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Tensor Product is distributive.

I get stuck at the proof:

If $T$ is a $p$-tensor and $S, U$ a $q$ tensor. Then I need to show that $$ T \otimes (S \cdot U) = (T \otimes U) \cdot (S \otimes U).$$ Denote $S(u_1, \ldots, u_q) \cdot U(w_1, \ldots, w_q) = S \cdot U (v_1, \ldots, v_q)$. Then \begin{eqnarray*} &&T \otimes (S \cdot U)(v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q})\\ & =& T (v_1, \ldots, v_p) \cdot (S \cdot U)(v_{p+1}, \ldots, v_{p+q})\\ & =& T (v_1, \ldots, v_p) \cdot (S (u_{p+1}, \ldots, u_{p+q}) \cdot U(w_{p+1}, \ldots, w_{p+q}))\\ &? =& (T \otimes U) \cdot (S \otimes U) \end{eqnarray*}

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    $\begingroup$ Normally, when we talk of the tensor product being distributive, it is over a group operation written as "+".... $\endgroup$
    – user14972
    Commented Jun 20, 2013 at 2:09

1 Answer 1

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Let $V$ be a vector space and $V^*$ the dual space. Suppose $\alpha,\beta \in V^*$. Let $\alpha \otimes \beta :V \times V \rightarrow \mathbb{R}$ be defined by $$ \alpha \otimes \beta (x,y) = \alpha(x)\beta(y) $$ for all $x,y \in V$. The proof that $\alpha \otimes (\beta + \gamma) = \alpha \otimes \beta + \alpha \otimes \gamma$ is as follows: let $x,y \in V$, $$ \begin{align} \alpha \otimes (\beta + \gamma)(x,y) &= \alpha(x)(\beta + \gamma)(y) \\ &= \alpha(x)[\beta(y) + \gamma(y)] \\ &= \alpha(x)\beta(y) + \alpha(x)\gamma(y) \\ &= \alpha \otimes \beta(x,y) + \alpha \otimes\gamma(x,y) \\ &= (\alpha \otimes \beta + \alpha \otimes\gamma)(x,y). \\ \end{align} $$ As $x,y$ were arbitrary the identity $\alpha \otimes (\beta + \gamma) = \alpha \otimes \beta + \alpha \otimes \gamma$ follows. The steps above are justified by the point-wise defined addition of multilinear mappings and the vector space structure of $V$.

A similar proof can be given for the tensor product of other tensors. I merely give the proof for dual-vectors, or one-forms if you prefer. Hope this helps. I took the liberty of replacing your $\cdot$ with $+$.

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  • $\begingroup$ Thank you very much. Very helpful. Thanks James. $\endgroup$ Commented Jun 20, 2013 at 3:27

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