Chain homotopy: Why du+ud and not du+vd? - MathOverflow

Chain homotopy: Why du+ud and not du+vd?

2009-12-15T11:12:19Z

When one wants to prove that a morphism $f_*$ between two chain complexes $\left(C_*\right)$ and $\left(D_*\right)$ is zero in homology, one of the standard approaches is to look for a chain homotopy, i. e., for a map $U_n:C_n\to D_{n+1}$ defined for every $n$ that satisfies $f_n=d_{n+1}U_n+U_{n-1}d_n$ for every $n$. However, this is not strictly necessary: For example, it is often enough to have two maps $U_n:C_n\to D_{n+1}$ and $V_n:C_n\to D_{n+1}$ defined for every $n$ that satisfy $f_n=d_{n+1}U_n+V_{n-1}d_n$ for every $n$. This way, when constructing $U_n$ and $V_n$, one doesn't have to care for them to "fit together", because each is used only one time.

However, at least my experience suggests that one does not gain much from this - when one tries to construct these $U_n$ and $V_n$, they turn out (after some simplification) to be the same.

My question is: What is the deeper reason behind this? Why do chain homotopies like to "fit together" although they don't need to?

I am sorry if this makes no sense...

EDIT: Thanks David, it seems I can't write a single absatz without a stupid mistake.

Answer by David Speyer for Chain homotopy: Why du+ud and not du+vd?

2009-12-15T12:59:18Z

EDITED, because I think I see the big picture now.

We have the following theorem: let $C$ and $D$ be complexes of projective objects. The following are equivalent: (a) the map $f: C \to D$ is the zero map in the derived category (b) there is a homotopy between $f$ and $0$.

In this theorem, the definition of homotopy is that $f=du+ud$. So my answer to your question is: In practice, if a map induces the zero map on homology, it is probably zero in the derived category. There do exist maps which are of the form $du+vd$, but are not of the form $du+ud$, see my other answer.

Paragraph about hereditary algebras deleted because I don't think it was quite right.

Some more elementary observations

(1) It is required that $f$ be a map of chain complexes, so $df=fd$. So we want $d(du+vd) =(du+vd)d$ or $dud=dvd$. This doesn't force $u=v$, but it is the easiest way to achieve it.

(2) There is a topological way of thinking of the condition $du+ud=f$, which I learned from Joel Kamnitzer. Let $I$ be the chain complex with $I_1=\mathbb{Z}$, $I_0 = \mathbb{Z}^2$ and the map $I_1 \to I_0$ given by $(1 \ -1)$. Let $\partial I$ be the subchain complex where $(\partial I)_0=I_0$ and $(\partial I)_i=0$ for all other $i$.

Then writing $f=du+ud$ is equivalent to finding a map $u:C \times I \to D$ such that, when we restrict to $C \times (\partial I)$, we have the map $f$ on one component and $0$ on the other. $I$ is the chain complex of the obvious triangulation of the unit interval. Thinking of $I$ as the unit interval, this really is a homotopy between $f$ and $0$. I can't think of an analogous geometric motivation for $f=du+vd$.

Answer by David Speyer for Chain homotopy: Why du+ud and not du+vd?

2009-12-15T14:34:59Z

Here is an example of a map of chain complexes which can be realized as $du+vd$ but not as $du+ud$.

Let $C$ and $D$ both be the chain complex

$$\cdots 0 \to \mathbb{Z}/p^2 \to \mathbb{Z}/p^2 \to 0 \to \cdots$$

where the nontrivial map is multiplication by $p$.

Map $C$ to $D$ by multiplication by $p$ in the first nontrivial degree, and by zero in the other nontrivial degree. This is $du+vd$, where $u$ is $1$ and $v$ is $0$ (in the only nontrivial degree).

On the other hand, this map cannot be written as $du+ud$. If we had such a representation, then $u$ would be multiplication by $a$ for some $a \in \mathbb{Z}/p^2$. The two vertical maps would then both be $pa$. In particular, it is impossible that one would be zero and the other not.

If we consider these complexes as objects in the derived category of $(\mathbb{Z}/p^2)$-modules, they are complexes of projective objects. So this is an example of a map of complexes which induces zero maps on cohomology but is not zero in the derived category.

What happens if we pass to a quotient category where maps like this are zero? I have no idea! Does anyone?

For completeness, I add an example of a map which induces zero on homology and can not be written as $du+vd$. Consider $$\begin{matrix} 0 & \to & \mathbb{Z} & \to & \mathbb{Z} \\ \downarrow & & \downarrow & & \downarrow \\ \mathbb{Z} & \to & \mathbb{Z} & \to & 0 \end{matrix}$$

where the nontrivial horizontal maps are multiplication by $p$, and the vertical map is the identity. The nontrivial homology is in different degrees in top and bottom, so the map on homology is zero. But any map of the form $du+vd$ would be $0$ modulo $p$ in the central column.

Answer by Mikhail Bondarko for Chain homotopy: Why du+ud and not du+vd?

2010-01-05T21:25:26Z

Amusingly, I have considered the category of complexes where maps of the form du+vd are killed in: Weight structures vs. $t$-structures; weight filtrations, spectral sequences, and complexes (for motives and in general), to appear in J. of K-theory, http://arxiv.org/abs/0704.4003, in section 3.1. My main observation is that the composition of two chain complex morphisms of form du+vd yield a morphism homotopic to zero; so you actually do not forget very much information. In particular, the projection functor from the homotopy category of complexes to this more coarse category is conservative i.e. a non-isomorphism does not become an isomorphism.

You may also find some motivation for introducing such a category in Remark 1.5.2 of my paper.

Answer by darij grinberg for Chain homotopy: Why du+ud and not du+vd?

2010-04-12T11:43:02Z

I have just found another article on this: Absolute Homology by Michael Barr.