TITLE: Understanding electric potential in the context of gravitational potential?
QUESTION [1 upvotes]: I'm trying to understand the concept of electric potential and the significance of an electric field being related to the gradient/derivative of an electric potential.
I found this question that explains how potential energy relates to force in the $-z$ direction - Is force the derivative of energy?
What would an analogous answer be for electric potential and force in one dimension?

REPLY [1 votes]: The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g).
Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg.  If you want to break it down further, I guess mg = Q/sec.
I'm not sure if this answers your question, but it at least gets things started.


TITLE: Spivak's Calculus, Chapter 5 "Limits", problem 41: If $c>0$, then $c^{1/n}$ approaches $1$ as $n$ > becomes very large.
QUESTION [0 upvotes]: The following is a problem from chapter 5, "Limits", of Spivak's Calculus

(a) For $c>1$, show that $c^{1/n}=\sqrt[n]{c}$ approaches $1$ as $n$ becomes very large. Hint: Show that for any $\epsilon>0$ we cannot
have $c^{1/n}>1+\epsilon$ for large $n$.

(b) More generally, if $c>0$, then $c^{1/n}$ approaches $1$ as $n$
becomes very large.

Note that the concept of a sequence hasn't been introduced at this point in the book. Only the basic definition of a limit of a function. I am looking for solutions that only use the latter concept.
My question regards the part $(b)$.
Here is the solution to part $(a)$.
Assume $c>1$ and $f(n)=c^{1/n}$
Bernoulli's Inequality says that
$$\forall \epsilon>0\ \forall n>0,\ 1<1+n\epsilon\leq (1+\epsilon)^n$$
Therefore
$$\lim\limits_{n\to \infty}(1+n\epsilon)=\infty\implies \lim\limits_{n\to\infty}(1+\epsilon)^n=\infty$$
This means that
$$\forall \epsilon>0\ \forall M>0\ \exists N>0\ \forall n, n>N \implies (1+\epsilon)^n>M$$
If we specify $M=c$ then
$$\forall \epsilon>0\ \exists N>0\ \forall n, n>N \implies (1+\epsilon)^n>c \implies 0<c^{1/n}-1<\epsilon$$
Now part $(b)$
Let $c>0$.
Assume $c^{1/n}>1+\epsilon$, for all $n\in\mathbb{N}$.
Then $c>(1+\epsilon)^n>1+n\epsilon$ by Bernoulli's Inequality.
But then $n<\frac{c-1}{\epsilon}$ for all $n \in \mathbb{N}$. But we can choose any natural number $n_1>\left \lfloor \frac{c-1}{\epsilon} \right \rfloor +1$ and have $n_1>\frac{c-1}{\epsilon}$.
Therefore, we have a contradiction, and can infer that
$$\forall \epsilon>0\ \exists M\ \forall n, n>M \implies 0<c^{1/n}<1+\epsilon$$
$$\implies -1<c^{1/n}-1<\epsilon\tag{1}$$
Note that if in this proof we had assumed $c>1$ then $(1)$ would be $0<c^{1/n}-1<\epsilon$, and this proof could have been used for part $(a)$. We assumed, however that $c>0$.
How do we complete this proof? Is it possible with this proof by contradiction? Is there another way?

REPLY [1 votes]: It's good to prove first the case (a) and then the case (b), since (b) follows from (a) immediately.
Proof of (b)

If $c>1$ than we are done.
If $c=1$ then $c^{1/n}=1\to 1$.
If $0<c<1$ then $1/c>1$ and from (a) we get $$\frac 1{c^{1/n}} = \left(\frac 1c\right)^{\frac 1n}\to 1\quad \text{ so }\quad c^{\frac 1n}=\frac{1}{\frac 1{c^{1/n}}}\to \frac 11 = 1.$$


text:
In this section we give an explicit formula for $\m$ in the affine case. This formula is equivalent to Macdonald's Constant Term conjecture \cite{Macdonald82}, which was proven by Cherednik \cite{Cherednik95}--the equivalence is explained in \cite{Macdonald03}. The formula we give in Theorem \ref{thm:affineformula} can be deduced from formulas which have appeared in the literature before. Our contribution is to phrase it in a uniform way for all affine types, in particular including $A^{(2)}_{2r}$.  Such a concrete formula is probably unavailable in indefinite Kac-Moody type, but \S \ref{sect:ordertwovanishing} below generalizes some properties of this formula to all types.

Consider an affine Kac-Moody root system $\Phi$. We specify a zeroth vertex in the Dynkin diagram as follows. Except in the case of $A^{(2)}_{2r}$, we choose $0$ to agree with \cite{Kac90}. In the case of $A^{(2)}_{2r}$, our vertex $0$ is Kac's vertex $r$. Note that these are the conventions of \cite{Beck-Nakajima04}, which are chosen to interact well with Drinfeld's loop presentation of the quantum affine algebra. The Dynkin diagram with vertex $0$ removed defines a finite root subsystem $\Phi_{\fin}$. Let $Q_\fin$ be the root lattice corresponding to this subsystem, and $Q^+_{\fin}$ its positive cone.

Recall that every imaginary root of $\fg$ is of the form $k \delta$, where $\delta$ is the minimal imaginary root. For each imaginary root $\eta$ we define $S(\eta) = \{ \beta \in  Q^+_{\fin} \suchthat \beta + \eta \in \Phi_{\real} \}$ and let:
\begin{align}
  \label{eq:6}
\m_\eta = \prod_{\beta \in S(\eta)} \frac{ (1 - t^{\height(\beta)} e^\eta)^2} { (1 - t^{\height(\beta)-1} e^\eta) (1 - t^{\height(\beta)+1} e^\eta) }
\end{align}

Then we have the following theorem, which is equivalent to Cherednik's resolution of the constant term conjecture.
\begin{Theorem}\label{thm:affineformula}\cite{Cherednik95}, \cite[3.12]{Macdonald03}, \cite[5.8.20]{macbook}
  \begin{align}
    \label{eq:7}
    \m = \prod_{\eta  \in \Phi^+_{\imag}} \m_\eta
  \end{align}
\end{Theorem}

For every affine Kac-Moody root system other than $A^{(2)}_{2r}$, the identity \eqref{eq:7} follows from \cite[3.12]{Macdonald03}. For the root system $A^{(2)}_{2r}$ we may can derive the constant term formula \cite[5.8.20]{macbook} from that of the root system $(C_r^{\vee},C_r)$ following the specialization choices of Sharma and Viswanath \cite[Section 3]{sharma-viswanath}.

\subsection{Divisibility by $(1-t)^2$}
The results of \S \ref{sect:ordertwovanishing} below are generalizations of some aspects of Theorem \ref{thm:affineformula} to arbitrary Kac-Moody type. Specifically, Theorem \ref{thm:roots_only} will assert that $\m$ can always be expressed as a product over imaginary roots, as in \eqref{eq:7}.

The shape of the factors $\m_{\eta }$ in \eqref{eq:6} motivates Theorem \ref{thm:order2vanishing}, which asserts that a certain polynomial is divisible by $(1-t)^2.$ Write $\m_{\eta }$ in product form, as in \eqref{eq:2}: $\m_{\eta }=\prod_{n=0}^{N_{\eta}}(1-t^n e^{\eta})^{-m(\eta, n)}$. By \eqref{eq:6}, the polynomial
$m_{\eta}(t)=\sum_{n=0}^{N_{\eta}}m(\eta, n)t^{n}$

is divisible by $(1-t)^2$. We see that $m_{\eta}(t) = (1-t)^2 \chi_{\eta}(t)$, where:
\begin{align}
  \label{eq:1}
  \chi_{\eta}(t) = \sum_{\beta \in S(\eta)} t^{\height(\beta)}
\end{align}

In indefinite type, it is no longer true that $\chi_{\eta}(t)$ is a polynomial with positive coefficients, but Conjecture \ref{conj:alternating} points to the possibility of some interesting positivity properties.

config: arxiv


TITLE: Divergence of a vector field defined by a single variable only
QUESTION [1 upvotes]: I understand how to take the divergence of a vector field defined by (x,y,z) but how can one take the divergence for the following:
$$\bar F = \bar a \times \bar r$$where $\bar a$ is a constant vector and $\bar r$ is $(bcos(\theta) , bsin(\theta) , c\theta)$, with b and c being positive constants and $\theta$ taking values between 0 to $\frac{5\pi}{2}$.
I have worked out the cross product to be
$$
\begin{pmatrix}
a_2c\theta - a_3b\sin(\theta)\\
a_3bcos(\theta) - a_1c\theta\\
a_1bsin(\theta) - a_2bcos(\theta)\\
\end{pmatrix}
$$
if we let $\bar a = (a_1,a_2,a_3)$
However, I am not sure how to proceed after this to work out the divergence, since there is no x, y, or z (apparently the answer turns out to be 0).
Thank you for your help.

REPLY [0 votes]: $\nabla \cdot (\overrightarrow{a}\times\overrightarrow{r})=\overrightarrow{r}\cdot (\nabla\times\overrightarrow{a})-\overrightarrow{a}\cdot(\nabla\times\overrightarrow{r})=-\overrightarrow{a}\cdot(\nabla\times\overrightarrow{r})$
for $\overrightarrow{r}$, $x=b\cos\theta, y=b\sin\theta, z=c\theta$, can be reparameterized as :
$x=b\cos(\cfrac{z}{c})$
$y=b\sin(\cfrac{z}{c})$
$z=z$
$\nabla \times\overrightarrow{r}=-\cfrac{b}{c}\Big(\cos(\cfrac{z}{c})\mathbf{i}+\sin(\cfrac{z}{c})\mathbf{j}\Big)$


text:
\begin{definition}[Definition:Metric System/Scaling Prefixes/kilo-]
'''kilo-''' is the  [[Definition:SI Units|Système Internationale d'Unités]] [[Definition:Metric Scaling Prefixes|metric scaling prefix]] denoting a multiplier of $10^3$.
=== [[Symbols:K/kilo-|Symbol]] ===
{{:Symbols:K/kilo-}}
\end{definition}

config: wiki


TITLE: Does "gerrymandering" matter?
QUESTION [4 upvotes]: In the United States, it is often a raw point discussing the issue of redistricting, or so unfavorably called, "Gerrymandering."
Background:  In the United States House of Representatives, each State is allotted a number of representatives based on the State's population.  Within the state, each of its representatives are assigned a geographical "district."  The point of contention comes with the fact that a states Governor may periodically get to redraw the districting lines.
My question is: mathematically, probabilistically, can this work?  My intuition tells me that if one line is moved over to improve the chances of one side winning one district, won't that just hurt the chances of winning the other?

REPLY [6 votes]: Imagine a state with 12 million inhabitants, and 6 districts. The voters are equally split with 6 million Democrats and 6 million Republicans. Note that each district contains two million voters, and it seems fair that there should be an even split in representation.
However, the districts are gerrymandered so that district one contains only 2 million Democrats and no Republicans. The other 4 million Democrats and 6 million Republicans are evenly distributed through the other districts.
Now there are five districts, each 60% Republican and 40% Democrat - as a result, we expect one Democratic district and 5 Republican districts. So gerrymandering significantly changed the result.
This is an extreme case, but if we were to even set it up so that one district contains 1.5 million Democrats with the remaining Democrats evenly distributed, we'd find that the remaining five districts have moderate Republican advantages, even though the first district is overwhelmingly Democratic.


text:
\begin{document}
\title{Gradient estimates for a nonlinear parabolic equation under Ricci flow}
\author{Shu-Yu Hsu\\
Department of Mathematics\\
National Chung Cheng University\\
168 University Road, Min-Hsiung\\
Chia-Yi 621, Taiwan, R.O.C.\\
e-mail:syhsu@math.ccu.edu.tw}
\date{June 25, 2008}
\smallbreak \maketitle
\begin{abstract}
Let $(M,g(t))$, $0\le t\le T$, be a n-dimensional complete noncompact
manifold, $n\ge 2$, with bounded curvatures and metric $g(t)$ evolving by
the Ricci flow $\frac{\partial g_{ij}}{\partial t}=-2R_{ij}$. We will extend
the result of L.~Ma and Y.~Yang and prove a local gradient estimate for
positive solutions of the nonlinear parabolic equation
$\frac{\1 u}{\1 t}=\Delta u-au\log u-qu$ where $a\in\R$ is a constant
and $q$ is a smooth function on $M\times [0,T]$.
\end{abstract}

\vskip 0.2truein

Key words: complete noncompact manifold, Ricci flow, local gradient estimate,
nonlinear parabolic equation\\

Mathematics Subject Classification: Primary 58J05, 58J35

\vskip 0.2truein
\setcounter{equation}{0}
\setcounter{section}{0}

\section{Introduction}
\setcounter{equation}{0}
\setcounter{thm}{0}

Recently there is a lot of studies on Ricci flow on manifolds by R.~Hamilton
and others \cite{H}, \cite{CLN}, because Ricci flow is a powerful tool in
analyzing the structure of manifolds. Let $(M,g(t))$, $0\le t\le T$, be
a n-dimensional complete noncompact manifold with metric $g(t)$ evolving
by the Ricci flow
\begin{equation}
\frac{\1 g_{ij}}{\1 t}=-2R_{ij}\quad\mbox{ in }M.
\end{equation}
$(M,g(t))$ is said to be a gradient Ricci soliton (\cite{H},\cite{T})
if there exists a constant $c\in\R$ such that
\begin{equation}
\mbox{Ric}=cg+D^2f\quad\mbox{ in }M
\end{equation}
where $D^2f$ is the Hessian of $f$. $(M,g(t))$ is a expanding gradient
soliton if $c<0$ and $(M,g(t))$ is a shrinking gradient soliton if $c>0$.
As observed by L.~Ma \cite{M} if we let $f=\log u$, then after some
computation using (1.2) we get
\begin{equation}
\Delta u+2cu\log u=(A_0-nc)u\quad\mbox{ in }M
\end{equation}
for some constant $A_0$. Then the study of the gradient soliton on $M$ is
reduced to the study of the properties of the solution $u$ of (1.3).
In \cite{M} L.~Ma using the technique of S.Y.~Cheng, P.~Li and S.T.~Yau
\cite{CY}, \cite{LY}, proved the local gradient estimates of positive
solutions of the equation
\begin{equation*}
\Delta u-au\log u-bu=0\quad\mbox{ in }M
\end{equation*}
where $a>0$ and $b\in\R$ are constants for complete noncompact manifolds
with a fixed metric and curvature locally bounded below. In \cite{Y}
Y.~Yang extended L.~Ma's result and proved the local gradient estimates of
positive solutions of the equation
\begin{equation*}
u_t=\Delta u-au\log u-bu\quad\mbox{ in }M\times (0,T]
\end{equation*}
where $a,b\in\R$ are constants for complete noncompact manifolds with a fixed
metric and curvature locally bounded below. In \cite{CTY}
A.~Chau, L.F.~Tam and C.~Yu, proved the local gradient estimates of
positive solutions of the conjugate heat equation
\begin{equation*}
u_t=\Delta^tu-qu\quad\mbox{ in }M\times (0,T]
\end{equation*}
where $q$  is a smooth function on $M\times [0,T]$ and $\Delta^t$ is the
Laplace operator with respect to the metric $g(t)$ for any complete
noncompact manifold $(M, g(t))$, $0\le t\le T$, with bounded curvature
and metric $g$ satisfying
\begin{equation}
\frac{\1 g_{ij}}{\1 t}=2h_{ij}\quad\mbox{ in }M\times (0,T]
\end{equation}
where $h(x,t)=(h_{ij}(x,t))$ is a smooth symmetric tensor on $M\times [0,T]$.
In this paper we will extend their results and prove the local gradient
estimates of positive solutions of the equation
\begin{equation}
u_t=\Delta^tu-au\log u-qu\quad\mbox{ in }M\times (0,T]
\end{equation}
where $a\in\R$ is a constant and $q$ is a smooth functions on $M\times [0,T]$
for any complete
noncompact manifold $(M, g(t))$, $0\le t\le T$, with bounded curvature
and metric $g$ satisfying (1.4).  In particular
our result holds for complete noncompact manifolds with metric evolving by
Ricci flow.

We start with some definitions. Let $\nabla^t$ and $\Delta^t$ be the
covariant derivative and Laplacian with respect to the metric $g(t)$.
When there is no ambiguity, we will drop the superscript and write $\nabla$,
$\Delta$, for $\nabla^t$, $\Delta^t$, respectively. For any $r>0$,
$x_0\in M$, $0<t_1\le t_0\le T$, let $B_r(x_0)$ be the
geodesic ball with center $x_0$ and radius $r$ with respect to the
metric $g(0)$.
For any $x_1,x_2\in M$, let $r(x_1,x_2)$ be the distance between $x_1$
and $x_2$ with respect to the metric $g(0)$.

\section{Local gradient estimates}
\setcounter{equation}{0}

In this section we will prove the local gradient estimates for positive
solutions of (1.5). We will let $(M, g(t))$, $0\le t\le T$, be a complete
noncompact manifold with metric $g$ satisfying (1.4) and
$$
|Rm|\le k_0\quad\text{ on }M\times [0,T]
$$
for some constant $k_0>0$ for the rest of the paper. Similar to \cite{CTY}
we will assume that
$q$ is a smooth functions on $M\times [0,T]$ and $h=(h_{ij})$ is a
smooth symmetric tensor on $M\times [0,T]$. We will also assume the following
for the metric $g$ and symmetric tensor $h(x,t)=(h_{ij}(x,t))$:

\begin{enumerate}\rm
\item[(A1)] $\|h\|, \|\nabla^th\|$ are uniformly bounded on $[0,T]$ where
the norms are taken on $M$.
\item[(A2)] $|q|,|\nabla^tq|, |\Delta q|$ are uniformly bounded
on $M\times [0,T]$.
\end{enumerate}

We also let $H=g^{ij}h_{ij}$ be the trace of $h=(h_{ij})$.

\begin{thm}
Let $u$ be a positive solution of (1.5). Then for any $\alpha>1$ and
$0<\delta<1$
there exists a constant $C_1>0$ depending on $k_0$, $\alpha$, $\delta$,
$a_+=\max (a,0)$ and the space-time uniform bound of $\|h\|, \|\nabla^th\|$,
$|q|$, $|\nabla^tq|$, $|\Delta^tq|$ such that
\begin{equation}
\frac{|\nabla u|^2}{u^2}-\alpha\frac{u_t}{u}-\alpha a\log u-\alpha q
\le\frac{n\alpha^2}{2(1-\delta)t}(1+ta_-)+C_1\quad\text{ in }M\times (0,T]
\end{equation}
where $a_-=-\min (a,0)$.
\end{thm}
\begin{proof}
We will use a modification of the technique of \cite{CTY} and \cite{Hs2} to
prove the theorem. Let $f=\log u$. Then by (1.5) $f$ satisfies
\begin{equation}
f_t=\Delta f+|\nabla f|^2-af-q.
\end{equation}
By \cite{LY} and \cite{CTY} $f$ satisfies
\begin{equation}
\left\{\begin{aligned}
(\Delta f)_t=&\Delta f_t-2h_{ij}f_{ij}-2h_{ik;i}f_k+\nabla H\cdot\nabla f\\
(|\nabla f|^2)_t=&2\nabla f_t\cdot\nabla f-2h(\nabla f,\nabla f)
\end{aligned}\right.
\end{equation}
where $h(\nabla f,\nabla f)=h_{ij}f_if_j$. Let
\begin{equation}
F(x,t)=t[|\nabla f|^2-\alpha (f_t+af+q)].
\end{equation}
By (2.3) and the Weitzenbock-Bochner formula \cite{A},
$$
\Delta (|\nabla f|^2)=2\nabla f\cdot\nabla (\Delta f)+2|\nabla^2f|^2
+2R_{ij}f_if_j,
$$
we have in normal coordinates
\begin{align}
t^{-1}\Delta F=&\Delta(|\nabla f|^2)-\alpha\Delta f_t
-\alpha a\Delta f-\alpha\Delta q\nonumber\\
=&2|\nabla^2f|^2+2\nabla f\cdot\nabla (\Delta f)+2R_{ij}f_if_j
-\alpha(\Delta f)_t-2\alpha h_{ij}f_{ij}-2\alpha h_{ik;i}f_k
\nonumber\\
&\qquad +\alpha\nabla H
\cdot\nabla f-\alpha a\Delta f-\alpha\Delta q.
\end{align}
Now by (2.2) and (2.4),
\begin{equation}
\Delta f=f_t+af+q-|\nabla f|^2
=\frac{1}{\alpha}\left(|\nabla f|^2-\frac{F}{t}\right)-|\nabla f|^2
=-\frac{1}{\alpha}\left((\alpha -1)|\nabla f|^2+\frac{F}{t}\right).
\end{equation}
Then by (1.4), (2.2), (2.4) and (2.6),
\begin{align}
&-\alpha(\Delta f)_t+2\nabla f\cdot\nabla(\Delta f)\nonumber\\
=&\frac{\1}{\1 t}\left(\frac{F}{t}+(\alpha -1)|\nabla f|^2\right)
+2\nabla f\cdot\nabla(f_t+af+q-|\nabla f|^2)\nonumber\\
=&\frac{F_t}{t}-\frac{F}{t^2}+2\alpha\nabla f\cdot\nabla f_t
+2\nabla f\cdot\nabla(af+q-|\nabla f|^2)-2(\alpha-1)h_{ij}f_if_j
\nonumber\\
=&\frac{F_t}{t}-\frac{F}{t^2}+2\alpha\nabla f\cdot\nabla\biggl (
\frac{1}{\alpha}|\nabla f|^2-\frac{F}{\alpha t}-af-q\biggr )
+2\nabla f\cdot\nabla(af+q-|\nabla f|^2)\nonumber\\
&\qquad -2(\alpha-1)h_{ij}f_if_j\nonumber\\
=&\frac{F_t}{t}-\frac{F}{t^2}-\frac{2}{t}\nabla f\cdot\nabla F
-2(\alpha -1)a|\nabla f|^2-2(\alpha -1)\nabla f\cdot\nabla q
-2(\alpha -1)h_{ij}f_if_j.
\end{align}
Hence by (2.5), (2.6) and (2.7),
\begin{align}
&t^{-1}(\Delta F-F_t+2\nabla f\cdot\nabla F)\nonumber\\
=&2\sum_{i,j}(f_{ij}^2-\alpha h_{ij}f_{ij})
+2R_{ij}f_if_j-(\alpha -1)a|\nabla f|^2-2(\alpha -1)\nabla f\cdot\nabla q
\nonumber\\
&\qquad -2(\alpha -1)h_{ij}f_if_j-\frac{F}{t^2}+a\frac{F}{t}
-2\alpha h_{ik;i}f_k+\alpha\nabla H\cdot\nabla f-\alpha\Delta q.
\end{align}
By (2.8) and Young's inequality, for any $0<\delta<1$
there exist constants $C_1>0$, $C_2>0$, $C_3>0$ and $C_4>0$
which depend on $a_+$ and such that
\begin{align}
&t^{-1}(\Delta F-F_t+2\nabla f\cdot\nabla F)\nonumber\\
\ge&2(1-\delta)\sum_{i,j}f_{ij}^2-C_1(|\nabla f|+|\nabla f|^2)
-C_2-\frac{F}{t^2}+a\frac{F}{t}\nonumber\\
\ge&\frac{2(1-\delta)}{n}(\Delta f)^2-C_3|\nabla f|^2
-C_4-\frac{F}{t^2}+a\frac{F}{t}.
\end{align}
By (1.4) and (A1) there exist constants $c_1>0$,
$c_2>0$, such that
\begin{equation}
\left\{\begin{aligned}
&c_1g_{ij}(x,0)\le g_{ij}(x,t)\le c_2g_{ij}(x,0)\quad\forall
x\in M, 0\le t\le T\\
&c_1g^{ij}(x,0)\le g^{ij}(x,t)\le c_2g^{ij}(x,0)\quad\forall
x\in M, 0\le t\le T.\end{aligned}\right.
\end{equation}
Let $x_0\in M$ and $r(x)=r(x,x_0)$. Then by (2.10) there exists a constant
$c_3>0$ such that
\begin{equation}
|\nabla^tr(x)|^2\le c_3\quad\mbox{ in }M\quad\forall 0\le t\le T.
\end{equation}
Let $\psi\in C^{\infty}([0,\infty))$, $0\le\psi\le 1$, be such that
$\psi (r)=1$ for all $0\le r\le 1$, $\psi (r)=0$ for all $r\ge 2$ and
$\psi''\ge -c_4$, $0\ge\psi'\ge -c_4$ and $\psi'{}^2/\psi\le c_4$ on
$[0,\infty)$ for some constant $c_4>0$. Let $R\ge 1$ and
$$
\phi (x)=\psi\left (\frac{r(x)}{R}\right).
$$
Then
\begin{equation}
\frac{|\nabla\phi|^2}{\phi}=\frac{|\nabla^tr(x)|^2}{R^2}
\cdot\frac{\psi'{}^2}{\psi}\le\frac{c_3c_4}{R^2}.
\end{equation}
By an argument similar to \cite{C} and \cite{LY} we may assume without
loss of generality that $\phi (x)$ and $r(x)$ are both smooth on $M$.
By (2.10), (2.11), the Laplacian comparison theorem \cite{SY} and an
argument similar to the proof of Lemma 1.3 of \cite{Hs1} there exists
a constant $c_5>0$ such that
\begin{align}
\Delta^tr(x)\le&c_5\quad\forall x\in M, r(x)\ge R, 0\le t\le T
\nonumber\\
\Rightarrow\qquad\Delta^t\phi=&\psi''\frac{|\nabla^tr(x)|^2}{R^2}+\psi'
\frac{\Delta^tr(x)}{R}
\ge-\frac{c_3c_4}{R^2}-\frac{c_4c_5}{R}
\ge -C_6(R^{-1}+R^{-2})
\end{align}
in $M\times [0,T]$ for some constant $C_6>0$. Suppose the function
$\phi F$ attains its maximum at $(x_0,t_0)$. If $(\phi F)(x_0,t_0)\le 0$
for any $R\ge 1$, then (2.1) holds in $M\times [0,T]$ and we are done.
Hence we may assume without loss of generality that there exists $R_1\ge 1$
such that for any $R\ge R_1$, $(\phi F)(x_0,t_0)>0$.
Then at $(x_0,t_0)$, $F_t\ge 0$,
\begin{equation}
\nabla (\phi F)=0\quad\Rightarrow\quad\phi\nabla F=-F\nabla\phi
\end{equation}
and by (2.2), (2.9), (2.12), (2.13) and (2.14),
\begin{align}
0\ge&\Delta (\phi F)\nonumber\\
\ge&\phi\Delta F+2\nabla\phi\cdot\nabla F+F\Delta\phi\nonumber\\
\ge&t_0\phi\frac{2(1-\delta)}{n}(\Delta f)^2-C_3t_0\phi
|\nabla f|^2-C_4t_0\phi-\frac{F}{t_0}\phi+a\phi F+\phi F_t
+2F\nabla f\cdot\nabla\phi\nonumber\\
&\qquad -2F\frac{|\nabla\phi|^2}{\phi}-C_6F(R^{-1}+R^{-2})\nonumber\\
\ge&t_0\phi\frac{2(1-\delta)}{n}(|\nabla f|^2-(f_t+af+q))^2-C_3t_0\phi
|\nabla f|^2-C_4t_0\phi-\frac{F}{t_0}\phi-a_-\phi F\nonumber\\
&\qquad -C_5F|\nabla f|\phi^{\frac{1}{2}}R^{-1}-C_6F(R^{-1}+R^{-2})
\end{align}
for some constant $C_5>0$. Let
$$
\sigma=\frac{\alpha^2-1}{\alpha^2},\quad
\delta_1=\frac{2(\alpha -1)}{\alpha^2}\quad
\mbox{ and }\tau=\frac{1}{2}(\sigma -\delta_1).
$$
Then by (2.15) and an argument similar to the proof on P.13--14 of \cite{CTY},
\begin{align}
0\ge&-\phi F[C_7t_0(1+\delta_1^{-1})(R^{-1}+R^{-2})+1+t_0a_-]-C_8(1+\tau^{-1})t_0^2
\nonumber\\
&\qquad
+\frac{2(1-\delta)}{n}(1-\sigma)(t_0\phi)^2(|\nabla f|^2-\alpha (f_t+af+q))^2
\nonumber\\
\ge&-\phi F[C_7t_0(1+\delta_1^{-1})(R^{-1}+R^{-2})+1+t_0a_-]
-C_8(1+\tau^{-1})t_0^2
+\frac{2(1-\delta)}{n}\alpha^{-2}(\phi F)^2
\end{align}
for some constant $C_7>0$ and $C_8>0$. Let $y=\phi F$ and $b=
C_7t_0(1+\delta_1^{-1})(R^{-1}+R^{-2})+1+t_0a_-$. Then by (2.16),
\begin{align}
&y^2-\frac{n\alpha^2b}{2(1-\delta)}y\le\frac{C_8n}{2(1-\delta)}
\alpha^2(1+\tau^{-1})t_0^2\nonumber\\
\Rightarrow\quad&\left(y-\frac{n\alpha^2b}{4(1-\delta)}\right)^2
\le\left(\frac{n\alpha^2b}{4(1-\delta)}\right)^2
+\frac{C_8n}{2(1-\delta)}\alpha^2(1+\tau^{-1})t_0^2\nonumber\\
\Rightarrow\quad&y\le\frac{n\alpha^2}{2(1-\delta)}
[C_7T(1+\delta_1^{-1})(R^{-1}+R^{-2})+1+Ta_-]
+C_9T.
\end{align}
where $C_9=\alpha\left(\frac{C_8n}{2(1-\delta)}\right)^{\frac{1}{2}}
(1+\tau^{-1})^{\frac{1}{2}}$. Then for any $x\in B_R(x_0)$,
$$
F(x,T)\le\frac{n\alpha^2}{2(1-\delta)}[C_7T(1+\delta_1^{-1})(R^{-1}+R^{-2})
+1+Ta_-]+C_9T
$$
Replacing $T$ by $t$ and letting $t\in (0,T]$ we have for any
$x\in B_R(x_0)$,
\begin{align}
&|\nabla f|^2-\alpha f_t-\alpha af-\alpha q
\le\frac{n\alpha^2}{2(1-\delta)}[C_7(1+\delta_1^{-1})(R^{-1}+R^{-2})+
\frac{1}{t}+a_-]+C_9\nonumber\\
\Rightarrow\quad&|\nabla f|^2-\alpha f_t-\alpha af-\alpha q
\le\frac{n\alpha^2}{2(1-\delta)t}(1+ta_-)
+C_9\quad\mbox{ as }R\to\infty\nonumber.
\end{align}
and the theorem follows.
\end{proof}

\begin{cor}
Let $(M, g(t))$, $0\le t\le T$, be a complete noncompact manifold with metric
$g$ evolving by the Ricci flow (1.1) and
$$
|\nabla^iRm|\le k_0\quad\text{ on }M\times [0,T]\quad\forall i=0,1
$$
for some constant $k_0>0$. Let $u$ be a positive solution of (1.5). Then
for any
$\alpha>1$ and $0<\delta<1$ there exists a constant $C_1>0$ depending on
$k_0$, $\alpha$, $\delta$, $a_+$, the space-time uniform bound of
$|q|$, $|\nabla^tq|$, $|\Delta^tq|$ such that
\begin{equation*}
\frac{|\nabla u|^2}{u^2}-\alpha\frac{u_t}{u}-\alpha a\log u-\alpha q
\le\frac{n\alpha^2}{2(1-\delta)t}(1+ta_-)+C_1\quad\text{ in }M\times (0,T].
\end{equation*}
\end{cor}

\begin{cor}
Let $(M, g)$ be a complete noncompact manifold with a fixed metric
$g$ satisfying
$$
|Rm|\le k_0\quad\text{ in }M
$$
for some constant $k_0>0$. Let $u$ be a positive solution of
$$
\Delta u-au\log u-qu=0\quad\text{ in }M
$$
where $q$ is a smooth function on $M$ with $|q|$, $|\nabla q|$
and $|\Delta q|$ uniformly bounded on $M$. Then there exists a  constant
$C_2>0$ such that
\begin{equation*}
u\le C_2\quad\mbox{ in }M\quad\mbox{ if }a<0
\end{equation*}
and there exists a  constant $C_3>0$ such that
\begin{equation*}
u\ge C_3\quad\mbox{ in }M\quad\mbox{ if }a>0.
\end{equation*}
\end{cor}
\begin{proof}
By Theorem 2.1 with $h_{ij}=0$, $\alpha=2$, $\delta=1/2$, there exists a
constant $C_1>0$ such that (2.1) holds for all $t>0$. Letting $t\to\infty$
in (2.1),
\begin{align*}
&\frac{|\nabla u|^2}{u^2}-2a\log u-2q\le 4na_-+C_1\nonumber\\
\Rightarrow\quad &-2a\log u-2q\le 4na_-+C_1\nonumber\\
\Rightarrow\quad &\left\{\begin{aligned}
&u\le e^{(4na_-+C_1+2\|q\|_{\infty})/2|a|}\quad\mbox{ in }M
\quad\mbox{ if }a<0\\
&u\ge e^{-(C_1+2\|q\|_{\infty})/2a}\qquad\quad\mbox{ in }M
\quad\mbox{ if }a>0\end{aligned}\right.
\end{align*}
and the corollary follows.
\end{proof}


text:
{\bf Problem.} There is a prime number which is a factor of every sum of three  consecutive integers. What is the number?
{\bf Level.} Level 3
{\bf Type.} Number Theory
{\bf Solution.} Let $n-1$, $n$, and $n+1$ be three consecutive integers. Their sum is $(n-1) + n + (n+1) = 3n$, which is always divisible by $\boxed{3}$, but not necessarily by any other prime.

set_name: MATH


TITLE: Numerability on subset of $\mathbb{R}$ with induced topology
QUESTION [0 upvotes]: A friend of mine has this statement, that I cannot establish wether it is true or false.
Let us consider $\mathbb{R}$ with the euclidean topology. Let us consider $A \subset \mathbb{R}$ such as the topology induced on it is the discrete topology, so $A$ is numerable.
Actually he states that the statement holds whenever $A$ is closed, but I ignore his proof.
I tried to demonstrate, recreating something like the Cantor set, that it is false proved that there exists a non separable set $A$ with that induced topology, but I cannot prove the existence of such $A$.
With discrete topology I mean that for every $x \in A$ there exists an open neighborhood of $x$, that I'll call $I(x)$ such that $I(x) \cap A = {x} $.
Any suggestions?

REPLY [1 votes]: For each $p\in A$ there is a $U_p$ such that $A \cap U_p = \{p\}$; the collection $\{U_p\}$ is uncountable. In each $U_p$ we can find an interval $V_p$ with rational endpoints containing $p$. Since also $A \cap V_p = \{p\}$, the $V_p$ are distinct and the collection $\{V_p\}$ is uncountable. But the collection of intervals $(q,r)$ with rational endpoints is in bijection with ordered pairs of rational numbers, of which there are only countably many.


TITLE: Show a ring has only one prime ideal
QUESTION [1 upvotes]: Let $p$ be an odd prime and $R = \mathbb{Z}/p^n\mathbb{Z}[x]/(x^p-1)$. I want to show that $(x-1,p)/(x^p-1)$ is the unique prime ideal of $R$. Indeed, consider the surjective homomorphism $\varphi: \mathbb{Z}/p^n\mathbb{Z}[x] \to \mathbb{Z}/p\mathbb{Z}$ given by $f(x) \mapsto f(1)  (\text{mod } p)$, we have $\ker(\varphi) = (x-1,p)$. By the first and the third isomorphism theorems, $(x-1,p)/(x^p-1)$ is a maximal ideal of $R$, and hence prime. However, I don't know how to show uniqueness. The hint says $(x-1)^p \equiv 0 (\text{mod $p$})$ in $R$, but I'm not sure how to use it.

REPLY [1 votes]: Let $P$ be a prime ideal of $R$.

You've shown that the ideal $(x-1,p)$ is a maximal ideal of $R$, hence to show that $P$ is unique, it suffices to show that $x-1\in P$ and $p\in P$.

If a product of finitely many elements of $R$ yields an element of $P$, then since $P$ is a prime ideal, it follows that at least one of the factors must be an element of $P$.

Since $p^n=0$ in $R$, we have $p^n\in P$, hence $p\in P$.

It remains to show $x-1\in P$.

Since $x^p-1=0$ in $R$, we have $x^p-1\in P$.

Applying the hint, since $p\in P$ and $x^p-1\in P$, the congruence
$$
(x-1)^p\equiv x^p-1\;(\text{mod}\;p)
$$
implies $(x-1)^p\in P$, hence $x-1\in P$.


TITLE: Where does the parameterization for a circle in 3D come from?
QUESTION [1 upvotes]: I was trying to draw circles in 3D in such a way that if I create a list of unit circles varying $\theta$ by some interval, and parameterizing those for $0≤\phi≤2\pi$, I would get a sphere.
$$x = \cos\phi\cos\theta \\
y = \sin\phi \\
z = \cos\phi\sin\theta$$
I stumbled upon a similar question and I'm wondering why it works. I understand of course why a circle would be parameterized in 2D by $(\cos\phi,\sin\phi)$ but I'm not sure why this parameterization tilts the unit circle - specifically how the z-coordinate makes sense.

REPLY [2 votes]: $2$-dimensional rotations are enough to understand all of this.

$3$-dimensional circles
You said you know that $(\cos\phi,\sin\phi)^\top$ describes a circle in $2$D and you want to consider many rotated copies of this circle so that it formes a sphere. The first thing we have to do is to go to higher dimensions. So when our circle is placed in the $xy$-plane of out $3$-dimensional space, it is given by
$$(\cos\phi,\sin\phi,0)^\top$$
where we just added a zero in the last component to show that it has no expanse into the $z$-direction.

$3$-dimensional rotation matrices
You said you know $2$-dimensional rotation matrices. Here is how to build a simple $3$-dimensional equivalent:
$$R(\phi)=\begin{pmatrix}
\cos \phi&\sin\phi\\
-\sin\phi &\cos\phi
\end{pmatrix}\quad\Rightarrow\quad R_z(\phi)=\begin{pmatrix}
\cos \phi&\sin\phi&0\\
-\sin\phi &\cos\phi&0\\
0&0&1
\end{pmatrix}.
$$
While in two dimensions there is no axis to rotate on (we rotate around points), in three dimensions we have to specify such an axis. This example is very simple. Our modified matrix just has another row/columns which shows that we completely ignore the point's $z$-coordinate. We rotate as if it where a $2$D point in the $xy$-plane. After rotation the point has the same $z$-coordinate as before the rotation (because of the $1$ in the $zz$-entry of the matrix).
So the matrix $R_z(\phi)$ rotates around the $z$-axis by an amount $\phi$. Here are the matrices which rotate around $y$- and $x$-axis respecitively (I think you can see why):
$$
R_y(\phi)=\begin{pmatrix}
\cos \phi&0&\sin\phi\\
0&1&0\\
-\sin\phi &0&\cos\phi
\end{pmatrix},\qquad R_x(\phi)=\begin{pmatrix}
1&0&0\\
0&\cos \phi&\sin\phi\\
0&-\sin\phi &\cos\phi
\end{pmatrix}.
$$

The rotated circle
Now we have our $3$D circle in the $xy$-plane around the $z$-axis. To make it rotate in such a way that it forms a sphere, we must rotate it around an axis other than $z$, e.g. the $x$-axis. So we take out circle-description $(\cos\phi,\sin\phi,0)^\top$ and we apply a rotation matrix $R_x(\theta)$:
$$
\begin{pmatrix}
1&0&0\\
0&\cos \theta&\sin\theta\\
0&-\sin\theta&\cos\theta
\end{pmatrix}\begin{pmatrix}\cos\phi\\\sin\phi\\0\end{pmatrix}
=\begin{pmatrix}\cos\phi\\\sin\phi\cos\theta\\-\sin\phi\sin\theta\end{pmatrix}.
$$
Well, this last vector might not be exactly what was given to you by the other answer, but this may be the case because they chose a circle in an other plane than the $xy$-plane or rotated around an other axis than $z$. You can play a bit with this idea to find the correct configuration. At least, this result shows structural similarities to the one given to you in the other answer.


TITLE: When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole-number quot...
QUESTION [2 upvotes]: When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole-number quotient with a remainder of 1.  How many distinct values are possible for this three-digit dividend?

I think that the answer to this the number of 3 digit numbers ending with 1 that are not primes, but I don't have a method to find that in a reasonable amount of time.

REPLY [1 votes]: As the dividend is a three digit number ending with $2$, it is:
$$10d+2 \quad (10 \leqslant d \leqslant 99)$$
I assume you do not include $1$ as a satisfying one-digit divisor, since it would be weird to say remainder $1$ when divided by $1$ (if this isn't the case, any three digit number ending with $2$ would work and the answer would simply be $99-9=90$). We now take the remaining single-digit numbers into consideration.
Any three digit dividend satisfying you properties would show that $10d+1$ is divisible by some single digit number from $2$ to $9$. Clearly, $2$, $4$, $6$ or $8$ cannot divide $10d+1$ as the latter is an odd number. Moreover, as $5$ divides $10d$, $5$ cannot divide $10d+1$. The only remaining single digit numbers are $3$,$7$ and $9$. If $9$ divides $10d+1$, then $3$ would divide $10d+1$ too, since $9=3^2$. Thus, any number satisfying our properties is of the form $10d+1$ such that atleast one of $3$ or $7$ divide $10d+1$.
If $3$ divides $10d+1$, we have:
$$10d+1=3(3d)+(d+1) \implies d+1 \text{ is divisible by $3$}$$
This means that $d$ leaves remainder $2$ when divided by $3$. The possibilities are $d=11,14,\ldots,98$ which is $30$ possibilities.
If $7$ divides $10d+1$ if and only if $7$ divides $5(10d+1)=50d+5$. We have:
$$50d+5=7(7d)+(d+5) \implies d+5 \text{ is divisible by $7$}$$
This means that $d$ leaves remainder $2$ when divided by $7$. The possibilities are $d=16,23,\ldots,93$ which is $12$ possibilities.
However, we cannot simply add $30+12=42$ and say that $42$ is the answer. This is because if $10d+1$ is divisible by both $3$ and $7$, we would have counted this number twice. Thus, we have to remove all the numbers divisible by $3$ and $7$.
For a number $10d+1$ to be divisible by $3$ and $7$, we need $d$ to leave remainder $2$ when divided by $3$ and $7$. This means $d-2$ leaves remainder $0$ when divided by $3$ or $7$. This shows that $21$ divides $d-2$. Thus, $d-2$ can take values $21,42,63,84$ and thus, $d$ can take values $23,44,65,86$. Since these four values were counted twice, we remove them once. This means we remove $4$ possibilities.
The final answer is $30+12-4=38$ such unique three digit dividends.
public class Main{
public static void main(String[] args){
    // S is the number of unique three digit dividends
    int S=0;
    // i is a possible three digit dividend
    for(int i=100;i<1000;i++){
        if(i%10==2){
            // j is a possible one digit divisor
            for(int j=2;j<10;j++){
                if(i%j==1){
                    S+=1; break;
                    }
                }
            }
        }
        System.out.println(S);
    }
}

The above is a simple (and naive) Java program to verify our answer. You can check that the output is:
38

verifying that our answer is indeed right!


TITLE: Example of an $\omega_1$ decreasing chain of dense semicontinua?
QUESTION [2 upvotes]: In his well-known paper Bellamy constructs an indecomposable continua with exactly two composants. The setup is as follows:
We have an inverse-system $\{X(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$ of metric indecomposable continua and retractions. For each $X(\alpha)$ a composant $C(\alpha) \subset X(\alpha)$ is specified and each $C(\alpha)$ maps into $C(\beta) $.
The inverse limit $X$ has exactly two composants. The first is the union
$\bigcup\{X(\beta): \beta < \omega_1\}$ where we identify $X(\beta)$ with the set of sequences $(x_\alpha) \in X$ with $x_\beta = x_{\beta+1} = x_{\beta+2} = \ldots . $ The second composant is the inverse limit $\{C(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$.
Observe there is no reason a priori for the second composant to be nonempty. However I do not believe an example is know.
My question is an easier one. Can you think of an example of a metric continuum $M$ and a $\omega_1$-indexed decreasing nest of dense semicontinua with empty intersection? We call the set $S \subset M$ a semicontinuum to mean for each $x,y \in S$ there exists a continuum $K$ with $\{x,y\} \subset K \subset S$.
If the second composant was empty the family $\{f^\alpha_0(C(\alpha)): \alpha < \omega_1\}$ would be such a nest for $M = X_0$.
If we index by $\omega$ instead an example is easy to come by. Let $M$ be the unit disc and $Q = \{q_1,q_2, \ldots\}$  and enumeration of the rational points on the boundary. Let $S(n)$ be formed by drawing the straight line segment from each element of $\{q_n,q_{n+1}, \ldots\}$ to each rational point of $(0,1/n) \times \{0\}$. Then add in $(0,1/n) \times \{0\}$ itself to make the space a semicontinuum.
Indexing by $\omega_1$ must somehow get around the fact that any $\omega_1$ decreasing nest of closed subsets of a metric continuum eventually stabilizes.
It feels like this would be easier if we assume the Continuum Hypothesis.

REPLY [1 votes]: I am not assuming any CH.

Let $\ \Omega\ := \{\alpha: \alpha<\omega_1\},\ $ and $\ i:\Omega\rightarrow [0;1]\ $ be injective and such that
$\ \Gamma\ :=\ i(\Omega)\ $ is condensed in $\ [0;1].\ $ More generally,
let $\ \Gamma_\alpha\ :=\ i([0;\alpha)).\ $ Also let $\ M:=[0;1]^2.\ $
Then
$$ S_\alpha\,\ :=\,\ (\Gamma\setminus\Gamma_\alpha)\times[0;1]
\ \cup\ [0;1]\times(\Gamma\setminus\Gamma_\alpha) $$
for $\ \alpha<\omega_1,\ $ is the required decreasing $\omega_1$-sequence which has empty intersection.

Condensed means that every non-empty open subset of $\ [0;1]\ $ has an uncountable intersection with $\ \Gamma$.


TITLE: Topologies for graphs
QUESTION [0 upvotes]: Some of the basic definitions in Graph theory made me wonder if there is by any chance a way to give a graph $G$ a topology, such that these definitions can be understood as versions of analogous definitions given in topology. For example, is there a topology for $G$ such that the definition of "connected graph" can be understood as equivalent to $G$ being a connected (or arc-connected) topological space?

REPLY [2 votes]: There is a notion of topological realization of a graph, basically given by taking points in $\Bbb R^n$ and gluing copies of the unit interval $[0,1]$ to it, wherever the graph has an edge. This results in a metric space.
Some results can be translated between this topological realization and the graph. For example the graph is connected, if and only if the realization is (path)-connected. Similarly the graph has a cycle, if and only if there is an embedding of a circle $\Bbb S^1$ into the realization etc.


TITLE: Is this probability inequality true?
QUESTION [0 upvotes]: This question may be simple, though I'm not managing to find an answer. Let $X$ and $Y$ be two dependent random vectors in in $\mathbb{R}^d$, with joint probability density $\mu(x,y)$ (with respect to the Lebesgue measure). For any subset $A \subset \mathbb{R}^d$ and vector $t \in \mathbb{R}^d$, define
$$
A+t=\{x+t=(x_1+t_1, \ldots, x_d+t_d): x \in A\}.
$$
Is it true that
$$
P(X-Y \in A, Y \in B) \leq \sup_{t \in B}P(X \in A+t)
$$
where $A$ and $B$ are measurable proper subset of $\mathbb{R}^d$? The inequality is trivially true if $X$ and $Y$ are independent:
$$
P(X-Y \in A, Y \in B) =\int_B \left[\int_{A+y} \mu(x|y)dx\right] \mu(y)dy\\
=\int_B \left[\int_{A+y} \mu(x)dx\right] \mu(y)dy\\
=\int_B P(X\in A+y)\mu(y)dy\\
\leq \sup_{y \in B}P(X\in A+y)
$$
where $\mu(x|y)$, $\mu(x)$ and $\mu(y)$ are the conditional density of $X$ given $Y=y$, the marginal density of $X$ and the marginal density of $Y$, respectively. What about the case where $X$ and $Y$ are dependent (i.e. $\mu(x|y)\neq \mu(x)$)?
EDIT 1 Here is my attempt:
$$
\mathbb{P}(X-Y \in A, Y\in B)= \int_{B}\int_{A+y}\mu(x,y)dxdy\\
\leq \sup_{t\in B}\int_{B}\int_{A+t}\mu(x,y)dxdy\\
=\sup_{t\in B}\int_{A+t}\int_B\mu(x,y)dydx\\
\leq \sup _{t\in B}\int_{A+t}\mu(x)dx\\
=\sup _{t\in B}\mathbb{P}(X \in A+t)
$$
but I have doubts about the second and third lines, I'm not sure they're correct. I pass from the third to the fourth line by using the fact that $\mu(x,y)$ is nonnegative and
$$
\int_B\mu(x,y)dy \leq \int_{\mathbb{R}^d}\mu(x,y)dy=\mu(x).
$$
EDIT 2 Are there conditions (different from independence) under which the inequaliy holds true? In the somewhat patological case $X=Y$ with probability one, as highlighted in the answer below, the inequality may not be true. But what if $X \neq Y$ with probability 1? Are there conditions under which, in such an istance, the inequality is satisfied?

REPLY [3 votes]: This is false:
Let the probability distribution be $P(X = 0, Y = 0) = P(X = 1, Y = 1) = \frac{1}{2}$ and let $A = \{ 0\}$, $B = \{ 0, 1\}$. Then the left-hand side of your inequality is $1$ while  the right-hand side is $\frac{1}{2}$.
If you want your densities to be continuous just convolve with some smooth highly concentrated function (both $X, Y$ and $A, B$).
Actually, if $X=Y$ with probability one and $A = \{0\}$, $B = \mathbb{R}$ then your inequality is false as long as distribution of $X$ is non-constant. For example, if $X$ is normal then left-hand side is $1$ while right-hand side is zero.


text:
{\bf Problem.} Let $\alpha$ and $\beta$ be conjugate complex numbers such that $\frac{\alpha}{\beta^2}$ is a real number and $|\alpha - \beta| = 2 \sqrt{3}.$  Find $|\alpha|.$
{\bf Level.} Level 4
{\bf Type.} Intermediate Algebra
{\bf Solution.} Let $\alpha = x + yi$ and $\beta = x - yi.$  Then from $|\alpha - \beta| = 2 \sqrt{3},$ $2|y| = 2 \sqrt{3},$ so $|y| = \sqrt{3}.$

Now, $\frac{\alpha}{\beta^2}$ is real.  Since $\alpha$ and $\beta$ are complex conjugates, $\alpha^2 \beta^2$ is real, so $\frac{\alpha}{\beta^2} \cdot \alpha^2 \beta^2 = \alpha^3$ is real.  But
\[\alpha^3 = (x + yi)^3 = (x^3 - 3xy^2) + (3x^2 y - y^3) i,\]so $3x^2 y - y^3 = 0,$ or $y(3x^2 - y^2) = 0.$  Since $|y| = \sqrt{3},$ $y \neq 0,$ so $3x^2 = y^2 = 3,$ which means $x^2 = 1.$  Therefore,
\[|\alpha| = \sqrt{x^2 + y^2} = \sqrt{1 + 3} = \boxed{2}.\]


text:
\section{Additional Notation}

A controller is a mapping $\pi: \mathcal{X}\times \{0,\ldots H-1\} \rightarrow
\mathcal{U}$.   Given a instantaneous cost function $c:\mathcal{X} \times
\mathcal{U}\rightarrow \R$, we define the cost (or the ``cost-to-go'' ) of a policy as:
\[
J^\pi(x;c,W) =  \Exp \left[ \sum_{h=0}^{H-1} c(x_h, u_h)  \Big|
  \pi, x_0=x, W
\right]
\label{eq: cost}
\]
where the expectation is under trajectories sampled under $\pi$
starting from $x_0$ in model parameterized by $W$.
The ``cost-to-go'' at state $x$ at time $h \in \{0,\ldots H-1\} $ is denoted by:
\[
J^\pi_h(x;c,W)  = \Exp \left[ \sum_{\ell=h}^{H-1} c(x_\ell, u_\ell)  \Big|
  \pi, x_h=x
\right]  .
\]

When clear from context, we let the episode $t$ index the
policy, e.g. we write $J^{t}(x;c)$ to refer to $J^{\pi^t}(x,c)$.
Subscripts refer to
the timestep within an episode and superscripts index the episode
itself, i.e. $\phi^t_h$ will refer to the random vector which is the
observed features during timestep $h$ within episode $t$.  We let
$\mathcal{H}_t$ denote the history up to the beginning of episode $t$.

Also, $\|x\|_M^2 := x^\top M x$ for  a vector $x$ and a matrix $M$.


text:
\begin{document}

\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{rem}[thm]{Remark}
\newtheorem{coro}[thm]{Corollary}
\newtheorem{conj}[thm]{Conjecture}
\newtheorem{de}[thm]{Definition}
\newtheorem{hyp}[thm]{Hypothesis}

\newtheorem{nota}[thm]{Notation}
\newtheorem{ex}[thm]{Example}
\newtheorem{proc}[thm]{Procedure}

\title{Rootless pairs of $EE_8$-lattices}
\author{Robert L.~Griess, Jr.}
\address[R.\,L.~Griess, Jr.]{
Department of Mathematics, University of Michigan,
Ann Arbor, MI 48109-1043,
USA}
\email{rlg@umich.edu}

\author{Ching Hung Lam}
\address[C.\,H. Lam]{
Department of Mathematics and
National Center for Theoretical Sciences,
National Cheng Kung University,
Tainan, Taiwan 701}

\email{chlam@mail.ncku.edu.tw}

\begin{abstract}
We describe a classification of
pairs $M, N$ of lattices isometric to $EE_8  := \sqrt 2  E_8$ such that the lattice $M + N$ is integral and rootless and such that the dihedral group associated to them has order at most 12. It turns out that most of these pairs may be embedded in the Leech lattice.  Complete proofs will appear in another article.  This theory of integral lattices has connections to vertex operator algebra theory and moonshine.
\end{abstract}

\maketitle

\section{Introduction}
Associated to a sublattice $S$ of an integral lattice $L$ is an orthogonal
transformation of order 2 on Euclidean space (defined to be $-1$ on $S$ and 1 on the annihilator of $S$). This involution leaves  $L$ invariant if $S$ satisfies the \textit{RSSD condition} ($2L \le S + ann(S )$) \cite{bwy, cbwy}.
An example of such an $S$ is $\sqrt 2 U$, where $U$ is an integral unimodular lattice, such as $EE_8:=\sqrt 2 E_8$.
A pair of RSSD sublattices gives a dihedral group in the isometry group of $L$.

In this article, we describe our classification of  pairs of $EE_8$-lattices
which span an integral and rootless lattice and whose associated involutions
(isometries of order 2) generate a dihedral group of order at most
12.
There are exactly $11$ such pairs.  Most can be realized inside the Leech lattice, which we denote by $\Lambda$.
We refer to an upcoming article  \cite{gl} for proofs and details.

This work may be considered purely  as a study of positive definite integral lattices.
Our real motivation, however, is the evolving theory of vertex operator algebras (VOA) and their automorphism groups.\footnote{AMS classifications: 20C10 (integral representations of finite groups); 11H55 (quadratic forms); 11F22 (relations to Lie algebras and finite groups).}

\section{Motivation from VOA theory}
The primary connection between the Monster and vertex operator
algebras was established in \cite{flm}.
In \cite{Mi}, Miyamoto gave a theory for defining involutions on a VOA.
Suppose that $(V, Y, \o, \allone)$ is a VOA. A degree $2$ element $e\in V$ is called a \textit{rational conformal vector of central charge $1/2$} if the
components of $Y(e,z)=\sum_{n\in \ZZ} e_n z^{-n-1}$
(The $e_i$ are in $End(V)$)
together with $Id_V$ span a Virasoro Lie algebra with central charge $\half$ and the subVOA of $V$ generated by $w$ is a simple VOA, so is isomorphic to simple Virasoro VOA $L(\half, 0)$.

The simple Virasoro VOA $L(\half, 0)$ has only three irreducible modules,
$L(\half, 0)$, $L(\half, \half )$, $L(\half, \frac 1{16})$, and the fusion rules give two formulas for an involution on $V$.

Since $L(\frac{1}{2},0)$ is rational, i.e., all $L(\frac{1}{2},0)$-modules are completely reducible, we can decompose $V:=V_0 \oplus V_{\half } \oplus V_{\sixteenth}$ into the direct sum of $V_c$, where
$V_c$ denotes the sum of all $L(\frac{1}{2},0)$-submodules of $V$ isomorphic to $L(\half, c)$.

\medskip

(1)  Define a linear map  $\tau_e:V \to V$ to be $1$ on $V_0 \oplus V_\half$ and $-1$ on $V_\sixteenth$. Miyamoto\,\cite{Mi} showed that $\tau_e$ is always an automorphism of $V$. This automorphism is called \textit{a Miyamoto involution} if $\tau_e$ has order $2$.

\medskip

(2) If $\tau_e=Id_V$, i.e., $V_{\sixteenth}=0$, then the linear map $\sigma_e$ defined by $1$ on $V_0$ and $-1$ on $V_\half$ is also an automorphism. It is called \textit{a Miyamoto involution of $\sigma$-type}.

\medskip

These involutions have been studied a lot.
Miyamoto proved \cite{Mi1}
the important result
that there is a bijection between the conjugacy class of $2A$
involutions in the Monster simple group and conformal vectors of
central charge $\half$ in the moonshine vertex operator algebra
$V^\natural$.

\subsection{Lattice type vertex operator algebras}
Recall that a lattice VOA $V_L$ associated with an even positive definite lattice $L$ is given by $V_L=S(\hat{\mathfrak{h}}^-) \otimes \CC [L]$, as a vector space, where
 $\mathfrak{h}:=\CC \otimes L$, $\hat{\mathfrak{h}}^-= \oplus_{n=1}^\infty t^{-n} \otimes \mathfrak{h}$,  $S(\hat{\mathfrak{h}}^-)$ is the symmetric algebra of
$\hat{\mathfrak{h}}^-$ and $\CC[L]=\mathrm{span}_\CC\{ e^\a| \a\in L\}$ is the group algebra of $L$.

Let $V_L$ be a lattice VOA.  It is common to study an involution on
$V$ which is a ``lift'' of the $-1$ map on $L$.
The standard lift of $-1$, called $\theta$, acts as $-1$ on $\hat{\mathfrak{h}}^-$ and takes
 $e^\a$ to $e^{-\a}$, for all $\a$.
The fixed point subVOA of $\theta$ is denoted $V_L^+$.

For VOAs of lattice type $V_L$ or $V_L^+$, there are two known explicit general formulas for rational conformal vectors of central charge $1/2$.

\medskip

\noindent {\it First formula: }

When $\a$ is a norm 4 vector of $L$,
\begin{equation}\labtt{aa1cv}
e=\sixteenth (t^{-1}\otimes \a)^2 \pm \fourth (e^\a +e^{-\a})
\end{equation}
is a rational conformal vector of central charge $1/2$.

\medskip

\noindent {\it Second formula: }

When $E$ is a sublattice of $L$, $E\cong EE_8$,
\begin{equation}\labtt{ee8cv}
e=\frac 18 q +\frac 1{32}\sum_{\a \in E/\{\pm 1\} } \varphi (\a) (e^\a + e^{-\a})
\end{equation}
is also  a rational conformal vector of central charge $1/2$,
(here $q$ is a sum $\sum_i  (t^{-1}\otimes u_i)^2$, where $u_i$ is an orthonormal basis of  $\CC\otimes E$;
$\varphi$ is a homomorphism $E\rightarrow \{\pm 1\}$).

\medskip

This second formula explains our interest in $EE_8$.

\subsection{Why rootless?}

If $L$ is rootless, $V_L^+$ has zero degree 1 term and so (by general VOA theory) the degree 2 part is a commutative algebra (generally nonassociative).  In this case, the automorphism group of $V_L^+$ is finite (not generally true for VOAs). Moreover, it is conjectured \cite{lsy} that the two kinds of conformal vectors in \refpp{aa1cv} and \refpp{ee8cv} will exhaust all the rational conformal vectors of
central charge $1/2$ in $V_L^+$ if $L$ is rootless. This conjecture was proved when $L$
is a $\sqrt{2}$ times a root lattice or the Leech lattice
\cite{lsy,ls} but it is still open if $L$ is a general rootless
lattice.   The results of this paper could help settle this
conjecture.

\subsection{Why dihedral groups of order at most 12? }
We may consider pairs of Miyamoto involutions and what they generate.  For a general VOA, this is an arbitrary dihedral group.
In the case of $V_L^+$ when $L$ is rootless,  the possibilities are limited to dihedral of order at most 12, by a recent theorem of Shinya Sakuma \cite{Sa}.  In fact, his
hypotheses are more general: the VOA need not be lattice type, but it should be defined over $\RR$ with a positive definite form, have no negative terms, the degree 0 component should be 1-dimensional, and degree 1 component should be 0-dimensional.  His proof is quite technical.

Even without the Sakuma theorem, one knows that
the restriction to
dihedral groups of order at most $12$ is relevant to a study of sporadic groups.  Consider the 2A class of involutions in the Monster \cite{gms} and classes within the Baby Monster, $Fi_{24}$, etc., discussed in \cite{gn}.

\section{The classification of integral, rootless $EE_8$-pairs}

In this section, we shall discuss our classification of $EE_8$-pairs. The following is our main theorem.

\begin{thm} Suppose that in Euclidean space, the integral lattice $L$ is the sum $M+N$ of sublattices, where $M\cong N \cong EE_8$, and $L$ is rootless. Let $t_M$ be the involution defined by $M$ and let $t_N$ be the  involution defined by $N$.  Assume that the dihedral group $\la t_M, t_N \ra$ has order at most 12.
Then the possibilities
for $L$ are
listed in Table \ref{NREE8SUM} and all these possibilities exist.  The lattices in Table \ref{NREE8SUM} are unique up to isometry of pairs $M, N$.  Except for $DIH_4(15)$, all of them embed as sublattices of the Leech lattice.
\end{thm}

\def\dih#1#2{DIH_{#1}(#2)}

\begin{table}[bht]
\caption{ \bf  NREE8SUMs: integral rootless lattices which are sums of
$EE_8$s}
\begin{center}
\begin{tabular}{|c|c|l|c|c|}
\hline
 Name & $\<t_M,t_N\>$ &Isometry type of $L$\,(contains)& $\dg L$ & In\,Leech?\cr
 \hline \hline
 $\dih{4}{12}$ & $Dih_4$ & $\geq  DD_4^{\perp 3}$ & $1^4 2^6 4^2$ & Yes\cr
 \hline
 $\dih{4}{14}$ & $Dih_4$ & $\geq AA_1^{\perp 2} \perp DD_6^{\perp 2}$ & $1^4 2^8 4^2$  &Yes\cr
 \hline
 $\dih{4}{15}$ & $Dih_4$ & $\geq AA_1\perp EE_7^{\perp 2}$ & $1^22^{14}  $  & No\cr
 \hline $\dih{4}{16}$ & $Dih_4$ & $\cong EE_8\perp EE_8$ & $2^{16}$ &   Yes\cr
 \hline $\dih{6}{14}$ & $Dih_6$ & $\geq AA_2\perp A_2\otimes E_6 $ & $1^7 3^3 6^2$ &Yes\cr
 \hline
 $\dih{6}{16}$ & $Dih_6$ & $\cong A_2\otimes E_8 $ & $1^8 3^8 $ &Yes\cr
 \hline
 $\dih{8}{15}$ & $Dih_8$ & $ \geq AA_1^{\perp 7}\perp EE_8$     & $1^{10}4^5 $&   Yes\cr
 \hline
 $\dih{8}{16, DD_4}$ & $Dih_8$ & $\geq DD_4^{\perp 2} \perp  EE_8$  & $1^8 2^4 4^4$  &Yes\cr
 \hline
 $\dih{8}{16, 0}$ & $Dih_8$ & $\cong BW_{16} $ & $1^8 2^8$ &Yes\cr
 \hline $\dih{10}{16}$ & $Dih_{10}$ & $\geq A_4\otimes A_4 $ & $1^{12}5^4$  &Yes\cr
 \hline $\dih{12}{16}$ & $Dih_{12}$ & $\geq AA_2\perp AA_2 $ & &\cr
 & & $\ \perp A_2\otimes E_6 $ & $1^{12}6^4 $  & Yes\cr
 \hline \hline
\end{tabular}
\medskip

$X^{\perp n}$ denotes the orthogonal sum of $n$ copies of the lattice $X$.
\end{center}
\label{NREE8SUM}
\end{table}

\begin{nota}
$A_1, \dots, E_8$ denote the root lattices of the corresponding type;

$AA_1$, $\dots$, $EE_8$ denote the lattices isometric to $\sqrt{2}$ times the lattices $A_1, \dots, E_8$;

$\dg L = \dual L / L$ is the discriminant group of $L$;

$\dih {r}s$ means a pair of $EE_8$ lattices $(M, N)$ such that $t_M$, $t_N$ generates a dihedral group of order $r$ and $rank(M+N)=s$;

$DIH_8(16, P)$ means $DIH_8(16)$ and that $P= ann_M(N)\cong ann_N(M)$.
\end{nota}

\subsection{Some highlights of the proof}\labtt{hlight}
Next, we shall discuss the main steps for the classification.
We go through cases $|t_Mt_N| = 2,3,4,5,6$.  Our respective analyses are called {\it $DIH_4$-theory, $DIH_6$-theory,
$DIH_{8}$-theory, $DIH_{10}$-theory,  $DIH_{12}$-theory.  }

In $\la t_M, t_N \ra$, let $g:=t_Mt_N$.   Then $\ZZ [\la g \ra]$ acts on $L$ and it acts on $J:=ann_L(Fix_L(g))$, where $Fix_L(g)$ denotes the set of all fixed points of $g$ in $L$.  The action is that of as a ring of integers in a number field when $|g|$ is prime.

The main idea is to determine possibilities for $Fix_L(g)$, $J$, $ann_M(N)$, $ann_N(M)$ and related sublattices.
Exhaustive case by case analysis gives a list of candidates.  In all cases, the candidates are proved unique, given certain things we deduce about their sublattices.

\medskip

First, we observe that $\la t_M, t_N \ra$ acts faithfully on $L$ and leaves invariant  $Fix_L(g)=M\cap N$.

\medskip

\noindent \textit{Case:} $|g|=2,3,5$.

In these cases,  we  determine all sublattices of $E_8$ which are direct summands and whose discriminant group  is an elementary abelian $p$-group for $p=2, 3, 5$.  Exhaustive case by case analysis gives a list of candidates. It turns out  $M\cap N$ is isometric to $\sqrt 2$ times one of these lattices.  In fact,   $M\cap N\cong 0, AA_1, AA_1\perp AA_1$, or $DD_4$ if $|g|=2$; $M\cap N\cong 0$ or $AA_2$ if $|g|=3$; and $M\cap N=0$ if $|g|=5$.

\medskip
Given $M\cap N$, we then analyse $J$ and its sublattices.

\medskip

When $|g|=2$, $\la t_M, t_N \ra$ is a four-group. Then  we have $M\cap J=ann_M(N)$, $N\cap J=ann_N(M)$ and $M\cap N \perp ann_M(N)\perp ann_N(M)$ is an index $2$ sublattice of $L$.  In this case, the isometry type of $L$ is uniquely determined by $M\cap N$.

\medskip

When $|g|=3$, we consider the $\ZZ [\la g \ra]$-submodule $K$ generated by $M\cap J$.  Then $K$ is a sublattice of $J$ and $K$ is isomorphic as a lattice to $A_2\otimes \frac 1{\sqrt 2}(M\cap J)$.  The possibilities for $M\cap J$ in this case are $EE6$ or $EE8$. Again, the isometry type of $L$  is uniquely determined by $M\cap N$.

\medskip

When $|g|=5$, $M\cap N=0$. We show that for any norm $4$ vector $\a\in L$, the $\ZZ [\la g \ra]$-submodule generated by $\a$ is isomorphic as a lattice to $AA_4$. In fact, we show that $L=M+N$ contains a sublattice $U$ isometric to the orthogonal sum of $4$ copies of $AA_4$ such that $M\cap U\cong N\cap U \cong AA_1^8$. The uniqueness of $L$ is then shown by explicit glueings.

\medskip

\noindent \textit{Case:} $|g|=4,6$.

In each of these cases, let $h:=g^2$. Then $(M, Mh)$ and $(N, Nh)$ are $EE_8$ pairs whose associated dihedral group has order $4$ or $6$. We then use the results for $Dih_4$ and $Dih_6$ to deduce the structures of $L$. It turns out that there is only one possible case for $|g|=6$ but $3$ different cases for $|g|=4$.

A proof that the candidates are really rootless is made easier by a magic tool.  Most candidates can be embedded in the Leech lattice by direct constructions (and use of a uniqueness result).    Since Leech is rootless, so is our candidate $L$.  See the last section for examples of embeddings.

\section{ Glauberman-Norton subgroups}

In \cite{gn},  Glauberman-Norton computed the centralizers in the Monster of a pair of 2A-involutions.
We hope to study these subgroups by promoting pairs $M, N$  of lattices from our list to involutions on the Moonshine VOA.
An interesting point is that the Glauberman-Norton subgroups  have ``Levi factors'' which look like index $2$ subgroups of a direct product of a pair of Weyl groups modulo centers.  Our theory may give a context for this by viewing such a subgroup as acting on the Moonshine VOA.   At this time, we can see actions of direct products of Weyl groups on large subVOAs of the Moonshine VOA,
in the following way.
The Moonshine VOA is a twisted lattice type VOA based on the Leech lattice.  The Leech lattice contains sublattices of types $AA_1$ or $EE_8$ which will give a concrete realization of each pair of $2A$-involutions in the Glauberman-Norton theory (by Formula \refpp{aa1cv} or Formula \refpp{ee8cv}).  Since there are examples of tensor products of lattices contained in our   $M+N$ when $|g|=3$ or $5$ \refpp{hlight}, we have actions of a finite group with quotient a direct product of Weyl groups
on a subVOA of lattice type.

\section{Examples of some pairs $M, N$ in the Leech lattice}

In this section,
we shall give a few examples of $EE_8$-pairs $M, N$ in the Leech lattice.
Because of space limitations, these are just for the $DIH_6$-theory.
First, we recall some notation for describing the Leech lattice and its isometries \cite{cs, Gr12}.

\begin{nota}
Arrange  the set $\Omega=\{1, \dots, 24\}$ into a $4\times 6$ array such that the six columns forms a sextet.
For each codeword in the Golay code $\mathcal{G}$, $0$ and $1$ are marked by a
blanked and non-blanked space, respectively, at the corresponding
positions in the array. For example, $(1^8 0^{16})$ is denoted by
the array
\[
\begin{array}{|cc|cc|cc|}
\hline *& *&\ &\ &\ &\ \\
* & *&\ &\ &\ &\ \\
* & *&\ &\ &\ &\ \\
* & *&\ &\ &\ &\ \\ \hline
\end{array}
\]

By the same notation, every vector in the Leech lattice $\Lambda$ can be written as the form
\[
X=\frac{1}{\sqrt{8}}\left[ X_1 X_2 X_3 X_4 X_5 X_6\right],\quad \text{
juxtaposition of column vectors}.
\]
For example,
\[
\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
 \hline 2  & 2& 0  & 0 & 0 & 0 \\
 2 &  2 &  0&  0& 0& 0\\
 2 &  2& 0& 0&  0 & 0 \\
 2 &  2& 0&  0& 0 & 0\\ \hline
\end{array}.
\]
\end{nota}

\begin{de}\labtt{defofxi}

Let
\[
A= \frac{1}2 \left [
\begin{matrix}
-1 & 1& 1& 1\\
1 & -1& 1& 1\\
1 & 1& -1& 1\\
1 & 1& 1& -1
\end{matrix}\right].
\]

Define a linear map $\xi:\Lambda \to \Lambda$  by  $ X\xi = AXD$, where
\[
X=\frac{1}{\sqrt{8}}\left[ X_1 X_2 X_3 X_4 X_5 X_6\right]
\]
is a vector in the Leech lattice $\Lambda$ and $D$ is the diagonal matrix
\[
\begin{pmatrix}
-1&0&0&0&0&0\\
0& 1&0&0&0&0\\
0& 0& 1&0&0&0\\
0&0&0 &1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0& 1\\
\end{pmatrix}.
\]
Recall that $\xi$ defines an isometry of $\Lambda$ (cf. \cite[p. 288]{cs} and \cite[p. 97]{Gr12}).

\end{de}

\begin{nota}
 Let $\mathcal{O}$ be an octad. Denote by $E(\mathcal{O})$ the sublattice of the Leech lattice $\Lambda$ which is supported on $\mathcal{O}$. Then $E(\mathcal{O})\cong EE_8$.
\end{nota}

Here are examples for the $DIH_6$ cases.

\subsubsection{$ |g|=3$.}\labtt{g=3} In this case, $M\cap N =0$ or
$AA_2$.

\medskip
\noindent $\bf \dih{6}{16}$

Let
\[
\mathcal{O}_1=
\begin{array}{|cc|cc|cc|}
\hline  * & * &  &  &  & \\
* & * &\ &\ &\ &\ \\
* & * &\ &\ &\ &\ \\
* & * &\ &\ &\ &\ \\ \hline
\end{array}
\]
be an octad and $M:=E(\mathcal{O}_1)\cong EE_8$. We choose a basis
$\{\b_1,\dots, \b_8\}$ of $M$, where
\begin{align*}
\b_1= \frac{1}{\sqrt{8}}\,
\begin{array}{|cc|cc|cc|}
\hline 4 & -4& 0  & 0 & 0 & 0 \\
0 &  0&  0&  0& 0& 0\\
 0 & 0&  0& 0&  0 & 0 \\
 0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, &&
 \b_2= \frac{1}{\sqrt{8}}\,
 \begin{array}{|rr|rr|rr|}
\hline 0  & 4& 0  & 0 & 0 & 0 \\
      -4 &  0&  0&  0& 0& 0\\
       0 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, \\

\b_3= \frac{1}{\sqrt{8}}\,
 \begin{array}{|rr|rr|rr|}
\hline 0  & 0& 0  & 0 & 0 & 0 \\
       4 &  -4&  0&  0& 0& 0\\
       0 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, &&
\b_4=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0  & 0& 0  & 0 & 0 & 0 \\
       0&  4 &  0&  0& 0& 0\\
       -4 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array},
\\

\b_5=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0  & 0& 0  & 0 & 0 & 0 \\
       0&  0 &  0&  0& 0& 0\\
       4 & -4&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array},  &&
\b_6=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0  & 0& 0  & 0 & 0 & 0 \\
       0 & 0&  0&  0& 0& 0\\
       0 & 4&  0& 0&  0 & 0 \\
       -4&  0& 0&  0& 0 & 0\\ \hline
\end{array}, \\

\b_7=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline -4  & -4& 0  & 0 & 0 & 0 \\
       0&  0 &  0&  0& 0& 0\\
       0 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, &&
\b_8=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline\ 2  &\ 2& 0  & 0 & 0 & 0 \\
       2 &  2 &  0&  0& 0& 0\\
       2 &  2& 0& 0&  0 & 0 \\
       2 &  2& 0&  0& 0 & 0\\ \hline
\end{array}.
\end{align*}

Let $N$ be the lattice generated by the vectors
\begin{align*}
&\a_1=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline\ \ 2 & -2&\ \ 2  & -2 &\ \ 2 &\ \, 0 \\
\ 0 & \ 0&\ 0&\ 0&\ 0&  2\\
\ 0 &  \ 0&\ 0&\ 0&\ 0 & 2 \\
\ 0 & \ 0&\ 0&\ 0&\ 0 &\ \, 2\\ \hline
\end{array}, &&
 \a_2=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0 &\ 2&\ 0&\ 2&  0 &\ 2 \\
      -2 &\ 0& -2&\ 0&\ 0 &\ -2 \\
     \ 0 &\ 0&\ 0&\ 0&\ 2 &\ -2\\
     \ 0 &\ 0&\ 0&\ 0&\ 0 &\ 0 \\ \hline
\end{array},\\

&\a_3=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline \ 0 &\  0&\ 0  &\ 0 &\ 0 & -2 \\
     \ \ 2 & -2 & \ \,2 & -2 & 2  &\ 0\\
       \ 0 &\ 0 &\ 0  &\ 0 & -2  &  0\\
       \ 0 &\ 0 &\ 0  &\ 0 & 2  &\ 0 \\ \hline
\end{array}, &
&\a_4=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline  \ 0 &\ 0  &   \ 0&    0&\ 0 &\ 2 \\
        \ 0 &\ \ 2&   \ 0&\ \ 2&  -2&\ \ 2\\
         -2 &\   0&    -2&\   0&\ 0 & \ 0\\
        \ 0 &  \ 0&   \ 0& \  0&\ 0 & -2 \\ \hline
\end{array},\\

&\a_5=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline   \ 0 &\ 0&  \ 0&\ 0& \ 0 &-2 \\
         \ 0 &\ 0&  \ 0&\ 0&\ 2& 0\\
       \ \ 2 & -2&\ \ 2& -2&\ 2 &0\\
         \ 0 &\ 0&\ 0&  0& -2 & 0 \\ \hline
\end{array}, &
&\a_6=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline \ 0 &  \ 0&\ 0 &  \ 0& 0   &\ 2 \\
       \ 0 &  \ 0& \ 0&  \ 0& 0   & -2\\
       \ 0 &\ \ 2&   0&\ \ 2&\ -2   &\ 2\\
        -2 &  \ 0&  -2&    0&\ 0 &\ \ 0 \\ \hline
\end{array},\\

&\a_7=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline     -2 &  -2&  -2& -2& \ \ 0 & -2 \\
           \ 0 &\ 0& \ 0&\ 0& -2& \ 0\\
           \ 0 &\ 0& \ 0&\ 0& -2 &\ 0\\
           \ 0 &\ 0&   0&\ 0& -2 &\ 0 \\ \hline
\end{array}, &
&\a_8=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline \ \ 1 &\ \ \, 1&\ \ 1  &\ \ \,1& \ -3 & 1 \\
       \ \ 1 &\ \ 1&\ \ 1  &\ \ 1& \  \ 1 & 1 \\
       \ \ 1 &\ \ 1&\ \ 1  &\ \ 1& \  \ 1 & 1 \\
       \ \ 1 &\ \ 1&\ \ 1  &\ \ 1& \  \ 1 & 1 \\ \hline
\end{array}.
\end{align*}
By checking the inner products, it is easy to shows that $N\cong EE_8$. Note that $\a_1, \dots, \a_7 $ are supported on octads and thus $N\leq \Lambda$.

In this case, $M\cap N=0$ and $\{\b_1, \b_2, \dots, \b_8, \a_1,\dots, \a_8\}$ is a basis of $L=M+N$. The Gram matrix of $L=M+N$ is given by

\[
\left[
\begin{array}{rrrr rrrr| rrrr rrrr}
4&-2&0&0&0&0&0&0&2&-1&0&0&0&0&0&0\\
-2&4&-2&0&0&0&-2&0&-1&2&-1&0&0&0&-1&0\\
0&-2&4&-2&0&0&0&0&0&-1&2&-1&0&0&0&0\\
0&0&-2&4&-2&0&0&0&0&0&-1&2&-1&0&0&0\\
0&0&0&-2&4&-2&0&0&0&0&0&-1&2&-1&0&0\\
0&0&0&0&-2&4&0&0&0&0&0&0&-1&2&0&0\\
0&-2&0&0&0&0&4&-2&0&-1&0&0&0&0&2&-1\\
0&0&0&0&0&0&-2&4&0&0&0&0&0&0&-1&2\\ \hline
2&-1&0&0&0&0&0&0&4&-2&0&0&0&0&0&0\\
-1&2&-1&0&0&0&-1&0&-2&4&-2&0&0&0&-2&0\\
0&-1&2&-1&0&0&0&0&0&-2&4&-2&0&0&0&0\\
0&0&-1&2&-1&0&0&0&0&0&-2&4&-2&0&0&0\\
0&0&0&-1&2&-1&0&0&0&0&0&-2&4&-2&0&0\\
0&0&0&0&-1&2&0&0&0&0&0&0&-2&4&0&0\\
0&-1&0&0&0&0&2&-1&0&-2&0&0&0&0&4&-2\\
0&0&0&0&0&0&-1&2&0&0&0&0&0&0&-2&4
\end{array}
\right]
\]

\medskip

By looking at the Gram matrix, it is clear that $L=M+N\cong A_2\otimes E_8$. The Smith invariant
sequence is $ 1111 1111 3333 3333$.

\medskip

\noindent $\bf \dih{6}{12}$

Let
\[
\mathcal{O}_2=\begin{array}{|cc|cc|cc|}
\hline \ & *&* &* &* &* \\
* &\ &\ &\ &\ &\ \\
* &\ &\ &\ &\ &\ \\
* &\ &\ &\ &\ &\ \\ \hline
\end{array}
\]
and  $M:=E(\mathcal{O}_2)$. Set $N:=M\xi $, where  $\xi$ is the
isometry defined in  \refpp{defofxi}. Consider
\begin{align*}
\g_1= \frac{1}{\sqrt{8}}\,
\begin{array}{|cc|cc|cc|}
\hline 0 & 0& 0  & 0 & 0 & 0 \\
 0 &  0&  0&  0& 0& 0\\
 -4 & 0&  0& 0&  0 & 0 \\
 4 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, &&
 \g_2= \frac{1}{\sqrt{8}}\,
 \begin{array}{|rr|rr|rr|}
\hline 0  & 0& 0  & 0 & 0 & 0 \\
      -4 &  0&  0&  0& 0& 0\\
       4 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, \\

\g_3= \frac{1}{\sqrt{8}}\,
 \begin{array}{|rr|rr|rr|}
\hline 0  & -4& 0  & 0 & 0 & 0 \\
       4 &  0&  0&  0& 0& 0\\
       0 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, &&
\g_4=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0  & 4& -4  & 0 & 0 & 0 \\
       0 &  0 &  0&  0& 0& 0\\
       0 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array},\\

\g_5=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0  & 0& 4  & -4 & 0 & 0 \\
       0 & 0&  0&  0& 0& 0\\
       0 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, &&
\g_6=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0  & 0& 0  & 4 & -4 & 0 \\
       0&  0 &  0&  0& 0& 0\\
       0 & 0&  0& 0&  0 & 0 \\
       0 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, \\

\g_7=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline 0  & 0& 0  & 0 & 0 & 0 \\
       0&  0 &  0&  0& 0& 0\\
       -4 & 0&  0& 0&  0 & 0 \\
       -4 &  0& 0&  0& 0 & 0\\ \hline
\end{array}, &&
\g_8=\frac{1}{\sqrt{8}}\,
\begin{array}{|rr|rr|rr|}
\hline\ 0  &\ 2& 2  & 2 & 2 & 2 \\
       2 &  0 &  0&  0& 0& 0\\
       2 &  0& 0& 0&  0 & 0 \\
       2 &  0& 0&  0& 0 & 0\\ \hline
\end{array}.
\end{align*}
Then $\{\g_1, \dots,\g_8\}$ is a basis of $M$ and $\{\g_1\xi ,
\dots,\g_8\xi \}$ is a basis of $N$.

\medskip

By the definition of $\xi$, it is easy to show that $\g_1\xi=-\g_1,
\g_2\xi=-\g_2$. Moreover, for any $\a\in M=E(\mathcal{O}_2)$, $\a
\xi$ is supported on $\mathcal{O}_2$ if and only if $\a\in
\mspan_\ZZ\{ \g_1, \g_2\}$.  Hence, $F=M\cap N=\mspan_\ZZ\{ \g_1,
\g_2\} \cong AA_2$. Then $ann_M(F)\cong ann_N(F)\cong EE_6$ and
$L=M+N$ is of rank $14$.

Note that
$\{\g_1, \g_2, \g_3, \dots, \g_8\}$ is a basis and $\{\g_1, \g_2, \g_3\xi, \dots, \g_8\xi \} $ is a basis of $N$. Therefore,
$$\{\g_1, \g_2\} \cup \{ \g_3, \dots, \g_8\}\cup \{\g_3\xi, \dots, \g_8\xi \} $$
is a basis of $L$ and the Gram matrix of $L$ is given by

\[
\left[
\begin{array}{cccccccccccccc}
  4&-2&0&0&0&0&0&0&0&0&0&0&0&0\\
  -2&4&-2&0&0&0&-2&0&-2&0&0&0&2&0\\
  0&-2&4&-2&0&0&0&0&0&1&0&0&-2&0\\
  0&0&-2&4&-2&0&0&0&1&-2&1&0&0&0\\
  0&0&0&-2&4&-2&0&0&0&1&-2&1&0&0\\
  0&0&0&0&-2&4&0&0&0&0&1&-2&0&0\\
  0&-2&0&0&0&0&4&-2&2&0&0&0&0&1\\
  0&0&0&0&0&0&-2&4&0&0&0&0&-1&-2\\
  0&-2&0&1&0&0&2&0&4&-2&0&0&0&0\\
  0&0&1&-2&1&0&0&0&-2&4&-2&0&0&0\\
  0&0&0&1&-2&1&0&0&0&-2&4&-2&0&0\\
  0&0&0&0&1&-2&0&0&0&0&-2&4&0&0\\
  0&2&-2&0&0&0&0&-1&0&0&0&0&4&2\\
  0&0&0&0&0&0&1&-2&0&0&0&0&2&4
  \end{array}
  \right]
\]
whose Smith invariant sequence is $1111\ 11111 \ 333 66$.

In this case,  $ann_M(F)+ann_N(F) \cong A_2\otimes E_6$ and thus $L$ contains a sublattice isometric to $AA_2 \perp (A_2\otimes E_6)$.


TITLE: Entirely "Bare-hands" proof that completely additive states are ultraweakly continuous
QUESTION [3 upvotes]: Originally I had asked this question: Positive Linear Functionals on Von Neumann Algebras
I got some responses that directed me to a variety of resources, some of which I could not understand because the terminology was not introduced within the confines of what is readable on googlebooks, or even within that book, and others were inaccessible to me. (I actually own 0 of the references there, but some were possible to find within the school reading room.  For the next while I will not have access to the reading room and so references will not be helpful unless they are google books readable.)  I decided that perhaps it is best for me to try to solve the problem myself, and so I broke it into 2 steps.

Every completely additive state is ultraweakly continuous. (Or equivalently weakly continuous on the unit ball.)
Every ultraweakly continuous state is of the form described in the above link.

I actually succeeded in doing 2., and I would like to see 1 done.
It would be most helpful for the proof to be elementary, not only because I'm a beginning student in the subject of operator algebras, but because I want to see if it's possible.  The statement seems simple enough.  For me, elementary means it uses only introductory facts for Von Neumann algebras such as the bicommutant theorem, the Borel functional calculus, Kaplansky density, comparison of projections and other things like that.  It may be possible to dig up a proof of this fact, or even an elementary one, by putting together many resources, but since MSE is partly about being a resource itself, as I understand, I hope you can keep your answer at least somewhat self-contained.  For a more precise definition of self-contained, what I mean is I know, and maybe it's even true that most people who would ask this type of question would know, everything before Ch. 7 in these notes.  The only problem is the proof of 7.1.7 is wrong, which is why I pursued my 2-step program above. (Dixmier can't be read from online)  Many thanks in advance.

REPLY [2 votes]: It looks like the proof in the notes you link to can be fixed by simply changing the line following Lemma 7.1.7 from
$$
| \phi(x p_\alpha) | \leq \phi(p_\alpha x^* x p_\alpha)^{1/2} \phi(p_\alpha)^{1/2} = \| x \xi_\alpha \| \phi(p_\alpha)^{1/2},
$$
to
$$
| \phi(x p_\alpha) | \leq \phi(p_\alpha x^* x p_\alpha)^{1/2} = \| x \xi_\alpha \|.
$$
Edit:  I guess the end of the proof will need to change a bit as well since it only gives $\| \eta_\alpha \| = 1$.  What does follow however is that $\phi$ is a limit of ultraweakly continuous linear functionals.  (Specificlly, $\phi$ is the limit of functionals $x \mapsto \phi(x \sum_{\alpha \in F} p_\alpha )$ where $F$ ranges over finite sets.)  Since the space of ultraweakly continuous linear functionals is closed it then follows that $\phi$ itself is also ultraweakly continuous.  This shows that (2) implies (4) in the notes linked to, and as mentioned in the notes we already have (4) implies (3) at this point.


TITLE: Triviality of a bundle given that the corresponding bundle of covering spaces is trivial
QUESTION [8 upvotes]: Let $F,\ E,\ B$ be connected manifolds and  $$F\hookrightarrow E\to B$$ be a fiber bundle. Suppose that $F$ is simply-connected and that $\pi_1(E)=\pi_1(B)= \mathbb Z_2$.
Now consider the universal covering manifolds $\tilde E$ and  $\tilde B$ of $E$ and $B$ respectively. Suppose that we have the following fiber bundle
$$F\hookrightarrow \tilde E\to \tilde B$$
and given that this bundle is trivial i.e. $\tilde E=F\times \tilde B$. Does this imply that also $E=F\times B$?

REPLY [3 votes]: No.  For instance, let $B=\mathbb{R}P^2$ and let $E$ be the $S^2$-bundle on $B$ classified by the map $B\to \mathbb{R}P^\infty=B\mathbb{Z}_2\to BO(3)$ where the first map is the inclusion and the second map is induced by the map $\mathbb{Z}/2\to O(3)$ sending a generator to $-I$.  This $S^2$-bundle is nontrivial since it is not orientable: going around a loop that generates $\pi_1(\mathbb{R}P^2)$, the orientation on the fiber gets reversed since $-I$ is orientation-reversing.  Since $F=S^2$ is simply connected, the long exact sequence on homotopy groups immediately implies $\pi_1(E)\cong\pi_1(B)\cong \mathbb{Z}_2$.
On the other hand, the universal cover bundle $\tilde{E}$ on $\tilde{B}$ is the bundle classified by the composition $\tilde{B}\to B\to B\mathbb{Z}_2\to BO(3)$.  But since $\tilde{B}$ is simply connected, the composition $\tilde{B}\to B\to B\mathbb{Z}_2$ is already nullhomotopic, and thus the classifying map of the bundle $\tilde{E}$ is nullhomotopic so it is a trivial bundle.


TITLE: How to develop $(n+1)^{p+1}-1=(p+1)\sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}$?
QUESTION [0 upvotes]: So, I am really sorry if this is a stupid question, but I am trying to solve a problem (Spivak $2-7$) for which there is a part where I have
$$
(k+1)^{p+1}-k^{p+1} = (p+1)k^p + \text{terms involving lower powers of k}.
$$
as in
$$
(k+1)^{p+1}-k^{p+1} = (p+1)k^p + Ak^{p-1} + Bk^{p-2}+...
$$
where the author develops this expression for k = $1,...,n$.
From which I get
$$
(n+1)^{p+1}-1=(p+1)\sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}
\tag{1}
$$
the latter part of the equation from which I stole from the author, but I understand that part, the issue is that the author gets:
$$
\frac{(n+1)^{p+1}}{p+1} = \sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}
\tag{2}
$$
My question is, how do we get from (1) to (2)? A
similar
question I have looked at derives this:
$$
\frac{(n+1)^{p+1}}{p+1} - \frac{1}{p+1}
\tag{3} =\sum_{k=1}^n k^p + \sum_{k=1}^n \mathcal{O}(k^r)
$$
(where $\mathcal{O}(k^r)$ represents the terms for $\sum_{k=1}^{n}k^r \text{, for r < p}$)
which seems more understandable than (2) as in (3) I see that the answerer just divided the identity by $(p+1)$, from which I also mean to ask, can you divide the terms in $\mathcal{O}(k^r)$ and still keep $\mathcal{O}(k^r)$? What I mean by this is that when the term $\mathcal{O}(k^r)$ represents $Ak^{p-1} + Bk^{p-2}+...$, can you still call $\frac{\mathcal{O}(k^r)$}{p+1}$, $\mathcal{O}(k^r)$? The reason I find (3) more understandable than (2) is that at least in (3), the "$-1$" term remains, where did the "$-1$" term go in (2)!?
I am very sorry if this question represents the utmost stupidity in me, and if the question is stupid in itself, thank you.

REPLY [0 votes]: Here's a way
to get the recurrence
for the power sums.
If
$s_m(n)
=\sum_{k=0}^n k^m
$
then
$\begin{array}\\
(n+1)^m
&=\sum_{k=0}^n ((k+1)^m-k^m)\\
&=\sum_{k=0}^n (\sum_{j=0}^{m-1}\binom{m}{j}k^j)\\
&= \sum_{j=0}^{m-1}\binom{m}{j}\sum_{k=0}^nk^j\\
&= \sum_{j=0}^{m-1}\binom{m}{j}s_j(n)\\
&= ms_{m-1}(n)+\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\\
\text{so}\\
s_{m-1}(n)
&=\dfrac1{m}\left((n+1)^m-\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\right)\\
&=\dfrac1{m}\left(\sum_{j=0}^m \binom{m}{j}n^j-\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\right)\\
&=\dfrac1{m}\left(n^m+mn^{m-1}+\sum_{j=0}^{m-2} \binom{m}{j}n^j-\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\right)\\
&=\dfrac1{m}\left(n^m+mn^{m-1}+\sum_{j=0}^{m-2} \binom{m}{j}(n^j-s_j(n))\right)\\
\end{array}
$
Immediately,
$s_m(n)
\lt \dfrac1{m+1}(n+1)^{m+1}
$.


TITLE: The 8 vectors to be made non-collinear
QUESTION [1 upvotes]: Consider the set of $8$ vectors $V=\{ai+bj+ck:a,b,c \in \{-1,1\}\}$. How can I choose three non-collinear vectors from $V$?
My try:
Let there be three vectors
\begin{align*}
V_1&=a_1i+b_1j+c_1k,\\
V_2&=a_2i+b_2j+c_2k,\\
V_3&=a_3i+b_3j+c_3k.
\end{align*}
Now for $V_1,V_2,V_3$ to be collinear determinant of their $x,y,z$ components should be zero. But what next? What conditions should I apply?

REPLY [1 votes]: If you visualize the set of vectors, you will notice that they form the vertices of a cube. Since no three vertices of the cube are collinear, any choice of $V_1,V_2,V_3$ as three disctinct vectos of the set will do.
Let $V_1,V_2,V_3$ be any three distinct vectors of the set $V$.
Note that $V_2-V_1$ is a nonzero vector with components $\in\{-2,0,2\}$. Likewise, $V_3-V_1$ and $V_3-V_2$ are nonzero vectors with components $\in\{-2,0,2\}$.
The only way for $V_3-V_1$ to be a multiple of $V_2-V_1$ which is equivalent to collinearity of $V_1,V_2,V_3$) is therefore that $V_3-V_1=\pm(V_2-V_1)$. But "$+$" leads to $V_3=V_2$, contrary to assumption. So $V_3-V_1=-(V_2-V_1)$ and hence $V_3-V_2=2\cdot (V_1-V_2)$ is a nonzero vector with components in $\{-4,0,4\}$. But we already know that its components are $\in\{-2,0,2\}$, so they must be zero, i.e. again $V_3=V_2$ contrary to assumption.


TITLE: What does this function converge to in distribution?
QUESTION [0 upvotes]: Suppose $D_i$ ~ $Bern(q)$. Let $\hat{q} = \frac{1}{N}\Sigma{D_i}$. Find the distribution of $\frac{N(\hat{q} - q)^2}{1-q}$ as N goes to infinity.
By Central Limit Theorem I have $dlim(\sqrt{N}(\hat{q} - q)) = N(0, q(1-q))$. Apply Continuous Mapping Theorem,  $N(\hat{q} - q)^2$ converges to $[N(0, q(1-q)]^2$. Is this Chi-squared distribution? If I divide by $q(1-q)$, what distribution would I get?

REPLY [0 votes]: As you already stated, by using the CLT, continuous mapping and Slutsky, you have that
$$
\frac{   \sqrt{n}( \hat{q} - q) }{\sqrt{q(1-q)} }\xrightarrow{D} Z,
$$
such that $\sim \mathcal{N}(0,1)$. Hence, by using the continuous mapping for $g(x) = x^2$ once again, you have that
$$
g \left( \frac{   \sqrt{n}( \hat{q} - q) }{\sqrt{q(1-q)} } \right) \xrightarrow{D} g(Z)=Z^2.
$$
One of the definitions of the $\chi^2$ distribution is a square of the standard normal random variable, thus
$$
g(Z) =Z^2\sim\chi_{(1)}^2.
$$