Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I got a question about resolutions of complexes, I just wanted to make sure I'm looking at them the right way. Let

$\cdots \rightarrow P^{-1} \rightarrow P^{0} \rightarrow X \rightarrow 0 \rightarrow \cdots$

be a projective resolution of $M$. $M$ can be considered as a complex

$\cdots \rightarrow 0 \rightarrow 0 \rightarrow X \rightarrow 0 \rightarrow 0 \rightarrow \cdots$,

and the resolution can be considered as a quasi-isomorphism of complexes

enter image description here

Now from what I know the resolution of a 'complex' is a Cartan-Eilenberg resolution, that is, a bicomplex:

enter image description here

But to be consistent with notation, shouldn't the resolution actually be considered the quasi-isomorphism $X^{\bullet} \rightarrow (SP)^{\bullet}$?

enter image description here,

where $(SP)^{n} = \oplus_{i + j = n}P^{i,j}$ is the diagonal complex. I just wanted to be clear about this.

share|improve this question
    
I guess it is context dependent, in the language of (triangulated) homotopy category, a projective resolution is, similar to what you said, a (right bounded) complex of projectives $P^\bullet$ with a quasi-isomorphism $P^\bullet \to X^\bullet$. In fact, the bicomplex resolution can be collapsed into such projective resolution by taking $P^n = \bigoplus_{i+j=n} P^{i,j}$ with (co)boundary map $d_h + d_v$, where $d_h$ and $d_v$ are differential in horizontal and vertical direction respectively (note, the summation needs to respect the sign convention at each component). –  Aaron Jan 1 '13 at 22:59
    
I hadn't seen this, thank you so much @Aaron. –  Richard Jennings Mar 1 '13 at 1:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.