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The tensor product $S\otimes_R T$ of $S$ and $N$ over $R$ is a module. A multilinear form $L:V^r \to R$ is called an $r$-tensor on $V$.

On the one hand a tensor is a function sending elements of $V^r$ to $R$. But a tensor product has no notion of a function sending elements from something to something. So why are tensor products called tensor products and why are tensors called tensors? How does one make the connection between tensor products and tensors? Is there a reason why they both have the word "tensor" in them even though one of them is a function but the other is not?

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See also…. – Hans Lundmark Jan 10 '12 at 7:09
up vote 5 down vote accepted

The vector space of all multilinear maps $V^r\to R$ is isomorphic (in a canonical way) to the $r$th tensor power $(V^*)^{\otimes r}=\underbrace{V^*\otimes V^*\otimes\cdots\otimes V^*}_{r \text{ times}}$ of the dual space $V^*$.

For example, suppose $B=\{e_i:1\leq i\leq n\}$ is a basis for $V$, $B^*=\{\phi_i:1\leq i\leq n\}$ the basis of $V^*$ dual to $B$, and $\phi:V\times V\to R$ a bilinear map. Then under the canonical isomorphism above, $\phi$ corresponds to the element of $V^*\otimes V^*$ given by $$\sum_{1\leq i,j\leq n}\phi(e_i,e_j)\phi_i\otimes\phi_j$$.

The abstract tensor product operation $\otimes$ is a, well, abstraction of the general ideal behind the study of multilinear maps.

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One way to characterize a tensor product is through its universal property.

For example, suppose $U,V,W$ are vector spaces over some field $\mathbb{F}$. Then a tensor product $U\otimes V \otimes W$ is a vector space (over $\mathbb{F}$) paired with a trilinear map $\cdot \otimes \cdot \otimes \cdot : U \times V \times W \to U \otimes V \otimes W$ (denoted $(u,v,w) \mapsto u\otimes v\otimes w$) such that any tri-linear map $f : U \times V \times W \to M$ uniquely factors through $U\otimes V\otimes W$.

That is: There exists a unique linear map $\bar{f}:U \otimes V \otimes W \to M$ such that $f(u,v,w)=\bar{f}(u\otimes v\otimes w)$.

It's not hard to show that such a vector space exists and that any two "tensor products" of $U$, $V$, and $W$ are canonically isomorphic. Thus we speak of "the" tensor product.

The whole business about factoring maps through the tensor product means that trilinear maps from $U\times V\times W$ are interchangeable with linear maps from $U\otimes V\otimes W$. This should begin to explain why the term tensor is used in these different contexts.

Now to elaborate on Mariano's answer. Running more with the Physics or Differential Geometry crowd you'll hear talk of $(r,s)$-tensors. These are multilinear maps with $r$ vector and $s$ dual vector inputs which output a scalar. So (using a universal property) this is no more than an element of $(V\otimes \cdots \otimes V \otimes V^* \otimes \cdots \otimes V^*)^*$ ($V$ occuring $r$ times and $V^*$ occuring $s$ times).

If $V$ is a finite dimensional vector space (which is usually the context for this use of "tensor"), we can do even better, such a multilinear map can be (canonically) identified with an element of $V^* \otimes \cdots \otimes V^* \otimes V \otimes \cdots \otimes V$ ($V^*$ occuring $r$ times and $V$ occuring $s$ times). Such an element looks like $$ \sum v^*_{i_1} \otimes \cdots \otimes v^*_{i_r} \otimes v_{j_1} \otimes \cdots \otimes v_{j_s}$$ where $v^*_{i_k} \in V^*$ and $v_{j_\ell} \in V$.

This element of $V^* \otimes \cdots \otimes V^* \otimes V \otimes \cdots \otimes V$ can acts like a multilinear map as follows $$ (w_1,\dots,w_r,w^*_1,\dots,w^*_s) \mapsto \sum v^*_{i_1}(w_1)\cdots v^*_{i_r}(w_r) \cdot w^*_1(v_{j_1})\cdots w^*_s(v_{j_s})$$

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I suppose the undefinded $M$ ist an arbitrary vector space over $\mathbb{F}$? – yippy_yay Jun 6 '15 at 15:45
Yes............. – Bill Cook Jun 6 '15 at 16:01

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