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Prove $L \otimes M = M \otimes L$ only if either $L=0$ or $M=0$

I saw this statement on Linear Algebra (2ed) written by Hoffman and Kunze. I can't figure out how to prove it. The multilinear forms $L$ (and $M$) are from $V^r$ (and $V^s$) into $K$ where $K$ is a commutative ring with identity and $V$ is a $K$-module. I think it is reasonable to exclude $L=M$.

Thanks.

Added by the crowd. Here's the relevant excerpt.

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Is $V$ an arbitrary $K$-module or a finite free $K$-module? (In fact, I wonder if your $V$ should be $K$.) –  KCd Jul 1 '12 at 2:25
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What if I take $V = K$, $r = s = 1$, $L = M$ the identity map $K \to K$? Then both of the resulting maps $K \times K \to K$ are just multiplication. Am I crazy? –  Dylan Moreland Jul 1 '12 at 2:42
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I think excluding $L=M$ is a bad idea. One should either figure out what question they actually meant to ask (or possibly that they did ask and you got the context wrong), or acknowledge that the statement is simply wrong, and replace it with a correct statement. –  Hurkyl Jul 1 '12 at 4:05
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There are definitely exceptions other than $L=M$. For example, $L=\lambda M$ for some scalar $\lambda$. Less trivially, if $L=M\otimes M$. What if we suppose that neither of $L$ or $M$ is a scalar multiple of a tensor power of the other? –  Jonas Meyer Jul 1 '12 at 6:51
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We also have to worry about things like $L=N\otimes N$, $M=N\otimes N\otimes N$. What if we suppose that no tensor power of $L$ is a scalar multiple of a tensor power of $M$, and vice versa? –  Jonas Meyer Jul 1 '12 at 6:57

2 Answers 2

up vote 6 down vote accepted

After such a long chain of (old) comments, a summary is in order. By definition, $L\otimes M$ is the function on $V^{r+s}$ such that $$(L\otimes M)(\alpha_1,\dots,\alpha_{r+s})=L(\alpha_1,\dots,\alpha_r)\,M(\alpha_{r+1},\dots,\alpha_{r+s}) \tag1$$ Similarly, $$(M\otimes L)(\alpha_1,\dots,\alpha_{r+s})=M(\alpha_1,\dots,\alpha_s)\,L(\alpha_{s+1},\dots,\alpha_{s+r}) \tag2$$ The following is a sufficient condition for the functions defined by (1) and (2) to be equal, pointed out in comments by Jonas Meyer.

$(*)\qquad$ $L$ and $M$ are scalar multiples of some tensor powers of a multilinear function $N$.

Since $(*)$ does not require either $L$ or $M$ to vanish, the statement in the book is false.

It hasn't been resolved whether $(*)$ is also a necessary condition for $L\otimes M=M\otimes L$.

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I've substantially revised my answer. –  Brian Rushton Jul 5 '13 at 17:12

Consider each linear function as a column vector in the dual space with some basis. Then the tensor product of $n$ such vectors can be visualized as a hyper matrix (or tensor, as it's usually described in physics) of dimension $n$, where each entry is the product of entries in each column vector. Two vectors $a,b$ tensored gives a matrix where each row is a multiple of the first vector, and the multiple of each row is given by the entries in the second column vector.

Tensoring such a matrix with a third tensor gives a cubical matrix where each horizontal slice is a multiple of the base matrix, and the multiples are given by the third column vector, etc.

Note that every single row (i.e. 1-dimensional subset) in the $i$th direction of the tensor hyper matrix is a multiple of the $i$th vector in the tensor product. Either every such row is zero (and the tensor is zero) or we can recover the $i$th vector in the tensor product up to a scalar factor.

This means that in non-zero tensor products, the multiplicands and their order are uniquely determined up to scalar products. Switching the order of tensoring as described in the problem is equivalent to taking an $r+s$ dimensional tensor permuting the order of the multiplicands by a cyclic permutation (I.e. shifting every index by $s$ and modding indices by $r+s$). If the tensor is invariant under such a cyclic permutation, then the multiplicands of the product tensor must be, up to a scalar factor,periodic with period $gcd(s,r+s)=gcd(s,r)$, i.e. the 2 vectors being tensored must be scalar multiples of tensor powers of the same tensor.

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