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For any given topological group $G$ we have Segal's construction/definition of $BG$. I'm recalling it in case the details turn out to be relevant.

Form the disjoint union of $G^n\times\Delta_n$ for $n\geq 0$ and identify points via $(d_i\cdot,\cdot)\sim (\cdot,\partial_i \cdot)$ where $\partial_i:\Delta_{n-1}\to\Delta_n$ are the face maps and $d_i :G^n\to G^{n-1}$ maps $(g_1,\ldots,g_n)\to (g_1,\ldots , g_i g_{i+1},\ldots,g_n)$ for $i\neq 0,n$. When $i=0$ (resp. $n$) the map $d_i$ is the projection on to the first (resp. last) $n-1$ factors.

On the other hand, one can consider $G$ as a topological category with one object and $G$ as its morphisms. One then considers the topological nerve $\mathcal{N}(G)$ which has one $0$-simplex, one $1$-simplex for each element of $G$, $2$-simplices are given by triangles labelled by $g_1,g_2,g_3$ such that the interior of the triangle corresponds to a path $h$ joining $g_1 g_2$ and $g_3$. Higher simplices are defined similarly. If one uses the natural gluing on this collection, one arrives at the geometric realization of $\mathcal{N}(G)$. Let's call it $\mathcal{B}G$. So here's my question, which may very well be known.

How are the two constructions $BG$ and $\mathcal{B}G$ related?

In fact, this prompts a more general question :

Given a topological category (in the sense of Segal) $\mathcal{C}$, i.e., $\mathit{Ob}$ and $\mathit{Mor}$ are topological spaces, one can form the topological nerve $\mathcal{N}(\mathcal{C})$, as explained above, and then take its realization. On the other hand, the usual nerve $N(\mathcal{C})$ of $\mathcal{C}$ is a simplicial object in $\mathbf{Top}$ and we can take its realization. How are these two realizations related?

I would love to know the answer to the above question for the usual notion of topological category too, viz., where only morphisms $\mathit{Mor}(x,y)$ is required to be a (compacty generated Hausdorff) space for any $x,y\in\mathit{Ob}$.

EDIT : May be I hadn't explained the topological nerve definition which Harry succintly does in his comment below. It is also the same definition given in Higher Topos Theory by J. Lurie.

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How is your "usual nerve" any different all from your topological nerve? –  David Carchedi Oct 29 '10 at 0:52
    
They're not the same. The 'topological nerve' looks like a homotopy version of the usual nerve, and contains it as a sub-simplicial space. For example, the space of 2-simplices for NG is G\times G, but the space of 2-simplices for \cal{N}G is (G x G) x_G G^I. But to be concrete, how are the 3-simplices formed? There are several choices I can think of. Also, how is the 'topological nerve' formed for general categories in Top? It's not a straightforward choice (Top-enriched categories are straightforward, though) –  David Roberts Oct 29 '10 at 1:13
    
Maybe I'm confused, what IS the definition of topological nerve, if its not just the enriched nerve? That's what I meant. The only definition of "topological nerve" I know is giving each set $N(C)_n$ coming from the nerve of the underlying category with the subspace topology inherited from $C_1^n$. So, if not this, what do you mean? –  David Carchedi Oct 29 '10 at 1:18
    
What's wrong with doing the following: Consider a topological group to be a Top-enriched category with one object, apply the total singular complex functor to each homspace, then take the homotopy-coherent nerve of this sSet-enriched category. –  Harry Gindi Oct 29 '10 at 1:41
    
By homotopy-coherent nerve, define $\mathcal{N}(C)_n:=Hom_{Cat_\Delta}(FU_\cdot([n]),C)$ where $FU_\cdot$ is the comonad resolution (that is, the bar construction for the free-category comonad). –  Harry Gindi Oct 29 '10 at 1:45

2 Answers 2

They're the same. You can construct the geometric realization of a simplicial space $X:\Delta^{op} \to Top$ by taking its co-end with the functor $\Delta \to Top$ which sends the $n$ to the "standard n-simplex" $\Delta^n$. Segal's construction is an explicit description of this co-end in the particular case that $X$ is the enriched nerve of a topological group.

There is something to be said however, since often this construction "isn't right". To be more specific, its really the "FAT geometric realization" which captures the correct homotopy type of a topological group/groupoid, in the sense of the existence of universal principal bundle over it, or, in a more abstract way, it is equal to the weak homotopy type of its corresponding topological stack. Under some mild conditions on your groupoid (e.g. the unit map being a neighborhood deformation retract), then you still get the right answer. That fat geometric realization is essentially what you get by keeping only the strictly increasing arrows in $\Delta$ when you form the co-end.

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Do we know if the fat realization agrees (up to weak homotopy equivalence) with the realization obtained from the coherent nerve? –  Chris Schommer-Pries Oct 29 '10 at 5:15
    
@Chris: I'm not sure, but, I am very interested if you find out! I'll ask around and likewise let you know. –  David Carchedi Oct 30 '10 at 16:04
    
@Somnath: Since you say you are a newbie with category theory, perhaps my answer is a little "high-browed". May I suggest you take a look at Section VII.2 of "Sheaves in Geometry and Logic", about tensoring functors (taking co-ends). An explicit description of how to compute such tensors is given. I suggest you work through yourself proving that the co-end I claim is in fact the same as Segal's construction, as doing this yourself will certainly make things much more clear for you than anything I can say. –  David Carchedi Oct 30 '10 at 16:10

The correct construction for a topological category is as follows:

If C is a topological category, we can replace it trivially with a simplicial category by taking the simplicial singular complex associated to each hom-space. By abuse of notation, we will call this functor $Sing$.

Now it suffices to give the answer for simplicial categories.

However, to find the classifying space of a simplicial category, we take its associated quasicategory by looking at the homotopy coherent nerve.

The homotopy coherent nerve is usually constructed formally as the adjoint of another functor called $\hat{FU}$, which is the extension of the bar construction $\bar{FU}$ for the associated comonad $FU:Cat\to Cat$ of the free-forgetful adjunction $U:Quiv\rightleftarrows Cat:F$.

Specifically, given any comonad based at $X$, we can form the bar construction, which gives us a functor from $X\to X^{\Delta^{op}}$. This is done by taking objects to be $F_k=F^{k+1}$ the degeneracies to be instances of the comultiplication map $s_i:F_k\to F_{k+1}=F^i\mu F^{k-i}:F^{k+1}\to F^{k+2}$ and faces given by the appropriate application of the counit (the idea is similar to the above, and I leave it as an exercise). In particular, we may take the whole simplicial object in $End(X)$, which gives us our functor $\bar{F}:X\to X^{\Delta^{op}}$

Back to our specific case, we resolve the comonad $FU:Cat\to Cat$ to a functor $\bar{FU}:Cat\to Cat_\Delta$ (since the resolution is trivial on objects , we can say this with a straight face). Restricting $\bar{FU}$ to $\Delta$, which can always be embedded as a full subcategory of $Cat$. By general abstract nonsense, any functor $X\to C$ where C is cocomplete lifts to a unique colimit preserving functor $Psh(X)\to C$ (since taking presheaves gives a "free" cocompletion). Standard notation suggests that we call this functor $\hat{FU}:sSet\to Cat_\Delta$, but following Lurie, we will call it $\mathfrak{C}$. In particular, this functor has a right adjoint called the homtopy coherent nerve, which we can compute as follows:

$$\mathcal{N}(C)_n:=Hom_{Cat_\Delta}(\mathfrak{C}(\Delta^n),C)$$.

for any simplicial category $C$.

Returning to your original case, $BC=\mathcal{N}(Sing(C))$ for a topological category $C$, and for a topological group, we need only notice that a topological group is identical to a one-object Top-enriched category, all of whose morphisms are invertible (something like this).

As for why this is the right definition, I fear I must refer you to Lurie's HTT. It relies on a proof of a certain Quillen equivalence, and alas, the margins are too small...

Edit: Alright, so the reason why they agree is covered in §4.2.4 of HTT, I'm pretty sure.

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and by topological category, you mean Top-enriched category, as opposed to a category in Top... –  David Roberts Oct 29 '10 at 4:05
    
I'm new to this categorical stuff but I think I got your homotopy coherent nerve definition which is used to define the topological nerve and your answer did help improve my understanding. But is it clear that at least in the example of $G$, the two constructions give the same thing? –  Somnath Basu Oct 29 '10 at 4:10
    
@Harry: Can you be more precise about where in Segal's paper he compares the realization of the coherent nerve of G to the realization of the "usual nerve" of G? –  Chris Schommer-Pries Oct 29 '10 at 4:27
    
@Chris: I was wrong, it didn't. Lurie shows it in HTT 4.2.4 –  Harry Gindi Oct 29 '10 at 4:44
    
I mean, I think so. I keep seeing statements that look like it, but then aren't. –  Harry Gindi Oct 29 '10 at 5:00

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