If I have a finite partially ordered set $K$, I can look at its derived category of finite dimensional representations $D(K)$. Note that $D(K^{op}) \simeq D(K)^{op}$ by linear duality.

But when do we have an equivalence $D(K) \simeq D(K^{op})$?

The kind of thing I have in mind is in this paper of Justin Curry: http://www.math.upenn.edu/~jucurry/papers/co_sheaf_dereq.pdf

When the poset comes from a finite cell complex, there is such a duality which interchanges the standard injectives with skyscrapers.

Does this depend only on the poset, and can this always be done?

  • $\begingroup$ The articles by Sefi Ladkani may be relevant, for instance arxiv.org/pdf/0705.0946v2.pdf $\endgroup$
    – F. C.
    Sep 26, 2012 at 20:25
  • $\begingroup$ What could it possibly depend on beside the poset? $\endgroup$ Sep 26, 2012 at 21:23
  • $\begingroup$ @Qiaochu: The fact that the poset came from a cell complex, giving the added data of dimensions of cells, orientations, ... $\endgroup$ Sep 26, 2012 at 21:30
  • $\begingroup$ @Sam: okay, I think I misinterpreted what you meant by "this." I interpreted it to mean "the existence of an equivalence $D(K) \cong D(K^{op})$." $\endgroup$ Sep 26, 2012 at 22:49

1 Answer 1


I don't know a very general answer. Your duality on cell complexes resembles Verdier duality and has a local nature, but some of these equivalences aren't like that. E.g. $K = (0 < 1 < 2)$ and $K^{op} = (2 < 1 < 0)$ are isomorphic as posets and so we get $D(K) = D(K^{op})$ from that, but I am pretty sure $K$ doesn't have a Verdier dualizing sheaf.

But here is an answer to a narrower question: when is there an equivalence $D(K) = D(K^{op})$ that is similar to the one on regular cell complexes? I mean an equivalence that is Verdier-like in the sense that it takes the standard injectives on $K$ to the skyscrapers on $K^{op}$, up to a shift.

For each $x \in K$, define $J_x:K \to \mathrm{Mod}_k$ by $$ J_x(w) = \begin{array}{cc} k & \text{if $w \leq x$}\\\ 0 & \text{otherwise} \end{array} $$ These are the indecomposable injective objects in the abelian category of functors $K \to \mathrm{Mod}_k$.

The $J_x$ have a simple Hom pattern: $\mathrm{Hom}(J_x,J_y) = k$ if $x \geq y$. All other Homs (and Exts) vanish. (They form an "exceptional collection" in $D(K)$.)

The simple objects are the skyscrapers $\delta_x$. You can compute the Homs and Exts between $\delta_x$ and $\delta_y$ by writing down an injective resolution of $\delta_y$ whose $p$th term is $$ \bigoplus_{y = y_0 \geq y_1 \geq y_2 \cdots \geq y_p} J_{y_p} $$ The differentials have degree $+1$. Then $\mathrm{Hom}(\delta_x,\text{that injective resolution})$ is a complex whose $p$th term is $$ \bigoplus_{y = y_0 \geq y_1 \geq y_2 \cdots \geq y_p = x} k $$ This is the cochain complex that computes something like the relative cohomology $H^*(N,\partial N)$ where $N$ is the nerve of the interval $ \{p \in K \mid x \leq p \leq y\} $ and $\partial N$ is the subcomplex of simplices that don't contain the edge $[x < y]$. So a necessary condition is for this $\partial N$ to be a homology sphere.

I think this condition is sufficient too, except you have to worry a little bit about the dimensions of those spheres. You have to be able to choose integers $d(x)$ for each $x \in K$ so that the dimension of that homology sphere is $d(y) - d(x)$ or something.

Somebody once told me that this condition on the intervals in posets has a standard name, maybe "Gorenstein star posets" but I am not sure I am remembering that right.

  • $\begingroup$ Awesome! Now I remember that you might have explained this to me before, but I had forgotten... $\endgroup$ Sep 27, 2012 at 1:57

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