t-structures on the derived category of finitely generated abelian groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:45:17Z http://mathoverflow.net/feeds/question/55930 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55930/t-structures-on-the-derived-category-of-finitely-generated-abelian-groups t-structures on the derived category of finitely generated abelian groups SGP 2011-02-19T00:00:07Z 2011-02-23T08:20:25Z <p>is it possible to explicitly parametrise all the t-structures on the derived category of finitely generated abelian groups?</p> http://mathoverflow.net/questions/55930/t-structures-on-the-derived-category-of-finitely-generated-abelian-groups/55967#55967 Answer by Sasha for t-structures on the derived category of finitely generated abelian groups Sasha 2011-02-19T08:33:47Z 2011-02-23T08:20:25Z <p>I guess the answer is the following. Take arbitrary subset $S$ of the set of all prime numbers. Let <code>$A_S = &lt;Z,\{Z/pZ\}_{p \not\in S},\{Z/pZ[-1]\}_{p \in S}&gt;$</code>. Then $A_S$ is a heart of the t-structure which is obtained by a simple tilting from the standard one. The claim is that any bounded t-structure is a shift of one of those.</p> <p>The main part of the proof is to show that all objects $Z/pZ$ and $Z$ are pure in any t-structure. For this use the fact that $Z$-modules have homological dimension 1 and that the objects from the heart of a t-structure don't have negative $Ext$'s between.</p> <p>EDIT: Let us show that $Z/pZ$ is pure. By shifting the t-structure we can assume that $Z/pZ \in D^{\le 0}$, but $Z/\pZ \not\in D^{\le 1}$. Then consider the triangle $$Z/pZ \to H^0(Z/pZ) \to \tau^{\le -1}(Z/pZ)[1].$$ Recall that any object in $D^b(Z)$ is a direct sum of $Z[i]$ and $Z/q^r Z[j]$ with $q$ prime. If there is a summand $M$ in $H^0(Z/pZ)$ such that $Hom(Z/pZ,M) = 0$ then $M$ is also a summand of $\tau^{\le -1}(Z/pZ)[1]$, hence $M \in D^{\le -1}[1] = D^{\le -2}$ and simultaneously $M \in D^0$ (since both subcategories are closed w.r.t. taking direct summands), so there is a nontrivial $Hom$ from $D^{\le -2}$ to $D^0$ which is impossible. </p> <p>So, all summands in $H^0(Z/pZ)$ should have a nontrivial $Hom$ from $Z/pZ$ into them. Thus we can only have $Z[1]$, $Z/p^rZ$ or $Z/p^rZ[1]$. Assume <code>$H^0(Z/pZ) = Z^a[1] + \oplus_i Z/p^{r_i}Z + \oplus_j Z/p^{s_j}Z[1]$</code> and the above triangle has form $$Z/pZ \to Z^a[1] + \oplus_i Z/p^{r_i}Z + \oplus_j Z/p^{s_j}Z[1] \to \tau^{\le -1}(Z/pZ)[1].$$ Now if we look at it in the standard t-structure we shall see that the third term has a $p$-torsion cohomology in degree 0 and a cohomology in degree $-1$ which is a sum of $Z^a$ and a $p$-torsion. Since any complex in $D(Z)$ is a sum of its cohomology, we conclude that $\tau^{\le -1}(Z/pZ)$ is a sum of $Z^a$, $p$-torsion and $p$-torsion shifted by $-1$. Once again, we cannot have same summands in $D^0$ and $D^{\le -1}$, hence the only possibility is that $H^0(Z/pZ) = Z/pZ$, so it is pure.</p> <p>A similar argument shows that $Z$ is also pure.</p> http://mathoverflow.net/questions/55930/t-structures-on-the-derived-category-of-finitely-generated-abelian-groups/56061#56061 Answer by Chris Brav for t-structures on the derived category of finitely generated abelian groups Chris Brav 2011-02-20T13:31:07Z 2011-02-20T13:31:07Z <p>The question and Sasha's answer can be generalized to a Noetherian ring $R$. The parametrization is in terms of functions from the integers to a specialization closed subset of $Spec R$. I do not have access to precise references at the moment, but see the paper 'Invariants of t-structures and classiﬁcation of nullity classes' by Don Stanley and further references therein.</p>