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In an abelian category $\mathcal A$ with enough projectives, we have the Yoneda pairing $$\operatorname{Ext}^p_{\mathcal A}(Y,Z)\otimes \operatorname{Ext}_{\mathcal A}^q(X,Y)\longrightarrow \operatorname{Ext}_{\mathcal A}^{p+q}(X,Z),$$ $$a\otimes b \mapsto a\smile b.$$ This pairing induces a graded algebra structure on $\operatorname{Ext}_{\mathcal A}^*(X,X)$ for any object $X$.

I'm interested in finding conditions on a given class $b\in \operatorname{Ext}_{\mathcal A}(X,X)^q$, $q>0$, ensuring that $$-\smile b\colon \operatorname{Ext}_{\mathcal A}^p(X,X)\longrightarrow \operatorname{Ext}_{\mathcal A}^{p+q}(X,X)$$ is an isomorphism for $p>0$ and surjective for $p=0$. I would appreciate (but I'm not restricted to) conditions related to a $q$-fold extension $$X\hookrightarrow P_q\rightarrow \cdots\rightarrow P_0\twoheadrightarrow X$$ representing $b$. Notice that this can only happen if $X$ is projective (in this case it's trivial) or of infinite projective dimension.

A sufficient condition would be that we can take all $P_i$ projective, $i=1,\dots,q$. I don't know if this condition is necessary. Of course, we can always take $P_i$ projective for $i=1,\dots, q-1$ (not including $q$), but what would then be (necessary and sufficient, if possible) conditions on $P_q$ so that the above property holds? Would it be enough that $P_q$ be of finite projective dimension?

I'm mostly interested in $\mathcal A=$ the category of bimodules over a $k$-algebra $R$ and $X=R$ as a bimodule, so $\operatorname{Ext}_{\mathcal A}^*(X,X)$ would be the Hochschild cohomology of $R$ if it is $k$-projective.

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Is $n$ the same as $q$? –  Sasha Feb 25 at 19:31
    
Yes, thanks, edited. –  Fernando Muro Feb 25 at 22:02
    
If you've already chosen $P_i$ projective for $i < q$, a necessary and sufficient condition is that $Ext(P_q,X) = 0$. –  Tyler Lawson Feb 26 at 4:47
    
@TylerLawson thanks! I think I can see that $\operatorname{Ext}_{\mathcal A}^n(P_q,X)=0$ for $n\geq q$ is sufficient. I can't even see necessity. Could you please elaborate a little bit your comment? I'd be glad to accept it as an answer. –  Fernando Muro Feb 26 at 8:46
    
@TylerLawson, sorry, I now see what you mean, I was so confused that was taking as covariant what is contravariant, still, I can just see sufficiency. –  Fernando Muro Feb 26 at 9:09

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