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Suppose that I have given you a bigraded vector space $V = \bigoplus_{i,j} V_{i,j}$. The first grading is a "homological" $\mathbb Z$-grading, and the second is an independent $\mathbb Z$-grading. Furthermore, I will supply a differential $\partial : V \to V$ (i.e. $\partial^2 = 0$). With respect to the homological grading, the differential does what you expect, lowering homological degree by $1$:
$$\forall i,\quad \partial : \bigoplus_j V_{i,j} \to \bigoplus_j V_{i-1,j}$$ You probably would expect that the differential preserves (say) the independent grading, but alas, with respect to this independent grading the differential is only filtered, which is to say "nonincreasing". Actually, I will guarantee to you a little bit more. Namely, there is some bound $b \in \lbrace 0,1,2,\dots\rbrace$ such that the differential $\partial$ does not decrease the "$j$"-degree by more than $b$: $$ \forall i\forall m, \quad \partial : V_{i,m} \to \bigoplus_{m-b \leq j \leq m} V_{i-1,j} $$

Finally, suppose that I guarantee to you that $\partial$ is exact, meaning that $\operatorname{im}(\partial) = \operatorname{ker}(\partial)$, so that all homology groups vanish.

My question is: is it possible to bound the degrees of a homotopy witnessing this exactness?

More specifically, recall that the best way to prove that a differential $\partial$ is exact is to demonstrate a homotopy $h : V \to V$ such that $\partial h + h\partial = 0$. As always, I will demand that $h$ raises the homological "$i$" degree by $1$.

Can such an $h$ be found which does not increase the "$j$" degree by more than, say, $b$?

If $b = 0$, then since I am working in vector spaces the answer is YES. Specifically, if $b=0$, then I can break my complex into a direct sum indexed by $j$, and for each $j$ the subcomplex $\partial : V_{\bullet,j} \to V_{\bullet-1,j}$ is exact, and therefore has a contracting homotopy.

Note that I do need to allow $h$ to increase degree in general. For example, if the only non-zero terms in $V$ are $$ V_{1,1} = V_{0,0} = \mathbb k = \text{ground field} $$ and $\partial$ is an isomorphism $V_{1,1} \to V_{0,0}$, then the only contracting homotopy is the inverse isomorphism $h : V_{0,0} \to V_{1,1}$, which increases "$j$"-degree by $1$.

On the other hand, $h$ cannot be bounded below. For example, consider the complex in which $$ V_{1,j} = V_{0,j} = \mathbb k \text{ for } 0 \leq j \leq n \text{ and } 0 \text{ otherwise} $$ and $$ \partial = \begin{pmatrix} 1 & -1 & \\ & 1 & -1 & \\ && 1 & \ddots & \\ &&& \ddots & -1 \\ &&&& 1 \end{pmatrix} $$ then $$ h = \partial^{-1} = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ & 1 & 1 & & 1\\ && 1 & \ddots & \vdots \\ &&& \ddots & 1 \\ &&&& 1 \end{pmatrix} $$

Similarly, if $\partial$ is bounded above, but not by $0$, then then answer is NO.

Finally, note that since $\partial$ is non-increasing, its $j$-degree-preserving piece is also a differential on $V$; specifically, it is the associated graded differential $\operatorname{gr}(\partial)$. If $\operatorname{gr}(\partial)$ is exact, then so is $\partial$, and moreover the answer to my question is YES, and in fact you can find an $h$ which is non-increasing. The example above of the invertible upper-triangular matrix is typical.

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1 Answer 1

up vote 4 down vote accepted

No. As you remark, in the special case in which the whole thing is concentrated in homological degrees $0$ and $1$ exactness means that the total map $\oplus_j V_{0,j}\to \oplus_j V_{1,j}$ is an isomorphism, and the question is what if any boundedness properties the inverse isomorphism has.

For any $N\ge 0$ you can make an example (closely related to one described in the question) in which the nontrivial spaces are the $V_{0,j}$ with $0\le j\le N$ and the $V_{1,j}$ with $1\le j\le N+1$, all $1$-dimensional, such that the inverse map takes the generator of $V_{0,j}$ to the sum of the generators of $V_{1,k}$ for $j+1\le N+1$.

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Good point. And I probably was close to thinking of something like that. In terms of matrices, what you're saying is: the matrix for $\partial$ looks like it has $1$s on the main diagonal and $-1$s on whichever diagonal raises degree (hence the inverse matrix fills in the triangle), but in fact the degrees are rigged so that the $-1$s are preserve degree and the $1$s lower it. –  Theo Johnson-Freyd Nov 5 '12 at 5:40
    
Yes, that's it. –  Tom Goodwillie Nov 5 '12 at 12:53

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