# Tensor Product is associative, distributive, not commutative.

Tensor Product is associative, distributive, not commutative.

Here is my attempt to show tensor product is associative, is it legit?

If $T$ is a $p$-tensor and $S$ a $q$ tensor, then $T \otimes S$ is a $p+q$ tensor: $$T \otimes S(v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q}) = T(v_1, \ldots, v_p) \cdot S(v_{p+1}, \ldots, v_{p+q}).$$ Now consider a third tensor, a $r$-tensor $U$. Then \begin{eqnarray*} & & (T \otimes S ) \otimes U (v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q}, v_{p+q+1}, \ldots, v_{p+q+r})\\ &= &(T \otimes S (v_1, \ldots, v_{p+q})) \cdot U(v_{p+q+1}, \ldots, v_{p+q+r})\\ &= &T (v_1, \ldots, v_{p}) \cdot S(v_{p+1}, \ldots, v_{p+q}) \cdot U(v_{p+q+1}, \ldots, v_{p+q+r})\\ &= &T (v_1, \ldots, v_{p}) \cdot (S(v_{p+1}, \ldots, v_{p+q}) \cdot U(v_{p+q+1}, \ldots, v_{p+q+r}))\\ &= &T (v_1, \ldots, v_{p}) \cdot (S \otimes U (v_{p+1}, \ldots, v_{p+q+r}))\\ &= &T \otimes (S \otimes U)(v_1, \ldots, v_{p+q+r}). \end{eqnarray*}

Distributivity - is this proof legit?

First note that tensor product distribute over addition. If $T$ is a $p$-tensor and $S, U$ a $q$ tensor. Then I need to show that $$T \otimes (S + U) = (T \otimes U) + (S \otimes U).$$

Then \begin{eqnarray*} &&T \otimes (S + U)(v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q})\\ & =& T (v_1, \ldots, v_p) \cdot (S + U)(v_{p+1}, \ldots, v_{p+q})\\ & =& T (v_1, \ldots, v_p) \cdot (S (v_{p+1}, \ldots, v_{p+q}) + U(v_{p+1}, \ldots, v_{p+q}))\\ & =& T (v_1, \ldots, v_p) \cdot S (v_{p+1}, \ldots, v_{p+q}) + T(v_1, \ldots, v_p) \cdot U(v_{p+1}, \ldots, v_{p+q})\\ & =& (T \otimes S) + (T \otimes U) \end{eqnarray*}

It is not commutative - is this legit?

\begin{eqnarray*} & & T \otimes S(v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q}) = T(v_1, \ldots, v_p) \cdot S(v_{p+1}, \ldots, v_{p+q})\\ &= &S(v_{p+1}, \ldots, v_{p+q}) \cdot T(v_1, \ldots, v_p) = S \otimes T(v_{p+1}, \ldots, v_{p+q}, v_{1}, \ldots, v_{p})\\ & \neq & S \otimes T(v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q}). \end{eqnarray*}

• Pay special attention to the indices in the very last line. – anon Jun 20 '13 at 0:45
• Presumably show that $\otimes$ distributes over addition of tensors. – anon Jun 20 '13 at 0:58
• The correct identity for "tensor distributes over dot product" would be $T \otimes (S \cdot U)=(T \otimes S) \cdot (T \otimes U).$ Why is distributivity helpful? Consider taking the tensor product of a tensor and a linear combination of tensors. – Loki Clock Jun 20 '13 at 1:03
• With the dot product, it shouldn't. – Loki Clock Jun 20 '13 at 1:28
• @MathSnail The law you're trying to prove there is not distributivity. Nor is it distributivity of the tensor product over addition - which would make the tensor product the multiplication of a ring over the vector space, if associativity also holds. Not that you couldn't try to prove it distributes over the dot product, but I'm not sure of the significance. – Loki Clock Jun 20 '13 at 1:55