When is the derived category of representations of a finite poset equivalent to its opposite? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:01:09Z http://mathoverflow.net/feeds/question/108191 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108191/when-is-the-derived-category-of-representations-of-a-finite-poset-equivalent-to-i When is the derived category of representations of a finite poset equivalent to its opposite? Sam Gunningham 2012-09-26T19:32:50Z 2012-09-26T23:50:41Z <p>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. </p> <p>But when do we have an equivalence $D(K) \simeq D(K^{op})$?</p> <p>The kind of thing I have in mind is in this paper of Justin Curry: <a href="http://www.math.upenn.edu/~jucurry/papers/co_sheaf_dereq.pdf" rel="nofollow">http://www.math.upenn.edu/~jucurry/papers/co_sheaf_dereq.pdf</a></p> <p>When the poset comes from a finite cell complex, there is such a duality which interchanges the standard injectives with skyscrapers. </p> <p>Does this depend only on the poset, and can this always be done?</p> http://mathoverflow.net/questions/108191/when-is-the-derived-category-of-representations-of-a-finite-poset-equivalent-to-i/108198#108198 Answer by David Treumann for When is the derived category of representations of a finite poset equivalent to its opposite? David Treumann 2012-09-26T23:50:41Z 2012-09-26T23:50:41Z <p>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. $K = (0 &lt; 1 &lt; 2)$ and $K^{op} = (2 &lt; 1 &lt; 0)$ 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.</p> <p>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.</p> <p>For each $x \in K$, define $J_x:K \to \mathrm{Mod}_k$ by $$J_x(w) = \begin{array}{cc} k &amp; \text{if w \leq x}\\ 0 &amp; \text{otherwise} \end{array}$$ These are the indecomposable injective objects in the abelian category of functors $K \to \mathrm{Mod}_k$.</p> <p>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 $D(K)$.)</p> <p>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 $[x &lt; y]$. So a necessary condition is for this $\partial N$ to be a homology sphere. </p> <p>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.</p> <p>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.</p>