What is a Tensor? Definition & Applications

[sal]

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Synge and Schild is something of a classic though it's now considered rather out of date.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

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]

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. dx^k is a differential but dx^k is not a differential. It is the gradient of a coordinate making it a basis 1-form.

Pete

 

[mathwonk]

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.

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

best regards,

roy

 

[sal]

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'm sure you are correct about Einstein's willingness to use the modern forms. I wish I knew when the modern machinery was invented -- certainly, the notion that a sensible definition of a vector could be something like

\sum_a v^a \frac{\partial}{\partial x^a}

is not something that would ever have occurred to me, and I can't believe it's something that's just been floating around forever. It seems like it must have been an invention (or discovery) that happened at some particular moment, probably in the last 7 decades -- but when? I have seen no mention of its origin in any textbook.

Until that was invented, as far as I know there was no rigorous basis for tensor calculus -- the infinitesimals were all they had, and that's why they were used so heavily.

 

[Hurkyl]

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]

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]

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]

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]

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 equation 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]

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]

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.

Best wishes

Pete

 

[robphy]

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.

By chance, are you talking about "jets"?

 

[mathwonk]

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]

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]

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

 

[Feynman]

A Tensor is a multilinear continus form like the scalar product

 

[sal]

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]

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]

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.

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]

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]

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]

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]

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]

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]

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]

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 a lot 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]

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]

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]

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!