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I read the following statement (equation 22) in "Monoidal 2-structure of bimodule categories" by Justin Greenough:

Let $\mathcal{C}$ be a finite tensor category (abelian k-linear rigid monoidal category with simple unit and finite dimensional Hom spaces). Let $\mathcal{M}$ and $\mathcal{N}$ be exact left module categories over $\mathcal{C}$.

We introduce the left $\mathcal{C}$-module structure in $\mathcal{M} \boxtimes \mathcal{N}$ (the Deligne' tensor product of $\mathcal{M}$ and $\mathcal{N}$) by:

$$X \otimes (M \boxtimes N) = (X \otimes M) \boxtimes N,$$ where $X \in \mathcal{C}$.

Then the equation 22 tells us that $$\underline{Hom}_{\mathcal{M} \boxtimes \mathcal{N}}(M \boxtimes N, S \boxtimes T) = \underline{Hom}_{\mathcal{M}}(M, S) \otimes \underline{Hom}_{\mathcal{N}}(N,T),$$ where $\underline{Hom}_{*}$ are internal hom for left $\mathcal{C}$ structure in $\mathcal{M} \boxtimes \mathcal{N}$, $\mathcal{M}$ and $\mathcal{N}$.

Now, let us consider the simple case:

let $\mathcal{C}$ be a unitary fusion category and $\mathcal{M} = \mathcal{N} = \mathcal{C}$. Then by the definition of internal Hom and the equation above, we have

$$Hom_{\mathcal{C} \boxtimes \mathcal{C}}(1 \boxtimes 1, X \boxtimes X^*) \cong Hom_{\mathcal{C}}(1, \underline{Hom}_{\mathcal{C} \boxtimes \mathcal{C}}(1 \boxtimes 1, X \boxtimes X^*))\\ \cong Hom_{\mathcal{C}}(1, \underline{Hom}_{\mathcal{C}}(1, X) \otimes \underline{Hom}_{\mathcal{C}}(1,X^*)), $$ where $1$ is the unit of $\mathcal{C}$ and $X$ is a simple object in $\mathcal{C}$ such that $X \ncong 1$ and $X^*$ is the left (or right) adjont of $X$.

Since $\underline{Hom}_{\mathcal{C}}(1, X) = X$ and $\underline{Hom}_{\mathcal{C}}(1, X^*) = X^*$, we have

$$\{0\} = Hom_{\mathcal{C}}(1,X) \otimes Hom_{\mathcal{C}}(1, X^*) \cong Hom_{\mathcal{C} \boxtimes \mathcal{C}}(1 \boxtimes 1, X \boxtimes X^*) \cong Hom_{\mathcal{C}}(1, X \otimes X^*) \neq \{0\}.$$

So it seems that we have a contradiction here. Can anyone tell me if I made a mistake somewhere? Does the equation (22) in "Monoidal 2-structure of bimodule categories" hold? Thank you in advance!

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  • $\begingroup$ Interesting question. The arXiv version of the referenced paper is arxiv.org/abs/0911.4979, where equation 22 is (8). It may also be worth noting that this concept of "internal hom" is not really internal, but rather enrichment in $\mathcal{C}$, such that the given $\mathcal{C}$-module structure becomes copowering. $\endgroup$ – Tobias Fritz Nov 21 '17 at 14:43
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That equation is not correct. You should be suspicious because the definition of the $\mathcal{C}$-module category structure on $\mathcal{M} \boxtimes \mathcal{N}$ doesn't use the $\mathcal{C}$-module category structure on $\mathcal{N}$, only the one on $\mathcal{M}$. So how could the $\mathcal{C}$-enriched hom of $\mathcal{N}$ show up?

The correct equation is:

$$ \underline{Hom}_{\mathcal{M} \boxtimes \mathcal{N}}(M \boxtimes N, S \boxtimes T) \cong \underline{Hom}_{\mathcal{M}}(M , S) \otimes {Hom}_{\mathcal{N}}( N, T)$$

Here the second factor is just the usual vector space valued hom in $\mathcal{N}$ (and we are using that $\mathcal{C}$ is naturally a $Vect$-module category).

With this corrected equation you no longer get a contradiction because it is $Hom_{\mathcal{C}}(1, X^*) = 0$ which shows up instead of $\underline{Hom}_{\mathcal{C}}(1, X^*) = X^*$.

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  • $\begingroup$ Ok, I totally agree with you. Then in the proof of Lemma 3.21, the author argue as $$Hom_{\mathcal{N}}(N, \underline{Hom}_{\mathcal{M}}(M,S) \otimes T) = Hom_{\mathcal{C}}(1, \underline{Hom}_{\mathcal{N}}(N, \underline{Hom}_{\mathcal{M}}(M, S) \otimes T)) = Hom_{\mathcal{C}}(1, \underline{Hom}_{\mathcal{M} \boxtimes \mathcal{N}}(M \boxtimes N, S \boxtimes T)) $$. So this can not be true either! right? $\endgroup$ – heller Nov 22 '17 at 0:51
  • $\begingroup$ You are correct. Lemma 3.21 is actually false. Here is an easy counter example: C is the tensor category Vect[G] of G-graded vector spaces for the finite group G. The forgetful functor to Vect, which forgets the grading, is a tensor functor. This makes Vect a C-module category. We let M = Vect as a right C-module category and N=Vect as a left C-module category. Then clearly $M \boxtimes N \cong Vect$, but a nice exercise shows that $$Fun_{Vect[G]}(M^{op}, N) \cong Rep(G) $$ is the category of $G$-representations. So the functor "I" in lem 3.21 is definitely NOT an equivalence in general. $\endgroup$ – Chris Schommer-Pries Nov 22 '17 at 14:35
  • $\begingroup$ Note that although Lem 3.21 is false, Thm 3.20 is (almost) true. (The almost is that the definition of $M^{op}$ is subtly wrong.) See Corollary 3.4.11, Remark 3.4.12, and footnote 3 in our paper. We spent a while confused about this too! $\endgroup$ – Noah Snyder Nov 25 '17 at 18:06

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