TITLE: Is there a higher homotopical spinor theory?
QUESTION [16 upvotes]: Gabriele Vezzosi and I have been musing on the following. Consider the standard double cover $\mathop{\rm Pin}_{n} \to \mathrm{O}_{n}$, whose kernel is $\mathbb Z/2\mathbb Z$. This allows to associate with every $\mathrm{O}_{n}$-principal bundle $E$ over a space $S$ a class in $\mathrm H^{2}(X, \mathbb Z/2\mathbb Z)$, which is well known to be the second Stiefel--Whitney class of $E$.
Now, suppose that we have given a simplicial group $G$, with a central simplicial subgroup $Z$ such that the quotient $G/Z$ is homotopy equivalent to a topological group $H$. Assume that $Z$ is homotopy equivalent, as a simplicial group, to an Eilenberg--MacLane group $K(A, m)$, where $m$ is a non-negative integer and $A$ an abelian group. Then, if we are not mistaken, we should obtain a map from $H$-principal bundles on $S$ to $\mathrm H^{m+2}(S, A)$, and thus a characteristic class for $H$-principal bundles: the group $G$ gives a homomorphism from $H$ to the classifying simplicial group $B\,K(A,m) = K(A, m + 1)$, which yields a homorphism $B\,H \to B\,K(A,m+1) = K(A, m + 2)$, and this should give the characteristic class.

Now, the question is: for each $m ≥ 0$, is there a simplicial group $\mathop{\rm Pin}_{n}^{(m)}$, with a surjective homomorphism $\mathop{\rm Pin}_{n}^{(m)} \to \mathrm{O}_{n}$, whose kernel is a central simplicial subgroup $K(\mathbb Z/2\mathbb Z, m)$, such that the associated characteristic class is the $(m+2)^{\rm nd}$ Stiefel--Whitney class?
If there isn't, is there some analogue in which $G$ is something less than a simplicial group (maybe an $H$-group, or something along that line)? We think this should be possible, but we'd be interested in an explicit construction, more than in an abstract existence theorem.

Ultimately, we'd like to have an analogous construction for simplicial schemes over a field, in order to study higher Hasse--Witt classes, in the sense of Jardine; but we were wondering if such a construction, in the simpler case of simplicial sets, is known to topologists.

REPLY [4 votes]: This seems to work for getting a central extension (even though it looks a little bit too simple). To be
specific I am using May's notation with $\newcommand{\hw}{\overline{W}}\hw$ for
the classifying space of a simplicial group and
$\newcommand{\kl}{\mathbf{G}}\kl$ for the Kan loop group. Start as Charles does
with the the map $\hw\mathbf{O}_n\rightarrow \hw\hw\mathrm{K}(\mathbb{Z}/2,m)$.
Applying the Kan loop construction to this map we get a map of simplicial groups
$\kl\hw\mathbf{O}_n\rightarrow \kl\hw\hw\mathrm{K}(\mathbb{Z}/2,m)$ and as $G$
and $\hw$ are adjoint we have a map of simplicial groups
$\kl\hw\hw\mathrm{K}(\mathbb{Z}/2,m)\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)$
and composing we get a map of simplicial groups
$\kl\hw\mathbf{O}_n\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)$. Now, there is an
exact sequence of abelian simplicial groups
$0\rightarrow\mathrm{K}(\mathbb{Z}/2,m)\rightarrow
H\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)\rightarrow0$ with $H$ contractible.
Pulling back this along
$\kl\hw\mathbf{O}_n\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)$ gives a central
extension $1\rightarrow\mathrm{K}(\mathbb{Z}/2,m)\rightarrow
H'\rightarrow\kl\hw\mathbf{O}_n\rightarrow1$ which has the right homotopy type.
Note that this goes more or less backwards in the following observations: If
$1\rightarrow Z\rightarrow G\rightarrow H\rightarrow1$ is a central extension,
then as multiplication $Z\times G\rightarrow G$ is a group homomorphism we get a
map $\hw Z\times\hw G\rightarrow\hw G$ which is an action of the simplicial
group $\hw Z$ on $\hw G$ making $\hw G\rightarrow\hw H$ a $\hw Z$-torsor giving
a classifying map $\hw H\rightarrow\hw\hw Z$.