TITLE: How to read this gas rate
QUESTION [0 upvotes]: Please see attached image.

So does this mean its 32.7 cents per therm? Or 32.07 cents per therm? I'm very confused. Also why are natural gas rates marketed in this fractional way and not in whole numbers. Thanks.

REPLY [1 votes]: We have
$$\frac {7}{10}=0.7$$
so the gas rate is $32.7$ cents per therm. One reason why the image would be written that way is because writing $32^{7/10}$ requires less space and ink than writing $32.7$.


text:
\section{BN-Pairs and the Weyl functor}

Before introducing the general concepts of building and BN-pair, we study the standard example of projective spaces (from the building point of view).\\

\subsection{Projective space}

Let $\mathbf{R}$ be a division ring (= skew field), let $n \in \mathbb{N}$, and let $V = V(n,\mathbf{R}) = \mathbf{R}^n$\index{$V(n,\mathbf{R})$} be the $n$-dimensional (left or right) vector space over $\mathbf{R}$. We define the {\em $(n - 1)$-dimensional (left or right) projective space}\index{projective!space} $\PG(n,\mathbf{R})$\index{$\PG(n,\mathbf{R})$} as being the set
\begin{equation}
(\mathbf{R}^n \setminus \{0\})/\sim,
\end{equation}
where the equivalence relation ``$\sim$'' is defined by (left or right) proportionality, with the subspace structure being induced by that of $V$. (When $n$ is not finite, similar definitions hold.) The choice of ``left'' or ``right'' does not affect the isomorphism class.
If $\mathbf{R} = \F_q$ is the finite field with $q$ elements ($q$ a prime power), we also write $\PG(n - 1,q)$\index{$\PG(n - 1,q)$} instead of
$\PG(n - 1,\F_q)$. Sometimes the notations $\hP^{n - 1}(\mathbf{R})$\index{$\hP^{n - 1}(\mathbf{R})$}, $\hP^{n - 1}(q)$\index{$\hP^{n - 1}(q)$}, $\mathbb{P}^{n - 1}(\mathbf{R})$\index{$\mathbb{P}^{n - 1}(\mathbf{R})$} and $\mathbb{P}^{n - 1}(q)$\index{$\mathbb{P}^{n - 1}(q)$} occur as well.

There is also a notion of {\em axiomatic projective space}\index{axiomatic!projective space}, which is defined to be an {\em incidence geometry} (defined later in this section) which is governed by certain axioms, which are (of course) satisfied by ``classical'' projective spaces over division rings. A truly remarkable thing is that Veblen and Young \cite{VY} showed that if the dimension $n - 1$ of such a space is at least three, it {\em is} isomorphic to some $\PG(n - 1,\mathbf{R})$. And this is well-known {\em not} to be true when the dimension is less than three. \\

\subsection{Representing spaces as group coset geometries}

Let $\hP$ be a finite\\
($n$-)dimensional projective space over some division ring $\mathbf{R}$.
Consider any $\mathbf{R}$-base $\hB$. Define a simplicial complex (in the next section to be formally defined, and called ``apartment''\index{apartment}) $\mC \equiv \mC(\hB)$, by letting it be the set of all possible subspaces of $\hP$ generated by
subsets of $\hB$. (Let it also contain the empty set.) Define a (maximal) ``flag''\index{flag} or {\em chamber}\index{chamber} in $\mC$ as a maximal chain (so of length $n + 1$) of subspaces
in $\mC$. Let $F$ be such a fixed flag.

Consider the special projective linear group
$K := \PSL_{n + 1}(\mathbf{R})$ of $\mathbf{P}$.
Then note that $K$ acts transitively on the pairs $(\mC(\hB'),F')$,
where $\hB'$ is any $\mathbf{R}$-base and $F'$ is a maximal flag in $\mC(\hB')$.

Put $B = K_{\mC}$ and $N = K_{F}$; then note the following properties:

\begin{itemize}
\item[(i)]
$\langle B,N \rangle = K$;
\item[(ii)]
put $H = B \cap N
\lhd N$ and $N/H = W$;
then $W$ obviously is isomorphic to the symmetric group $\mathbf{S}_{n + 1}$ on $n + 1$ elements. Note that a presentation of $\mathbf{S}_{n + 1}$ is:
\begin{equation}
\langle   s_i \vert s_i^2 = \1, (s_is_{i + 1})^3 = \1, (s_is_j)^2 = \1,     i, j \in \{1,\ldots,n + 1\}, j \ne i \pm 1          \rangle.
\end{equation}
\item[(iii)]
$Bs_iBwB \subseteq BwB \cup
Bs_iwB$ whenever $w \in W$ and $i\in\{1,2,\ldots,n + 1\}$;
\item[(iv)]
$s_iBs_i \ne B$ for all $i\in\{1,2,\ldots,n + 1\}$.
\end{itemize}

(Here, expressions such as $BwB$ mean $B\widetilde{w}HB$, where $\widetilde{w}$ is any representant of $\widetilde{w}H = w$.)

Now let $K \cong \PSL_{n + 1}(\mathbf{R})$ be as above, and suppose that $B$ and $N$ are groups satisfying these properties. Define a geometry $\mB_{(K,B,N)}$\index{$\mB_{(K,B,N)}$} as follows.

\begin{itemize}
\item[(B1)]
Its elements are left cosets in $K$ of the groups $P_i$ which properly contain $B$ and are different from $K$, $i = 1,\ldots,n + 1$;
\item[(B2)]
two elements $gP_i$ and $hP_j$ are incident if they intersect nontrivially.\\
\end{itemize}

\begin{proposition}
$\mB_{(K,B,N)}$ is isomorphic to $\PG(n,\mathbf{R})$.
\end{proposition}

\medskip
\subsubsection{Low dimensional cases}

For dimension $n = 1$, our definition of axiomatic space doesn't make much sense.
Here we rather {\em start} from a division ring $\mathbf{R}$, and define $\hP$, the {\em projective line} over $\mathbf{R}$, as being the set $(\mathbf{R}^2 \setminus \{0\})/\sim$,
where $\sim$ is defined by (left) proportionality. So we can write

\begin{equation}
\hP = \{(0,1)\} \cup \{(1,\ell) \vert \ell \in \mathbf{R}\}.
\end{equation}
Now $\PSL_2(\mathbf{R})$ acts naturally on $\mathbf{P}$; in fact, we have defined the projective line as a permutation group equipped with the natural doubly transitive action
of $\PSL_2(\mathbf{R})$.
Defining a geometry as we did for higher rank projective spaces,
through the ``$(B,N)$-pair structure''
of $\PSL_2(\mathbf{R})$, one obtains the same notion of projective line.\\

Restricting to finite fields, we obtain the following very simple

\begin{proposition}
A finite projective line has $q + 1$ points, for some prime power $q$.
\end{proposition}

The $2$-dimensional case is different, still. Here, other than in the $1$-dimensional case, one obtains a nontrivial geometry; the axioms now boil down to just demanding that each two different points are incident with precisely one line, that, dually, any two distinct lines intersect in precisely one point, and there exist four distinct points of which no three are on the same line. So we need not require additional algebraic structure in order to have interesting objects. Here we cannot say much about the order of the plane a priori.

\subsubsection{Representation by diagram}

We represent the presentation of $\mathbf{S}_{n + 1}$ as above
in the following way (this will be explained in more detail in the next section):

\bigskip

$\mathbf{A}_{n + 1}$: \begin{tikzpicture}[style=thick, scale=1.5]
\foreach \x in {1,2,3,5,6}{
\fill (\x,0) circle (2pt);}

\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (3.5,0);
\draw (4.5,0) -- (5,0);
\draw (5,0) -- (6,0);
\draw (4,0) node {$\dots$} ;

\end{tikzpicture} \hspace{0.5cm} ($n\geq 0$)\\

(The number of vertices is $n + 1$ | each vertex corresponding to an
involution in the generating set of involutions | and
we have an edge between vertices $s_i$ and $s_j$
if and only if $\vert j-i\vert=1$.)\\

\subsection{Simplicial complexes}

 Recall that a (combinatorial) {\em simplicial complex}\index{simplicial complex} is a pair $(\mS,Y)$, where $Y$ is a set and $\mS \subseteq 2^Y$, such that $Y \in \mS$ and
\begin{equation}
U \subseteq V \in \mS \Longrightarrow U \in \mS.
\end{equation}
We are ready to introduce buildings. We will not provide each result with a specific reference | rather, we refer the reader to \cite{AbBr}.\\

\subsection{Combinatorial definition}

A {\em chamber geometry}\index{chamber!geometry} is a geometry
\begin{equation}
\Gamma = (\mC_1,\mC_2,\ldots,\mC_j,\I)
\end{equation}
 of rank $j$  (so $\Gamma$ has $j$ different
kinds of elements and $\I$ is an incidence relation between the elements such that no two elements belonging to the
same $\mC_i$, $1 \leq i \leq j$, can be incident) so that the simplicial complex $(\mC,X)$, where $\mC = \cup_{i = 1}^{j}\mC_i$
and $S \subseteq \mC$ is contained in $X$ if and only if every two distinct elements of $S$ are incident, is a chamber
complex (as in, e.g., \cite{POL}).
A {\em building}\index{building} $(\mC,X)$ is a thick chamber geometry $(\mC_1,\mC_2,\ldots,\mC_j,\I)$
of rank $j$, where $\mC = \cup_{i = 1}^j\mC_i$,
together with a set $\mA$ of thin chamber subgeometries, so that:

\begin{itemize}
\item[(i)]
every two chambers are contained in some element of $\mA$;
\item[(ii)]
for every two elements $\Sigma$ and $\Sigma'$ of $\mA$ and every two simplices $F$ and $F'$, respectively contained in $\Sigma$ and
$\Sigma'$, there exists an isomorphism $\Sigma \mapsto \Sigma'$ which fixes all elements of both $F$ and $F'$.
\end{itemize}

If all elements of $\mA$ are finite, then the building is called {\em spherical}\index{spherical!building}. Elements of $\mA$ are called {\em apartments}\index{apartment}. \\

\subsection{Coxeter groups and systems}

We need to introduce the notions of ``Coxeter system'' and ``Coxeter diagram''.\\

\subsubsection{Coxeter groups}
A {\em Coxeter group}\index{Coxeter!group} is a group with a presentation of type

\begin{equation}
\langle s_1,s_2,\ldots,s_n \vert (s_is_j)^{m_{ij}} = \1 \rangle,
\end{equation}

\noindent
where $m_{ii} = 1$ for all $i$, $m_{ij} \geq 2$ for $i \ne j$, and $i, j$ are natural numbers bounded above by the natural number $n$. If $m_{ij} = \infty$, no relation of the form
$(s_is_j)^{m_{ij}}$ is imposed. All generators in this presentation are involutions. The natural number $n$ is the {\em rank}\index{rank} of the Coxeter group.

A {\em Coxeter system}\index{Coxeter!system} is a pair $(W,S)$, where $W$ is a Coxeter group and $S$ the set of generators defined by the presentation.

Recall that a {\em dihedral group}\index{dihedral group} of {\em rank}\index{rank} $n$, denoted $\mathbf{D}_n$\index{$\mathbf{D}_n$}, is the symmetry group of a regular $n$-gon in the real plane. \\

\subsubsection{Coxeter matrices}

A square $n \times n$-matrix $(M)_{ij}$ is a {\em Coxeter matrix}\index{Coxeter!matrix} if it is symmetric and defined over $\mathbb{Z} \cup \{\infty\}$, has only $1$s on the diagonal, and has $m_{ij} \geq 2$ if $i \ne j$.
Starting from a Coxeter matrix $(M)_{ij}$, one can define a Coxeter group $\langle s_1,s_2,\ldots,s_n \vert (s_is_j)^{m_{ij}} = \1 \rangle$, and conversely.
Different Coxeter systems can give rise to the same Coxeter group, even if the rank is different. \\

\subsubsection{Coxeter diagrams}

Let $(W,S)$ be a Coxeter system. Define a graph, called ``Coxeter diagram''\index{Coxeter!diagram}, as follows. Its vertices are the elements of $S$. If $m_{ij} = 3$, we draw a single edge between $s_i$ and $s_j$; if $m_{ij} = 4$, a double edge, and if $m_{ij} \geq 5$, we draw a single edge with label $m_{ij}$. If $m_{ij} = 2$, nothing is drawn.
If the Coxeter diagram is connected, we call $(W,S)$ {\em irreducible}\index{irreducible!Coxeter diagram}. If it has a finite number of vertices, we call $(W,S)$ {\em spherical}\index{spherical!Coxeter diagram}.\\

The irreducible, spherical Coxeter diagrams were classified by H. S. M. Coxeter \cite{HSMC}; the complete list is the following.

\bigskip

$\mathbf{A}_n$\index{Coxeter!diagram!of type $\mathbf{A}_n$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {1,2,3,5,6}{
\fill (\x,0) circle (2pt);}

\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (3.5,0);
\draw (4.5,0) -- (5,0);
\draw (5,0) -- (6,0);
\draw (4,0) node {$\dots$} ;

\end{tikzpicture} \hspace{0.5cm} ($n\geq 1$)\\

$\mathbf{B}_n = \mathbf{C}_n$\index{Coxeter!diagram!of type $\mathbf{C}_n$}\index{Coxeter!diagram!of type $\mathbf{B}_n$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {1,2,3,5,6}{
\fill (\x,0) circle (2pt);}

\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (3.5,0);
\draw (4.5,0) -- (5,0);
\draw (5,0.035) -- (6,0.035);
\draw (5,-0.035) -- (6,-0.035);
\draw (4,0) node {$\dots$} ;

\end{tikzpicture}\hspace{0.5cm} ($n\geq 2$)\\

$\mathbf{D}_n$\index{Coxeter!diagram!of type $\mathbf{D}_n$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {1,2,3,5,6}{
\fill (\x,0) circle (2pt);}

\fill (5,1) circle (2pt);

\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (3.5,0);
\draw (4.5,0) -- (5,0);
\draw (5,0) -- (6,0);
\draw (5,0) -- (5,1);
\draw (4,0) node {$\dots$} ;

\end{tikzpicture}\hspace{0.5cm}  ($n\geq 4$)\\

$\mathbf{E}_n$\index{Coxeter!diagram!of type $\mathbf{E}_n$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {0,2,3,4}{
\fill (\x,0) circle (2pt);}

\fill (2,1) circle (2pt);

\draw (0,0) -- (.5,0);
\draw (1.5,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (4,0);
\draw (2,0) -- (2,1);
\draw (1,0) node {$\dots$} ;

\end{tikzpicture}
\hspace{0.5cm}  ($n=6,7,8$)\\

$\mathbf{F}_4$\index{Coxeter!diagram!of type $\mathbf{F}_4$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {0,1,2,3}{
\fill (\x,0) circle (2pt);}

\draw (0,0) -- (1,0);
\draw (2,0) -- (3,0);
\draw (1,0.035) -- (2,0.035);
\draw (1,-0.035) -- (2,-0.035);

\end{tikzpicture}\\

$\mathbf{H}_3$\index{Coxeter!diagram!of type $\mathbf{H}_3$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {0,1,2}{
\fill (\x,0) circle (2pt);}

\draw (0,0) -- (1,0);
\draw (1,0) -- (2,0);
\draw (1.5,.25) node {$5$} ;

\end{tikzpicture}\\

$\mathbf{H}_4$\index{Coxeter!diagram!of type $\mathbf{H}_4$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {-1,0,1,2}{
\fill (\x,0) circle (2pt);}
\draw (-1,0) -- (0,0);
\draw (0,0) -- (1,0);
\draw (1,0) -- (2,0);
\draw (1.5,.25) node {$5$} ;

\end{tikzpicture}\\

$\mathbf{I}_2(m)$\index{Coxeter!diagram!of type $\mathbf{I}_2(m)$}: \begin{tikzpicture}[style=thick, scale=1.3]
\foreach \x in {1,2}{
\fill (\x,0) circle (2pt);}

\draw (1,0) -- (2,0);
\draw (1.5,.25) node {$m$} ;

\end{tikzpicture}\hspace{0.5cm}  ($m\geq 5$)\\

\medskip
\subsection{Incidence geometries}

 Having met some standard examples of incidence geometries, we now
 introduce general incidence geometries\index{incidence!geometry} in a formal way. These objects will be our way to approach
 combinatorial geometries in the present chapter.

 An {\em incidence geometry}\index{incidence!geometry} or  {\em Buekenhout-Tits geometry}\index{Buekenhout-Tits!geometry} consists of a set $X$ of {\em objects}\index{objects}
provided with a symmetric relation $\I$ called {\em incidence}\index{incidence}
and a surjective function

\begin{equation}
t:X \longrightarrow I
\end{equation}
that assigns a {\em type}
to each object, such that two objects of the same type are never incident.
The set $I$ is the set of {\em types}\index{type}. The cardinality $|I|$
is called the {\em rank}\index{rank!of incidence geometry} of the geometry.

Denote the geometry by $\Gamma = \Gamma(X,\I,I,t)$.

If the rank is two, we also speak of a {\em point-line geometry}\index{point-line geometry} (the assignment
\begin{equation}
\I\ \ \longrightarrow\ \  \{\mbox{point},\mbox{line} \}
\end{equation}
being bijective but arbitrary).

\subsubsection{Geometries as incidence graphs}

It will be particularly important in the $\mathbb{F}_1$-context
 to see incidence geometries as a kind of graph. An extra feature which comes in handy is that essentially (or better: usually) the automorphism group $A$ of the
 geometry is the same as the automorphism group $B$ of the associated graph. In any case, $A \leq B$, and if
 $[B : A] \ne 1$, then this quantity is a measure for the number of types of objects that play the same role.\\

An incidence geometry can be viewed as a multipartite graph $\Gamma$
with vertex set $X$ and partition $\{X_i \mid i \in I\}$
(with $X_i = t^{-1}(i)$), with incidence taken as adjacency.\\

The geometry $\Gamma$ is called {\em connected}\index{connected incidence geometry} when the graph $\Gamma$
is connected. The graph without vertices is not connected:
a connected graph has precisely one connected component, while the graph
without vertices has no connected component.

A {\em flag}\index{flag} $F$ in $\Gamma$ is a clique, which is by definition a complete subgraph. No two elements of a flag
have the same type. The {\em rank}\index{rank!of flag} of $F$ is $|t(F)|$ (that is, $|F|$).
The {\em corank}\index{corank of a flag} of $F$ is $| I \setminus t(F)|$.
The {\em residue}\index{residue} ${\rm Res}(F)$ (also written $\Gamma_F$) is the geometry
with set of objects $Y = \{y \in X \setminus F \mid F \cup \{y\}
~\mbox{\rm is a flag}\}$, incidence inherited from $\Gamma$,
set of types $I \setminus t(F)$, and type function inherited from $\Gamma$.
The geometry $\Gamma$ is called {\em residually connected}\index{residually connected} when every residue
of rank at least two is connected (and hence nonempty), and every residue
of rank one is nonempty.\\

\medskip
\subsection{BN-Pairs and buildings}

A group $G$ is said to have
a {\em BN-pair}\index{BN-pair} $(B,N)$, where $B, N$ are subgroups of $G$, if:

\begin{itemize}
\item[(BN1)]
$\langle B,N \rangle = G$;
\item[(BN2)]
$H = B \cap N
\lhd N$ and $N/H = W$ is a Coxeter group with distinct generator set of involutions
$S = \{ s_j \vert j \in J \}$;
\item[(BN3)]
$BsBwB \subseteq BwB \cup
BswB$ whenever $w \in W$ and $s\in S$;
\item[(BN4)]
$sBs \ne B$ for all $s\in S$.
\end{itemize}

The group $B$, respectively $W$, is a {\em Borel subgroup}\index{Borel subgroup},
respectively the {\em Weyl group}\index{Weyl!group}, of $G$. The  quantity $\vert S\vert$
is called the \emph{rank}\index{rank!of BN-pair} of the BN-pair. If $W$ is a finite group, the BN-pair is {\em spherical}\index{spherical!BN-pair}. It is {\em irreducible}\index{irreducible!BN-pair} if the corresponding Coxeter system is.
Sometimes we call $(G,B,N)$ also a {\em Tits system}\index{Tits system}.\\

\begin{remark}
{\rm
Asking that $W$ is a Coxeter group is in fact redundant; by the other axioms and the fact that $S$ consists of involutions, it is not hard to show that $W$ {\em must be} a Coxeter group, and that $S$ is uniquely determined as the set of elements in $W^\times$ for which
\begin{equation}
B \cup BsB
\end{equation}
is a group.
}
\end{remark}

\subsubsection{Buildings as group coset geometries}

To each Tits system $(G,B,N)$ one can associate a building $\mB_{(G,B,N)}$ in a natural way, through a group coset construction. For that reason we introduce the standard {\em parabolic subgroups}\index{parabolic subgroup}; these are just the proper subgroups of $G$ which properly contain $B$. Let $I \subset J$, and define
\begin{equation}
W_I := \langle s_i \vert i \in I \rangle \leq W.
\end{equation}

Then
\begin{equation}
P_I := BW_IB
\end{equation}
is a subgroup of $G$ which obviously contains $B$, and vice versa it can be shown that any standard parabolic subgroup has this form.

We are ready to introduce $\mB_{(G,B,N)}$.

\begin{itemize}
\item[(B1)]
\textsc{Elements}: (or ``subspaces'' or ``varieties'') are elements of the left coset spaces $G/P_I$, $\emptyset \ne I \subset J \ne I$.
\item[(B2)]
\textsc{Incidence}: $gP_I$ is incident with $hP_L$, $I \ne L$, if these cosets intersect nontrivially.
\end{itemize}

The {\em rank}\index{rank!of a building} of $\mB_{(G,B,N)}$ is the rank of the BN-pair.
The building $\mB_{G,B,N}$ is {\em spherical}\index{spherical!building} when the BN-pair $(B,N)$ is; note that this is in accordance with the aforementioned synthetic definition of ``spherical building'' (taken that
there is already a BN-pair around). It is {\em irreducible}\index{irreducible!building} when $(B,N)$ is irreducible.\\

\subsubsection{$G$ as an automorphism group}

The group $G$ acts as an automorphism group, by multiplication on the left, on $\mB_{(G,B,N)}$.  The kernel  $K$ of this action is the biggest normal subgroup of $G$ contained in $B$, and is equal to
\begin{equation}
K=\bigcap_{g\in G}B^g.
\end{equation}

For the sake of convenience, suppose $J$ is the finite set $\{ 1,2,\ldots,n\}$, $n \in \mathbb{N} \setminus \{0\}$.
The group $G/K$ acts faithfully on $\mB_{(G,B,N)}$ and the stabilizer of the flag
\begin{equation}
F = \{P_{\{1\}},P_{\{1,2\}},\ldots,P_J\}
\end{equation}
is $B/K$. If $K = \{\1\}$, we say that the Tits system is {\em effective}\index{effective Tits system}.\\

Let $\Sigma$ be an apartment of $\mB_{(G,B,N)}$, and let its elementwise stabilizer be $E$; then $NE$ is the global
stabilizer of $\Sigma$.
We can write

\begin{equation}
E = \bigcap_{w\in W}B^w.
\end{equation}

The next theorem sums up several properties.

\begin{theorem}[\cite{AbBr,Titslect}] \label{Titssystem} Let $(G,B,N)$ be a Tits system with Weyl group $W$. Then the geometry $\mB_{(G,B,N)}$
is a Tits building. Setting
\begin{equation}
K=\bigcap_{g\in G}B^g\mbox{ and } E=\bigcap_{w\in W}B^w, \end{equation}
we have that $G/K$
acts naturally and faithfully by left translation on
$\mB_{(G,B,N)}$. Also, $B$ is the stabilizer of a unique flag $F$
and $NE$ is the stabilizer of a unique apartment containing $F$,
and the triple $(G/K,B/K,NE/K)$ is a Tits system
associated with $\mB_{(G,B,N)}$.
Moreover, $G/K$ acts transitively on the sets $(A,F')$, where $A$ is an apartment and $F'$ is a maximal flag (chamber) in $A$.
\end{theorem}

The Tits system $(G,B,N)$ is called
\emph{saturated}\index{saturated} precisely when
$N=NE$, with $E$ as above. Replacing $N$ by $NE$, every Tits
system is ``equivalent'' to a saturated one.\\

\subsubsection{Bruhat decomposition}

Let $G$ be a group with a spherical, saturated, effective BN-pair $(B,N)$. Then the ``Bruhat decomposition''\index{Bruhat decomposition} tells us that

\begin{equation}
G = BWB = \coprod_{w \in W}BwB,
\end{equation}
where $W = N/(B \cap N)$ is the Weyl group. Note that with $I \subset J$, we also have

\begin{equation}
P_I = BW_IB = \coprod_{w \in W_I}BwB.
\end{equation}

\subsubsection{Classification of BN-pairs}

 If the rank of an abstract spherical building is at least $3$, Tits showed in a celebrated work
 \cite{Titslect} that it is always associated to a BN-pair in the way explained above, and this deep observation led him eventually to classify all spherical BN-pairs of rank $\geq 3$ (cf. \cite[11.7]{Titslect}).

So Tits realized a far reaching generalization of the Veblen-Young theorem for spherical buildings, which roughly could be formulated as follows.

\begin{theorem}[Classification of spherical buildings | Tits \cite{Titslect}]
An
irreducible spherical building of rank at least $3$ arises from a simple algebraic group (of relative rank at least $3$) over an arbitrary division ring.\\
\end{theorem}

\bigskip
\begin{tabular}{c | c}
 & \\
\mbox{\textsc{Projective}}\ \ \mbox{\textsc{spaces}} &\mbox{\textsc{Buildings}}\\
 & \\
\hline \\
\mbox{\textsc{Veblen-Young:}}   & \mbox{\textsc{Tits:}} \\
 & \\
\mbox{dim} $\geq 3$:  & \mbox{rank} $\geq 3$: \\
\mbox{vector}\ \ \mbox{spaces} &\mbox{BN-pairs;}\ \ \mbox{simple}\ \ \mbox{algebraic} \ \ \mbox{groups} \\
\mbox{over} \ \ \mbox{division} \ \ \mbox{rings}   & \mbox{over} \ \ \mbox{division} \ \ \mbox{rings}\\
 & \\
 \mbox{dim} $2$:  & \mbox{rank} $2$: \\
 \mbox{axiomatic}\ \ \mbox{projective}\ \ spaces &\mbox{generalized}\ \ \mbox{polygons} \\
  & \\
   & \\
\end{tabular}

\medskip
\subsection{The rank $2$ case}

Combinatorially, a  {\em generalized $n$-gon}\index{generalized!$n$-gon} ($n \geq 3$) is a point-line geometry $\Gamma = (\mP,\mB,\I)$
for which  the following axioms are satisfied:

\begin{itemize}
\item[(i)]
$\Gamma$ contains no ordinary $k$-gon (as a subgeometry), for $2 \leq k < n$;
\item[(ii)]
any two elements $x,y \in \mP \cup \mB$ are contained in some ordinary $n$-gon (as a subgeometry) in
$\Gamma$;
\item[(iii)]
there exists an ordinary $(n + 1)$-gon (as a subgeometry) in $\Gamma$.
\end{itemize}

A {\em generalized polygon}\index{generalized!polygon} (GP)\index{GP} is a generalized $n$-gon for some $n$.

By (iii), generalized polygons have at least three points per line and three lines per point.  The generalized $3$-gons are precisely
the projective planes. A geometry $\Gamma$ which satisfies (i) and (ii) is a {\em weak generalized $n$-gon}\index{weak generalized $n$-gon}.
If (iii) is not satisfied for $\Gamma$, then $\Gamma$ is called {\em thin}\index{thin}.
Otherwise, it is called {\em thick}\index{thick}. Sometimes we will speak of ``thick (respectively thin)
generalized $n$-gon'' instead of ``thick (respectively thin) weak generalized $n$-gon''.
Generalized polygons were introduced by Tits in his triality paper \cite{Tits};  the basic reference is \cite{POL}.\\

It is not hard to show that once (iii) is also satisfied for a weak generalized polygon, there are constants $s$ and $t$ such that
each point is incident with $t + 1$ lines, and each line is incident with $s + 1$ points. If the polygons were to be classical (that is, {\em Moufang}\index{Moufang} \cite{POL}), then there are division rings $\K$ and $\mathbb{L}$ such that $s + 1 = \vert \K \vert + 1$ and $t + 1 = \vert \mathbb{L} \vert + 1$.
If (iii) is not satisfied, it can be shown that each line is incident with precisely $1 + 1$ points, so that thin generalized polygons are the polygons over $\Fun$.
And thin generalized $n$-gons are nothing else than ordinary $n$-gons. We will come back to this in more detail.

Note that there are many equivalent definitions for the notion of generalized polygon\footnote{Think, e.g., of the definition of generalized $3$-gon and of axiomatic projective plane given in the introduction to this chapter.}, but the present one is very natural in the characteristic one context | we only use
$\Fun$-polygons to describe the axioms.\\

The relation between buildings and generalized polygons,
as observed by Tits in \cite{Titslect} (see also \cite{POL}, \S 1.3.7, of which the notation is used), is as follows:

\begin{itemize}
\item[(S)]
{\em Suppose $(\mC,X)$, $\mC = \mC_1 \cup \mC_2$, is a spherical building of rank $2$. Then $\Gamma = (\mC_1,\mC_2,\I)$ is a
generalized polygon. Conversely, suppose that $\Gamma = (\mP,\mB,\I)$ is a generalized polygon, and let $\mF$ be the set of its flags.
Then $(\mP \cup \mB, \emptyset \cup \{ \{v\} \vert v \in \mP \cup \mB \} \cup \mF)$ is a chamber geometry of rank $2$.
Declaring the thin chamber geometry corresponding to any ordinary subpolygon an apartment, we obtain a spherical
building of rank $2$.}
\end{itemize}

\medskip
\subsubsection{Duality}

Interchanging the role of points and lines, that is, applying the map
\begin{equation}
\Gamma = (\mP,\mB,\I)\ \  \overset{D}\longrightarrow\ \  \Gamma^D = (\mB,\mP,\I),
\end{equation}
we obtain the (point-line) {\em dual}\index{dual} of $\Gamma$. It is also a GP (with the parameters switched).

\medskip
\subsubsection{Polygons as graphs}

Let $\mS = (\mP,\mB,\I)$ be a generalized $n$-gon. The {\em (point-line) incidence graph}\index{incidence!graph} $(V,E)$ of $\mS$ is defined by taking $V = \mP \cup \mB$, where an edge is drawn between vertices if the corresponding elements in $\mS$ are incident; $(V,E)$ then is a bipartite graph of diameter $n$ and girth $2n$. Vice versa, such graphs
define GPs.

Let the graph corresponding to $\mS$ be denoted by $\Gamma$.
We
call $(x_0,\ldots,x_k)$ a {\em (simple) path}\index{path} if the $x_i$ are pairwise distinct and $x_i$ is adjacent to $x_{i+1}$
for $i = 0,\ldots,k - 1$. The natural graph theoretic distance function on $\Gamma$ is denoted
by ``$\mathrm{d}$'' or sometimes ``$\mathrm{d}_n$''\index{$\mathrm{d}$}\index{$\mathrm{d}_n$}. The set of elements at distance $i$ from some element $x \in \Gamma$ is
denoted by $\Gamma_i(x)$\index{$\Gamma_i(x)$}. Elements at distance $n$ are called {\em opposite}\index{opposite}.

\subsection{$\mathbb{F}_1$-Buildings and the Weyl functor}

We now define the Weyl functor, and
describe some of the examples Tits mentioned in \cite{anal}.\\

\subsubsection{The Weyl functor\index{Weyl!functor}}

Note again that since $\mathbb{F}_1$ expresses the idea of an Absolute Arithmetic, it is clear that the buildings of a certain prescribed type $\mathbf{T}$ over $\mathbb{F}_1$ should be present in any thick building of the same type $\mathbf{T}$.\\

Motivated by the properties which a building over $\mathbb{F}_1$ of type $\mathbf{T}$ should have, we are ready to define such geometries (and their groups) in general.
Let  $\mB = (\mC_1,\mC_2,\ldots,\mC_j,\I)$ be a thick building
of rank $j$ and type $\mathbf{T}$ (given by one of the Coxeter diagrams below), and let $\mA$ be its set of thin chamber subgeometries.
Suppose $(B,N)$ is a saturated effective BN-pair associated to $\mB$; its Weyl group $W$ is a Coxeter group defined by one of the Coxeter graphs below.

\begin{proposition}
A building of rank $j$ and type $\mathbf{T}$ defined over $\mathbb{F}_1$ is isomorphic
to any element of $\mA$. Its automorphism group is isomorphic to the  Coxeter group $W$.
\end{proposition}

\bigskip

$\mathbf{A}_n$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {1,2,3,5,6}{
\fill (\x,0) circle (2pt);}

\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (3.5,0);
\draw (4.5,0) -- (5,0);
\draw (5,0) -- (6,0);
\draw (4,0) node {$\dots$} ;

\end{tikzpicture} \hspace{0.5cm} ($n\geq 1$)\\

$\mathbf{C}_n$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {1,2,3,5,6}{
\fill (\x,0) circle (2pt);}

\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (3.5,0);
\draw (4.5,0) -- (5,0);
\draw (5,0.035) -- (6,0.035);
\draw (5,-0.035) -- (6,-0.035);
\draw (4,0) node {$\dots$} ;

\end{tikzpicture}\hspace{0.5cm} ($n\geq 2$)\\

$\mathbf{D}_n$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {1,2,3,5,6}{
\fill (\x,0) circle (2pt);}

\fill (5,1) circle (2pt);

\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (3.5,0);
\draw (4.5,0) -- (5,0);
\draw (5,0) -- (6,0);
\draw (5,0) -- (5,1);
\draw (4,0) node {$\dots$} ;

\end{tikzpicture}\hspace{0.5cm}  ($n\geq 4$)\\

$\mathbf{E}_n$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {0,2,3,4}{
\fill (\x,0) circle (2pt);}

\fill (2,1) circle (2pt);

\draw (0,0) -- (.5,0);
\draw (1.5,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (4,0);
\draw (2,0) -- (2,1);
\draw (1,0) node {$\dots$} ;

\end{tikzpicture}
\hspace{0.5cm}  ($n=6,7,8$)\\

$\mathbf{F}_4$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {0,1,2,3}{
\fill (\x,0) circle (2pt);}

\draw (0,0) -- (1,0);
\draw (2,0) -- (3,0);
\draw (1,0.035) -- (2,0.035);
\draw (1,-0.035) -- (2,-0.035);

\end{tikzpicture}\\

$\mathbf{H}_3$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {0,1,2}{
\fill (\x,0) circle (2pt);}

\draw (0,0) -- (1,0);
\draw (1,0) -- (2,0);
\draw (1.5,.25) node {$5$} ;

\end{tikzpicture}\\

$\mathbf{H}_4$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {-1,0,1,2}{
\fill (\x,0) circle (2pt);}
\draw (-1,0) -- (0,0);
\draw (0,0) -- (1,0);
\draw (1,0) -- (2,0);
\draw (1.5,.25) node {$5$} ;

\end{tikzpicture}\\

$\mathbf{I}_2(m)$: \begin{tikzpicture}[style=thick, scale=1.2]
\foreach \x in {1,2}{
\fill (\x,0) circle (2pt);}

\draw (1,0) -- (2,0);
\draw (1.5,.25) node {$m$} ;

\end{tikzpicture}\hspace{0.5cm}  ($m\geq 5$)\\

\subsubsection{Rank $2$ case}

Generalized $n$-gons over $\mathbb{F}_1$ are ordinary $n$-gons, and their automorphism groups are dihedral groups $\mathbf{D}_n$ \cite{anal}.
It follows that the corner stones of the spherical buildings of rank at least $3$ over $\mathbb{F}_1$ are the ordinary $n$-gons with $n = 3, 4, 6, 8$ (since these gonalities are the only ones which do occur in the corresponding thick buildings by \cite{FeHi}). Still, it is important to note that in the rank $2$ examples, {\em all} positive integer values for $n$ occur (except $n = 0,1,2$).

\subsubsection{Quadrics}

We give one final explicit example | it concerns quadrics. \\

Let $n \in \mathbb{N}^\times$.
A {\em quadric}\index{quadric over $\F_1$} of projective dimension $2n$ or $2n + 1$ over $\mathbb{F}_1$ is a set $\mQ$ of $2(n + 1)$ points arranged in pairs $x_0,y_0,x_1,y_1,\ldots,x_n,y_n$, and its subspaces are
the subsets not containing any couple $(x_i,y_i)$. The Witt index of the so defined quadrics is $n$. The quadrics in dimension $2n$ have the further property that the maximal singular subspaces ($n$-spaces consisting of $n + 1$ points) are partitioned in two types, namely those containing an even number of points of $\{a_0,a_1,\ldots,a_n\}$ and those containing an odd number. Automorphisms are permutations of the set $\mQ$ which preserve the given pairing in the $2n$-dimensional case.\\

\bigskip
\subsubsection{Trees as $\mathbb{F}_1$-geometries}

If we allow the value $\infty = n$ in the definition of generalized $n$-gon, we obtain a point-line geometry $\Gamma$ without closed paths, such that any two points or lines are contained in a path  without end points.
So $\Gamma$ becomes a tree (allowing more than $2$ points per line) without end points. Its apartments are paths without end points, and the Weyl group is an infinite dihedral group (generated by the reflections about two different adjacent vertices of such an apartment). So in this setting, a generalized $\infty$-gon over $\mathbb{F}_1$ is a tree of valency $2$  without end points.

\begin{figure}
\begin{center}
\begin{tikzpicture}[style=thick, scale=2]
\foreach \x in {1,2,3,4}{
\fill (\x,0) circle (2pt);}
{\fill (0,-0.5) circle (2pt);}
{\fill (5,-0.5) circle (2pt);}
{\fill (-0.5,-1.5) circle (2pt);}
{\fill (5.5,-1.5) circle (2pt);}
\draw (1,0) -- (2,0);
\draw (2,0) -- (3,0);
\draw (3,0) -- (4,0);
\draw (4,0) -- (5,-0.5);
\draw (0,-0.5) -- (-0.5,-1.5);
\draw (1,0) -- (0,-0.5);
\draw (5,-0.5) -- (5.5,-1.5);
\draw[dashed] (-0.5,-2.5) -- (-0.5,-1.5);
\draw[dashed] (5.5,-2.5) -- (5.5,-1.5);

\end{tikzpicture}
\caption{Generalized $\infty$-gon over $\mathbb{F}_1$.}
\end{center}
\end{figure}

Consider, for instance, $G = \mathbf{SL}_2(\mathbb{F}_q((t^{-1})))$. Then $G$ has a BN-pair $(B,N)$, where

\begin{equation}
B = \{  \left(\begin{array}{cc} a & b\\ c & d   \end{array}\right) \in  \mathbf{SL}_2(\mathbb{F}_q[[t^{-1}]])  \vert c \equiv 0\mod{t^{-1}}\},
\end{equation}
and $N$ is the subgroup of $G$ consisting of elements with only $0$ on the diagonal or only $0$ on the antidiagonal.
Its Weyl group is an infinite dihedral group generated by

\begin{equation}
s_1 =  \left(\begin{array}{cc} 0 & -1\\ 1 & 0   \end{array}\right)\ \ \mathrm{and}\ \ s_2 =  \left(\begin{array}{cc} 0 & -t\\ 1/t & 0   \end{array}\right).
\end{equation}

The corresponding building (defined in the same way as before) is a generalized $\infty$-gon with $q + 1$ points per line and $q + 1$ lines per point.
Its apartments are exactly the trees we introduced earlier in this section.\\

 \bigskip
\subsection{Generalizations}

One notes that it is possible to relax the BN-pair axioms and still get a meaningful theory, and a Weyl functor.
For example, let us fix a certain category of groups $\mathbf{C}$, and let us
consider groups $G$ with subgroups $C, E$ such that

\begin{itemize}
\item[(GBN1)]
$\langle C,E \rangle = G$;
\item[(GBN2)]
$H = C \cap E
\lhd E$ and $E/H$ is isomorphic to an object of $\mathbf{C}$;
\item[(GBN3)]
(to be filled in appropriately).
\end{itemize}

Now we fix a set of groups $\mC$ with the property that each of its elements is a proper subgroup of $G$ which properly contains $C$, and
construct a geometry $\Gamma = \Gamma(C,E;\mathbf{C},\mC)$ as follows.

\begin{itemize}
\item[(GB1)]
\textsc{Elements}: are elements of the left coset spaces $G/P$, $P \in \mC$.
\item[(GB2)]
\textsc{Incidence}: $gP$ is incident with $hP'$, $P \ne P'$, if these cosets intersect nontrivially.
\end{itemize}

With $\mathbf{C}$ = category of Coxeter groups and $\mC$ = all proper subgroups properly containing $C$, and (GBN3) replaced by
the Bruhat decomposition axioms, we obtain a BN-pair $(C,E)$.

Defining (maximal) flags as before, the reader notes the following:

\begin{proposition}
$G$ acts by left translation as an automorphism group of $\Gamma$. This action is transitive on the maximal flags of the geometry.
\end{proposition}

The Weyl functor for this category of group data would be

\begin{equation}
\underline{\mA}: G \longrightarrow E/H,
\end{equation}
and on the geometrical side, it should send the geometry to the geometry induced by the left coset spaces $E/H$ ($= \{eH \vert e \in E\}$).\\

\quad\textsc{Question}\quad
{\em Find good candidates for $\mathbf{C}$ and $\mC$, and
formulations for {\rm (GBN3)} such that the Weyl images have precisely $2$ points per line.}\\

It is precisely this kind of question which we will be considering in the next section from the synthetic side. (The game that we play there is first
to imagine what the Weyl images should be | certain thin geometries which are fixed objects of the functor we want to define | and build, using
certain prescribed axioms, the general geometries ``defined over $\Fun$'' using the Weyl geometries as bricks. One could play the same game here:
imagine what the geometries induced on $E/H$ should be, and ask a similar question.)\\

config: arxiv


TITLE: Does $A^{-1} = A_0^{-1} + O_p(n^{-1})$ where $A = A_0 + O_p(n^{-1})$?
QUESTION [2 upvotes]: Let $A$ be a random matrix that converges in probability to a constant matrix $A_0$ at the rate $O_p(n^{-1})$:
$$
A = A_0 + O_p(n^{-1}).
$$
I want to know if $A^{-1} = A_0^{-1} + O_p(n^{-1})$?
Let $I$ be the identity matrix. Here is my attempt at proving it, which I'm not sure is valid mainly because I don't know if we are allowed to define a series expansion in the form of $O_p$ variables.
$$
\begin{align}
A^{-1} &= (A_0 + O_p(n^{-1}))^{-1} \\
&= (A_0(I + A_0^{-1}O_p(n^{-1}))^{-1} \\
&= (A_0(I + O_p(n^{-1}))^{-1} \\
&= (I + O_p(n^{-1}))^{-1}A_0^{-1}.
\end{align}
$$
Now we perform a Neumann series expansion which gives
$$
\begin{align}
(I + O_p(n^{-1}))^{-1}
&= \sum_{k=0}^\infty (-1)^{k} (O_p(n^{-1}))^{k} \\
&= I - O_p(n^{-1}) \\
&= I + O_p(n^{-1}).
\end{align}
$$
I have assumed here that $(O_p(n^{-1}))^0$, is this true?
Then we substitute this into $(I + O_p(n^{-1}))^{-1}A_0^{-1}$ to get
$$
\begin{align}
A^{-1} &= (I + O_p(n^{-1}))A_0^{-1} \\
&= A_0^{-1} + O_p(n^{-1})A_0^{-1} \\
&= A_0^{-1} + O_p(n^{-1}).
\end{align}
$$
In the deterministic case, if we wanted to invert the matrix $(I+K)$ we need to have $||K|| < 1$ to ensure convergence. So does this mean that in the probabilistic case, to invert the matrix $(I + O_p(n^{-1}))$ we need to have $||O_p(n^{-1})|| < 1$? So the norm of an $O_p(n^{-1})$ term is required to be less than one..does this phrase even make sense?
It seems an $O_p(n^{-1})$ term will be increasingly likely to have a value less than $1$ as $n \to $\infty$, since the probability is becoming more concentrated around zero, but I don't think it can ever be fully guaranteed to be below one?
So does $A^{-1} = A_0^{-1} + O_p(n^{-1})$ really hold and if it does is my proof valid? If not, what is a valid proof?

REPLY [1 votes]: Given a stochastic $A$ and fixed $A_0$ of size $d\times d$ such that $A=A_0+O_p(n^{-1})$, we can show $A^{-1}=A_0^{-1}+O_p(n^{-1})$ by noticing that matrix inverse is differentiable, so by local linearity rates are preserved.$\DeclareMathOperator\vect{vec}$
Consider the vector-valued vector-input function $f$ over flattened invertible matrices where $ f(\vect X)=\vect X^{-1}$. This vector-valued function is differentiable on the open set of invertible matrices, such that, refering to the indices $ij, kl$ from the unflattened matrices,
$$
\partial_{ij}f_{kl}(\vect X)=-(X^{-1})_{jk}(X^{-1})_{li}\,.
$$
For any convex combination $A_t$ of $A_0,A$, we have on the event $A_t$ is invertible,
$$\|Df(\vect A_t)\|\le \|Df (A_t)\|_F\le \|\vect Df (A_t)\|_1=\|\vect A_t^{-1}\|_1^2\le d^2\|A_t^{-1}\|_F^2\,.$$
Then since $\|A-A_t\|_F=O_p(n^{-1})$ by the assumption, as the other answer shows with Neumann series expansion, we have $A_t^{-1}$ exists whp and converges entrywise in probability to $A^{-1}$, so $\|A_t^{-1}\|_F=O_p(\|A^{-1}\|_F)=O_p(1)$.
Thus whp $\|Df(\vect A_t)\|$ is bounded by some constant $M$ and by the mean value inequality on $f$, $\|A_0^{-1}-A^{-1}\|_F\le M\|A_0-A\|_F$, which implies the conclusion.


TITLE: finding expected value and variance using conditional expected value and variance
QUESTION [0 upvotes]: Let $X \sim G(p)$ and $Y|X \sim B(j,p)$, I need to express $E[Y]$ and $Var(Y)$ using $p$
first I tried expressing $E[Y]$:
$$E[Y|X=j]=\sum_{i=0}^{j}i\cdot\binom{j}{i}p^i(1-p)^{j-i}=\sum_{i=0}^{j}i\frac{j!}{i!(j-i)!}p^i(1-p)^{j-i}=jp\sum_{i=0}^j\frac{(j-1)!}{(i-1)!(j-i)!}p^{i-1}(1-p)^{j-i}=jp\sum_{i=0}^{j}\binom{j-1}{i-1}p^{i-1}(1-p)^{j-i}=jp(p+1-p)^{j-1}=jp$$
hence
$$E[Y]=E[E[Y|X]]=\sum_{j=0}E[Y|X=j]p_x(j)=\sum_{j=0}jp(1-p)^{j-1}p=p^2\sum_{j=0}j(1-p)^{j-1}$$
But I don't know how to reach a solution for this last series, and get an answer for $E[Y]$

I then tried expressing $Var(Y)$
Let $Y_i$ be an indicator, therefore $Y=\sum Y_i$ so
$$Var(Y|X)=Var(\sum_{i=1}^{j}Y_i|X=j)=\sum_{i=1}^j Var(Y_i)=jp(1-p)$$
I already know that $E[Y|X]=jp$ so all I need is to calculate
$$Var(Y)=E[Var(Y|X)]+Var(E[Y|X])=E[jp(1-p)]+Var(jp)$$
but I have no idea how I can calculate this either
In both cases I followed the same steps that were in the examples I got, so I fail to understand what did I do wrong, this is an introduction exercise so it shouldn't be too complicated, yet I am at a complete loss

REPLY [1 votes]: First question

But I don't know how to reach a solution for this last series

$$\sum_{j=0} j(1-p)^{j-1}$$
set $(1-p)=q$ and observe that
$$jq^{j-1}=\frac{d}{dq}q^j$$
...the sum of the derivative is the derivative of the sum...the resulting sum is easy because it is a geometric series....and the solution easy follows

Second question
Your text is the following
$X\sim G(p)$ and $Y|X=j\sim Bin (j,p)$
Thus using the variance decomposition you get, as you found
$$\mathbb{V}[Y]=\mathbb{V}[jp]+\mathbb{E}[jpq]=p^2\cdot\frac{q}{p^2}+pq\frac{1}{p}=2(1-p)$$

EDIT: I think it is easy for you to rewrite the text in the following way
$X\sim G(p)$
$Y|X\sim Bin(x,p)$
Thus it is easy to realize that, i.e.
$$E(x pq)=pqE[X]=pq\frac{1}{p}=q$$


TITLE: Restriction of scalars is special case of internal hom
QUESTION [2 upvotes]: Since restriction of scalars is right adjoint to extension of scalars, which is just a tensor product, it must be (isomorphic to) the internal hom. I'm having trouble showing this is the case:
Let $f:R\rightarrow S$ be an arrow in $\mathsf{Rng}$. Let $M$ be a left $S$-module. How can I prove its restriction of scalars to $R$, denoted by $\bar M$, is (naturally isomorphic to) the internal hom $\mathsf{hom}(R,M)$? I can't even think of the isomorphism...
I am also confused because the general tensor hom adjunction for monoidal categories is an adjunction of endofunctors, and here, it seems $-\otimes S$ should somehow be adjoint to $\mathsf{hom}(R,-)$...

REPLY [3 votes]: There is a more general tensor-hom adjunction for bimodules. It is important, because if the rings are not commutative, the tensor product does not stay in a fixed category of module over a ring. More precisely, you need a right $B$-module $M$ and a left $B$-module $N$ to form the tensor product $M\otimes_B N$. And this is not a $B$-module anymore. However, if $M$ or $N$ (or both) carry another ring structure, then so does $M\otimes_B N$. Hence if $N$ is a $(B,C)$-bimodule, then we have for any ring $A$ a functor
$$\cdot\otimes_B N : (A,B)-\operatorname{bimod}\rightarrow(A,C)-\operatorname{bimod}.$$
Similarly, if $N$ and $P$ are right $C$-modules, then $\operatorname{Hom}_C(N,P)$ is not a $C$-module anymore. However, if $N$ carries a left $B$-module structure, then $\operatorname{Hom}_C(N,P)$ carries a right $B$-module structure, and if $P$ carries a left $D$-module structure, so does $\operatorname{Hom}_C(N,P)$. So in general, we have a functor for any ring $D$
$$\operatorname{Hom}_C(N,\cdot):(D,C)-\operatorname{bimod}\rightarrow(D,B)-\operatorname{bimod}$$
In this context, the tensor-hom adjunction is the following : let $M$ be a $(A,B)$-bimodule, $N$ a $(B,C)$-bimodule and $P$ a $(D,C)$-bimodule, then :
$$\operatorname{Hom}_C(M\otimes_B N,P)=\operatorname{Hom}_B(M,\operatorname{Hom}_C(N,P))$$
natural as $(D,A)$-bimodules in $M,N,P$.
The extension/restriction of scalar adjunction is indeed a special case of the tensor-hom. Let $f:R\rightarrow S$ a morphism of rings. Consider $S$ as a left $R$-module, and right $S$-module. Then for any right $S$-module $N$, $\operatorname{Hom}_S(S,N)$ is a right $R$-module and
$$ \operatorname{Hom}_S(M\otimes_R S,N)=\operatorname{Hom}_R(M,\operatorname{Hom}_S(S,N))$$
for any right $R$-module $M$.
Finally, it is easy to show that $\operatorname{Hom}_S(S,N)=N$ as right $R$-module.


TITLE: Why do bubbles initially stick to the side or the bottom of a glass while boiling?
QUESTION [10 upvotes]: When boiling water, I've noticed that bubbles will appear to grow at the bottom of the glass for a period of time and then rise.  I've read that bubbles form from nucleation centers but I don't know the specifics of the nucleation dynamics.  Is there a possible scenario where the bubbles, while still appearing to adhere to the bottom of the glass, are in an equilibrium where the bottom par of the bubble is being formed by water evaporating into gas, but at the top of the bubble, which might be slightly cooler, the gas is redissolving into liquid phase? If the two rates are the same then the bubble would appear stationary.  At some point the water heats up enough to have rate of evaporation greater than dissolving throughout the entire bulk of the liquid and the bubble will survive its transit to the top of the liquid.  I think that falling drops experience a dynamic that is th opposite of this.
A couple other thoughts that I've considered:

I know that boiling occurs when the vapor pressure of water is greater than atmospheric pressure.  I believe that this condition is achieved even when the bubble is stationary since the bubble is displacing the surrounding bulk water and air.
I think that carbonated beverages can have bubbles stick to the side of a glass, but I suspect that the reason for this is different but i don't have an intuition about why carbon dioxide bubbles in a cold drink might stick, since the beverage is cold.

REPLY [19 votes]: The bubbles are already on the surface, they are just too small to see with the naked eye.
Wetting a surface, even at room temperature, results in tiny gas/vapor bubbles at defect sites due to surface tension. For example, surface tension prevents water from seeping into tiny crevices (on the order of microns).
These tiny gas pockets expand when heated, and eventually you can see them. They were on the surface the entire time, they just expanded. They stay on the surface because surface tension pulls down and balances the upward buoyant force.
If you keep adding more energy, however, the gas in the bubble will expand. Eventually the bubble will eject from the surface because the surface tension scales inversely with bubble radius, so the force holding it back decreases. Furthermore, as the bubble increases in volume at the surface, it gains an appreciable buoyant force that overcomes surface tension. At this point, the bubble rises.
You can actually superheat water above the boiling point if you have a surface that has small enough defects, since this makes it more difficult for gas bubbles to be trapped when the surface is wetted.
Anyways, the bubbles seem to stick to the sides of the container because they were always there to begin with, thanks to surface tension. You only see them when higher temperatures cause the gas inside them to expand.

REPLY [10 votes]: It is expensive, in energy terms to, make a surface like the surface of a bubble.
By sticking to the walls or bottom of a container some of the surface is free, so it is energeticly favorable to stay partly attached until enough gas forms that the ratio of surface area to volume of gas reaches some limit (which depends on surface tension and the nature of the liquid and surface)


TITLE: Define a conic section relative to its vertices
QUESTION [0 upvotes]: What is the equation of a conic section illustrated below. Points $A$ and $B$ are the vertices, separeted by a distance of $1$, and point $C$ is an arbitrary point on the conic. Point $A$ is also the center of a circle which passes through $O$, the center of the conic. Point $C'$ is the special case of point $C$ where the conic intersects that circle.
The equation should show the relationship between the lengths of segments $\overline{AC}$ and $\overline{BC}$, and a third variable which defines the shape of the conic. For example, in the special case where the conic is a circle (no third variable used), the equation is: $\overline{AC}^2 + \overline{BC}^2 = 1$. For the degenerate case where the conic is an infinitely thin ellipse which becomes a line segment (no third variable necessary) the equation is: $\overline{AC} + \overline{BC} = 1$. For the degenerate case of an infinetely tall ellipse which becomes two parallel lines, the equation is $\overline{AC}^2 + 1= \overline{BC}^2$ or  $\overline{BC}^2 + 1= \overline{AC}^2$. For the case of the degenerate hyperbola where the solution is two opposed rays, the equation is $\overline{AC} - \overline{BC} = 1$ or $\overline{BC} - \overline{AC} = 1$
I had asked the question nearly this way in my previous StackExchange post, and it was properly answered, but the answer was two equations - one for a hyperbola and one for an ellipse. The "third variable" had to go to infinity for the degenerate case of an infinitely tall ellipse which becomes parallel lines. I am therefore including suggestions for the third variable so that the answer is a single equation. Here are three fairly obvious suggestions for the third variable:
The measure of angle $\angle BAC'$
The length of segment $\overline{BC'}$
The length of segment $\overline{OC'}$    (I'm thinking that an isosceles triangle might keep the math simpler.)
I've assumed that the distance between $A$ and $B$ is 1, though that may not be optimum. And for the nitpickers, the vertices shown here are actually co-vertices.

REPLY [0 votes]: In an answer to the previous related question, writing $a = AC$ (length of the segment $\overline{AC}$)
and $b = BC$ (length of the segment $\overline{BC}$), there are separate equations for the ellipse and for the hyperbola,
$$
\left(h^2 - \frac14\right) (a^2 - b^2)^2 + \frac12(a^2 + b^2) = h^2 + \frac14
$$
and
$$
\left(h^2 + \frac14\right) (a^2 - b^2)^2 - \frac12(a^2 + b^2) = h^2 - \frac14,
$$
respectively.
In addition, there is the case of two vertical lines through the points $A$ and $B$,
which has the equation
$$
(a^2 - b^2 - 1)(a^2 - b^2 + 1) = 0,
$$
which is equivalent to
$$
(a^2 - b^2)^2 = 1.
$$
Note that all of these equations can be written in the form
$$
p (a^2 - b^2)^2 + (q - p)(a^2 + b^2) = q \tag 1
$$
where $p$ and $q$ are real constants. So if we want to come up with a single equation that covered all three cases, we can clearly do it if we get to choose $p$ and $q$ independently at will.
Some other observations: for the hyperbola, $p > q,$ for the ellipse, $p < q,$ and for the vertical lines, $p = q.$ For the extreme cases of the degenerate hyperbola and ellipse, $p = -q.$
But note that only the ratio between $p$ and $q$ really matters, because
if we multiply both $p$ and $q$ by the same non-zero factor we get an equation with the same solutions for $a$ and $b$ as before.
If we want to come up with a single equation where we select just a single parameter value to determine the shape of the graph, we need formulas for $p$ and $q$ in terms of that parameter that allow all of the necessary ratios to be achieved.

Any of the parameters in the question should work, but let's try the parameter
$\beta = BC'$ (the length of the segment $\overline{BC'}$).
The figure tells us that at one of the points on the desired curve we have
$a = \frac12$ and $b = \beta.$
If we plug those values into the generic formula in Equation $(1)$, we have
$$
p \left(\left(\tfrac12\right)^2 - \beta^2\right)^2 + (q - p)\left(\left(\tfrac12\right)^2 + \beta^2\right) = q.
$$
Multiply everything by $4$ so we don't have to deal with powers of $\frac12,$
expand $(1 - 4\beta^2)^2$, collect all the $p$ terms on one side and all the $q$ terms on the other, and we get something equivalent to
$$ (12 - 16 \beta^2) q = (3 + 24 \beta^2 - 16 \beta^4) p. $$
An easy way to make this true for any value of $\beta$ is to set
\begin{align}
p &= 12 - 16 \beta^2, \\
q &= 16 \beta^4 - 24 \beta^2 - 3.
\end{align}
Making these substitutions for $p$ and $q$ in Equation $(1)$ and simplifying the difference $q - p$ we get:
$$
(12 - 16 \beta^2) (a^2 - b^2)^2 + (16 \beta^4- 8 \beta^2 - 15)(a^2 + b^2)
 = 16 \beta^4 - 24 \beta^2 - 3. \tag2
$$
If we let $\beta$ range through all values between $\frac12$ and  $\frac32$ (that is, all the distances from $B$ to any point on the circle around $A$),
we get the full range of ellipses and hyperbolas,
with a circle for $\beta = \frac{\sqrt3}2$ (where $\beta$ is the leg opposite $A$ in a right triangle $\triangle AC'B$)
and parallel lines for $\beta = \frac{\sqrt5}2$ (where $\beta$ is the hypotenuse of a right triangle $\triangle BAC'$).

To use the measure of the angle $\angle BAC'$ as the single parameter, let that angle be $\theta.$
Then at the point $C'$, we have $a = \tfrac12$ (again) and
$b^2 = \left(\tfrac12\right)^2 + 1^2 - 2\cdot \tfrac12\cdot 1\cdot \cos\theta
 = \tfrac54 - \cos\theta$
by the Law of Cosines.
Making these substitutions for $a$ and $b^2$ in Equation $(1)$, we get
$$
p \left(\cos\theta - 1\right)^2 + (q - p)\left(\tfrac32 - \cos\theta\right) = q
$$
Expanding and collecting terms in $p$ and $q$, we can arrive at the equivalent equation
$$
 q (1 - 2 \cos\theta) = p (1 + 2 \cos\theta - 2\cos^2\theta),
$$
from which we can find suitable values for $p$ and $q$:
\begin{align}
p &= 1 - 2 \cos\theta, \\
q &= 1 + 2 \cos\theta - 2\cos^2\theta.
\end{align}
Putting these values into Equation $(1)$ and simplifying, we have
$$
(1 - 2 \cos\theta) (a^2 - b^2)^2 + (4 \cos\theta - 2\cos^2\theta)(a^2 + b^2)
 = 1 + 2 \cos\theta - 2\cos^2\theta. \tag3
$$
We get the full set of ellipses and hyperbolas by letting $\theta$ range from $0$ to $\pi,$ corresponding to all possible locations of the point $C'$ along the upper half of the circle about $A.$
The circle occurs at $\theta = \frac\pi3$
and the two parallel lines occur at $\theta = \frac\pi2.$
It should also be possible to derive Equation $(3)$ by substituting
$\beta = \tfrac54 - \cos\theta$ in Equation $(2).$

Let $d = OC'$, the length of the segment $\overline{OC'}$.
Using the definition of the angle $\theta$ from above, we can write
$d = \sin\frac\theta2.$
Therefore the double-angle formula for the cosine
(interpreting $\theta$ as double the angle $\frac\theta2$) implies that
$\cos\theta = 1 - 2\sin^2\frac\theta2 = 1 - 2d^2.$
If we make this substitution for $\cos\theta$ in Equation $(3),$ we get
$$
(1 - 2 (1 - 2d^2)) (a^2 - b^2)^2 + (4 (1 - 2d^2) - 2(1 - 2d^2)^2)(a^2 + b^2)
 = 1 + 2 (1 - 2d^2) - 2(1 - 2d^2)^2,
$$
which simplifies to
$$
(4 d^2 - 1) (a^2 - b^2)^2 + (2 - 8 d^4)(a^2 + b^2) = 1 + 4 d^2 - 8 d^4. \tag 4
$$
For this formula we get the full range of ellipses and hyperbolas by letting $d$ range between $0$ and $1,$ which is the range of possible lengths of chords of the circle about $A$. The circle occurs at $d = \frac12$ and the two parallel lines occur at $d = \frac{\sqrt2}{2}.$

Any of Equations $(2)$, $(3)$, or $(4)$ can be converted into a conventional equation of a conic in one of the usual positions in Cartesian coordinates by assuming that $A$ has coordinates $\left(-\frac12,0\right)$
and $B$ has coordinates $\left(\frac12,0\right).$
If the coordinates of an arbitrary point $C$ on the conic are $(x,y)$,
then $a^2 = \left(x + \frac12\right)^2 + y^2$
and $b^2 = \left(x - \frac12\right)^2 + y^2.$
As shown in the previous answer,
this implies that $(a^2 - b^2)^2 = 4x^2$ and $a^2 + b^2 = 2x^2 + 2y^2 + \tfrac12.$
We can substitute these expressions for $(a^2 - b^2)^2$ and $a^2 + b^2$
in any of the Equations $(2)$, $(3)$, or $(4)$
in order to obtain an equation in $x$ and $y$ that can be plotted in Cartesian coordinates. For example, Equation $(4)$ is equivalent to
$$
(4 d^2 - 1) 4x^2 + (2 - 8 d^4)\left(2x^2 + 2y^2 + \tfrac12\right)
 = 1 + 4 d^2 - 8 d^4,
$$
which is equivalent to
$$
4 d^2 (1 - d^2)x^2 + (1 - 4d^4)y^2 = d^2(1 - d^2).
$$

There are two special cases that still do not work out quite the way I think the question was looking for, namely the "degenerate ellipse" and the "degenerate hyperbola."
For example, using Equation $(4)$ we would expect to obtain the degenerate ellipse when $C' = O$ (that is, $d = 0$) and the degenerate hyperbola when $C'$ is exactly opposite from $O$ on the circle (that is, when $d = 1$).
But when $d = 0,$ Equation $(4)$ becomes
$$
-(a^2 - b^2)^2 + 2(a^2 + b^2) = 1,
$$
whereas when $d = 1$ the equation is
$$
3(a^2 - b^2)^2 - 6(a^2 + b^2) = -3.
$$
But the second equation is exactly equivalent to the first one.
In fact, both of these equations are equivalent to
$$
((a - b)^2 - 1)((a + b)^2 - 1) = 0,
$$
although it takes some algebra to get there.
When the first factor is zero, that is, when $(a - b)^2 = 1,$ you get a point on the "degenerate hyperbola,"
and when the second factor is zero, that is, when $(a + b)^2 = 1,$ you get a point on the "degenerate ellipse."
I do not think it is possible for one formula to continuously transform from the formula of the "degenerate ellipse" through the usual ellipses and hyperbolas to the formula of the "degenerate hyperbola."
One reason is that if you make the substitution
$a^2 = \left(x + \frac12\right)^2 + y^2$, $b^2 = \left(x - \frac12\right)^2 + y^2$
into such a formula, you should get a formula for the Cartesian coordinates of the conic. In order to be a correct formula for the non-degenerate conics, this formula must be a quadratic polynomial in $x$ and $y.$ But there is no quadratic polynomial in $x$ and $y$ that produces a line segment or two opposite rays.
One reason for that is that there is no way to produce such figures by intersecting an infinite double circular cone with an infinite plane.
(For this same reason, neither figure is usually considered to be an actual conic,
even in generalized contexts where we consider two parallel lines to be a conic.)
What you can produce by intersecting an infinite double circular cone with an infinite plane is a single line in the plane.
This happens when the vertex of the cone is on the plane and the plane is tangent to one of the circular cross-sections of the cone.
And a single line through the points $A$ and $B$ is what is produced as the graph of the equation when either $d = 0$ or $d = 1.$
If you really want to include in your taxonomy the non-conic curves that you call a "degenerate ellipse" and a "degenerate hyperbola," I think the best policy to set them aside as special cases to be handled by separate equations.
These two equations can be combined into the general equation of the line through $A$ and $B,$ which is a case of the general equation for all the other figures,
but to have a general equation that suddenly reduces to one of these two separate equations exactly when the parameter reaches the appropriate end of its range,
I think you would need to resort to at least one inelegant discontinuous function defined by cases, which would be a lot uglier than just defining the cases that you want in the obvious way. (One equation for $d = 0,$ one for $d = 1,$ and another for every other value of $d.$)


TITLE: Converting powers of 3 into powers of 2?
QUESTION [0 upvotes]: I'm stuck on a problem. I have a term $3^m$ where $m$ is an integer $> 0$. and I want to represent it as $2^m + a$ however I don't want to keep the $a$. I am looking for a formula to represent $a$ using numbers and the variable $m$. Is this possible to do?
Thanks

REPLY [0 votes]: This is more a collection of information than a formula.  It seems tough to get an elementary function on this.
We can get a generating function for this.
There's a lot of information on this on Sloan's Online Encyclopedia of Integer Sequences, A001047.
It looks like you have:
$$3^m = 2^m + a$$
Also, I believe you want to represent $a$ as a formula of $m$.
This can be done by a generating function.  The way this works is that we use induction.  We consider what we'd get with $3^{m-1}$, and then calculate what we'd get with $3^m$.
First, we want to figure out the difference between $3^m$ and $2^m$.  We realize that we can visualize $3^m$ as a "hypercube"; that is, an $m$-dimensional cube with sides of length 3.  Now $2^m$ is an $m$ dimensional cube with sides of length 2.  So the difference between the larger cube and the smaller cube is adding length one to each side of the smaller cube.
We do this by first adding $3^{m-1}$.  This gives us part of what we need.  Then we can add twice the previous case (remember, we're doing this by induction).  That will give us the required difference.
Now, we can construct a generating function to give us the answer.  If you're interested in a very, very simple introduction, you can download a free copy of Wilf's Generatingfunctionology here.
THE GENERATING FUNCTION
We state this as recursively as:
$$a_{m+1} = 2 a_m + 3^{m-1}$$
This translates into the generating function:
$$
\begin{align}
\sum_{m=1}^\infty{a_{m+1}x^m} &= 2 \cdot \sum_{m=1}^\infty{a_{m}x^m} + \frac{3}{(1-3x)} \\
\frac{A(m)-1}{x} &= 2A(m) + \frac{3}{(1-3x)}\\
A(m) &= \frac{1}{1-5x+6}
\end{align}
$$
The series proceeds like:
$$1, 5x, 19x^2, 65x^3, 211x^4, \dots$$


TITLE: Probability that the sum of 40 numbers from $X$ ~ $Unif(-1/2,1/2)$ is greater then 3?
QUESTION [0 upvotes]: There are $40$ independent random variables, $X_i$ ~ $Unif(\frac{-1}{2},\frac{1}{2})$ for $i \in \{1,2,...,40\}$. What is the probability that their sum is greater then 3?
My attempt at a solution is definitely incorrect.
By letting $Y = X_1+X_2+...+X_{40}$ all we will have to do is find $P(Y>3)$, but thats the same evaluating $1 - P(Y\leq3)$. This can be solved since we know that the pdf is $f(x)=\frac{1}{1/2-(-1/2)}=1$. Putting this all together we get:
$$P(Y>3)=1-P(Y \leq 3)=1-\int_{-20}^{3} 1 dy$$
I got a lower bound of -20 since the lowest value you could pick everytime is $-1/2$ so just multiple it by 40 to get a minimum sum value. The problem is that this evaluates to a negative value. more specifically $-22$.
Can someone tell me what my mistake is?

REPLY [0 votes]: Let $S_{40}=\sum\limits_{i=}^{40} X_i$ where $X_i\sim U(-1/2,1/2)$ with mean $\mu=0$ and variance $\sigma^2=1/12$ Then $$Z=\frac{S_{40}-0}{\sqrt{40}(1/12)}\sim \mathcal{N}(0,1)$$ approximately by the Central Limit Theorem. Thus $$P(S_{40}>3)=P\left(\frac{S_{40}-0}{\sqrt{40}(1/12)}> \frac{3}{\sqrt{40}(1/12)}\right)$$
which  is then computable via the Normal table