Finding a chain map.

Question

Here is the question I am trying to solve:

Let $C_2 = \langle T| T^2 = 1 \rangle$ be the group of order 2. Let $R = \mathbb Z$ or $R = \mathbb F_2.$ Let $\mathcal{W}$ be the $C_2$-equivariant cellular chains of $S^{\infty}$ with two cells, $e_n$ and $Te_n,$ of each dimension $n \geq 0,$ and boundary $d(e_{2k + 1}) = (1 - T)e_{2k}$ and $d(e_{2k}) = (1 + T)_{e_{2k} - 1}.$ This gives the resolution $$0 \leftarrow \mathcal{W}_0 \leftarrow \mathcal{W}_1 \leftarrow \dots $$ with each $\mathcal W_n = \langle e_n \rangle \cong R[C_2]$ and homology

$$ H_i(\mathcal W)= \begin{cases} R, & i = 0\\ 0, & i > 0 \end{cases} $$

The tensor product $\mathcal{W} \otimes \mathcal{W}$ is the chain complex with $\bigoplus_{i + j = n}\mathcal{W}_i \otimes \mathcal{W}_j$ in degree $n$ and differential $d(x \otimes y) = dx \otimes y + (-1)^{|x|} x \otimes dy.$ Let $C_2$ act diagonally on $\mathcal{W} \otimes \mathcal{W}: T(x \otimes y) = (Tx) \otimes (Ty).$

$(1)$ With $R = \mathbb F_2,$ find a $C_2$-equivariant chain map $\Delta: \mathcal{W} \to \mathcal{W} \otimes \mathcal{W}$ with $\Delta (e_0) = e_0 \otimes e_0.$

$(2)$ Do the same with $R = \mathbb Z.$

My thoughts and questions:

0- why $(1-T)e_0 = e_0 \otimes e_0 - Te_0 \otimes Te_0$? can someone show me the details of the calculation please?

a- Equivariant in this context means $\mathcal{W}$ and $\mathcal{W} \otimes \mathcal{W}$ are $R[C_2]$-modules, and $\Delta$ should be an $R[C_2]$-module homomorphism. The idea of the exercise is to show that $\tau \Delta \neq \Delta$

b- I got a hint to work $e_1$ and $e_2$ first but I do not know the details of doing this. I am guessing I should use the given two maps $d(e_{2k + 1}) = (1 - T)e_{2k}$ and $d(e_{2k}) = (1 + T)_{e_{2k} - 1}$ but still how?

c- I also got the following diagram (maybe it contain mistakes):

d- I also, do not know how to use the given differential in my solution to $1$ and $2.$ maybe this is required for solving the following question (which is also given as #3):

With $R = \mathbb F_2 or \mathbb Z,$ find a chain homotopy $\Delta_1: \mathcal{W}_n \to (\mathcal{W} \otimes \mathcal{W})_{n + 1}$ between $\Delta$ and $\tau \Delta.$ That is $d \Delta_1 + \Delta_1 d = \Delta - \tau \Delta.$ Here, $\tau(x \otimes y) = (-1)^{|x||y|} y \otimes x.$

Could someone show me the details of the solution of $(1)$ & $(2),$ please? How can I find the $4 \times 8$ matrix over $\mathbb Z$ for $e_1$? How can I find the $8 \times 12$ matrix for $e_2,$ can someone give me more details please?

Answer 1

This exercise is essentially asking for a chain-level approximation to the diagonal map $S^\infty \to S^\infty \times S^\infty$. There is probably a geometric way to write down a Alexander-Whitney map, but let's use your hint and attack the problem algebraically by hand at least for the first few generators.

We are given that $\Delta(e_0) = e_0 \otimes e_0$. By equivariance, we have $\Delta(Te_0) = Te_0 \otimes Te_0$, so we know what the map $\Delta$ does on the $0$th level. In fact, $\Delta(Te_k) = T \Delta(e_k)$, so it is enough to define $\Delta$ on $e_k$ for each $k$ and then $\Delta$ will automatically be $C_2$-equivariant.

The next question to ask is what is $\Delta(e_1)$? The only condition we have is that $\Delta$ is a chain map. This means that whatever $\Delta(e_1)$ is, it has to satisfy $d \Delta(e_1) = \Delta(d e_1)$. We can compute the right hand side: $$\begin{align*} d \Delta(e_1) &= \Delta(d e_1) \\ &= \Delta(e_0 - Te_0) \\ &= \Delta(e_0) - \Delta(Te_0) \\ &= e_0 \otimes e_0 - Te_0 \otimes Te_0. \end{align*}$$ So we need to find something whose differential is $e_0 \otimes e_0 - Te_0 \otimes Te_0$.

Let's compute a few differentials: let's try $e_0 \otimes e_1$ first: $$d(e_0 \otimes e_1) = e_0 \otimes (e_0 - Te_0) = e_0 \otimes e_0 - e_0 \otimes Te_0.$$

Ok, this gives us the $e_0 \otimes e_0$ term in $d\Delta(e_1)$, but we still need $-Te_0 \otimes Te_0$ and a way to cancel out $-e_0 \otimes Te_0$. But note that $$e_0 \otimes Te_0 - Te_0 \otimes -Te_0 = (e_0 - Te_0) \otimes Te_0 = de_1 \otimes Te_0 = d(e_1 \otimes Te_0).$$ Therefore, we have $$d(e_0 \otimes e_1 + e_1 \otimes Te_0) = e_0 \otimes e_0 - Te_0 \otimes Te_0.$$ So let's set $\Delta(e_1) = e_0 \otimes e_1 + e_1 \otimes Te_0$.

We can play the same game for $\Delta(e_2)$, using our new definition of $\Delta(e_1)$ (and thus $\Delta(Te_1)$). We have $$\begin{align*} d \Delta(e_2) &= \Delta(de_2) \\ &= \Delta(e_1 + Te_1) \\ &= \Delta(e_1) + \Delta(Te_1) \\ &= (e_0 \otimes e_1 + e_1 \otimes Te_0) + (Te_0 \otimes Te_1 + Te_1 \otimes e_0) \end{align*}$$

Let's compute more differentials; let's try $e_0 \otimes e_2$: $$d(e_0 \otimes e_2) = e_0 \otimes(e_1 + Te_1) = e_0 \otimes e_1 + e_0 \otimes Te_1.$$

The $e_0 \otimes e_1$ is what we want, but again we have a junk term $e_0 \otimes Te_1$. We see that we want the $e_1 \otimes Te_0 + Te_0 \otimes Te_1$ terms eventually, so why not kill two birds with one stone and use $$d(e_1 \otimes Te_1) = (e_0 - Te_0) \otimes Te_1 - e_1 \otimes (Te_0 - e_0) = e_0 \otimes Te_1 - Te_0 \otimes Te_1 - e_1 \otimes Te_0 + e_1 \otimes e_0.$$ This leaves us with $e_1 \otimes e_0 + Te_1 \otimes e_0$, which we recognize as $d(e_2 \otimes e_0)$. So putting this altogether, we have $$d(e_0 \otimes e_2 - e_1 \otimes Te_1 + e_2 \otimes e_0) = e_0 \otimes e_1 + e_1 \otimes Te_0 + Te_1 \otimes e_0 + Te_0 \otimes Te_1 = d(\Delta(e_2)).$$ So we can set $\Delta(e_2) = e_0 \otimes e_2 - e_1 \otimes Te_1 + e_2 \otimes e_0$.

We have not shown that these are the only choices for $\Delta(e_1)$ and $\Delta(e_2)$, and you might be worried that we'll run into an obstruction if this is not the right choice as we continue to lift this map up the chain complex; however, there is a result in homological algebra about lifting maps from free resolutions to acyclic ones that guarantees that this procedure will succeed.

You should try to continue and compute $\Delta(e_3)$, $\Delta(e_4)$, $\ldots$ until you spot a pattern.

In this case, one possible diagonal approximation is given by $$\Delta(e_k) = \sum_{i=0}^k (-1)^{i(k-i)} e_i \otimes T^i e_{k-i}.$$ It's tedious, but I leave it to you to actually check this it is a chain map. You can ignore the signs if you work mod $2$.

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Editing to answer questions in comments.

1. We can compute $\tau \Delta$ directly, using the formula for $\Delta$ in my answer: $$\tau \Delta (e_k) = \sum_{i=0}^k T^i e_{k-i} \otimes e_i,$$ and simply observe that $\tau \Delta \neq \Delta$.
2. Equivariance means $T \circ \Delta = \Delta \circ T$, where $T$ acts as indicated on $\mathcal{W}$, and diagonally on $\mathcal{W} \otimes \mathcal{W}$.
3. Normally, one uses the method of acyclic models to create these higher diagonal approximations, which is slick but does not give explicit formulae. Alternatively, using a similar method as above, you can deduce $\Delta_1(e_0) = 0$, $\Delta_1(e_1) = e_1 \otimes e_1$, $\ldots$. I think that in general, $$\Delta_1(e_k) = (-1)^{k+1} \sum_{\substack{0 \leq i < j \leq k \\ j - i \text{ odd}}} e_{k-j+i+1} \otimes T^i e_{j-i}$$ but I don't have very high confidence about the signs (which depend in part on how you define a chain homotopy), so you should double check. You could probably extract the right signs from Steenrod's original paper "Products of Cocycles and Extensions of Mappings". Another place to look is this more recent paper by Blumfiel, Medina-Mardones, and Morgan (see Theorem 3.2), but they seem to work mod $2$ only.