10
$\begingroup$

I am working through Weibel's Introduction to Homological Algebra, and am having trouble finding examples for Exercise 1.2.6

  1. A second quadrant double complex $C$ with exact columns such that $\mathrm{Tot}^{\prod}(C) $ is acyclic but $\mathrm{Tot}^{\oplus}(C)$ is not
  2. A second quadrant double complex C with exact row such that $\mathrm{Tot}^{\oplus}(C)$ is acyclic but $\mathrm{Tot}^{\prod}(C)$ is not
  3. A double complex (in the entire plane) for which every row and every column is exact, yet neither $\mathrm{Tot}^{\prod}(C)$ nor $\mathrm{Tot}^{\oplus}(C)$ is acyclic

I have tried just playing around with examples, but none seem to work. E.g. for part 1, I am struggling to even find $\mathrm{Tot}^{\oplus}$ acyclic with exact rows.


I am aware that this is a similar question referring to the same problem, but I did not want to edit that question as the spirit of that question is whether or not parts 1 and 2 are incorrect (this misunderstanding is quickly cleared up in the comments, and hence, there are no answers to that question).

$\endgroup$

1 Answer 1

14
$\begingroup$

I would butcher the indexes horribly anyway, so let me not indicate them; when one draws the examples, hopefully the arguments will be clear even without it.

Also looking at Weibel, it seems that his arrows go down and to the left, and also the squares anti-commute, so I will try to mimic that.

For 1:

On the main diagonal going through the second quadrant, put $\mathbb{Z}$ everywhere. Do the same on the diagonal right above it in the second quadrant. Put $0$ everywhere else. Let all the maps that have chance to be nonzero be the identity $1: \mathbb{Z} \rightarrow \mathbb{Z}.$ (Note that this is anti-comutative double complex since every square will contain $0$ as initial or terminal vertex.)

A sketch looks somewhat like this:

\begin{array} & \dots &&&& \\ & \vdots &&&\\ &\mathbb{Z}& \leftarrow & \mathbb{Z}&&\\ &&&\downarrow \\ &&&\mathbb{Z}& \leftarrow & \mathbb{Z}& \\ &&&&&\downarrow \\ &&&&& \mathbb{Z} \end{array} Here all the indicated arrows are identities and all the other spots are $0$.

Then the product totalization has only two non-zero terms, and in between them the map \begin{align*} \alpha: \mathbb{Z}^{\times \mathbb{Z}_{\geq 0}} &\longrightarrow \mathbb{Z}^{\times \mathbb{Z}_{\geq 0}} \\ (x_n)_{n \geq 0} &\longmapsto (x_n+x_{n-1})_{n \geq 0} \;, \end{align*} (with the convention that $x_{-1}=0$, so the first component of the image is just $x_0$).

Similarly, the sum totalization has the only nonzero arrow given as \begin{align*} \beta: \mathbb{Z}^{\oplus \mathbb{Z}_{\geq 0}} &\longrightarrow \mathbb{Z}^{\oplus \mathbb{Z}_{\geq 0}} \\ (x_n)_{n \geq 0} &\longmapsto (x_n+x_{n-1})_{n \geq 0} \;. \end{align*}

Note that $\alpha,$ and consequently $\beta,$ are both injective: If $(x_n)_{n \geq 0}$ is mapped to $0$, we have the series of equations \begin{align*} x_0&=0\\ x_1+x_0&=0\\ x_2+x_1&=0\\ \dots \end{align*} and thus, recursively one gets all $x_n$ equal to $0$.

For surjectivity of $\alpha,$ proceed similarly, namely given any element in the image $(y_n)_{n \geq 0}$, to find the preimage is to solve the system of equations \begin{align*} x_0&=y_0\\ x_1+x_0&=y_1\\ x_2+x_1&=y_2\\ \dots \end{align*} which you are always able to do (recursive rule $x_0=y_0,\; x_n=y_n-x_{n-1}$). However, note that this also shows that $\beta$ is not surjective: If $(y_n)_{n \geq 0}$ is chosen as $(1, 0, 0, 0, \dots)$, then the recursive formula above gives you $x_n=(-1)^n$. Thus, although $(y_n)$ has only finitely many nonzero components, the preimage does not.

For 2:

Similarly, populate the main diagonal and this time the one below it in the second quadrant by $\mathbb{Z}$'s, and put $0$ everywhere else. Again put identities everywhere that you can, and zero aps everywhere else.

The sketch changes only by the fact that the zigzag now starts horizontally and not vertically, that is, we have

\begin{array} & \dots &&&&&& \\ & \vdots &&&&&\\ &\mathbb{Z}& \leftarrow & \mathbb{Z}&&&&\\ &&&\downarrow && \\ &&&\mathbb{Z}& \leftarrow & \mathbb{Z}& && \\ &&&&&\downarrow &&\\ &&&&& \mathbb{Z} & \leftarrow & \mathbb{Z} \end{array}

Again, the totalizations are supported in two consecutive degrees only, and the non-trivial map is given this time by \begin{align*} \alpha: \mathbb{Z}^{\times \mathbb{Z}_{\geq 0}} &\longrightarrow \mathbb{Z}^{\times \mathbb{Z}_{\geq 0}} \\ (x_n)_{n \geq 0} &\longmapsto (x_n+x_{n+1})_{n \geq 0} \;, \end{align*} \begin{align*} \beta: \mathbb{Z}^{\oplus \mathbb{Z}_{\geq 0}} &\longrightarrow \mathbb{Z}^{\oplus \mathbb{Z}_{\geq 0}} \\ (x_n)_{n \geq 0} &\longmapsto (x_n+x_{n+1})_{n \geq 0} \;. \end{align*} for product and direct sum totalizations, resp. In this case, $\alpha$ is not injective: an element of the kernel is for example $(1, -1, 1, -1, \dots)$. However, from the fact that in the sum totalization, each element has $x_n=0$ for all big enough $n$, one can deduce that $\beta$ is injective (If an element $(x_n)$ is mapped to $0$, take $n$ such that $x_n=0$ and recursively show that $x_{n-1}=x_{n-2}= \dots = x_0=0$).

One can show that both $\alpha$ and $\beta$ are surjective: For $\alpha$, given $(y_n)_n$ in the target you can set $x_0=0$ and then use the recursive formula $x_{n+1}=y_n-x_n$ to define the rest. In the case of $\beta,$ one has to go backwards: Given $(y_n)_n$ in the target, let $y_N$ be the last nonzero entry. Then set $x_{N+1}=x_{N+2}=\dots =0,$ and work out the first $N+1$ entries $x_n$ by the formula $x_n=y_n-x_{n+1}.$ In the end, $(x_n)_n$ is the (unique) preimage to $(y_n)_n$.

So in conclusion, $\beta$ is an isomorphism, hence the sum totalization is acyclic, while $\alpha$ is not, and the product totalization has some homology (given by kernel of $\alpha$).

For 3:

In the last case - you guessed it - one does the same but indefinitely to both sides. That is, populate two neighboring full diagonals by $\mathbb{Z}$'s and put identities between them.

The totalizations have, again, two nonzero terms, and in between them the analogous maps $\alpha, \beta$: \begin{align*} \alpha: \mathbb{Z}^{\times \mathbb{Z}} &\longrightarrow \mathbb{Z}^{\times \mathbb{Z}} \\ (x_n)_{n} &\longmapsto (x_n+x_{n+1})_{n} \;, \end{align*} \begin{align*} \beta: \mathbb{Z}^{\oplus \mathbb{Z}} &\longrightarrow \mathbb{Z}^{\oplus \mathbb{Z}} \\ (x_n)_{n} &\longmapsto (x_n+x_{n+1})_{n} \;. \end{align*}

This time, however, note that $\alpha$ is still not injective (for similar reasons as in part 2) and that $\beta$ is still not surjective (more or less for the same reason as in 1) - see what happens if one tries to hit an element of the form $(\dots, 0, 0, 1, 0, 0, \dots)$).

$\endgroup$
1
  • 1
    $\begingroup$ Nice! Really well explained too! $\endgroup$ Apr 25, 2020 at 7:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .