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Sasha Patotski
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Where toshould I search for resolutions?

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Sasha Patotski
  • 1.3k
  • 1
  • 11
  • 21

Where to search for resolutions?

In my research, to test some conjectures or just to illustrate some facts, I often need to compute some explicit examples of derived functors (in the sence of Quillen's model categories). Mainly I work in the category of differential non-negatively graded algebras over some field $k$ of characteristic zero (denote this category $DGA^+_k$). I will regard that differentials have degree $-1$.

It seems like the main problem when computing these functors is to find an appropriate cofibrant resolution. To be more precise, for an algebra $A\in DGA^+_k$ a map $\phi\colon R\twoheadrightarrow A$ is a cofibrant resolution if $\phi$ is surjective in each degree quasi-isomorphism and $R$ is semi-free. Here $R\in DGA^+_k$ is called semi-free if it's underlying graded algebra $R_\#$ is free.

It's not a big deal to construct some resolution. For example, the cobar-bar adjunction gives a resolution $\Omega B(A)\twoheadrightarrow A$. But this resolution is "huge". So the problem is how to construct resolutions that are "small".

I know a couple of examples. For exapmle, the (commutative) polynomial algebra $A=k[x,y]$ has a resolution in $DGA^+_k$ of the form $R=k\langle x,y,t\rangle$, where $\deg x=\deg y=0$, $\deg t=1$ and $d(t)=xy-yx=[x,y]$.

Algebra $A=k[x,y,z]$ has resolution of the form $R=\langle x,y,z;\xi,\lambda,\theta;t\rangle$ with $\deg x,y,z=0$, $\deg \xi,\lambda,\theta=1$ and $\deg t =2$. The differential is defined by $d(\xi)=[y,z],d(\lambda)=[x,y],d(\theta)=[z,x]$ and $d(t)=[x,\xi]+[y,\theta]+[z,\lambda]$.

So my questions are the following.

  1. Do you know any other nice examples of algebras with simple resolutions? What are resolutions of symmetric algebras, matrix algebras?

  2. How can you come up with such nice resolutions? I understand that probably there is no general recipe, but maybe there are some techniques, or hints, how to do that.

I hope this question is appropriate to ask here.

Thanks a lot for your help!