# The tensor product of two monoidal categories

Given two monoidal categories $$\mathcal{M}$$ and $$\mathcal{N}$$, can one form their tensor product in a canonical way?

The motivation I am thinking of is two categories that are representation categories of two algebras $$R$$ and $$S$$, and the module category of the tensor product algebra $$R \otimes_{\mathbb{C}} S$$.

Also, if $$\mathcal{M}$$ and $$\mathcal{N}$$ are assumed to be braided monoidal, can we tensor the braidings?

The book Tensor Categories discusses, with many variations, the details of Robert McRae's answer. Just like for vector spaces, there are a number of related but inequivalent "tensor products" of linear categories, with the choice dependent on the types of linear categories considered: sufficiently finite dimensional (vector spaces / categories) have only one reasonable tensor product, but the "algebraic" tensor product of infinite-dimensional objects can be completed in various ways. Locally finite abelian categories is a particularly good choice.

Once you have made such a choice, tensoring (braided) monoidal structures is typically easy. You probably will need to require that the monoidal structure is "continuous" for however you chose to complete your tensor products. Again, see the Tensor Categories book for details.

• Another nice reference is Ignacio Lopez Franco, "Tensor products of finitely cocomplete and abelian categories" (arxiv.org/abs/1212.1545) which carefully defines and compares various notions of tensor product. Jul 31, 2019 at 8:01

If your categories are locally finite abelian, I think you are looking for the Deligne tensor product of $$\mathcal{M}$$ and $$\mathcal{N}$$. The Deligne tensor product $$\mathcal{M}\boxtimes\mathcal{N}$$ does inherit braided monoidal structure from $$\mathcal{M}$$ and $$\mathcal{N}$$ if these are braided monoidal.

• What happens if the categories are not locally finite? Jul 9, 2019 at 20:44
• For example, $k$-linear and finite dimensional. Jul 9, 2019 at 20:45

If you're acquainted with $$\infty$$-categories you might also be interested in the tensor product of presentable $$\infty$$-categories: it's a real tensor product, in that it provides an equivalence $$\begin{array}{cr} C_1\times\dots\times C_n \to D & \text{cocont.}\\\hline C_1\otimes\dots\otimes C_n \to D \end{array}$$ between the colimit-preserving functors from $$C_1\times\dots\times C_n$$ to $$D$$ and the functors from the tensor product to $$D$$.

There's plenty of reasons why "multilinear" gets categorified in "cocontinuous"; also, presentable categories all are cartesian (because they admit finite products, and all other shapes of limits for that matter), so they are not the most general monoidal categories, but their tensor product is fairly well understood.

Hope this helps!

• There is no need to jump to the $\infty$-categorical setting for the question. Feb 23, 2020 at 19:26
• I didn't say there was; but perhaps it is possible to back-engineer the infty-definition to get the 1-definition Feb 23, 2020 at 23:10
• Sure. And this tensor product has been known long before $\infty$-categories were popular. Feb 23, 2020 at 23:27
• Thank god categories were popular way before $\infty$-stuff ;) Feb 24, 2020 at 19:08

No, there is not tensor product. You need to make further assumptions that could give you several potentially distinct tensor products.

You can always form the product category $$M\times N$$. It is a monoidal category. Any sensible tensor product will be a quotient category of $$M\times N$$.

For instance, in the Deligne tensor product, mentioned by McRae, you assume that $$M$$ and $$N$$ are abelian, $$k$$-linear. Then $$M\times N$$ is $$k$$-bilinear and you can ask for the universal quotient $$M\otimes_k N$$ such that any $$k$$-bilinear monoidal functor $$M\times N\rightarrow P$$ factors via a $$k$$-linear monoidal functor $$M\otimes_k N\rightarrow P$$. Such thing clearly exists: you dont even need to assume that the categories are abelian but you need $$k$$-linearity! However, $$M\otimes_k N$$ is not necessarily abelian, even if $$M$$ and $$N$$ are such. If it is abelian, then it is Deligne tensor product.

Overall, you need some "balancing" conditions to form a tensor product. In the case of $$k$$-linear categories, the balancing is performed by $$Vec(k)$$, the category of vector spaces that acts on both $$M$$ and $$N$$. There are several ways you can do "balancing" but only some of them preserve monoidal and/or braided structure. Others will lose it.

For instance, the following wabbity balancing is cool. For this I need a third monoidal category $$C$$ and two monoidal functors $$C\rightarrow M$$ and $$C\rightarrow N$$. By a $$C$$-balanced functor I understand a bifunctor $$M\times N \rightarrow D$$, together with an equivalence of two resulting trifunctors $$M\times C \times N \rightarrow D$$. I claim that there exists a monoidal category $$M\otimes_CN$$ with the universal $$C$$-balanced bifunctor $$M\times N \rightarrow M\otimes_C N$$, with two caveats:

• It is not longer braided, you need some conditions for that.
• It is no longer a category :-)) The issue is that your hom-s will be proper classes, unless you put some smallness condition, for instance, $$C$$ being small will do.