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\begin{document}
\title[Instances of the Baues problem]
{On some instances of the generalized Baues problem}
\author[Victor Reiner]{Victor Reiner}
\email{reiner@math.umn.edu}
\address{School of Mathematics\\
University of Minnesota\\
Minneapolis, MN 55455}
\keywords{cyclic polytopes, generalized Baues problem,
Spanier-Whitehead duality}
\thanks{Author partially supported by Sloan Foundation and
University of Minnesota McKnight-Land Grant Fellowships.}
\begin{abstract}
We present an approach applicable to certain instances of
the generalized Baues problem of Billera, Kapranov, and Sturmfels.
This approach involves two applications of Alexander/Spanier-Whitehead
duality. We use this to show that the generalized
Baues problem has a positive answer for the surjective map of cyclic polytopes
$C(n,d) \rightarrow C(n,2)$ if $n < 2d+2$ and $d \geq 9$.
\end{abstract}
\maketitle
\centerline{Draft version- April 1998.}
\centerline{Not to be submitted for publication, just for the fun of it!}
\section{Introduction}
The generalized Baues problem (GBP) of Billera, Kapranov, and Sturmfels
\cite[\S 3]{BilleraKapranovSturmfels} asks whether
a certain poset associated to an affine surjection
$\pi:P \rightarrow Q$ of polytopes has the
homotopy type of a sphere, when the poset is endowed with
a standard topology. Although it is known that
this question has a negative answer in general, there are many
interesting special cases for which the answer is known or
conjectured to be positive. For motivation and a survey of general
results on the GBP, see \cite{Reiner}.
The purpose of this paper is to outline an approach to the GBP
under certain conditions on the polytope $P$ and the map $\pi$.
We apply this approach to positively answer the GBP in the
case of the natural surjection of a cyclic polytope $C(n,d)$
onto the cyclic polygon $C(n,2)$ if $n < 2d+2$
and $d \geq 9$. The approach uses Alexander (or more strongly,
Spanier-Whitehead) duality twice, in order to work with posets that
are more tractable than the original. This approach was partly
inspired by the special result on the GBP obtained in
\cite[Theorem 1.2]{ALRS}. It is also similar to a double-usage
of Alexander duality occurring in work of Stanley
\cite[Lemma 2.8]{Stanley} in a somewhat different context.
\section{The approach}
We first introduce subdivisions and the Baues poset $\BauesPQ$.
Let $\pi: P \rightarrow Q$ be an affine surjection of polytopes $P,Q$,
of dimensions $d',d$ respectively. Denote by $V$ the vertex set of
$P$, and say $V$ has cardinality $n$.
Let $\A$ be the point set $\pi(V)$, and we assume for ease of exposition
that $\A$ also has cardinality $n$, that is, no two vertices of $P$ have
the same image under $\pi$; this assumption is not essential for our results. Note that $Q$ is the convex hull $\conv(\A)$.
A {\it subdivision} of $\A$ is a collection of pairs $\{(A_\alpha,Q_\alpha)\}$
where
\begin{enumerate}
\item[$\bullet$] $A_\alpha$ are subsets of $\A$,
\item[$\bullet$] each $Q_\alpha$ is the convex hull of $A_\alpha$ and
is $d$-dimensional,
\item[$\bullet$] the union of the $Q_\alpha$ covers $Q$,
\item[$\bullet$] for any $\alpha, \beta$, the intersection
$Q_\alpha \cap Q_\beta$ is a face $F$ (possibly empty) of both
$Q_\alpha$ and $Q_\beta$, and $A_\alpha \cap F = A_\beta \cap F$
\end{enumerate}
The set of subdivisions of $\A$ is ordered by {\it refinement} in the
following fashion:
$\{(A_\alpha,Q_\alpha)\} \leq \{(A'_\beta,Q'_\beta)\}$ if and only
if for every $\alpha$ there exists some $\beta$ with
$A_\alpha \subseteq A'_\beta$
(and hence also $Q_\alpha \subseteq Q'_\beta$). We will denote
by $\omega(\A)$ the poset of subdivisions of $\A$ ordered by
refinement.
Certain subdivisions of $\A$ called {\it coherent} subdivisions
are singled out by a geometric property which we will not explain
here; see \cite[Chapter 7]{GKZ} or \cite{Lee-triangulations}.
However, we will use the following
fundamental theorem of Gelfand, Kapranov, and Zelevinsky
\cite[Chapter 7, Theorem 2.4]{GKZ}
about the subset of coherent subdivisions, which
partly explains their importance:
\begin{theorem}
\label{secondary-polytopes}
The subposet of coherent subdivisions inside $\omega(\A)$
is isomorphic to the poset of faces
of an $(n-d-1)$-dimensional convex polytope, called the {\it secondary
polytope} $\Sigma(\A)$.
\end{theorem}
\noindent
In particular, $\omega(\A)$ contains a top element $\hat{1}$
corresponding to the very coarse subdivision $\{(Q,\A)\}$.
Our approach to the GBP will only apply in situations where all
subdivisions of $\A$ are coherent. This is a serious restriction,
although there are examples known to have this property- see
\cite{Lee-triangulations}. In particular, when $\A$ is the set of vertices
of a convex polygon in the plane, all of its subdivisions are coherent.
In this paper, when we talk about the topology of a poset,
we are implicitly identifying the poset with the geometric realization of
its {\it order complex}, that is the simplicial complex of chains
in the poset \cite[\S 9]{Bjorner}. With this understanding, if we
assume that all subdivisions of $\A$ are coherent, then the poset
$\omega(\A)-\hat{1}$ triangulates the sphere $\sphere^{n-d-2}$.
This follows because $\omega(\A)-\hat{1}$
is the face poset of the boundary complex of the
$(n-d-1)$-dimensional secondary polytope $\Sigma(\A)$, and
hence triangulates the barycentric subdivision of this complex.
A subdivision of $\A$ is $\pi$-{\it induced} if each of the
sets $\A_\alpha$ has the property that the corresponding subset
$V_\alpha \subset V$ is the set of vertices of a proper face of $P$.
The Baues poset $\BauesPQ$ is defined to be the subposet of
$\omega(\A)$ consisting of the $\pi$-induced subdivisions of $\A$.
With this defined, we can now phrase the GBP:
\begin{question}
{\bf (The generalized Baues problem}) \cite[\S 3]{BilleraKapranovSturmfels}\\
Is the Baues poset $\BauesPQ$ homotopy
equivalent to the sphere $\sphere^{d'-d-1}$?
\end{question}
This problem has been resolved positively in many special cases,
in particular when
$d=1$ \cite{BilleraKapranovSturmfels} or when $d'-d \leq 2$
\cite{RambauZiegler}, but has a negative answer in general
\cite{RambauZiegler}; see \cite{Reiner} for a survey of these results.
For our approach to the GBP, in addition to the Baues poset $\BauesPQ$,
we will consider its complement $\nonBauesPQ:= \omega(\A) - \BauesPQ - \hat{1}$.
Identifying a face of $P$ with the set of its vertices, we let $\faces(P)$
denote the poset of proper non-empty faces of $P$, considered as an induced subposet of
the Boolean algebra $2^V$. We will also consider its complementary
subposet $\nonfaces(P):=2^V - F(P) - \{V\}$.
For two topological spaces $X, Y$, we will write $X \homeo Y$,
$X \approx Y$, and $X \she Y$ resp. to mean that $X$ is {\it homeomorphic},
{\it homotopy equivalent}, or {\it stably homotopy equivalent} to $Y$, respectively.
Recall that $X \she Y$ means that there exists some nonnegative integer
$p$ such that their $p$-fold suspensions are homotopy equivalent,
that is, $\susp^p X \approx \susp^p Y$. In particular, when
$X \she Y$ they share the same integral homology groups in each dimension,
due to the homotopy invariance of homology groups and
the suspension isomorphism
$H_i(X, \integers) \cong H_{i+1}(\susp X, \integers)$.
The following lemma is the crux of our approach to the GBP.
\begin{lemma}
\label{main-lemma}
Using the notations already established, assume
that $\pi: P \rightarrow Q$ satisfies the two hypotheses that
\begin{enumerate}
\item[$\bullet$] all subdivisions of $\A$ are coherent, and
\item[$\bullet$] $\nonBauesPQ \approx \nonfaces(P)$.
\end{enumerate}
Then
$$
\BauesPQ \she \sphere^{d'-d-1}.
$$
\end{lemma}
\begin{proof}
Spanier-Whitehead duality \cite{SpanierWhitehead} asserts that for every
subcomplex $X$ of a CW-sphere $\sphere^m$, there is a another
CW-complex $D_m X$ having the same homotopy type as $\sphere^m - X$,
and the stable homotopy type of $D_m X$ is determined
by the stable homotopy type of $X$. In other words, if
$A \she B$, then $D_m A \she D_m B$.
We will use the fact that when a poset
$\Lambda$ triangulates $\sphere^m$, any subposet
$X \subseteq \Lambda$ has the property that the order complexes of
$X, \Lambda-X$ are deformation retracts of each other's complements
within $\sphere^m$ \cite[Lemma 4.7.27]{BLSWZ}.
Hence $D_m X \she \Lambda-X$ in this situation. With this in mind, we have
the following chain of stably homotopy equivalences, which are explained
below:
\begin{equation*}
\begin{align*}
\BauesPQ &\she D_{n-d-2} \nonBauesPQ \\
&\she D_{n-d-2} \nonfaces(P) \\
&\she D_{n-d-2} D_{n-2} \faces(P) \\
&\she D_{n-d-2} D_{n-2} \sphere^{d'-1} \\
&\she D_{n-d-2} \sphere^{n-d'-2} \\
&\she \sphere^{d'-d-1}
\end{align*}
\end{equation*}
The first line is an application
of Spanier-Whitehead duality to the subspaces
$$
\BauesPQ, \nonBauesPQ \quad \hookrightarrow \quad
\omega(\A)-\hat{1} \homeo \sphere^{n-d-2}
$$
where we have used the assumption that all subdivisions of
$\A$ are coherent to conclude that
$\omega(\A)-\hat{1} \homeo \sphere^{n-d-2}$.
The second line comes
from our assumption that $\nonBauesPQ \approx \nonfaces(P)$.
The third is another application of Spanier-Whitehead duality,
this time to
$$
\faces(P),\nonfaces(P) \quad \hookrightarrow \quad
2^V - \{ \varnothing, V \} \homeo \sphere^{n-2}.
$$
The fourth comes from the fact that $\faces(P)$ triangulates
the boundary of $P$, a $d'$-dimensional polytope. The last
two lines follow from the fact that $D_m \sphere^{k} \she \sphere^{m-k-1}$.
\end{proof}
\begin{remark} \rm \ \\
The conclusion of Lemma \ref{main-lemma} is stronger than
the assertion which follows from a double usage of
Alexander duality, namely that
$\BauesPQ$ has the same integral homology groups as $\sphere^{d'-d-1}$,
but is weaker than the desired conclusion of the
GBP, i.e. that $\BauesPQ \approx \sphere^{d'-d-1}$.
However, it is not as weak as it might seem at first glance,
as we now explain.
First, since the GBP is known to hold whenever $d'-d \leq 2$
\cite{RambauZiegler}, we may
assume that $d'-d \geq 3$, and hence $\sphere^{d'-d-1}$ is simply-connected.
The following well-known lemma, whose proof we include for completeness,
says that when the conclusion
of Lemma \ref{main-lemma} holds, we only need to check whether the fundamental
group of $\BauesPQ$ is trivial:
\begin{lemma}
\label{Whitehead}
Let $X$ be a $CW$-complex with $X \she \sphere^k$ for some $k \geq 2$,
and with $X$ simply connected. Then $X \approx \sphere^k$.
\end{lemma}
\begin{proof}
Since $X \she \sphere^k$, we know that $X$ has the same integral
homology groups as $\sphere^k$. Since $X$ and $\sphere^k$ are
simply-connected, the Hurewicz Theorem \cite[p. 397]{Spanier}
says that they have the same homotopy groups. In particular,
$\pi_k(X) \cong \pi_k(\sphere^k) \cong \integers$, so there
is a map $f:\sphere^k \rightarrow X$ whose homotopy class corresponds
under these isomorphisms to $1 \in \integers$. It follows from
the definition of the Hurewicz homomorphism that $f$ induces
an isomorphism between the $k$-dimensional homology groups of $\sphere^k$ and
$X$. Since both $X, \sphere^k$ have all other homology groups trivial,
an application of a Whitehead theorem \cite[p. 399]{Spanier} says
that $f$ induces an isomorphism between all the homotopy groups
of $\sphere^k$ and $X$. But then since $X$ is a CW-complex,
another Whitehead theorem \cite[p. 405]{Spanier} implies that
$f$ induces a homotopy equivalence between $\sphere^k$ and $X$.
\end{proof}
\end{remark}
\section{Applying the main lemma}
In order to apply Lemma \ref{main-lemma}, we need tools to
compare the homotopy type of the posets $\nonfaces(P)$ and $\nonBauesPQ$.
Our approach will be to find good coverings of spaces homotopy equivalent
to these posets, and then compare the nerves of these covers.
For the remainder of the paper, we will assume that
\begin{enumerate}
\item[$\bullet$]
$P$ is a simplicial polytope, i.e. that its boundary faces are
all simplices, and
\item[$\bullet$]
$\A$ has only coherent subdivisions.
\end{enumerate}
Because of our assumption that $P$ is simplicial, the poset $\nonfaces(P)$
forms an order filter in the Boolean algebra $2^V$, and hence is dual or
oppposite to the face poset of a simplicial complex $\Delta$. We
can therefore replace $\nonfaces(P)$ by $\Delta$ up to homeomorphism. Every
minimal nonface $N$ of $P$ corresponds to a maximal face $F_N$ of $\Delta$,
and we let $\F=\{F_N\}$ be the covering of $\Delta$ by these maximal faces.
This is a good covering in the sense that any intersection
$\cap_{i=1}^r F_{N_i}$ is either empty or contractible, and note
that the latter happens if and only if $\cup_{i=1}^r N_i \subsetneq V$.
Hence by the usual Nerve Lemma \cite[(10.6)]{Bjorner} one can replace
$\nonfaces(P)$ by the $\nerve(\F)$ up to homotopy equivalence.
We summarize our conclusions in the following proposition:
\begin{proposition}
\label{non-faces-nerve}
Assuming $P$ is simplicial, $\nonfaces(P) \approx \nerve(\F)$.
Here $\nerve(\F)$ has vertices indexed by the minimal
non-faces of $P$, and a face for each collection $N_1,\ldots,N_r$
of minimal faces with $\cup_{i=1}^r N_i \subsetneq V$.
\end{proposition}
\begin{remark} \rm \ \\
One might ask whether the assumption that $P$ is simplicial
is important in the previous covering/nerve construction.
Even without assuming that $P$ is simplicial, one can cover
the order complex of the
poset $\nonfaces(P)$ by the order complexes of the subposets $P_N$
where $P_N$ is the set of non-faces of $P$ which contain $N$, and
$N$ ranges over the minimal non-faces of $P$. On the other hand,
this turns out not to always be a good cover. For example, let
$P$ be a triangular prism whose two triangular faces have vertices
labelled $a,b,c$ and $a',b',c'$ in the obvious way, so that $a,a'$
span an edge, as do $b,b'$ and $c,c'$. Then the minimal non-faces
$N_1=ab', N_2=a'b$ have the property that $P_{N_1}, P_{N_2}$
are each contractible, but they intersect in a poset with only two
incomparable elements $\{aa'bb'c, aa'bb'c'\}$ which is neither contractible
nor empty.
\end{remark}
We wish to also replace $\nonBauesPQ$ by something homotopy equivalent,
and to do this, we need to review the notion of the secondary fan
associated to $\A$.
Without loss of generality, assume the points of $\A$ affinely
span $\reals^d$. Identify $\A$ with a $(d+1) \times n$ matrix whose
columns give the coordinates of the points in $\A$ with an extra $(d+1)^{st}$
coordinate equal to $1$ appended to each. Any matrix $(n-d-1) \times n$
matrix $\A^*$ whose row space coincides the nullspace of $\A$ is
called a {\it Gale transform} of $\A$, and we regard $\A^*$ as a configuration
of $n$ points in $\reals^{n-d-1}$ which are its columns. If $a$ is a
point given by some column of the matrix $\A$, let $a^*$ be the point
given by the corresponding column of $\A^*$. Given a subset
$A^* \subset \A^*$, let $\cone(A^*)$ denote the set of
nonnegative linear combinations of the element in $a^*$.
The {\it secondary fan} of $\A$ is the common refinement of all
cones $\cone(A^*)$ for $A^* \subset \A^*$. It turns out that
the secondary fan is the normal fan to the secondary polytope $\Sigma(\A)$.
\begin{theorem}\cite{BilleraFillimanSturmfels}
\label{secondary-fan}
The poset of coherent subdivisions of $\A$ is dual (or opposite) to the
the poset of non-zero cones in the secondary fan of $\A$.
Specifically, a coherent subdivision uses a pair $(A,Q)$
with $A =\{a_1,\ldots,a_r\} \subset \A$
if and only if the corresponding cone of the secondary fan
lies in the relative interior of $\cone(\A^*-\{a^*_1,\ldots,a^*_r\})$.
\end{theorem}
As a consequence, in the case of interest for us when all subdivisions
of $\A$ are coherent, the poset of subdivisions of $\A$ is the
face poset of a regular cell complex homeomorphic to $\sphere^{n-d-2}$,
namely the decomposition of the unit sphere in $\reals^{n-d-1}$ by
the cones of the secondary fan of $\A$. Call this cell complex $K$,
and let $K'$ be the subspace which is the
union of all cells of $K$ indexed by elements
in $\nonBauesPQ$. Although $K'$ need not in general be a subcomplex of $K$,
by \cite[Lemma 6.2]{ALRS} it is homotopy equivalent to $\nonBauesPQ$.
We wish to find a good cover of $K'$. For any minimal non-face
$N$ of $P$, let $K_N$ be the union of all cells of $K$
indexed by subdivisions $\{(A_\alpha,Q_\alpha)\}$ of $\A$ which use
some $A_\alpha$ containing $N$. Then we have a covering $\E=\{K_N\}$
of $K'$ by letting $N$ range over the minimal non-faces of $P$.
Then $\nerve(\E)$ of this covering will have vertices indexed
by the minimal non-faces of $P$, just like $\nerve(\F)$ did.
\begin{proposition}
\label{non-Baues-nerve}
Assuming $P$ is simplicial and all subdivisions of $\A$ are
coherent, $\nonBauesPQ \approx \nerve(\E)$.
A collection $N_1,\ldots,N_r$ of minimal faces indexes a
face of $\nerve(\E)$ if and only if there exists a single
proper subdivision $\{(A_\alpha,Q_\alpha\}$ of $\A$ having some
$A_{\alpha_i}$ containing $N_i$ for each $i$.
\end{proposition}
\begin{proof}
We already have seen that we can replace $\nonBauesPQ$ by the
space $K'$ up to homotopy equivalence. So it suffices to check
that $\E$ is a good covering of $K'$. It is easy to check
from the correspondence between coherent subdivisions and cones
given in Theorem \ref{secondary-fan} that each $K_N$ corresponds
to a convex union of cones in the secondary fan. Therefore,
any intersection $\cap_{i=1}^r K_{N_i}$ corresponds to a convex union
of cones (possibly the $0$ cone), and hence is either empty or
contractible as a subspace of the spherical complex $K$.
The last assertion in the Proposition follows from the last
assertion in Theorem \ref{secondary-fan}.
\end{proof}
By Lemmas \ref{non-faces-nerve}, \ref{non-Baues-nerve} we can
compare the nerves of the two covers $\E, \F$ in place of
comparing the homotopy types of $\nonfaces(P)$ and
$\nonBauesPQ$.
\begin{proposition}
With the above notation, there is an inclusion of simplicial
complexes
$$
i:\nerve(\F) \quad \hookrightarrow \quad \nerve(\E)
$$
\end{proposition}
\begin{proof}
Both nerves have vertex sets indexed by the minimal non-faces
of $P$. Assume $N_1,\ldots,N_r$ are minimal non-faces which span
a face of $\nerve(\F)$, that is $N:= \cup_{i=1}^r N_i \subsetneq V$.
Then there always exist proper subdivisions of $\A$ whose
restriction to $\conv(N)$ contains only the pair $(N,\conv(N))$,
that is leaving $\conv(N)$ completely unsubdivided.
For example, one can use the {\it pulling} construction of
Lee \cite{Lee-triangulations} on the remaining vertices $V-N$.
Therefore $N_1,\ldots,N_r$ span a face of $\nerve(\E)$.
\end{proof}
Finally, we apply these results in a concrete situation, relating to
cyclic polytopes. The {\it cyclic polytope} $C(n,d)$ is the
convex hull of any $n$ points on the {\it moment curve}
$\{(t,t^2,\ldots,t^d)\}$ in $\reals^d$. Although this polytope
depends upon the choice of the $x_1$-coordinates $t_1 < \ldots < t_n$
of the points on the moment curve, much of the combinatorics of these
polytopes does not depend upon this choice. In particular, Gale's
Evenness Criterion \cite[Theorem 0.7]{Ziegler} describes the face lattice
of $C(n,d)$ independent of this choice.
The map $\pi: \reals^{d'} \rightarrow \reals^d$ which forgets the last
$d'-d$ coordinates clearly restricts to a surjection
$\pi:C(n,d') \rightarrow C(n,d)$, and one can show that the
set of subdivisions of $C(n,d)$ and also the subset of $\pi$-induced
subdivisions are independent of the choice of the $t_i$.
Subdivisions and the GBP for these maps between
cyclic polytopes have been studied
a great deal in the recent past
\cite{ALRS, EdelmanReiner, Rambau, EdelmanRambauReiner, RambauSantos},
and in \cite[Conjecture 19]{Reiner} we conjectured that the GBP always
has a positive answer for these maps $\pi:C(n,d') \rightarrow C(n,d)$.
This is known to be true in the following cases:
\begin{enumerate}
\item[$\bullet$] $d=1$ \cite{BilleraKapranovSturmfels}
\item[$\bullet$] $d'-d \leq 2$ \cite{RambauZiegler}
\item[$\bullet$] $n=d'+1$ \cite{RambauSantos}
\item[$\bullet$] $n=d'+2$ and $d=2$ \cite{ALRS}.
\end{enumerate}
We will consider the case where $d=2$ (and rename $d'$ by $d$ for ease
of notation), i.e. $\pi:C(n,d) \rightarrow C(n,2)$.
Note that $P=C(n,d)$ is always a simplicial polytope, and
it is well-known \cite{Lee-associahedron, Lee-triangulations}
that all triangulations of $Q = C(n,2)$ are coherent.
Therefore our approach will apply whenever we can say something
about the inclusion $i: C(n,d) \hookrightarrow C(n,2)$.
\begin{lemma}
\label{nerve-isomorphism}
Consider the map $\pi: C(n,d) \rightarrow C(n,2)$, and let $\E, \F$
be the coverings of $\nonfaces(P),\nonBauesPQ$
described above. Then whenever $n < 2d+2$,
the map
$$
i:\nerve(\F) \hookrightarrow \nerve(\E)
$$ is an isomorphism.
\end{lemma}
\begin{proof}
Let $C(n,d)$ have vertex set $[n]:=\{1,2,\ldots,n\}$,
where $i$ denotes the vertex $(t_i,i_i^2,\ldots,t_i^d)$.
We give the proof only when $d$ is even, since the description of
the minimal non-faces of $C(n,d)$ is slightly simpler in this case.
The case when $d$ is odd is similar.
One can check using Gale's Evenness Criterion that if $d$ is even,
the subsets of $[n]$ which index minimal non-faces of
$C(n,d)$ are exactly the subsets of cardinality $\frac{d}{2}+1$ which
contain no two consecutive residues modulo $n$.
We already know that $i$ is an injective simplicial map,
so we need only show it is surjective. In other words, we must show that if
$N_1,\ldots,N_r$ are minimal non-faces of $C(n,d)$
whose projections into $C(n,2)$ each lie inside some polygon of a
fixed proper subdivision $S$, and if $\cup_{i=1}^r N_i = [n]$,
then $n \geq 2d+2$.
Because $S$ is a proper subdivision of the convex $n$-gon $C(n,2)$,
we can find two subpolygons $Q_1, Q_2$ used in $S$, each of
whose boundaries contain only one interior edge of the polygon.
Since $\cup_{i=1}^r N_i = [n]$, for each $j=1,2$, the polygon $Q_j$ must contain
a collection of projections of the $N_i$ which cover almost all of
its vertices, missing at most two vertices of $Q_j$ (namely the two vertices
on the boundary edge of $Q_j$ which forms an interior edge of $C(n,2))$.
By the description of minimal non-faces of $C(n,d)$ given in the
first paragraph, this implies each of $Q_1,Q_2$ must have at least $d+2$
vertices, so $Q$ must have at least $2d+2$ vertices.
In other words, $n \geq 2d+2$.
\end{proof}
From Lemma \ref{main-lemma} we immediately deduce the following.
\begin{corollary}
\label{stable-Baues}
$\omega(C(n,d) {\overset{\pi}{\rightarrow}} C(n,2)) \she \sphere^{d'-3}$
whenever $n <2d+2$.\qed
\end{corollary}
In light of Corollary \ref{stable-Baues} and Lemma \ref{Whitehead},
we are particularly interested in knowing when $\BauesPQ$ is
simply connected.
\begin{lemma}
\label{simply-connected}
$\omega(C(n,d) {\overset{\pi}{\rightarrow}} C(n,2))$
is simply connected whenever $d \geq 10$.
\end{lemma}
\begin{proof}
We begin by observing that whenever $P$ is simplicial and $\A$ has
only coherent subdivisions, the Baues poset $\BauesPQ$ is actually
the face poset of a regular cell complex. To see this, note that by
Theorem \ref{secondary-polytopes}
the poset $\omega(\A)$ of all subdivisions of $\A$ is the face poset of the
secondary polytope $\Sigma(\A)$, which is a regular cell complex.
One can then check using Theorem \ref{secondary-fan} that since $P$ is
simplicial, the subposet $\BauesPQ$ is an order ideal in $\omega(\A)$,
and hence indexes the cells of some regular cell subcomplex $L$
of the secondary polytope $\Sigma(\A)$.
Recall that the fundamental group of a regular cell complex can
be computed in terms of its $2$-skeleton alone. Therefore it suffices
for us to show that when $d \geq 10$, the $2$-skeleton of the cell complex
$L$ is simply connected.
In fact, we claim that $L$ has the same $2$-skeleton as $\Sigma(\A)$
when $d \geq 10$, due to the neighborliness of cyclic polytopes.
Gale's Evenness Criterion implies $C(n,d)$ is a
$\lfloor d/2 \rfloor$-{\it neighborly} polytope
\cite[Corollary 0.8]{Ziegler}, meaning that every subset of
its vertices having cardinality $\lfloor d/2 \rfloor$ or less
spans a boundary face. It is then easy to describe the
$0$-cells, $1$-cells, $2$-cells in $\Sigma(\A)$
(the {\it associahedron}), and analyze when they are
$\pi$-induced from the surjection $\pi:C(n,d) \rightarrow C(n,2)$:
\begin{enumerate}
\item[$\bullet$] $0$-cells of $\Sigma(\A)$ correspond to triangulations of $C(n,2)$, and since these subdivisions are made up of polygons with at
most $3$ vertices, they will all be $\pi$-induced if $d \geq 6$.
\item[$\bullet$] $1$-cells of $\Sigma(\A)$ correspond to subdivisions
of $C(n,2)$ having mostly triangles and exactly one quadrangle, so they
will all be $\pi$-induced if $d \geq 8$.
\item[$\bullet$] $2$-cells of $\Sigma(\A)$ correspond to subdivisions of $C(n,2)$ having mostly triangles, and either two quadrangles or one pentagon,
so they will all be $\pi$-induced if $d \geq 10$.
\end{enumerate}
We therefore conclude that when $d \geq 10$, the fundamental group of\\
$\omega(C(n,d) {\overset{\pi}{\rightarrow}} C(n,2)$ coincides with
that of $\Sigma(\A) \homeo \sphere^{n-4}$, which is simply connected
since $n~\geq~d~\geq~10$.
\end{proof}
\begin{remark} \rm \ \\
\label{Christos-remark}
C. Athanasiadis (personal communication) has shown that
$\BauesPQ$ is also simply-connected when $d=9$, and hence
one can weaken the hypotheses of the previous Lemma \ref{simply-connected}.
He observes that any $2$-cell in the associahedron $\Sigma(\A)$
which is missing from the subcomplex $L$ considered
in the proof of Lemma \ref{simply-connected} must be pentagonal. He
then gives a clever inductive argument for why the boundary
of each such pentagonal $2$-cell is null-homotopic within $L$.
\end{remark}
From the Lemma \ref{simply-connected}, Remark \ref{Christos-remark},
Corollary \ref{stable-Baues}, and Lemma \ref{Whitehead} we immediately deduce
\begin{corollary}
\label{GBP}
The GBP has a positive answer for $\pi: C(n,d) \rightarrow C(n,2)$
when $n < 2d+2$ and $d \geq 9$.
\end{corollary}
\section{Questions}
\begin{enumerate}
\item Can one do a more detailed analysis
the $2$-skeleton of $\omega(C(n,d) {\overset{\pi}{\rightarrow}} C(n,2)$,
along the lines of Remark \ref{Christos-remark}, thereby further
weakening the hypothesis on $d$ in Lemma \ref{simply-connected}?
\item Can one show that the inclusion of nerves
as in Lemma \ref{nerve-isomorphism} is sometimes a homotopy equivalence
when it is not an isomorphsim, thereby weakening the hypothesis that
$n < 2d+2$ in Lemma \ref{nerve-isomorphism}?
\item One might think of applying our method to the projections
$\pi:C(n,d') \rightarrow C(n,d)$ when $d > 2$.
It was shown in \cite{ALRS} that the only other non-trivial
special cases where $Q=C(n,d)$ has only coherent subdivisions are
$(n,d) = \{(7,3),(8,3),(8,4)\}$. Bearing in mind that the GBP
in this situation is already settled positively in the
cases $d=1$ or $d'-d \leq 2$ or $n=d'+1$,
this means that there is only one remaining case with $d \geq 3$ where
our method might apply, namely $\pi:C(8,6) \rightarrow C(8,3)$.
However, it turns out that this case is also covered by
\cite[Theorem 1.2]{ALRS}, hence there seems to be nothing new to be proved
for projections of cyclic polytopes by this method when $d > 2$.
\item As a step in the proof of Lemma \ref{main-lemma} we observed that
$$\nonfaces(P) \she D_{n-2}\sphere^{d'-1} \she \sphere^{n-d'-2}.$$
This suggests the following stronger question:
\begin{question}
For any $d$-dimensional polytope $P$ with $n$ vertices, is
the poset $\nonfaces(P)$ homotopy equivalent to $\sphere^{n-d-2}$,
not just stably homotopy equivalent?
\end{question}
\noindent
We suspect that the answer is ``Yes", and that the geometry of {\it Gale
diagrams} \cite[Lecture 6]{Ziegler} can be used to prove this,
but have not been able to carry this out.
We also suspect that the answer is ``No" if instead
we only look at the poset of non-faces in some regular cell complex
homeomorphic to $\sphere^{d-1}$, rather than the non-faces
of a convex polytope.
\end{enumerate}
\section{Acknowledgments}
The author thanks Don Kahn for helpful conversations regarding
Spanier-Whitehead duality, and Christos Athanasiadis for the content
of Remark \ref{Christos-remark}.
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