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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?

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Crossposted at MSE: math.stackexchange.com/questions/405108 –  Zhen Lin May 28 '13 at 21:26
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Whether or not this identification is okay really depends on what purpose you need tensors for. For things like multi-linear algebra over the reals or complexes, for subjects like relativity, this is perfectly good. –  Ryan Budney May 28 '13 at 21:41
    
As Ryan Budney said, the definition is OK in the case of finite-dimensional vector spaces over a field (if I remember correctly, this is how Halmos defines it). However, it fails miserably if anything is infinite-dimensional or we work over a ring. Moreover, I find the definition extremely unintuitive -- more so than the classical construction as a quotient of the free module on pairs (which describes exactly how one computes in a tensor product). –  darij grinberg May 29 '13 at 0:18
    
I think that the point is that the (much preferable to me) definition of the tensor product as the quotient of the free module on pairs has a slight pedagogical disadvantage in the case of finite-dimensional vector space as a quotient of two infinite-dimensional ones. To many students it may look much scarier than the definition via bilinear maps. But it really depends on one's purpose and way of thinking, which one to prefer. –  Vladimir Dotsenko May 29 '13 at 8:53
    
This makes use of the universal property of the tensor product that it linearizes bilinear mappings, and of reflexivity of finite dimensional vector spaces over a field. In view of this it is perfectly okay, IMHO.I would even define it as: $V\otimes W = L_2(V\times W, k)^\ast$. –  Peter Michor May 29 '13 at 9:35
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