5
$\begingroup$

$\newcommand\Mod[1]{#1\text{-Mod}}$Does any one have a reference on a explicit equivalence between $$\Mod{A\otimes_\mathbb{C} B} \cong \Mod A\boxtimes \Mod B?$$

The proof in "Tensor Categories EGNO" uses the universal property of $\boxtimes$, but I would like to see an explicit functor that defines an equivalence.

In BaKi definition 1.1.15 there is a explicit description of $\mathcal{C}_1\boxtimes \mathcal{C}_2$ when $\mathcal{C}_1$, $\mathcal{C}_2$ are additive categories over $k$. Namely,

  • $\operatorname{Ob}(\mathcal{C}_1\boxtimes \mathcal{C}_2)=$ finite sums of the form $\bigoplus X_i\boxtimes Y_i$, where $X_i\in \operatorname{Ob}(\mathcal{C}_1)$, $Y_i\in \operatorname{Ob}(\mathcal{C}_2)$.
  • and $\operatorname{Hom}_{\mathcal{C}_1\boxtimes \mathcal{C}_2}(\bigoplus X_i\boxtimes Y_i,\bigoplus X_j'\boxtimes Y_j')= \bigoplus \operatorname{Hom}(X_i,X'_j)\otimes \operatorname{Hom}(Y_i,Y'_j)$.

Using this description I wanted to construct the equivalence.

I tried showing that \begin{align*} F:\Mod A\boxtimes \Mod B &\rightarrow \Mod{A\otimes_\mathbb{C} B}\\ X\boxtimes Y&\mapsto X\otimes_\mathbb{C}Y \end{align*} is full, faithul, and essentially surjective.

But I got stuck on trying to show that this functor is essentially surjective. There is what seems to be a counter example, shown on MathStack (Link).

Any help/suggestions would be greatly appreciated.

I currently only need the case when the modules are finite dimensional. But dont want to be restricted to finite dimensional algebras. Namely I am working with finite dimensional modules over $U(\mathfrak{g}_1\oplus \mathfrak{g}_2)\cong U(\mathfrak{g}_1)\otimes U(\mathfrak{g}_2)$.

Edit (More context): I was really trying to use the Relative Tambara Tensor product introduced in (Tam01), and saw a sentence in (DSS18) (First Paragraph), that when using $\mathcal{C}=\operatorname{Vect}$ it agrees with Deligne's Tensor Product (though I could have misunderstood).

The Relative Tambara Tensor Product has a similar description of objects and morphisms, but also has much more relations.

$\endgroup$
1
  • 1
    $\begingroup$ For your particular problem it’s probably better to rewrite that as comodules for the corresponding quantized function algebra which will be better behaved under tensor product. I think this is discussed in one of the Ben Zvi-Brochier-Jordan papers $\endgroup$ Mar 12 at 15:38
6
$\begingroup$

The explicit definition you give is not the Deligne tensor product, it’s what’s often called the “naive” tensor product. The naive tensor product typically won’t be abelian. The Deligne tensor product is universal for right exact bifunctors, and there’s nothing about right exactness in the naive tensor product.

There is no description of the Deligne tensor product that’s more explicit than the theorem you’re looking at, more general Deligne tensor products are explicitly constructed by taking colimits of this construction for subcategories.

$\endgroup$
1
  • $\begingroup$ Hello Noah, thanks for the reply! I actually wanted to work with the relative Tambara Tensor Product which had a similar description. And in your paper "The Balanced Tensor Product of Module Categories" with Douglas and Schommer-Pries, I saw a sentence saying that the relative Tambara Tensor product using Vect is equivalent to the Deligne Tensor Product. $\endgroup$ Mar 12 at 17:40

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.