Is the derived category of abelian groups a subcategory of the stable homotopy category? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:45:42Z http://mathoverflow.net/feeds/question/120235 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120235/is-the-derived-category-of-abelian-groups-a-subcategory-of-the-stable-homotopy-ca Is the derived category of abelian groups a subcategory of the stable homotopy category? name 2013-01-29T17:37:01Z 2013-01-29T23:37:21Z <p>An extension of the Dold-Kan equivalence gives an adjunction between the stable homotopy category and the (unbounded) derived category of abelian groups $SH \rightleftarrows D(Ab)$.</p> <p>Question 1: Is the right adjoint $D(Ab) \to SH$ faithful?</p> <p>Question 2: If not, is there a class of objects on which it is faithful (for example compact objects).</p> http://mathoverflow.net/questions/120235/is-the-derived-category-of-abelian-groups-a-subcategory-of-the-stable-homotopy-ca/120262#120262 Answer by Gregory Arone for Is the derived category of abelian groups a subcategory of the stable homotopy category? Gregory Arone 2013-01-29T21:24:56Z 2013-01-29T23:37:21Z <p>I think the answer to Question 1 is positive. Think of $SH$ as the homotopy category of modules over the sphere spectrum $S$. The category $D(Ab)$ is equivalent to the homotopy category of modules over the Eilenberg-Mac Lane spectrum $HZ$. Your adjunction is equivalent to the adjunction between $S$-modules and $HZ$ modules, where the right adjoint is pullback along the natural map of ring spectra $S\to HZ$, and the left adjoint is the functor $M\mapsto HZ\wedge M$.</p> <p>Your question is equivalent to this: given $HZ$-modules $M, N$, is the map</p> <ul> <li>$[M, N]_{HZ}\to [M, N]_S$</li> </ul> <p>injective? By adjunction </p> <p>$[M,N] = [HZ \wedge M, N]_{HZ}$ </p> <p>and the map * is induced by the map $HZ\wedge M \to M$. I claim that the last map is a split surjection in the homotopy category of $HZ$-modules. </p> <p><b> Edited </b> to account for Fernando's comment.</p> <p>Since every $HZ$-module splits (non-naturally) as a wedge sum of Eilenberg Maclane modules It is enough to check this claim when $M=HA$, in which case it is an easy calculation. The homotopy groups of $HZ\wedge HA$ are isomorphic to the homology groups of $HA$. By Huriewicz theorem, this is $A$ in dimension zero. Using the general splitting result again, it follows that $HA$ is a summand of $HZ\wedge HA$ in the category of $HZ$-modules.</p> <p>Therefore * is injective.</p> http://mathoverflow.net/questions/120235/is-the-derived-category-of-abelian-groups-a-subcategory-of-the-stable-homotopy-ca/120268#120268 Answer by Fernando Muro for Is the derived category of abelian groups a subcategory of the stable homotopy category? Fernando Muro 2013-01-29T22:42:50Z 2013-01-29T22:42:50Z <p>I've found the following somewhat intricate way of answering Q1 in the affirmative. Any complex in $D(Ab)$ quasi-isomorphic to a graded abelian group. Hence, it is enough to consider complexes concentrated in a single degree. Given an abelian group $A$ and $n\in\mathbb Z$, let $(A,n)$ be the abelian group $A$ concentrated in degree $n$. For simplicity, I will use the same notation for the Eilenberg-MacLane spectrum $\Sigma^nHA$. In the derived category we have, $$D(Ab)((A,n),(B,n))=\operatorname{Hom}(A,B),$$ $$D(Ab)((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$ $$D(Ab)((A,n),(B,m))=0\text{ otherwise}.$$ In the stable homotopy category we have the stable Eilenberg-MacLane groups $$SH((A,n),(B,m))=H^{m+k}(A,n+k;B),\quad k>>0.$$ It is well known, since E-ML's "On the groups..." (Annals) that $$SH((A,n),(B,n))=\operatorname{Hom}(A,B),$$ $$SH((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$ and that the functor $D(Ab)\rightarrow SH$ is the identity on the previous morphism sets. Hence we are done. The groups $SH((A,n),(B,m))$ are however non-trivial for $m>n+1$, in general.</p>