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Short question. Suppose we have vector spaces $V_1,V_2,V_3,V_4$ and a linear map

$f: V_1\otimes V_2 \to V_3 \otimes V_4$.

Are there always linear maps $f_1: V_1 \to V_3$ and $f_2: V_2 \to V_4$, such that

$f\simeq f_1 \otimes f_2$?

If yes, why is it?

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Even a linear map $V\otimes V\to V\otimes V$ needn't be of the form $A\otimes B$, e.g. $v\otimes w\mapsto w\otimes v$; this example indicates the possibility of linear maps "mixing things" across the $\otimes$ symbol.

The embedding ${\rm hom}(V_1,V_3)\otimes\hom(V_2,V_4)\hookrightarrow\hom(V_1\otimes V_2,V_3\otimes V_4)$ is easily seen to be a surjection if $V_1,V_2,V_3,V_4$ are all finite-dimensional, simply by comparing dimensions, but there will necessarily be elements in $\hom\otimes\hom$ that are not expressible as pure tensors, except in the very special circumstances of $\dim V_1=\dim V_3=1$ or $\dim V_2=\dim V_4=1$.

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No, this is not always the case, for example if $k$ is a field then consider the map $k \otimes k \to k^2 \otimes k^2$ defined by $1 \otimes 1 \mapsto (1, 0) \otimes (1, 0) + (0, 1) \otimes (0, 1)$.

Another way to see this is that if these are spaces over the finite field $\mathbb F_p$ and $V_1, V_2, V_3, V_4$ have dimensions $a, b, c, d$ respectively then $\hom(V_1, V_3)$ has $p^{ac}$ elements, $\hom(V_2, V_4)$ has $p^{bd}$ elements, and $\hom(V_1 \otimes V_2, V_3 \otimes V_4)$ has $p^{abcd}$ elements but there are $p^{ac + bd}$ maps in $\hom(V_1, V_3) \times \hom(V_2, V_4)$ and hence $p^{ac + bd - 1}$ maps of the form $f_1 \otimes f_2$.

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  • $\begingroup$ Ok.I see. Does it work, if we restrict to the case $f: V_1 \otimes V_2 \to V_1 \otimes V_3$? $\endgroup$
    – Bobby
    Jan 12 '15 at 5:05
  • $\begingroup$ @whacka: I missed a factor of $p$ but otherwise the counting argument is fine. The pure tensors are essentially enumerated by the direct product and the counting argument is counting the difference between the set of elements in the tensor product and the set of pure tensors in the tensor product. As you say yourself, the difference between the two is the true culprit. $\endgroup$
    – Jim
    Jan 12 '15 at 5:54
  • $\begingroup$ @Bobby: No, it still doesn't work. It doesn't even work if we restrict ourselves to the case $V \otimes V \to V \otimes V$. $\endgroup$
    – Jim
    Jan 12 '15 at 6:01
  • $\begingroup$ Ah I didn't realize that's what you were doing. $\endgroup$
    – whacka
    Jan 12 '15 at 6:03

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