The "grassmannian" of a simplicial complex

Question

This question is mainly a reference request – I have a construction which seems natural, so I am quite convinced it should be standard, but I don't know what it is called.

Take an $n$ dimensional simplicial complex $X$ and define a new simplicial complex $X_{k}$ in the following way – the vertices of $X_{k}$ will be $k$ dimensional simplices of $X$ and two such vertices will be connected by an edge if the two corresponding $k$ simplices in $X$ are contained in a $k+1$ simplex in $X$. This is the skeleton of $X_{k}$ and complete it by gluing higher dimensional simplices whenever they are formed (so you get an $n-k$ simplicial complex).

My questions – what is the standard name for this and where can I find a reference on? Moreover, is it true that whenever $X$ is contractible then $X_k$ is simply connected (and even that the homologies of order < $n-k$ vanish)?

Answer 1

The claims about simple connectivity and homology vanishing in the last paragraph is false.

For a failure of simple connectivity, let $X$ be the two dimensional simplicial complex with vertex set $\{ a,b,c,d,e \}$ and maximal faces $abe$, $bce$, $cde$, $ade$. This is contractible: It is a solid square subdivided into four triangles.

Then $X_1$ retracts onto the circle whose vertices are $ae$, $be$, $ce$ and $de$.

For a failure of homology vanshing, let $X$ be the three dimensional simplicial complex with vertex set $\{ a,b,c,d, p,q,r,s,x \}$ and maximal faces $abpx$, $bpqx$, $bcqx$, $cqrx$, $cdrx$, $drsx$, $adsx$, $apsx$. So we have triangulated a cylinder and coned it from the point $x$. Since there is a cone point, the complex is contractible.

Each tetrahedron of $X$ contributes a hollow octahedron to $X_1$, and they are glued together into a ring. So $X_1$ is homotopic to a wedge of $S^1$ and eight $S^2$'s, and in particular $H^1(X_1) \neq 0$.

Answer 2

What you describe is similar to something well-known that does satisfy your condition (ie, has the same homotopy as $X$ in dimensions less than $n-k$). I'm speaking of the dual complex to the dual $(n-k)$-skeleton of your simplicial complex. Let me call this $X^{\ast (n-k)\ast}$ (which is short for $((X^\ast)^{(n-k)})^\ast$). This is not a simplicial complex in general, only a cone complex. It has one vertex for each $k$-simplex; vertices corresponding to those $k$-simplices that belong to the same $(k+1)$-simplex are connected by an "edge" (that is, a cone over those vertices - notice that there can be more than two of them); and so on. You can find the discussion of dual complexes in standard texts on PL topology, but cone complexes as such are not clearly exposed there; see references on cone complexes in my answer here.