Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

[edited!] Given a group $\pi$ and an integer $n>1$, what are examples of Eilenberg-Maclane spaces $K(\pi, n)$ that can be constructed as "known" manifolds? (or if not a manifold, say some space people had a pre-existing desire to study before $K(\pi,n)$ spaces were identified as being of interest)

Constructing $K({\bf Z}, 2)$ as ${\bf CP}^{\infty}$ is the only example I know - but there must be more out there.

I'm interested in concrete examples (like the one above) that could, e.g., be given in a Topics grad course for topology students. They seem to be scarse, so it would be nice to know what was known.

Note: I've excluded $n=1$ because most people know examples (or can figure them out) in this case.

share|improve this question
11  
In what sense is $\mathbb{C} \mathbb{P}^{\infty}$ a manifold? –  Pete L. Clark Oct 29 '10 at 2:17
1  
I believe there's a theorem to the effect that they will not be finite-dimensional manifolds, so one necessarily needs to consider Frechet, Banach etc. manfolds. –  David Roberts Oct 29 '10 at 2:38
9  
Given a nice inclusive definition of "manifold" that allows some examples, what would be an example of a weak homotopy type that is not represented by a manifold? –  Tom Goodwillie Oct 29 '10 at 2:46
13  
Any countable simplicial complex is homotopy-equivalent to a Hilbert manifold. The idea is to inductively embed the skeleton into the Hilbert cube in such a way that you have a regular neighbourhood, making your simplicial complex homotopy-equivalent to the open regular neighbourhood -- and since it's open in Hilbert space it's a manifold. –  Ryan Budney Oct 29 '10 at 2:49
9  
There's a lovely paper of Kodama and Michor (2006) where they show that the component of $Imm(S^1,\mathbb R^2)/Diff^+(S^1)$ corresponding to the the figure-8 immersion has that homotopy-type of a $K(\mathbb Z,2)$. Here $Imm(S^1,\mathbb R^2)$ denotes immersions of $S^1$ in the plane, and we're modding out by orientation-preserving reparametrizations. –  Ryan Budney Oct 29 '10 at 4:37
show 11 more comments

5 Answers 5

Let BTOP and BPL be the classifying spaces of topological/PL-sphere bundles and $TOP/PL$ the homotopy fiber of the map $BPL \to BTOP$. The $TOP/PL$ is a model for a $K(\mathbb{Z}/2\mathbb{Z},3)$ by Kirby and Siebenmann. This identifies a third cohomology class as obstruction to get a PL-structure on a topological sphere bundle.

share|improve this answer
    
Ah, I had thought there was some result like this, but had only very vague recollections about it. Are there any proofs known other than Kirby and Siebenmann's? Are there any people working on this kind of thing in this decade? All the results I know in this direction are quite "old". –  Romeo Oct 30 '10 at 15:26
1  
I think, in the book by Madsen and Milgram are some results of this sort. And I don't really know, what you mean, but while there's certainly less activity in the PL-world today than a few decades ago, at least topological manifolds are a topic, quite a few people are still working on. And the quoted result is surely essential to compare the topological, PL and smooth world. –  Lennart Meier Oct 31 '10 at 21:12
2  
I was just wondering who was carrying on the Kirby-Siebenmann, Ranicki, et all torch in the 21st century. A lot of topology grad students I know these days have never really heard the word "PL"... –  Romeo Nov 4 '10 at 0:50
    
@Lennart: What is a reference for this neat fact? Thanks! –  David Carchedi May 20 '13 at 13:06
    
@David: I think, it is in the Kirby-Siebenmann book "Foundational Essays on Topological Manifolds...". –  Lennart Meier May 20 '13 at 17:36
show 2 more comments

Following up on Dai's answer, one can go a step further since $P U(H)$ is obviously a group. So if we can find a contractible space on which it acts freely, the quotient will be the next level up (namely, a $K(\mathbb{Z},3)$.

Such a space can be constructed as follows: take our favourite (separable, though that's not necessary) Hilbert space, $H$, and consider $HS(H)$, the space of Hilbert-Schmidt operators on $H$. This is isomorphic to the Hilbert tensor product $H^* \widehat{\otimes} H$ so is a Hilbert space. Its unitary group is thus contractible. The group $U(H)$ acts on $HS(H)$ by conjugation, and once we divide out by the centre this becomes free. Thus $P U(H)$ acts on $U(HS(H))$ freely and so the quotient is a $K(\mathbb{Z},3)$.

However, as $P U(H)$ does not act centrally on $U(HS(H))$, the iteration stops here.

share|improve this answer
    
That is very nice! –  Andreas Thom Oct 29 '10 at 7:37
    
Very cool, thanks for the details. –  Romeo Oct 30 '10 at 15:27
add comment

If $M$ is a hyperfinite type $I\!I\!I_1$ factor, then (at least conjecturally), its group of outer automorphisms is a $K(\mathbb Z,3)$.

This is based on the following three properties of that von Neumann algebra:
• The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$.
• The group of unitaries in $M$ is contractible.
• The automorphism group of $M$ is contractible (conjectural).

To see that $Out(M)\cong K(\mathbb Z,3)$, apply the long exact sequence of homotopy groups to the following two fiber sequences: $$ U(Z(M)) \to U(M) \to Inn(M) $$ $$ Inn(M) \to Aut(M) \to Out(M) $$


As a consequence, we also get that $BOut(M)\cong K(\mathbb Z,4)$.

I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $Out(M)$ as the automorphism group of a naturally occurring mathemtical object.

share|improve this answer
    
This sounds interesting, but is new to me - any beginning references to recommend? –  Romeo Nov 19 '10 at 19:07
    
Popa, Sorin; Takesaki, Masamichi; The topological structure of the unitary and automorphism groups of a factor. ams.org/mathscinet/search/… –  André Henriques Nov 19 '10 at 21:33
    
How far are we from knowing that $Aut(M)$ is contractible? Is it just tricky, or do people have no idea how to do it? –  David Roberts Oct 2 '13 at 6:00
1  
I enjoyed that talk you gave at MSRI. Is that stack you build really a $K(\mathbb{Z},4)$ or is that also conjectural as in the first part of the answer? Or is it the stack that is the real deal, and we don't really know about the space? –  David Roberts May 28 at 8:06
1  
Out(M) is not a space: it's a sheaf on Top (and if you force it to be a space, then it acquires the coarse topology). Any Sheaf on Top has an associated homotopy type (via its singular complex), and that's what I claim to be a K(Z,4). Whether or not I should keep the adjective "conjectural" is debatable: I think that I can prove all the claims that I'm making, but I haven't worked out all the details yet... –  André Henriques May 28 at 8:16
show 1 more comment

The following example appears in the definition of twisted $K$-theory.

Let $H$ be an infinite dimensional separable Hilbert space over $\mathbb{C}$. Since the unitary group $U(H)$ is contractible, the projective unitary group $PU(H)= U(H)/S^1$ has the homotopy type of $K(\mathbb{Z},2)$. The fact that $BPU(H)\simeq K(\mathbb{Z},3)$ and the fact that $PU(H)$ acts on the space of Fredholm operators $\mathrm{Fred}(H)$ are essential in the definition of twisted $K$-theory.

share|improve this answer
    
Argh, was too slow in my comment to the question. –  David Roberts Oct 29 '10 at 5:38
    
Very nice. Is there a place you recommend for an exposition on this? –  Romeo Oct 29 '10 at 6:28
1  
Atiyah and Segal's paper on Twisted K-theory is where I would start reading about this. –  Loop Space Oct 29 '10 at 6:58
    
The first one... –  David Roberts Oct 29 '10 at 10:11
add comment

There is a very nice model of $K(\mathbb Z,n)$ which is given by the free abelian topological group on the pointed space $(S^n,\star)$, let us call that $F(S^n,\star)$. An element in $F(S^n,\star)$ is given by a finite set of points in $S^n \setminus \lbrace\star\rbrace$ such that each point in this finite carries a non-zero integer as a label with the obvious addition. The topology is more subtle to describe and made in such a way that $F(S^n,\star)$ is an abelian topological group, the inclusion $S^n \subset F(S^n,\star)$ is continuous and $\star=0$ in $F(S^n,\star)$.

Though, I am not sure whether $F(S^n,\star)$ is an infinite-dimensional manifold (I think not), it is still pretty regular being a topological group and a CW-complex at the same time.

This is all very classical and was studied in detail in

Dold, Albrecht; Thom, René, Quasifaserungen und unendliche symmetrische Produkte., Ann. of Math. (2) 67 1958 239–281.

share|improve this answer
3  
More generally, if M is any Moore space (many of these are finite dimensional manifolds!) then taking the geometric realization of the free abelian group on the singular simplices of M will give you the corresponding Eilenberg-MacLane space. –  Saul Glasman Oct 29 '10 at 13:13
1  
Or you could take symmetric prodcuts of $S^n$ labelled by elements of any abelian group $A$: that produces $K(A,n)$. But is that "geometric"? –  Johannes Ebert Oct 29 '10 at 14:00
    
@Saul: Wow, nice, I've never seen that construction before. Where does it come from? –  Romeo Oct 30 '10 at 15:28
    
@Andreas, cool, thanks, this kind of thing is perfect (and definitely wouldn't have found that on my own). Never read a paper of Dold before... –  Romeo Oct 30 '10 at 15:30
add comment

Your Answer

 
discard

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.