# Question on resolutions for arbitrary chain complexes.

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).

• 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. – Louis A Feb 16 '13 at 18:59
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
(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

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.
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. – Louis A Feb 17 '13 at 15:25