Tensor product of two arbitrary vectors an arbitrary tensor?

  • #1
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I am trying to show that if (C^ab)(A_a)(B_b) is a scalar for arbitrary vectors A_a and B_b then C^ab is a tensor.

I want to take the product of the two vectors then use the quotient rule to show that C^ab must then be a tensor. This lead to the question of whether or a not the product of two arbitrary vectors is again a completely arbitrary tensor. I guess this boils down to asking whether or not every tensor can be generated by a tensor product of two vectors.
 

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  • #2
andrewkirk
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What is the 'C^ab' to which you refer?
If it is a function, then the result does not hold, because it could be a nonlinear function, and tensors are linear.
If it's not a function, then what do you mean by the symbol string '(C^ab)(A_a)(B_b)'?

You'll probably find it easier to make your question clear by using latex. Use double # symbols to start and end in-line latex, and double $ symbols to start and end display latex (ie formulas on a line of their own).

The answer to your last question is No. Most (2 0) tensors cannot be expressed as the product of two vectors but are instead a linear combination of sums of up to ##n^2## such vector products, where ##n## is the dimension of the underlying vector space..
 
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  • #3
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Ahh, thank you for the answer. Two in fact. I did not know you can just LaTeX directly into this forum. $$C^{ab}$$ was left unknown to conclude through the quotient rule that it is indeed a tensor. The way I understood the quotient rule, as stated from multiple sources, is that we can determine $$C^{ab}$$ is tensor just by knowing that $$A_a and B_b$$ are arbitrary and that $$C^{ab}B_{b}A_{a}$$ equals a scalar.
 
  • #4
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Ahh, thank you for the answer. Two in fact. I did not know you can just LaTeX directly into this forum. $$C^{ab}$$ was left unknown to conclude through the quotient rule that it indeed was a tensor. The way I understood the quotient rule, as stated from multiple sources, is that we can determine $$C^{ab}$$ is tensor just by knowing that $$A_a and B_b$$ are arbitrary and that $$C^{ab}B_{b}A_{a}$$ equals a scalar.
Which part of andrewkirk's answer didn't you understand?
 
  • #5
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Which part of andrewkirk's answer didn't you understand?
Well I was in the process of clarifying what I meant by the symbol string '(C^ab)(A_a)(B_b)'.
 
  • #6
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Well I was in the process of clarifying what I meant by the symbol string '(C^ab)(A_a)(B_b)'.
I can be a tensor and linear in ##A_a## and ##B_b##, if ##C^{ab}## is a rank 1 matrix. But without further restrictions it is in general no tensor.
 
  • #7
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I can be a tensor and linear in ##A_a## and ##B_b##, if ##C^{ab}## is a rank 1 matrix. But without further restrictions it is in general no tensor.
Could you elaborate? Is my understanding of the quotient rule wrong? If we have ##C^{ab}A_a## with arbitrary tensor ##A_a## and this product transforms like a tensor under a change of coordinate basis then ##C^{ab}## is a tensor. Is that correct?

It should also be pointed out that I am using summation/index/Einstein notation.
 
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  • #8
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A matrix ##C^{a,b}## of rank 1 means you can write each column vector as a multiple of the first (as long as it isn't all 0 but some has to be different from 0). That is ##C^{a,b} = (A_1 \vec B , ... , A_n \vec B) = (A_1,...,A_n)⊗B = \vec A ⊗ \vec B.## And on the other hand each tensor ##\vec A ⊗ \vec B## can be written as such a matrix if you chose a basis, i.e. a coordinate system.
In general, given ##C^{a,b}## a matrix, you will have ##C^{a,b} = \sum^{a,b=n}_{a,b=1} \vec A_a ⊗ \vec B_b##.

And as andrewkirk has said: if ##C^{a,b}## represents a nonlinear function, it will have nothing to do with tensors. I don't really understand what your ##C^{a,b}## is and why you write a tensor just ##A_a##. Which tensor? ##\vec v_1 ⊗ \vec v_2 ⊗\vec v_3 ⊗\vec v^*_1 ⊗ \vec v^*_2 ⊗\vec v^*_3##?
 
  • #9
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How can any textbook justify something like this then?
upload_2016-2-22_18-57-47.png

I've seen it like this and other similar forms in other books. It seems like it doesn't matter what "T" here is as long as A and B are tensors and that A is invariant under coordinate transformations.
 
  • #10
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Let me think about it to get it straight. In this basic form I have to assume that ##A## and ##B## are of the general form ##\vec v_1 ⊗ ... ⊗ \vec v_m## with ##\vec v_i ∈ V## , ##\dim V = n## and ##T## is a linear function. (Unfortunately "Quotient Rule" doesn't tell me something in this context. Maybe a physicist is more familiar with it.)
 
  • #11
andrewkirk
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How can any textbook justify something like this then?
View attachment 96310
I've seen it like this and other similar forms in other books. It seems like it doesn't matter what "T" here is as long as A and B are tensors and that A is invariant under coordinate transformations.
I think ##T## is supposed to be an array of scalars, where the scalars change when one changes basis. Such an array may be a tensor, or a non-tensor, with the Christoffel symbols being examples of the latter. The way the Christoffel symbols change upon basis change is not the same as how tensor representations change, because Christoffel symbols are not tensors.

I think what the book is doing is using ##BT## to denote the coordinate-based contraction (aka 'inner product') of the representation of tensor ##B## in basis ##C## with the basis ##C##-representation of the array ##T## of scalars. What that means is to take the outer product of the two arrays and then contract on designated pairs of indices or, if no pairs are designated, on pairs in matching positions.
The conclusion is that, if the result of doing that is the same as the basis-##C## representation of tensor ##A##, and that this result holds regardless of the choice of basis ##C##, then ##T## is a tensor.

It would be an instructive - though possibly laborious and messy - exercise to do this where ##T## denotes the set of Christoffel symbols, and observe that the equation fails when one changes basis.

Here's a link, although I found that example, as with others I skimmed, still rather opaque.
 

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