18
$\begingroup$

A cospectrum (in the context of homotopy theory) is defined to be a sequence of spaces $X_0, X_1, \ldots, X_n, \ldots, $ equipped with maps $X_{n+1}\to \Sigma X_n$, for each $n$. So cospectra are similar to spectra, except that the structure maps point in the opposite direction.

Cospectra were first introduced by Elon Lima in 1959. I learned about them form Browder's classic paper on the Kervaire invariant problem. Browder seems to make rather essential use of this construction. I had never seen the concept before, so tried to search the literature on cospectra. I found virtually nothing. There are a couple of papers developing some properties of cospectra, but they did not seem to lead to any further activity. As far as I can see, practically no one else investigated cospectra or used them for anything.

So we have a definition that is a more or less natural variant of a very influential one. A definition that was used once in an important paper, and nowhere else. I find this curious, therefore I want to ask

Question 1 Is there a reasonable way to rewrite Browder's proof without cospectra?

According to my limited understanding, the reason for introducing cospectra is that they provide an alternative approach to Spanier-Whitehead duality. Lima's motivation was to define Spanier-Whitehead dual of spaces more general than finite CW complexes. As far as I could make out, Browder's motivation was similar. He needed to have a notion of Spanier-Whitehead dual that was well-behaved for non-finite spectra, and cospectra seem to do the job for him. Therefore I wonder if the same could be accomplished using the duality between spectra and pro-spectra, or maybe by just using finite approximations to spectra.

If question 1 does not have an obvious positive answer, then there is a natural follow up:

Question 2 How come no one else found use for cospectra? Is there some good mathematics lying that way, waiting to be discovered?

Improve this question
$\endgroup$
7
  • $\begingroup$ Does Lima produce any natural examples of cospectra? $\endgroup$
    – Tyrone
    Commented Feb 9, 2021 at 13:47
  • 2
    $\begingroup$ I tend to think of a general spectrum, in the sense of a sequence of spaces with maps $X_n \to \Omega X_{n+1}$ , as a non-fibrant stand-in for an $\Omega$-spectrum (i.e. spectrum where those maps are equivalences); we keep the category of spectra around because it's a convenient place to construct suspension spectra. But the only $\Sigma$-cospectrum (i.e. cospectrum where the maps $X_{n+1} \to \Sigma X_n$ are equivalences) is contractible. So I don't know where to place cospectra. This may just be a shortcoming of my conceptual apparatus, but I imagine many others are in the same boat. $\endgroup$
    – Tim Campion
    Commented Feb 9, 2021 at 13:58
  • 3
    $\begingroup$ I'll have to dig out my notes, but I'm pretty sure prespectra can be excised from Browder's paper without working too much (at least I think I did so for an expository talk on the material a few years ago). This does not mean they cannot be useful, but I don't think they have much relevance in this particular example. $\endgroup$
    – Denis Nardin
    Commented Feb 9, 2021 at 15:22
  • $\begingroup$ I came across CW cospectra by Hikida which describes how to get a homology theory from a cospectrum. $\endgroup$
    – Tim Campion
    Commented Feb 9, 2021 at 15:22
  • 2
    $\begingroup$ A joint paper of mine had cause to use cospectra (in another category) and I wrote Brayton Gray to ask about the status of his previous work, which is referenced in a few places. He sent a nice response that he gave some lectures about it but never wrote it up, and that not much was within the range of calculation (I believe some of it got subsumed by the Bousfield-Kuhn functor). $\endgroup$
    – Tyler Lawson
    Commented Feb 9, 2021 at 16:50

1 Answer 1

Reset to default
7
$\begingroup$

Brayton Gray had a nice preprint about cospectra and unstable $v_n$-periodic homotopy. This was maybe 20 years ago, and I can't find any online version. To confirm my memory that this existed, by searching the web, I also found mention of this in notes by Neil Strickland.

I likely have a paper version in my office, which I am not visiting very often right now. But maybe I, or someone else, can reconstruct his basic idea.

Improve this answer
$\endgroup$
4
  • 6
    $\begingroup$ I think that Brayton Gray's cospectra are different; they are related to the colimit of $[\Omega^kX,\Omega^kY]$ as $k\to\infty$, and I do not think that that interacts with the construction in this question. (But I also think that Brayton's version is very interesting and that it deserves to be investigated further.) $\endgroup$
    – Neil Strickland
    Commented Feb 9, 2021 at 16:57
  • 5
    $\begingroup$ Ah, I bet Gray's cospectra have maps $\Sigma X_{n+1} \rightarrow X_n$. And in appropriate $v_n$-local categories, there are objects that one can think of as infinite suspensions (analogous to infinite loopspaces). $\endgroup$
    – Nicholas Kuhn
    Commented Feb 9, 2021 at 17:33
  • $\begingroup$ I had completely forgotten about the very surprising point in Nicholas Kuhn's comment, but it turns out he told me this before! The point being that the Bousfield-Kuhn functor has a left adjoint (as we know now, it's in fact monadic, given by the "Lie algebra" monad), and the left adjoint carries in the "spectrum" structure from the (stable) category of $K(n)$-local spectra. $\endgroup$
    – Tim Campion
    Commented Feb 10, 2021 at 1:42
  • 1
    $\begingroup$ Just to add a reference to the pile. The type of cospectra with maps $\Sigma X_{n+1} \to X_n$ also appear in "Stable frames in model categories" by Lenhardt. If my memory recalls correctly, Lenhardt uses cospectra to show that the homotopy category of any stable model category is a module over the stable homotopy category, and that (symmetric) spectra (of simplicial sets) is initial among stable model categories. (There's a nice account of this story in "Foundations of Stable Homotopy Theory" by Barnes and Rotizheim.) $\endgroup$
    – Niall Taggart
    Commented Apr 16, 2021 at 8:05

You must log in to answer this question.

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