What is a Tensor? Definition & Applications

[mathwonk]

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]

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.


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]

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]

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.

 

[mathwonk]

Interestingly, I did not receive a single request for copies of my notes on algebra.

 

[sal]

Interestingly, I did not receive a single request for copies of my notes on algebra.

I suspect the tiny handful of group members who have a clue what you've been talking about are already up to their eyebrows in algebra texts.

Are your notes online somewhere? I'm not sure I could follow them, but I'd be interested in seeing what you said about exterior products, at the least.

 

[mathwonk]

you encourage me to make my notes web ready!

 

[fizixx]

dang

No. None of that has anything to do with tensors. At least not according to any definition that I've seen. For a definition see

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

Pete

pmb_phy...

Ummm...look at the original post. The user asked for a basic definition of tensors, not some horrid conflagration such as that on that website. I thought mathwonk did a fine job of getting the idea off to a good start...and I thought a 'good job' to mathwonk was in order. He/she tried and even qualified his/her own offering. You tried too, but instead of posting to the original question you posted to mathwonk, and went overboard in my opinion. I certainly respect your post tho. You probably know far more about tensors than I do, but I know an appropriate post when I see one.

Kindly, and respectfully...
fiz~

 

[fizixx]

good explanations

mathwonk...I enjoy your explanations. I know so little about this topic. You're humble, but you are confident in what you know, and you have good arguments to back up what you say.

pmb... hope your back gets better. You have some good thoughts as well.