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Consider a $\mathbb{Z}$-graded chain complex $A^{\bullet}$, I know that a bounded below complex is one such that $A^i = 0$ for $i$ sufficiently small, and a bounded above complex is one such that $A^i = 0$ for sufficiently large $i$, every bounded above (or below) complex has a Cartan-Eilenberg resolution and every unbounded complex has a resolution too (according to Spaltenstein and other authors).

I wanted to ask:

  • Does this apply to unbounded complexes that are not $\mathbb{Z}$-graded? Or finite complexes? Arbitrary chain complexes? An $\mathbb{N}$-graded complex for example? Do they too have resolutions?

I just wanted to know as I don't think I know the reason as to why boundedness is such a key thing, I know that the classical construction of resolutions for complexes works only for bounded complexes but I don't know why or where it fails for unbounded complexes.

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$\mathbb N$-graded complexes are bounded below. I don't understand the rest of questions, eg what do you mean by 'arbitrary chain complex'? You don't seem to mean unbounded, since you already know that case. –  Fernando Muro Feb 16 '13 at 18:17
    
1) What do you want to know that's not answered in Spaltenstein? 2) What is an example of a complex that is not ${\mathbb Z}$ - graded? –  Steven Landsburg Feb 16 '13 at 18:24
    
An example could be a complex starting at $0$: $A^0 \rightarrow A^1 \rightarrow A^2 \rightarrow A^3 \rightarrow \cdots$, with NO ZEROS to the left, where $A^0$ is not zero. –  Samuel Mf Feb 16 '13 at 18:59
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Fernando's point is that if you add the zeroes, you can construct an injective resolution for that, which also has zeroes on the left, which you can then remove. –  Mariano Suárez-Alvarez Feb 16 '13 at 20:51
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(Keeping a calmed tone, by the way, goes a long way to getting a useful MO experience) –  Mariano Suárez-Alvarez Feb 16 '13 at 20:55
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1 Answer

up vote 1 down vote accepted

Start with a (possibly bounded) sequence of maps satisfying $dd=0$. Per Fernando's comment, you can always add an infinite number of zeroes on the left and/or right to create a ${\mathbb Z}$-graded complex. You can then build a Cartan-Eilenberg resolution of that ${\mathbb Z}$- graded complex. Your question (I think) is whether this Cartan-Eilenberg resolution can be truncated to give a Cartan-Eilenberg-like resolution of your original sequence (i.e. a resolution whose coboundaries and cohomology are resolutions of your original sequence's coboundaries and cohomology). The answer is yes, because (thinking of your original sequence as a row) the C-E construction puts a column of zeros wherever your row has a zero --- and throwing away columns of zeros can't change the coboundaries and cohomologies of the rows.

(This assumes that you define "cohomology" in the obvious way at the beginning and end of your sequence --- you haven't actually told us what your definition is.)

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Thank you! That's so obvious, I got so confused there, that's what I thought, I had this sequence of objects and didn't think about the grading, I just saw that there were no $0's$ to the left or right and thought of it as an unbounded complex, and that's why I got confused when I was trying to use Spaltenstein's construction, and thought at first 'is this even a complex?', didn't see that you could just do that and that's why I asked, but now it's clear. –  Samuel Mf Feb 17 '13 at 15:25
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