Tensor Product: Basis for Higher Order Tensors and Its Proof

In summary, the tensor product of two tensors of lower rank forms the basis for ANY tensor of higher order. This can be proven by looking at the transformation properties of tensors and using the kronecker delta. There are 16 components in total for a tensor of rank (0,2) because there are 4 choices for the basis vector in each slot, giving a total of 4x4=16 components. However, understanding this concept and the kronecker delta may be difficult for high schoolers studying tensor analysis.
  • #1
Terilien
140
0
How can we prove that the tensor product between two tensors of lower rank forms the basis for ANY tensor of higher order? also WHY is it it true?

ANY TENSOR of higher order.
 
Last edited:
Physics news on Phys.org
  • #2
The tensor product of two tensors is a tensor of higher order. What do you mean by "forms the basis"?
 
  • #3
most physics books say that the tensor product between two tensors is the most general higher order tensor. how can we prove this?
 
  • #4
Terilien said:
most physics books say that the tensor product between two tensors is the most general higher order tensor. how can we prove this?
I find that an odd saying. Could you please reference several physics texts which say this so that the likelyhood that I'll have one of them will be good? Thanks

Best wishes

Pete
 
  • #5
pmb_phy said:
I find that an odd saying. Could you please reference several physics texts which say this so that the likelyhood that I'll have one of them will be good? Thanks

Best wishes

Pete

a first course in general relativity.
 
  • #6
Terilien said:
a first course in general relativity.
That one I have. What page should I turn to?

Pete
 
  • #7
It's on page 71.
 
  • #8
Terilien said:
most physics books say that the tensor product between two tensors is the most general higher order tensor. how can we prove this?

If you look at page 71 you'll see that Schutz says quite explicitly

The most general [itex](0,2)[/itex] tensor is not a simple outer product, but it can always be represented as a sum of such tensors.

This contradicts what you've said above. In fact, Schutz goes to great lengths to explain how and why the most general (0,2) tensor must be a sum of tensor product terms. What, specifically, is the difficulty that you're having?
 
  • #9
Ah that was just a simple misunderstanding. however there is another section i have a hard time with. It's on page 70. It's the part where they talk about the absis of the gradient one form. i don't quite understand what's being done. could you guide me through it step by step?

I'm finding tensor analysis to be quite difficult actually. Is this normal for high schoolers studying the subject?

to be more precise:

I'm having a hard time with the section on page 70 where he talks about basis one forms for the gradient vector.

To be more precise here's a quote" note that the index aappears as a supercript in the denominator and as a subscript on the right hand side. As we have seen this is consistent eith the transformation properties of the expression.

In particular we have:

Then he introduces a symbol that I don't understand at all. I don't understand the conclusion after that.

My other problem on page 71 deals with the fact that he says that "since each index has four values there are 16 components". could you explain that in more detail?
 
Last edited:
  • #10
Terilien said:
Ah that was just a simple misunderstanding. however there is another section i have a hard time with. It's on page 70. It's the part where they talk about the absis of the gradient one form. i don't quite understand what's being done. could you guide me through it step by step?

I'm finding tensor analysis to be quite difficult actually. Is this normal for high schoolers studying the subject?
I wouldn't worry if you're in high school and finding tensor analysis quite difficult!

to be more precise:

I'm having a hard time with the section on page 70 where he talks about basis one forms for the gradient vector.

To be more precise here's a quote" note that the index aappears as a supercript in the denominator and as a subscript on the right hand side. As we have seen this is consistent eith the transformation properties of the expression.

In particular we have:

Then he introduces a symbol that I don't understand at all.
He introduces [tex]x^{\alpha}_{, \beta}\equiv \delta^{\alpha}_{\beta}[/tex] [sorry, I don't know how to offset the indices like in the text]. Note he is using , to denote partial derivative, so the LHS is [tex]\frac{\partial x^{\alpha}}{\partial x^{\beta}}[/tex] Do you know what the kronecker delta is?
I don't understand the conclusion after that.
The conclusion comes from looking at (3.12). Do you understand this equation?

My other problem on page 71 deals with the fact that he says that "since each index has four values there are 16 components". could you explain that in more detail?

alpha and beta can take the values {0,1,2,3}, and so the tensor [itex]f_{\alpha \beta}[/itex] has components f_00, f_01, f_02, ..., f_33. There are 16 in total.
 
  • #11
i'm not sure if i understand 3.12 properly. I'm not too well versed in the kronecker delta. I will give it a shot though.

I think it means that the output can only equal the correspong components multiplied together IF the basis one form applied to the basis vectors equal some identity map?

i'm not sure. I've always found the kronecker delta somehwat confusing.

Do the sixteen componets simply follow from linearity? sorry it's kind of weird to me. I know that the componets are the outputs for every basis vector, but still. How does one expand the mapping in such a way that we can show that there are 16 components.

i know that if you apply a one form on a basis vector you get the sum of a0b0.

Sorry about all this I'm just very eager, and i don't want to ruin it by misinterpreting anything.
 
Last edited:
  • #12
The kronecker delta is defined as (im switching to latin indices, as they're easier to type!) [tex]\delta^a_b=\left\{\begin{array}{cc}1,&\mbox{ if } a=b \\1, & \mbox{ if } a\neq b \end{array}\right\}[/tex], so, (3.12) simply says that a basis oneform [itex]\tilde{\omega}^a[/itex] acting on a basis vector [itex]\vec{e}_b[/itex] is 0 if [itex]a\neq b[/itex] and is 1 if a=b. Comparing this with [itex]x^{\alpha}_{, \beta}\equiv \delta^{\alpha}_{\beta}[/itex] gives us the result that the basis one form is [itex]\tilde{d}x^a[/itex].

Now, to explain the 16 components. Let f be the arbitrary tensor. The components [itex]f_ab[/itex] of this tensor are obtained by contracting f with the basis vector ea in the first slot, and eb in the second slot. We have 4 choices for each basis vector, and so there are 4x4=16 components (since for each choice of basis vector in slot one, we have 4 choices for the basis vector in slot 2).
 
  • #13
the sixteen component thing seems so obvious now. so what your saying is that the most general possible 0,2 tensor has sixteen components? That makes sense.

however i don't understand the significance of the kronecker delta thing. that well at least. Oh wait.

I'm still having a hard time with the kronecker delta thing and 3.12 and the gradient thing. what does [itex]x^{\alpha}_{, \beta}\equiv \delta^{\alpha}_{\beta}[/itex] mean again? how do we know that it equals the kronecker delta? I'm not clear on its significance.

Sorry I'm not used to learning this way.
 
Last edited:
  • #14
It might be helpful to consider the following when trying to understand the Kronecker delta. Suppose that you have some system of coordinates [itex]x^a[/itex], where [itex]a=0,1,2,\ldots,m[/itex]. Now consider the partial derivative

[tex]\frac{\partial x^a}{\partial x^b} = x^a_{,b}[/itex]

Schutz tells you that [itex]x^a_{,b}=\delta^a_{\phantom{a}b}[/itex]. To see why this is so, consider a simple example where [itex]a=0,1[/itex]. Then [itex]x^a_{\phantom{a},b}[/itex] can be represented as a matrix:

[tex]x^a_{\phantom{a},b} =
\left(
\begin{array}{cc}
\frac{\partial x^0}{\partial x^0} & \frac{\partial x^0}{\partial x^1} \\
\frac{\partial x^1}{\partial x^0} & \frac{\partial x^1}{\partial x^1}
\end{array}
\right) =
\left(
\begin{array}{cc}
1 & 0 \\ 0 & 1
\end{array}
\right) = \delta^{a}_{\phantom{a}b}
[/tex]

The reason that, for example, [itex]\partial x^0/\partial x^1 = 0[/itex] while [itex]\partial x^0/\partial x^0=1[/itex] should be obvious. If it isn't, note that our coordinates are supposed to be independent, so if you differentiate [itex]x^0[/itex] with respect to [itex]x^1[/itex] you will get zero, while differentiating [itex]x^0[/itex] with respect to [itex]x^0[/itex] will give you 1. So it should be easy to see why the definition of the Kronecker delta allows you to write this. Extending things to a scenario where you have [itex]m[/itex] coordinates is then trivial:

[tex]
x^a_{\phantom{a},b}
=
\left(
\begin{array}{cccc}
\frac{\partial x^0}{\partial x^0} & \frac{\partial x^0}{\partial x^1} &
\cdots & \frac{\partial x^0}{\partial x^m} \\
\frac{\partial x^1}{\partial x^0} & \frac{\partial x^1}{\partial x^1} &
\cdots & \frac{\partial x^1}{\partial x^m} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{\partial x^m}{\partial x^0} & \frac{\partial x^m}{\partial x^1} &
\cdots & \frac{\partial x^m}{\partial x^m}
\end{array}
\right)
=
\left(
\begin{array}{cccc}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{array}
\right)
= \delta^{a}_{\phantom{a}b}.
[/tex]
 
Last edited:
  • #15
Ok I get 3.12 but am still having trouble withthe material at the beginning of page 70. it would be nice if someone pmed me, but posting here would be OK. I'm really sorry btw. These are the few topics I've been having trouble with. i rarely ask things online.

Please try to epxlian it to me as clearly as possible! This is the last thing i can't understand in his tensor analysis section!
 
Last edited:
  • #16
Terilien said:
I'm finding tensor analysis to be quite difficult actually. Is this normal for high schoolers studying the subject?
Its probably one of the most difficult branches of math that there are. Graduate students have trouble with that math too. So if you're having trouble then you're normal. :biggrin:

Pete
 
  • #17
Ok so right now my main problems are equation 3.24, which i'd like explained in more detail and the basis for the gradient one forms.
 
Last edited:
  • #18
high schoolers? uh, no it is not normal for high schoolers to find it difficult, because most high schoolers are prudent enough to leave the topic for 4-5 years later. do you understand trig well? plane geometry? solid geometry? logic? algebra of polynomials? probability? matrices? calculus of one and several variables? topology?

if not, my suggestion is leave the tensors alone.
 
  • #19
as to the original question, why are monomials a basis for all polynomials? (these are products of coordinates.) similarly, products of linear forms give basis for all multilinear functions of any degree.
 
  • #20
mathwonk said:
high schoolers? uh, no it is not normal for high schoolers to find it difficult, because most high schoolers are prudent enough to leave the topic for 4-5 years later. do you understand trig well? plane geometry? solid geometry? logic? algebra of polynomials? probability? matrices? calculus of one and several variables? topology?

if not, my suggestion is leave the tensors alone.


Everything there but topology.
 
  • #21
Terilien said:
Ok so right now my main problems are equation 3.24, which i'd like explained in more detail

Well, the tensors [itex]\tilde{\omega}^{ab}[/itex] are introduced and defined in (3.23) such that the tensor f can be written as the sum of components of f multiplied by these new tensors omega. From (3.21) we see that the components of f (fab)can be obtained by contracting f with unit vectors ea in the first slot, and eb in the second (as I discussed above). Using this fact and (3.23) we obtain [tex]f_{cd}=\bold{f}(\vec{e}_c,\vec{e}_d)=f_{ab}\tilde{\omega}^{ab}(\vec{e}_c,\vec{e}_d)[/tex]. Now, on the left hand side we have the cd'th component of f, and on the right is the ab'th component multiplied with [itex]\tilde{\omega}^{ab}(\vec{e}_c,\vec{e}_d)[/itex]. So, we see for the left to equal the right, we must have that [itex]\tilde{\omega}^{ab}(\vec{e}_c,\vec{e}_d)=\delta^a_{ \phantom{a}c}\delta^b_{ \phantom{a}d}[/itex], which is (3.24)
 
  • #22
Terilien said:
Everything there but topology.
If that's the case then I wouldn't worry about it. I myself don't know topology and my probability is only enough to work in quantum mechanics.

It appears that all your questions were answered before I was able to get to them. Is there anything else I can help with? How much do you know about special relativity and physics?

Pete
 
  • #23
pmb_phy said:
If that's the case then I wouldn't worry about it. I myself don't know topology and my probability is only enough to work in quantum mechanics.

It appears that all your questions were answered before I was able to get to them. Is there anything else I can help with? How much do you know about special relativity and physics?

Pete

I know some university level classical dynamics, special relativitivistic kinematics and dynamics. I also know a bit of electromagnetism.

That stuff was all pretty easy, though the complete lack of rigour that physicists use when explaining gauss's law is insulting.


I know enough physics to manage with some things things, though it would be nice to know more electromagnetism.

I don't know much probability and hope to do as little as possible.

Right now i just don't understand equation 3.19. everything else is fine.
 
Last edited:
  • #24
Terilien said:
I know enough physics to manage with some things things, though it would be nice to know more electromagnetism.
Right now I'm refreshing myself in EM. I'm using Ohanian's book "Classical Electrodynamics - 2nd Ed.". I highly recommend this book.
Right now i just don't understand equation 3.19. everything else is fine.
As I said, it was simply the introduction of new notation. Instead of writing out the entire expression of the partial of [itex]\phi[/itex] with respect to x[itex]\alpha[/itex] you can just express that as [itex]\phi[/itex],[itex]\alpha[/itex]. Its merely a simpler way of expressing it for the ease of reading. Without a lot of the notational shortcuts that we use in this area of math the equations would look more hellish than they already are. :biggrin:

Pete
 
  • #25
Terilien said:
Everything there but topology.

*is frightened of you*
 
  • #26
anyway , I'm ok now, so i guess this thread oughta be closed.
 
  • #27
forgive my scepticism, but if you understahnd all that stuff, why is gradf perpendicular to a level set of f?

and what does greens theorem say and why is it true? (in a rectangle, say.)

and how do you prove a continuous function has a maximum on a closed bounded interval?
 
  • #28
mathwonk said:
forgive my scepticism, but if you understahnd all that stuff, why is gradf perpendicular to a level set of f?

and what does greens theorem say and why is it true? (in a rectangle, say.)

and how do you prove a continuous function has a maximum on a closed bounded interval?


green's theorem says that the circulation is around an interval is equal to the sum of iinfintessimal circulations within the area. Say that you have a square, you can find the cxirculation around it through very simple methods. then put another square beside it and calculate it's circulation. For obivous reasons(related to orientation of the sides), the sum of the circulations will be the circulation around both squares. since any area can be apporximated by squares we can approximate the circulation around the boundary of the region. If we take the limit as the number of squares approach infinity we get the total circulation around the boundary. if you want me to give a better explanation just ask.

I'm kind of offended, though I undersrtand your skepticism.

As for the second thing, I've nver seen such a proof but it seems very easy to prove in an informal way.

now if your wondering why I'm having difficulty with tensors, i guess I'm stupid.

Edit: i did not notice the first question, it's very easy to prove. Take a level surface and differentiate both sides with respect to t(where t can represent the paremter of a arbitrary cruve through the surface) using implicit differentiation. you get that the dot product between an arbitrary tangent vector and the gradient of the level surface equals zero meaning that they are orthogonal.


I'm actually quite offended, but again I understand your skepticism. i get this kind of treament from most professionals i meet. Unfortunately i haven't met many.
 
Last edited:
  • #29
Terilien said:
green's theorem says that the circulation is around an interval is equal to the sum of iinfintessimal circulations within the area. Say that you have a square, you can find the cxirculation around it through very simple methods. then put another square beside it and calculate it's circulation. For obivous reasons(related to orientation of the sides), the sum of the circulations will be the circulation around both squares. since any area can be apporximated by squares we can approximate the circulation around the boundary of the region. If we take the limit as the number of squares approach infinity we get the total circulation around the boundary. if you want me to give a better explanation just ask.

I'm kind of offended, though I undersrtand your skepticism.

As for the second thing, I've nver seen such a proof but it seems very easy to prove in an informal way.

now if your wondering why I'm having difficulty with tensors, i guess I'm stupid.

(One of) the reasons he's sceptical is that you're having difficulty understanding what the Kronecker delta is. I have to say that I'm a bit sceptical too, particularly given that the definition of the Kronecker delta is pretty much as simple and straightforward as mathematics can get. I also found your quip about a "lack of rigour" when explaining Gauss' law to be pretty strange.

Then again, I may be completely wrong.
 
  • #30
Ah ok I understand. Yes it's actually very simple now. It's just that when I was introduced to it, it was this weird alien symbol who's significance i could not grasp.

I know I'm stupid. I have no illusions about that.

Luckily I understand it now.

As for gauss' law for inverse square laws, it was just a that a few treatments made statements that were too general.

Now I feel like I'm on trial. i hate it when this happens.

I will admit though that I have difficulty learning in an ordered and sequential manner. learning disabilities at work!

what must I explain now? stokes? gauss's? change of variables? this is exactly why I would never ask questions in the past.
 
Last edited:
  • #31
I'm not asking you to prove yourself and I'm certainly not saying that you're stupid. Regardless of how old you are and regardless of your background, the fact that you're asking questions proves that you're not thick.

Seriously, if you're in high school and you've been able to understand and absorb this much physics and mathematics then you should do very well if you choose to study physics at university. I would however suggest that you look at a book such as Chern, Chen, and Lam to learn differential geometry before trying to learn too much about general relativity.

Oh, and complex analysis is an absolute must. I'm still amazed at how many undergrads I see that don't understand it at all.

Terilien said:
what must I explain now? stokes? gauss's? change of variables? this is exactly why I would never ask questions in the past.

You'd better get used to explaining yourself. Not only is it the best way to actually learn something, if you ever end up in grad school you'll do little else.
 
Last edited:
  • #32
coalquay404 said:
I'm not asking you to prove yourself and I'm certainly not saying that you're stupid. Regardless of how old you are and regardless of your background, the fact that you're asking questions proves that you're not thick.

Seriously, if you're in high school and you've been able to understand and absorb this much physics and mathematics then you should do very well if you choose to study physics at university. I would however suggest that you look at a book such as Chern, Chen, and Lam to learn differential geometry before trying to learn too much about general relativity.

Oh, and complex analysis is an absolute must. I'm still amazed at how many undergrads I see that don't understand it at all.

I've never found a resource on complex analysis, I am VERY interested in ANY introduction to rigorous mathematics.chern chem and lam? what's the book like? does it start with gaussian geometry or full blown generality?

I'd actually prefer to study mathematics outside of physics but i can't find any non physics books that treat advanced topics with a conversational tone.

"for what will it profit a man to know the formal definition of a submanifold, yet lose sight of the it's true meaning?"

Most mathematics books, start with something like, definition 1A.

"A "insert fancy term here" is a bijective mapping between two surjective Y spaces".

I know that's not anything specific, but I think you get the point. I'd love to major in physics, but trouble with things like school are getting in the way. sometimes it takes a while for me to get over my fear of studying a certain topic. I'm generally very afraid of not understanding something.

Oh well I'm getting a tad sentimental. My problem is that i generally keep my thoughts on topics to myself because I'm very sensitive to criticism.

"More articulate in person"

I'm not trying to be whiny. I'm willing to learn formal mathematics, I just like seeing WHY. intuition is the key to mathematics and science.

I'm actually extremely new to anything formal in the field of math. I'm used to thinking of a problem and finding the answer.
 
Last edited:
  • #33
Terilien said:
I've never found a resource on complex analysis, I am VERY interested in ANY introduction to rigorous mathematics.chern chem and lam? what's the book like? does it start with gaussian geometry or full blown generality?

"Chern, Chen, and Lam" refers to this book. (For some reason it's far more expensive in the US store than here in the UK, even taking exchange rates into account.) S. S. Chern was possibly the greatest geometer of the last century and had an uncanny knack for explaining geometry in a perfectly correct but chatty style. This book is probably the best one (in my opinion) from which to learn geometry if you don't feel like getting bogged down in Bourbaki-style rigour. He begins with multilinear algebra and continues through Riemannian geometry, Lie groups, and fibre bundles. It really is excellent.

Terilien said:
I'd actually prefer to study mathematics outside of physics but i can't find any non physics books that treat advanced topics with a conversational tone.

This is known as the curse of Bourbaki. Probably the best way to learn mathematics in a "conversational tone" is by doing just that: talking to people. Books are great as references, but you'll probably end up learning a lot more from other people.

Terilien said:
"A "insert fancy term here" is a bijective mapping between two surjective Y spaces".

I know that's not anything specific, but I think you get the point. I'd love to major in physics, but trouble with things like school are getting in the way. sometimes it takes a while for me to get over my fear of studying a certain topic. I'm generally very afraid of not understanding something.

Oh well I'm getting a tad sentimental. My problem is that i generally keep my thoughts on topics to myself because I'm very sensitive to criticism.

Again, if you want to learn, get used to being criticised. You can learn a lot from books but you'll learn the most from talking to people who already know the things you're interested in. Being criticised (sometimes even viciously) is a great way to sharpen your ideas. If you can spot the difference between somebody launching an ad hominem attack and somebody who's picking your ideas apart, you can spot the difference between good criticism and bad criticism. And then you'll be half way towards getting a PhD.
 
  • #34
I've always wanted to meet a math professor of some sort. Unfortunately I have no idea how to.
 
  • #35
Terilien said:
I've always wanted to meet a math professor of some sort. Unfortunately I have no idea how to.

It's pretty simple. If you live near to a university, email the maths department and ask if they have any outreach programs or any staff members who are willing to talk to prospective students. I can (almost) guarantee you that you'll get a good response and that you'll have a chance to talk to somebody who's doing a job that you may one day like to have. Mathematicians and physicists are really receptive to people who show an interest in the field so try to get in contact with someone.
 
<h2>1. What is a tensor product?</h2><p>A tensor product is a mathematical operation that combines two or more tensors to create a new, higher-order tensor. It is often used in multilinear algebra and is denoted by the symbol ⊗.</p><h2>2. How is the basis for a higher-order tensor determined?</h2><p>The basis for a higher-order tensor is determined by taking the tensor product of the basis vectors for each individual tensor. For example, the basis for a third-order tensor would be the tensor product of the basis vectors for three individual tensors.</p><h2>3. What is the significance of the tensor product in higher-order tensors?</h2><p>The tensor product is significant because it allows for the representation of higher-order tensors in terms of lower-order tensors, making it easier to perform calculations and manipulate these complex objects.</p><h2>4. Can you provide an example of a tensor product and its proof?</h2><p>One example of a tensor product is the cross product of two vectors in three-dimensional space. The proof for this can be shown using the properties of the tensor product, such as distributivity and associativity, along with the definition of the cross product itself.</p><h2>5. How is the tensor product used in real-world applications?</h2><p>The tensor product has many real-world applications, particularly in physics and engineering. It is used in the study of electromagnetism, quantum mechanics, and fluid dynamics, among others. It is also used in computer science for tasks such as image and signal processing.</p>

1. What is a tensor product?

A tensor product is a mathematical operation that combines two or more tensors to create a new, higher-order tensor. It is often used in multilinear algebra and is denoted by the symbol ⊗.

2. How is the basis for a higher-order tensor determined?

The basis for a higher-order tensor is determined by taking the tensor product of the basis vectors for each individual tensor. For example, the basis for a third-order tensor would be the tensor product of the basis vectors for three individual tensors.

3. What is the significance of the tensor product in higher-order tensors?

The tensor product is significant because it allows for the representation of higher-order tensors in terms of lower-order tensors, making it easier to perform calculations and manipulate these complex objects.

4. Can you provide an example of a tensor product and its proof?

One example of a tensor product is the cross product of two vectors in three-dimensional space. The proof for this can be shown using the properties of the tensor product, such as distributivity and associativity, along with the definition of the cross product itself.

5. How is the tensor product used in real-world applications?

The tensor product has many real-world applications, particularly in physics and engineering. It is used in the study of electromagnetism, quantum mechanics, and fluid dynamics, among others. It is also used in computer science for tasks such as image and signal processing.

Similar threads

  • Differential Geometry
Replies
2
Views
467
  • Differential Geometry
Replies
9
Views
2K
  • Differential Geometry
Replies
8
Views
2K
  • Differential Geometry
Replies
3
Views
3K
  • Linear and Abstract Algebra
Replies
10
Views
239
  • Quantum Physics
Replies
11
Views
1K
Replies
2
Views
2K
  • Differential Geometry
Replies
3
Views
1K
Replies
2
Views
722
  • Differential Geometry
Replies
1
Views
1K
Back
Top