# What's the relationship between B Aut(G) and B Aut(BG) for a (discrete) group G?

(Some of the notational choices I"m about to make might be iffy; I'm happy to take suggestions for improvements.)

Let $G$ be a (discrete) group. Think of it as an object in the $2$-category of small categories; then it has an automorphism $2$-group $\text{Aut}(G)$, and this automorphism $2$-group has a classifying space $B \text{Aut}(G)$ given, for example, by taking the geometric realization of the nerve. On the other hand, let $X$ be a space with (weak) homotopy type $BG$. Under mild conditions $\text{Aut}(X)$, the group of homeomorphisms from $X$ to itself, is a topological group, which also has a classifying space $B \text{Aut}(X)$. (Edit: The comments below strongly suggest I should instead be talking about the topological monoid of homotopy equivalences from $X$ to itself, so let's do that instead.)

Are these classifying spaces (weakly) homotopy equivalent?

• Are you sure you don't mean homotopy equivalences from X to itself instead of homeomorphisms? – Tom Goodwillie Jan 31 '14 at 2:09
• @Tom: oh, hmm. Will the corresponding classifying spaces have different homotopy types in general? I really know nothing about these groups. – Qiaochu Yuan Jan 31 '14 at 2:11
• Think of $G$ trivial and $X\sim BG$ a closed $1$-disk. – Tom Goodwillie Jan 31 '14 at 2:47
• But if $Aut(X)$ means the topological monoid of homotopy equivalences $X\to X$ then it has a classifying space in some sense -- anyway, a delooping -- whose homotopy type depends only on that of $X$ under mild assumptions. And if $X$ is an Eilenberg-MacLane space $BG$ for discrete $G$ then the homotopy type of $BAut(X)$ will depend on $G$ in the way you are thinking of, I believe. I don't know $2$-categories and $2$-groups well enough to really know what I'm talking about, but this is probably your $BAut(G)$. – Tom Goodwillie Jan 31 '14 at 2:54
• Its $\pi_1$ is the outer automorphisms of $G$ (cokernel of ordinary group homomorphism $G\to Aut(G)$) and its $\pi_2$ is the center of $G$ (the kernel of the same), and the other $\pi_n$ are trivial. – Tom Goodwillie Jan 31 '14 at 2:54

The 2-nerve of the automorphism 2-group of the groupoid $(G\Rightarrow *)$ is indeed homotopy equivalent to $BAut(BG)$.
$$\text{1-nerve of the 2-group of automorphisms of (G\Rightarrow *)}$$
is isomorphic, not just homotopy equivalent, as a group object in simplical sets to the $$\text{simplicial set of automorphisms of NG},$$ where $NG$ denotes the simplicial nerve of the group $G$.
Since $NG$ is a fibrant simplicial set, its simplicial set of self-maps is homotopy equivalent to the topological space of self-maps of $BG:=|NG|$. That equivalence restricts to an equivalence between the self-maps of $NG$ that are homotopy equivalences and the self-maps of $|NG|$ that are homotopy equivalences. Now, one notes that a self-map of $NG$ is a homotopy equivalence iff it is an isomorphism. Therefore, the $$\text{simplicial set of automorphisms of NG}$$ is homotopy equivalent to the topological monoid of $$\text{self-homotopy equivalences of |NG|}.$$ Moreover, that equivalence is induced by a map of monoids from the geometric realization of the former to the latter.
To get the equivalence that you asked about, apply $B$ to both sides and use the fact that $B$ commutes with geometric realizations'', that is, there is a commutative diagram (up to invertible natural transformation) of functors $$\begin{matrix} \text{group objects in simplcial sets} & \stackrel B \to & \text{simplicial sets}\\ \downarrow |-| & & \downarrow |-| \\ \text{topological groups} & \stackrel B \longrightarrow & \text{topological spaces} \end{matrix}$$