2
$\begingroup$

It seems like tensor product of modules over a ring and tensor product of algebras over a ring are studied somewhat independently, is it possible to see these tensor product as one single phenomenon ? For example a tensor product of non-unitary algebras over a ring : modules been then non-unitary algebras with null multiplication, algebras are non-unitary algebras having a unit (but the embedding of category is not fully faithful, that should be a problem I think). Anyone know how to do that ?

This question is motivated by the following : is it possible de interpret the the tensor product of modules as the coproduct in a richer category (my guess : non-unitary algebras) ? My final motivation is : how can be define in general a tensor product "indexed by" any category ? While the tensor product of a finite number of modules can be seen as a tensor product over a finite set, what is a good notion of tensor product indexed by an infinite set ? Or a general category ?

$\endgroup$
2
$\begingroup$

There are many threads here so an answer can't be completely satisfying. But evidently tensor products are defined generically for modules. If those modules happen to carry other structure then the result might also carry that structure. Say if the modules happen to be rings then the tensor product is a ring. I am not aware of any introduction to tensor products that meaningfully separates the development of tensors of modules from tensors of rings so to the first point, I think perhaps this is a missunderstanding of the references you have in mind. They may be assuming some earlier treatment where a common foundation was made clear already and are just now specializing for convenience. But all tensor products do in fact depend on one unified definition of tensor product (as having a universal mapping property from bilinear to linear) as given by Hassler Whitney in the 1930's. See for example Rotman Section 8.4

Rotman, Joseph J., Advanced modern algebra, Upper Saddle River, NJ: Prentice Hall/Pearson Education. xv, 1012 p., append. 27 p. (2002). ZBL0997.00001.

Categorical definitions of $\otimes$ have been given in the context of symmetric monoidal categories. However be aware the obvious model is $\otimes_A$ where $A$ is commutative. Non-commutative $A$ wont quite fit that model nor will multiple $A'$s. In particular $U_1\otimes_{A_1}U_2\otimes_{A_2} U_3$ isn't even well-defined. Does one mean $(U_1\otimes_{A_1}U_2)\otimes_{A_2} U_3$, in which case $A_2$ acts on $U_1\otimes_{A_1} U_2$, or does it mean $U_1\otimes_{A_1}(U_2\otimes_{A_2} U_3)$ where now $A_1$ acts on $(U_2\otimes_{A_2} U_3)$? Most authors intend a third idea, which is that $U_2$ is an $(A_1,A_2)$-bimodule and thus $U_1\otimes_{A_1}U_2$ becomes an $A_2$-module and $U_2\otimes_{A_2} U_3$ an $A_1$-module; so, some form of associativity works. But one sees the defects in thinking of such a definition as an $n$-ary product. So even the concept of a tensor over a set as suggested needs shoring up before further generalizations. This in part is why the axioms of symmetric monoidal categories assume a single tensor product.

It turns out that if you form $U\otimes_{\Delta} V$ for any set $\Delta$ operating linearly on $U$ and $V$ (formally there is a function $\Delta\to \mathrm{End}(U)\times \mathrm{End}(V)^{op}$), then in fact you can effectively "close" $\Delta$ to an associative ring, an adjoint algebra $A(\Delta)$. And $U\otimes_{A(\Delta)} V=U\otimes_{\Delta} V$ -- equal not just isomorphic. So in this sense for a binary tensor product any creative set of operations will result in the same as just using associative algebras; see Theorem 2.11 in the below reference.

Brooksbank, Peter A.; Wilson, James B., Groups acting on tensor products., J. Pure Appl. Algebra 218, No. 3, 405-416 (2014). ZBL1300.20032.

There is a recent generalization of that result for $n$-ary tensor products, but the result is that the closure of the operators you tensor over is a Lie algebra, not an associative algebra. In short, tensoring over categories -- whatever that should mean -- probably wont get you what you want. You need a Lie structure, and categories don't have that.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.