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.
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$\begingroup$ You should read what wikipedia has to say about the ring structure on Ext: en.wikipedia.org/wiki/…. It seems like whenever Yoneda products and products given by the "pick a resolution" method (see link) both exist, then they'll be the same (probably by an Eckman-Hilton argument?). In this case it should be graded-commutative. Right? $\endgroup$– Dylan WilsonCommented Jul 27, 2010 at 6:39
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$\begingroup$ This seems right; but it does not apply in several interesting cases (for example, the case of sheaves). $\endgroup$– AngeloCommented Jul 27, 2010 at 6:44
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3$\begingroup$ Not at all. For $R$ a local commutative ring and $A$ its residue field, $\mathrm{Ext}(A,A)$ can be computed by a resolution but it is ver seldom graded commutative. It is always the enveloping algebra of a graded Lie algebra and the Lie algebra may contain for instance large free subalgebras. For instance for $R=k[x_1,...,x_n]/(x_1,...,x_n)^2$ the Lie algebra is free on $n$ generators (in degree $1$). $\endgroup$– Torsten EkedahlCommented Jul 27, 2010 at 8:53
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2$\begingroup$ As Torsten points out, graded commutativity of Ext(AA) will not hold for every object. However, it will hold when A is the unit object in an appropriate monoidal category, by an Eckman-Hilton argument. An important example of this is Hochschild cohomology of a ring or dga R, which is defined as Ext(RR) in the category of R-bimodules. To show this is graded commutative, you should think of R as the unit in the monoidal derived category of bimodules and apply Eckmann-Hilton. Here, you should think of the derived category as being enriched over graded vector spaces by summing up all Exts, or $\endgroup$– Chris BravCommented Jul 27, 2010 at 10:19
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2$\begingroup$ ... even better, think of the derived category as enriched over cochain complexes (think of it as a dg category). Then you get that Hochschild cochains has two commuting products (Yoneda and tensor product), so is an $E_{2}$-algebra. Note that by Eckmann-Hilton, an $E_{2}$-algebra in graded vector spaces is just a graded commutative algebra. $\endgroup$– Chris BravCommented Jul 27, 2010 at 10:22
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2 Answers
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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 $2$.
answered Jul 28, 2010 at 8:52
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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 $0$.
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.
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 $A$?
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$\begingroup$ A simple module is cyclic (1-generated). $\endgroup$ Commented Jul 28, 2010 at 19:56
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2$\begingroup$ You need some (finiteness...) hypothesis on $R$ for his to be true: take $k=\mathbb C$, $R=\mathbb C(t)$, and $A$ the $1$-dimensional $R$-vector space. $\endgroup$ Commented Jul 29, 2010 at 21:10