text:
\begin{document}

\title{LVMB manifolds and simplicial spheres}
\author{Jerome Tambour}

\maketitle

\begin{abstract}
\noindent
LVM and LVMB manifolds are a large family of examples of non kähler manifolds. For instance, Hopf manifolds and Calabi-Eckmann manifolds can be seen as LVMB manifolds. The LVM manifolds have a very natural action of the real torus and the quotient of this action is a simple polytope. This quotient allows us to relate closely LVM manifolds to the moment-angle manifolds studied (for example) by Buchstaber and Panov. The aim of this paper is to generalize the polytopes associated to LVM manifolds to the LVMB case and study its properties. In particular, we show that it belongs to a very large class of simplicial spheres. Moreover, we show that for every sphere belonging to this class, we can construct a LMVB manifold whose associated sphere is the given sphere. We use this latter result to show that many moment-angle complexes can be endowed with a complex structure (up to product with circles).
\end{abstract}

\vskip 4mm
\section*{Introduction}
It is not easy to construct non kähler compact complex manifolds. The simplest example is the well know Hopf's manifold ($\cite{H}$, 1948), which gives a complex structure on the product of spheres $S^{2n+1}\times S^1$ as a quotient of $\C^n\backslash\{0\}$ by the action of a discrete group. The Hopf's manifold has many generalizations: firstly, by Calabi and Eckmann $\cite{CE}$ who give a structure of complex manifold on any product of spheres (of odd dimension). Then by Santiago Lopez de Medrano, Alberto Verjovsky ($\cite{LdM}$ and $\cite{LdMV}$) and Laurent Meersseman $\cite{M}$. In these last generalizations, the authors obtain complex structures on products of spheres, and on connected sums of products of spheres, also constructed as a quotient of an open subset in $\C^n$ but by the action of a nondiscrete quotient this time. These manifolds are known as \emph{LVM manifolds}.

\vskip 4mm

The construction in $\cite{M}$ has (at least) two interesting features: on the one hand, the LVM manifolds are endowed with an action of the torus $(S^1)^n$ whose quotient is a simple convex polytope and the combinatorial type of this polytope characterizes the topology of the manifold. On the other hand, for every simple polytope P, it is possible to construct a LVM manifold whose quotient is $P$.

\vskip 4mm

In $\cite{B}$, Frédéric Bosio generalize the construction of $\cite{M}$ emphasing on the combinatorial aspects of LVM manifolds. This aim of this paper is to study the \emph{LVMB manifolds} (i.e. manifolds constructed as in $\cite{B}$) from the topological and combinatorial viewpoints. In particular, we will generalize the associated polytope of a LVM manifold to our case and prove that this generalization belongs to a large class of simplicial spheres (named here \emph{rationally starshaped spheres}).

\vskip 4mm

In the first part, we briefly recall the construction of the LVMB manifolds as a quotient of an open set in $\C^n$ by an holomorphic action of $\C^m$. In the second part, we study \emph{fundamental sets}, the combinatorial data describing a LVMB manifold and its connection to pure simplicial complexes\footnote{In this paper, the expression "simplicial complex" (or even "complex") stands for "abstract simplicial complex" and a "geometric simplical complex" is a geometric complex whose cells are (straight) simplices in some Euclidean space $\R^n$.}. Mainly, we show that an important property appearing in $\cite{B}$ (the $SEU$ property) is related to a well-known class of simplicial complexes: the pseudo-manifolds. In the same part, we also introduce the simplicial complex associated to a LVMB manifold and we show that this complex generalizes the associated polytope of a LVM manifold. In the third and forth parts, we are mainly interested in the properties of this complex and we show that the complex is indeed a simplicial sphere. To do that, we have to study another action whose quotient is a toric variety closely related to our LVMB manifold. This action was already studied in $\cite{MV}$ and $\cite{CFZ}$ but we need a more thorough study. Finally, in the fifth part, we make the inverse construction: starting with a rationally starshaped sphere, we construct a LVMB manifold whose associated complex is the given sphere. Using this construction, we show an important property for moment-angle complexes: up to a product of circles, every moment-angle complex arising from a starshaped sphere can be endowed with a complex structure of LVMB manifold.

\vskip 4mm

To sum up, we prove the following theorems:

\vskip 2mm

\begin{thm} \label{thmprinc} Let $\cal N$ be a LVMB manifold. Then its associated complex $\cP$ is a rationally starshaped sphere. Moreover, if $\cal N$ is indeed a LVM manifold, $\cal P$ can be identified with (the dual of) its associated polytope.
\end{thm}

\vskip 2mm

\begin{propnonnum}
Every rationally starshaped sphere can be realized as the associated complex $\cal P$ for some LVMB manifold.
\end{propnonnum}

\vskip 2mm

\begin{thm} Up to a product of circles, every moment-angle complex arising from a starshaped sphere can be endowed with a complex structure of LVMB manifold.
\end{thm}

\subsection*{Notations}
In this short section, we fix several notations which will be used throughout the text:

\begin{itemize}
	\item We put $I_z=\{\ k\in\{1,\dots ,n\}\ /\ z_k\neq 0\ \}$ for every $z$ in $\C^n$.
	\item $\D$ is the closed unit disk in $\C$ and $S^1$ its boundary.
	\item Moreover, $exp$ will be the map $\C^n\rightarrow (\C^*)^n$ defined by $exp(z)=(e^{z_1},\dots ,e^{z_n})$ (where $e$ is the usual exponential map of $\C$).
	\item The characters of $(\C^*)^n$ will be noted $X_n^m$: $ X_n^m(z)=z_1^{m_1}\dots z_n^{m_n}$.
	\item And the one-parameter subgroups of $(\C^*)^n$ will be noted $\lambda_n^u$: $\lambda_n^u(t)=(t^{u_1},\dots,t^{u_n})$.
	\item $<,>$ is the usual \textbf{non hermitian} inner product on $\C^n: <z,w>=\sum_{j=1}^nz_jw_j$
	\item We will identify $\C^m$, as a $\R$-vector space, to $\R^{2m}$ via the morphism
	\[z\mapsto (Re(z_1),\dots,Re(z_n),Im(z_1),\dots,Im(z_n))\]
	\item We set $Re(z)=(Re(z_1),\dots,Re(z_n))$ and $Im(z)=(Im(z_1),\dots,Im(z_n))$, so \[z=Re(z)+iIm(z)=(Re(z),Im(z))\]
	\item In $\R^n$, Conv(A) (resp. pos(A)) is the convex hull (resp. the positive hull) of a subset A.
	\item For every $v$ in $\R^n$, we set $\widetilde{v}=(1,v)\in\R^{n+1}$.
\end{itemize}

\vskip 4mm
\section{Construction of the LVMB manifolds}

In this section, we briefly recall the construction of LVMB manifolds, following the notations of $\cite{B}$. Let $M$ and $n$ be two positive integers such that $n\geq M$. A \emph{fundamental set} is a nonempty subset $\cal E$ of $\cP(\{1,\dots,n\})$ whose elements are of cardinal $M$. Elements of $\cal E$ are called \emph{fundamental subsets}.

\vskip 4mm

\begin{rmq} For practical reasons, we sometimes consider fundamental sets whose elements does not belong to $\{\ 1,\dots,n\ \}$ but to another finite set (usually $\{\ 0,1,\dots, n-1\ \}$).
\end{rmq}

\vskip 4mm

Let $P$ be a subset of $\{1,\dots,n\}$. We say that $P$ is \emph{acceptable} if $P$ contains a fundamental subset. We set $\cal A$ as the set of all acceptable subsets. Finally, an element of $\{1,\dots,n\}$ will be called \emph{indispensable} if it belongs to every fundamental subset of $\cal E$. We say that $\cal E$ is of type $(M,n)$ (or $(M,n,k)$ if we want to emphasize the number $k$ of indispensable elements).

\vskip 4mm

\begin{exempl}\label{exemple} For instance, $\cE=\{\ \{1,2,5\},\{1,4,5\},\{2,3,5\},\{3,4,5\}\ \}$ is a fundamental set of type $(3,5,1)$.
\end{exempl}

\vskip 4mm

Two combinatorial properties (named $SE$ and $SEU$\footnote{for Substitute's Existence and Substitute's Existence and Uniqueness} respectively, cf. $\cite{B}$) will be very important in the sequel:
\[
\forall\ P\in\cE,\hskip 2mm \forall\ k \in\{1,\dots,n\},\hskip 2mm \exists\ k'\in P;\hskip 2mm (P\backslash\{k'\})\cup\{k\}\in\cE
\]

\vskip 4mm

\[
\forall\ P\in \cE, \hskip 2mm \forall\ k\in\{1,\dots,n\}, \hskip 2mm \exists!\ k'\in P; \hskip 2mm (P\backslash\{k'\})\cup\{k\}\in\cE
\]

\vskip 7mm

Moreover, a fundamental set $\cal E$ is \emph{minimal} for the $SEU$ property if it verifies this property and has no proper subset $\widetilde\cE$ such that $\widetilde\cE$ verifies the $SEU$ property. Finally, we associate to $\cE$ two open set $\cS$ in $\C^n$ and $\cV$ in $\PP^{n-1}$ defined as follow:

\vskip 2mm

\[\cS=\{\ z\in \C^n/\  I_z\in \cA\ \}\]

\vskip 4mm

The open $\cal S$ is the complement of an arrangement of coordinate subspaces in $\C^n$. We also remark that, for all $t\in\C^*$, we have $I_{tz}=I_z$ so the following definition is consistant (where $\PP^{n-1}$ is the complex projective space of dimension $n-1$):

\vskip 2mm

\[\cV=\{\ [z]\in \PP^{n-1}/\  I_z\in \cA\ \}\]

\vskip 4mm

\begin{exempl} In \ref{exemple}, $\cE$ verifies the $SEU$ property and the set $\cS$ is

\[
\cS=\{\ (z_1,z_2,z_3,z_4,z_5)\ /\ (z_1,z_3)\neq0,\ (z_2,z_4)\neq0,\ z_5\neq0\ \}\ \simeq\ \left(\C^2\backslash\{0\}\right)^2\times\C^*
\]

\end{exempl}

\vskip 4mm

Now, we suppose that $M=2m+1$ is odd and we fix $l=(l_1,\dots,l_n)$ a family of elements of $\C^m$. We can define an action (called \emph{acceptable holomorphic action}) of $\C^*\times\C^m$ on $\C^n$ defined by:

\vskip 2mm

\[
(\alpha,T)\cdot z=(\ \alpha e^{<l_1,T>}z_1,\ \dots,\ \alpha e^{<l_n,T>}z_n\ )\hskip 4mm \forall\ (\alpha,T,z) \in \C^*\times\C^m\times\C^n
\]

\vskip 4mm

\begin{rmq} If $m=0$, the previous action is just the classical one defining $\PP^{n-1}$.
\end{rmq}

\vskip 4mm

In the sequel, we focus only on families $l$ such that for every $P\in \cE$, $(l_p)_{p\in P}$ spans $\C^m$ (seen as a $\R$-affine space). In this case, we said that $(\cE,l)$ is a \emph{acceptable system}. If $\alpha\in\C^*$, $T\in\C^m$ and $z\in\C^n$, we have $I_{(\alpha,T).z}=I_z$, so $\cal S$ is invariant for the acceptable holomorphic action. Moreover, the restriction of this action to $\cal S$ is free (cf. $\cite{B}$, p.1264).

\vskip 4mm

As a consequence, we note $\cal N$ the orbit space for the action restricted to $\cal S$. Notice that we can also consider an action of $\C^m$ on $\cal V$ whose quotient is again $\cal N$. We call $(\cE,l)$ a \emph{good system} if $\cal N$ can be endowed with a complex structure such that the natural projection $ \cS\rightarrow{\cal N}$ is holomorphic. Such a manifold is known as a \emph{LVMB manifold}. Since a quotient of a holomorphic manifold by a free and proper action (cf. $\cite{Hu}$, p.60) can be endowed with a complex structure, we only have to check the last property. Here, according to $\cite{Hu},\ p.59$, we define a proper action of a Lie Group $G$ on a topological space $X$ as a continuous action such that the map $G \times X \rightarrow X \times X$ defined by $(g,x)\ \mapsto (g\cdot x,x)$ is proper. Notice that in our case, the group $G$ is not discrete. If $G$ is a discrete group which acts freely and properly on a space $X$, then the action is properly discontinuous.

\vskip 4mm

According to $\cite{B}$, we make the following definition:

\vskip 3mm

\begin{defn} Let $(\cE,l)$ be a acceptable system. We say that it verifies the \emph{imbrication condition} if for every $P,Q$ in $\cal E$, the convex hulls $Conv(l_p,p\in P)$ and $Conv(l_q,q\in Q)$ does not have disjoint interiors.
\end{defn}

\vskip 4mm

In $\cite{B}$, the following fundamental theorem is proved:

\vskip 2mm

\begin{thm}[$\cite{B}$, p.1268] \label{thmBosio1} A acceptable system is a good one if and only if $(\cE,l)$ verifies the $SE$ property and the imbrication condition.
\end{thm}

\vskip 4mm

\begin{rmq} In $\cite{B}$, it is also proved that a good system is minimal for the $SEU$ property.
\end{rmq}

\vskip 4mm

Finally, a \emph{LVM manifold} is a manifold constructed as in $\cite{LdMV}$ or $\cite{M}$. We don't explain here the whole construction of the LVM manifolds. The only thing we need here is that is a special case of LVMB. Indeed, we have the following theorem (and we use this theorem as a definition of LVM manifolds):

\vskip 2mm

\begin{thm}[$\cite{B}$, p.1265] \label{LVM} Every LVM manifold is a LVMB manifold. To be more precise, let $\cal O$ be the set of points of $\C^m$ which are not in the convex hull of any subset of $l$ with cardinal $2m$. Then, a good system $(\cE,l)$ is the good system of a LVM manifold if and only if there exists an unbounded component $O$ in $\cal O$ such that $\cal E$ is exactly the set of subsets $P$ of $\{1,\dots,n\}$ with $(2m+1)$ for cardinal such that $O$ is contained in $Conv(l_p,p\in P)$.
\end{thm}

\vskip 4mm

\begin{exempl} We come back to \ref{exemple}. If we set $l_1=l_3=1$, $l_2=l_4=i$ and $l_5=0$, then the imbrication condition is fulfilled and $(\cE,l)$ is a good system. As a consequence, \ref{thmBosio1} and \ref{LVM} imply that $\cN$ can be endowed with a structure of a LVM manifold. In $\cite{LdMV}$, the LVM manifolds constructed from a good system of type $(3,n,k)$ are classified up to diffeomorphism. Here, $\cE$ has type $(3,5,1)$ and we have $\cN\approx S^3\times S^3$. Notice that we can also use the theory of moment-angle complex (cf.the last section of this text) to do this calculation.
\end{exempl}

\vskip 4mm

To conclude this section, we recall how is constructed the polytope associated to a LVM manifold $(\cE,l)$. It is clear that the natural action of $(S^1)^n$ on $\C^n$ preserves $\cS$ and that this action commutes with the holomorphic action. So, we have an induced action of $(S^1)^n$ on $\cal N$. Up to a translation of the $l_j$ (which does not change the quotient $\cN$), \ref{LVM} allow us to assume that $0$ belongs to $Conv(l_1,\dots,l_n)$ (this condition is known as \emph{Siegel's condition}) and in this case, the quotient $P$ of the action of $(S^1)^n$ on $\cN$ can be identified with:

\[
P=\left\{\ (r_1,\dots,r_n)\in(\R_+)^n\ /\ \Sigma_{i=1}^n r_jl_j=0,\ \Sigma_{i=1}^n r_j=1\ \right\}
\]

\vskip 4mm

This set is clearly a polytope and it can be shown that it is simple (cf. \cite{BM}, lemma $0.12$). The polytope $P$ is called the \emph{polytope associated} to the LVM manifold $\cN$.

\vskip 4mm

\begin{exempl} For the previous example, we have to perform a translation on the vectors $l_j$ with the view to respect the Siegel's condition. For example, $\cN$ is also the LVM manifold associated with the good system $(\cE,\lambda)$ where $\lambda_1=\lambda_3=\frac{3}{4}-\frac{i}{4}$, $\lambda_2=\lambda_4=-\frac{1}{4}+\frac{3i}{4}$ and $\lambda_5=-\frac{1}{4}-\frac{i}{4}$. A calculation shows that the polytope $P$ is the square

\[
P=\left\{\ \left(\frac{1}{4}-r_3,\frac{1}{4}-r_4,r_3,r_4,\frac{1}{2}\right)\ /\ r_3,r_4\in [0,\frac{1}{4}]\ \right\}
\]

\end{exempl}

\vskip 4mm
\section{Fundamental sets and associated complex}

In this section, we will briefly study the fundamental sets introduced in $\cite{B}$ and defined above and construct a simplicial complex whose combinatorial properties reflect the geometry of a LVMB manifold. Here, $\cE$ is a fundamental set of type ($M,n$). The integer $M$ is not supposed to be odd. For the moment, our aim is to relate the above properties to some classical ones of simplicial complexes.

\vskip 4mm

Let us begin with some vocabulary. \emph{Faces}, or \emph{simplices} are subsets of a simplicial complex. If a complex $K$ is \emph{pure-dimensional} (or simply \emph{pure}), the simplices of maximal dimension $d$ are named \emph{facets} and the faces of dimension 0 (resp. $d-1$) are the \emph{vertices} (resp. the \emph{ridges}) of the complex.

\vskip 4mm

The first important definition in this paper is the following:

\vskip 2mm

\begin{defn} Let $\cal E$ be a fundamental set of type $(M,n)$. The \emph{associated complex} of $\cal E$ is the set $\cal P$ of subsets of $\{1,\dots,n\}$ whose complement (in $\{1,\dots,n\}$) is acceptable.
\end{defn}

\vskip 4mm

As we will see later, this complex is the best choice for a combinatorial generalization of the associated polytope of a LVM manifold.

\vskip 4mm

First properties of $\cP$ are the following:

\vskip 2mm

\begin{prop} Let $\cal E$ be a fundamental set of type $(M,n,k)$. Then, its associated complex $\cal P$ is a simplicial complex on $\{1,\dots,n\}$. Moreover, $\cal P$ is pure-dimensional of dimension $(n-M-1)$ and has $(n-k)$ vertices. Its vertices are precisely the non-indispensable elements of $\{1,\dots,n\}$ for $\cal E$ and its facets are exactly the complements of the subsets of $\cal E$.
\end{prop}

\vskip 4mm

\begin{demo} $\cal P$ is obviously a simplicial complex. Moreover, the maximal simplices of $\cal P$ are exactly the complements of minimal subsets of $\cal A$, that are fundamental subsets. The latters have the same number $M$ of elements, so every maximal simplex of $\cal P$ has $n-M$ elements. Finally, an element $j\in\{1,\dots,n\}$ is a vertex of $\cal P$ if and only if $\{1,\dots,n\}\backslash\{j\}$ contains a fundamental subset, that is $j$ is not indispensable. $\square$
\end{demo}

\vskip 4mm

\begin{exempl} The complex $\cP$ associated to the fundamental set of \ref{exemple} is the complex with facets $\{ \{1,2\},\{2,3\},\{3,4\},\{1,4\}\}$. So, $\cP$ is the boundary of a square.
\end{exempl}

\vskip 4mm

\begin{rmq} Conversely, every pure complex can be realized as the associated complex of a fundamental set: let $\cal P$ be a pure-dimensional simplicial complex on the set $\{1,\dots,v\}$ with dimension $d$ and $v$ vertices. Then, for every integer $k$, there exists $M,n$ two integers and $\cal E$ a fundamental set of type $(M,n,k)$ whose associated complex is $\cal P$. If $k$ is fixed, this fundamental set $\cal E$ is unique.
\end{rmq}

\vskip 4mm

Moreover, the $SEU$ property can be expressed as a combinatorial property of $\cP$:

\vskip 2mm

\begin{prop} \label{facettes} $\cal E$ verify the $SEU$ property if and only if

\[
\forall\ Q\in \cP_{max},\ \forall\ k\in \{1,\dots,n\},\ \exists!\ k'\notin Q;\ (Q\cup\{k'\})\backslash\{k\}\in\cP_{max}
\]

where $\cP_{max}$ is the set of facets of $\cal P$.
\end{prop}

\vskip 4mm

\begin{demo} First, we assume that $\cE$ verifies the $SEU$ property. Let $Q$ be a facet of $\cP$ and $k\in\{1,\dots,n\}$. Then $P=Q^c$ belongs to $\cE$. The $SEU$ property implies that there is $k'$ in $P$ (so $k'\notin Q$) such that $P'=(P\backslash\{k'\})\cup\{k\}$ belongs to $\cE$. As it was claimed, $P'^c$ is also a facet of $\cP$. Moreover, we obviously have $P'^c=(Q\cup\{k'\})\backslash\{k\}$. Finally, if  $Q''=(Q\cup\{k''\})\backslash\{k\}$ is a facet of $\cal P$ with $k''\neq k'$ and $k''\notin Q$, then, we would have $(P\backslash\{k'\})\cup\{k\}$ in  $\cE$, which contradicts the $SEU$ property. The proof of the converse is analogous. $\square$
\end{demo}

\vskip 4mm

\begin{cor} \label{crete} Let $\cal E$ be a fundamental set. Then its associated complex $\cal P$ verify the $SEU$ property if and only if every ridge of $\cal P$ is contained in exactly two facets of $\cal P$.
\end{cor}

\vskip 4mm

\begin{demo} To begin, we assume that $\cE$ verifies the $SEU$ property. Let $Q$ be a ridge of $\cP$. By definition, $Q$ is included in a facet $P$ of $\cP$. We put $P=Q\cup\{k\}, k\in\{1,\dots,n\}\backslash Q$. By \ref{facettes}, there exists $k'\notin P$ (and so $k\neq k'$) such that $P'=(P\cup\{k'\})\backslash\{k\}$ is a facet of $\cP$. We have $P'=Q\cup\{k'\}$ so $Q$ is contained in at least two facets of $\cP$. Let assume that $Q$ is contained in a third facet $P''=Q\cup\{k''\}$. However, in this case, we would have $P''=(P\cup\{k''\})\backslash\{k\}$, which contradicts \ref{facettes}.
\\Conversely, let $Q$ be a facet of $\cP$ and $k\in\{1,\cdots,n\}$. If $k\in Q$, then $P=Q\backslash\{k\}$ is a ridge of $\cP$ and by hypothesis, $P$ is  contained in exactly two facets $Q_1$ and $Q_2$. One of them, say $Q_1$, is $Q$. The other is $Q_2$ and we have $Q_2=P\cup\{k'\}$. Then we have $k'\notin Q$ (on the contrary, we would have $Q_2=Q=Q_1$) and $Q_2=(Q\cup\{k'\})\backslash\{k\}$. Moreover, if $Q_3=(Q\cup\{k''\})\backslash\{k\}$ is another facet of $\cP$ with $k''\notin Q$, so $Q_3$ contains $P$ and by hypothesis, $Q_3=Q_2$ (i.e $k''=k'$). If $k\notin Q$, we remark that the element $k'\in\{1,\dots,n\}$ such that $Q'=(Q\cup\{k'\})\backslash\{k\}$ is a facet of $\cP$ is $k'=k$. Indeed, if $k'=k$, then $Q'=Q$ is a facet of $\cP$. And if $k'\neq k$, then $k\notin Q\cup\{k'\}$ and, as a consequence, $Q'=Q\cup\{k'\}$ is not in $\cP$. $\square$
\end{demo}

\vskip 4mm

\begin{defn} Let $\cE$ be a fundamental set of type $(M,n)$. We define the (unoriented) graph $\Gamma$ stating that its vertices are fundamental subsets of $\cal E$ and two vertices $P$ and $Q$ are related by an edge  if and only if there exist $k\notin P,k'\in P$ such that $Q=(P\backslash\{k'\})\cup\{k\}$. Equivalently, we relate two subsets of $\cal E$ if and only if they differ exactly by one element. $\Gamma$ is called \emph{replacement graph} of $\cal E$.
\end{defn}

\vskip 4mm

\begin{prop} Let $\cal E$ be a fundamental set of type $(M,n)$ which verifies the $SEU$ property. Then, there exists an integer $p\in\N^*$, and fundamental sets $\cE_j$ of type $(M,n)$ which are minimal for the $SEU$ property and such that $\cal E$ is the disjoint union $\displaystyle{\bigsqcup_{j=1}^p\cE_j}$.
\end{prop}

\vskip 4mm

\begin{demo} We proceed by induction on the cardinal of $\cal E$. If $\cal E$ is minimal for the $SEU$ property, then there is nothing to do. Let assume that it is not the case: there exists a proper subset $\cE_1$ of $\cal E$ which is minimal for the $SEU$ property. We put $\overline\cE$ for its complement $\cE\backslash\cE_1$. It is obvious that $\overline\cE$ is a fundamental set (of type $(M,n)$). We claim that $\overline\cE$ verifies the $SEU$ property. Let $P$ be an element of $\overline\cE$ and $k\in\{1,\dots,n\}$. If $k\in P$, then, putting $k'=k$, we have that $(P\backslash\{k'\})\cup\{k\}=P$ is an element of $\overline\cE$. It is the only choice (for $k'$) since $P$ is in $\cal E$ and $\cal E$ verifies the $SEU$ property. Let assume now that $k$ is not in $P$. Since $P$ is an element of $\cal E$, there exists exactly one $k'\in P$ such that $P'=(P\backslash \{k'\})\cup\{k\}$ is an element of $\cal E$, too. We claim that $P'$ cannot be in $\cE_1$. Indeed, if it was the case, since $\cE_1$ is minimal for the $SEU$ property, there would exist exactly one $k''\in P'$ such that $P''=(P'\backslash\{k''\})\cup\{k'\}\in\cE_1$. But $P=(P'\backslash\{k\})\cup\{k'\}$ is in $\cE$ and $k\in P'$. So, $k''=k$ and $P''=P$. As a consequence, $\overline\cE$ is a fundamental set of type $(M,n)$ which verifies the $SEU$ property with cardinal strictly smaller than $\cal E$. Applying the induction hypothesis on $\overline\cE$, we have the  decomposition of $\cal E$ we were looking for. $\square$
\end{demo}

\vskip 4mm

\begin{rmq} The decomposition of the previous proposition induces a decomposition of the vertex set of $\Gamma$. In the proof, we show that an element of  $\cE_j$ is related only to others elements in $\cE_j$. Consequently, each set $\cE_j$ is the vertex set of a connected component of $\Gamma$. This also implies that this decomposition is unique up to the order. We call \emph{connected components} of $\cal E$ the sets  $\cE_j$.
\end{rmq}

\vskip 4mm

\begin{cor} \label{min} Let $\cal E$ be a fundamental set of type $(M,n)$ and $\Gamma$ its replacement graph. We assume that $\cal E$ verifies the $SEU$ property. Then, the following assertions are equivalent:

\begin{enumerate}
	\item $\cal E$ is minimal for the $SEU$ property.
	\item $\cal E$ has only one connected component.
	\item $\Gamma$ is connected.
\end{enumerate}
\end{cor}

\vskip 4mm

\begin{defn} Let $K$ be a simplicial complex. $K$ is a \emph{pseudo-manifold} if these two properties are fulfilled:
\begin{enumerate}
	\item every ridge of $K$ is contained in exactly two facets.
	\item for all facets $\sigma,\tau$ of $K$, there exists a chains of facets $\sigma=\sigma_1,\dots,\sigma_n=\tau$ of $K$ such that $\sigma_i\cap\sigma_{i+1}$ is a ridge of $K$ for every $i\in\{0,\dots,n-1\}$.
\end{enumerate}
\end{defn}

\vskip 4mm

For instance, every simplicial sphere is a pseudo-manifold. More generally, a triangulation of a manifold (that is a simplicial complex whose realization is homeomorphic to a topological manifold) is also a pseudo-manifold. Now, the proposition below shows that the notion of pseudo-manifold is exactly the combinatorial property of $\cP$ which characterizes the fact that $\cE$ is minimal for the $SEU$  property:

\vskip 2mm

\begin{prop} \label{pseudomanifold} Let $\cal E$ be a fundamental set with $n>M$. Then, $\cal P$ is a pseudo-manifold if and only if $\cal E$ is minimal for the $SEU$ property.
\end{prop}

\vskip 4mm

\begin{demo} On the one hand, let assume that $\cal E$ is minimal for the $SEU$ property. This implies that every ridge of $\cal P$ is contained in exactly two facets  (cf. \ref{crete}). Moreover, let $\sigma,\tau$ be two distinct facets of $\cal P$. So, $P=\sigma^c$ and $Q=\tau^c$ are two fundamental subsets. By minimality for the $SEU$ property, $\Gamma$ is connected (cf. \ref{min}). Consequently, there exists a sequence $P_0=P,P_1,\dots ,P_r=Q$ of fundamental subsets such that $P_i$ and $P_{i-1}$ differ by exactly one element. We note $R_i$ the acceptable subset $P_{i-1}\cup P_i$ with $M+1$ elements. Its complement $R_i^c$ is thus a face of $\cal P$ with $n-M-1=dim(\cP)$ elements. If we put $\sigma_i=P_i^c$, we have $R_i^c=\sigma_{i-1}\cap\sigma_i$ so $\sigma_0=\sigma,\dots ,\sigma_r=\tau$. Consequently, $\cal P$ is a pseudo-manifold.
\\On the other hand, we assume that $\cal P$ is a pseudo-manifold. Then, thanks to \ref{crete}, $\cal E$ verifies the $SEU$ property. Moreover, $\cal E$ will be minimal for this property if and only if $\Gamma$ is connected (cf. \ref{min}). Let $\sigma,\tau$ be two distinct elements of $\cal E$. Then $\sigma^c$ and $\tau^c$ are facets of $\cal P$. Since $\cal P$ is a pseudo-manifold, there exists a sequence $\sigma_0=\sigma^c$, $\sigma_1,\dots,\sigma_r=\tau^c$ of facets of $\cal P$ such that for every $i$, $\sigma_i$ and $\sigma_{i-1}$ share a ridge of $\cal P$. This means that $\sigma_i^c$ and $\sigma_{i-1}^c$ are fundamental subsets $\cal E$ which differ only by an element, and consequently, $\sigma_i^c$ and $\sigma_{i-1}^c$ are related in $\Gamma$. This implies that $\Gamma$ is connected and $\cal E$ is minimal for the $SEU$ property. $\square$
\end{demo}

\vskip 4mm

\begin{rmq} The case where $n=M$ corresponds to $\cP=\{\emptyset\}$. It is not a pseudo-manifold since the only facet is $\emptyset$ and it doesn't contains any simplex with dimension strictly smaller.
\end{rmq}

\vskip 4mm

Finally, we prove the following proposition which is the motivation for the study of the associated complex:

\vskip 2mm

\begin{prop} Let $(\cE,l)$ be a good system associated to a LVM manifold and $P$ its associated polytope. Then the associated complex $\cal P$ of $\cal E$ is combinatorially equivalent to the border of the dual of $P$.
\end{prop}

\vskip 4mm

\begin{demo} Since $(\cE,l)$ is a good system associated to a LVM manifold, there exists a bounded component $O$ in $\cal O$ such that $\cal E$ is exactly the set of subsets $Q$ of $\{1,\dots,n\}$ with $(2m+1)$ for cardinal such that $O$ is included in the convex hull of $(l_q,q\in Q)$. Up to a translation, (whose effect on the action is just to introduce an automorphism of $\C^m\times\C^*$ and so does not change the action, cf. $\cite{B}$), we can assume that $\cal E$ is exactly the set of subsets $P$ of $\{1,\dots,n\}$ with $(2m+1)$ for cardinal such that the convex hull of $(l_p,p\in P)$ contains $0$. In this setting, according to the page 65 of $\cite{BM}$, the border of $P$ is combinatorially characterized as the set of subsets $I$ of $\{1,\dots,n\}$ verifying \[I \in P \Leftrightarrow 0\in Conv(l_k,k\in I^c)\]
So, $I$ is a subset of  $P$ if and only if $I^c$ is acceptable, i.e. $I\in \cP$.
\\As a consequence, from a viewpoint of set theory, $P$ and $\cal P$ are the same set. We claim that the order on these sets $P$ and $\cal P$ are reversed. On the one hand, the order on $\cal P$ is the usual inclusion (as for every simplicial complex). On the other other, we recall the order on the face poset of $P$ given in $\cite{BM}$:
\\every $j-$face of $P$ is represented by a $(n-2m-1-j)-$tuple. So, facets of $P$ are represented by a singleton and vertices by a $(n-2m-2)-$tuple. A face represented by $I$ is contained in another face represented by $J$ if and only if $I\supset J$. So, combinatorially speaking, the poset of $P$ is $(\cP,\supset)$, which prove the claim. Finally, the poset for the dual $P^*$ is $(\cP,\subset)$, and the proof is over. $\square$
\end{demo}

\vskip 4mm

\vskip 4mm
\section{Condition $(K)$}

Consequently, $\cal P$ is exactly the object we are looking for to generalize the associated polytope of a LVM manifold to the case of LVMB manifolds. What we have to do now is to study its properties. Our main goal for the moment is to prove the following theorem:

\vskip 2mm

\begin{thm} Let $(\cE,l)$ be a good system of type $(2m+1,n)$. Then $\cal P$ is a simplicial $(n-2m-2)-$sphere.
\end{thm}

\vskip 4mm

\begin{rmq} The theorem is trivial in the LVM case since the associated complex $\cal P$ is a polytope.
\end{rmq}

\vskip 4mm

To prove the previous theorem, we have to focus on good systems which verify an additional condition, called condition $(K)$: there exists a real affine automorphism  $\phi$ of $\C^m=\R^{2m}$ such that $\lambda_j=\phi(l_j)$ has coordinates in $\Z^{2m}$ for every $j$. For instance, if all coordinates of $l_i$ are rationals, then $(\cE,l)$ verify condition (K). Note that the imbrication condition is an open condition. As a consequence, it is sufficient to prove the previous theorem for good system verifying the condition $(K)$. Indeed, since $\Q^n$ is dense in $\R^n$, a good system $(\cE,l)$ which does not verify the condition $(K)$ can be replaced by a good system which verifies the condition, with the same associated complex $\cP$.

\vskip 4mm

The main interest of condition $(K)$ stands in the fact that we can associate to our holomorphic acceptable action an algebraic action (called \emph{algebraic acceptable action}) of $(\C^*)^{2m+1}$ on $\C^n$ (or an action of $(\C^*)^{2m}$ on $\PP^{n-1}$):

\vskip 4mm

Let $(\cE,l)$ be a fundamental set of type $(2m+1,n)$ verifying condition $(K)$. We set $l_j=a_j+ib_j,a_j,b_j\in\Z^m$ for every $j$ and $a_j=(a_j^1,\dots,a_j^m)$. We can define an action of $(\C^*)^{2m+1}$ on $\C^n$ setting:

\vskip 2mm

\[
\left(u,t,s\right)\cdot z=\left(u \ t_1^{a_1^1}\dots t_m^{a_1^m}\ s_1^{b_1^1}\dots s_m^{b_1^m}\ z_1,\dots, u\ t_1^{a_n^1}\dots t_m^{a_n^m}\  s_1^{b_n^1}\dots s_m^{b_n^m}\ z_n\right)\ \forall u\in\C^*,t,s\in\left(\C^*\right)^m, z\in{\C}^n
\]

\vskip 4mm

Using the notation $X_{2m+1}^{\widetilde{l_j}}$ for the character of $(\C^*)^{2m+1}$ defined by $\widetilde{l_j}=(1,l_j)$, we can resume the formula describing the acceptable algebraic action by:

\vskip 2mm

\[t \cdot z=\left(X_{2m+1}^{\widetilde{l_1}}(t) z_1,\dots, X_{2m+1}^{\widetilde{l_n}}(t)z_n \right)\ \forall t\in (\C^*)^{2m+1},\ z\in\C^n
\]

\vskip 4mm

It is clear that $\cal S$ is invariant by this action. So we can define $X$ as the topological orbit space of $\cal S$ by the algebraic action. As for the holomorphic acceptable action, we can define an action of $\left(\C^*\right)^{2m}$ on $\cal V$ whose quotient is also $X$. In $\cite{CFZ}$, proposition 2.3, it is shown that the holomorphic acceptable action of $\C^m$ on $\cal V$ can be seen as the restriction of the algebraic acceptable action to a closed cocompact subgroup $H$ of $(\C^*)^{2m}$. As a consequence, we can define an action of $K=(\C^*)^{2m}/H$ on $\cal N$ whose quotient can be homeomorphically identified with $X$.

\vskip 4mm

The principal consequence of this result is the following:

\vskip 2mm

\begin{prop} $X$ is Hausdorff and compact.
\end{prop}

\vskip 4mm

\begin{demo} Let $p:{\cal N}\rightarrow X$ be the canonical surjection. $K$ is a compact Lie group so $p$ is a closed map (cf. $\cite{Br}$, p.38). Consequently, $X$ is Hausdorff. Finally, since $p$ is continuous and $\cal N$ is compact, we can conclude that $X$ is compact. $\square$
\end{demo}

\vskip 4mm

Another important consequence for the sequel of the article is that the algebraic action on $\cal S$ (or $\cal V$) is closed. Moreover, since every complex compact commutative Lie group is a complex compact torus (i.e. a complex Lie group whose underlying topological space is $(S^1)^n$, cf. $\cite{L}$, Theorem 1.19), we can say that $K$ is a complex compact torus.

\vskip 4mm

Using an argument of $\cite{BBS}$, we show that:

\vskip 2mm

\begin{prop} $t\in(\C^*)^{2m}$ is in the stabilizer of $[z]$ if and only if $\forall i,j\in I_z$, we have
\[
X_{2m+1}^{l_i}(t)=X_{2m+1}^{l_j}(t)
\]
\end{prop}

\vskip 4mm

\begin{demo} Let us fix some linear order on $\Z^{2m}$. Up to a permutation of the homogenous coordinates of $\PP^{n-1}$, we can assume that  $l_j\leq l_{j+1}$ for every $j\in\{1,\dots ,n-1\}$ (notice that such a permutation is an equivariant automorphism of $\PP^{n-1}$). We set $j_0=min(I_z)$ the smallest index of nonzero coordinates of $z$. Then, for every $t$ which stabilizes $[z]$, we have\[[z]=t.[z]=[X_{2m+1}^{l_j}(t)z_j]=[0,\dots ,0,X_{2m+1}^{l_{j_0}}(t)z_{j_0},\dots ,X_{2m+1}^{l_n}(t)z_n]\]
\\So $[z]=[0,\dots ,0,z_{j_0},\dots ,X_{2m+1}^{l_n-l_{j_0}}(t)z_j]$. In particular, we have $X_{2m+1}^{l_n-l_{j_0}}(t)z_j=z_j$ for every $j$. If $j\in I_z$, then  $X_{2m+1}^{l_n-l_{j_0}}(t)=1$. $\square$
\end{demo}

\vskip 4mm

\begin{rmq} In $\cite{BBS}$, it is shown that $[z]$ is a fixed point for the algebraic acceptable action if and only if $\forall i,j\in I_z$, we have $l_i=l_j$.
\end{rmq}

\vskip 4mm

Consequently, we have:

\vskip 2mm

\begin{prop} Every element of $\cal V$ have a finite stabilizer for the algebraic acceptable action.
\end{prop}

\vskip 4mm

\begin{demo} First, we recall that an element of $\cal V$ has at least $(2m+1)$ nonzero coordinates and amongs this coordinates, there are $(2m+1)$ coordinates such that $(l_p)_{p\in P}$ spans $\C^m$ as a real affine space. Up to a permutation, we can assume that $(1,2,\dots ,2m+1)$ is contained in  $I_z$ and in $\cal E$. In this case, we have $X_{2m+1}^{l_j-l_{2m+1}}(t)=1$ for every $j=1,\dots ,2m$ and every $t$ in the stabilizer of $[z]$. We put $L_j=l_j-l_{2m+1}$. Writing $t_j=r_je^{2i\pi\theta_j}$, we get that $r=(r_1,\dots ,r_{2m+1})$ and $\theta=(\theta_1,\dots ,\theta_{2m+1})$ verify the following systems: \[M.ln(r)=0,\ M.\theta\equiv0\ [1]\]

\vskip 4mm

where $M=(m_{i,j})$ is the matrix defined by $m_{i,j}=L_j^i$ and $ln(r)=(ln(r_1),\dots,ln(r_{2m+1}))$. The system being acceptable, $(l_1,\dots ,l_{2m+1})$ spans $\C^m$ as a real affine space, which means exactly that the real matrix M is inversible.
\\Consequently, $ln(r)=0$ (i.e $|t_j|=1$ pour tout j) and $\theta\equiv0\ [det(M^{-1})]$. So, $r_je^{2i\pi\theta_j}$ can take only a finite number of values. As a conclusion, the stabilizer of $[z]$ is finite, as it was claimed. $\square$
\end{demo}

\vskip 4mm

\begin{rmq} An analogous proof shows that the stabilizer of $z\in \cS$ is finite, too.
\end{rmq}

\vskip 4mm
\section{Connection with toric varieties}

In this section, our main task is to recall that $X$ is a toric variety and to compute its fan. When the fan $\Sigma$ is simplicial, we can constructed a simplicial complex $K_\Sigma$ as follow: we note $\Sigma(1)$ for the set of rays of $\Sigma$ and order its elements by $x_1,\dots,x_n$. Then, the complex $K_\Sigma$ is the simplicial complex on $\{1,\cdots,n\}$ defined by:

\[
\forall\ J\subset\{1,\cdots,n\},\hskip 3mm (\ J\in K_\Sigma \Leftrightarrow\ pos(x_j,j\in J)\in \Sigma\ )
\]

\vskip 4mm

This complex $K_\Sigma$ is \emph{the underlying complex} of $\Sigma$. We recall a theorem which will be very important in the sequel:

\vskip 2mm

\begin{prop} \label{sphere} Let X be a normal separated toric variety and $\Sigma$ its fan. We suppose that $\Sigma$ is simplicial. Then, the three following assertions are equivalent:
\begin{enumerate}
\item $X$ is compact.
\item $\Sigma$ is complete in $\R^n$.
\item $\Sigma$'s underlying simplicial complex is a $(n-1)$-sphere.
\end{enumerate}
\end{prop}

\subsection{Toric varieties}

To begin with, it is clear that $\cS$ and $\cV$ are separated normal toric varieties. As it is explained in $\cite{CLS}$, we can associate to a separated normal toric variety with group of one-parameter subgroups $N$ a fan $\Sigma$ in the real vector space $N_\R=N\otimes\R$ whose cones are rational with respect to the lattice $N$\footnote{a cone $\sigma$ is said to be rational for $N$ if there a family $S$ of elements of $N$ such that $\sigma=pos(S)$. Moreover, if $S$ is free in $N_\R$, we say that $\sigma$ is simplicial}. So, we can compute the fan associated to $\cal S$:

\vskip 2mm

\begin{prop} \label{eventail} Let $(e_i)_{i=1}^n$ be the canonical basis for $\R^n$. Then, the fan describing $\cal S$ in $\R^n$ is
\[
\Sigma(\cS)=\{\ pos(e_i,i\in I)/\ I \in \cP\}
\]
\end{prop}

\vskip 4mm

\begin{demo} To compute the fan of a toric variety, one has to calculate limits for its one-parameter subgroups. The embedding of $(\C^*)^n$ in $\cal S$ is the inclusion and the one-parameter subgroups of $(\C^*)^n$ have the form  $\lambda^m(t)=(t^{m_1},\dots,t^{m_n})$ with $m=(m_1,\dots,m_n)\in\Z^n$.
\\So, the limit of $\lambda^m(t)$ when t tends to $0$ exists in $\C^n$ if and only if $m_i\geq 0$ for every $i$. In this case, the limit is $\epsilon=(\epsilon_1,\dots,\epsilon_n)$ with $\epsilon_j=\delta_{m_j,0}$ (Kronecker's symbol).
\\Of course, the limit has to be in $\cal S$, which implies for $I_\epsilon$ to be acceptable. But $I_\epsilon=\{j/m_j=0\}$ so the condition means exactly that $\{j/m_j>0\}$ belongs to $\cal P$. $\square$
\end{demo}

\vskip 4mm

\begin{exempl} For the fundamental set $\cE=\left\{\{1,2,5\},\{1,4,5\},\{2,3,5\},\{3,4,5\}\right\}$ of \ref{exemple}, we have ${\cS}=\left\{z/(z_1,z_3)\neq 0, (z_2,z_4)\neq 0, z_5\neq 0 \right\}$. As a consequence, the fan $\Sigma(\cS)$ is the fan in $\R^5$ whose facets are the $2-$dimensional cones $pos(e_1,e_2), pos(e_1,e_4), pos(e_2,e_3)$ and $pos(e_3,e_4)$ (where $e_1,\dots,e_5$ is the canonical basis of $\C^5$).
\end{exempl}

\vskip 4mm

\begin{rmq}
We can also easily compute orbits of $\cal S$ for the action of $(\C^*)^n$. For $I\subset\{1,\dots,n\}$, we set $O_I=\{z\in\C^n/I_z=I^c\}$. Then if $z\in\cal S$, its orbit is $O_{I_z^c}$.
\\So, we can describe $\cal S$ as the partition: $\cS=\displaystyle{\bigsqcup_{I\in\cP} O_I}$

\vskip 4mm

Moreover, in the orbit-cone correspondance between $\cal S$ and $\Sigma(\cS)$ (cf. $\cite{CLS}$ ch.3), $O_{I}$ corresponds to $\sigma_I=pos(e_j/j\in I)$.
\end{rmq}

\vskip 4mm

\begin{rmq} In quite the same way, we can show that the fan of $\cal V$ is $\Sigma(\cV)=\{pos(e_i,i\in I)/I\in \cP\}$ in $\R^{n-1}$, with $(e_1,\dots,e_{n-1})$ defined as the canonical basis of $\R^{n-1}$ and $e_n=-(e_1+\dots+e_{n-1})$.
\end{rmq}

\vskip 4mm

In $\cite{CLS}$, it is proven that $X$ is an irreducible separated compact normal toric variety. In the next section, we will detail the construction with the view to identify its group of one-paramater subgroup and the structure of its fan.

\vskip 4mm

\begin{exempl} For the good system $(\cE,l)$ with $\cE=\{\{1,2,5\},\{1,4,5\},\{2,3,5\}\{3,4,5\}\}$ and $l_1=l_3=1$, $l_2=l_4=i$ and $l_5=0$, the algebraic action is

\vskip 2mm

\[(\alpha,t,s)\cdot z=(\alpha t z_1,\alpha s z_2,\alpha t z_3,\alpha s z_4,\alpha z_5)\]

\vskip 2mm

Using the automorphism of $\left(\C^*\right)^3$ defined by $\phi(\alpha,t,s)=(\alpha,\alpha t,\alpha s)$, we can see that the quotient $X$ of the algebraic action is also the quotient of $\cS$ by the action defined by

\vskip 2mm

\[(\alpha,t,s)\cdot z=(t z_1,s z_2,t z_3,s z_4,\alpha z_5)\]

\vskip 2mm

so $X$ is the product $\PP^1\times\PP^1$ and the projection $\cN\rightarrow X$ comes from the usual definition of $\PP^2$ as a quotient of $S^3.$
\end{exempl}

\vskip 4mm

To conclude this section, let $f:(\C^*)^{2m+1}\longrightarrow (\C^*)^n$ be the map defined by

\vskip 2mm

\[
f(u,t,s)=(X_{2m+1}^{\widetilde{l_1}}(u,t,s),\cdots,X_{2m+1}^{\widetilde{l_n}}(u,t,s))
\]

\vskip 2mm

The algebraic acceptable action on $\C^n$ is just the restriction to $Im(f)$ of the natural action of the torus $(\C^*)^n$ on $\C^n$.

\vskip 4mm

\begin{prop} $Ker(f)$ is finite.
\end{prop}

\vskip 4mm

\begin{demo} Let $t$ be in $Ker(f)$. Then $X^{\widetilde{l_j}}(t)=1$ for every $j$. We set $t=(r_1e^{2i\pi\theta_1},\dots,r_{2m+1}e^{2i\pi\theta_{2m+1}})$ and $l_j=(l_j^1,\dots,l_j^{2m})$ and we have $r_1r_2^{l_j^1}\dots r_{2m+1}^{l_j^{2m}}=1$ for every $j$. That means that we have $A.v=0$ where $A=(a_{i,j})$ is the real matrix defined by $a_{i,j}=\widetilde{l_i}^j$ and $v$ is the vector $v=(ln(r_j))$. Since $l_1,\dots l_n$ affinely spans $\R^{2m}$, we can extract an inversible matrix $\widetilde{A}$ of order $2m+1$ from $A$ and we can deduce that $v=0$. This means that $|t_j|=1$ for every $j$. Moreover, if $\theta=(\theta_1,\dots,\theta_{2m+1})$, we get that $\widetilde{A}.\theta$ is an element of $\Z^{2m+1}$. Consequently, $\theta$ is an element of $\frac{1}{\delta}\Z^n$, where $\delta=det(\widetilde{A})$. Finally, we can deduce that $e^{i\theta_j}$ is a $\delta-$th root of unity for every $j$. Particularly, there are only a finite number of elements in the kernel of $f$. $\square$
\end{demo}

\vskip 4mm

\begin{rmq} Generally, $f$ is not injective. For instance, if we consider $\cE=\{\{1,2,4\},\{2,3,4\}\}$ and $l_1=1,l_2=i,l_3=p,l_4=-1-i$, where $p$  is a nonzero positive integer. $(\cE,l)$ is a good system and for $p=4$, $f$ is not injective. Note that for $p=3$, $f$ is injective.
\end{rmq}

\vskip 4mm

\begin{defn} We define $T_N$ as the quotient group $\left(\C^*\right)^n/Im(f)$
\end{defn}

\vskip 4mm

We recall that $\left(\C^*\right)^n$ is included in $\cal S$ as an dense Zariski open subset. $(\C^*)^n$ is invariant by the action of $(\C^*)^{2m+1}$. This implies that $T_N$ can be embedded in $X$ as an dense open subset. Moreover, the action of $(\C^*)^n$ on $\cal S$ commutes with the action of $(\C^*)^{2m+1}$, so the action of $T_N$ on itself can be extended to an action on $X$. In $\cite{CLS}$, it is proven that $X$ is an irreducible separated compact normal algebraic variety.

\subsection{The algebraic torus $T_N$}

We note $F:\C^{2m+1}\rightarrow\C^n$ the linear map defined by \[F\left(U,T,S\right)=\left(U+<a_j,T>+<b_j,S>\right)_j\]

\vskip 4mm

The matrix of $F$ has $(1,l_j)$ as j-th row so $F$ has maximal rank. So, $F$ is injective. The family $f_j=F(e_j)$, with $(e_1,\dots ,e_{2m+1})$ the canonical basis of $\C^{2m+1}$ is a basis for $Im(F)$. We can notice that each $f_j$ has integer coordinates. We complete this basis to a basis $(f_1,\dots ,f_n)$ of $\C^n$ with integer coordinates. Next, we define the map $G:\C^n\rightarrow\C^{n-2m-1}$ by linearity and
$G(f_j)=\left\{\begin{array}{cl}
										0 	 & j\in\{1,\dots,2m+1\}\\
										g_j  & otherwise\\
							\end{array}
				\right.$

(with $(g_{2m+2},\dots ,g_n)$ the canonical basis of $\C^{n-2m-1}$)

\vskip 4mm

It is clear by construction that the following sequence is exact:

\[ \xymatrix{
				0 \ar[r] & \C^{2m+1} \ar[r]^F & \C^n \ar[r]^G & \C^{n-2m-1} \ar[r] & 0
						}
\]
Moreover, we have \[F(\Z^{2m+1})\subset \Z^n,\ G(\Z^n)\subset \Z^{n-2m-1}\]

\vskip 4mm

Let $t\in\left(\C^*\right)^{2m+1}$ and $T$ be some element of $\C^{2m+1}$ such that $t=exp\left(T\right)$. We put $g\left(t\right)=exp\left(G\left(T\right)\right)$. The previous remark has for consequence that $g$ is well defined. Moreover, we have:

\vskip 2mm

\begin{prop} $g$ is a group homomorphism and the following diagram is commutative:

\[
\xymatrix{
    \left(\C^*\right)^{2m+1} \ar[r]^-{f}  & \left(\C^*\right)^n \ar[r]^-{g}  & \left(\C^*\right)^{n-2m-1} \\
    \C^{2m+1} \ar@{->>}[u]_-{exp}  \ar@{^{(}->}[r]^-{F} &  \C^n \ar@{->>}[u]_-{exp} \ar@{->>}[r]^-{G} & \C^{n-2m-1} \ar@{->>}[u]_-{exp}
  }
\]

\end{prop}

\vskip 4mm

Finally, we can prove that:

\vskip 2mm

\begin{prop} $g$ is surjective and $Ker(g)=Im(f)$
\end{prop}

\vskip 4mm

\begin{demo} The surjectivity of $g$ is clear (since $G$ and $exp$ are surjective).
\\So, we have only to show that $Ker(g)=Im(f)$.
By the construction of $F$ and $G$ and by commutativity of the previous diagram, we have for every $t=exp(T)$, $g\circ f(t)=g\circ f\circ exp(t)=exp\circ G\circ F(T)=exp(0)=1$ so $Im(f)\subset Ker(g)$.
\\Conversely, let $t$ belong to $Ker(g).$ We put $t=exp(T)$, for some $T\in \C^n$ and we have $1=g(t)$ so $exp(G(T))=1$.
\\As a consequence, $G(T)\in 2i\pi\Z^{n-2m-1}$, i.e. \[G(T)=(2i\pi q_{2m+2},\dots ,2i\pi q_n)=2i\pi q_{2m+2}g_{2m+2}+\dots 2i\pi q_ng_n=G(2i\pi q_{2m+2}f_{2m+2}+\dots 2i\pi q_nf_n)\]

\vskip 3mm

So $T-(2i\pi q_{2m+2}f_{2m+2}+\dots 2i\pi q_nf_n)\in Ker(G)=Im(F)$. We can write that \[T=\lambda_1f_1+\dots\lambda_{2m+1}f_{2m+1}+2i\pi q_{2m+2}f_{2m+2}+\dots 2i\pi q_nf_n\]

\vskip 3mm

Finally, $T=F(\lambda_1e_1+\dots\lambda_{2m+1}e_{2m+1})+2i\pi(q_{2m+2}f_{2m+2}+\dots q_nf_n)$, which implies that \[t=exp(F(\lambda_1e_1+\dots\lambda_{2m+1}e_{2m+1}))=f(exp(\lambda_1e_1+\dots\lambda_{2m+1}e_{2m+1}))\in Im(f)\ \square \]
\end{demo}

\vskip 3mm

Particularly, $T_N=(\C^*)^n/Im(f)$ is isomorphic to $(\C^*)^{n-2m-1}$ (and, as it was claimed in the previous section, $X$ is a toric variety). We note $\overline{g}$ for the isomorphism between $T_N$ and $(\C^*)^{n-2m-1}$ induced by $g$.
\vskip 4mm

\begin{defn} We note $\lambda_T^u$ for the one-parameter subgroup of $T_N$ defined by $\lambda_T^u=\overline{g}^{-1}\circ \lambda_{n-2m-1}^u$. Since  $\overline{g}$ is an isomorphism, every one-parameter subgroup of $T_N$ has this form.
\end{defn}

\vskip 4mm

\textbf{Notation:} Let $\phi:T_1\rightarrow T_2$ be a group homomorphism between two algebraic torus $T_1$ and $T_2$. We will note $\phi^*$ for the morphism between the groups of one-parameter subgroup of $T_1$ and $T_2$ induced by $\phi$: $\phi^*(\lambda)=\phi\circ\lambda$.

\vskip 4mm

The group of one-parameter subgroup of $\left(\C^*\right)^n$ is $\{\lambda_n^u/u\in\Z^n\}$ which we will identify with $\Z^n$ via $u\leftrightarrow \lambda_n^u$. Via this map, the map $F$ and $G$ are exactly the morphisms induced by $f$ and $g$ (respectively):

\vskip 2mm

\begin{prop} With the above identification, we have $F=f^*$ and $G=g^*$. Precisely:
\begin{enumerate}
\item For every $v$ in $\Z^{2m+1}$, $f\circ\lambda_{2m+1}^v$ is the one-parameter subgroup $\lambda_n^{F(v)}$ of $\left(\C^*\right)^n$.
\item For every $v$ in $\Z^n$, $g\circ\lambda_n^v$ is the one-parameter subgroup $\lambda_{n-2m-1}^{G(v)}$ of $\left(\C^*\right)^{n-2m-1}$.
\end{enumerate}
\end{prop}

\vskip 4mm

\begin{demo} Let us take some $t\in\C^*$.
\\1) We have $\lambda_{2m+1}^v(t)=(t^{v_1},...,t^{v_{2m+1}})$ so \[f\circ\lambda_{2m+1}^v(t)=(t^{v_1+v_2a_j^1+\dots+v_{m+1}a_j^m+\dots+v_{2m+1}a_j^m})_j=(t^{<v,\widetilde{l_j}>})_j=\lambda_n^{F(v)}(t)\]

2) Let us pick some $T\in\C^n$ such that $t=exp(T)$. We put $w=G(v)=(w_1,...,w_{n-2m-1})$. If we note $g_j$ (resp. $G_j$) for the coordinates functions of $g$ (resp. $G$) in the canonical bases, we can easily verify that $g_j\circ exp=exp\circ G_j$ and that $w_j=G_j(v)$ for every $j$.
\\Next, we have $\lambda^v_n(t)=(t^{v_1},\dots,t^{v_n})=(e^{Tv_1},\dots,e^{Tv_n})$, so $g\circ\lambda^v_n(t)=g(e^{Tv_1},\dots,e^{Tv_n})=exp\circ G(Tv_1,\dots,Tv_n)=exp(G_1(Tv),\dots,G_{n-2m-1}(Tv))=(e^{G_1(Tv)},\dots,e^{G_{n-2m-1}(Tv)})$.
\\Finally, $g\circ\lambda^v_n(t)=(e^{TG_1(v)},\dots,e^{TG_{n-2m-1}(v)})=(e^{Tw_1},\dots,e^{Tw_{n-2m-1}})=\lambda_{n-2m-1}^w(t)$. $\square$
\end{demo}

\vskip 4mm

We would like to identify the group $N$ of one-parameter subgroups of $T_N$ with some lattice in $\C^n/Im(F)$. A natural candidate is described as follow: Let $\Pi$ be the canonical surjection $\C^n\rightarrow \C^n/Im(F)$ and $\overline{G}:\C^n/Im(F)\rightarrow\C^{n-2m-1}$ be the linear isomorphism induced by $G$. Notice that $\overline{G}$ is also a $\Z-$module isomorphism between $\Z^{n-2m-1}$ and $\Pi(\Z^n)$. If $u$ belongs to $\Z^n$, we set $\lambda_T^{\Pi(u)}$ for the one-parameter subgroup $\lambda_T^{G(u)}$ (notice that this definition make sense since $Im(F)=Ker(G)$). As a consequence, we have: $N=\{\lambda_T^{\Pi(u)}/u\in\Z^n\}$ which can be identified with $\Pi(\Z^n)=\Z^n/Im(F).$

\vskip 3mm

Now, we can define an "exponential" map between $\C^n/Im(F)$ and $T_N$: we define $exp:\C^n/Im(F)\rightarrow T_N$ by setting $exp(\Pi(z))=\pi\circ(exp(z))$ for every element $z$ in $\C^n$. The fact that $Ker(g)=Im(f)$ implies that this map is well defined.(Alternatively, we can define this exponential as the map $\overline{g}^{-1}\circ exp\circ\overline{G}$). By construction, we have $\pi\circ exp=exp\circ\Pi$ and $exp\circ\overline{G}=\overline{g}\circ exp$. Moreover, for every $v\in\Z^n$, $\pi\circ\lambda_n^v=\lambda_T^{\Pi(v)}$, that is to say that $\Pi=\pi^*$.

To sum up, we have the following commutative diagram:
\label{suitesexactes}
\[
\xymatrix{
		0 \ar[r] & \C^{2m+1} \ar@{^{(}->}[rr]^F \ar[rd]^F \ar@{->>}[dd]^{exp} & & \C^n \ar@{->>}@/_2pc/[ll]^{\widetilde{F}} \ar@{->>}[rr]^G \ar@{->>}[rd]^\Pi \ar@{->>}[dd]^{exp} & & \C^{n-2m-1} \ar[r] \ar@{->>}[dd]^{exp} \ar@{^{(}->}@/_2pc/[ll]^{\widetilde{G}} & 0\\
   	& & Im(F) \ar@{^{(}->}[ru]^i \ar@{->>}[dd] & & \C^n/Im(F) \ar[ru]_{\overline{G}} \ar@{->>}[dd]^{exp}  \\
    & (\C^*)^{2m+1} \ar@{-}[r]^f \ar@{->>}[rd]^f & \ar[r] & (\C^*)^n \ar@{-}[r] \ar@{->>}[rd]^\pi & \ar@{->>}[r]^-{g} & (\C^*)^{n-2m-1} \ar[r] & 1\\
    & & Im(f) \ar@{^{(}->}[ru]^i & & T_N \ar[ru]_{\overline{g}} \ar@{=}[d] \\
    & & & & (\C^*)^n/Im(f)
}
\]

\subsection{Study of the projection $\pi$}

In this last section, we are in position to prove the first part of \ref{thmprinc}. Let $\Sigma$ denote the fan in $N_\R=N\otimes\R$ associated to $X$ (this fan exists since $X$ is separated and normal). In order to use \ref{sphere}, we have to prove that $\Sigma$ is simplicial. According to $\cite{CLS}$, we just have to prove that $X$ is an orbifold.

\vskip 4mm

As it was shown previously, the holomorphic acceptable action on $\cal S$ is free and the algebraic action on $\cal S$ has only finite stabilizers. Consequently, every stabilizer for the action of $K$ on $\cal N$ is finite. So, $X$ is the quotient of the compact variety $\cal N$ by the action of a compact Lie group $K$ and every stabilizer for this action is finite.

\vskip 4mm

Finally, since it is a continuous map defined on a compact set, we can claim that  the map

\[\begin{array}{rccl} \phi_n: & K & \longrightarrow  &{\cal N} \\ & h & \mapsto & h\cdot n  \end{array}\]

is proper for every element $n$ in $\cal N$. By Holmann's theorem (cf. $\cite{O}$, \S 5.1), we get that $X$ is indeed an orbifold. Taking advantage of \ref{sphere}, we have proved that $K_\Sigma$, the underlying complex of the fan $\Sigma$, is a $(n-2m-2)-$sphere.

\vskip 4mm

Now, the following proposition will prove \ref{thmprinc}:

\vskip 2mm

\begin{thm} $K_\Sigma$ and $\cal P$ are isomorphic simplicial complexes.
\end{thm}

\vskip 4mm

\begin{demo} Fisrt of all, $\pi$ is by construction a toric morphism. According to $\cite{CLS}$, this implies that $\pi$ is equivariant (for the toric actions of $\left(\C^*\right)^n$ and $T_N$), so $\pi$ sends an orbit $O_I$ in $\cal S$ to an orbit in $X$. We set $\widetilde{O_I}$ to be the unique orbit in $X$ containing $\pi(O_I)$. Moreover, the map $\pi^*$ (identified with $\Pi$) from $\R^n$ into $N_\R$ preserves the cones.

\vskip 4mm

We can easily show that $\cal S$ and $X$ have exactly the same number of orbits, i.e that the quotient of $X$ by its torus $T_N$ is in bijection with the quotient of $\cal S$ by its torus $\left(\C^*\right)^n$. Since $\pi$ is surjective, we get that the assignation  $O_I\rightarrow\widetilde{O}_I$ (induced by $\pi$) is bijective. As a consequence, $\Pi$ induces a bijection between the cones of $\Sigma(\cS)$ and those of $\Sigma$

\vskip 4mm

If $\sigma$ is a cone belonging to $\Sigma(\cS)$, we note $O(\sigma)$ the orbit in $\cal S$ associated to $\sigma$ (cf $\cite{CLS}$, ch.3). Particularly, $O(\sigma_I)=O_I$. We will also note in the same way the orbits in $X$. Moreover, the image of the cone $\sigma$ by $\Pi$ will be noted $\widetilde{\sigma}$. So, we have $\widetilde{O(\sigma)}=O(\widetilde{\sigma})$.
\\Still from $\cite{CLS}$, $\pi$ preserves the partial order of faces: if $\tau$ is a face of $\sigma$ in $\Sigma(\cS)$, then $\widetilde{\tau}$ is a face of $\widetilde{\sigma}$.

\vskip 4mm

On the other hand, a slight modification of the proof of the fact that $(\C^*)^n/Im(f)$ (i.e. $\pi\left(O_\emptyset\right)$) is isomorphic to $\left(\C^*\right)^{n-2m-1}$ shows that $\pi(O(\sigma))$ is isomorphic to $(\C^*)^{n-2m-1-dim(\sigma)}$. At the level of cones, this means that cones of $\Sigma(\cS)$ with the same dimension are sent to cones of $\Sigma$ with the same dimension. In particular, $\Pi$ sends rays to rays. This last property means that $\Pi$ induces a bijection between vertices of $\cal P$ and vertices of $K_\Sigma$, bijection which we also note $\Pi$.
\\It is clear by what preceeds that this very last map is an simplicial complex isomorphism. $\square$
\end{demo}

\vskip 4mm

As a consequence, $\cal P$ is indeed a simplicial sphere. More precisely, what we have is a particular type of simplicial spheres, namely spheres which are underlying simplicial complex of some complete fan:

\vskip 2mm

\begin{defn} Let $K$ be a $d$-sphere. $K$ is said to be rationally starshaped if there exists a lattice $N$ in $\R^{d+1}$, a point $p_0\in N$ and a realization\footnote{The fact that a $d-$sphere has a realization in $\R^{d+1}$ is an open question (cf. $\cite{MW}$, \S 5).} $|K|$ for $K$ in $\R^{d+1}$ such that every vertex of $|K|$ belongs to $N$ and every ray emanating from $p_0$ intersects $|K|$ in exactly one point. The realization $|K|$ is said to be \emph{starshaped} and we say that $p_0$ belongs to the \emph{kernel} of $|K|$.
\end{defn}

\vskip 4mm

\begin{cor} Let $(\cE,l)$ be a good system verifying $(K)$. Then its associated complex $\cal P$ is a rationally starshaped sphere.
\end{cor}

\vskip 4mm

\begin{demo} We have already seen that $\cal P$ is a simplicial sphere combinatorially equivalent to $K_{\Sigma}$. If we put $\Sigma(1)=\{\rho_1,\dots,\rho_v\}$ for the distincts rays of $\Sigma$ and $u_1,\dots,u_v$ for generators of $\rho_1,\dots,\rho_v$ (respectively) in $N$, then the geometric simplicial complex $C$ whose simplexes are $Conv(u_i,i\in I)$ for $I$ in $\cal P$ is obviously a realization of $\cal P$ in $\R^{n-2m-1}$ with rational vertices. The point $0$ is in the kernel of $C$ so $\cal P$ is rationally starshaped. $\square$
\end{demo}

\vskip 4mm
\section{Inverse construction}

\subsection{The construction}

In this section, we give a realization theorem: for every rationally starshaped sphere, there exists a good system whose associated sphere is the given one. In $\cite{M},\ p.86$, the same kind of theorem is proven for simple polytopes and LVM manifolds. One of the main interests of this theorem is that it gives us a clue to answer to an open question: does there exist a LVMB manifold who has not the same topology than a LVM manifold? The idea is to use this theorem to construct a LVMB manifold from a rationally starshaped sphere $\cP$ which is not polytopal and that this manifold has a particular topology. For example, we expect that the LVMB manifold coming from the Brückner's sphere or the Barnette's one (the two $4$-spheres with $8$ vertices which are not polytopal) are good candidates.

\vskip 4mm

Let $\cal P$ be a rationally starshaped $d$-sphere with $v$ vertices (up to a simplicial isomorphism, we will assume that these vertices are $1,2,\dots,v$). So, there exists a lattice $N$ and a realization $|\cal P|$ of $\cal P$ in $\R^{d+1}$ all of whose vertices belong to $N$. We can assume that $0$ is in the kernel of $|\cal P|$ and that $N$ is $\Z^{d+1}$.

To begin, we suppose that $v$ is even and we put $v=2m$. We note $\cal E$ the set defined by \[\{P\in\{1,...,v+d+1\}/ P^c\ is\ a\ facet\ of\ \cP\}\] and $\cE_0=\{\{0\}\cup P / P\in \cE\}$ We also note $A$ the matrix whose columns are $x_1,\dots,x_v$. We label its rows $x^1,\dots,x^{d+1}$, and finally, we have:

\vskip 4mm

\begin{thm} If $e_1,\dots,e_v$ is the canonical basis of $\R^v$, then $(\cE_0,(0,e_1,\dots,e_v,-x^1,\dots,-x^{d+1}))$ is a good system of type $(v+1,d+v+2)$ whose associated complex is $\cal P$.
\end{thm}

\vskip 4mm

First, since $\cal P$ is a $d-$sphere, it is a pure complex whose facets have $d$ elements. As a consequence, every subset in $\cE_0$ has $v+1$ elements. So, $\cE_0$ is a fundamental set of type $(v+1,v+d+2)$ (as usual, we will note the set of its acceptable subsets $\cA_0$). Notice that, since every facet of $\cal P$ has vertices in $\{1,\dots,v\}$, $\cE_0$ has $d+2$ indispensable elements. Moreover, by definition, $\cE_0$'s associated complex is clearly $\cal P$. Since $\cal P$ is a sphere, hence a pseudo-manifold, \ref{pseudomanifold} implies that $\cE_0$ is minimal for the SEU property. We can also notice that the same is true for $\cal E$ (with set of acceptable subsets $\cal A$).

\vskip 4mm

Secondly, we have to check that the vectors $0,e_1,e_2,\dots, e_v,-x^1,\dots,-x^{d+1}$ "fits" with $\cE_0$ to make a good system. We put $\rho_j=pos(x_j)$ for the ray generated by $x_j,j=1,\dots,v$. We also note  $\Sigma(\cP)$ the fan defined by $\Sigma(\cP)=\{pos(x_j,j\in I)/I\in\cP\}.$ Then, by definition, $\Sigma(\cP)$ is a simplicial fan whose underlying complex is $\cal P$. As a consequence, $\Sigma(\cP)$ is rational with respect to $N$ and complete (since $\cal P$ is a sphere, cf. \ref{sphere}). We set $X$ for the compact normal toric variety associated to $\Sigma(\cP)$. Following $\cite{Ha}$, we will construct $X$ as a quotient of a quasi-affine toric variety by the action of an algebraic torus.

\vskip 4mm

In what follows, $(e_j)_{j=1,\dots,N}$ is the canonical basis of $\C^N$ (for any $N$). Let $\widetilde{\Sigma}$ be the fan in $\R^{v+d+1}$ whose cones are $pos(e_j,j\in J)$, $J\in\cal P$. Obviously, $\widetilde{\Sigma}$ is a noncomplete simplicial fan whose underlying complex is $\cal P$, too. We note $\widetilde{X}$ the (quasi-affine) toric variety associated to this fan. Then, the open set $\cS=\{z\in \C^{v+d+1}/I_z\in \cA\}$ is exactly the set $\widetilde{X}$. Indeed, the computation of the proof of \ref{eventail} shows that the fan of $\cS$ is the fan in $\R^{v+d+1}$ whose rays are generated by the canonical basis and whose underlying complex is $\cal P$. Thus, $\cS$ and $\widetilde{X}$ have exactly the same fan, so they agree. We can also observe that $\cS_0=\C^*\times \cS=\{(z_0,z)\in\C^{v+d+2}/I_{z_0,z}\in \cA_0\}$ have the same fan, but seen in $\R^{v+d+2}$.

\vskip 4mm

Finally, we define the map $f:(\C^*)^v\rightarrow(\C^*)^{v+d+1}$ by $f(t)=(t,X_v^{-x^1}(t),\dots,X_v^{-x^{d+1}}(t))$. Note that $t=(X_v^{e_1}(t),\dots,X_v^{e_v}(t))$. According to $\cite{Ha}$, $X$ is the quotient of $\widetilde{X}=\cS$ by the restriction of the toric action of $(\C^*)^{v+d+1}$ restricted to $Im(f)$. It's a geometric quotient because $\widetilde{\Sigma}$ is simplicial.

\vskip 4mm

Considering $l_0=0,l_1=e_1,\dots,l_{2m}=e_{2m},-x^1,\dots,l_{v+d+2}=-x^{d+1}$ as elements of $\C^m$, we can define an holomorphic action of $(\C^*)\times \C^m$ on $\cS_0$ by setting

\[
(\alpha,T)\cdot (z_0,z)=(\alpha e^{<l_j,T>}z_j)_{j=0}^{v+d+2}\ \forall \alpha\in\C^*,t\in \C^m, (z_0,z)\in\cS_0
\]

\vskip 4mm

It is clear that $(\cE_0,l)$ verifies $(K)$. Moreover, the algebraic acceptable action associated to this system has $X$ for quotient. Indeed, a computation shows that this algebraic action is defined by \[(\alpha,t)\cdot (z_0,z)=(\alpha z_0,f(t)\cdot z)\  \forall \alpha\in\C^*,t\in(\C^*)^{2m}, (z_0,z)\in\cS_0\]
If we note $\cV_0=\{[z_0,z]/(z_0,z)\in\cS_0\}$ the "projectivization" of $\cS_0$, then the orbit space for the algebraic action is the quotient of $\cV_0$ by the action defined by

\[
t \cdot [z_0,z]=[z_0,f(t)\cdot z]\  \forall t\in(\C^*)^{2m}, [z_0,z]\in\cV_0
\]

\vskip 4mm

But $\cV_0=\{[1,z]/z\in \cS\}$ is homeomorphic to $\cal S$ so the claim is proved.

\vskip 4mm

The only thing we have to check is that $(\cE_0,l)$ is a good system, that is the orbit space $\cal N$ for the holomorphic action is a complex manifold. We have seen that $X$ is a geometric quotient so the action of $\left(\C^*\right)^{2m}$ is proper (see, for example, $\cite{BBCMcG}$, p.28). As a consequence, the action of $\C^m$ is proper too, and since there is no compact subgroups in $\C^m$ (except $\{0\}$ of course), its action is free. Finally, the action is proper and free so $\cal N$ can be endowed with a structure of complex compact manifold.

\vskip 4mm

\begin{rmq}
1) Since $\cal N$ is a complex compact manifold, the imbrication condition is fulfilled (cf. \ref{thmBosio1}). This gives also another proof for the fact that the SEU property is also fulfilled.
\\2) The transformation of $(x_1,\dots,x_v,e_1,\dots,e_{d+1})$ to $(e_1,\dots,e_v,-x^1,\dots,-x^{d+1})$ is called a linear transform of $(x_1,\dots,x_v,e_1,\dots,e_{d+1})$ (see $\cite{E}$ for example). In $\cite{M}$, the construction of a LVM manifold starting from a simple polytope used a special kind of linear transform called Gale (or affine) transform (see. $\cite{E}$)
\end{rmq}

\vskip 4mm

Now, we suppose that $v=2m+1$ is odd. The construction is very similar. However, we have an additional step: we define an action of $(\C^*)^{v+1}$ on $\cS_0$ by \[(t_0,t)\cdot (z_0,z)=(t_0z_0,f(t)\cdot z)=(X^{e_0}(t)z_0,X^{e_1}(t)z_1,\dots,X^{-x^{d+1}}(t)z_{v+d+1})\] where $f$ and the action of $Im(f)$ are defined as above, and $e_0$ is the first vector of $\C^{v+1}$ with coordinates $(z_0,z_1,\dots, z_{v+1})$. The orbit space for this last action is still $X$. The rest is as above: we define a fundamental set $\cE_*=\{\ \{-1\}\cup E\ /\ E\in \cE_0\ \}$ and we have:

\vskip 4mm

\begin{thm} If $e_0,\dots,e_v$ is the canonical basis of $\R^{v+1}$, then $(\cE_*,(0,e_0,e_1,\dots,e_{v-1},-x^1,\dots,-x^{d+1}))$ is a good system of type $(v+2,d+v+3)$ whose associated complex is $\cal P$.
\end{thm}

\subsection{LVMB manifolds and moment-angle complexes}

In this section, we follow closely the definitions and notations of $\cite{BP}$. We use the previous result of the above section to show that many moment-angle complexes with even dimension can be endowed with a complex structure as a LVMB manifold.

\vskip 4mm

\begin{defn} Let $K$ be a simplicial complex on $\{1,\dots,n\}$ with dimension $d-1$. If $\sigma\subset\{1,\dots,n\}$, we put
\[
C_\sigma=\{t\in[0,1]^n/t_j=1\ \forall j\notin \sigma\}
\]

and

\[
B_\sigma=\{z\in\D^n/|z_j|=1\ \forall j\notin \sigma\}
\]

\vskip 2mm

The moment-angle complex associated to $K$ is \[\cZ_{K,n}=\bigcup_{\sigma\in K} B_\sigma\]
\end{defn}

\vskip 4mm

\begin{exempl} For instance, if $K$ is the border of the $n$-simplex, $\widetilde{\cZ}_K$ is the sphere $S^{2n-1}$ (cf. \cite{BP}).
\end{exempl}

\vskip 4mm

In $\cite{BP},$ lemma $6.13$, it is shown that if $K$ is a simplicial sphere, then $\cZ_{K,n}$ is a closed manifold. Moreover, in $\cite{B}$, to prove \ref{thmBosio1} (p.1268 in $\cite{B}$), Bosio introduces the set $\widehat{M_1'}$ defined by

\[
\widehat{M_1'}=\{\ z\in\D^n/\ J_z \in \cA\ \}
\]

\vskip 2mm

where $J_z=\{\ k\in\{1,\dots,n\}/\ |z_j|=1\ \}$. This set is the quotient of $\cS$ by the restriction of the holomorphic to $\R_+^*\times\C^m$ and as a consequence, $\cN$ is the quotient of $\widehat{M_1'}$ by the \emph{diagonal action} of $S^1$ defined by:

\vskip 2mm

\[
e^{i\theta}\cdot z=(\ e^{i\theta}z_1,\dots,e^{i\theta}z_n\ )\hskip 3mm \forall \theta\in\R,\ z=(z_1,\dots,z_n)\in \widehat{M_1'}\subset \C^n
\]

\vskip 4mm

\begin{prop}\label{complexe} We have $\widehat{M_1'}=\cZ_{\cP,n}$
\end{prop}

\vskip 4mm

\begin{demo} Indeed, we have:
\[\begin{array}{rcl}
\widehat{M_1'} & = & \displaystyle{\bigcup_{\sigma\in\cP} \{z\in\D^n/ \forall j\notin\sigma, |z_j|=1 \}} \\
 		    			 & = & \displaystyle{\bigcup_{\tau\in\cA}\{z\in\D^n/\tau\subset I_z\}} \\
 		    			 & = & \{z\in \D^n/J_z\in\cA\} \hskip 20mm \square \\
\end{array}\]
\end{demo}

\vskip 4mm

\begin{prop} \label{complexe2} Let $(\cE,l)$ be a good system with type $(2m+1,n,k)$ and $\cN$ the LVMB associated to this system. If $k>0$, then $\cN$ is homeomorphic to a moment-angle complex.
\end{prop}

\vskip 4mm

\begin{demo} For the clarity of the proof, we assume that $n$ is indispensable. Let $\cP$ be the sphere associated to $\cE$ and $\cS$ the open subset of $\C^n$ whose quotient by the holomorphic action is $\cN$. According to \ref{complexe}, the quotient $\widehat{M_1'}=\cS/(\R_+^*\times \C^m)$ can  be identified with $\cZ_{\cP,n}$.

\vskip 4mm

Let $\phi$ be the map defined by

\[
\begin{array}{rcccl}
\phi & : & \cZ_{\cP,n} & \rightarrow & \C^{n-1} \\
	   &	 & z					 & \mapsto 		 & \left(\frac{z_1}{z_n},\dots, \frac{z_{n-1}}{z_n} \right)
\end{array}
\]

Since $n$ is indispensable, we have $|z_n|=1$ for every $\cZ_{\cP,n}$ so $\phi$ is well defined. Moreover, $\phi$ is continuous and a simple calculation shows that $\phi$ is invariant for the diagonal action and $\phi(\cZ_{\cP,n})=\cZ_{\cP,n-1}$. We claim that if $\phi(z)=\phi(w)$, then $z$ and $w$ belong to the same orbit for the diagonal action. Indeed, if $\phi(z)=\phi(w)$, we have

\[
\left(\frac{z_1}{z_n},\dots,\frac{z_{n-1}}{z_n}\right)=\left(\frac{w_1}{w_n},\dots,\frac{w_{n-1}}{w_n}\right)
\]

\vskip 4mm

We have $|z_n|=|w_n|=1$, so we put $\frac{z_n}{w_n}=e^{i\alpha}$ and we have $z=e^{i\alpha}w$.

\vskip 4mm

As a consequence, $\phi$ induce a map $\overline{\phi}:\cN\rightarrow \cZ_{\cP,n-1}$ which is continuous and bijective. Actually, this is an homeomorphism since the inverse map $\phi^{-1}$ is the continuous map

\[
\begin{array}{rcccl}
\phi^{-1} & : & \cZ_{\cP,n-1} & \rightarrow & \cN \\
	       	&   & z	 				 		& \mapsto 	  & [(z,1)]
\end{array}
\]

where $[(z,1)]$ denotes the equivalence class of $(z,1)\in \cZ_{\cP,n}$ of the diagonal action.
\end{demo}

\vskip 4mm

\begin{cor}\label{complexe3} Let $\cP$ be a rationally starshaped sphere. Then there exists $N\in\N^*$  such that $\cZ_{\cP,N}$ can be endowed with a complex structure as LVMB manifold.
\end{cor}

\vskip 4mm

\begin{demo} Since $\cP$ is a rationally starshaped sphere, there exists a good system $(\cE,l)$ with type $(2m+1,n,k)$ whose associated complex is $\cP$ (cf. the previous subsection). Moreover, our construction of $(\cE,l)$ implies that $k>0$. So, as stated in \ref{complexe2}, $\cN$ is homeomorphic to $\cZ_{\cP,n-1}$. So, we put $N=n-1$ and we endow $\cZ_{\cP,N}$ with the complex structure induced by this homeomorphism.
\end{demo}

\vskip 4mm

\begin{rmq} Let $N_0$ be the smallest integer $N$ as in \ref{complexe3}. Then, for every $q\in\N$, $\cZ_{\cP,N_0+2q}$ can also be endowed with a complex structure and then we have $\cZ_{\cP,N_0+2q}=\cZ_{\cP,N_0}\times(S^1)^{2q}$. Indeed, let $\Lambda$ be the matrix whose columns are the vectors of the good system $(\cE,l)$ given in the proof of \ref{complexe3}. We put $\widetilde{\cE}=\{\ P\cup\{N_0+1,N_0+2\}/\ P \in \cE \}$ and we define $\lambda_1,\dots,\lambda_{n+2}$ as the columns of the matrix

\[
\begin{pmatrix}
\Lambda 		& 0  & 0 \\
-1 \dots -1 & 1  & 0 \\
-1 \dots -1 & -1 & 1
\end{pmatrix}
\]

Then it is easy to show that $(\widetilde{\cE},(\lambda_1,\dots,\lambda_{n+2}))$ is a good system and that we have $\cZ_{\cP,N_0+2}=\cZ_{\cP,N_0}\times(S^1)^2$
\end{rmq}

\bibliographystyle{short}
\bibliography{versionanglaise}

\end{document}

config: arxiv


TITLE: Implementation details of implicit arguments
QUESTION [7 upvotes]: How do the implicit arguments work in proof assistants such as Agda or Coq? Specifically, how are the blanks filled in? What kind of resolution algorithm is used?
Are there any papers written about implicit arguments?

REPLY [2 votes]: I don’t think there is any detailed paper description of this for Coq, however there is one for Matita which as far as I know is quite close to how Coq does things.
Unification is taken as a black box there, if you want to open that box, the best I know of in the Coq ecosystem is UniCoq and its accompanying paper. Again, it is not what actually happens in the internals of Coq, but it should be close enough to satisfy some curiosity.


TITLE: Solving a set of equations containing sums of logarithms
QUESTION [2 upvotes]: I'm trying to solve a set of $J+1$ equations for variables $x,y_1,\ldots,y_J$. The equations are as follows:
\begin{align*}
\sum_j\sum_i\frac{1}{a_{ij}}\log\left(\frac{a_{ij}}{x+y_j}\right)&=MN\\
\left(N-\sum_i\frac{1}{a_{ij}}\log\left(\frac{a_{ij}}{x+y_j}\right)\right)y_j&=0,\,\,j=1,\ldots,J
\end{align*}
where each $a_{ij}>0$ and $x>0, y_j\ge0$. The logarithms are base $e$. Note that we cannot divide the second set of equations by $y_j$ because some may be zero. I've tried rearranging the equations using $\log\left(\frac{a_{ij}}{x+y_j}\right) = \log(a_{ij}) - \log(x+y_j)$ but it doesn't seem to help. Any ideas?

Attempt:
Using properties of logarithm and exponential functions, I've managed to write the equations as
\begin{align*}
\prod_j\prod_i\left(\frac{a_{ij}}{{x+y_j}}\right)^{1/a_{ij}}&=\exp(MN)\\
\prod_i\left(\frac{a_{ij}}{{x+y_j}}\right)^{y_j/a_{ij}}&=\exp(Ny_j),\,\,j=1,\ldots,J
\end{align*}
This looks nicer, but it doesn't appear to lead anywhere.
An alternate approach:
Based on N74's comments, by defining $z_j = \log(x+y_j)$, we can alternatively write the equations as
\begin{align*}
\sum_j\sum_i\left[\frac{\log(a_{ij})}{a_{ij}}-\frac{z_j}{a_{ij}}\right]&=MN\\
\left(N-\sum_i\left[\frac{\log(a_{ij})}{a_{ij}}-\frac{z_j}{a_{ij}}\right]\right)(\exp(z_j)-x)&=0
\end{align*}
Defining $A = N-\sum_i\frac{\log(a_{ij})}{a_{ij}}$, $B = \sum_j\sum_i\frac{\log(a_{ij})}{a_{ij}}-MN$, and $c_j=\sum_i\frac{1}{a_{ij}}>0$, we have
\begin{align*}
\sum_jc_jz_j &= B\\
(A+c_jz_j)(\exp(z_j)-x)&=0,\,\,j=1,\ldots,J
\end{align*}
This is where I run into trouble.

REPLY [1 votes]: Using your definitions:
 $$z_j = \log(x+y_j)$$ $$A = N-\sum_i\frac{\log(a_{ij})}{a_{ij}}$$ $$B = \sum_j\sum_i\frac{\log(a_{ij})}{a_{ij}}-MN$$ $$c_j=\sum_i\frac{1}{a_{ij}}>0,$$ we want to solve:
$$\begin{align*}
\sum_jc_jz_j &= B\\
(A+c_jz_j)(\exp(z_j)-x)&=0,\,\,j=1,\ldots,J
\end{align*}.$$
To make null a product we need that one of the factors is null, so we have a solution when $z_j=\log x$ and another solution for $z_j=-{A\over c_j}$, for each $j$ in $[1,J]$.
Let's discuss some solutions.

All $y_j$ are null:
In this case $z_j=\log x$  so $$\sum_jc_j\log x = B$$ $$x=\exp {B \over \sum_jc_j}$$
Only $y_{j^*}$ is not null:
In this case $z_j=\log x$  for $j\neq j_*$ and $c_{j^*}z_{j^*}=-A$ so $$\sum_{j\neq j^*}c_j\log x = B+A$$ $$x=\exp {B+A \over \sum_{j\neq {j^*}}c_j}$$ $$y_{j^*}=\exp {-A \over c_{j^*}}-x$$
All $y_j$ are not null:
In this case $c_jz_j = -A$ so the first equation becomes $$-JA=B.$$ This equation doesn't depend on the unknowns, so the system is either impossible (${B\neq -JA}$), or undetermined ($x$ cannot be fixed).

Obviously you can find all the other solutions by just fixing how many $y_j$are not null.


TITLE: How to evaluate integral $\frac{1}{2 \pi i}\int \limits_{c-i \infty}^{c+i \infty} \frac{ds}{s(1-q^{1-s})}\text{?}$
QUESTION [3 upvotes]: How to evaluate the integral $$\frac{1}{2 \pi i}
\int \limits_{c-i \infty}^{c+i \infty} \frac{ds}{s(1-q^{1-s})}\text{?}$$ I tried with Perron's formula but I couldn't solve it. The result of the integral is $\frac{1}{2}$. Can someone help please?!

REPLY [0 votes]: For some $q > 1$ let
$$f(x) = \sum_{n \ge 0,q^n < e^x} q^n$$
For $x = n \log q$ we take the mean value of the left and right limit, that is
$$f(n \log q) = \frac{f(\epsilon+n\log q)+f(-\epsilon+n\log q)}{2}$$
Thus $f(0) = 1/2$

For $\Re(s ) > 1$
$$F(s) = \frac{1}{1-q^{1-s}} = \sum_{n=1}^\infty q^n q^{-sn} = s \int_0^\infty f(x) e^{-sx}dx$$
By inverse Fourier/Laplace/Mellin transform (or Perron's formula), for $c > 1$ and $x \ne n \log q$
$$f(x) = \frac{1}{2i\pi} \int_{c-i \infty}^{c+i\infty} \frac{F(s)}{s} e^{sx}ds$$
With $x=0$
$$f(0) = \frac12= \frac{1}{2i\pi} \int_{c-i \infty}^{c+i\infty} \frac{F(s)}{s} ds= \frac{1}{2i\pi} \int_{c-i \infty}^{c+i\infty} \frac{1}{s(1-q^{1-s})} ds$$
And the RHS is analytic in $q$, so we can extend by analytic continuation to $q \in \mathbb{C}$, and the same for $c$.


TITLE: Defining the inverse of the Bott map
QUESTION [3 upvotes]: I hope this isn't too narrowly focused.  I have a question concerning the inverse of the Bott map as defined in Atiyah's paper, Bott Periodicity and the Index of Elliptic Operators.  On page 122 he defines it as the composition
$$K^{-2}(X) \to K(S^2 \times X) \stackrel{\text{index} \bar\partial}{\longrightarrow} K(X).$$
My question is: what is the first map?  At first I thought it was
$$
K^{-2}(X) = \tilde K(S^2 \wedge X^+) \to \tilde K(S^2 \times X^+) \to \tilde K(S^2\times X) \subset K(S^2 \times X)
$$
where the first map is the pullback of the projection and the second is the pullback of the inclusion.  But from the 6 term exact sequence, for this first map to be injective (which it needs to be for this to be correct) we need
$$
K(S^2 \times X^+, S^2 \times X) \to \tilde K(S^2 \times X^+)
$$
to be the zero map which doesn't seem correct since $S^2 \times X^+/S^2 \times X$ is identifiable with $S^2 \sqcup \{pt\}$ and if we pullback a non-trivial bundle over this space to $S^2 \times X^+$ it seems like it won't trivialize.
Thanks!

REPLY [1 votes]: Actually I think I found a better way to look at the map, which shows that it is injective.  Consider the pair $(S^2 \times X, \{pt\} \times X)$.  Then since $\{pt\} \times X$ is a retract of $S^2 \times X$, the 6-term sequence in $K$-theory splits to give a short exact sequence
$$
0 \to K(S^2 \times X, \{pt\} \times X) \to K(S^2 \times X) \to K(\{pt\} \times X).
$$
But $(S^2 \times X) / (pt \times X) \simeq S^2 \wedge X^+$, giving the desired map.


TITLE: Projecting a Positive Matrix
QUESTION [0 upvotes]: If a matrix is positive-definite, is its projection on a subspace positive-definite as well?
The question concerns linear operators in general. Let $O$ be a positive-definite operator and $P$ the operator that projects into a given subspace. Is $P\circ O \circ P$ also positive-definite?

REPLY [0 votes]: If $P$ is symmetric (orthogonal projector), then $POP$ is non-negative definite since $a' POP a = (Pa)'O (Pa) \geq 0$.

REPLY [0 votes]: Under the assumption of positive-definiteness of $O$, we cannot guarantee that $POP$ is positive-definite.  For example, consider
$$
O=\begin{pmatrix}
1 & 0 \\
0 & 2
\end{pmatrix} , \qquad
P=\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}
$$
However we can show a slightly weaker statement:

If $O$ is positive-semidefinite and $P$ is a projection, then $POP$ is positive-semidefinite.

To see this, let $V$ be an inner product space with inner product $\langle\cdot,\cdot\rangle$, let $O\in L(V)$ be positive-semidefinite, and let $P$ be a projection (i.e. a self-adjoint idempotent).  Then for any $x\in V$ we have
$$ \langle x,POPx\rangle=\langle P^*x,OPx\rangle =\langle Px,OPx\rangle=\langle(Px),O(Px)\rangle\geq 0.$$
and thus $POP$ is positive-semidefinite.

EDIT  In fact,the same argument shows a slightly stronger statment.  Since idempotency of the projection wasn't necessary, we have

If $O$ is positive-semidefinite and $T$ is self-adjoint, then $TOT$ is positive-semidefinite.


TITLE: Poisson Process maths question
QUESTION [2 upvotes]: I'm not quite sure how to tackle this question,
The number of customers arriving
to a bank and a supermarket each independently follow a Poisson process with intensity
3 and 5 customers per hour, respectively. What is the probability that the first customer will be
to the bank ?
Any help/guidance is appreciated

REPLY [1 votes]: hint
The time $T_B$ for a customer to arrive at the bank is exponential with mean $1/3$ and the time $T_S$ for one to arrive at the supermarket is an exponential with mean $1/5$, independent from $T_B.$ You just need to compute $P(T_B< T_S).$


TITLE: Prove $z = \frac {5^{125 }- 1}{5^{25} - 1} $ is composite.
QUESTION [0 upvotes]: We need to prove that $z = \cfrac {5^{125 }- 1}{5^{25} - 1} $ isn't a prime number . This form comes out to be the sum of a geometric progression with the first term $1$ and ratio $5^{25}$, with five terms. So it reduces to the form:
$$z=1 + 5^{25} + 5^{50} + 5^{75} + 5^{100} $$
Now you could present it as a difference of two squares, but that turns out to be a tedious task.
Any help after this? Thanks a lot!

REPLY [1 votes]: The number $x^4+x^3+x^2+x+1$ is the product of two values of $x^2+3x+1 \pm (x+1) \sqrt{5x}$
Since $x = 5^{25}$ makes $\sqrt{5x}=5^{13}$, it has a solution.
This has been known since the time of K.F.Gauss.
Cunningham Project
The number $b^n-1$ represents a number of $n$ digits, each $b-1$, in base b, and thus the divisors correspond to numbers with period of $n$.  For example, with $b=10, n=7$, we get the number $9999999$.  A good deal of number theory has gone into this.
One can find that there is a unique factor $^bA_n$ for every value of b and n, and that any pair are coprime, unless $p \mid n$ and $m/n$ is a power of $p$.  Apart from these and sevenites, the list of factors of $^bA_n$ is just the list of primes with this as a period.
Primes that have a period of n, by Fermat's little theory, are of the form $p=mx+1$.  While this greatly reduces the number of primes to check, it really is not enough.
Gauss's quadratic law tells us if the period n divides p an even or odd number of times, and for most values of n, this halves the number of primes to search.  But where the base is of the form $b=xy^2$, then the values of either x (for 1 mod 4), or 2x (for 2, 3 mod 4), have an algebraic factorisation of the kind given above.
So, for example, where b=120, x=30, and we would find that ^{120}A_{120a+60} has an algebraic factorisation, that is, odd numbers times 2*30.  For b=10, x=10, and we find that ^{10}A_{40a+20} has an algebraic factorisation.
The Cunningham project is to find the factors of these A values, as far as technology permits, for b=2-10, 12.  I have done similar runs for base 120.  The tables contain some unimaginately large numbers, and more than half of them contain a value larger than $\sqrt{A}$.
The number in this example is $^5A_{125}$, which passes the second test (ie algebraic and gaussian factorisation), but the rest is an uninspiringly large number best left to utilities like factor.
For $b^n-a^n$, the same general rules hold, except that the divisors of the base are $ab$, and $abx$ as square defines the value of $x$.  So $3^n-2^n$ is a sixth-square, and like all sixth-squares (6, 24, 54, ...), it follows that $A_n$ has a factorisation if $n$ is an twelve times an odd number.


TITLE: Grothendieck's Galois theory without finiteness hypotheses
QUESTION [24 upvotes]: This is motivated by the discussion here. The usual definition of the etale fundamental group (as in SGA 1) gives automatic profiniteness: Grothendieck's formulation of Galois theory states that any category with a "fiber functor" to the category of finite sets satisfying appropriate hypotheses is isomorphic to the category of finite, continuous $G$-sets for a well-defined profinite group $G$ (taken as the limit of the automorphism groups of Galois objects, or as the automorphism group of said fiber functor). In SGA1, the strategy is to take finite etale covers of a fixed (connected) scheme with the fiber functor the set of liftings of a geometric point.
One problem with this approach is that $H^1(X_{et}, G)$ for a finite group $G$ (which classifies $G$-torsors in the etale topology) is not isomorphic to $\hom(\pi_1(X, \overline{x}), G)$ unless $G$ is finite. Indeed, this would imply that the cohomology of the constant sheaf $\mathbb{Z}$ would always be trivial, but this is not true (e.g. for a nodal cubic).
However, Scott Carnahan asserts on the aforementioned thread that the "right" etale fundamental group of a nodal cubic should be $\mathbb{Z}$, not something profinite.
How exactly does this work? People have suggested that one can define it as a similar inverse limit of automorphism groups, but is there a similar equivalence of categories and an analogous formalism for weaker "Galois categories"? (Perhaps one wants not just all etale morphisms but, say, torsors: the disjoint union of two open immersions might not be the right candidate.) I'm pretty sure that the finiteness is necessary in the usual proofs of Galois theory, but maybe there's something more.

REPLY [14 votes]: In "The pro-étale topology for schemes" (http://arxiv.org/abs/1309.1198), Bhatt and Scholze introduce the pro-étale fundamental group which seems to give a good answer to your question (see, e.g., Theorem 1.10).  The pro-étale fundamental group also compares well against the usual étale fundamental group and the "SGA3 étale fundamental group".


TITLE: Show that the discrete metric can not be obtained from $X\neq\{0\}$
QUESTION [1 upvotes]: If $X \neq \{ 0\}$ is a vector space. How does one go about showing that the discrete metric on $X$ cannot be obtained from any norm on $X$?
I know this is because $0$ does not lie in $X$, but I am having problems. Formalizing a proof for this.
This is also my final question for some time, after this I will reread the answers, and not stop until I can finally understand these strange spaces.

REPLY [1 votes]: Suppose$(X,d)$ is a vector space with the discrete metric.
If$X$ is a normed space, then for ∀ $x≠y ∈ X$,$α≠0 ∈K, αx≠αy$.So then $\lVert\alpha x-\alpha y\rVert = 1$
However, by metric and norm properties,
$$\lVert\alpha x-\alpha y\rVert =\lVert\alpha(x-y)\rVert = \lvert\alpha\rvert\,\lVert x-y\rVert\le\vert \alpha \vert.1=\alpha,$$ a contradiction!!


TITLE: Is there a name for the quotient group of isomorphic permutations?
QUESTION [0 upvotes]: Suppose we have finite tuple $v$ of two (repeating) elements $x$ and $y$, the first occurring $m$ times, the second $n$ times.
Suppose we have a permutation $\pi \in S_{n+m}$. Then there are multiple permutations $\pi'$, such that $\pi(v) = \pi(v')$ (namely, those permutations that permute the $x$ and $y$ elements among themselves). For given $m$ and $n$ we can define an equivalence relation ~ on the permutations $\pi\in S_{n+m}$, such that $\pi ~\pi'$ if $\pi(v) = \pi'(v)$.
Is there a special name for the quotient group (or another group isomorphic to it)? I understand that this group will depend on $m$ and $n$.

REPLY [3 votes]: As indicated in the comments, you probably mean $S_{n+m}$. Now, let $A$ be the set of all tuples of length $n + m$ with $m$ copies of $x$ and $n$ copies of $y$. There is an action of $S_{n + m}$ on $A$, which is $S_{n+m} \times A \longrightarrow A$ via $(\pi, v) \mapsto \pi(v)$. Now, to this group action we can associate the stabilizer of $v$, denoted $G_v = \{\pi \in S_{n+m} : \pi(v) = v\}$. This is a subgroup of $S_{n + m}$.
Now, the equivalence relation you defined was $\pi \sim \pi'$ if $\pi(v) = \pi'(v)$. This is equivalent to $\pi^{-1} \pi'(v) = v$, which is precisely the definition of $\pi^{-1} \pi' \in G_v$. So $\pi \sim \pi'$ if and only if $\pi^{-1} \pi' \in G_v$. The latter condition is precisely the equivalence relation defining the cosets $S_{n+m}/G_v$.
This is therefore the quotient you are looking for, but it is not necessarily a group. Well any finite set can easily be equipped with some group structure, but you probably mean $[\pi] [\pi'] = [\pi \pi']$, where the brackets denote the equivalence class. In other words, to rightly call this a quotient group would mean that there is a group structure on $S_{n+m}/G_v$ such that the quotient map $S_{n + m} \longrightarrow S_{n+m}/G_v$ is a group homomorphism. This is only possible if $G_v \subseteq S_{n + m}$ is a normal subgroup. However, it is not hard to compute that $\pi G_v \pi^{-1} = G_{\pi(v)}$. Thus, $G_v$ is normal in $S_{n+m}$ if and only if $G_v = G_{\pi(v)}$ for all $\pi \in S_{n+m}$. This is true in some trivial cases like $n = 0$ or $m = 0$ but it need not be true in general. So in general, this quotient is not a quotient group.
There is still a sort of multiplication on this quotient: the action of $S_{n+m}$ on it. While the product $[\pi] [\pi'] = [\pi \pi']$ is not well defined, the action $\pi [\pi'] = [\pi \pi']$ is. In coset notation, this is $\pi(\pi' G_v) = (\pi \pi') G_v$. So we cannot multiply two elements of the quotient, but we can left multiply by elements of $S_{n+m}$.
I'd like to point out another useful fact in analyzing this quotient. So far we have shown that $S_{n+m}/{\sim} = S_{n+m}/G_v$ and observed its action from $S_{n+m}$. There is a useful isomorphism (of sets with an $S_{n+m}$ action) of the latter object with the orbit of $v$. The orbit is defined $S_{n+m} v = \{\pi(v) : \pi \in S_{n + m}\} \subseteq A$. The isomorphism is described by $S_{n+m} / G_v \longrightarrow S_{n+m} v$ taking the coset $\pi G_v \mapsto \pi(v)$. That this is a well defined isomorphism is called the orbit stabilizer theorem.


TITLE: Direct Limit of Both Rings and Their Modules
QUESTION [1 upvotes]: Suppose we have a directed set $\langle I,\leq\rangle$, with a direct system $\langle A_i,f_{ij}\rangle$ of rings and a direct system $\langle M_i,g_{ij}\rangle$ of abelian groups, such that each $M_i$ is an $A_i$ module via $h_i:A_i\times M_i\to M_i$. Then suppose that these actions are compatible with the direct systems, so $$g_{ij}(h_i(a,m))=h_j(f_{ij}(a),g_{ij}(m))$$
  Then do we have an action $$\mathop{\lim_{\longrightarrow}}A_i\times\mathop{\lim_{\longrightarrow}}M_i\to\mathop{\lim_{\longrightarrow}}M_i$$ which is determined by these systems?

Here is what I have tried so far:
Let $$\begin{align*}A&=\mathop{\lim_{\longrightarrow}}A_i\\
M&=\mathop{\lim_{\longrightarrow}}M_i\end{align*}$$
Then does the following define an action $h:A\times M\to M$?

Any $a\in A$, $m\in M$ must have some representative $a_i\in A_i$, $m_j\in M_j$. If $i\leq j$, set $a_j=f_{ij}(a_i)$, if $j\leq i$ then set $m_i=g_{ji}(m_j)$. In either case, we may now assume we have $a_k\in A_k$ and $m_k\in M_k$. Then define $h(a,m)$ to be $h_k(a_k,m_k)$ modulo the equivalence relation on the direct limit.

If so, is there a "nice" way of showing this? I'm struggling at the moment fiddling with the definitions directly, I was hoping for something possibly using the universal properties of colimits, but since we are working in two different categories I can't immediately see how to do this.
Any help would be much appreciated.

REPLY [2 votes]: You can work in a monoidal category if you add some assumptions on how $\otimes$ behaves with respect to direct limits. In the specific situation of rings and modules, you want to work in the monoidal category $\mathbf{Ab}$ with tensor product the usual tensor product $\otimes_{\mathbb Z}$.
Indeed in this situation a ring is just a monoid in this category, and modules over this ring are just modules over this monoid.
So your question becomes : let $(C,\otimes)$ be a monoidal category, $(A_i)$ a directed system of monoids, $(M_i)$ a directed system such that $M_i$ is an $A_i$-module in a compatible way, that is, the following diagrams commute whenever $i\leq j$ : $$\require{AMScd}\begin{CD}A_i\otimes M_i @>>> M_i\\
@VVV @VVV \\
A_j\otimes M_j @>>> M_j\end{CD}$$
Assume that both direct limits (which are actually colimits) exist, let's call them $A,M$. Under what conditions is there an $A$-module structure on $M$ that is compatible with our direct system ?
First of all, one has to see when $A$ carries a monoid structure. That is, we want a map $A\otimes A\to A$. It's not clear how to do that, except if $\otimes$ commutes with direct colimits. In this case, what we want amounts to a map $\varinjlim_{(i,j)\in I\times I} A_i\otimes A_j \to A$, and then note that $\Delta = \{(i,i)\in I^2\}$ is cofinal in $I^2$ so we just need a map $\varinjlim_i A_i\otimes A_i\to A$ and this is easy to find, and it's easy to check that we get a monoid in the end.
So with this extra assumption, $A$ has  a monoid structure. When can we then get an $A$-module structure on $M$ ? Well our assumption is enough : we want a map $A\otimes M\to M$, which again by the extra assumption boils down to a map $\varinjlim_i A_i\otimes M_i \to M$, which is again easy to find, and to check that this does provide a module structure on $M$; which is quite obviously compatible with the module structures on $M_i$.
Now how does this relate to your example ? Well as I said, a ring is nothing but a monoid in $(\mathbf{Ab},\otimes_\mathbb Z)$, and over such a ring a module is well just a module.
Moreover, since everything is nice and algebraic, one can check that the various colimits over a directed system computed in the various categories all agree (that is, if you compute $\varinjlim_i A_i$ in $\mathbf{Ring,Ab, Set}$, you get the same result up to a forgetful functor; and if you compute $\varinjlim_i M_i$ in $\mathbf{Ab,Set}$, the same thing happens), therefore there is no need to worry about that.
Finally, $\otimes_\mathbb Z$ in $\mathbf{Ab}$ does preserve directed colimits, in fact it preserves all colimits, as it is a left adjoint, so the extra assumption is satisfied.
You could probably also do this directly in $\mathbf{Set}$, because $\times$ does commute with directed colimits for the same reasons, but it would be longer and more tedious because a ring is not "just a monoid in $(\mathbf{Set},\times)$", you have more things to write down.
This is the (a) categorical approach, but of course in such a situation it can also be nice to have a more down-to-earth description - and here it's particularly easy to find :
if you have $x\in A, m\in M$, then $x$ is in the image of $A_i$ for some $i$, $m$ in the image of $M_j$ for some $j$; take $k\geq i,j$ then $x$ is in the image of $A_k$, $m$ in the image of $M_k$, you can then compute $xm$ in $M_k$ and take its image in $M$. You easily check that this doesn't depend on any of the choices made because of the compatibility condition (this is exactly what we're doing when defining the module structure abstractly via the colimit - the thing I haven't checked is that everything is compatible, but that's easy)


TITLE: Complex number under Sqrt
QUESTION [0 upvotes]: Is it possible that $\sqrt{a+ i \: b} $ could not be written as one of the below forms?
$ i \sqrt{-a-i\:b} $
or
$-i \sqrt{-a-i\:b}$
$a$ and $b$ are real constants.

REPLY [1 votes]: $\displaystyle \sqrt{a+ i \: b}=\sqrt{-i^2(a+ i \: b)}=\sqrt{i^2} \cdot \sqrt{-(a+ i \: b)}=i\sqrt{-a-ib}$
Because: $\sqrt{i^2}=\sqrt{(\sqrt{-1})^2}=\sqrt{-1}=i$
Update: Thanks to Watson's comment as we can also find here we can't always separate the square roots but here we can.
If $|Arg(i^2)+Arg(a+ib)|≥π$, then $\sqrt{i^2 (a+ib)}=\sqrt {i^2} \sqrt {a+ib}$ does not hold,
but we have: $|Arg(i^2 \in \Bbb R)+Arg(a+ib)|=|0+Arg(a+ib)|=|Arg(z)|\leqπ$, So it is ok since always an argumnet of a complex number lies in the specified range.


TITLE: How to find maximum value of trig function?
QUESTION [0 upvotes]: How to find maximum value of this:
$$y = 5\sin x - 12\cos x$$
And I am more intrested in solving process, rather than answer. I know the answer. I am familiar with derivatives, not so good, but as I found prime of this function:
$$y' = 5\cos x+12\sin x$$
but I dont know what to do next.
P.S Excuse me if grammar was poor, not a native english speaker.

REPLY [0 votes]: Hint: This is equivalent to solving a system of two equations in two unknowns, $5a+12b=0$, and $a^2+b^2=1$.


TITLE: Are planes in $3$-dimensions two-dimensional?
QUESTION [4 upvotes]: Are planes in $3$-dimensions two-dimensional?

The reason I ask is because mathematically the $xy$-plane exists in $3$D space but appears to be $2$D, but how can something $2$D be in $3$D space? I therefore conclude that a plane in $3$D space must have a thickness that is infinitesimal. Isn't this a contradiction, though, since planes have a thickness of that of a point and since a point has no thickness neither does the plane?

REPLY [5 votes]: The first thing to mention when talking about the term dimension is that in mathematics there is an exact definition of this term. So, when we speak about dimensions, the first thing to do is to define what dimension means. For the kind of geometry you use, the definition of the term dimension from a mathematical field called linear algebra is the appropriate one. In this sense, dimension is a property of so-called vector spaces.

Let's start by defining the term vector space (I will only consider vector spaces over $\mathbb{R}$ here to simplify things): The set
$V=\{(x_1,x_2,\ldots,x_n)\mid x_1,x_2,\ldots,x_n\in\mathbb{R}\}$
is a vector space of dimension $n$. Please note: Not every vector space is of this form!
Remark: From school you may know $\mathbb{R}^2$ (which is a vector space of dimension 2) and $\mathbb{R}^3$ (which is a vector space of dimension 3).
The interesting question is: Given a subset $U$ of $V$, what is its dimension? We can only define what the dimension of $U$ is if $U$ itself is a vector space. $U\subseteq V$ is a vector space, if and only if there exist $v_1,v_2,\ldots,v_m\in V$ such that
$U=\{a_1v_1+a_2v_2+\ldots+a_mv_m\mid a_1,a_2,\ldots,a_m\in\mathbb{R}\}$
Now two cases can occur: Either $U=V$ holds or $U\subset V$ holds. For $U=V$, the dimension of $U$ is obviously $n$. For $U\subset V$, we still have to define what dimension means.
For defining the term dimension, we first have to define what linearly independent means: A set $W=\{w_1,w_2,\ldots,w_m\},W\subseteq V$ is called linearly independent, if the equation
$a_1w_1+a_2w_2+\ldots+a_mw_m=\mathbf{0}\quad(a_i\in\mathbb{R})$
only holds for $a_1=a_2=\ldots=a_m=0$.
For example: The set $\{(1,0,0),(0,1,0),(3,2,0)\}$ is not linearly independent, because
$3\cdot (1,0,0)+2\cdot (0,1,0)-(3,2,0)=(0,0,0)$
On the other hand, the set $\{(1,0,0),(0,1,0)\}$ is linearly independent, because
$a_1\cdot (1,0,0)+a_2\cdot (0,1,0)=(0,0,0)$ implies $a_1=a_2=0$.
Now we can define what dimension means:
$U=\{a_1v_1+a_2v_2+\ldots+a_mv_m\mid a_1,a_2,\ldots,a_m\in\mathbb{R}\}$
has dimension $m$ if and only if $\{v_1,v_2,\ldots,v_m\}$ is a linearly independent set.
Example (continuation of the above example):
$\{a_1(1,0,0)+a_2(0,1,0)+a_3,(3,2,0)\mid a_1,a_2,a_3\in\mathbb{R}\}$
is the same vector space as
$\{a_1(1,0,0)+a_2(0,1,0)\mid a_1,a_2\in\mathbb{R}\}$;
its dimension is 2 (because $\{(1,0,0),(0,1,0)\}$ is linearly independent).
Remark: Given a vector space $U$, its dimension is unique.
Remark: If $n$ is the dimension of $V$, $m$ is the dimension of $U$ and $U\subseteq V$, then $m\leq n$ holds.
If $U\subseteq V$ is a vector space and $v_0\in V$,
$U^\prime=v_0+U=\{v_0+a_1v_1+a_2v_2+\ldots+a_mv_m\mid a_1,a_2,\ldots,a_m\in\mathbb{R}\}$
is called an affine vector space. The dimension of $U'$ is defined as the dimension of $U$.

In the above points, we have defined the term dimension for vector spaces. What remains to be done is to show how vector spaces are related to basic geometry:

Any point can be represented by an affine vector space $\{v_0\}\subseteq V$. Therefore, points have dimension 0.
Any line can be represented by an affine vector space $\{v_0+a_1v_1\mid a_1\in\mathbb{R}\}$, where $v_0\in V$ and $v_1\in V$ are given. Therefore, lines have dimension 1. (The set contains the points that lie on the line.)
Any plane can be represented by an affine vector space $\{v_0+a_1v_1+a_2v_2\mid a_1,a_2\in\mathbb{R}\}$, where $v_0\in V$, $v_1\in V$ and $v_2\in V$ are given and $v_1,v_2$ are linearly independent. (The set contains the points that lie in the plane.)

Please note: There is no limit on how big a dimension can be; for example, you can also define vector spaces of dimension 1000.