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Let's say that I have a three dimensional tensor $A = ((B \times_1 C_1) \times_2 C_2) \times_3 C_3$. where $B$ is $n \times n \times n$ tensor, $C_i$ are $n \times n$ matrices and $A$, as a result, is $n \times n \times n$ tensor. The notation $\times_k$ means tensor product on dimension $k$ of the tensor.

  1. Now, when I compute $(((A \times_1 D) \times_2) E \times_3 F)$, where $D$, $E$ and $F$ are $n \times n$ matrices is it the same like computing:

    $G = ((B \times_1 (C_1D) \times_2 (C_2 E)) \times_3 (C_3F)$

  2. If the answer to the above is positive, what if I want to multiply each of the three dimensions of $G$ by a tensor $K$ of dimension $n \times n \times n \times n$ (i.e. compute

    $((B \times_1 (C_1D \times_{1,2} K) \times_2 (C_2 E \times_{1,2} K)) \times_3 (C_3F \times_{1,2} K)$

Is there a way to do it by multiplying directly some dimensions of $G$ by $K$ without "having access" to $C_i$,$D$,$E$ and $F$, but just having $G$?

For example, does tensor product follow distributivity, and therefore the above equals:

$(((B \times_1 (C_1D ) \times_2 (C_2 E )) \times_3 (C_3F ) ) \times_{2,3} K$



By "tensor product", I mean for example that $B \times_1 C_1$ is a three dimensional tensor, such that $[B \times_1 C_1]_{ijk} = \sum_l B_{ljk} C_{li}$.

Unfortunately, I am not sure how to define $\times_{1,2}$ exactly as a product, but it basically means that if $K$ is an $n\times n \times n \times n$ tensor, i.e. a function $R^{n \times n} \rightarrow R^{n \times n}$, then $C_1 D \times_{1,2} K = K(C_1 D)$.

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I have a parsing problem with the phrase "tensor product on dimension $k$ of the tensor". Can you give one actual example to illustrate what you're asking? –  KCd May 17 '12 at 5:48
Is for example $(B \times_1 C_1)_{ijk} = \sum_\alpha B_{\alpha jk}(C_1)_{\alpha,i}$? –  martini May 17 '12 at 6:08
@martini, yes, exactly. –  kloop May 17 '12 at 12:09
@KCd: I made some edits to reflect what I meant. –  kloop May 17 '12 at 12:20
What you call a "three-dimensional" tensor is usually called a "rank 3" tensor (on an $n$-dimensional space), and your "tensor product on dimension $k$ of the tensor" is usually called contraction on the $k$th index. Look up this terminology and use it. Using conventional notation, you begin in $(V^{\otimes 3}) \otimes (V \otimes V^*) \otimes (V \otimes V^*) \otimes (V \otimes V^*)$ and contract 3 times, so the answer is again in $V^{\otimes 3}$. Let $c_i \colon V^{\otimes 3} \otimes V^* \rightarrow V^{\otimes 2}$ be contraction with the $i$th factor of $V^{\otimes 3}$. (contd.) –  KCd May 17 '12 at 14:37
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