# Product of 3rd rank tensor with squared vector

## Main Question or Discussion Point

Greetings,

can somebody show me how to calculate such a term?

P= X E² where X is a third order tensor and E and P are 3 dimensional vectors.

Since the result is supposed to be a vector, the square over E is not meant to be the scalar product. But the tensor product of E with itself yields a matrix, and a 3 rank tensor times a matrix cannot be a vector, can it?
According to a book, the result for the components is Pi=∑j,kXi,j,kEjEk

Regards

andrewkirk
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It is really bad practice to write it that way - as you have pointed out. It causes unnecessary confusion.

However, what they probably mean is (XE)E. That is, contract the tensor with the vector E to get a second order tensor. Then contract it with E again to get a first order tensor, which is a vector. The formula you wrote gives that in terms of coordinates.

Also, what they wrote is ambiguous, as just writing XE2 or even (XE)E doesn't tell us against which index we should contract the vector. This is clarified in the coordinate version, which tells us they are contracting them against the 2nd and 3rd indices, but that is not apparent from the question, unless your text has specified a convention about which indices one should assume one is contracting post-written vactors against..

Can you specify what 'contracting' means in that context? I don't see how you can combine a 3rd order tensor and a vector to get a matrix. In a lesser dimension, this would be equivalent to somehow multiplying a matrix and a scalar to get a vector ?

andrewkirk
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Contracting a 3rd order, 3-dimensional tensor with components ##X_{ijk}## on the 2nd index with a vector ##\vec E=(E^1,E^2,E^3)## is an operation whose result is a 2nd order tensor (ie matrix) ##\mathbf Y## with components ##Y_{ik}=\sum_{s=1}^3 X_{isk}E^s##. Contracting on a different index - the r-th index (r = 1 or 3) uses the same formula but with the s moved to the r-th position in the subscripts of ##X##.

Ok, and what is the meaning of this contraction? Is there an analogon for normal matrices?

andrewkirk
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Ok, and what is the meaning of this contraction? Is there an analogon for normal matrices?
Yes. Contraction of a tensor with a vector is analogous with pre-multiplication by a row vector if it is on the first index, and with post-multiplication by a column vector if it is on the last index. For a matrix, which is a second order tensor and hence has only two indices, that covers all possibilities. But for a higher order tensor there are additional indices - the middle ones - on which one can contract, and that has no analog with matrix multiplication.

WWGD
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Ok, and what is the meaning of this contraction?
There is a natural isomorphism between ## T^k_{j+1} (V) ## ( mixed tensors ) for V finite-dimensional and the space of multilinear maps:
##(V^{*} \times V^{*} \times......V^{*}) \times (V \times V\times\times V....\times V ) \rightarrow V ##
((k,j) factors respectively; parentheses not needed, my choice) . See, e.g. p 12/29 in
http://www.maths.ed.ac.uk/~aar/papers/leeriemm.pdf
EDIT: See (2.3) in p.13/30.

Last edited:

Yes. Contraction of a tensor with a vector is analogous with pre-multiplication by a row vector if it is on the first index, and with post-multiplication by a column vector if it is on the last index. For a matrix, which is a second order tensor and hence has only two indices, that covers all possibilities. But for a higher order tensor there are additional indices - the middle ones - on which one can contract, and that has no analog with matrix multiplication.
So do I understand it right like this?. How do I contract a matrix with a vector? Let's say at the first index. Now I just multiply my row vector from the left with my matrix and I get another row vector? And on the second index, I just carry out multiplication from the right with a column vector?
So contracting a 3rd order tensor with a vector yields a 2nd order tensor, and if I contract this with a vector, I get a 1st order tensor, so the dimension gets reduced everytime?

There is a natural isomorphism between ## T^k_{j+1} (V) ## ( mixed tensors ) for V finite-dimensional and the space of multilinear maps:
##(V^{*} \times V^{*} \times......V^{*}) \times (V \times V\times\times V....\times V ) \rightarrow V ##
((k,j) factors respectively; parentheses not needed, my choice) . See, e.g. p 12/29 in
http://www.maths.ed.ac.uk/~aar/papers/leeriemm.pdf
EDIT: See (2.3) in p.13/30.
Sorry, I don't know anything about the 'pure maths' field of tensors, I don't think the pure mathematical description is understandable for me. Still, thanks for the answer.

WWGD
andrewkirk
Homework Helper
Gold Member
So do I understand it right like this?. How do I contract a matrix with a vector? Let's say at the first index. Now I just multiply my row vector from the left with my matrix and I get another row vector? And on the second index, I just carry out multiplication from the right with a column vector?
So contracting a 3rd order tensor with a vector yields a 2nd order tensor, and if I contract this with a vector, I get a 1st order tensor, so the dimension gets reduced everytime?
Yes that's all correct.

Having written that, I realise that my formula in post 2 of (XE)E is still ambiguous, because it does not specify on which indices we do the contractions - it could be on any of 1&2, 2&3 or 3&1. The formula you wrote with subscripts and superscripts specifies it's 2&3, but the formulas without sub/superscripts - both theirs and mine - fail to distinguish between the three possible interpretations.

Yes that's all correct.

Having written that, I realise that my formula in post 2 of (XE)E is still ambiguous, because it does not specify on which indices we do the contractions - it could be on any of 1&2, 2&3 or 3&1. The formula you wrote with subscripts and superscripts specifies it's 2&3, but the formulas without sub/superscripts - both theirs and mine - fail to distinguish between the three possible interpretations.
Alright, thank you very much :)