4
$\begingroup$

According to http://ocw.mit.edu/courses/mathematics/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/lecture-notes/lecture2.pdf. , if $X$ is any topological space, then its cochain complex $C^* := C^*(X)$ has a (homotopy) symmetric multiplication

$$D_2(C^*) := (C^* \otimes C^* \otimes E(\mathbb{Z}/2))_{\mathbb{Z}/2} \rightarrow C^*$$

where $E(\mathbb{Z}/2)$ is a homological model for a contractible space with free $\mathbb{Z}/2$-action.

pre-Question: how can I describe concretely this symmetric multiplication? (For instance, if I pick a cellular model for $X$ then how do I write it down?)

If $X$ is a topological space with $G$-action, then we can form its homotopy quotient $X/G := (X \times EG)/G$, where $EG$ is a contractible space with free $G$-action. At the level of cochains, we have $C^*(X/G) = C^*(X)^{hG} := Hom(C_*(EG), X)^G$. So I should get a symmetric multiplication

$$ D_2(C^{hG}) \rightarrow C^{hG}$$

Question: more generally, if $C$ is any chain complex with symmetric multiplication $D_2(C) \rightarrow C$, is there an induced multiplication $D_2(C^{hG}) \rightarrow C^{hG}$? What is it?

$\endgroup$
5
  • $\begingroup$ For the prequestion, take a look at this: mathoverflow.net/questions/210540/… . For the second question: I am sure it does if, instead of just a multiplication you work with E_∞-structures (the forgetful functor from E_∞-algebras to chain complexes preserves homotopy limits being a right Quillen adjoint). I suspect it is true also for just the symmetric power but I need to think about it more. $\endgroup$ Aug 29 '16 at 23:59
  • $\begingroup$ Thanks! My intuition/guess was that this map should exist for formal reasons, but the short story is that I used it sort of carelessly in a calculation and got a nonsensical answer, so now I'm trying to make sure I understand every step meticulously. I think it is a tiny bit subtle to say what this map is, i.e. given two maps $EG \rightarrow X$ and an element of $C_*(EZ/2)$ how exactly to put this together into another map $EG \rightarrow X$ in an equivariant way. I have an example in mind which is messy to say here but which I would be happy to elaborate on over e-mail, if you would like. $\endgroup$
    – user84144
    Aug 30 '16 at 0:07
  • $\begingroup$ Also, while I greatly respect these higher-categorical notions, and have at least heard enough "slogans" about E_{infty} algebras to appreciate a little what they mean, I would prefer as much as possible to adhere to a pedestrian point of view here... I'd be really happy to write down, even for a simple cell complex, how to concretely write down a multiplication $D_2(C) \rightarrow C$ in terms of bare-bones generators. $\endgroup$
    – user84144
    Aug 30 '16 at 0:19
  • 2
    $\begingroup$ For the prequestion: I'm sure this map is induced by the diagonal $X\to X\times X\to (X\times X)\times_{\mathbb{Z}/2} E\mathbb{Z}/2$. So to describe what it does on cochains would require a cellular approximation of the diagonal, possibly with some extra equivariance properties. Anyway, some relevant references (none of which do exactly what you want) are Hatcher 4.L, Steenrod and Epstein's "Cohomology operations" and Bruner-May-McClure-Steinberger's "H_\infty ring spectra and their applications" $\endgroup$
    – Mark Grant
    Aug 30 '16 at 8:13
  • $\begingroup$ Mark, what you say sounds very sensible but I have to admit that I am confused by the gradings. If you look at Lurie's notes, the chain complex he takes for $E(\mathbb{Z}/2)$ is a free resolution of the cohomology of a point, so for instance it is concentrated in negative degrees, so the cochain complex of $X \times X \times E\mathbb{Z}/2$ wouldn't be what I call about $C \otimes C \otimes E(\mathbb{Z}/2)$). That's one of the reasons I am confused about this business! $\endgroup$
    – user84144
    Aug 30 '16 at 17:54
5
$\begingroup$

The construction of the map $$D_2(C^*)\rightarrow C^*$$ goes back at least to N. E. Steenrod. In his beautiful paper "Products of cocycles and extensions of mappings", Ann. of Math. (2) 48 (1947), 290–320, he gives a very explicit combinatorial description of this map.

Take for $E\mathbb{Z}/2$ the chain complex whose i-chains ($i\geq 0$) are generated by the regular reprensation of the symmetric group $\mathbb{Z}/2=\{e,\tau\}$ i.e we have $$(E\mathbb{Z}/2)_i=\mathbb{Z}<e_i,\tau.e_i>$$ together with $de_i=e_{i-1}+(-1)^{i}\tau.e_{i-1}$.

$e_i$ corresponds to the $i$-th cup-i product. N. E. Steenrod gives in his paper an explicit formula for the action of the cup-i product on cochains.

This formulas can be found in many papers and many works where they were extended to an $E_{\infty}$-structure on singular cochains. One can cite:

  • C. Berger and B. Fresse,"Combinatorial operad actions on cochains", Math. Proc. Cambridge Philos. Soc. 137 (2004), no. 1, 135–174.

  • J. P. May: "A general algebraic approach to Steenrod operations", The Steenrod Algebra and its Applications (Proc. Conf. to Celebrate N. E. Steenrod’s Sixtieth Birthday, Battelle Memorial Inst., Columbus, Ohio, 1970), Lecture Notes in Mathematics, Vol. 168, Springer, Berlin, 1970, pp. 153–231.

  • J. McClure and J. Smith : "Multivariable cochain operations and little n-cubes", preprint arXiv:math.QA/0106024(2001).

$\endgroup$
3
  • $\begingroup$ This is great! Do these also address the question of whether such a structure automatically descends to a symmetric product on the homotopy invariants of a chain complex? $\endgroup$
    – user84144
    Aug 30 '16 at 15:59
  • $\begingroup$ You're welcome. Can you be more precise? $\endgroup$
    – David C
    Aug 30 '16 at 16:12
  • $\begingroup$ So above in the question I ask if a symmetric multiplication on $C$ automatically induces one on $C^{hG}$, which coincides with the one coming from topology via $C^{hG} = C^*(X/G)$ when $C = C^*(X)$, and if so how to describe it. $\endgroup$
    – user84144
    Aug 30 '16 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.