Let $$ \otimes:R\times R\rightarrow W $$ $$ f:R\times R\rightarrow R~~,~f(X,Y)=XY $$ $\otimes$ is tensor product, $W$ is a vector space, and $f$ is a bilinear map. As I know , we need to find a linear map $g:W\rightarrow R$ such that $f=g\circ \otimes$ . Because I can define $$ g(X\otimes Y) = f(X,Y) ~~,~~ X,Y\in R $$ then $W$ is $$ W=\{XY | X,Y\in R \} =R $$ Is this is a right example about tensors?

I am fuzzy with tensor products. So I try to make a specific example. But I am not sure whether there are mistakes.


2 Answers 2


Your question clearly is:

$$ g(X\otimes Y) = f(X,Y) ~~,~~ X,Y\in R $$ then $W$ is $$ W=\{XY | X,Y\in R \} =R $$ Is this is a right example about tensors?


(I'll assume $R = \mathbb R$)

This is a wrong statement of a valid example about tensors, but can be fixed by considering formal subtleties. $W$ is not equal to $R$, it is actually isomorphic to $R$. The basic reason is that you chose $W$ to be the codomain of a generic tensor product you denote with $\otimes$. But of course you can specify you're taking the $\otimes$ map to be the ordinary multiplication map in $R$ directly, but that's not strictly the abstract tensor product formulation.

Let us clarify this.

What you denote by $W$ is the space of linear combinations of the image of a generic universal bilinear map $\otimes$. $W$ is usually denoted $R\otimes R$. You can think of it as:

$$W = R\otimes R = \left\{\sum X\otimes Y,~~~X,Y\in R \right\}.$$

We usually omit these sums since most things of interest are linear on $W$ or defined as to be linear, so we can focus on defining nontrivial rules at each component of the sum (simple or pure tensor products).

The full map $g$ is constructed as a linear extension of the rule $$g(X\otimes Y) = f(X,Y) = XY$$ which takes $$t = \sum X\otimes Y \in R\otimes R$$ to $$g(t) = g\left(\sum X\otimes Y\right) = \sum g(X\otimes Y) = \sum f(X,Y) = \sum XY.$$

But $t = \sum X\otimes Y = \sum XY (1\otimes 1) = (\sum XY) 1\otimes 1$ since $\otimes$ is bilinear. Hence, to map to your notation as much as possible,

$$ W = \left\{\sum X\otimes Y\right\} = \left\{\left(\sum XY\right) (1\otimes 1)\,|\,X,Y\in R\right\}, $$ which is isomorphic to the set you mentioned which is $R$.


I think this is what you're after:

If $R$ is a ring, then the map $f : R \times R \to R$ given by $f(x, y) = x y$ is $R$-bilinear, so it induces a linear map $\tilde{f}:R\otimes_R R \to R$ given by $\tilde{f}(x \otimes y) = f(x, y) = x y$.

In general, tensor products of modules (or, more concretely, vector spaces) turn bilinear maps $f : A \times B \to C$ into linear maps $\tilde{f} : A\otimes B \to C$ via the formula $\tilde{f}(x\otimes y) = f(x, y)$.

  • $\begingroup$ beginners often miss that defining $\tilde f$ by $\tilde{f}(x \otimes y) = f(x, y) = x y$ in fact means $\tilde{f}$ is the linear extension of this rule to all of $A\otimes B$, so that the full $\tilde f$ acts on sums of simple tensor products and yields sums of $R$-products $\endgroup$
    – rfabbri
    Mar 1, 2018 at 14:27

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