I'm a novice to homotopical algebra, but I've found myself confronted with it by necessity and have some basic questions...

I'm going to consider chain complexes over a field $F := \mathbb{F}_2$. Given a chain complex $C$, I'm interested in two operations:

  • the ``homotopy Sym'', where I form $(C \otimes C \otimes E (\mathbb{Z}/2))_{\mathbb{Z}/2}$ where the $\mathbb{Z}/2$ acts diagonally on the tensor product. I'll call this $hSym^2 C$.
  • if $C$ has a $\mathbb{Z}$-action, then I can form ``homotopy quotient'' $C/\mathbb{Z}$, which is $(C \otimes E \mathbb{Z})_{\mathbb{Z}}$, with $\mathbb{Z}$ acting diagonally. I don't know if there is "official" notation for this; I'll just call it $hC/\mathbb{Z}$.

(Edit: I thought I wrote this down but must have deleted it accidentally; $E G$ is a projective $F[G]$-resolution of the complex $F$ in degree $0$, which is supposed to represent a point; thus $EG$ is morally to be the chain complex of some contractible space on which $G$ acts freely.)

So my question is about how the composition of these two operations in either order are related. If $C$ has a $\mathbb{Z}$-action, then I think $hSym^2 C$ still has a $\mathbb{Z}$-action, so I could form

$$ h( hSym^2 C )/\mathbb{Z} $$

or I could do things in the opposite order:

$$ hSym^2 (hC/\mathbb{Z}). $$

Based on naive intuition about how ordinary quotients work, I guess that there should be an induced map

$$ h( hSym^2 C )/\mathbb{Z} \rightarrow hSym^2 (hC/\mathbb{Z}) $$

  1. Is this right?
  2. And if it is, then is the above map an (edit:quasi-)isomorphism? (I guess probably not in general)
  3. How can I understand this map explicitly? For instance, if I choose an explicit model for $E \mathbb{Z}/2$ and $E \mathbb{Z}$, like the standard ones that spit out $\mathbb{RP}^{\infty}$ and $S^1$, then I should in principle be able to write it down explicitly, but I'm confused about how that goes.
  • 1
    $\begingroup$ What is $E(\mathbf{Z})$? $\endgroup$
    – user97187
    Aug 14 '16 at 4:04
  • $\begingroup$ @user84144 What you call $hC/\mathbb{Z}$ is called homotopy fixed points and usually denoted $C^{h\mathbb{Z}}$ (and in fact it behaves more like fixed points than quotients). Similarly, your homotopy Sym is usually called a restricted power and denoted $D_2(C)$ (although this notation might be much more common in homotopy theory than in homological algebra) $\endgroup$ Aug 14 '16 at 9:10
  • 2
    $\begingroup$ @DenisNardin The homotopy fixed points is $\hom(E{\mathbb Z}, C)^{\mathbb Z}$. I am guessing/hoping that $(C\otimes E{\mathbb Z})^{\mathbb Z}$ is really meant to be $(C\otimes E{\mathbb Z})_{\mathbb Z}$, since with is what homotopy quotient usually means. If I am not mistaken, the invariants are always trivial in this case. $\endgroup$ Aug 14 '16 at 9:57
  • $\begingroup$ @GregoryArone Uh you're right. I am sorry for the mix-up. I agree that it is probably $(C\otimes E\mathbb{Z})_{\mathbb{Z}}$ $\endgroup$ Aug 14 '16 at 10:21
  • $\begingroup$ Thanks for the corrections. I have edited to (hopefully!) fix things. $\endgroup$
    – user84144
    Aug 14 '16 at 16:34

If $C$ has an action of $\mathbb Z$, then $C\otimes C$ has an action of the wreath product ${\mathbb Z}\wr {\mathbb Z}/2$. This is the split group extension $1\to {\mathbb Z}\times{\mathbb Z}\to {\mathbb Z}\wr {\mathbb Z}/2\to {\mathbb Z}/2\to 1$ associated with the natural action of ${\mathbb Z}/2$ on ${\mathbb Z}\times {\mathbb Z}$. Inside this group you have the subgroup ${\mathbb Z}\times {\mathbb Z}/2$, where ${\mathbb Z}$ is the diagonal subgroup of ${\mathbb Z}\times {\mathbb Z}$.

With this notation, $h(hSym^2C)/{\mathbb Z}$ is the homotopy quotient of $C\otimes C$ by ${\mathbb Z}\times {\mathbb Z}/2$, while $hSym^2(hC/{\mathbb Z})$ is the homotopy quotient of $C\otimes C$ by ${\mathbb Z}\wr {\mathbb Z}/2$. So your map can be understood as the quotient map $$[C\otimes C\otimes E({\mathbb Z}\times {\mathbb Z}/2)]_{{\mathbb Z}\times {\mathbb Z}/2}\to [C\otimes C\otimes E({\mathbb Z}\wr {\mathbb Z}/2)]_{{\mathbb Z}\wr {\mathbb Z}/2}$$ associated with the group inclusion ${\mathbb Z}\times {\mathbb Z}/2\hookrightarrow {\mathbb Z}\wr {\mathbb Z}/2$. In particular, the map is not an isomorphism, or even a chain homotopy equivalence, except in the most trivial case.

You can make this map pretty explicit by choosing nice chain level models for $E\mathbb Z$, $E{\mathbb Z}/2$ and $E{\mathbb Z}\wr {\mathbb Z}/2$. As you say, a good model for $E{\mathbb Z}/2$ is given by cellular chains on $S^{\infty}$, with the standard cell structure that has two cells in each dimension. To get a nice model for $E{\mathbb Z}$, think of $\mathbb Z$ acting on $\mathbb R$ by translation. Equip $\mathbb R$ with a cell structure with a zero cell for every integer $n$, and a $1$-cell for every interval $[n, n+1]$. The cellular chain complex gives you a nice small model for $E\mathbb Z$. Since you said you wanted to work with ${\mathbb F}_2$ coefficients, it has the form ${\mathbb F}_2[{\mathbb Z}]\leftarrow {\mathbb F}_2[{\mathbb Z}]\leftarrow 0 \cdots $, where the boundary homomorphism is determined by $[n, n+1]\to [n+1]-[n]$. Once you have models for $E{\mathbb Z}$ and $E{\mathbb Z}/2$, the chain complex $E{\mathbb Z}\otimes E{\mathbb Z}\otimes E{\mathbb Z}/2$ gives a model for ${\mathbb Z}\wr {\mathbb Z}/2$.

  • $\begingroup$ Thanks, this seems like exactly what I was looking for! Unfortunately I won't get a chance to think this through for a couple days. I assume that I can figure this out by writing things out, but is it known if the two compositions $C -> h(hSym^2 C)/\mathbb{Z}$ are homotopic (I mean the same in homology)? $\endgroup$
    – user84144
    Aug 14 '16 at 16:57
  • $\begingroup$ @user84144 A word of caution: homotopic and the same in homology are two very different properties for maps of complexes (the first means that there is a chain homotopy connecting them) $\endgroup$ Aug 14 '16 at 19:11
  • $\begingroup$ What are "the two compositions $C\to h(hSym^2C)/{\mathbb Z}$"? Are you referring to two natural homomorphisms between these chain complexes? What are they? I am not sure I can see even one. $\endgroup$ Aug 14 '16 at 20:03

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