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I am wondering if there are "elementary" ways to check for $n$-connectedness of simplicial complexes in terms of simple conditions on vertices.

For example

Let $X$ be the curve complex on a surface $S$ (the exact type of curve complex does not matter). That is, the vertices of $X$ correspond to curves on $S$. A $p$-simplex of $X$ consists of $p+1$ distinct, disjoint curves.

I am hoping to get conditions like the following examples:

$X$ is connected iff given a curve $c$, and any other curve $\gamma$, there is a sequence of curves $\gamma = \gamma_1, \ldots \gamma_n = c$ such that $\gamma_i$ is disjoint from $\gamma_{i-1}$ and $\gamma_{i+1}$. ie. This is just the saying there is a path from any vertex to the vertex corresponding to $c$.

For $1$-connected, it might look something like this: $X$ is $1$-connected iff for a fixed curve $c$, given an edge in $(a,b)$ in $X$, ie two disjoint curves, there is a sequence of disjoint curves (edges) $(a,b) = (a_1,b_1), \ldots ,(a_n,b_n)$ such that

  1. EITHER ($a_i = a_{i+1}$ and $b_{i+1}$ is disjoint from both $a_{i-1}$ and $b_{i-1}$) OR ($b_i = b_{i+1}$ and $a_{i+1}$ is disjoint from both $b_{i-1}$ and $a_{i-1}$)

  2. $b_n = c$ OR $a_n = c$

For a picture of what the $1$-connected argument is trying to achieve imagine a closed path and slowly filling in with triangles. For each edge $(a,b)$ we get paths $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ which bound a disk (the disjointness conditions ensure we can fill in with triangles). Actually, this doesn't quite work completely, since edges $(a,b)$ and $(b,c)$ might give rise to different $b$ paths, so one has to ensure that there is a disjointness condition for adjacent vertices also.

So can anyone point me in the right direction if such conditions exist? While I am not sure about the exact form of the answer I want, it would be good if it were something concrete like my above examples to work with. Or could it just be that this is this not really the right question to consider for connectivity arguments?

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I'm not quite sure what exactly you are looking for. An elementary way for proving $n$-connectedness results is the usage of $n$-connected carriers, see for example and the references therein. – j.p. May 15 '11 at 18:03
up vote 2 down vote accepted

I don't know whether this is useful in practice, but the straightforward generalization of your 0-connectedness criterion to a 1-connectedness criterion goes as follows.

Any 1-cycle of the form $[(a_0, a_1), (a_1, a_2), \ldots, (a_k, a_0)]$ can be transformed into the trivial (empty) 1-cycle by a sequence of the following moves:

  • $[\ldots, (a, b), (b,c),\ldots] \to [\ldots, (a,c),\ldots]$ if $a$, $b$ and $c$ are mutually disjoint (i.e. they correspond to a 2-cell in the complex)

  • $[\ldots, (a,c),\ldots] \to [\ldots, (a, b), (b,c),\ldots]$ if $a$, $b$ and $c$ are mutually disjoint

  • $[\ldots, (a, b), (b,b'),\ldots] \to [\ldots, (a, b),\ldots]$ if $b$ and $b'$ are parallel (remove degenerate 1-cell)

  • $[\ldots, (a, b), \ldots] \to [\ldots, (a, b),(b,b'),\ldots]$ if $b$ and $b'$ are parallel (insert degenerate 1-cell)

  • $[(a, b), (b,c), (c, a)] \to [\;\;]$ (empty) if $a$, $b$ and $c$ are mutually disjoint

The basic idea is that a triangulated disk is "shellable" in some sense. There should be a similar criterion for general $k$-connectedness (in terms of $k$-dimensional Pachner moves on triangulations of the $k$-sphere), but I haven't thought carefully about the shellability issues in higher dimensions.


I think that allowing degenerate simplices (i.e. parallel disjoint curves) means that one doesn't need to worry about shellability.

In case it wasn't clear what I meant by Pachner moves, here they are for $k=2$. The context is a triangulation of the 2-sphere where the vertices are labeled by curves and the curves at the vertices of a $j$-simplex ($j=1,2$) are mutually disjoint.

  • Two adjacent triangles $(a,b,c),(b,c,d)$ can be replaced by $(a,d,b), (a,d,c)$ if $a$ and $d$ are disjoint.

  • A single triangle $(a,b,c)$ can be replaced by three adjacent triangles $(a,b,d),(a,c,d),(b,c,d)$ if $d$ is disjoint from $a$, $b$ and $c$ (and we can also do the reverse move, replacing three triangles by one).

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Thanks Kevin, this is the type of answer I am looking for. I'll need to think a bit more for the $k$-connected case. The reason this might be useful to me is that for the complex I am working with, the only nice description of what the simplices are, are through disjointness properties. However, because given a simplex, it is easy to find candidate vertices (with other conditions I might like) that are connected to it, I can then run this sort of argument. I am however not sure if this is the "best" way to do this. – TriThang Tran May 17 '11 at 0:59

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