19
$\begingroup$

Recall that two $k$-algebras $A, B$ are Morita equivalent iff their categories of left modules are equivalent. However, this relation turns out to be rather fine and one introduces a coarser equivalence relation of derived Morita equivalence by using (bounded) derived categories of modules, along with their triangulated structure.

(Note that I am no expert in this matters and I was mostly exposed to this viewpoint through algebraic geometry, where one instead works with bounded derived categories of coherent sheaves on a variety.)

However, as I understand it, the derived category of an algebra arises as a homotopy category of the stable model category of chain complexes. The latter may be seen as presenting a homotopy theory (ie. an $(\infty, 1)$-category), for example through the process of simplicial localization (and probably also some more direct, dg-theoretic methods?).

One could then say that two algebras are "higher derived Morita equivalent" if their $(\infty, 1)$-categories of (bounded?) complexes are equivalent as higher categories. The question is as follows: What can we say about this new equivalence relation? How far is this relation from derived Morita equivalence? How far is it from ordinary Morita equivalence?

I have no intuition about this and I can imagine answers that completely equate "higher derived Morita equivalence" with either of these two, although it would be probably most interesting if it was somewhere between them.

Note that one can imagine that somehow the derived category of an algebra remembers all the "higher homotopy", as it happens to be the case for some other homotopy categories of stable model categories. For example, in "The stable homotopy category is rigid" by S. Schwede it is proven that any stable model category $\mathcal{C}$ that satifies $ho(\mathcal{C}) \simeq \mathcal{SHC}$ (as triangulated categories, where the latter is the stable homotopy category) is in fact Quillen equivalent to model category of spectra, so they present the same homotopy theory.

I ask the question since I am currently studying higher categories and this led me to wonder what is their possible strength as invariants of other mathematical objects.

$\endgroup$

2 Answers 2

24
$\begingroup$

Derived Morita equivalence is the same as higher derived Morita equivalence. Clearly, $\Leftarrow$ is obvious, and $\Rightarrow$ follows from Theorem 2.6 in Dugger-Shipley's 'K-theory and derived equivalences' Duke Math. J. 124 (2004), no.3, 587--617. This is surprising at a first glance, it follows from the fact that algebras are concentrated in degree $0$. The analogous result for DG-algebras is false, compare 'A curious example of triangulated-equivalent model categories which are not Quillen equivalent' Algebraic and Geomtric Topology 9 (2009), no. 1, 135-166, by the same authors. There're still some interesting open questions connected to this, though.

$\endgroup$
0
17
$\begingroup$

Fernando's answer is excellent, but I can't resist mentioning what is perhaps the simplest counterexample to a generalization to your question. As Fernando says, there are counterexamples if you generalize from rings concentrated in degree 0 to dg-algebras. These examples are somewhat complicated, but if you generalize further to $A_\infty$ ring spectra, there is a very easy example.

Fix a prime $p$ and an integer $n>0$ and consider the Morava K-theory spectrum $K(n)$, which can be given an $A_\infty$ structure. The homotopy groups $\pi_*K(n)=\mathbb{F}_p[v_n^{\pm1}]$ are a graded field, and it follows that every $K(n)$-module is free (a wedge of suspensions of $K(n)$). In particular, the homotopy category of $K(n)$-modules is semisimple, and is actually equivalent (as a triangulated category) to the category of graded $\mathbb{F}_p[v_n^{\pm1}]$-vector spaces. This category is also equivalent to the homotopy category of $H\mathbb{F}_p[v_n^{\pm1}]$-modules (or equivalently, dg-modules over the graded ring $\mathbb{F}_p[v_n^{\pm1}]$). However, the corresponding $(\infty,1)$-categories are not equivalent. Indeed, the space of endomorphisms of a simple $K(n)$-module is $\Omega^\infty K(n)$, while the space of endomorphisms of a simple $H\mathbb{F}_p[v_n^{\pm1}]$-module is $\Omega^\infty H\mathbb{F}_p[v_n^{\pm1}]\simeq \prod_i K(\mathbb{F}_p,2i(p^n-1))$. These spaces have isomorphic homotopy groups, but are otherwise quite far from being homotopy equivalent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.