A 1991 paper of Lewis, title “Is there a convenient category of spectra?” proves that there is no category $\mathrm{Sp}$ satisfying the following desiderata$^1$:

  1. There is a symmetric monoidal smash product $\wedge$;
  2. There is an adjunction $\Sigma^\infty\colon\mathrm{Top}_*^\mathrm{CGWHaus}\rightleftarrows\mathrm{Sp}:\Omega^\infty$;
  3. The sphere spectrum $\mathbb{S}$ is a unit for $\wedge$;
  4. There is either a natural transformation $$(\Omega^\infty E)\wedge(\Omega^\infty F)\Rightarrow\Omega^\infty(E\wedge F)$$ or a natural transformation $$\Sigma^\infty(E\wedge F)\Rightarrow(\Sigma^\infty E)\wedge(\Sigma^\infty F).$$ Furthermore, these natural transformations are required to commute with the unity, commutativity, and associativity isomorphisms of $\mathrm{Top}_*^\mathrm{CGWHaus}$ and $\mathrm{Sp}$;
  5. There is a natural weak equivalence $\Omega^\infty\Sigma^\infty X\xrightarrow{\cong}\varinjlim(\Omega^n\Sigma^nX)$.

Question: The theorem is proved for (ordinary) categories. But what happens for $\infty$-categories? Is the $\infty$-category of spectra defined in Higher Algebra “convenient”?

The $\infty$-category $\mathrm{Sp}$ satisfies (in an appropriate sense) [1] and [3] (HA, and [2] (HA What about [4] and [5]?

$^1$Which I am partly paraphrasing from the 2017 Talbot notes and partly paraphrasing from Lewis.

  • 5
    $\begingroup$ Yes of course. $\Sigma^{\infty}$ is symmetric monoidal, $\Omega^{\infty}$ is lax symmetric monoidal, and $\Omega^{\infty}\Sigma^{\infty}X$ is equivalent to $\mathrm{colim} \Omega^n\Sigma^nX$. Also your footnote "2" is incorrect- the loops-infinity suspension-infinity adjunction is just an adjunction, not an equivalence. $\endgroup$ – Dylan Wilson Feb 9 at 2:24
  • 2
    $\begingroup$ Some references: 4b) follows from the universal properties: $\mathsf{Spaces}_*$ is the unit object in the symmetric monoidal $\infty$-category of presentable pointed $\infty$-categories, and $\mathsf{Sp}$ is a pointed, presentably symmetric monoidal $\infty$-category so the essentially unique colimit-preserving functor from $\mathsf{Spaces}_*$ determined by $S^0$ has an essentially unique refinement to a symmetric monoidal functor. (cf. HA.4.8.2). 4a) right adjoints to symmetric monoidal functors are automatically lax monoidal (HA. And 5) can be proven directly or use HA. $\endgroup$ – Dylan Wilson Feb 9 at 2:42
  • $\begingroup$ @DylanWilson Thank you very much for the great pointers! (I have deleted the footnote.) $\endgroup$ – Théo de Oliveira Santos Feb 9 at 3:34

My colleague Dylan answered first (I keep telling him not to spend too much time on this toy :) but I both agree and disagree with his "Yes of course". The same words are used with different meanings and implications in the point-set and infinity category setting. So the word "convenient" has correspondingly different meanings! With the meaning of words in the infty category world, Yes means yes. But the heart of Lewis's argument is that in the point-set world the automorphisms of the unit object of a symmetric monoidal topological category give a point-set level commutative topological monoid. He argues that 1-5 imply that the unit component of $QS^0$ is equivalent to an honest commutative topological monoid, which would imply that it is equivalent to a product of Eilenberg-Moore spaces, which it is not. There is no corresponding contradiction in the infty category world. (Dylan, ok?)

Edit: I'll answer comments in order and add a bit of math to my previous answer. Dylan, enjoy yourself and I'll see you when you get back from Vancouver. Mike, Gaunce's paper was published in 1991, when Jacob was 13 years old, so long before $\infty$ categories were born. Harry, you have a choice. The Lewis-May category of spectra satisies 2, 3, part of 4, and 5, but not 1. The EKMM category of S-modules satisfies 1, 3 (but non-cofibrantly, as Dan says), a version of 2, but not 5. I want to expand a bit on 1. In fact, there is a notion of a graded monoidal symmetric monoidal category, never published but known since the 1970's, and the external version of the category of Lewis-May spectra is symmetric monoidal in that sense. The point of EKMM is to internalize the external smash product while retaining as much as possible of the good connection with spaces. Diagram spectra (symmetric and orthogonal) use a more elementary internalization of a symmetric monoidal graded category, sacrificing the close relationship with spaces. The paper ``Diagram spaces, diagram spectra and spectra of units'', https://msp.org/agt/2013/13-4/p01.xhtml, of John Lind shows how best to relate spectra and spaces starting from different models of symmetric monoidal categories of spectra (symmetric monoidal in the good old-fashioned sense of course :).

  • 11
    $\begingroup$ Of course I agree that $\infty$-categorical concepts differ from their point-set counterparts- and that Lewis's theorem is indeed a theorem ;) I was just answering the question that was asked! And alright, I'll get off the toy... but it's the weekend! Can I at least watch a movie or something? $\endgroup$ – Dylan Wilson Feb 9 at 3:25
  • 7
    $\begingroup$ Would you say that it was known even at the time of Lewis's theorem that "there is no corresponding contradiction in the $\infty$-category world", at least in some informal sense predating formal definitions of "$\infty$-category"? Or did that realization emerge later? $\endgroup$ – Mike Shulman Feb 9 at 4:57
  • 1
    $\begingroup$ Professor May, which one of these five properties was violated by EKMM (or simplicial symmetric spectra)? Only the on-the-nose symmetry of the smash product? $\endgroup$ – Harry Gindi Feb 9 at 5:42
  • 4
    $\begingroup$ @HarryGindi Elemndorf wrote a paper about this: 5) fails. See books.google.com/… $\endgroup$ – Dan Ramras Feb 9 at 6:24
  • 3
    $\begingroup$ Actually, in the last paragraph of his 3-page paper, I see that Elmendorf says that there's a different choice for the adjoint pair $(\Omega^\infty, \Sigma^\infty)$ in which 5) holds but 3) fails. Somehow the key underlying weirdness of S-modules, I think, is that the unit for the smash product is not cofibrant. $\endgroup$ – Dan Ramras Feb 9 at 6:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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