4
$\begingroup$

Let $R$ be a commutative ring and let $\mathrm{Mod}_R$ be the category of (left) $R$-modules.

Question: Is it true that the categories $\mathcal{Z}(\mathrm{Mod}_R)$ and $\mathrm{Mod}_R$ are equivalent?

I have read the previous claim in a couple of places but without any proof or reference, and I cannot really find one. In general if $\mathcal{C}$ is a braided category, then there is a fully faithful functor $\mathcal{C} \hookrightarrow \mathcal{Z}(\mathcal{C})$, but in general this is not an equivalence. Furthermore, it seems to me that the functor $\mathrm{Mod}_R \hookrightarrow \mathcal{Z}(\mathrm{Mod}_R)$ has little chance to be an equivalence since $\mathrm{Mod}_R$ is symmetric monoidal and the center is (highly) braided.

Improve this question
$\endgroup$
2
  • $\begingroup$ I believe it is indeed true that the Drinfeld center of the abelian category $Mod_R$ is equivalent to its Drinfeld center (I don't know a reference off the top of my head though). To see the non-triviality of the braiding one has to look at the analogous derived construction. $\endgroup$
    – Sam Gunningham
    Commented Mar 7, 2021 at 15:30
  • 5
    $\begingroup$ The relevant derived statement is that $Z(D(X)) = D(LX)$ where $D(X)$ is the dg derived category of quasi-coherent sheaves on a scheme $X$ and $LX$ is its derived loop space. If $X=Spec(R)$, $D(LX)=Sym(\mathbb{L}_R[1])-dgmod$ where $\mathbb{L}_R$ is the cotangent complex. Note that the heart of the $t$-structure on this category is just the abelian category $R-mod$. See e.g. arxiv.org/pdf/0805.0157.pdf for a discussion of this sort of thing. $\endgroup$
    – Sam Gunningham
    Commented Mar 7, 2021 at 15:32

1 Answer 1

Reset to default
6
$\begingroup$

Let $(X,\Phi)$ be an object of the Drinfeld center.

We'd like to prove that $(X,\Phi)$ is isomorphic to $(X,$ standard symmetry isomorphism $)$, which would prove the equivalence as you say we already have fully faithfulness.

Compose the isomorphism $\Phi$ with the inverse of the standard symmetry isomorphism, to get a natural map $X\otimes_R - \to X\otimes_R-$.

Now, evaluating in $R$ gives an automorphism $f:X\to X$, and, by naturality and the fact that $X\otimes_R Y$ is generated by pure tensors, it follows that in general, the morphism $X\otimes_R Y\to X\otimes_R Y$ is given by $f\otimes_R \mathrm{id}_Y$

(here I'm using something specific to $\mathrm{Mod}_R$, I'm not sure how general it could actually be in the context of symmetric monoidal categories)

In particular, this means that $\Phi $ is actually given by $x\otimes y\mapsto y\otimes f(x)$. Moreover, note that the "associativity" of $\Phi$ imposes $f = f^2$, and because $f$ is an automorphism, it implies $f= \mathrm{id}_X$.

Therefore $\Phi$ is actually the standard symmetry isomorphism.

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ As pointed out in the comments below the question, if you go to a more homotopical setting, this will be related to topological Hochschild homology of $R$ with certain coefficients; namely if you look at the proof above, we're essentially looking at $X$ viewed as an $R$-bimodule and maps $X\to X$; for $X= R$ this is exactly topological Hochschild cohomology of $R$; in general this will be something like $Map_R(THH(R;X), X)$ or $Map_R(X,THC(R;X))$ $\endgroup$
    – Maxime Ramzi
    Commented Mar 7, 2021 at 16:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .