Realization of a bisimplicial set

Question

Hello,

I just need some clarification (or a good reference) for the definition of the realization of a bisimplicial set, this is what i have when $X$ is a bisimplicial set its realization is

$\cup_n X_n \times \Delta[n]$ subject to the following equivalence relation $(d_ix,p) \sim (x,d_ip), (x,p) \in X_n \times \Delta[n-1]$

$(s_ix,p) \sim (x,s_ip), (x,p) \in X_{n-1} \times \Delta[n]$

My question is since the realization of a bisimplicial set is a simplicial set what are the k-simplicies $|X|_k$ are? is it correct that the $|X|_k=\cup_n (X_{n,k} \times \Delta[n]_k)/\sim$ but then what is $\sim$ in this case?

Any help is appreciated.

Answer 1

There are three ways to realize a bisimplicial set, but they all give the same result in the sense that the resulting spaces are functorially homeomorphic.

1) For each fixed n you have a "horizontal" simplicial set where the $k$-simplices are $X_{n,k}$. Realize this to get a space $|X|_n$. These now fit together to form a simplicial space where the space of $n$-simplices is $|X|_n$. Now realize that space.

2) Do the same with $n$ and $k$ reversed.

3) Take the $k$-simplices to be $X_{k,k}$.

You can find a proof that these all give the same result starting around the middle of page 86 in Quillen's paper on Higher Algebraic K-Theory. (Quillen works with simplicial spaces and you want simplicial sets, but a simplicial set is just a simplicial space in which each $X_{n,k}$ has the discrete topology.)

Answer 2

A remix with sampling from all the answers and comments. (Should just be a comment)

A bisimplicial set is a contravariant functor $\Delta^{op}\times\Delta^{op}\to Sets$. Now, you can view a set as a discrete space, as a discrete simplicial set, as a simplicial space in two ways, or as a discrete bisimplicial set. Next consider covariant functors on $\Delta\times\Delta$ sending the object $([n],[k])$ to the product $\Delta[n]\times\Delta[k]$. Now this product of simplices has a model in a number of different categories and taking coends you gives different "realizations" of the original bisimplicial set. The geometric realization is the coend that takes $\Delta{n}\times\Delta{k}$ to be the product of simplices.. There are two ways for one factor of $\Delta{n}\times\Delta{k}$ to be a space and the other factor a simplicial set, giving a simplicial space. The corresponding coends are the simplicial spaces obtained by realizing vertically or realizing horizontally. If you view $\Delta{n}\times\Delta{k}$ as the product simplicial set then the coend is the diagonal of the original bisimplicial set, so one can reasonably refer to the diagonal as the simplicial set realization. Finally, if you take $\Delta{n}\times\Delta{k}$ to be the bisimplicial set that is the external product of two simplicial sets then the coend is the original bisimplicial set.

Next a few general comments addressing questions that weren't asked, why is the geometric realization useful and why all these coends? A simplicial set, a simplicial space or a bisimplicial set is an example of a diagram of spaces. Any diagram of spaces has a homotopy colimit characterized by its homotopy invariant properties. A geometric realization is of interest only when it agrees with the homotopy colimit. This is always the case for simplicial and bisimplicial sets but for simplicial spaces there are additional conditions. Let $D\colon I\to Spaces$ be a diagram of spaces and let $ * \colon I^{op}\to Spaces $ be the constant contravariant functor with value a one point space. Then the coend is the colimit. Let $ * ^{cof} $ be a suitably cofibrant replacement of $*$ in the categroy of $I^{op}$ spaces and assume that $D$ is suitably cofibrant then the coend is the homotopy colimit. There are a bunch of technialities because there are many choices for model structures on digram category. More generally each of the coends from paragraph 2 is trying to be a homotopy colimit.

Answer 3

View $X$ as a simplicial object

$X: \Delta^{op} \rightarrow SSet$

either vertically or horizontally and consider the Yoneda inclusion

$F: \Delta \rightarrow SSet, [n] \mapsto \Delta[n]$

You obtain a functor $\Delta \times \Delta^{op} \rightarrow SSet, ([n], [k]) \mapsto F([n]) \times X([k])$. The realization of $X$ is the coend of this functor. See http://ncatlab.org/nlab/show/bisimplicial+set

Of course, if you realize the realization further to a space, you obtain Steven's realizations since the geometric realization commutes with coends.

Answer 4

I think the easiest way to see it is that :

$|X|_k=\cup_n (X_{n,k} \times \Delta[n]_k)/\sim$ subject to the following equivalence relation $(d_ix,p) \sim (x,d_ip), (x,p) \in X_{n,k} \times \Delta[n-1]_k$

$(s_ix,p) \sim (x,s_ip), (x,p) \in X_{n-1,k} \times \Delta[n]_k$