# Notion of finite dimensional simplicial space

I was wondering, what a $N$-dimensional simplicial space $X$ should be. Of course the degeneracy maps force the spaces to be nonempty in high dimensions. Currently I have two different versions and i am wondering whether they are equivalent:

1) There are no nondegenerate simplexes above dimension $N$.

2) Let $\Delta|_N$ be the full subcategory of $\Delta$ consisting of the objects $[0],\ldots ,[N]$. The inclusion induces a restriction functor $R$ from $\Delta-$spaces to $\Delta|_N$-spaces, which has a left adjoint $L$. There is a canonical map $L(R(X))\rightarrow X$. $X$ should be called $N$-dimensional, iff this map is an isomorphism,i.e. a homeomorphism on each object of $\Delta$.

To make the question more precise: $2)\Rightarrow 1)$ can be easily seen, as the map $L(R(X))\rightarrow X$ cannot hit any nondegenerate simplex above dimension $N$ by construction. The other way round: Given $1)$, then all the maps occuring in the natural transformation are surjective and continuous. Injectivity should follow from the relations in $\Delta$. So why is the inverse map of sets continuous?

(In the category of simplicial sets (and not spaces) both notions should be equivalent using the same argumentation.)

• Is your map $X\to LR(X)$ going the right way? If L is left adjoint then surely the map goes from $LR(X)$ to $X$ (I think: 'Free group on underlying set goes to group'!) The map is then, geometrically, the inclusion of the N-skeleton. That I suspect from later on is a typo. I think you will find the answer if you describe L explicitly. Commented Feb 12, 2010 at 18:41
• oops sry. you are right . Commented Feb 12, 2010 at 18:49

Interesting question!

I'm cautiously optimistic that 1) $\Rightarrow$ 2). As you say, if $X$ satisfies 1), then $L(R(X))\to X$ is a continuous bijection. (Because it's a statement about point-sets, and the thing is true for simplicial sets.)

If $N=0$, it should be easy: $L(R(X))$ is the constant simplicial space, whose value at each [k] is $X_0=X([0])$. The canonical map $L(R(X))\to X$ is the one which at degree $k$ is given by the map $s:X_0\to X_k$ defined by the composite of degeneracy operators. But we know that the composite $X_0\to X_k\to X_0$ is the identity, where $d:X_k\to X_0$ is a composite of face operators. So if $s$ is a bijection, $d$ is its continuous inverse.

For general $N$, let $Y=L(R(X))$. The functor $R$ also has a right adjoint, which I'll call $M$. Let $Z=M(R(X))$. Just as the space $Y_k$ looks like a colimit of a certain diagram of the spaces $X_0,\dots,X_N$, the space $Z_k$ looks like a limit of a certain diagram of these spaces.

There are canonical maps $Y=L(R(X)) \to X \to M(R(X))=Z$. I would like to claim that the composite $f:Y\to Z$ gives a homeomorphism of $Y$ onto its image. If you can prove this, that will give your result, since a continuous inverse to $Y\to X$ will be given by $X\to f(Y)\approx Y$.

• It may help for comparison with known results to us the skeleton and coskelton functor terminology here. Commented Feb 12, 2010 at 21:54
• Good point. Specifically, L(R(X)) is the Nth "skeleton" of X, and M(R(X)) is the Nth "coskeleton" of X. Commented Feb 13, 2010 at 0:08
• @ Henrik As 'dimension' could be ambiguous, perhaps some sort of name such as N-skeletal might be better for this concept. My worry is that if the spaces of simplices have themselves a dimension there is a risk of confusion. As a 'for instance', if we start with a bisimplicial set and realise geometrically in one direction, then the question still makes sense, yet in that `realised' direction no restriction is being made. Commented Feb 13, 2010 at 10:09
• i agree. The dimension of a simplicial space or set really depends on the chosen functor and not only on the geometric realisation. But i have some hope, that for good spaces of simplices the dimension agrees with some dimension of the realisation. Currently I believe, that if the space of simplices is totally disconnected Hausdorff, then the "dimension" of the simplicial space agrees with the topological dimension of its realisation. Commented Feb 13, 2010 at 21:51
• There are results on profinite simplicial spaces, but not ones relating to dimension of a realisation. I have looked at profiniteness from various angles but generally it seemed not that good an idea (for what I wanted to do) to take geometric realisations. You may have a good reason to want to do that but it does seem a strange thing to me. The category of profinite simplicial spaces is very nice for doing lots of things, and I never had the desire to form some geometric realisation of one of them. Commented Feb 15, 2010 at 17:01

OK I checked, how the adjoint functors looks like. Given any $\Delta|_N$ simplicial space $X$. To define $L(X)$, we have to extend $X$ to the whole category $\Delta$. I am just telling, what $L(X)$ does on $[N+1]$. Then you keep extending the functor in the same way:

$L(X)([N+1]):=(0,\ldots,N)\times X([N])/\sim$, where the equivalence relation is given by $(j,s_k(x))\sim (k+1,s_j(x))$ for $0\le j\le k\le N,x\in X[N-1]$. The $i$-th degeneracy map is induced by the inclusion of the i-th summand. Using the relations in $\Delta$ one can also define the face maps.

The right adjoint functor is given by $M(X)([N+1]):= ( (x_0,\ldots,x_{N+1})|\partial_ix_j=\partial_{j-1}x_i\mbox{ for } 0 \le i < j \le N+1 )\subset \prod_{i=0}^{N+1}X[N]$. The face maps are just the projections and one can define the degeneracy maps using the relations in $\Delta$.

So let $X$ be a $\Delta$-space. The natural transformation is given by $L(R(X))([N+1])\rightarrow M(R(X))([N+1])\qquad (i,x)\mapsto (\partial_0 s_i(x),\ldots,\partial_{N+1} s_i(x))$.

Using the relations in $\Delta$ one can show, that this map is injective. So the remaining question is, whether this map is an open map (considered as a map onto the image).

I just wanted to write a comment, that I could show the upper equivalence conly for simplicial sets and simplicial Hausdorff spaces (but this answer is too long for a comment).

This question still makes sense in other categories replacing 1) by

1') The span of all degenerate subobjects $s_0(X([N-1])),\ldots,s_{N-1}(X([N-1]))$ is the whole of $X([N-1])$.

For example it fails in the category of simplicial groups. There are two simplicial groups $X$ and $Y$ , whose restrictions to $\Delta|_1$ agree. But if you consider the span of all degenerate elements in $X([2])$ resp. $Y([2])$ those spans are not isomorphic. Consider the simplicial set $S$ with one 0-simplex $a$ and 1 nondegenerate 1-simplex $b$. Define $X([k]):=Y([k]):=\mathbb{Z}(S([k])$ for $k=0,1$. Then you can define $X([2])$ as $\mathbb{Z}(a,s_0(b),s_1(b))$ and $Y([2])$ as $\mathbb{Z}(a)\times F(s_0(b),s_1(b))$, where $F(s_0(b),s_1(b))$ denotes the free group in $s_0(b),s_1(b)$.

The face and degeneracy maps in $S$ induce in both cases well defined face and degeneracy maps. Hence this shows, that $X$ and $Y$ give a counterexample.