My understanding is that in differential geometry one typically defines a 2-tensor on V (an $\mathbb{R}$ vector space) as a multilinear map from $V \times V \rightarrow \mathbb{R}$. The collection of all such maps is called $V \otimes V$.

Thus each element of $V \otimes V$ is a multilinear map. In algebra one defines $M \otimes N$ for $R$ modules (assume commutative with identity) as a quotient module of a (very large) free module. The algebra construction corresponds to the differential geometry construction in the case of vector spaces.

Are there any results or theorems on $M$ and $N$ which indicate when it is possible to identify a basic element $m_{1} \otimes n_{1}$ of $M\otimes N$ with a multilinear map $M\times N \rightarrow R$? Is it true that if every basic element $m_{1} \otimes n_{1}$ induces a multilinear map $M\times N \rightarrow R$ then $M \otimes N$ is a free?

  • $\begingroup$ One doesn't define tensor products via free modules etc., one defines them via their universal property, which immediately gives the connection to multilinear maps (here: bilinear). What you describe is just one possible construction of the tensor product. Concerning your question: $m_1 \otimes n_1$ has nothing to with a bilinear map $M \times N \to R$. $\endgroup$ – Martin Brandenburg Sep 26 '14 at 18:46
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    $\begingroup$ The definition of $V \otimes W$ as the space of bilinear maps $V \times W \to \mathbb{R}$ is unfortunate. One should define tensor products via universal properties, for modules and then vector spaces in particular (these are no separate cases!). If $V,W$ are finite free, then $V \otimes W \cong \hom(V^* \otimes W^*,R) \cong \mathrm{Bilin}(V^* \times W^*,R)$. The other definition ("differential geometry") ignores dualization. It makes it even impossible to define a map $V \otimes W \to V' \otimes W'$ from maps $V \to V'$, $W \to W'$, which is a minimal requirement of a natural construction. $\endgroup$ – Martin Brandenburg Sep 26 '14 at 18:53
  • $\begingroup$ Each $m\otimes n$ is a tensor, so they do have something with bilinear maps to do. $\endgroup$ – Lehs Sep 26 '14 at 18:58
  • $\begingroup$ @Martin I know it is best to define the tensor product via a universal property. However, it is the case for vector spaces that one has the fact that tensors can be considered multilinear maps in a natural way. Are there conditions in which a basic element of the tensor product of two modules can be considered a bilinear map in a natural way other than the case that M and N are free R-modules? $\endgroup$ – sykh Sep 26 '14 at 20:33
  • $\begingroup$ @Martin Pertaining to the last sentence in your second comment. I don't see why this is true in case of vector spaces. Maps $V \rightarrow V'$ and $W \rightarrow W'$ would induce homomorphisms on the tensor product. As far as I can see but I'm not knowledgeable in this area. The "differential geometry" definition only works for vector spaces and not modules. If that is your point then I understand it. $\endgroup$ – sykh Sep 26 '14 at 20:42

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