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I Tensor Product

  1. Feb 4, 2017 #1
    We've been learning about tensor products. In particular, we've been looking at index notation for the tensor products of matrices like these:
    ##
    \left( \begin{array}{cc}
    a_{11} & a_{12} \\
    a_{21} & a_{22} \end{array} \right)##
    And
    ##
    \left( \begin{array}{cc}
    b_{11} & b_{12} \\
    b_{21} & b_{22} \end{array} \right)##

    In index notation the tensor product is ##(a \bigotimes b)_{ijkl} = a_{ij}b_{kl}##. And apparently, there's an expression in terms of these indices which tells you which row and which column ##a_{ij}b_{kl}## will be on. So I tried the row number being ##ik## and the column number being ##kl## and that worked up to the third row and column, then it didn't any more. I've written out the matrix and I've been staring at it, and I cannot see what the expression for row and column number is. Does anyone know it? I'd really appreciate someone showing me, because now it's going to bother me for ages. Thanks for any help! :)
     
    Last edited: Feb 4, 2017
  2. jcsd
  3. Feb 4, 2017 #2
  4. Feb 4, 2017 #3
    If it does then I can't find it, or more probably I didn't understand it. In a normal matrix, of course, the row is just given by ##i## and the column by ##j## and there is an equivalent idea for tensors but it's not at all easy; I thought for the entry ##a_{ij}b_{kl}## it would be in row ##ik## and column ##jk## and it isn't; nor is it in row ##ikl## or row ##ijl## and I just can't find a general expression for the location of some element ##a_{ij}b_{kl}##. Sorry if it's obvious, I did read through the Wikipedia page but we started the tensor product a few days ago and most of what the article says is a bit beyond me.

    I'm only really interested in square matrices, which I hoped would make things simpler. Not sure it does though!
     
  5. Feb 4, 2017 #4
    Ah. Unless row and column in a tensor mean something different to in a matrix. Is the tensor product of two 2x2 matrix treated as having 4 rows and columns or just 2? I was treating it as having 4.
     
  6. Feb 4, 2017 #5

    fresh_42

    Staff: Mentor

    You can write each matrix ##A = (a_{kl}) = \Sigma_i \,v_i \otimes w_i\;##. So if ##B = (b_{mn}) = \Sigma_j \,v'_j \otimes w'_j## is another matrix, then ##A \otimes B = \Sigma_i \, \Sigma_j \, v_i \otimes w_i \otimes v'_j \otimes w'_j \,##. This is a four dimensional cube, a matrix of matrices, i.e. entries ##(a_{ij} \cdot B) = (a_{ij} \cdot (b_{mn}))\, ##.
     
  7. Feb 4, 2017 #6
    Ah ok. Can I split it into parts? In the first quadrant, top left, the row and column are simply given by the indices of B. Then for the rest it's the ##i+k## for the row and ##j+l## for the column?
     
  8. Feb 4, 2017 #7

    fresh_42

    Staff: Mentor

    Top left is the matrix ##a_{11}\cdot B##, the ##a_{11}## multiple of the entire matrix ##B##.
    Each entry is a multiple of the entire matrix ##B##, namely ##a_{ij}B##

    The hierarchy goes:
    grade 0 - scalars ##c##
    grade 1 - vectors ##\vec{v}##
    grade 2 - matrices ##\vec{v} \otimes \vec{w}##
    grade 3 - cubes ##\vec{u}\otimes \vec{v}\otimes \vec{w}##
    ##\ldots##

    However it rarely makes sense to think of tensors in such a coordinate representation.

    It is far more helpful to think of them in terms of properties. E.g. a Lie multiplication in a Lie algebra ##\mathfrak{g}## can be written as
    $$ [X,Y] = \sum_{i=1}^m \, u_i(X)\cdot v_i(Y) \cdot w_i \; \textrm{ and } \; u_i \otimes v_i \otimes w_i \in \mathfrak{g}^* \otimes \mathfrak{g}^* \otimes \mathfrak{g}$$
     
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