# Cohomology of Double Complex (extreme case)

Let $K^{p,q}$ be a double complex with ascending indices, i.e. the differentials are $d':K^{p,q} \rightarrow K^{p+1,q}$ and $d'':K^{p,q} \rightarrow K^{p,q+1}$. Suppose that $K^{p,q}=0$ if $p<0$ or $q<0$ and that $H^n(K^{p \cdot})=0$ when $n>0$ for any $p$. Define $X^p=ker(K^{p,0} \rightarrow K^{p,1})$. Then $X$ is a complex with differential induced by $d'$. How can we show that $H^n(K) = H^n(X)$?

I am interested in an argument that does not involve any spectral sequence theory. Reference: Matsumura, Commutative Ring Theory, p. 277.

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Consider the inclusion $X\to K$ and let $L$ be the quotient, so that we have a short exact sequence of first-quadrant double complexes $$0\to X\to K\to L\to 0$$

Now show by hand that $L$ is exact. Suppose $n>0$ and we have an $n$-cocycle in $L$, which is a sequence $(x_0,\dots,x_n)$ with $x_i\in L^{i,n-1}$ for each $i$ such that $d'x_i=d''x_{i+1}$ if $0\leq i<n$, $d''x_0=0$ and $d'x_n=0$. In particular, $x_0$ is a cocycle for the vertical differential. The hypothesis on exactness implies there is a $y_0$ such that $d''y_0=x_0$. Then $d''(d'y_0-x_1)=d'd''y_0-d''x_1=d'x_0-d''x_1=0$, and $d'y_0-x_1$ is a cocycle for the vertical differental. Exactness implies there is a $y_1$ such that $d''y_1=d'y_0-x_1$, so that $x_1=d''y_1-d'y_0$.

In this way we go down the $(n-1)$th diagonal. When we reach the last step, though, we have to use the fact that $L$ is constructed in a specific way.

It is the sort of boring argument that is best done in detail by no one other that by yourself!

(And which spectral sequences perfectly package...)

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The useful property of $L$ is that its columns are exact, and that is enough to do this. –  Mariano Suárez-Alvarez Feb 18 '13 at 17:08
Thanks a lot, in my efforts i had not considered taking the quotient :) I would appreciate some clarification: the $n$-cocycle of $L$ is just an element of $L^n$, right? Now $L^n = K^n / Kerd''_{n,0}$. But $Kerd''_{n,0}$ is just a submodule of $K^{n,0}$. So i don't see how we get this relation $d'x_i = d''x_{i+1}$ on the entries of the cocycle. What am i missing? Also, when you say that $L$ is exact, what exactly do you mean? –  Manos Feb 18 '13 at 17:19
«$L$ is exact» means «the associated simple complex is exact». As for your 1st question: an $n$-cochain $x$ in the complex $L$ is an element of $L^{0,n}\oplus L^{1,n-1}\oplus\dots\oplus L^{n,0}$, so can be written as a tuple $x=(x_0,\dots,x_n)$. Now write what exactly $dx=0$ means. –  Mariano Suárez-Alvarez Feb 18 '13 at 17:49
Professor, in studying your answer i have the following question: the map $L^{0,n} \rightarrow L^{0,n+1}$ is given by $x+Kerd''_{0,n} \mapsto d''_{0,n}x + Kerd''_{0,n+1}$, right? Now since $H^q(K^{p \cdot})=0$ for any $q>0$, this means that $Kerd''_{0,n+1} = Imd''_{0,n}$. Hence $L^{0,n} \rightarrow L^{0,n+1}$ is the zero map. What am i doing wrong? –  Manos Feb 19 '13 at 17:03
Your hypothesis is that $H^q(K^{p\bullet})$ is zero only when $q>0$. Be careful :D –  Mariano Suárez-Alvarez Feb 19 '13 at 18:14