14
$\begingroup$

Cobordism genera can often be refined to $E_\infty$-orientations in the sense of Ando-Blumberg-Gepner-Hopkins-Rezk:

1) the mod 2 Euler characteristic $MO\to H\mathbb{F}_2$;

2) the $\widehat A$-genus $MSpin\to KO$ (Ando-Hopkins-Rezk, Joachim);

3) the Todd genus $MSpin^c\to K$ (Joachim);

4) the Witten genus $MString\to tmf$ (Ando-Hopkins-Rezk).

Now, the signature (and Arf invariant, etc.) lifts to an $\mathbb{L}$-theory orientation for $PL$-bundles (e.g. Ranicki), and $MPL$ and $\mathbb{L}$ are both $E_\infty$-ring spectra.

My question is the following: is it known whether this orientation has an $E_\infty$ refinement? If the answer is yes, I would appreciate a reference.

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Welcome to Mathoverflow! I have no idea what the answer to this is, but it seems hard. We'd have to know something about, like... the units of L-theory...For KO and tmf we knew stuff about the K(n)−localizations and used Rezk′s magic juice; do we know anything about the K(n)−localizations of L-theory? This sounds hard. But also like buckets of fun. $\endgroup$
    – Dylan Wilson
    Commented May 17, 2013 at 19:17
  • 9
    $\begingroup$ I am unfamiliar with L-theory. Nevertheless, I came across a recent article on the arxiv which seems related: "Commutativity properties of Quinn spectra" (arxiv.org/abs/1304.4759). It states in remark 1.4 that the Sullivan-Ranicki orientation from $MSTop$ to L-theory is a ring map of symmetric ring spectra, which I assume to mean a map of associative/$A_\infty$ monoids. Immediately before that remark, it is also stated that the authors are unaware of any previous result on the multiplicativity of the symmetric signature. Perhaps this article and the references therein will be helpful. $\endgroup$
    – Ricardo Andrade
    Commented May 18, 2013 at 2:52
  • 4
    $\begingroup$ @Dylan: if I remember rightly, $L[1/2]$ is equivalent to an Eilenberg-MacLane spectrum, whereas $L_{(2)}$ is equivalent to $kO_{(2)}$, so the $K(n)$-localizations are not hard. However, these are not $E_\infty$ equivalences (or at least not obviously so). $\endgroup$
    – Neil Strickland
    Commented May 19, 2013 at 17:11
  • 4
    $\begingroup$ Neil: yes, $L$ is well-understood, but you have it backwards: actually, $L[1/2]=KO[1/2]$; $L_{(2)}$ is EM. A simple check: $L$ is 4-periodic, while $KO_{(2)}$ is only 8-periodic. $\endgroup$
    – Ben Wieland
    Commented May 20, 2013 at 20:08
  • $\begingroup$ @Ricardo: I think your comment answers the question. Thanks! $\endgroup$
    – Ben Knudsen
    Commented Nov 9, 2013 at 17:21

1 Answer 1

Reset to default
7
$\begingroup$

[Since my comment above appears to have been helpful, I am repeating it here.]

I must admit I am unfamiliar with L-theory. Nevertheless, I came across a recent article on the arXiv which is related: Commutativity properties of Quinn spectra by Gerd Laures and James McClure. It states in remark 1.4 that the Sullivan–Ranicki orientation from $MSTop$ to L-theory is a ring map of symmetric ring spectra, which I assume to mean a map of associative or $A_\infty$ monoids. Immediately before that remark, it is also stated that the authors are unaware of any previous result on the multiplicativity of the symmetric signature.

Improve this answer
$\endgroup$

You must log in to answer this question.

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