors

One of the specific examples wikipedia cites of a tensor, is a homogeneous polynomial of degree two. There is also a short tutorial on this forum, in the thread "Math Newb wants to know whats a tensor", by chroot, where the scalar product is cited as an example of a rank 2 covariant tensor, not rank zero, (actually he calls it type (0,2), or rank 2 covariant and rank zero contravariant).

I also gave an explicit calculation in the thread "tensor product" that shows how to use the modern definition of a tensor to compute the tensor product as matrices.

If we get together on this, it will be a major success for both of us, but not worth a backache.

best regards,

roy

pmb_phy

Aug11-04, 03:45 PM

Hi again

Yep. Back problems suck big time. Due to that sitting and typing too long I had to be rushed to the hospital in an ambulance. I've experienced levels of pain (from sitting here typing too long) that I never knew even existed, and I know pain since I've had 8 bone marrow biopsies. I have no intention of letting that happen again.

Meanwhile, if you happen to have A short course in General Relativity, J. Foster, J.D. Nightingale then see section 1.10. It will clarify the difference between contravariant vectors and 1-forms (aka covariant vectors). In Euclidean space you can get away with igoring the difference, but not in general. A 1-form maps vectors to real numbers. They are distinct objects from vectors (but they are related). That is why I have been emphasizing the placement of the indices. In some areas, such as mechanics, one uses only Cartesian tensors and this distinction never arises. A Cartesian tensor is an example of an affine tensor which is different than, but somewhat similar to, a tensor.

Note that it is common practice to use what Foster and Nightingale phrase as follows on page 45
[quote]
From here onwards we shall adopt a much used convention which is to confuse a tensor with its components. This allows us to refer simply to the tensor Tab, rather than the tensor with components Tab.

g is literally the tensor while gab is literally the components. They are defined by gab = g(ea,eb).

Pete

mathwonk

Aug11-04, 04:27 PM

Hi again Pete,

I admire your tenacity in this forum! It inspires me after my own last weeks "procedure" near the area I use for sitting.

I discovered that this whole hash about competing languages has already been discussed at unbelievable length in the thread "intro to differential forms, started by lethe, and subsequently largely deleted by him, over a flap about use of more informal language.

Check out posts #50,.... in that thread to see some of the same discussions we have been having about up or down indices. Of course they were talking about the case of anticommutative covariant tensors, or differential p - forms, rather than general tensors.

It appears the original tutorial posted by lethe still exists at another site, namely

http://www.sciforums.com/showthread.php?t=20843&page=1&pp=20

He gives the full monty discussion there of the language I was advocating above. In particular, he does define general tensors on the way to defining skew commutative ones

And thanks for the reference to Foster and Nightingale.

Maybe I can still learn some relativity!

best wishes,

roy

mathwonk

Aug11-04, 04:52 PM

PS

It seems one confusion is that although (the components of) a covariant tensor are subscripts, the basic co - tensors themselves are apparently written as superscripts.

i.e. gjk for the components as opposed to dx^j dx^k, for the basic guys.

E.g. the basic tensor dx^1dx^1 written in components,

would just be the matrix (gjk),

where all the gjk equal zero except g11 = 1.

Similarly, the coordinates (components) of a contravariant tensor are written as super, and the indices on the actual basic tensors are written as subs.

i.e. a^j as opposed to ej.

That way when we conjoin the components of a contra -tensor with the components of a co - tensor, it does mean to contract and get a number,

e.g. summ gjk h^(jk)

But when we conjoin the components gjk say of a co tensor, with the symbols for the basic co tensors, it only means to sum them up as numbers times co - tensors, so it is still a co tensor.

e.g. summ gjk dx^j dx^k, is a rank 2 cotensor.

So here are two versions of the same object

classical covariant 2 tensor: gjk

modern version of same covariant 2 tensor: summation gjk dx^j dx^k.

classical version of contravariant 2 tensor h^(jk)

modern version of same contravariant 2 tensor: summation h^(jk) ej ek.

Then (summation gjk dx^j dx^k) acts on: (summation h^(jk) ej ek),

by contracting their components: summation gjk h^(jk). this is a number.

How does this seem?

mathwonk

Aug11-04, 04:55 PM

PPS:
In some sense the modern point of view has made the wonderful contribution of doubling the number of indices!!

I.e. the modern way of writing that last contraction would be:

(summation gjk dx^j dx^k) (summation h^(rs) er es)

= big summation (gjk dx^j dx^k) (h^(rs) er es)

= big summation (gjk h^(rs)) (dx^j dx^k)(er es)

= big summation (gjk h^(rs)) (dx^j(er))(dx^k(es))

= big summation (gjk h^(rs)) kronecker^(jk),(rs)

[this last because dx^jdx^k (ej ek) = 1, and all other pairings

dx^jdx^k (er es) are zero]

= summation gjk h^(jk).

But no one would often do this I hope.

I admit that if one understands the indices on the components, there is never any need for the basic tensors, but it seems almost to throw out the baby and keep the bathwater, to an index - free guy like me.

I admit too the indices are too complicated for me. I even wrote an algebra book once, including treating tensors in a coordinate free way, and actually wrote out the tensor product of matrix as a consequence of these definitions, but it was a terrifying experience.

Just for laughs, I confess that to me the tensor product, Atensor(blank), is actually "the unique right exact functor on R - modules that commutes with direct sums and takes value A on the field R of real numbers", but I would not readily say that here, if we weren't good friends by now!

Perhaps that reveals why I am having such a hard time understanding classical tensors though.

Peace,

roy

mathwonk

Aug11-04, 05:10 PM

By the way, in your nice post 17 above, you distinguish tensors from components by writing the tensors as bold, e.g. in the last line, you write "e suba (bold)" for a basic contravariant 1 tensor.

In that same spirit, on the site

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

the symbols dx^j in equation (1) should be bold, since they are the entirely analogous basic covariant 1-tensors.

The fact that they are not bold, leads to the confusion that this expression denotes a 0 - tensor instead of a 2 - tensor,

i.e. equation (1) is a sum of, scalar multiples of, pairwise products of, basic 1 tensors,

hence it is a homogeneous "polynomial" of degree 2 in the basic 1 tensors,

i.e. a 2-tensor.

Does that seem believable?

pmb_phy

Aug11-04, 06:35 PM

In that same spirit, on the site

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

the symbols dx^j in equation (1) should be bold, since they are the entirely analogous basic covariant 1-tensors.

The components of tensors are in italics.

The fact that they are not bold, leads to the confusion that this expression denotes a 0 - tensor instead of a 2 - tensor,

That expression is a tensor of rank 0. If you notice, it is the contraction of a second rank covariant tensor with two rank 1 tensors. Such a contraction is always a tensor of rank zero. Why do you keep calling the interval a dstensor?

i.e. equation (1) is a sum of, scalar multiples of, pairwise products of, basic 1 tensors,

hence it is a homogeneous "polynomial" of degree 2 in the basic 1 tensors,

i.e. a 2-tensor.

Does that seem believable?
Nope. Sorry.

Pete

mathwonk

Aug11-04, 07:43 PM

Pete,

This is the basic point you are missing in my opinion. I cannot make it any clearer though than I have in post #19 above. If that doesn't do it, I have no more to add.

Of course I could always be wrong! (and I frequently am.)

have a good one, my friend.

roy

mathwonk

Aug11-04, 09:17 PM

Pete,

Here is a good looking reference for notes on relativity that uses both indices and the conceptual approach, by a clear expert. If it makes sense to you, you do not need to concern yourself with anything I have been saying. He also recommends the famous book Gravitation?, by Misner, Thorne, and Wheeler, which apparently also uses the modern approach.

These are notes of Sean Carroll at Institute for Theoretical Physics at Santa Barbara, UCAL.

http://xxx.lanl.gov/abs/gr-qc/9712019

best,

roy

PS: I was going to quit after 100 posts but this is kind of addictive! I must stop soon though, as I have to go to work.

pmb_phy

Aug12-04, 08:58 AM

Pete,

Here is a good looking reference for notes on relativity that uses both indices and the conceptual approach, by a clear expert.

Yep. I've seen them and have read part of them several years ago and during the last few years as reference. I've also read MTW as well as many other GR and tensor analysis texts.

And I understand what you were saying. What I was saying is that it is incorrect.

I've been trying to find the notes I used when I first studied GR but the site is not working right now. It will be under http://arcturus.mit.edu/~edbert when it does come up.

Pete

mathwonk

Aug12-04, 11:32 AM

Hi Pete, another day another dollar.

Well, let's try again.

I believe that since x is a function, its differential dx is a covariant 1-tensor, i.e. a section of the cotangent bundle, because it acts on a tangent vector, via directional differentiation, and spits out a number.

Similarly, the tensor product of dx^j and dx^k is a covariant tensor of rank 2, because it acts on a pair of tangent vectors and spits out a number, namely the product of their jth and kth coordinates in the x - coordinate system. so dx^j dx^k is a covariant 2 tensor, i.e. a section of (T^) tensor(T^).

Now you are saying that gjk is also a 2 tensor. well, from what you have told me, it is called a 2 tensor as abuse of language. But what is the actual 2 tensor it is shorthand for? I understand it to be shorthand for the covariant 2 tensor:
summation gjk dx^j dx^k.

That would make it also a section of (T^)tensor(T^),

hence not a candidate for contraction against another such bird.

So here is the crucial point. You must contract tensors of opposite variance.

I,e, summation gjk dx^j dx^k is not actually a contraction.

The problem to me seems to be that if one thinks any expression with some indices up and others down is a contraction, one gets in trouble.

I.e. you are mixing two different languages here. the notation dx^j always stands for a section of the cotangent bundle, namely the differential of x^j. Hence dx^j really is a tensor, not just the components of one, i.e. it is not abuse of language to call dx^j a tensor.

if you want to contract two tensors, they have to be written in the same way, not one as an actual tensor and the other as components of a tensor.

If you want to contract gjk with some thing, it has to be with something like h^(jk), which would be the components of a contravariant 2 tensor like

summation h^(jk) ej ek.

The problem is that the components of a tensor transform opposite to the way the basis elements of the tensor transform.

Thus the basic covariant tensors are written with indices up, while their components are written with indices down.

Thus you do contract components having opposite indices, but a sum of components and basis elements with opposite placed indices is not a contraction.

Lets be more concrete.

What bundle do you think dx^j dx^k is a section of? and what bundle do you think gjk represents a section of?

You canno, contract them unless your answer is that they are sections of mutually dual bundles, but to me that would be hard to bring in line with any accepted notation or usage.

roy

mathwonk

Aug12-04, 11:45 AM

Pete,

Let me give an example from your own post 17. I will write G for "bold g" and

ea for "e lower a".

Then you say that G is the tensor, and gab = G(ea,eb) are the components.

That is correct. But another way to write the tensor G is as:

G = summation gab dx^a dx^b.

In fact that is the meaning of the statement that "gab are the components of G".

Here is how it looks written out fully:

ea and eb are basic contravariant 1 tensors,

and dx^a and dx^b are the dual basic covariant 1 tensors.

I.e. dx^a(ea) = 1, and dx^a(eb) = 0 for any b different from a.

Any way of acting on the tensors ea,eb can be expressed in terms of these simplest possible actions. Thus any tensor G can be expressed as a linear combination of dx's.

I.e. there are some numbers gab such that G = summation gab dx^a dx^b.

Those numbers gab are computed by evaluating both sides

on the basic contravariant tensors (ea,eb).

One obtains:

G(ea,eb) = summation gjk dx^j dx^k (ea,ab) = gab,

because almost all terms dx^j dx^k (ea,ab) in the sum are zero

except dx^a dx^b (ea,ab) = 1.

I.e. because G and summation gab dx^a dx^b are both (the same) cotensor,

they can both be evaluated on the basic contra-tensors (ea,eb), and yield the same answer.

peace.

mathwonk

Aug12-04, 02:51 PM

Dear Pete,

One last contribution. if e1,...,en, are basis vectors for R^n, i.e. e1 = (1,0,...,0),

e2 = (0,1,0,...,0), .....,en = (0,....,0,1), (where ej means e "subscript j")

Then any vector V can be written as a linear combination of these basis vectors with number coefficients.

In the notational convention taught in the books you have referred me to, for instance Sean Carroll, these number coefficients are written with superscripts,

thus to say the vector V has coefficients a^1,....,a^n, means we can write V as

V = summation a^j ej = (a1,....,an).

By your logic, because these have oppositely placed indices this would be a contraction, and hence a 0 - tensor, i.e. a number, whereas I hope we agree this is clearly a 1 - tensor, i.e. a plain old vector.

According to Carroll, the only significance iof these objects having oppositely placed indices is that the object obtained by summing them is an invariant object, independent of coordinates. In particular here it is a vector.

By duality with your example in post #17, where you find the coeeficients or components of G, by evaluating G on the basic vectors ej, here we could also find the coefficients, i.e. components of V, by evaluating V on the basic covectors dx^j, or if you like, by evaluating those covectors on V.

(Carroll uses a notation like theta^j for dx^j.)

I.e. V(dx^j) = dx^j(V) = a^j. THIS is a contraction, between a contratensor V and a cotensor dx^j, and yields a number, a^j.

Oh well, I'm sure we agree in some sense, since we both seem to be getting along successfully using the subject, but we just do not understand what each other are saying.

Hope to meet you some day and have a nice glass of wine. That always helps me clear my head and open my mind.

roy

sal

Aug12-04, 03:04 PM

Pete asked me for another opinion on this one, so I'm going to stick my oar in...

First, I have to say that all you guys seem to be arguing about is notation, and IMHO the only "correct" notation is notation that's either completely universal and already known to the reader, or notation that's explained in the text. The only "incorrect" notation is notation that's so confusing, nonstandard, or self-contradictory that the reader can't figure it out.

Ain't none of that latter stuff here, that I can see. :smile:

Pete,
Let me give an example from your own post 17. I will write G for "bold g" and ea for

e_a.

Then you say that G is the tensor, and gab = G(ea,eb) are the components.

That is correct. But another way to write the tensor G is as:

G = summation gab dx^a dx^b.
Almost but not quite.

If you want to be nit-picky squeaky-clean correct, and you want to follow standard conventions here, you need to use a tensor product on the right, and you didn't do that. You should write

G = \sum g_{ab} dx^a \otimes dx^b

Otherwise you've got an ordinary product, which is a very different thing.

What you actually wrote was identical to what Pete had in equation (1) in http://www.geocities.com/physics_world/ma/intro_tensor.htm. That is not an equation between tensors. Rather, it is an equation in terms of infinitesimals. (See the end of this post for more on the line element.)

They're "physicist's infinitesimals", which have been in common use clear back to Einstein's papers. They're a shorthand for taking a limit. If you reduce the power to 1 and divide through by dt, poof, you have a derivative.

In fact that is the meaning of the statement that "gab are the components of G".
Yup, you can write it as a tensor equation in terms of the basis covectors, you can write it as a line element in terms of infinitesimals, you can write it as matrix (since it's rank 2), or you can just write it as a single bold-faced letter. They all mean the same thing, as long as your reader is on the same wavelength you're on!

Back on the subject of the basis covectors, you said someplace that they should be bold. Well, that's one convention. Not everybody does that. Furthermore, some people also use a tilde overtop to indicate that they're covectors, and not just infinitesimals, or even vector gradients.

And some people never use the word "covector" at all and would say that my post was riddled with nonsense as a result -- they'd call them "dual vectors" or even "basis 1-forms".

I'm going to snip the rest of the post I was replying to, and comment on a few other items from earlier in the thread.

I believe that since x is a function, its differential dx is a covariant 1-tensor, i.e. a section of the cotangent bundle, because it acts on a tangent vector, via directional differentiation, and spits out a number.

As I said, that's one way of using the symbol "dx", or more commonly, dx, or even
\tilde{d}x .
It's not the only way to use dx; another very common use is as an infinitesimal, and in fact that's typically how the "line element" is written.

Going on,
Now you are saying that gjk is also a 2 tensor. well, from what you have told me, it is called a 2 tensor as abuse of language. But what is the actual 2 tensor it is shorthand for? I understand it to be shorthand for the covariant 2 tensor:
Calling it any kind of "2-tensor" is indeed an abuse of the language. A "2-tensor", most often, is a tensor on a 2-dimensional space; what you're talking about here is a rank 2 tensor.

Written as g_{ab}, it's actually a rank (0,2) tensor, but since it can be converted trivially into a rank (1,1) tensor or a rank (2,0) tensor, it's also perfectly reasonable to just refer to the whole "package" of 3 related tensors as a "rank 2 tensor".

I.e. you are mixing two different languages here. the notation dx^j always stands for a section of the cotangent bundle, namely the differential of x^j.
As I said, if that's a tensor you're talking about, then yeah, it's a section of the cotangent bundle. But just as often, it's an infinitesimal rather than a tensor.

In the land of physics, it's just not accurate to say it "always" stands for a 1-form.

Here is a good looking reference for notes on relativity that uses both indices and the conceptual approach, by a clear expert.
Check this reference again, and look at page 25, formula 1.95. It is exactly what Pete had. I quote:

"A more natural object is the line element, or infinitesimal interval:
ds^{2} = \eta _{\mu \nu}dx^{\mu} dx^{\nu} "

So as I said to start with, Pete's talking about an infinitesimal line element, you're talking about covectors, and you're both using the same notation. It's incorrect to say the notation is only used for one or the other -- it's used for both, and if it's not clear from the context which is meant, you need to spell it out.

I hope this helps, at least a little, with the confusion...

mathwonk

Aug12-04, 04:44 PM

Well it does help, because I thought "infinitesimals" went out with Newton. What do they mean to you? I also took Carroll's first chapter which you cite, as an imprecise conversational verson of the material before it gets precise.

At least I made it possible for you to understand what I meant by dx^j by defining it so you could tell I meant it is a differential.

if you read some of Pete's posts however you will see that he himself said that in his cited equation (1) that dx^j dx^k was 2 tensor, and also that gjk was a 2 tensor, and that therefore the combination summation gjk dx^j dx^k was a contraction to a 0 tensor.

Here is a quote from his post #22:

"That expression is a tensor of rank 0. If you notice, it is the contraction of a second rank covariant tensor with two rank 1 tensors. Such a contraction is always a tensor of rank zero."

So Pete never said his equation denoted infinitesimals, rather that it was a contraction of rank 2 tensors. do you agree with that? that is all I was puzzled by.

sal

Aug12-04, 05:24 PM

Well it does help, because I thought "infinitesimals" went out with Newton.
No indeed! They're alive and well in the physics community.

Mathematicians sometimes refer to them as "Physicist's sloppy infinitesimals".

What do they mean to you?

In general they're shorthand for a limit process. Older (pre-1950) books on tensor calculus used them exclusively. It was only in the last half of the 20th century that it became really common to use the formal definition of a tangent vector as a partial (path) derivative and cotangent vector as the dual of that, rather than just talking about an "infinitesimal displacement".

For example, Einstein, "On the Electrodynamics of Moving Bodies" (1905), p44 in the Dover edition "The Principle of Relativity", in the course of deriving the Lorentz transform, says

"Hence, if x' be chosen infinitesimally small..."

and he goes on from there, using derivatives and "infinitesimals".

Synge and Schild, "Tensor Calculus" (c) 1949, p. 9 in the Dover edition, in section 1.3, discussing contravariant vectors and tensors, say

"...These two points define an infinitesimal displacement or _vector_ PQ..."

Synge and Schild is something of a classic though it's now considered rather out of date.

I don't know when the machinery for handling tangent vectors rigorously was invented, but most of the seminal physics in relativity was developed without it.

I also took Carroll's first chapter which you cite, as an imprecise conversational verson of the material before it gets precise.
That's the most common way of writing the line element that I've seen, and it's done in terms of infinitesimals.

With enough effort you can define rigorous "differential" functions in one dimension and work it out that way, but if you do, you're just wallpapering over the original meaning which was a relationship among infinitesimals.

At least I made it possible for you to understand what I meant by dx^j by defining it so you could tell I meant it is a differential.

if you read some of Pete's posts however you will see that he himself said that in his cited equation (1) that dx^j dx^k was 2 tensor, and also that gjk was a 2 tensor, and that therefore the combination summation gjk dx^j dx^k was a contraction to a 0 tensor.

Here is a quote from his post #22:

"That expression is a tensor of rank 0. If you notice, it is the contraction of a second rank covariant tensor with two rank 1 tensors. Such a contraction is always a tensor of rank zero."

So Pete never said his equation denoted infinitesimals, rather that it was a contraction of rank 2 tensors. do you agree with that? that is all I was puzzled by.
I don't see how that can be correct.

The expression is a representation of a tensor. I would hesitate before saying g_{ab} dx^a dx^b is actually a contraction, because that would make it a single number (just as Pete said) and I don't see how to apply that to anything.

Pete may have a different notion as to what the terms in the line element representation of the metric mean, but what I've described here is, I believe, fully consistent with what's on his website.

mathwonk

Aug12-04, 06:08 PM

It has dawned on me that physicists may be willing to use something logically nonsensical just because Einstein did so, and achieved correct results.

I think if the modern version of differential geometry had been around in 1900 then Einstein would have used it instead.

I agree of course that physicists are really using their intelligence and intuition, rather than mathematiocal rigor, which is why they so seldom go astray.

I love the recent story of the puzzle as to how many rational cubic curves lie on a general quintic hypersurface in complex 4 space. The mathematicians, by brute force computation had one answer, and the physicists by relating the problem to one in quantum gravity or something, had a different prediction which popped out of a recursion formula and a differential eqaution they thought applicable.

Of course the physicists were actually right, and it led to a whole industry in enumerative algebraic geometry.

We mathematicians are merely trying to formulate precisely the intuitions physicicts seem blessed with because of their familiarity with nature. We are at a big disadvantage here.

But we do not seem to argue as hopelessly as some theoretical physicists do, because we do eventually make clear what we are saying.

My error in my previous long harangue, was not to ask precisely what Pete meant by his notation, and not to say precisely what I meant by it.

That was what I meant when I expressed confidence we would agree at some level, once we understood each other properly.

I have almost never heard a disagreement that was not found to be based on different interpretations of the same words being used.

pmb_phy

Aug12-04, 07:37 PM

if you read some of Pete's posts however you will see that he himself said that in his cited equation (1) that dx^j dx^k was 2 tensor,..

Caution is required here. So long as you know that I said that dx^j are the components of a vector then we're all set. I don't mind the shorthand statement that dx^k is a vector though but I'm never sure what you mean by it.

So Pete never said his equation denoted infinitesimals, rather that it was a contraction of rank 2 tensors.
Ummm .. scuse me, but I did say that in
http://www.geocities.com/physics_world/ma/intro_tensor.htm
right above Eq. (1), i.e.

The arc length, dl, between two closely spaced points on a curve is given in Cartesian coordinates, by ...

unless you didn't know that the arc length was an infinitesimal? Usually one doesn't need to state that explicitly since the notation speaks for itself (hence the purpose of notation). dl is an infinitesimal and Eq. (1) gives the square of dl.

Use caution when using notation. dxk is a differential but dxk is not a differential. It is the gradient of a coordinate making it a basis 1-form.

Pete

mathwonk

Aug12-04, 09:09 PM

sal

Aug12-04, 10:47 PM

Hurkyl

Aug12-04, 11:14 PM

Well, for funsies, I know of three ways that "infinitessimals" can be made rigorous.

One is algebraically; something with a power equal to zero. I thought it was the coolest thing when I saw the algebraic definition of the derivative of a polynomial:

f'(x) is the unique polynomial satisfying:
f(x + h) = f(x) + h f'(x) (mod h^2)

Another is though nonstandard analysis; use the hyperreals which are extremely similar to the reals, but there are positive hyperreals smaller than any positive real. (and thus called infinitessimal)

And another which I came across (I think) at Baez's site, which I don't fully understand, is through some clever logic. There's a way to consistently define an "arrow" whose head and tail are at the same point, yet they are different. These arrows can then be used naturally as infinitessimal vectors.

mathwonk

Aug12-04, 11:43 PM

Quote:
"Well, for funsies, I know of three ways that "infinitessimals" can be made rigorous.

One is algebraically; something with a power equal to zero. I thought it was the coolest thing when I saw the algebraic definition of the derivative of a polynomial:

f'(x) is the unique polynomial satisfying:
f(x + h) = f(x) + h f'(x) (mod h^2)"

Actually this algebraic definition of differentials is equivalent to the usual theory of one forms, so defines the same thing, not something different or more exotic.

The idea is that a one form is "a function on the diagonal vanishing once, modulo those vanishing twice".

If M is the ideal of polynomial functions on K^n vanishing at the origin, then the cotangent space or dual tangent space at the origin of K^n, is M/M^2, which if you think about it is all linear polynomials, i.e. it is all polynomials vanishing at the origin modulo those,vanishing to degree higher than one. Taylor's theorem shows this makes sense also for smooth functions.

Then to globalize this concept, consider the injection of X into XxX as the diagonal, where X is some manifold, or affine space, or scheme.

If I is the ideal of functions vanishing on the diagonal, then I/I^2 is the module of differentials on the diagonal, which is isomorphic to X. So I/I^2 is the module of one forms on X.

But this is merely an algebraic way to recover in algebraic geometry, the theory of one - forms from differential geometry.

Hurkyl

Aug13-04, 07:15 AM

A nifty thing about the algebraic approach is that you don't have to let h^2 be zero; you can use another power. e.g.

f(x + h) = f(x) + f'(x) h + f''(x) h^2/2 (mod h^3)

mathwonk

Aug13-04, 11:57 AM

yes! That idea gives you the "Taylor series" for an element of a local ring on a scheme.

In fact, a point of a scheme is a manifold point (non singular), apparently when the Taylor series in this sense is unique for eadch function. (At least that is true at a non singular point.)

The Taylor series is an element of the direct sum of the quotients M^r/M^(r+1), for all r>=0.

Since this was (I believe) Fermat's attitude to calculus, maybe Fermat invented schemes!

mathwonk

Aug13-04, 12:03 PM

In general, we can define order of contact of a line with a curve at a point (a, f(a)) by which power of x-a divides the difference between the eqution of the line and the equation of the curve.

A tangent line is one for which the order of contact is >= 2, and an infllectional tangent one for which it is >= 3, etc.

Then (I believe this is Descartes' method), we can do differential calculus for all polynomials this way just by algebra.

So what we teach in college about limits to define derivatives, is of course unnecessary except for the transcendental functions.

Of course we all know this, but the poor students still have to struggle with those hard ideas when these easy ones were good enough for Fermat, Descartes, etc...

I believe there are some books around that do calculus this way for polynomials and probably also for analytic functions (those given by power series.) I tried it once but it is hard to swim upstream against established practice.

mathwonk

Aug13-04, 12:09 PM

I once heard it suggested that a scheme structure could be considered as some differential equations on a manifold such that a function was considered zero at a point, or along a sub manifold, not just if its values were zero there but if also it satisfied the differential equation.

For instance the subscheme of the line, defined by the equation x^2 = 0, instead of x=0, should pick out the ideal of functions generated by x^2, i.e. functions vanishing at 0 along with their first derivatives. But I have never seen this worked out fully.

pmb_phy

Aug15-04, 10:55 AM

Right you are Pete.

My apologies. I am not too up on infinitesimals and it must have gone right by me. In general I want to thank you (and the others here) for your extraordinary patience with me.

Glad to help. I have infinite patience with people who are pleasant and you've been nothing less than very pleasant.

This is a great site for learning, and I will recommend it to my students too!

I didn't know that you were a teacher. What do you teach? College level etc?

By the way, there are different definitions of tensors used in areas such as general relativity according to different views. I'm more familiar with that view used by such authors as Ohanian, Misner Thorne and Wheeler, Foster and Nightingale, Schutz etc. and less familiar with that used by Wald. If youy want to learn about Wald's view on tensors then sal is the expert here. I'd pick his brain on Wald if I were you and wanted to learn both. I plan on learning Wald myself but I'm in no hurry. But as I dom sal will be the person I turn to for help. He's very good at math from what I've seen.

One last point - Regarding vectors and 1-forms, i.e. covariant and contravariant vectors. As I mentioned above these are different animals. They belong to dual spaces. If you're familiar with quantum mechanics they you may be familiar with bras and kets. bras and kets belong to dual spaces also and they too are different animals. :smile:

Best wishes

Pete

robphy

Aug15-04, 06:55 PM

mathwonk

Aug15-04, 10:35 PM

Yes I teach college level math from beginning calc to graduate algebraic geometry.

I have also taught Euler's theorem V-E+F = 2, to second graders, using cardboard polyhedra with colored faces. One of my 2nd grade students became an aerospace engineer!

I am a professional algebraic geometer, but always pitifully ignorant of physics.

Ironically I have been an invited lecturer at the International Center for Theoretical Physics in Trieste. But it seems they wanted to pick my brains about Riemann surfaces.

Just between us, physics rocks!

This site is amazing for the generosity you all display in answering absolutely any question, for essentially anyone. Congratulations. I have really learned a lesson in patience from you that I hope to bring to my teaching.

I do not know whether jets answer the question I posed or not. The problem is to find a differential equation that captures the same information as an arbitrary ideal in a ring, or at least in a polynomial ring.

I hope to meet you guys some time.

best,

roy

PS: the semester begins now, so if I do not answer for awhile, I am preoccupied with grading! (as you well know).

sal

Aug15-04, 10:54 PM

If you want to learn about Wald's view on tensors then sal is the expert here. I'd pick his brain on Wald if I were you and wanted to learn both. I plan on learning Wald myself but I'm in no hurry. But as I dom sal will be the person I turn to for help. He's very good at math from what I've seen.

I am a professional algebraic geometer...

I'm flattered by Pete's opinion, and I like to think I'm a bright guy, but after all I'm a programmer, not a mathematician ... and I think I know the difference between a "professional" and a "wannabe", even a "wannabe" who maybe "couldabeen" if he hadn't gotten distracted from math by computers his junior year in college...

If you want to grok Wald's views on math just look through his appendices; he lays it all out there. Save for his "abstract index" notation, his approach seems very standard to me, much like what I recall of Warner's "Foundations of Differentiable Manifolds", for instance.

mathwonk

Aug15-04, 11:04 PM

Thanks, I read Warner about 25 years ago, so its the physics I want to learn.

Feynman

Aug18-04, 11:11 AM

A Tensor is a multilinear continus form like the scalar product

sal

Aug18-04, 02:26 PM

A Tensor is a multilinear continus form like the scalar product
If you're going to post as "Feynman" at least try to be a little accurate, OK?

A "form", as the term is generally used, is a covariant tensor field, mapping points on a manifold into covariant tensors at each point; it is not a tensor per se.

What's more, contravariant tensors, of which vectors are the archetypical example, are also tensors (obviously) and they're not forms (obviously), because they're not covariant.

In general, a covariant tensor is a multilinear map from vectors to the real numbers. A covector is a covariant tensor of one argument. A contravariant tensor is a vector, or a more general multilinear map from covectors into the real numbers. A mixed tensor is a multilinear map from some number of vectors and some number of covectors into the real numbers. The tangent space at each point on a manifold is the collection of tangent vectors at that point. The cotangent space at a particular point on a manifold is the collection of covectors acting on tangent vectors at that point. A tensor field is a mapping from a manifold to the tensor algebra defined at each point.

There are a number of other ways of defining tensors and tensor fields but they all end up in the same place.

If you can't get your accuracy level up, maybe you should change your handle. (Or maybe you should change it anyway -- it's disrespectful, IMHO.)

mathwonk

Aug19-04, 10:37 AM

In the spirit of keeping a hot thread going, here's another possibly provocative suggestion: there is no such thing as "a tensor".

I.e. "tensor" is a verb, not a noun. Given any ring R and any two R modules M,N, one can form their tensor product over R, M(tensor(R))N.

The special case where R is the real numbers, and M =T is the tangent space to a manifold at a point, and N = T* is the dual or cotangent space, yields the example T(tensor(R))T*, whose elements are called by some authorities here, (1,1) tensors (at a point).

Then one can speak of fields of these objects, i.e. locally families of such elements parametrized by open sets U of the manifold, which is (equivalent to) a function from U to T(tensor(R))T*.

Then on larger open subsets of the manifold, such as the whole manifold, one must form a union with identifications of these products, and this leads to coordinate changes with various rules.

These special types of coordinate changes are themselves endowed with the name "tensor" in these posts, but they are really a very special aspect of a very special, but important, example.

In general one can tensor together any two modules, or any two bundles, such as a normal bundle and the exterior algebra of a cotangent bundle, or really anything, even objedcts that are not bundles, i.e. not locally products.

How about them bananas? Any bites?

pmb_phy

Aug19-04, 11:48 AM

In the spirit of keeping a hot thread going, here's another possibly provocative suggestion: there is no such thing as "a tensor".

That would certainly make life easier. :rofl:

I.e. "tensor" is a verb, not a noun.

That would be a re-definition/different use of the term "tensor". Ever notice that in a dictionary that words can have multiple meanings? Same in math at times. The term tensor, as commonly used in tensor analysis/differential geometry, is just as much a noun as is the term "vector". Especially since a vector is an example of a tensor.

Given any ring R and any two R modules M,N, one can form their tensor product over R, M(tensor(R))N.

That refers to the tensor product and not the tensor itself. This use of the term "tensor" is not the same use of the term "tensor" in differential geometry.

Pete

mathwonk

Aug19-04, 12:13 PM

Since this line of thought has proved so wildly popular I will push it further:

Why would one take the previous point of view? The answer may be in the fact that many geometric surfaces of interest are not manifolds, i.e. are not actually smooth objects which are locally parametrizable by euclidean space, hence local coordinates are not available.

I.e. in that event, even the familiar tangent and cotangent bundles are actually not products. In fact one definition of a manifold, after defining the intrinsic cotangent bundle, is that the (possibly singular) variety is a manifold if and only if the cotangent bundle is locally a product.

This means one cannot define even tangent and cotangent vectors in the simple intuitive way that has been used for manifolds. I.e. we think usually first of "what is a vector at a point"? Then we think of a field of vectors on an open coordinate set, and finally we introduce changes of coordinates for different but overlapping open sets.

Then for tensors, we use the same procedure, passing from pointwise, to local, to global. I.e. we go back to a point and define a tensor at a point, then take products of that eucldiean tensor space with an open coordinate set, and finally ask how the coefficients or components of the tensor change as we change coordinates.

But what if there is no possibility of introducing local coordinates near a certain "bad" point? i.e. a singular point? such as near the origin of a cone. Then the solution in algebraic and analytic geometry is to do the local version first, (taking account of the fact that "local" does not mean "locally trivial"), and after that the pointwise version. And tensor products play a crucial role, even in the definiton of vectors and covectors.

Here is what comes out:
From a certain strange point of view, the tangent bundle to a manifold X, is the same as the normal bundle to the diagonal in the product XxX. (This is familiar in differential topology, where the euler characteristic of a manifold is defined sometimes as the self intersection of the diagonal, i.e. via Hopf's theorem, as the number of zeroes of a general vector field.) Thus the conormal bundle to the diagonal is the cotangent bundle to X.

Now the advantage of this point of view, since we have not defined cotangents or tangents yet, is that conormal bundles are more basic than tangent bundles! I.e. the conormal bundle to the diagonal in XxX, is just the family of functions vanishing on the diagonal, modulo those vanishing twice!

To see what this has to do with derivatives, note the usual difference quotient defining a derivative, i.e. [f(x) - f(a)]/(x-a). See there, that denominator is a function of two variables, a and x, hence a function on the product XxX. Moreover note that when x=a, the numerator is zero, so "deltaf" is a function on the product XxX which vanishes on the diagonal.

However as a derivative, or a differential, df is not consiered zero unless it vanishes twice on the diagonal, i.e. unless after dividing out by the first order zero, i.e. by x-a, we still get zero. Now in algebra we just divide by x-a straightaway, and we can define the derivative of f at a, as the value of the actual algebraic quotient

[f(x)-f(a)]/(x-a), at a. That is how fermat and descartes took derivatives, or found tangents. But in analysis we must take a limit to evaluate this, the usual Newton definition of the derivative.

So to sum up, the cotangent bundle of X is by definition, locally the quotient of the ideal I of functions on XxX which vanish on the diagonal, modulo I^2, those vanishing twice.

If we let C(X) be the ring of functions on X, then it turns out that the ring of functions on XxX is locally C(X)[tensor]C(X), and I is the kernel of the multiplication map C(X)[tensor]C(X)-->C(X). Then the cotangent bundle of X is locally I/I^2. This ism true even at singular, i.e. non manifold, points.

Now this is all more or less true in algebraic and analytic geometry (plus or minus my inherent inaccuracy and ignorance), but I have not checked it for C infinity functions, as my notation suggests here. maybe Hurkyl would be interested in trying this out along with his investigation of general schemes and their smooth analogues.

To pass back to the point wise situation, one defines the pointwise cotangent space as the (pointwise) localization of the module I/I^2, at the point p, and this is done, guess what? by tensoring I/I^2 with the field of coinstants at the point.

I.e. T*(p) = (I/I^2)[tensor(C(X))]R, where R say is the field of real numbers, (and the tacitly assumed homomorphism from C(X) to R, is just evaluation at p).

So tensors have a huge variety of uses.

peace,

I hope this does not kill the interest of this thread for good. Just ignore what I said here if you like.

pmb_phy

Aug19-04, 01:13 PM

Why would one take the previous point of view? The answer may be in the fact that many geometric surfaces of interest are not manifolds, i.e. are not actually smooth objects which are locally parametrizable by euclidean space, hence local coordinates are not available.

Please clarify. Why do you think non-smooth surfaces are necessarily not locally parametrizable?

I.e. in that event, even the familiar tangent and cotangent bundles are actually not products.

I don't see that those bundles are even defined for anything but a manifold.
[qoute]
In fact one definition of a manifold, after defining the intrinsic cotangent bundle, is that the (possibly singular) variety is a manifold if and only if the cotangent bundle is locally a product.
[/quote]
Sorry dude but you lost me.

Pete

mathwonk

Aug19-04, 01:36 PM

The definition of the cotangent bundle of a not necessarily amnifold, is as I gave it above:

" If we let C(X) be the ring of functions on X, then it turns out that the ring of functions on XxX is locally C(X)[tensor]C(X), and I is the kernel of the multiplication map C(X)[tensor]C(X)-->C(X).

Then the cotangent bundle of X is locally I/I^2. This is true even at singular, i.e. non manifold, points."

As a simple example of trying to define tangent spaces geometrically for a non manifold, consider a cone, the zero set of x^2 + y^2 - z^2 = 0, and suppose that is our space.

then near the origin it is not locally like any open set in euclidean space, rather it looks like the union of two discs with their centers identified.

So in this simple case we could locally parametrize each disc separately with the proviso that the two centers are the same point.

But there are much much more complicated non manifolds imaginable, such the common zeroes of any polynomial in several variables at all.

Even in the simple case of the cone above we must decide what we mean by a vector tangent to the cone at the origin.

Now if tangency means order of contact more than one, then every line though the origin is tangent to the cone at the origin, since the equation of the cone has second order vanishing when restricted to any such line, because it begins with terms of order 2.

But such vectors are not "tangent" to the cone in the sense that they occur as velocity vectors for any curve in the cone. So the velocity vector definition would give a different concept of tangent vector for a cone than would the order of vanishing definition.

In fact for this cone the order of vanishing definition gives a three dimensional tangent space at the origin of an essentially two dimensional object, the cone. This is unfamiliar from manifold theory.

On the other hand the velocity vector to a curve definition, gives not a vector space at all, but another copy of the cone itself, a two dimensional object but not a linear object. So which one to use?

I.e. for a manifold essentially all different definitons one can think of for tangent vectors or cotangents vectors agree, but not for non manifolds.

When one uses the "order of contact" definition of a tangent space them one gets this phenomenon that the dimension of the tangent space jumps up at a non manifold point, and that causes the family or bundle of tangents not to be a product near there.

When one uses the velocity vector definition, one gets a non linear object at a non manifold point and that again causes the tangent spaces not to be locally trivial near there. There is no way out, if we stick to the desire to have a locally trivial family of objects, i.e. one parametrizable by an open subset of euclidean space.

So we use both concepts, order of contact definition gives the best linear approximation to our space, but possibly of alrger dimension. The velocity vector definition gives the right dimension, and a good approximation to the space, but possibly not linear.

Then a point is manifold point if these two agree, i.e. if the linear object has the right dimension, or equivalently if the correct dimensional object is linear.

The definition above, i.e. I/I^2, is the bundle dual to the order of contact definition of tangent bundle, because locally it gives m/m^2, the maximal ideal of functions vanishing at the point, modulo those vanishing twice.

Note that at the origin of the plane, this gives the vector space of linear polynomials, i.e. the space of all polynomials beginnig with terms of degree at least one, modulo those beginning with therms of degre at least two, i.e. linear polyonimals, i.e. the cotangent space at the origin.

sal

Aug19-04, 02:37 PM

I.e. "tensor" is a verb, not a noun.
That would be a re-definition/different use of the term "tensor"
Actually, in both colloquial and technical English, verbing a noun is perfectly acceptable.

"I wolfed down dinner and rushed out to shoe my horse so I could cart some things I'd already boxed up into town, but I was caught speeding and they booked me for it."

Nouning a verb is possible too but most verbs were nouned so long ago that it's hard to find recognizable examples. "Going for a run", "joining a sing around the campfire", "making a good throw" come to mind.

For an example of a noun which was verbed and then subsequently adjectived, consider "fish": I see a fish. I fish for it... I install a phone wire. I need to fish it through the wall. I do a neat, fast job of fishing the wire, and my boss says, "That was a good fish job".

In conclusion, if tensor is a noun, then it's surely a verb too, and conversely.

mathwonk

Aug19-04, 02:42 PM

to be a little more precise, I/I^2 is the local C(X) module of covector fields, rather than the "bundle" of cotangent spaces.

The covector fields are the reason for defining the cotangent spaces in the first place, so the fields are more important than the spaces, but if one wants to recover the actual points in the family of tangent spaces say, there is an algebraic way to do this. As a set of points, I believe (over the complex numbers) the tangent "bundle" would be (locally) the maximal ideal spectrum of the symmetric tensor algebra on the C(X) module I/I^2, whatever that means.

mathwonk

Aug19-04, 02:45 PM

can we make adjectives too? like "tensor than thou?"

or at least "tensor than necessary?"

Sometimes, to paraphrase Ghostbusters, I feel like "I've been tensored!"

pmb_phy

Aug19-04, 03:51 PM

Actually, in both colloquial and technical English, verbing a noun is perfectly acceptable. And when you do so you change the meaning of the word.

mathwonk - interesting stuff but alot of it I don't follow. The language I either don't recognize or have forgotten from non-use. When I'm allowed to sit here for extended times then we'll have to chat more.

Glad to have you aboard.

Pete

mathwonk

Aug19-04, 06:56 PM

Thank you Pete! it is very friendly and fun here.

I want to see if I understand enough of the mumbo jumbo i was parroting to give an example.

Lets try to reproduce the algebraic differential one forms, i.e. (0,1) tensors?, on the line (dual vector fields, things you integrate over paths).

The ring of algebraic functions on the line is C[t], polynomials in one variable. The ring of functions on the plane is C[x,y] which happens to be the tensor product as algebras, of the two polynomials rings C[x], and C[Y].

Then the diagonal embedding of the line into the plane, and restriction of polynomials from the plane, to the diagonal, correponds to the map C(X)tensorC(Y) = C[X,Y]-->C[t] taking f(X,Y) to f(t,t), I guess, what else? Then the kernel I, of this map contains things like f(x)-f(Y), which in tensor notation would be represented by
f(x)(tensor)1 - 1(tensor)f(Y), but so what.

Notice this object f(x)-f(Y), looks a lot like "deltaf", the numerator of a derivative.

I.e. this is a function vanishing on the diagonal. Now to get "df" out of this, we just consider it as an element of the quotient object I/I^2, i.e. just decree that such a thing is zero if it is a product of two such things. Now this is a little esoteric, but I beg to be given the benefit of the doubt since a derivative is indeed the second order value of a function, so it is zero if the function vanishes "doubly".

To prove algebraically that this gadget is what it should be, we define a more plebeian version, by simply taking all symbols df, for all f in C[t], then we take all linear combinations of products of form g df, for various g's and f's, and we call such a thing a differential one form.

But of course we have to have some relations, so we mod out by (i.e. consider to be zero) all such linear combinations of form d(f+g) - df - dg, and d(cf) - cdf, and d(fg) - fdg - gdf.

Then we really do have the space of differential one forms on the line. i.e. sums of thignjs like g df, with the usual relations. I suppose also we can show in this case that they all can be written uniquely as actually just g(x)dx, for some g. For more general, uglier spaces, especially non manifolds, this is not true.

Now I claim on good authority that I/I^2 is isomorphic to this module of differential one forms. To show it we have to have a map between them and show the map is an isomorphism.

Well just send df in the module of one forms, to f(X)-f(Y), in I, i.e. to delta f, or rather go ahead and send it further to the equivalence class of f(X)-f(Y) in I/I^2, i.e. send df to "df"!

this defines a map from the differential one forms to the space I/I^2.

It can be shown by someone with better algebra skills than mine that this is an isomorphism of C[t] modules.

well that was pretty wimpy, but i claim it is a sketch of an example of showing that I/I^2, in the case of the line embedded as the diagonal of the plane, really gives the expressions of form: summation gj dfj, i.e. fields of covariant tensors, i.e. differential one forms.

Now on objects that are not manifolds, i.e. that have singularities, these modules of fields are not locally trivial, hence are not sections of local product bundles. I suppose you can still take higher order tensor powers of them but to me it becomes a little hard to understand what you are getting.

You can see I run out of gas pretty quick after basic rank one tensors.

best,

roy

Hurkyl

Aug19-04, 08:51 PM

I didn't understand it until I looked at a nice simple case.

On S, the set of real valued functions on R^n differentiable around the origin, we have the operator d(.)(0) that takes a function to a cotangent vector at the origin.

Additive constants don't matter, so we can strip off the zeroeth order terms, leaving us with the subset I of functions in S that are zero at the origin.

(Note that I is an ideal of the ring S!)

Two functions of I evaluate to the same vector under d(.)(0) if and only if they have the same linear terms. In other words, if their difference consists only of second order terms and higher.

This set is precisely I^2: the set of sums of things of the form p q where p and q are in I.

For example,

f(x, y) = ax^2 + bxy + cy^2 + ...
= x (ax + by + ...) + y (cy + ...)

So, if we take the set, I, of all functions in S zero at the origin, and we mod out by I^2, the set of everything with a double zero at the origin, then d is a bijection between I/I^2 with R*^n.

As an example, let's take S to be the set of all polynomials in x and y.
Let f(x, y) = x^3 + 3xy^2 + 7xy - 3x - 7y

f is an element of I, since f(0, 0) = 0
Now, df = [3x^2 + 3y^2 + 7y - 3] dx + [6xy + 7x - 7] dy
so df(0, 0) = -3 dx - 7 dy = <-3, -7>

Also, consider g(x, y) = -3x - 7y. Then dg(0, 0) = <-3, -7> also.

Now, take (f-g)(x, y) = x^3 + 3xy^2 + 7xy. This can be written as
(f-g)(x, y) = x (x^2 + 3y^2 + y)

So, f - g is a product of two functions in I (and thus a sum of things that are a product of two functions in I), thus f - g is in I^2.

This confirms the earlier observation that if two functions have the same image under d(.)(0), then they differ by something in I^2.

So, the result is that the map, d(.)(0) : I/I^2 --> R*^n : f --> df(0) is an isomorphism!

This means, considering the vector space I/I^2 is just as good as considering R*^n of traditional differential forms.

I/I^2 has an advantage of being a purely algebraic construction, thus it can be used to define "differential forms" on things where we can't ordinarily talk about differentiation.

Hurkyl

Aug19-04, 11:35 PM

Wee, writing that up has helped me understand more.

Setting S to be some nice space of functions, like real functions analytic at the origin, we have:

S corresponds to all functions analytic at the origin. In other words, it consists of all functions that are given by a power series about the origin.

I is the ideal of all functions zero at the origin. It is all functions of S given by power series with no constant terms.

I^2 is the ideal of things that are sums of things of the form i*j where i and j are both in I. In this case, it is all power series without any constant or linear terms.

...

I^n is the ideal of things that are sums of things of the form i1*i2*...*in where all of the i_m are in I. It is all power series with terms only of degree n or more.

Now, because I is the set of all power series with no constant terms, if we mod out by I, we eliminate all terms with degree 1 or more. In particular,
f = g (mod I)
iff f(0) = g(0).

Similarly, because I^2 is the set of all power series with no constant or linear terms, if we mod out by I^2 we eliminate all terms with degree 2 or more. Thus,
f = g (mod I^2)
iff f(0) = g(0) and f'(0) = g'(0)

And so on.

In particular, if we just the ideal I^n and we mod out by I^(n+1), we're left with terms of degree exactly n: terms of lesser degree don't exist in I^n, and terms of greater degree are in I^(n+1) and thus equivalent to zero.

(Here, I'm setting I^0 = S)

So, if we interpret I^n / I^(n+1) as a vector space over R, then we get a nice thing. S/I is isomorphic R. I/I^2 is isomorphic to the space of linear forms. I^2/I^3 is isomorphic to the space of all (homogenous) quadratic forms. I^3/I^4 to cubic forms, et cetera.

As an example of using these for fun and profit, let's compute the Maclaurin series for 1/(2+x):

In the case where S is simply analytic functions in x, we have that I = (x); the set of all multiples of x. (such as x e^x), I^2 = (x^2), ..., I^n = (x^n)

The constant term lives in (a space isomorphic to) S/I. So, we have:
1/(2+x) = f(x) (mod x)
1 = f(x) (2+x) (mod x)
1 = 2 f(x) (mod x)
1/2 = f(x) (mod x)

So, the constant term is 1/2. (notice that 1/2 + x, or 1/2 + x e^x, or anything similar is fine; we are only finding the constant term, we don't care about the higher order terms)

So we've chosen 1/2 as the constant term, we can find the linear term. Actually, for funsies, let's use 1/2 + x as the constant term and see what happens:

1/(2+x) = [1/2 + x] + f(x) (mod x^2)
1 = [1/2 + x](2+x) + f(x)(2+x) (mod x^2)
1 = 1 + (5/2)x + f(x) (2 + x) (mod x^2)
-(5/2)x = f(x) (2 + x) (mod x^2)
We can see that f(x) = -(5/4)x satisfies this equation. (As expected, f(x) is an element of I) So, if we add our constant and linear terms, we get
1/(2+x) = [1/2 + x] - (5/4)x (mod x^2)
= 1/2 - (1/4)x (mod x^2)

Which is exactly what we expect the first two terms to be.

We can apply everyone's favorite algebraic trick to this too:

(2+x) (1/2 - x/4 + x^2/8 - x^3/16 + x^4/32)
= 1 - x/2 + x^2/4 - x^3 / 8 + x^4 / 16
+ x/2 - x^2/4 + x^3/8 - x^4/16 + x^5/32
= 1 + x^5/32
= 1 (mod x^5)

Giving us that 1/(2+x) = 1/2 - x/4 + x^2/8 - x^3/16 + x^4/32 (mod x^5)

As Mathwonk was pointing out, the key is that none of this involves any analysis whatsoever. We don't need derivatives, or even a topology!

As an example, let's take S to be the set of all functions continuous at the origin.

Then I is the set of all functions continuous and zero at the origin.

Then I^2 is... *drumroll*... the set of all functions continuous and zero at the origin!

Proof: let f(x) be continuous and zero at the origin. Take g(x) = |f(x)|^(1/2) and h(x) = sign(f(x)) g(x)

It's easy to see that both g and h are continuous at the origin, and that f = g*h. Therefore, any continuous function zero at the origin is also a product of two continuous functions zero at the origin, so that I^2 = I.

So, when we try to take a look at all linear forms, by looking at the space I/I^2, we find that the only linear form is zero!

And this is exactly what it should be, since our knowledge of ugly continuous functions tells us that the only reliable approximations of continuous functions in general are their evaluations!

mathwonk

Aug21-04, 10:24 PM

Hi, I'm back after the first day of school. Fortunately the other kids liked me enough not to take my lunch money.

Thus I am emboldened again to define "A TENSOR". I notice some dork with my same handle has maintained there is no such thing as a tensor, since "to tensor" is a verb.

But to paraphrase Bill Murray again, "I have been tensored therefore I am a tensor".

I.e. one can perhaps accept both uses of the word, properly restricted.

Thus:

Basic object: manifold X with a differentiable structure.

derived structure: tangent bundle T= T(X),
(family of tangent spaces Tp, at points p of X)

second derived structure: cotangent bundle T*
(family of dual tangent spaces T*p).

Operation: tensor product of bundles, yielding new bundles:

T(tensor)T(tensor)T(.....)T(tensor)T*(tensor)T*(.. ...)T*.

with r factors of T and s factors of T*.

Then a section of this bundle (drumroll), i.e. a function with domain X and value at p an element of Tp(tensor)Tp(....)T*p(tensor)T*p(....)T*p,

is called a tensor of type (r,s).

how are them peaches?

Hurkyl

Sep17-04, 06:24 PM

Notice this object f(x)-f(Y), looks a lot like "deltaf", the numerator of a derivative.

I checked out Hartshorne again to look at this stuff again; it introduces a construction of relative differential forms just as you have here... but it is missing this important sentence which explains what's happening. :smile:

I have a question though; I'm happy enough with polynomial rings, because we can just lump all of the generators together as you describe, but I want to make sure I have it right in the general case, since I can't find a definition of the tensor product of algebras anywhere.

When B is an algebra over A, I (mostly) understand the B-module BxB (where I'm using x for the tensor product over A)... to make it into an algebra, do we just define fxg * pxq as (fp)x(gq)?

mathwonk

Sep18-04, 07:04 AM

sure, why not? consult e.g.:

zariski samuel, p.179,
atiyah macdonald page 30,
lang, algebra, second edition, pages 576, 582.

also notes from my 1997 course math 845:
(where my fonts did not reproduce well.)

Categorical Sums of Commutative Rings and Algebras
As an extension of the ideas of the section above on base change, consider what happens if both modules in a tensor product are rings, hence R-algebras, rather than just R-modules. Let S, T be R-algebras, i.e. let ring maps ƒ:R-->S, ¥:R-->T be given, and form the R-module S.tens(R).T.

(which denotes the tensor product of the R modules S and T.)

This is both an S-module and a T-module, but we claim it is also a ring, and an R-algebra. The multiplication is the obvious one, i.e. (aºb)(sºt) = asºbt.

(where the little circle denotes the tensor product of two elements.)

Claim: This gives an associative, distributive operation, with identity 1º1. First we check it gives a well defined R-bilinear operation:
The function (SxT)x(SxT)-->StensT, taking ((a,b),(s,t)) to asºbt gives, for each fixed value of (s,t), a bilinear map on SxT, hence induces a linear map (StensT)x{(s,t)}-->StensT. The induced pairing (StensT)x(StensT)-->StensT is also bilinear in the second variable for each fixed element of StensT, hence induces a map (StensT)x(StensT)-->StensT, which is linear in each variable. Hence our propsed multiplication is well defined and R-bilinear.
Since (1º1)(sºt) = sºt, the element 1º1 acts as an identity on a set of generators, hence also everywhere. Similarly, (s0ºt0)(s1s2ºt1t2) = (s0s1s2ºt0t1t2) = (s0s1ºt0t1)(s2ºt2), so the product is associative on generators. Since these expressions are linear in each quantity siºti, associativity holds for all elements.
Since the R-module structures on S,T are by means of the maps ƒ:R-->S, and ¥:R-->T, the following elements of StensT = Stens(R)T are equal: r(xºy) = (rxºy) = (ƒ(r)x)ºy = xº(¥(r)y) = (xºry). Thus there is a unique R-algebra structure on StensT defined by the map R-->StensT, taking r to r(1º1) = 1rº1 = ƒ(r)º1 = 1º¥(r) = 1ºr1. Since for a,b in R, rab(1º1) = ƒ(a)º¥(êb) = (ƒ(a)º1)(1º¥(b)) = (a(1º1))(b(1º1)), and (a+b)(1º1) = a(1º1) + b(1º1), and 1ÿ(1º1), this is indeed a ring map.
Remark: With the understanding given above of the notation, we may write simply rº1 for r(1º1) = ƒ(r)º1 = 1º¥(r).

This simple construction yields a nice conclusion:
Theorem: Any two R-algebras R-->S, R-->T, have a direct sum in the category of R-algebras, namely: Stens(R)T.

etc....

mathwonk

Sep18-04, 07:55 AM

summary of my notes content:
Graduate Algebra, Main results

843: Main idea: Counting principle for a group acting on a set: the order of the group. equals the product of the order of the subgroup fixing a point, times the order of the orbit of that point.
Main theorems:
1) Sylow theorems on existence of p-subgroups of finite groups,
2) simplicity of An,
3) Jordan Holder theorem on existence and uniqueness of set of simple quotients for a finite group,
4) classification theorem: all finite abelian groups are products of cyclic groups.
5) Galois` theorem that a solvable polynomial has a ``solvable`` Galois group (i.e. the Galois group has an abelian normal tower), and an example of a polynomial whose Galois group is A5, hence has no abelian normal tower, thus an example of a polynomial with no solution formula by radicals.

844: Main idea: The Galois group of relative automorphisms of a simple field extension, is determined by the way the minimal polynomial of the generator factors in successive partial extensions.
1) Gauss` theorem that polynomial rings over a ufd are ufds`s,
2) existence of root fields for polynomials,
3) Hilbert`s basis theorem that a polynomial ring over a noetherian ring is noetherian,
4) the theorem of the primitive element (a finite separable extension is simple),
5) the fundamental theorem of Galois theory, (in a finite normal separable field extensions intermediate fields correspond one - one with subgroups of the Galois group, and the order of the Galois group equals the degree of the extension)
6) the converse of Galois` theorem, i.e. (over a field of characteristic zero) a polynomial is solvable if its Galois group has an abelian normal tower,
7) Cardano`s formulas for explicitly solving cubics and quartics using the structure of an abelian normal tower for the Galois group.

845: Main idea: Diagonalizing a matrix.
1) Theorem on existence of decomposition of a finitely generated module over a pid into a product of cyclic modules, and a procedure for finding it over a Euclidean ring (from a presentation).
2) Application to existence and uniqueness of rational canonical form for any matrix over a field, (a special representative for the conjugacy class of an element of a matrix group), [i.e. an endomorphism T of a k vector space V is equivalent to a structure of k[X] module on V, and the rational canonical form of T is equivalent to a decomposition of V as sum of standard cyclic submodules], and
3) of Jordan form for any matrix over a field in which the characteristic polynomial factors into linear factors.
4) Spectral theorems, (sufficient conditions for a matrix to be diagonalizable, especially into mutually ``orthogonal`` components),
5) multilinear algebra including tensor products (construction of a universal bilinear map out of AxB),
6) exterior products, duality, and the formula for the exterior powers of a direct sum.