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I am aware that many books on differential geometry define tensors as multilinear maps. Namely $$ V\otimes W := L_2(V^*\times W^*,\Bbb F) $$ I am also aware that this space is isomorphic to the tensor product in the finite dimensional case, but I am wondering if it is a good idea to think of tensor products as multilinear maps. Is there any reason why one would like to make this identification in the finite dimensional case? Or does this definition come from the idea that students new to the subject may have an easier time with this less abstract definition?

Thanks

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    $\begingroup$ But, even with the more standard algebraic definition, the key feature is the universal mapping property, which is oh-so-close to this. But you are correct that double-duality gets tangled in there. $\endgroup$ – Ted Shifrin May 28 '13 at 20:50
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    $\begingroup$ Could you elaborate please? I am interested as well. Thanks $\endgroup$ – Nicolas Bourbaki May 28 '13 at 20:55
  • $\begingroup$ What is the alternative definition of tensor you are referring to? I only know of the definition of the tensor product as identified with multilinear maps. $\endgroup$ – Christopher A. Wong May 28 '13 at 23:36
  • $\begingroup$ $V\otimes W = (\mbox{free product of $V$ and $W$})/(\mbox{a certain ideal})$? $\endgroup$ – Neal May 29 '13 at 1:34
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    $\begingroup$ The definition of $V\otimes W$ as $L_2 \left(V^\ast \times W^\ast, \mathbb F\right)$ is really just a trick to avoid talking of universal properties and huge free modules, which works as long as one is content with finite-dimensional spaces. I don't like this trick, as it also comes at the cost of clarity (but YMMV). The free-module-quotient construction, once one gets used to it a bit, looks very natural: it is almost precisely the straightforward formalization of the motivation, which is "formal linear combinations of terms of the form $v \otimes w$ modulo distributivity". $\endgroup$ – darij grinberg Jul 26 '14 at 19:00
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Most maths concepts (or objects) can be defined in many ways. A good rule (IMHO) is to use the definition which requires a minimal number of other concepts, and/or a minimal number of steps to be derived. And other possible definitions often become corollaries.

The following definition essentially requires the notions of vector spaces and linear forms and is pretty straightforward. It does not use the tensor product operation, which can be defined afterwards in that case.

Let $V$ be a vector space over a field $\Bbb F$. The set of linear forms of $V$ (linear maps from $V$ to $\Bbb F$) is called $V^*$.
A [p,q] tensor is a multilinear map from $(V^*)^p\times V^q=V^*\times V^*\times ...\times V\times V$ to $\Bbb F$ (also called a multilinear form since it has values in $\Bbb F$).
From that you can show that:

  • a scalar is a [0,0] tensor
  • a vector is a [1,0] tensor (but a [1,0] tensor is not necessarily a vector unless $V$ has finite dimension)
  • a covector (i.e. a linear form which is by definition an element of $V^*$) is a [0,1] tensor
  • a linear map (from $V$ to $V$) is a [1,1] tensor
  • etc.
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