# Making sense of defining tensor products $\bigotimes V$

This question is making sense of a definition. Suppose $R$ is a commutative ring.

How does one make a meaningful definition of $\bigotimes_1^k V_i$ where $V_i$ are $R$-modules? I know of the $k-$ary product definition. But I do not like it since it does not build up from the original operation $\otimes$, which can be regarded as a bifunctor.

We could show there is a well defined isomoprhism, $(A \otimes B) \otimes C \cong A \otimes (B \otimes C)$, $\tau_{A,B,C}: (a \otimes b) \otimes c \mapsto a \otimes (b \otimes c)$.

In spirit of this proof. Let us define the left-associated form of $k$ tensor products, $(\cdots (V_j \otimes V_{j+1}) \cdots \otimes V_{k-1}) \otimes V_k := L_j^k(V)$ for $j <k$.

So I wish to define $\bigotimes_1^k V_i$ as $L_1^k(V)$. How does one justify the "correctedness" of this definition?

My sketchy attempt:

Suppose we have a $n$-ary representation of $V_i$ with some parathensization, denoted $X$, by inductive hypothesis $$X \cong (L_1^m(V) ) \otimes (L_{m+1}^n(V)) \cong (L_1^m(V)) \otimes ((L_{m+1}^{n-1} (V) ) \otimes V_n)$$ by inductive hypothesis and the $3$ case, $$(L_1^m(V) \otimes L_{m+1}^{n-1}(V)) \otimes V_n \cong (L_1^{n-1}(V)) \otimes V_n \cong L_1^n(V)$$

This reasoning seems unnatural, somehow the isomorphisms are obscured. That is, we could have chosen any isomoprhism $\tau_{A,B,C}: (A \otimes B) \otimes C \rightarrow A \otimes (B \otimes C)$.

Is there a categorical explanation of all of this?

• Are you fine when defining the $k$-fold cartesian product of sets as $(\dots(X_1 \times X_2) \times X_3)\dots X_{k-1})\times X_k$? Because the same problem appears: $(\prod_1^k X_i) \times (\prod_k^n X_i)$ is not equal to $\prod_1^n X_i$, it is only in bijection with it. And you might not even notice it most of the time, taking element $(x_1,\dots,x_n)$ in it, without even thinking about the definition of such a tuple. If this is ok for you, then so should your problem too. (The formal justification is Hurkyl's answer.) – Pece Jul 28 '18 at 17:44

There is the general notion of a monoidal category; it is a category equipped with:

• a unit object $I$
• a bifunctor $\otimes$
• a canonical natural isomorphism between any two ways to paranthesize an expression and/or insert/remove copies of $I$

with the property that any way of combining the canonical natural isomorphisms via composition or tensor product gives you another canonical natural isomorphism.

As an example, there are canonical natural isomorphisms

$$\mu : I \otimes ((A \otimes B) \otimes C) \to A \otimes ((B \otimes I) \otimes C)$$ $$\nu : ((A \otimes B) \otimes C) \to A \otimes ((B \otimes I) \otimes C)$$ $$\omega : I \otimes (A \otimes ((B \otimes I) \otimes C)) \to A \otimes ((B \otimes I) \otimes C)$$

We can construct another natural isomorphism parallel to $\mu$ via $\omega \cdot (I \otimes \nu)$; these are guaranteed to be the same natural isomorphism.

This coherence property means its safe to pick whatever parenthesization you like; if you ever need to use two different parenthesizations there's a canonical way to do the reinterpretation.

In fact, we can even say something stronger: every monoidal category turns out to be monoidally equivalent to a strict one — one where the tensor product is actually an associative operation and where tensoring with $I$ is the identity functor. (and for all of the coherence morphisms, we've chosen the identity transformations to be the canonical isomorphisms)

There's a big theorem that says all of the coherence properties of a monoidal category can be expressed just by using the associator, the two unitors, and a handful of identities relating them; definitions of monoidal category you're likely to encounter will be in that form.

To me, the tensor product of modules $M_1,\ldots, M_k$ over a commutative ring $R$ is a module that represents the functor of $R$-multilinear maps from $M_1\times\cdots\times M_k$. This means that there is a natural correspondence between $R$-multilinear maps $M_1\times\cdots \times M_k\to N$ and $R$-module homomorphism: $\bigotimes_1^k M_i\to N$.

Write $\textrm{Mult}(M_1,\ldots,M_k;N)$ for the $R$-module of multilinear maps from $M_1\times\cdots \times M_k$ to $N$. Then $$\textrm{Mult}(M_1,\ldots,M_k;N)\cong \textrm{Hom}\left(\bigotimes_1^{k-1}M_i,\textrm{Hom}(M_k,N)\right) \cong\textrm{Bil}\left(\bigotimes_1^{k-1}M_i,M_k;N\right)$$naturally in $N$. This means that $$\bigotimes_1^{k-1}M_i\otimes M_k\cong\bigotimes_1^k M_i$$ as both represent the same functor. Iterating this gives your left-associated form. But you don't need to do this, you could work from the right, or in some more arbitrary manner instead. The point is, that all these representations of the $k$-fold tensor product represent the same representable functor and so they are all canonically isomorphic.

• Somewhat merging the two answers, look up Leinster's unbiased monoidal categories. – Berci Jul 29 '18 at 22:50