Yeah true, I just meant for G-symmetric spectra there

@Bogdan for what it's worth, you need to work with "symmetric G-spectra" to make this true (because G-symmetric spectra is Hausmann's name for a model, where commutative monoids are equivalent to commutative monoids in orthogonal G-spectra)

I am slightly confused now. Is the main point that being a commutative algebra in these other categories is not the same as being an $E_{\infty}$-algebra?

In LMS and in symmetric G-spectra (as Tom says) commutative algebras are the same as "naive" E_oo algebras

Marc Hoyois told me once that it's only with the black magic of orthogonal specctra that commutative algebras are richer (and I thought that this sentence was pretty badass)

ok. interesting. I am either having deja vu or remembering that Marc Hoyois quote.

@SeanTilson Strictly commutative rings in orthogonal G-spectra are a model for "G-commutative" ring spectra. They are basically E_∞-rings that come equipped with additional "norm maps" N^GR→R. Essentially you can multiply not only a set of elements of R but also a G-set of them. Computationally this expresses itself in the homotopy groups forming a Tambara functor.

Has anyone developed a coherent nerve for simplicially enriched bicategories? I'm fine assuming that all the $2$-morphisms are invertible.

I guess one way is to first strictify, then apply the coherent nerve to the morphism categories to obtain a simplicial category, then apply the coherent nerve again.

Hi all! any idea about how to compute the orientation character for the classifying space of the dihedral group? Since I've not found a geometric description of BD_{2n} (n even) it's hard to guess whether the loops preserve of reverse the orientation. Are there any other ways to figure it out? My aim is to identify the orientation cover, which for other reason can be either: two copies of BD_2n, BD_2(n/2), or BZ_n. But I've no clue how to show which one is the right one

@LuigiM The orientation cover for what? I don't think BD_{2n} has a notion of "local orientation"

I doubt this is a coherent concept. Surely thinking of BZ/2 as a union of even- or odd-dimensional projective planes might convince you of this

From a manifoldy perspective note that the orthogonal group of a Hilbert space is contractible, so you can't even get a decent notion of determinant map from O(H) to Z/2

I see your point, and I thank you for that.I'm trying to compute the integer homology of a Thom space of a non-orientable bundle, and I was told that Thom iso holds if one makes use of twisted coefficients. But then, in order to compute them my only idea was to use the l.e.s.involving the orientation double cover

No, the orientation double cover won't help you here (among other things because it doesn't exist). What you want is the covering space that trivializes the local system, but I am not sure how easy it is to figure it out without more details.

mmh, ok. Problem is I'm struggling finding some references where I can understand what I really need. Thanks anyway!