5
$\begingroup$

If $G$ is a discrete group, recall that a (naive) $G$-spectrum consists of based $G$-spaces $E_n$ together with based $G$-maps $\Sigma E_n \to E_{n+1}$, where we give the suspension coordinate the trivial action. A standard example is to take an abelian group $A$ with a $G$-action, i.e., a $G$-module structure. Then the Eilenberg-Mac Lane spectrum $hA$ has the structure of a $G$-spectrum. Notice that taking path components of $hA$ recovers $A$ together with its module structure.

Suppose we are given a extension of groups $$ 0 \to A \to E \to G \to 1 $$ where $A$ is any abelian group. Then we have an associated $G$-module structure (the action is given by lifting elements of $G$ to $E$ and then conjugating).

So we have an operation $$ \lbrace \text{extensions of } G \text{ by } A \rbrace \qquad \mapsto \qquad \lbrace G\text{-module structures on } A \rbrace . $$ The problem is to now fill in the diagram $$ \lbrace \quad\qquad\text{ ?}\qquad \quad \rbrace \quad \mapsto \qquad \lbrace G\text{-action on a spectrum } E \rbrace $$ $$ \downarrow\qquad\qquad \qquad \qquad\qquad \qquad \downarrow $$ $$ \lbrace \text{extensions of } G \text{ by } A\rbrace \qquad \mapsto \qquad \lbrace G\text{-module structures on } A\rbrace . $$ What is the appropriate notion of extension of $G$ by a spectrum?

After all, a spectrum is something akin to "an abelian group up to homotopy."

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ A possible short + lazy answer by turning a classification into a definition: An extension is classified by an element of $H^2(G,A)$. You can give $K(A,2)$ a $G$-action, form the associated bundle $K(A,2) \to Y \to BG$, and then this $H^2$ can be identified with the set of space of sections of $Y \to BG$. Now replace $K(A,2)$ with $B^2 X$ for $X$ an infinite loop space. (A more modern thinker than me might say something about extensions of the $\infty$-groupoid $BG$ by a symmetric monoidal $\infty$-groupoid.) $\endgroup$
    – Tyler Lawson
    Commented Jan 18, 2013 at 6:00
  • 1
    $\begingroup$ Yes, I was aware of the short and lazy answer---but it's not very satisfying. $\endgroup$
    – John Klein
    Commented Jan 18, 2013 at 11:47
  • $\begingroup$ John, it might help if you try to indicate what kind of answer you are hoping for. My first thought is something along the lines of a fiber bundle over $BG$ with fibers $A$'' (a bundle of spectra). But it is a variation of the same short and lazy'' answer. $\endgroup$
    – Gregory Arone
    Commented Jan 18, 2013 at 12:19
  • $\begingroup$ Greg, I'm confused by your answer. I think a bundle over $BG$ with fibers $A$ is really equivalent to specifying a $G$-action on the spectrum $A$ (i.e., the homotopy category of $G$-spectra is equivalent to the homotopy category of fiberwise spectra over $BG$). An "extension," whatever that might be, should have more structure than that: if we fix the given $G$-action on $A$, the extensions that induce the given $G$-action (up to some notion of equivalence) should be in bijection with $H^2(G;A)$, as Tyler points out. $\endgroup$
    – John Klein
    Commented Jan 18, 2013 at 13:41
  • $\begingroup$ @Greg: here's another way to see this classically. Let $A$ be a $G$-module. Then an extension gives a map $BG \to B\text{Aut}(BA)$, where $\text{Aut}(BA)$ is the monoid of unbased self-equivalences of $BA$. A $G$-module structure on $A$ is really a map $BG \to B\text{Aut}(A)$, where $\text{Aut}(A)=$ automorphisms of the abelian group. Applying $H_1$ to an automorphism gives a map $B\text{Aut}(BA) \to B\text{Aut}(A)$. I guess I'm looking for a version of this map in spectra: the hofiber of this map is identified with $B^2A$. By obstruction theory the lifts are classified by $H^2(G;A)$. $\endgroup$
    – John Klein
    Commented Jan 18, 2013 at 13:54

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .