Let X, Y and B be $E_\infty$ spaces, and let $p: X \rightarrow Y$ and $f: B \rightarrow Y$ be $E_\infty$ maps. We can ask for the space of lifts of f across p, that is the space of $E_\infty$ maps $g: B \rightarrow X$ such that $pg = f$.

Q1: What spectral sequences or other technology exists for computing the (homotopy groups of the) space of such E-infinity lifts?

Any $E_\infty$ lift $g: B \rightarrow X$ provides a lift of the map of commutative monoids $\pi_0 B \rightarrow \pi_0 Y$ across the map $\pi_0 X \rightarrow \pi_0 Y$.

Q2: Do all the techniques for computing E-infinity lifts in effect require that you first solve this $\pi_0$ lifting problem in commutative monoids, and begin an obstruction calculation from there, or are there techniques that solve both the $\pi_0$ problem and the E-infinity problem 'simultaneously' and perhaps in a way that eases both computations?

I am particularly interested in merely knowing if there exists an $E_\infty$ lift, thus might have asked the seemingly more basic question:

Q1': When is there an E-infinity lift of an E-infinity map?

I think the obstruction groups answering Q1' are liable to come packaged in the answers to Q1/Q2, but if there are separate techniques for the existence question, that would also be helpful.

Remark: I could imagine that one answer to Q1 involving relative Andre-Quillen cohomology might be extracted from Goerss-Hopkins, Moduli Problems for Structured Ring Spectra, but perhaps there are more elementary means. I'd be very interested for answers to Q1 along those or especially other lines of thought, and any ideas about Q2. Thanks!

  • 3
    $\begingroup$ John Francis points out below that this reduces to a problem in stable homotopy theory when X is grouplike, or is nearly so. I'm wondering if anybody has ever computed something like this, where X isn't grouplike, and B isn't something simple (like a free E-infinity space). I can't think of any such case, though that may just reflect my ignorance. $\endgroup$ – Charles Rezk Feb 17 '10 at 20:46
  • 2
    $\begingroup$ My best guess for finding something like that was to look through John Rognes' paper on topological logarithmic structures, where a log structure is basically some kind of map of E-infinity spaces to the multiplicative monoid of $\Omega^\infty R$. My coarse read on it (and my memory of comments he's made) seems to be that finding appropriate maps of E-infinity spaces (to give the right kinds of logarithmic structures) when $R = ku$ or the like is a difficult problem. I don't know of any systematic analysis of these mapping spaces even when X is the zero-component of $\Omega^\infty R$! $\endgroup$ – Tyler Lawson Feb 17 '10 at 21:13

If $Y$ and $B$ are grouplike, then the question, of course, immediately reduces to the case of spectra: The map $X\rightarrow B$ factors through the group completion $\Omega B X$ of $X$, so you can then deloop everything to get the corresponding spectra. Once you're in the setting of connective spectra, there are spectral sequences available, the most obvious being the one based on the Postnikov filtration. (This does start with figuring out the $\pi_0$ case, then inductively working through the higher homotopy groups, so affirmative to Q2. I'm not, unfortunately, optimistic about your hope in Q2.)

You can still reduce to the case of spectra with the more modest assumption that the induced map $\pi_0 K \rightarrow \pi_0 \Omega B K$ is an injection, for $K = B, Y$. In this case, you can group complete everything, and then try to solve the problem for the corresponding spectra. An $E_\infty$ lift $X\rightarrow Y$ over $B$ will be the same as a $E_\infty$ lift of the corresponding group completions satisfying an additional $\pi_0$ condition, that the image of $\pi_0 X\rightarrow \pi_0\Omega BX \rightarrow \pi_0 \Omega B Y$ lies in $\pi_0 Y$.

Perhaps you can extend this approach to deal with more general monoids, by writing them as extensions ${\rm Fiber} \rightarrow Y \rightarrow \Omega B Y$. (But you can't be interested in those, can you?)

  • $\begingroup$ Thanks, it's good to keep in mind that to be in spectra-land, it's sufficient for the group completion to be injective. Unfortunately, in my cases I expect the group completion to be a brutal operation on $\pi_0$. I'll contemplate whether some carefully chosen submonoid might have better properties. $\endgroup$ – cdouglas Feb 24 '10 at 0:12

The issue with the underlying monoid seems to complicate everything, in a similar way to how Postnikov decompositions are complicated by the $\pi_1$ issue. In that case, the common technique is to fix $G = \pi_1$ and consider the Postnikov tower as operating in the category of spaces over $K(G,1)$ rather than in the ordinary category of spaces.

So I don't see a lot of hope immediately for getting the $\pi_0$ problem out of the way at the same time; it seems like it colors the whole problem.

Once you've decided on a lifting $\pi_0 B \to \pi_0 X$, though, you can fix the underlying monoid because then you're reduced to studying lifts $B \times_{\pi_0 Y} \pi_0 X \to X$ over $\pi_0 X$.

If you then fix $M = \pi_0 X$ then there's certainly some kind of obstruction theory, but the problem is identifying the obstruction classes as coming from something cohomological that you can actually calculate. It seems to me that one should study the symmetric monoidal category of "spaces over $M$", with product having fibers $$ (X \star Y)_m = \coprod_{m' m'' = m} (X_{m'} \times Y_{m''}) $$ (which is some kind of left Kan extension), and try to get some handle on it.

Even when $M = \mathbb{N}$ the bookkeeping gets complicated. Then you're studying "graded $E_\infty$ spaces" and your obstruction theory will land in something like cohomology with coefficients in the relative homotopy groups of $Y$ over $B$, but you're taking cohomology of the "derived indecomposables" in your $E_\infty$ space. The zero'th space of derived indecomposables of an $E_\infty$ space $B$ over $\mathbb{N}$ is the topological Andr\'e-Quillen homology object of $B_0$. Even if $B_0$ is trivial, then the zero'th derived indecomposable space is trivial, the first is $B_1$, and the next is the homotopy cofiber of the squaring map $(B_1 \times B_1)_{h\mathbb{Z}/2} \to B_2$.

Based on this kind of futzing around I am led to believe that your obstructions may possibly occur in the relative topological Andre-Quillen cohomology of $\Sigma^\infty_+ B$ over $\Sigma^\infty_+ M$ with coefficients in the relative homotopy of $X$ over $Y$. But the problem seems very difficult for a general monoid $M$.

(Especially evidenced by the fact that Charles Rezk hasn't popped in here with an answer yet.)

  • $\begingroup$ I was waiting for someone with more rep-points than me to respond. $\endgroup$ – Charles Rezk Feb 17 '10 at 20:42
  • $\begingroup$ Thanks, glad to have the perspective on how relative TAQ enters, even if you confirmed my fears about the difficulty of the $\pi_0$ problem. $\endgroup$ – cdouglas Feb 24 '10 at 0:16

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.