When is the Yoneda product graded commutative? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:34:27Z http://mathoverflow.net/feeds/question/33476 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33476/when-is-the-yoneda-product-graded-commutative When is the Yoneda product graded commutative? Tom Price 2010-07-27T05:57:40Z 2010-07-29T20:45:08Z <p>Sometimes, given an object A in an Abelian category, the Yoneda product on Ext(A, A) is graded-commutative, for example in cases where it coincides with the cup-product in singular cohomology. Are there any nice theorems about when the Yoneda product is graded-commutative in general? Thanks in advance.</p> http://mathoverflow.net/questions/33476/when-is-the-yoneda-product-graded-commutative/33633#33633 Answer by Torsten Ekedahl for When is the Yoneda product graded commutative? Torsten Ekedahl 2010-07-28T08:52:53Z 2010-07-28T08:52:53Z <p>I move this to a more proper answer to discuss some subtle points of the question. The Eckman-Hilton argument (or more concrete calculations) shows, as Chris points out, that $\mathrm{Ext}(A,A)$ is commutative when $A$ is the unit for a monoidal category. The subtleties appear when we consider for instance the ring $R=k[x]/(x^2)$ for $k$ a field and $A=k$. Then $A$ has a uniform resolution $\dots\xrightarrow{x}R\xrightarrow{x}R\xrightarrow{x}R\to k\to 0$ giving $\mathrm{Ext}^i(A,A)=k$ for all $i$. Using the definition of the Yoneda product in terms of maps of resolutions we get that $\mathrm{Ext}(A,A)$ is the polynomial ring on $\mathrm{Ext}^1(A,A)$. This is graded commutative only when the characteristic is $2$ (and then it is not graded commutative in the strict sense of the square of odd degree elements being zero). However, it is exactly in characteristic $2$ that $R$ is the affine algebra of a finite group scheme (with $x\mapsto x\otimes1+1\otimes x$ as coproduct) with $k$ the unit for the associated monoidal structure on the category of $R$-modules. Hence we have a monoidal reason for the $\mathrm{Ext}$-algebra being graded commutative in characteristic. On the other hand we have a uniform description of the $\mathrm{Ext}$-algebra in all characteristics which just happens to fulfil the definition of being graded commutative in characteristic $2$.</p> http://mathoverflow.net/questions/33476/when-is-the-yoneda-product-graded-commutative/33695#33695 Answer by Graham Denham for When is the Yoneda product graded commutative? Graham Denham 2010-07-28T18:30:43Z 2010-07-28T18:30:43Z <p>Another starting point is to think of ${\rm Ext}(A,A)$ as the derived endomorphism ring of the object $A$ and recall Schur's lemma. If $A$ is a finitely-generated simple module over a ring $R$, then ${\rm Hom}_R(A,A)$ is a division algebra. For example, if $R$ is a $k$-algebra over an algebraically closed field $k$, then ${\rm Hom}_R(A,A)$ is isomorphic to $k$ (so, in particular, it is commutative.) Via Freyd-Mitchell embedding, this should give some idea what to expect in degree $0$.</p> <p>Going back the question, then, the examples one might have in mind are categories of modules over a group ring or enveloping algebra of a graded Lie algebra: in these cases, ${\rm Ext}(k,k)$ is group- or Lie algebra cohomology, respectively, and has a graded-commutative cup product, where $k$ is the trivial module.</p> <p>Perhaps there is a suitable "semisimplicity" hypothesis one could impose on the category so that ${\rm Ext}(A,A)$ is graded-commutative for all simple objects $A$?</p>