MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a connected space $B$, is there always some space $X$ with $B \simeq \mathbf{B}(\mathrm{Aut}(X))$?

Here by space I mean simplicial set, by $\mathrm{Aut}(X)$ I mean the simplicial monoid of auto-equivalences of $X$ (not just strict automorphisms), and by $\mathbf{B}$ I mean the classifying space of this; but I’d also be interested in answers for other reasonable interpretations of these terms.

Edit: A little background — this simply arose out of curiosity, not out of any desired application. Most of the people I’ve mentioned it to have a strong first impulse that the answer should be “no”, but none of us have been able to substantiate this.

share|cite|improve this question
If $B$ is of the form $\Sigma Y$ then $\Omega B$ is a free $A_{\infty}$-space. That means that $Aut(X)$ is a free $A_{\infty}$-space. Which gives a hint that the clam should be wrong. – Ilias Amrani Nov 15 '13 at 22:40
Maybe it's a matter of taste, but this seems like an awkward phrasing to me. I'd rather first say "Is every topological group equivalent to the automorphism group of a space?" which sounds like a natural question; and only then clarify with the technical details of what I mean by "topological group" and "automorphism group of a space." – Ben Wieland Nov 18 '13 at 20:39
Or "Does every connected homotopy type occur as a component of the space of all spaces?" – Tom Goodwillie Nov 19 '13 at 12:58
The rational homotopy version of this question is a well-known open problem. See here for instance: – Mark Grant Sep 11 '15 at 9:19
I think you can solve the problem locally using MatanP's nice answer and Kan-Thurston theorem in the following sense. First, for any connected space $X$, there is a discrete group $G$ such that $BG\rightarrow X$ induces an isomorphism in homology with local coefficients. Second, by MatanP's argument you get that for any connected space $X$ there is a map $BAut(BS)\rightarrow X$ which induces an isomorphism in homology with local coefficients. – Ilias Amrani Sep 28 '15 at 20:49
up vote 11 down vote accepted

Here is the $1$-type case. I assume all spaces are of the homotopy type of CW. Let me write $haut(X)$ (resp. $haut_*(X)$) for the monoid of self-equivalences (resp. pointed ones) to avoid posible confusion with the group-theoretic notation. These spaces have the correct homotopy type by our assumption.

From All Groups are Outer Automorphism Groups of Simple Groups by Droste, Giraudet and Göbel, for every discrete group $G$ we can write $G\cong out(S)$ where $S$ is simple. Recall the exact sequence $0\to Z(S)\to S\to aut(S)\to out(S)\to 0$. Since (thanks to Ricardo Andrade for pointing this) the $S$ appearing in the theorem is in fact non-abelian, $Z(S)=0$.

Looping down the universal fibration with fiber $BS$ we have the homotopy principal fiber sequence $S\to haut_*(BS)\to haut(BS)$ and thus $haut_*(BS)//S\simeq haut(BS)$. But $haut_*(BS)\simeq aut(S)$ and $S\to aut(S)$ is injective so the homotopy quotient $haut_*(BS)//S$ is the (ordinary) quotient $aut(S)/S\cong out(S)\cong G$. Hence, $G\simeq haut(BS)$.

share|cite|improve this answer
Fantastic — this is just the sort of result I was hoping to hear, and definitely worth being an answer not a comment. The last paragraph can be streamlined a little by using the general fact that $\mathrm{haut}(BG)$ is a 1-type with $\pi_0 = \mathit{Out}(G)$ and $\pi_1 = Z(G)$. – Peter LeFanu Lumsdaine Nov 18 '13 at 15:04
I am afraid I do not follow the statement "If $S$ is abelian we are done, so we can assume $S$ has trivial center." at the end of the second paragraph. If $S$ is abelian and non-trivial, then $\operatorname{haut(BS)}$ has non-trivial $\pi_1$. Am I missing something? – Ricardo Andrade Nov 18 '13 at 21:05
On the other hand, that does not really matter, since the article "All Groups are Outer Automorphism Groups of Simple Groups" (by Droste, Giraudet and Göbel) mentioned above actually proves that any group is isomorphic to the outer automorphism group of an infinite simple group $S$; $S$ is, in particular, non-abelian, and thus has trivial centre, as required for the argument. – Ricardo Andrade Nov 18 '13 at 21:05
If $S$ is (simple) abelian, then $G\cong out(S)\cong aut(S)\simeq haut(S)$ where $S$ is thought of as a discrete space. – MatanP Nov 18 '13 at 21:19
Dear @MatanP: For a discrete space $S$, $\operatorname{haut}(S)$ is just the group of self-bijections of $S$ (with the discrete topology), i.e. a symmetric group. The only group $S$ for which this coincides with $\operatorname{aut}(S)$ is the trivial group. – Ricardo Andrade Nov 18 '13 at 22:32

If you are willing to relax your wish from having answer of the form $\text{Aut}(X)$ to the more general group-like topological monoid $\text{Aut}_B(E)$ for a suitable fibration $E \to B$, then the answer is yes. Here $\text{Aut}_B(E)$ is the self-homotopy equivalences of $E$ covering the identity map if $B$. Here's why:

We can assume that $B$ is a CW complex. Let $G \to E \to B$ be a choice of universal principal bundle on $B$, where $G$ is a suitable topological group (here $E$ is contractible, so $B \simeq BG$).

Then $\text{Aut}_B(E) \simeq G$ as topological monoids. Hence, $B \simeq BG \simeq B\text{Aut}_B(E)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.