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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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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. –  Ted Shifrin May 28 '13 at 20:50
Could you elaborate please? I am interested as well. Thanks –  Nicolas Bourbaki May 28 '13 at 20:55
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. –  Christopher A. Wong May 28 '13 at 23:36
$V\otimes W = (\mbox{free product of $V$ and $W$})/(\mbox{a certain ideal})$? –  Neal May 29 '13 at 1:34
The tensor product is defined by it's universal mapping property. What you are referring to is an explicit construction of this. –  Nicolas Bourbaki May 29 '13 at 8:10

1 Answer 1

At least, over the reals $\Bbb{R}$ and in the finite dimensional case it is possible to find a satisfactory answer.

By taking vector space over the reals: $V={\rm span}\{b_1,...,b_n\}$ and $W={\rm span}\{c_1,...,c_m\}$, then define $T$ in $V\otimes W$ as the formal linear combination $$T=\sum_{\mu}\sum_{\nu}T^{\mu\nu}b_{\mu}\otimes c_{\nu}$$ -which everyone should abbreviate- as $T=T^{\mu\nu}b_{\mu}\otimes c_{\nu}$, it determined by some arrange $T^{ij}$ of scalars.

So, to associate to some multilinear ambit we are lead naturaly to map $$ V^*\times W^*\ \stackrel{T}\longrightarrow\ {\Bbb{R}},$$ appliying for covectors $f\in V^*$ and $g\in W^*$ an assignment $(f,g)\to T(f,g)$ is given by \begin{eqnarray*} T(f,g)&=&T^{sr}b_s\otimes c_r(f,g)\\ &=&T^{sr}f(b_s)g( c_r)\\ &=&T^{sr}f_sg_r\\ &=&\sum_s\sum_rT^{sr}f_sg_r\\ \end{eqnarray*} where we had simplfied $f(b_s)=f_s$ and $g(c_r)=g_r$.

Note that the number $T^{sr}f_sg_r$ is nothing more than the bilinear form: $$[f_1,...,f_n] \left(\begin{array}{ccccc} T^{11}&T^{12}&...&T^{1,n}\\ T^{21}&T^{22}&...&T^{2,n}\\ \vdots\\ T^{m1}&T^{m2}&...&T^{mn} \end{array}\right) \left(\begin{array}{c} g_1\\ \vdots\\ g_n \end{array}\right)= fTg^{\top}. $$

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The reals have nothing to do with this. This works for finite free modules over any commutative ring. –  darij grinberg Jul 26 at 18:58
At least to gain intuition, my friend –  janmarqz Jul 26 at 19:13

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