8
$\begingroup$

I have been asked to provide at reference for the fact that if $X\to B(BG)$ (where $G$ is the stable auto-homotopy-uquivalences of spheres - or $BG=Pic(S)$ in some peoples terminology) then the Thom spectrum associated to the based loop map

$$ \Omega X \to BG=Pic(S) $$

is a ring-spectrum. I was looking around for a good reference, but couldn't find one that was very clean cut (the paper is about something in symplectic geometry - so I would like the reference to be precise). I also don't know where this originally appeared - so I would prefer the original reference if it is clear or a reference that points back to the original.

Improve this question
$\endgroup$
12
  • $\begingroup$ A small correction: if $G$ is the stable auto-homotopy equivalences of spheres (what's usually called $F$ or $GL_1(\mathbb{S})$) then $BG$ is only the connected component at the identity of $Pic(\mathbb{S})$. Of course this does not change anything important. $\endgroup$
    – Denis Nardin
    Commented Aug 11, 2016 at 12:07
  • $\begingroup$ Are you are saying that $Pic(S)=\mathbb{Z}\times BG$? $\endgroup$
    – Thomas Kragh
    Commented Aug 11, 2016 at 15:06
  • 1
    $\begingroup$ (cont) The analogy to keep in mind is that, for a general $R$-module $M$, $\mathrm{Aut}(M)$ need not be commutative, but when $M$ is free of rank one, $\mathrm{Aut}(M)=R^\times$ and so it is commutative. $\endgroup$
    – Denis Nardin
    Commented Aug 12, 2016 at 6:32
  • 1
    $\begingroup$ @PeterMay Obviously I meant $E_\infty$ (that is commutative up to infinitely many coherences). This is obviously in contrast with the traditional terminology that you are referring to, but it is gaining traction and I think it is more intuitive. What you called commutative I usually call strictly commutative, to distinguish it (they are, as you note, very different concepts). I am sorry for any confusion that this somewhat non-standard choice of words may have caused. $\endgroup$
    – Denis Nardin
    Commented Aug 13, 2016 at 19:42
  • 1
    $\begingroup$ @PeterMay I think this terminology for commutative monoids is forced upon you by the one on commutative ring spectra if you want the group of units of a commutative ring spectrum to be a commutative monoid (and I at least do). That said this, as all conventions, is mostly a matter of personal taste. $\endgroup$
    – Denis Nardin
    Commented Aug 13, 2016 at 20:38

3 Answers 3

Reset to default
11
$\begingroup$

I think that the original paper is

Mahowald - Ring spectra that are Thom complexes

My favorite reference for the multiplicative properties of Thom spectra is

Antolín-Camarena, Barthel - A simple universal property of Thom ring spectra

Improve this answer
$\endgroup$
10
$\begingroup$

The late Gaunce Lewis's 1978 PhD thesis ``The stable category and generalized Thom spectra'' proved (as a special case) that the Thom spectra of $F$ and its oriented version $SF$ (alias $GL_1(S)$ or $SL_1(S)$) are $E_{\infty}$ ring spectra, the most highly structured kind of ring spectrum. The published version is Chapter IX of Springer LNS 1213, http://www.math.uchicago.edu/~may/BOOKS/equi.pdf, which was written so as to be as independent as possible of the previous chapters.

Improve this answer
$\endgroup$
3
$\begingroup$

Along the great references cited above, I think

Mahowald, Mark; Ray, Nigel A note on the Thom isomorphism. (English) Zbl 0469.55007 Proc. Am. Math. Soc. 82, 307-308 (1981).

is also a good reference. Theorem 1 of this paper, proved as Corollary 3, is what you are after. I must add that Theorem 1 of this paper was originally proved in "Mahowald. Ring spectra that are Thom complexes" that is cited above.

I also like to add that Thom spectrum of a map from a loop space into $BG$ is not necessarily a ring spectrum! The statement is that the Thom spectrum of any $H$-map, in particular loop map, is a ring spectrum.

Improve this answer
$\endgroup$

You must log in to answer this question.

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