Let $A$ and $B$ be $k$-algebras. And for convenience let's say $k$ is a field and both $A$ and $B$ are finite-dimensional.

A well known theorem independently discovered by Eilenberg and Watts states that every $k$-linear, right exact, cocontinuous functor $F: A\mathsf{-Mod}\to B\mathsf{-Mod}$ is of the form $M\otimes_A-$ for some bimodule ${}_B M_A$ (in fact $M=F(A)$).

A theorem of Rickard (with some refinements by other people like Keller) also states that if there is an equivalence $D^b(A) \to D^b(B)$ there is also an equivalence of the form $X\otimes_A^\mathbb{L}-$ for some $X\in D^b(B\otimes_k A^{op})$.

A reasonable question to ask is then:

Is every equivalence $D^b(A) \to D^b(B)$ itself of the form $X\otimes_A^{\mathbb{L}}-$ for some bounded complex of bimodules $X$?

or more generally:

Is every exact functor $D^b(A) \to D^b(B)$ of triangulated categories which commutes with all direct sums (plus maybe some other reasonable condition) of the form $X\otimes_A^\mathbb{L}-$ for some bounded complex of bimodules $X$ ?

  • 3
    $\begingroup$ Let me mention that the result you want is true if, instead of an equivalence of triangulated categories you ask for an equivalence of stable ∞-categories (this is a theorem by Schwede and Shipley, although they talk about Quillen equivalences of stable model categories). There are probably also dg variants if you're more comfortable with those. $\endgroup$ – Denis Nardin Jun 11 '18 at 15:58
  • 1
    $\begingroup$ Since I'm not fluent in $\infty$-categorial language, what does that mean for the category of chain complexes (which I guess is the appropriate $\infty$-category here) ? I would guess it is something like "everything I want holds up to homotopy (of homotopies of ...)", say "there ex. transformations $F\leftrightarrow X\otimes^L-$ that are not necessarily natural themselves, but make the appropriate diagram commute up to homotopy, and are not necessarily inverse to each other, but the compositions are homotopic to identities". Would that be true? $\endgroup$ – Johannes Hahn Jun 11 '18 at 16:03
  • 1
    $\begingroup$ @JohannesHahn The correct statement would be the following. You have $\infty$-categories $\mathcal D^b(A)$ and $\mathcal D^b(B)$, whose homotopy categories are $D^b(A)$ and $D^b(B)$, respectively. Suppose your equivalence $D^b(A) \to D^b(B)$ is actually induced by an equivalence $\mathcal D^b(A) \to \mathcal D^b(B)$. Then it is of the form you want, i.e. it is given by tensoring with a bounded complex. $\endgroup$ – Dan Petersen Jun 11 '18 at 16:57
  • $\begingroup$ Concretely, this means that one needs to prove that the functor $F$ can be upgraded to a construction which is in addition "homotopy coherent". $\endgroup$ – Dan Petersen Jun 11 '18 at 17:02
  • 1
    $\begingroup$ @JohannesHahn The point is that an equivalence of stable ∞-cats (/dg cats/your favorite enhancement of $D^b(-)$) is more data than just an equivalence of triangulated cats. As Dan Peterson is saying, the theorem is saying that if your equivalence can be lifted to a functor of ∞-cats (/...) then it is of the form you want. $\endgroup$ – Denis Nardin Jun 11 '18 at 17:03

Your first question seems to be a famous open question due to Rickard, see https://academic.oup.com/jlms/article-abstract/s2-43/1/37/888935?redirectedFrom=PDF .

It is known in several special cases, such as for hereditary algebras, see http://home.ustc.edu.cn/~xwchen/Personal%20Papers/A%20note%20on%20standard%20equivalence.pdf .

A very recent approach can be found here: https://www.sciencedirect.com/science/article/pii/S000187081830207X .

  • 1
    $\begingroup$ In particular, I find it worth pointing out that Rickard showed that every equivalence is of this form on objects. $\endgroup$ – Julian Kuelshammer Jun 11 '18 at 16:43

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.