Math Newb Wants to know what a Tensor is

  • #26
mathwonk
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Ohmigosh, I worked for over an hour on how to picture a covector and a differential form, and the web connection just shut down and lost it all.

Well, that spares you a long post.

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

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

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

This lets us picture at least the level sets of covectors within the vector space it self.
 
  • #27
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Okay I've read this post and the other refering to tensors, and followed the links, and I'm still lost.

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

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

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

Sorry if it seems like I'm asking a lot, but I would really like a deeper insight into GR and cosmology and seem to be getting no where. Anyway I'll leave it there coz I'm feeling rather stupid, right now. :wink:
 
  • #28
robphy
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These aren't definitions... just ways one can think about things. (What I am saying is nothing new... you'll probably find these in the various threads on this topic.)

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

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

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

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

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

Hopefully, this helps get you started. If so, maybe the other posts in these various threads are more digestible.
 
  • #29
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I'm a bit confused now about the commutivity (if such a thing exists) of tensors. I think part of my confusion stems from my lack of familiarity with this summation notation.

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

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

Does [tex] \Lambda^{\mu}_{\ \rho} H^{\rho\sigma}a_{\nu}\Lambda^{\nu}_{\sigma} [/tex] give the same thing as [tex] \Lambda^{\mu}_{\ \rho}\Lambda^{\nu}_{\sigma} H^{\rho\sigma}a_{\nu} [/tex] ?
 
  • #30
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Yes! Because the summation convention requires you multiply elements with the same (dummy-)index (one sub- and one super-scripted) and sum over all allowed values of the dummy index. So it does not matter in what order the tensors appear; simply because you multiply normal numbers for wich ab=ba !

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

Where the last equality comes from the fact that because e.g. [itex]A_0 B^0 = B^0 A_0 [/itex]. For they are just numbers.
 
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  • #31
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Ohhh.... excellent. Thanks for the explanation, it's all starting to make sense now. :smile:

EDIT:

Oops, I had just one more question.

What the heck is [tex]T^{\mu\nu} [/tex]? How exactly is that different from [tex]T^\mu_{\ \nu}[/tex]? Are they basically the same thing or are we talking about differences in terms of contravariant and covariant?
 
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  • #32
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These are different, and indeed the difference is that the latter has a co- and a contra-variant component while the firts has two contravariant components. You can relate the two by the metric: [itex]T^a_b=g_{bc} T^{ac}[/itex] (This is called 'lowering an index'). In Euclidean space these are the same because the metric is the identity matrix, but in a general Riemannian space the components can be different.
 
  • #33
mathwonk
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chroot, are you actually a cigar?
 
  • #34
gvk
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chroot said:
I'm not too clear on the geometrical meaning myself, though I can go through the motions and manipulate tensor expressions just fine.

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

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

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

- Warren
First of all, covariant and contravariant vectors are not different vectors. They represent ONE VECTOR (an arrow :-) in two different coordinate systems (dual, or reciprocal, or skew, or...coordinates). The reciprocal system is equally satisfactory for representing vectors, but 'contravariant' vector looks exactly the same as 'covariant'. So "visualize" them as ONE tangent arrow (toothpick) if you wish. Two paralell blades, probably, mean direct and reciprocal coordinate planes, which may have complement scale or orientation, but, of course, should be parallel (no less no more).
Secondly, any quantity that we wish to define, be it scalar, vector, or tensor, must be independent of the special coordinate system. We shell adopt this as fundamental principal. However, its representation will depend on the particular system.
 
  • #35
chroot
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The difference between a vector and a dual vector is not just a change of coordinate system, gvk. Vectors and dual vectors live in entirely different spaces, and are certainly not the same vector.

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

- Warren
 
  • #36
jcsd
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gvk said:
First of all, covariant and contravariant vectors are not different vectors. They represent ONE VECTOR (an arrow :-) in two different coordinate systems (dual, or reciprocal, or skew, or...coordinates). The reciprocal system is equally satisfactory for representing vectors, but 'contravariant' vector looks exactly the same as 'covariant'. So "visualize" them as ONE tangent arrow (toothpick) if you wish. Two paralell blades, probably, mean direct and reciprocal coordinate planes, which may have complement scale or orientation, but, of course, should be parallel (no less no more).
Secondly, any quantity that we wish to define, be it scalar, vector, or tensor, must be independent of the special coordinate system. We shell adopt this as fundamental principal. However, its representation will depend on the particular system.
I think what may of confused you is that for a Euclidean vector space the contravariant and covariant vectors are the 'same' (i.e. in any given frame a pair of dual vectors have the same compoents) as the compoents of the metric are simply the compoents of an identity matrix.

But in general a vector and it's one-form belong to different spaces, infact the dual space of some linear vector space S is the set of all linear functions that map a vector in S to some (in genral complex) number and it also constitutes a linear vector space in it's own right. Further the components of a pair of dual vectors Aν and Aν in the same frame are not in general the same (also belonging to differnet spaces they have different bases).
 
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  • #37
mathwonk
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a "vector" is the derivative of a curve, i.e. of a map from R^1 to R^n, whereas a covector (dual vector) is the opposite, the gradient of a function from R^n to R^1. (at one point)

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

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


(Unfortunately covectors are only covariant in differential geometry, they are contravariant in category theory, algebraic topology, and the rest of mathematics.)
 
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  • #38
gvk
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chroot said:
The difference between a vector and a dual vector is not just a change of coordinate system, gvk. Vectors and dual vectors live in entirely different spaces, and are certainly not the same vector.

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

- Warren
No, not only this I'm trying to say, Warren.
You completely forgot who asked "what a tensor is?" and what is the level of this person. He/she is a student of High School! If you will teach them in such manner, they never understand anything about vector and tensor calculus at all, and will never ask you again.
It is the same as you start to teach them QM before they learn anything about classical mechanics. But educational sequence in math plays even more important role than in natural sciences. Here everything should be sequential and well understood in elementary manner.
So, they first need understand that "a vector is an arrow or column", and "a tensor is a matrix" and what are the properties of these notions in Euclidean space. This is most important, because they may use the tensors in the future just to caclulate stress and deformations in materials. :smile:
 
  • #39
mathwonk
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i cannot agree with you less, gvk. this high school, student is also a student of multivariable calculus at UCLA. As such I think he is capable of understanding what things mean, not just the symbols used for them.
 
  • #40
chroot
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gvk said:
You completely forgot who asked "what a tensor is?" and what is the level of this person.
You were replying to a question of mine, not the original poster's.
He/she is a student of High School! If you will teach them in such manner, they never understand anything about vector and tensor calculus at all, and will never ask you again.
Purposefully simplifying material for pedagogical purposes does not give you license to say things that are patently false. You could say "vectors and dual vectors are closely related, and can be thought of as different "representations" of the same fundamental thing -- that would be fine. Saying that vectors and dual vectors are related through a coordinate transformation, on the other hand, is quite wrong, and would probably do more harm to the student's understanding than good.

- Warren
 
  • #41
gvk
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chroot said:
You were replying to a question of mine, not the original poster's.
Purposefully simplifying material for pedagogical purposes does not give you license to say things that are patently false. You could say "vectors and dual vectors are closely related, and can be thought of as different "representations" of the same fundamental thing -- that would be fine. Saying that vectors and dual vectors are related through a coordinate transformation, on the other hand, is quite wrong, and would probably do more harm to the student's understanding than good.
- Warren

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


And I did not tell that "vectors and dual vectors are related through a coordinate transformation". This is your interpretation.
Read carefully: "covariant and contravariant vectors are not different vectors. They represent ONE VECTOR (an arrow :-) in two different coordinate systems". I meant, of course, euclidean case.
Is that 'patently false'?
 
  • #42
gvk
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mathwonk said:
i cannot agree with you less, gvk. this high school, student is also a student of multivariable calculus at UCLA. As such I think he is capable of understanding what things mean, not just the symbols used for them.
What are you talking about?
I have a son in High School of MN, and have 30 yr. experience of teaching (in other country), and can tell you that US school's education in math is about 1.5-2,5 years behind the Europe's. Moreover, it's 1yr. behind what I learned at the same age as son many many years ago.
At the age 14 we knew all elementary and trigonometric functions, algebra (linear and quadratic equations), geometry on the plane and much much more. Now, in 10th grade, they start to learn so called 'integrated math' which has only 1/4 what i mentioned.
How do you expect they will to know the term "rank" of matrix or "homogeneous" transformation at the end of HS?
I guess, it is impossible with this level of math education, even a guy attended multivariable calculus at UCLA, he still missed a lots of math.
 
  • #43
chroot
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gvk,

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

- Warren
 
  • #44
mathwonk
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You do not seem to realize, gvk, that not all high schools in the US are the same low level. You are of course correct that the average high school and even most high schools, are at a low level, but there are some high schools which are different. And even at average high schools there are students who are different.

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

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

Your comments may be correct for a generic case but do not seem to apply to the student we are discussing.
 
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  • #45
gvk
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Could you please give me a little bit more detail about those high schools (city, state). I never heard about such unique cases here.
Sorry for digressions in posts. When I read a reply from StonedPanda and it sounded to me as a discouragment:
"I definitely picked the wrong name for this forum!"
 
  • #46
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So then, I begin to realize that the low level is a worldwide problem not only in Spain happens. Well, this is a very poor consolation, but... still is
 
  • #47
gvk
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chroot said:
A vector and its dual do not even exist in the same vector space, even when the spaces are Euclidean.
- Warren
Euclidean vector space with metric is coincided with its dual, so it is the same, R = R*.
The most students learn from elementary examples and only later move to the general definitions. It is not "my pedagogy", it's a natural way of learning. (I am sure you were on the same way too. By the way, in what age and where did you learn about dual space and covarience? Highschool or college?). However, in more general cases, you are right.
 
  • #48
chroot
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gvk said:
Euclidean vector space with metric is coincided with its dual, so it is the same, R = R*.
The most students learn from elementary examples and only later move to the general definitions. It is not "my pedagogy", it's a natural way of learning. (I am sure you were on the same way too. By the way, in what age and where did you learn about dual space and covarience? Highschool or college?). However, in more general cases, you are right.
I continue to disagree. A vector space and its dual space are not the same, even when the two spaces are both Euclidean. Just because two objects have the same components, for example, does not mean they are really the same. You still need to use the metric to convert between them -- it just happens that the metric is the identity matrix.

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

- Warren
 
  • #49
mathwonk
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gvk, try to understand what chroot is saying. there is big difference between saying there is an isomorphism between two spaces and saying they are the same.

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

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

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

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

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

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

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


best wishes,

PS: Some good high schools in US include the Paideia School in Atlanta, Andover and Exeter in New Hampshire I believe, Brooklyn technical high school and Bronx High shool of science in New York. Other students at average high schools attend university while still in hogh school. My older son e.g. easily exceled in calculus classes at Georgia Tech before being accepted to both Stanford and Harvard. I believe the high school student who was the top performer in my university level algebra class was from Gainesville high in Georgia. Westminster in Atlanta is also very strong, and there are many others. New Trier high school in Winnetka? Illinois is very famous.
 
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  • #50
mathwonk
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here is an example of when two different vector spaces can be regarded as almost the same: let V be any vector space over the real scalars, and V* = Hom(V,R) = its dual space, the space of all linear functions from V to R. Then let V** = (V*)*

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

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


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

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

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

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

confusing isn't it?

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

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

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

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

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

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


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

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