The thermal behaviors of oxygen-related complexes in boron doped Czochralski Silicon (Cz-Si) wafers at 450°C and 800°C were investigated using Fourier transform infrared spectroscopy (FTIR) and Hall mobility measurements. Activation of thermal donors (TDs) at 450°C leads to a decrease of both mobility and majority carrier concentration using the four point probes configuration of Van Der Pauw. It was found that annealing at 450°C would possibly affect the electronic properties of the Si wafers via the formation of interstitial dioxygen defects (IO2i), which exhibit an IR absorption band positioned at 545 cm–1. A strengthening of the IR bands peaking at around 1595 cm–1, 1667 cm–1, 1720 cm–1 and 1765 cm–1 occurs at 450°C, while they disappear at 800°C. At high temperatures, the precipitation of interstitial oxygen becomes predominant over all other oxygen-related reactions. The dynamic of oxygen-thermal donor generation-annihilation in Cz-Si involving the formation of small oxygen clusters is discussed.


*Jamie Foxx surely wasn't looking for trouble when he and some friends went to dinner at Catch Restaurant on Saturday, but that's what he got served.
At that point, according to eyewitnesses, the guy lunged at Jamie. The accounts vary, but one witness says he actually made contact with the actor, pulling him down onto a table. The witness goes on to say Jamie turned the guy around, put him in a choke hold and took him down.
The unknown dude and his crew were kicked out of the restaurant, reports TMZ.
OK, that was bad news for Jamie Foxx, but we've also got good news for Jamie.
It seems Showtime has picked up Foxx's series, "White Famous," starring former Saturday Night Live member Jay Pharaoh.
As we reported earlier, the project is based on an original idea by Foxx and is about the actor/comedian's real-life experiences coming up in the entertainment business. Tom Kapinos wrote the script for the series. Tim Story directed the pilot and will executive produce.
The series will also star Megalyn Echikunwoke as Floyd's ex-girlfriend, Sadie, the love of his life and the mother of his son. Utkarsh Ambudkar (The Mindy Project) plays Malcolm, Floyd's frustrated agent who wants his client to do whatever it takes to make it big.
As of this posting, it has not yet been determined when the series will premiere.


Here are some excellent ways you can get your free service and have your technician or spa benefit at the same time.
●Before and After Photos. If you have fine lines, wrinkles, hyper-pigmentation, acne or even eyebrows in sore need of reshaping, offer to let the establishment use your before and after picture in their brochure's or advertising in exchange for free or deeply discounted service.
●Research. If your spa is introducing a new product or procedure, you can volunteer to be the subject of their research and offer a generous testimonial to help boost their sales with the new product or treatment.
● Trade. Do you have a product or service you can trade? Can you trade photography, printing, dog walking or even babysitting for your favorite technician. Can you paint, repair or decorate? Do you have access to fresh flowers? Do you sale something that an individual or even the company may benefit from that you could trade for services?
●Refer a Friend. Many companies have a referral incentive, offering discounts for every referral that you supply. Word of mouth is a great, cost effective way to build a business and those who refer their friends should be rewarded. If you like the experience and the service you receive at a particular spa or salon, and if they have a reward system in place, tell your friends, cousins, aunts uncles, sisters, brothers and co-workers.
●Advertise. Do you have a newsletter or a blog? Do you have access to a company bulletin board where you work? Do you own a business where you could display brochures? These are all ways to help your favorite salon advertise with little or no cost to them. Make sure the salon that you are working with asks the , " Where did you hear about us," question so that they can track how many people you are actually bringing in to them.
If you are a frequent patron at a certain spa or salon, getting free or discounted service might be easier than you think. Be creative, recognize your unique talents and abilities that you have to share. You never know what doors you can open until you ask.


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The article on Home Diagnostic Kits in the September 2004 issue of Pharmacy Times should have said that the FDA has cleared home testing kits for consumer use.
The article on Fluoroquinolones in the September issue should have referred to nalidixic acid as a quinolone.
An error in editing resulted in misinformation in the article, Statins and Diabetes (October 2004). The table should have said that there are currently 6 statins on the market. Ezetimibe (Zetia) and the combination product ezetimibe/simvastatin (Vytorin) should not have been included.


WASHINGTON (AP) — For a couple of months, Stephen Strasburg "stirred up the baseball world," in his words.
Now the Washington Nationals' right-hander is embarking on the "boring" process of working his way back from reconstructive elbow surgery.
While waiting for that, Strasburg has been doing "some sort of workout every single day" to stay in shape.
The No. 1 pick in the 2009 draft made an electrifying debut for the Nationals on June 8, striking out 14 Pittsburgh Pirates in seven innings while earning the victory. He went 5-3 with a 2.91 ERA and 92 strikeouts in 12 starts before getting injured last month.
Strasburg has been spending time studying toward a degree at San Diego State, which he left after his junior year. He wound up signing a record $15.1 million, four-year contract with the Nationals.
His rehabilitation schedule has been outlined in detail, including a specific date when he should be able to throw his next professional pitch, if all goes well.
He spoke to several people who had the same Tommy John surgery that he did, including Nationals teammate Jordan Zimmermann — who is back with the team now — and St. Louis Cardinals pitcher Chris Carpenter.
One person Strasburg hasn't chatted with lately? Washington manager Jim Riggleman.
"I've called him a couple times and you don't hear back from him. (Pitching coach Steve) McCatty's called him. He calls back during the game. That way, he knows you're not going to be able to pick up and answer," Riggleman said with a smile.


Step 1: Find suitable object to place on little brother's head.
Step 2: Stand back and admire handiwork. Then point and laugh.
Step 3: Now wait and watch as little brother tries to get it off.
Step 4: Take a closer look to make sure he is still not enjoying this.
Step 5: If after completing Step 4, you realize your plan backfired (and your brother is now enjoying himself), change tactics completely.
Implement new motto : "If you can't beat 'em, join 'em."


text:
Home » Frequently asked questions » Resources » Humor » Humor, 2014
October 2014 Humor
A Halloween Tale
Police work can be entertaining as well as dangerous. Recently, a female sheriff's deputy arrested Patrick Lance, a 22 year old white male, who was fornicating with a pumpkin in the middle of a field at night. The next day, at the Gwinnet County (GA) courthouse, Lance was charged with lewd and lascivious behaviour, public indecency and public intoxication.
The suspect explained that he was passing a pumpkin patch on his way home from a drinking session when he decided to stop. He explained: " as there was no one around for miles, or at least I thought there was no one around" he stated.
Lance went on to say that he pulled over to the side of the road, picked out a pumpkin that he felt was appropriate to his purpose, cut a hole in it, and proceeded to satisfy his pressing need. 'Guess I was really into it, y'know?' he commented with evident embarrassment.
In the process of doing the deed, Lance failed to notice an approaching sheriff's car and was unaware of his audience until Deputy Brenda Taylor approached him. 'It was an unusual situation, that's for sure,' said Deputy Taylor.
'I walked up to Lawrence and he's just humping away at this pumpkin.' Deputy Taylor went on to describe what happened when she approached Lance ... 'I said: 'Excuse me sir, but do you realize that you're having sex with a pumpkin?'
He froze and was clearly very surprised that I was there, and then he looked me straight in the face and said: 'A pumpkin? Shit, is it midnight already?'
The court (and the judge) could not contain their laughter. Lance was found guilty only of public intoxication, fined $10. and sent on his way. The Washington Post wrote an article describing this as "The best come-back line ever.
[OK, prolly didn't happen, but it's a good October joke anyway.]
Humor - Summaries
‹ September 2014 Humor up November 2014 Humor ›

redpajama_set_name: RedPajamaCommonCrawl


you might want to know that you can save your knees and do it on a table.
I learned this method from my friend, Sally, some years ago.
It's made my least favorite part of the quilt-making process a pleasant afternoon or evening.
Here's how it looks when you're ready to start pinning.
And a used 6 foot folding banquet table.
You can also use your dining table, as long as you protect it with a mat.
Tape two toothpicks in a cross to mark the middle of the table.
You can also mark the middle of the sides and ends of the table with toothpicks and tape.
Use painter's tape if you're concerned about damaging the table surface.
Fold your backing fabric in quarters and iron at the center corner.
Clip your backing to the table.
Iron the center corner on your batting and quilt top.
Then place them over the clipped backing.
You will be able to feel where the center lines of the table are because of the toothpicks.
Pin this section of the quilt together.
leaving a row of the pinned area on the table.
Peel away the quilt top and batting.
Clip the backing to the table and carefully smooth the batting and top over it.
Repeat until the whole quilt is pinned.
When finished, I fold the table and store it under the guest room bed.
the perfect time to baste my corduroy quilt!
Some quilts take many inspirations to find their way.
a gift from Patricia at Okan Arts, a few years ago.
I pulled coordinating fabrics from my stash, and the red hopped in unexpectedly.
Cultural Fusion Quilts, page 55, and wanted to make it.
for a class with her in the summer of 2017.
Sujata very sweetly helped me lay out the design in class.
Lattice Quilt 64" x 64"
Then, Sujata got us going with the UandU Quilt Along at Basket Full of Scraps.
for the upteenth time, and saw his Lattice Quilt on page 34.
It jogged my memory of the unsewn quilt blocks from Sujata's class.
A few tweeks and a few hours later, the quilt top was done!
Many thanks to Patricia, Sujata and Roderick for your inspiration.
It's been a long time since I've posted on this blog.
I've missed reading your stories and writing my own.
and less time on everything else, including social media.
I've still found a little time to sew, and I've posted my quilts on Instagram.
IG posting is much quicker, but it just isn't as satisfying.
that make me feel good, and this is where they happen.
Briefly, here are a few things I've been doing since I last posted.
I'll come back and tell the stories later, one by one.
I'm still working on these EPP stars.
It's my travel project, and I've been finding fun photo opportunities on the road.
I'm still using up orphan blocks and using Word Play words in quilts.
and shown in Gwen Marston's Liberated String Quilts book.
I made a wool quilt, with wool from many sources, old and new.
I contemplated what to do with all of these quilts. This is only half the stack!
They all went back into the closet.
And I found this heart reaching out to me on the path to my backyard.
Thank you for being here and welcoming me back.
Actually, it happened months ago, but I didn't have decent photos until now. Meanwhile, it's been to two shows and it's been slept under a few times. You can find the piecing story here.
"Spare Parts" 86" x 86"
It was quilted by the very talented Sharon Tucker.
I just love what she did with it!
Here's "Spare Parts" with me at the Sisters Outdoor Quilt Show last weekend.
"Cowgirl Dreams" came too. And she sold!
It's the first time I've put a price tag on a quilt.
It makes me very happy to know that some cowgirl will be sleeping under it soon.
That's Mary Keasler from Tennessee with me.
Here's Mary with her fantastic "Sunday Best" quilt.
the best thing about it is the community involvement.
It appears that EVERYONE comes out to help.
Even the fire fighters help hang the show quilts early in the morning.
the friendly people were still working.
and these are only SOME of the many beauties we saw in the show!
"Cowgirl Dreams" 80 x 80"
just in time for the local guild show in March.
This is one that I can honestly say, "The fabric made me do it."
For the story and more fabric close ups, click here.
I decided to quilt it simply, in order to show off the print most clearly.
After stitching in "every stinkin' ditch," I cut a template and marked the curves.
Then I used the walking foot, which was pretty easy.
On the rope, well, I had to follow it, right?
This part I did with free motion quilting.
A few nights before the show, I washed it, feeling quite happy.
and some, not all, of the ivory fabrics had turned pink!
Luckily, a few hours soak in the bathtub with Oxyclean removed all the pink.
It also destroyed the fake buckskin fringe on the border.
Darn! I cut it off carefully and got it to the show on time.
It's been a long, wet winter here in Seattle with no opportunities for outdoor photos.
so hubby and I got outside with quilts and camera.
and have finished a handful of projects.
I'll post them one by one over the next few weeks.
First up, the "Haha" quilt top is complete! It finished at 60" x 72".
Here's a photo from my sewing table as I made the blocks.
The back will be this fabric salvaged from old curtains found on eBay.
I'm looking forward to quilting it with machine and hand stitches.
Instagram has also been a fun way to share and keep up with many of you.
If you haven't already, you can find me there @niftyquilts.
I've started a new project that will take me well into the new year.
It's using mostly my hubby's old shirts, along with a few stash fabrics.
I saw this photo on Barbara Brackman's blog back in 2014.
I love a quilt with a good sense of humor!
But the story of this quilt isn't funny at all.
It's told in this book.
Anna Horn made the quilt in 1917-1918.
Her son, Albert Horn, was away at war (WWI) at the time.
when Albert left to serve his country.
Anna comforted herself by stitching the initials that she and her son shared.
Albert survived the war and the quilt stayed in the family for generations.
It was eventually given to his great granddaughter, Ashley Horn.
Wishing you all laughter, comfort, love and creativity in 2017!
Today, and every day, I appreciate you, my "blog family."
Here's to more creativity and sharing for all of us in 2017!
It all started in 2013, when Quiltdiva Julie sent me a box of crazy pieced vintage blocks.
There were 24 in all, and they came from a quilt top she'd bought on Ebay.
The seller said it came from Ohio.
I had saved this photo on my computer screen for some years before that.
I don't remember the source, but it became my inspiration.
I sewed the blocks together and cut out the diamond shapes for the star.
There were just enough leftover for the border.
Then I bought some embroidery floss.
I learned a few stitches, and embroidered all the seams.
Eventually, it was ready for a background.
I picked this spring green solid, because it was the only thing that seemed to work.
I wasn't excited about the color, but thought I'd get over it.
Well, I didn't. I just didn't like all that green!
I decided to remove it and find another background.
Several hours of stitch-picking later, I was ready for auditions.
I tried practically everything I had on the shelf.
A few dear friends suffered with me and my auditions for WEEKS!
Then I went to the Houston Quilt Festival and found what I was looking for!
It's one of the new woven plaids from Kaffe Fassett.
The top is about 87" square, plenty big for a queen size bed.
I finished it last week.
THEN, just yesterday, I went to the Farmer's Market, like I do every Sunday.
The woman farmer I always buy from gave me this!
She said her great-grandmother had made it from old family clothes.
What a treasure! I tried to talk her out of giving it away.
But she said she had no room for it.
So I gladly brought it home.
this one was sewn all on one sheet with embroidery.
Now I have TWO vintage scrap tops to quilt and to love!
Bill Volckening found where the original inspiration quilt came from.
The quilt was shown in the European Patchwork show in France in 2012.
Katell Renon's quilt group in France was also inspired by the original.
Do go and see it here!


Wow - that's a big update. And congrats on the one year no-soda mark! My co-worker opened a can of Coke today and I just laughed because I immediately started salivating at the sound - shows you how long it's been since I've had a Coke, and it's not even been close to one year! You rock!
I just grabbed Karin's coke and drank it down... YUM. But, I really am not going right back to bad habits. Still mostly drinking water and some juice!


This year marks 20 years since Advantage Plastics was founded in 1999, continual development and growth thought out the past 20 years has been exciting, enjoyable and satisfying but this couldn't have been done without our fantastic customers, so thank you for making our success possible.
With our mission being all products produced are to be of excellence quality and design right through to delivery, Careful and dedicated detail to service will ensure customer satisfaction and growth with retaining long term relationships. We certainly do look forward to working with you thought out this year and for the next 20 years.


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\begin{document}

\title{Classification of $(1{,}2)$-reflective anisotropic hyperbolic lattices of rank $4$}

\author{Nikolay V. Bogachev}

\address{Moscow Institute of Physics and Technology;}

\address{Lomonosov Moscow State University;}

\address{Adyghe State University.}

\email{nvbogach@mail.ru}

\begin{abstract}
A hyperbolic lattice is called \textit{$(1{,}2)$-reflective} if its automorphism group
is generated by $1$- and $2$-reflections up to finite index.
In this paper we prove that the fundamental polyhedron of a $\mathbb{Q}$-arithmetic
cocompact reflection group in the three-dimensional Lobachevsky space contains an
edge such that the distance between its framing faces is small enough.
Using this fact we obtain a
classification of $(1{,}2)$-reflective anisotropic hyperbolic lattices of rank $4$.
\end{abstract}

\maketitle

\textbf{Keywords:}
reflective hyperbolic lattices, roots, reflection groups, fundamental polyhedra, Coxeter polyhedra.

\tableofcontents

\section{Introduction}
\label{s1}

\subsection{Preliminaries}
\label{ss1.1}

By definition, a \textit{quadratic lattice} is a free Abelian group with a non-degenerate integral
symmetric bilinear form
called an \textit{inner product}.
A quadratic lattice $L$ is
called \textit{Euclidean}
if its inner product is positive definite and it is called \textit{hyperbolic}
if its inner product is a form of signature $(n, 1)$.
A lattice $L$ is said to be \textit{isotropic} if the corresponding quadratic form represents zero,
otherwise it is said to be \textit{anisotropic}.

A lattice $L$ is said to be \textit{even} if $(x,x) \in 2 \mathbb{Z}$ for all $x \in L$,
otherwise it is said to be \textit{odd}. Any odd lattice $L$ contains a unique even sublattice
of index $2$ formed by all its vectors whose lengths have even squares.

Let $L$ be a hyperbolic lattice. We shall assume that it is embedded into the
\textit{Minkowski space} $V=L \otimes \mathbb{R}=\mathbb{E}^{n, 1}$,
and we shall take one of the connected components of the hyperboloid
\begin{equation}
\label{eq1}
\{x \in \mathbb{E}^{n, 1}\mid (x, x)=-1\}
\end{equation}
as a \textit{vector model} of the $n$-\textit{dimensional (hyperbolic) Lobachevsky space} $\mathbb{L}^n$
with the metric $\rho$ given by the formula $\cosh\rho(v,w)= -(v,w)$.
In this case, the group $\mathrm{Isom}(\mathbb{L}^n)$ of
motions of the Lobachevsky space is a subgroup $O'(V)$ of index $2$ of the pseudo-orthogonal group $O(V)$ and
it consists of all
transformations leaving invariant each connected component of the hyperboloid \eqref{eq1}.
The \textit{planes} in the vector model of the Lobachevsky space are non-empty
intersections of the hyperboloid with subspaces of $V$. The \textit{points at infinity} in this
model correspond to isotropic one-dimensional subspaces of $V$.

A primitive vector $e$ of a quadratic lattice $L$ is
called a \textit{root} or, more precisely, a
$k$-\textit{root}, where $k=(e,e) \in \mathbb{N}$, if $2(e, x) \in
k\mathbb{Z}$ for all $x \in L$. (For $k \leqslant 2$ the last condition is fulfilled automatically.)
Every root $e$ defines an \textit{orthogonal reflection} (called a $k$-\textit{reflection} if $(e,e)=k$)
in the space
$L \otimes \mathbb{R}$
$$
\mathcal{R}_e\colon x \mapsto x-\frac{2(e, x)}{(e, e)} e,
$$
which preserves the lattice $L$. In the hyperbolic case, $\mathcal{R}_e$ defines a reflection with respect to
the hyperplane
$$
H_e=\{x \in \mathbb{L}^n \mid (x, e)=0\}
$$
in the space $\mathbb{L}^n$ called a \textit{mirror} of the reflection $\mathcal{R}_e$.

Suppose that $\mathcal{O}(L)$ is the group of automorphisms of a lattice $L$. It is known that
$$
\mathcal{O}'(L)=\mathcal{O}(L) \cap \mathcal{O}'(V)
$$
is a discrete group of motions of the Lobachevsky space, and
its fundamental polyhedron has finite volume (see Venkov's paper~\cite{Ven37}). We denote by $\mathcal{O}_r(L)$,
$\mathcal{O}^{(2)}_r(L)$, and $\mathcal{O}^{(1{,}2)}_r(L)$ the subgroups of
$\mathcal{O}'(L)$ generated by all
reflections, all  $2$-reflections, and all  $1$- and $2$-reflections in~$\mathcal{O}'(L)$, respectively.

The lattice $L$ is said to be \textit{reflective}, $2$-\textit{reflective}, or $(1{,}2)$-\textit{reflective}
if the subgroup $\mathcal{O}_r(L)$,
$\mathcal{O}^{(2)}_r(L)$, or $\mathcal{O}^{(1{,}2)}_r(L)$, respectively,
has finite index in~$\mathcal{O}'(L)$. The lattice $L$ is reflective, $2$-reflective, or $(1{,}2)$-reflective
if and only if the fundamental polyhedron of the group $\mathcal{O}_r(L)$,
$\mathcal{O}^{(2)}_r(L)$, or $\mathcal{O}^{(1{,}2)}_r(L)$, respectively, has finite volume in the Lobachevsky space
$\mathbb{L}^n$.

It is obvious that every finite extension of a $2$-reflective
(or $(1{,}2)$-reflective) hyperbolic lattice is also a
$2$-reflective (respectively, $(1{,}2)$-reflective) lattice. We also
note  that every $2$-reflective hyperbolic lattice
is $(1{,}2)$-reflective.

Discrete groups generated by reflections were determined by H.\,S.\,M.~Coxeter. He classified them on spheres
$\mathbb{S}^n$ and in Euclidean spaces~$\mathbb{E}^n$ in~1934 (see \cite{Cox34}). A systematic
study of reflection groups in Lobachevsky spaces was initiated by Vinberg
in~1967 (see \cite{Vin67}). In the same paper, an  arithmeticity criterion was given for discrete groups
generated by reflections, new methods  of studying hyperbolic
reflection groups were suggested and various examples of such groups were  constructed.
In~1972, Vinberg proposed an algorithm
(see \cite{Vin72}, \cite{Vin73}) that, given a lattice $L$, enables
one to construct the fundamental polyhedron of the group $\mathcal{O}_r (L)$ and determine thereby
the reflectivity of the lattice $L$. It is known that cocompact
(with a compact or, equivalently, bounded fundamental polyhedron)
discrete reflection groups (as well as arithmetic reflection groups) do not exist in the Lobachevsky spaces of
dimension
$\geqslant 30$ (Vinberg, 1984, see \cite{Vin84}).
It was also proved that there are no reflective hyperbolic lattices of rank
$n+1 > 22$ (F.~Esselmann, 1996, see\ \cite{Ess96}). In~2007, V.\,V.~Nikulin
(see\ \cite{Nik07}) finally proved that, up to conjugacy, there are
only finitely many maximal arithmetic reflection groups (see also \cite{ABSW2008}).
These results give hope that all maximal arithmetic hyperbolic
reflection groups and, in particular, reflective hyperbolic lattices can be classified.

V.\,V.~Nikulin (1979, 1981 and 1984, see \cite{Nik79,Nik81b,Nik84}) classified all
$2$-reflective hyperbolic lattices of
rank not equal to $4$, and after that in~2000\ (see~\cite{Nik00}) he found all reflective hyperbolic lattices of rank $3$
with square free discriminants. In~1998, (see \cite{Vin98} and \cite{Vin07}) E.\,B.~Vinberg
classified all $2$-reflective hyperbolic lattices of rank $4$.
Subsequently, D.~Allcock (2011, see \cite{All12}) classified at all
all reflective hyperbolic lattices of rank $3$.

In 1989--1993 in the papers~\cite{Sch89,SW92,Wal93}  R.~Scharlau and C.~Walhorn
presented a list of all maximal groups of the form $\mathcal{O}_r(L)$, where $L$
is a reflective isotropic hyperbolic lattice of rank $4$ or $5$.
A similar result was obtained in 2017 in the dissertation of I.~Turkalj (see  \cite{Tur17})
for lattices of rank $6$.

Finally, the author of this paper announced a classification of all maximal
$(1{,}2)$-reflective anisotropic hyperbolic lattices of rank $4$ (2016 and
2017, see \cite{Bog16} and \cite{Bog17}). A more complete history of the problem
 can be read in the recent survey of M.~Belolipetsky~\cite{Bel16}.

\subsection{Notation}
\label{ss1.2}
We introduce some notation:

1)~$[C]$ is a quadratic lattice whose inner product in some
basis is given by a symmetric matrix $C$;

2)~$d(L) := \det C$ is the discriminant of the lattice $L = [C]$;

3)~$L \oplus M$ is the orthogonal sum of the lattices $L$ and $M$;

4)~$[k]L$ is the quadratic lattice obtained from $L$ by multiplying all inner
products by $k \in \mathbb{Z}$;

5)~$L^*=\{x \in L \otimes \mathbb{Q} \mid \forall\, y \in L \ \ (x, y) \in
\mathbb{Z}\}$~ is the \textit{adjoint} lattice.

Let $P$ be a compact acute-angled polyhedron in $\mathbb{L}^3$ and
let $E$ be some edge of it.
We denote by $F_1$ and $F_2$ the faces of the polyhedron $P$
containing the edge $E$. Let
 $u_3$ and $u_4$ be the unit outer
normals to the faces $F_3$ and $F_4$ containing the vertices of the edge $E$,
but not the edge itself.

\begin{definition}
\label{d1.1}
The faces $F_3$ and $F_4$ are called the framing edges of the edge $ E $,
and the number $|(u_3, u_4)|$ is its width.
\end{definition}

We associate with the edge $E$ the
set $\overline \alpha = (\alpha_{12}, \alpha_{13}, \alpha_{23}, \alpha_{14}, \alpha_{24})$,
where $\alpha_{ij}$ is the angle between the faces $F_i$ and $F_j$.

\subsection{Main results}
\label{ss1.3}
The main results of this paper are the following two assertions, the second
of which is proved with the help of the first one.

\begin{theorem}
\label{t1.1}
The fundamental polyhedron of every $\mathbb{Q}$-arithmetic cocompact
 reflection group in $\mathbb{L}^3$ has an edge of width less than $4.14$.
\end{theorem}

Recall that a $\mathbb{Q}$-arithmetic reflection group is any finite index subgroup of a group
of the form $\mathcal{O}' (L)$. It can be cocompact only in the case where the lattice $L$ is anisotropic.

In fact, a stronger result is obtained. Namely, it is proved that
there is an edge of width
$ t_{\overline \alpha}$, where
$t_{\overline \alpha} \leq 4.14$ is a number depending on the set $\overline \alpha$ of
dihedral angles around this edge (see Theorem~\ref{t2.3}).

\begin{theorem}
\label{t1.2}
Any $(1{,}2)$-reflective anisotropic hyperbolic lattice of rank $4$ over $\mathbb{Z}$ is either isomorphic
to
$[-7] \oplus [1] \oplus [1] \oplus [1]$ or
$[-15] \oplus [1] \oplus [1] \oplus [1]$,
or to an even index $2$ sublattice of one of them.
\end{theorem}

Actually, these lattices are $2$-reflective (see \cite{Vin07}).

The author hopes that the \emph{method of the outermost edge} (see \S\,\ref{s2}) employed
in this paper
will be applicable for classifying all reflective anisotropic hyperbolic lattices of rank $4$.

\medskip

The author regards  as his pleasant duty to express his deep gratitude to E.\,B.~Vinberg
for posing the problem and also for some
ideas, valuable advice, help, and attention.

\section{The method of the outermost edge and proof of Theorem~\ref{t1.1}}
\label{s2}

\subsection{Nikulin's method}
\label{ss2.1}
In this paper we shall use \textit{the method of the outermost edge},
which is a modification of \emph{the method of narrow parts of polyhedra}, applied by
V.\,V. Nikulin in his papers \cite{Nik80} and \cite{Nik00}.

\begin{definition}
\label{d2.1}
A convex polyhedron in the space $\mathbb{L}^n$ is the intersection of finitely many halfspaces
such that it has non-empty interior.
A generalized convex polyhedron is the intersection of  a family of halfspaces (possibly infinite)
such that any ball intersects only finitely many their boundary hyperplanes.
\end{definition}

\begin{definition}
\label{d2.2}
A generalized convex polyhedron is called acute-angled if all its dihedral angles do not exceed $\pi/2$.
A generalized convex polyhedron is called a Coxeter polyhedron if all its dihedral angles have the form
$\pi/k$, where $k \in \{2, 3, \dots, \infty\}$.
\end{definition}

It is known that fundamental domains of discrete reflection groups are generalized \textit{Coxeter polyhedra}
(see the papers \cite{Cox34},
\cite{Vin85}, and \cite{VS88}).

Here and throughout  by faces of a polyhedron we mean its $(n-1)$-dimensional faces.
The Gram matrix of a system of vectors $v_1, \dots, v_k$ will be denoted by $G(v_1, \dots, v_k)$.

In his earlier works (see Lemma~3.2.1 in~\cite{Nik80} and the proof
of Theorem~4.1.1 in~\cite{Nik81b}) V.\,V.~Nikulin proved the following
assertion\footnote{We present this assertion in a form convenient  for us, although
it was not formulated in this way anywhere.}.

\begin{theorem}
\label{t2.1}
Let $P$ be an
acute-angled convex polyhedron of finite volume in~$\mathbb{L}^n$.
Then there exists a face $F$ such that
$$
\cosh \rho (F_1, F_2) \le 7,
$$
for any faces $F_1$ and $F_2$ of $P$ adjacent to $F$, where
$\rho{(\,\cdot\,, \cdot\,)}$ is the metric in the Lobachevsky space\footnote{In Nikulin's papers, the squares of the lengths of the normals to faces are $2$, therefore, in his works
there is a bound $(\delta, \delta') \leq 14$.}.
\end{theorem}

In the proof of this assertion, the face $F$ was chosen as the outermost face
from some fixed point $O$ inside the polyhedron $P$. Notice that
this theorem enables us to bound at once the absolute value of the inner product of
outer normals to faces adjacent to the face $F$. Indeed, if $F_1$ and
$F_2$ intersect or are parallel, this value is equal to the cosine of
the dihedral angle between these faces, and if these faces diverge, then it
equals the hyperbolic cosine of the distance between them.

\subsection{The method of the outermost edge}
\label{ss2.2}
We have the following corollary of Theorem~\ref{t2.1}.

\begin{proposition}
\label{p2.1}
Each compact (i.\,e., bounded) acute-angled polyhedron
$P \subset \mathbb{L}^3$ contains an edge of width not greater than $7$.
\end{proposition}

\begin{proof}
Following V.\,V.~Nikulin (see\ \cite{Nik81b}), we consider an interior point $O$ in $P$. Let $E$ be
\textit{the outermost\footnote{In an acute-angled polyhedron,
the distance from the interior point to the face (of any dimension) is equal to the
distance to the plane of this face.}
edge from it}, and let $F$ be a face containing this edge.
Let $E_1$, $E_2$ be disjoint\footnote{Note that we consider the case where the framing
faces are divergent, since otherwise the absolute value of the inner product does not exceed one.}
edges of this face coming out from different vertices of $E$.

Let  $O'$ be the projection of  $O$ onto the face $F$. Note that
$O'$ is an interior point of this face, since otherwise the point
$O$ would lie outside of some dihedral angle adjacent to  $F$ (because the polyhedron $P$ is acute-angled).
Further, since $E$ is
the outermost edge of the polyhedron for $O$, then
$$
\rho(O, E) \ge \rho(O, E_i), \qquad i=1, 2.
$$
It follows from this and  the three
perpendiculars theorem that the distance between the point~$O'$
and the edge $E$ is not
less than the distance between this point and any other edge of the face $F$.
This means that the edge $E$ is the outermost edge to the point $O'$
inside the polygon $F$ and we can use Theorem~\ref{t2.1}.

Further, let $F_3$ and $F_4$ be the faces (with unit outer normals $u_3$ and $u_4$,
respectively) of the polyhedron $P$ framing the
outermost edge $E$ and containing the edges $E_1$ and $E_2$, respectively.
Clearly, the distance between the faces is not greater than the distance
between their edges.
Therefore,
$$
-(u_3,u_4)=\cosh \rho(F_3, F_4) \le \cosh \rho(E_1, E_2) \le 7.
$$
The proposition is proved.
\end{proof}

Let now $P$ be the fundamental polyhedron of the group
$\mathcal{O}_r(L)$ for an anisotropic hyperbolic lattice $L$
of rank $4$. The lattice $L$ is reflective if and only if the polyhedron
$P$ is compact (i.\,e., bounded).

Let $E$~be an edge (of the polyhedron $P$) of width not greater than $t$. By
Propositon~\ref{p2.1} we can ensure that $t \leq 7$ (if we take the outermost
edge from some fixed point $O$ inside the polyhedron $P$). Let $u_1$, $u_2$
be the roots of the lattice $L$ that are orthogonal to the faces containing
the edge $E$ and are the outer normals of these faces.
Similarly, let $u_3$, $u_4$ be the roots corresponding to the framing
faces. We denote these faces
by $F_1$, $F_2$, $F_3$, and $F_4$, respectively. If $(u_3, u_3) = k$, $(u_4, u_4) = l$, then
\begin{equation}
\label{eq2}
|(u_3, u_4)| \leq t\sqrt{kl} \leq 7 \sqrt{kl}.
\end{equation}

Since we solve the classification problem for $(1{,}2)$-reflective lattices,
we have to consider the fundamental polyhedra of arithmetic
groups generated by $1$- and $2$-reflections. In this case we are given bounds
on all elements of the matrix $G(u_1, u_2, u_3, u_4)$,
because all the faces $F_i$ are pairwise intersecting, excepting,
possibly, the pair of faces $F_3$ and $F_4$. But if they do not intersect,
then the distance between
these faces is bounded by inequality~\eqref{eq2}.
Thus, all entries of the matrix $G(u_1, u_2, u_3, u_4)$ are integer and bounded, so
there are only finitely many possible matrices
$G(u_1, u_2, u_3, u_4)$.

The vectors $u_1, u_2, u_3, u_4$ generate some sublattice $L '$
of finite index in the lattice $L$. More precisely, the lattice $ L $ lies between
the lattices $L'$ and $(L')^*$, and
$$
[(L')^* : L']^2=|d(L')|.
$$
Hence it follows that $|d(L')|$ is divisible by $[L:L']^2$. Using this, in each case we shall
find  for a lattice
$L'$ all its possible extensions of finite index.

To reduce the enumeration of matrices $G(u_1, u_2, u_3, u_4)$ we shall use some
additional considerations enabling us to get sharper
bounds on the number $|(u_3, u_4)|$ than in inequality~\eqref{eq2}.

\subsection{Bounds for the length of the edge
$E$ for a compact acute-angled polyhedron in $\mathbb{L}^3$}
\label{ss2.3}
In this subsection $P$ denotes a compact acute-angled polyhedron in the three-dimensional
Lobachevsky space $\mathbb{L}^3$.

Keeping the assumptions and notation of the previous sections, we denote the vertices of the edge $E$ by
$V_1$ and $V_2$. The dihedral angles between the faces $F_i$ and $F_j$ will be denoted by
$\alpha_{ij}$.

Let $E_1$ and $E_3$ be the edges of the polyhedron $P$ outgoing
from the vertex $V_1$ and let $E_2$ and $E_4$ be the edges outgoing from $V_2$ such that the edges
$E_1$ and $E_2$ lie in the face $F_1$. The length of the edge $E$ is
denoted by $a$, and the plane angles between the edges
$E_j$ and $ E $ are denoted by $\alpha_j$ (see Figure~\ref{fig1}).

\begin{figure}[!htp]
\begin{center}
\includegraphics{gen-edge.png}
\caption{The outermost edge}
\label{fig1}
\end{center}
\end{figure}

\begin{theorem}
\label{t2.2}
The length of the outermost edge satisfies the inequality
$$
a < \operatorname{arcsinh} \biggl(\frac{\operatorname{tanh} (\ln (\cot ({\alpha_{12}}/{4})))}
{\tan ({\alpha_3}/{2})}\biggr)+\operatorname{arcsinh} \biggl(\frac{\operatorname{tanh} (\ln (\cot
({\alpha_{12}}/{4})))}{\tan ({\alpha_4}/{2})} \biggr).
$$
\end{theorem}

\begin{proof}
Denote by $O_1$ and $O_2$ the orthogonal projections of the point $O$ onto the faces $F_1$ and
$F_2$, respectively. By the theorem of three perpendiculars, both points fall
under the projection onto this edge $E$ on the same point $A$, which is the projection
of the point $O$ onto this edge. Due to the fact that the polyhedron $P$ is acute-angled, the point
$A$ is an inner point of the edge~$E$.

Thus, we get a flat quadrilateral $AO_1 O O_2$, in which
$\angle A = \alpha_{12}$ (the dihedral angle between the faces $F_1$ and $F_2$),
$\angle O_1=\angle O_2=\pi/2$, $AO_1=a_1$, $AO_2=a_2$ (see\ Figure~\ref{fig2}).

\begin{figure}[!htp]
\begin{center}
\includegraphics[scale=1.4]{perp-edge.png}
\caption{A quadrilateral $AO_1 O O_2$}
\label{fig2}
\end{center}
\end{figure}

In the limiting case where the point $O$ is a point at infinity, the dihedral
angle~$\alpha_{12}$ is composed of the so-called \textit{angles of parallelism}
$\Pi(a_1)$ and~$\Pi(a_2)$. In our case $O \in
\mathbb{L}^3$, therefore,
$$\alpha_{12} < \Pi(a_1)+\Pi(a_2)=2 \mathop{\rm arctan}\nolimits(e^{-a_1}) + 2 \mathop{\rm arctan}\nolimits(e^{-a_2}).$$

Denote by $V_1 I$, $V_2 I$, $V_1 J$, $V_2 J$ the bisector of angles $\alpha_1$,
$\alpha_2$, $\alpha_3$, $\alpha_4$, respectively. Let $h_I$ and $h_J$ be
the distances from the points $I$ and $J$ to the edge $E$. Without loss of generality
 we can assume that
$h_J \leqslant h_I$.

Since the edge $E$ is the outermost edge for the point $O$, we have
\begin{equation}
\label{eq3}
\rho(O_1, E) \leq \rho(O_1, E_1), \rho(O_1, E_2), \qquad
\rho(O_2, E) \leq \rho(O_2, E_3), \rho(O_2, E_4).
\end{equation}
Then it is clear that $h_J \leq h_I \leq a_1$, $h_J \leq a_2$, since
inequalities~\eqref{eq3} imply that the points $O_1$ and $O_2$ lie inside flat
angles vertical to the angles $V_1 I V_2$ and $V_1 J V_2$, respectively (the scan of
faces around the edge $E$ is represented in Figure~\ref{fig3}).

\begin{figure}[!htp]
\begin{center}
\includegraphics[scale=1.9]{len-edge-1.png}
\caption{The scan}
\label{fig3}
\end{center}
\end{figure}

We have $\Pi(a_1), \Pi(a_2) \leq \Pi(h_J)$. It follows that
$\mathop{\rm arctan}\nolimits(e^{-h_J}) > \alpha_{12}/4$. Thus,
$$
h_J < \ln \biggl(\cot \biggl(\frac{\alpha_{12}}{4} \biggr)\biggr).
$$

We introduce the notation $A_0 := \operatorname{tanh} \bigl(\ln(\cot({\alpha_{12}}/{4}))\bigr)$. Then
$\operatorname{tanh} h_J < A_0$. Let $H$ be the projection of the point $J$ onto the edge $E$ and let
$x=\rho(H,V_1)$.

From the right triangles $V_1 J H$ and $V_2 J H$ we find
\begin{equation}
\label{eq4}
\operatorname{tanh} h_J=\tan \biggl(\frac{\alpha_4}{2}\biggr) \sinh(a-x) =
\tan \biggl(\frac{\alpha_3}{2} \biggr) \sinh x,
\end{equation}
which implies that
$$
\sinh x=\frac{\operatorname{tanh} h_J}{\tan ({\alpha_3}/{2})} < \frac{A_0}{\tan ({\alpha_3}/{2})},
\qquad \sinh (a-x) < \frac{A_0}{\tan ({\alpha_4}/{2})}.
$$
Hence,
$$
a=x+(a-x) <\operatorname{arcsinh} \biggl(\frac{A_0}{\tan ({\alpha_3}/{2})}\biggr)
+ \operatorname{arcsinh} \biggl(\frac{A_0}{\tan ({\alpha_4}/{2})}\biggr),
$$
which completes the proof.
 \end{proof}

\subsection{The proof of Theorem~\ref{t1.1} and bounds on  $|(u_3,
u_4)|$}
\label{ss2.4}
Let a polyhedron $P$ be the fundamental polyhedron of a
$\mathbb{Q}$-arithmetic cocompact reflection group in the three-dimensional
Lobachevsky space and let $E$ be the outermost edge of the polyhedron
$P$. Consider the set of unit outer normals  $(u'_1, u'_2, u'_3,
u'_4)$ to the faces $F_1$, $F_2$, $F_3$, $F_4$. Note that this vector system is
linearly independent. Its Gram matrix is
$$
G(u'_1, u'_2, u'_3, u'_4) =
\begin{pmatrix}
1 & -\cos \alpha_{12} & -\cos \alpha_{13} & -\cos \alpha_{14}\\
-\cos \alpha_{12} & 1 & -\cos \alpha_{23} & -\cos \alpha_{24}\\
-\cos \alpha_{13} & -\cos \alpha_{23} & 1 & -T\\
-\cos \alpha_{14} & -\cos \alpha_{24} & -T & 1
\end{pmatrix},
$$
where $T=|(u'_3, u'_4)|=\cosh \rho(F_3, F_4)$ in the case where the faces $F_3$ and
$F_4$ diverge. Recall that otherwise $T \leq 1$, and we do not need to consider this case separately.

Let $(u^*_1, u^*_2, u^*_3, u^*_4)$ be the basis dual to the basis $(u'_1,
u'_2, u'_3, u'_4)$. Then $u^*_3$ and $u^*_4$ determine the vertices $V_2$ and $V_1$ in the Lobachevsky space.
Indeed, the vector $v_1$ corresponding to the point
$V_1 \in \mathbb{L}^3$ is uniquely determined (up to scaling)  by the conditions
$(v_1, u'_1)=(v_1, u'_2)=(v_1, u'_3)=0$. Note that
the vector $u_4^*$ satisfies the same conditions. Therefore, the vectors $v_1$ and
$u^*_4$ are proportional, hence,
$$
\cosh a=\cosh \rho(V_1, V_2)=-(v_1, v_2)=-\frac{(u_3^*,u_4^*)}
{\sqrt{(u_3^*,u_3^*)(u_4^*,u_4^*)}}.
$$

It is known that $G(u^*_1, u^*_2, u^*_3, u^*_4)=G(u'_1, u'_2, u'_3, u'_4)^{-1}$,
whence it follows that $\cosh a$ can be expressed in terms of the algebraic complements
$G_{ij}$ of the elements of the matrix $G=G(u'_1, u'_2, u'_3, u'_4)$:
$$
\cosh a=-\frac{(u_3^*,u_4^*)}{\sqrt{(u_3^*,u_3^*)(u_4^*,u_4^*)}} =
\frac{G_{34}}{\sqrt{G_{33} G_{44}}}.
$$

Denote the right-hand side of the inequality from Theorem~\ref{t2.2} by $F(\overline
\alpha)$, where $\overline \alpha=(\alpha_{12}, \alpha_{13}, \alpha_{23},
\alpha_{14}, \alpha_{24})$; then this theorem implies that $\cosh
a < \cosh F(\overline \alpha)$. It follows that
\begin{equation}
\label{eq5}
\frac{G_{34}}{\sqrt{G_{33} G_{44}}} < \cosh F(\overline \alpha).
\end{equation}
For every $\overline \alpha$, in this way we obtain  a linear inequality
with respect to the number $T$.

\begin{lemma}
\label{l2.1}
The following relations are true:

{\rm(i)}~$\alpha_{12}+\alpha_{23}+\alpha_{13} > \pi,
\quad \alpha_{12}+\alpha_{24}+\alpha_{14} > \pi$;

{\rm(ii)}
\begin{alignat*}{2}
\cos \alpha_1&=\frac{\cos \alpha_{23}+\cos \alpha_{12} \cdot \cos \alpha_{13}}
{\sin \alpha_{12} \cdot \sin \alpha_{13}}, &\qquad
\cos \alpha_2&=\frac{\cos \alpha_{24}+\cos \alpha_{12} \cdot \cos \alpha_{14}}
{\sin \alpha_{12} \cdot \sin \alpha_{14}},
\\
\cos \alpha_3&=\frac{\cos \alpha_{13}+\cos \alpha_{12} \cdot \cos \alpha_{23}}
{\sin \alpha_{12} \cdot \sin \alpha_{23}}, &\qquad
\cos \alpha_4&=\frac{\cos \alpha_{14}+\cos \alpha_{12} \cdot \cos \alpha_{24}}
{\sin \alpha_{12} \cdot \sin \alpha_{24}}.
\end{alignat*}
\end{lemma}

\begin{proof}
To prove both parts of the lemma, we intersect each trihedral angle
with the vertices $V_1$ and $V_2$ by spheres centered at these points.
In the intersection we obtain spherical triangles the angles  of
which are the dihedral angles $\alpha_{ij}$, and the lengths of their edges
are the flat angles $\alpha_k$. This  implies at once the first assertion, and
the second one follows from the dual cosine-theorem for these triangles (see, for
example, \cite[p.\,71]{AVS88}). The lemma is proved.
\end{proof}

\begin{figure}[!htp]
\begin{center}
\includegraphics[scale=1.2]{bog04.pdf}
\caption{Coxeter diagram of the edge with angles
$(\pi/6, \pi/2, \pi/2, \pi/2, \pi/2)$}
\label{fig4}
\end{center}
\end{figure}

It is known that the dihedral angles of the fundamental polyhedron of an arithmetic
hyperbolic reflection group with a ground field $\mathbb{Q}$ can equal only
$\pi/2$, $\pi/3$, $\pi/4$, and $\pi/6$.

It is easy to verify that, taking into account Lemma~\ref{l2.1},\,(i), there are exactly
$44$ different (up to numbering) sets of angles $\overline \alpha$.
For each such set  $\overline \alpha$ inequality~\eqref{eq5} gives some
bound $T < t_{\overline \alpha}$.

\begin{table
\begin{center}
\caption{Coxeter diagrams of the outermost edge}
\vskip1mm
\includegraphics[scale=1.2]{tab.pdf}
\label{tab}
\end{center}
\end{table}

 For solving the $44$ linear inequalities a program was compiled
 in the computer algebra system Sage\footnote
The Sage Developers, \textit{the Sage Mathematics Software System
(Version~7.6)}, SageMath, \href{http://www.sagemath.org}{http:/\!/www.sagemath.org}, 2017},
its code is available on the Internet\footnote
N.~Bogachev, \textit{Method of the outermost edge/bounds},
\href{https://github.com/nvbogachev/OuterMostEdge/blob/master/bounds.sage/}
{https:/\!/github.com/nvbogachev/\allowbreak OuterMostEdge\allowbreak
/blob/master/bounds.sage/}, 2017}.

The obtained results are presented in Table~\ref{tab} in the form of a set of
Coxeter diagrams for the faces  $F_1$, $F_2$, $F_3$, $F_4$. The faces $F_3$ and $F_4$ will be
connected by a dotted line, and the whole diagram will be signed by the relevant
bound: $T < t_{\overline \alpha}$.

An example of how an edge diagram looks like for~$\overline \alpha=(\pi/6, \pi/2,
\pi/2, \pi/2, \pi/2)$ see in Figure~\ref{fig4}. In this figure we see that
$t_{(\pi/6, \pi/2, \pi/2, \pi/2, \pi/2)}=2.87$.

Thus, we have proved the following theorem.

\begin{theorem}
\label{t2.3}
The fundamental polyhedron of every $\mathbb{Q}$-arithmetic cocompact
 reflection group in~$\mathbb{L}^3$ contains an edge of width less than $t_{\overline
\alpha}$, where $t_{\overline \alpha}$ is the number (depending on the set $\overline
\alpha$), specified in Table~\ref{tab}. Moreover,
$$
\max_{\overline \alpha} \{t_{\overline \alpha}\}=t_{(\pi/4, \pi/2, \pi/3,
\pi/3, \pi/2)}= 4.14.
$$
\end{theorem}

The numbers given in Table~\ref{tab} were calculated on a computer with accuracy
up to eight decimal places. The table shows them rounded up to the nearest hundredth, which is
quite enough for our purposes. \textit{Note: the numbering of the faces on each diagram is
the same as in the Figure~\ref{fig2}.}

Let $E$ be the edge (chosen in Theorem~\ref{t2.3}) of width less than  $t_{\overline
\alpha}$ for some~$\overline \alpha$ in the fundamental polyhedron $P$ of the group
 $\mathcal{O}_r (L)$, and let $(u_3, u_3)=k$, $(u_4, u_4)=l$. Then
\begin{equation}
\label{eq6}
|(u_3, u_4)| < t_{\overline \alpha} \cdot \sqrt{kl}.
\end{equation}
Note that this bound is much better than estimate~\eqref{eq2}.

\section{Quadratic lattices}
\label{s3}

In this section we give some necessary information about
indefinite quadratic lattices. For more details, see, for example,~\cite{Kas78} or \cite{Vin84}.

Let $A$ be a principal ideal ring. A \textit{quadratic $A$-module} is a
free $A$-module of finite rank equipped with a non-degenerate
 symmetric bilinear form
with values in $A$, called an \textit{inner product}. In particular,
a quadratic $\mathbb{Z}$-module is called
a \textit{quadratic lattice}.
We denote by $[C]$ the standard module $A^n$ whose
inner product is defined by a Gram matrix $C$.
\textit{A quadratic space} is a free $\mathbb{F}$-module of finite rank over $\mathbb{F}$.

The determinant of the Gram matrix of a basis of a module $L$  is called
a \textit{discriminant} $d (L)$ of the quadratic $A$-module $L$.
It is defined up to multiplication by an element of $(A^*)^2 $ ($A^*$
denotes the group of invertible elements of the ring $A$) and can be regarded as
an element of the semigroup $A /(A^*)^2$.
A quadratic $A$-module $L$ is called \textit{unimodular} if $d(L) \in A^*$.

A nonzero vector $x \in L $ is called \textit{isotropic} if $(x, x) = 0$.
A quadratic
module $L$ is called \textit{isotropic} if it contains at least one
isotropic vector,
otherwise $L$ is called \textit{anisotropic}.

Since $(\mathbb{Z}^*)^2 = {1}$, the discriminant $d (L)$ of a
quadratic lattice $L$
is  an integer number. The unimodularity of a quadratic lattice $L$ is
equivalent to the property that $L$ coincides with its \textit{conjugate} lattice
$$
L^* = \{x \in L \otimes \mathbb{Q} \mid (x, y)
\in \mathbb {Z} \ \  \forall y \in L \}.
$$

For a lattice $L$, the invariant factors of the Gram matrix of a basis of $L$
are called \textit {invariant factors} of the lattice $L$.
Every quadratic lattice $L$ generates the quadratic real vector
space
$L_{\infty} = L \otimes \mathbb{R}$ and, for any prime~$p$, the quadratic $\mathbb{O}_p$-module
$L_p = L\,{\otimes}\,\mathbb{O}_p$, where $\mathbb {O}_p$ is the ring
of $p$-adic numbers.
The signature of
 the lattice $ L $ is defined as the signature of the space $ L_{\infty}$.
 It is obvious that if two quadratic
lattices $L$ and $M$ are isomorphic, then
they have the same signature and $ L_p \simeq M_p $ for any prime $p$.
The converse is also true under the following conditions:

(i) $L$ is indefinite;

(ii) for any prime $p$, the lattice $L$ has two invariant factors
divisible by  the same power of $p$.

The structure of quadratic $\mathbb{O}_p$-modules
can be described as follows.
Each such module
$L_p$ admits the \textit{Jordan decomposition}
$$
L_p = L_p^{(0)} \oplus [p] L_p^{(1)} \oplus [p^2] L_p^{(2)} \oplus \cdots,
$$
where all $L_p^{(j)}$ are unimodular quadratic $\mathbb{O}_p$-modules.
These unimodular modules are determined by $L$ uniquely up to an isomorphism, unless
$p \not = 2$. In the case $p = 2$ the rank and the parity of each such module are uniquely
determined by $L$.

\begin{proposition}[{\rm(See, for example, \cite{Vin84})}]
\label{p3.1}
If $L$ is a maximal quadratic lattice that is not contained
in any other quadratic lattice, then
$$
L_p=L_p^{(0)} \oplus [p] L_p^{(1)}
$$
for all primes $p\mid d(L)$.
\end{proposition}

\begin{definition}
\label{d3.1}
Let $a, b \in \mathbb{Q}^*_p$. Set
$$
(a, b)_p =
\begin{cases}
1 & \text{if the equation } ax^2 + by^2 = 1 \text{ has a solution in }\mathbb{Q}^*_p,
\\
-1 & \text{otherwise.}
\end{cases}
$$
The number $(a, b)_p $ is called the \textit{Hilbert symbol}.
\end{definition}

It is known that the group $ \mathbb{Q}^*_ p / (\mathbb{Q}^*_ p)^2$ can be
regarded as a vector space over $\mathbb{Z}_2 = \mathbb{Z} / 2 \mathbb{Z}$
of rank $2$  for $p \not = 2$
(respectively, of rank $3$ for $p = 2$).
The Hilbert symbol is a non-degenerate symmetric bilinear
form on this vector space.

Recall the definition of a Hasse invariant for an arbitrary quadratic
space $V$ over the field $\mathbb{Q}_p$. Let $a_1, \dots, a_m$ be the squares of lengths
of vectors of some orthogonal basis of the space $V$.

\begin{definition}
The number
$$
\varepsilon_p (f) = \prod_{i <j} (a_i, a_j)_p
$$
is called the Hasse invariant of the quadratic space $V$.
\end{definition}

The following assertions are known.

\begin{theorem}[{\rm See, for example, \cite[Lemma~2.6]{Kas78}}]
\label{t3.1}
A quadratic space $V$ of rank $4$ over the field $\mathbb{Q}_p$
is anisotropic if and only if the following conditions hold:

{\rm(1)}~$d(V) \in (\mathbb{Q}^*_p)^2$;

{\rm(2)}~$\varepsilon_p (V)=-(-1, -1)_p$.
\end{theorem}

\begin{theorem}[{\rm Strong Hasse principle, see, for example,
\cite[Theorem~1.1]{Kas78}}]
\label{t3.2}
A quadratic space $V$ over the field $\mathbb{Q}$ is isotropic if and only if
$V \otimes \mathbb{Q}_p$ is isotropic for all $p$, including $\infty$.
\end{theorem}

\begin{theorem}[{\rm Weak Hasse principle, see, for example,
\cite[Theorem~1.2]{Kas78}}]
\label{t3.3}
Two rational quadratic spaces are isomorphic over $\mathbb{Q}$ if and only if
they are  isomorphic over $\mathbb{R}$ and over all $\mathbb{Q}_p$.
\end{theorem}

\section{Methods of testing a lattice for $(1{,}2)$-reflectivity}
\label{s4}

\subsection{Vinberg's algorithm}
\label{ss4.1}

As it was said before, in 1972, Vinberg suggested an algorithm of constructing the fundamental
polyhedron of a hyperbolic reflection group. This algorithm is theoretically applicable to
any hyperbolic reflection group, but practically it is efficient only for groups of the
form $\mathcal{O}_r (L)$ (and also of the form $\mathcal{O}^{(2)}_r (L)$, etc.).

In this subsection we describe Vinberg's algorithm, following \cite{Vin72} and \cite{Vin73}.
We pick a point $v_0 \in \mathbb{L}^n$, which we shall call a
\textit{a basic point}. The fundamental domain $P_0$ of its stabilizer
$\mathcal{O}_r(L)_{v_0}$ is a polyhedral cone in $\mathbb{L}^n$. Let
$H_1, \dots , H_m$ be the sides of this cone and let $a_1, \dots , a_m$ be the
corresponding outer normals. We define the half-spaces
$$
H_k^-=\{x \in \mathbb{E}^{n,1}\mid (x, a_k) \le 0 \}.
$$
Then $P_0=\bigcap_{j=1}^m H_j^-$.

There is the unique fundamental polyhedron $P$ of the group
$\mathcal{O}_r(L)$ contained in~$P_0$ and
containing the point $v_0$. Its faces containing $v_0$ are formed by the cone faces
$H_1, \ldots, H_m$. The other faces $H_{m+1}, \ldots $ and the corresponding outer normals
 $a_{m+1} , \ldots $ are constructed by induction. Namely, for $H_j$ we take a mirror
 such that the root $a_j$ orthogonal to it satisfies the conditions:

1)~$(v_0, a_j) < 0$;

2)~$(a_i, a_j) \le 0$ for all $i < j$;

3)~the distance $\rho(v_0, H_j)$  is minimal subject to constraints~1) and~2).

The lengths of the roots of a lattice $L$ satisfy the following condition.

\begin{proposition}[{\rm(Vinberg, see \cite[Proposition~24]{Vin84})}]
\label{p4.1}
The squares of lengths
of roots in the quadratic lattice $L$ are divisors
of the doubled last invariant factor of $L$.
\end{proposition}

The fundamental polyhedron of a group $\mathcal{O}_r (L)$ is a Coxeter polyhedron and is
determined by its Coxeter diagram. A polyhedron is of finite volume if and only if it is
the convex span of a finite number of usual points or points at infinity of the space $\mathbb{L}^n$.
Every vertex of a finite-volume Coxeter polyhedron corresponds either to an elliptic subdiagram
or rank $n$ (usual vertices) of its Coxeter diagram or to a parabolic subdiagram
or rank $n-1$ (vertices at infinity). Thus, according to the Coxeter diagram, one can determine
whether a polyhedron has a finite volume. For more details, see, for example, Vinberg's papers
 \cite{Vin67} and \cite{Vin85}.

Some efforts to implement Vinberg's algorithm by using a computer have been made since
the 1980s, but they all dealt with particular lattices, usually with an orthogonal basis.
Such programs are mentioned, e.g., in the papers of Bugaenko (1992, see \cite{Bug92}),
Scharlau and Walhorn (1992, see \cite{SW92}),
Nikulin (2000, see \cite{Nik00}), and Allcock (2011, see \cite{All12}). But the programs
themselves have not been published; the only
exception is Nikulin's paper, which contains a program code for lattices of several different special
forms. The only known implementation published along with a detailed documentation is
Guglielmetti's 2016
program\footnote{see\ \href{https://rgugliel.github.io/AlVin}{https:/\!/rgugliel.github.io/AlVin}},
processing hyperbolic lattices with an orthogonal basis with square-free invariant
factors over several ground fields.
Guglielmetti used this program in his thesis (2017, see \cite{Gug17}) to classify reflective hyperbolic lattices
with an orthogonal basis with small lengths of its elements.
His program works fairly well in all dimensions in which
reflective lattices exist.

In this paper, we use the program created in 2017 by the author jointly with
A.\,Yu.~Perepechko. This program is available on the Internet
(see \cite{VinAlg2017}), and one can find its detailed description in \cite{BP18}.

\subsection{The method of ``bad'' reflections}
\label{ss4.2}

If we can construct the fundamental polyhedron of the group $O_r (L)$
for some reflective lattice $L$, then it is easy to determine whether it is $(1{,}2)$-reflective.
One can consider the group $\Delta$ generated by the $k$-reflections for $k > 2$ (we shall call them
``bad'' reflections)
in the sides of the fundamental polyhedron of the group $O_r (L)$.
The following lemma holds (see~\cite{Vin07}).

\begin{lemma}
\label{l4.1}
A lattice $L$ is $(1{,}2)$-reflective if and only if it is reflective and the group
$\Delta$ is finite.
\end{lemma}

Actually, to prove that a lattice is not $(1{,}2)$-reflective,
it is sufficient to construct only some part of the fundamental polyhedron containing
an infinite subgroup generated by bad reflections.

\section{Short list of candidate-lattices}
\label{s5}

\subsection{Plan for finding a short list}
\label{ss5.1}
Let $P$ be the fundamental polyhedron of the group $\mathcal{O}^{(1{,}2)}_r (L)$
for an anisotropic hyperbolic lattice $L$ of rank $4$.
This lattice is $(1{,}2)$-reflective if and only if $P$ is compact.
By Theorem~\ref{t2.3},
every such polyhedron contains an edge $E$ of width less than
 $t_{\overline \alpha}$, where $t_{\overline \alpha} \leq 4.14$ is the number
 depending on the set $\overline \alpha$ of dihedral angles around this edge.

Let $u_1$, $u_2$, $u_3$, $u_4$ be the roots of the lattice $L$ that are the outer
normals to the faces $F_1$, $F_2$, $F_3$, $F_4$, respectively. These roots generate
some sublattice
$$
L'=[G(u_1, u_2, u_3, u_4)] \subset L.
$$

Note that the elements of the Gram matrix $G(u_1, u_2, u_3, u_4)$
can assume only finitely many different values.
Namely, the diagonal elements can equal only~$1$ or $2$,
and the absolute values of the remaining elements $g_{ij}$ must be
strictly less than $\sqrt{g_{ii} g_{jj}}$, excepting $g_{34}=(u_3,
u_4)$, whose absolute value is bounded by $t_{\overline \alpha} \sqrt{(u_3,u_3) (u_4,
u_4)}$.

Thus, we obtain a finite list of matrices $G(u_1, u_2, u_3, u_4)$.
We pick in these matrices the ones that define anisotropic lattices, and
after that we find all their possible extensions.

In order to select only anisotropic lattices, we use a computer program%
\footnote{N.~Bogachev, \textit{Method of the outermost edge/is\_anisotropic},
\href{https://github.com/nvbogachev/OuterMostEdge/blob/master/is\_anisotropic}
{https:/\!/github.com/\allowbreak nvbogachev/\allowbreak OuterMostEdge/\allowbreak
blob/\allowbreak master/is\_anisotropic}, 2017},
based on the results and methods formulated in~\S\,\ref{s3} of this paper.

We split the list of all anisotropic lattices into isomorphism classes and
take only one representative of each class.  So, now we obtain
a substantially shorter list of anisotropic lattices that are pairwise non-isomorphic.

After that we find all their finite-index extensions
and verify the resulting list of candidate-lattices on $(1{,}2)$-reflectivity
using the methods described in~\S\,\ref{s4}.

\subsection{Short list of candidate-lattices}

The final program that  creates a list of numbers $t_{\overline \alpha}$ and
then, using
this list, displays all Gram matrices $G(u_1, u_2, u_3, u_4)$, is also
available on the Internet%
\footnote{N.~Bogachev, \textit{Method of the outermost edge/CandidatesFor12Reflectivity},
\href{https://github.com/nvbogachev/OuterMostEdge/blob/master/Is_equival}
{https:/\!/github.com/\allowbreak nvbogachev/\allowbreak
OuterMostEdge/blob/master/Is\_equival}, 2017}.

As the output we obtain matrices  $G_1$--$G_{7}$, for each of which we find all corresponding
extensions.

To each Gram matrix $G_k$ in our notation, there corresponds a lattice $L_k$ that can have some
other extensions. For each new anisotropic lattice
(non-isomorphic to any previously found lattice) we introduce the notation
$L(k)$, where $k$ denotes its number:
{\allowdisplaybreaks
\begin{align*}
G_{1} &=
\begin{pmatrix}
1 & 0 & 0 & -1\\
0 & 2 & 0 & -1\\
0 & 0 & 1 & -2\\
-1 & -1 & -2 & 2
\end{pmatrix}, \qquad L_1 \simeq [-7] \oplus [1] \oplus [1] \oplus [1] := L(1),
\\
G_{2} &=
\begin{pmatrix}
1 & -1 & 0 & 0\\
-1 & 2 & 0 & -1\\
0 & 0 & 1 & -4\\
0 & -1 & -4 & 2
\end{pmatrix}, \qquad L_2 \simeq [-15] \oplus [1] \oplus [1] \oplus [1] := L(2),
\\
G_{3} &=
\begin{pmatrix}
1 & 0 & 0 & -1\\
0 & 2 & -1 & 0\\
0 & -1 & 2 & -3\\
-1 & 0 & -3 & 2
\end{pmatrix}, \qquad L_3 \simeq [-3] \oplus [5] \oplus [1] \oplus [1] := L(3),
\\
G_{4} &=
\begin{pmatrix}
2 & 0 & 0 & -1\\
0 & 2 & -1 & -1\\
0 & -1 & 1 & -2\\
-1 & -1 & -2 & 2
\end{pmatrix}, \qquad L_4 \simeq [-23] \oplus [1] \oplus [1] \oplus [1] := L(4),
\\
G_{5} &=
\begin{pmatrix}
2 & -1 & 0 & -1\\
-1 & 2 & -1 & 0\\
0 & -1 & 1 & -4\\
-1 & 0 & -4 & 2
\end{pmatrix}, \qquad L_5 \simeq [-55] \oplus [1] \oplus [1] \oplus [1] := L(5),
\\
G_{6} &=
\begin{pmatrix}
2 & 0 & 0 & -1\\
0 & 2 & 0 & -1\\
0 & 0 & 2 & -3\\
-1 & -1 & -3 & 2
\end{pmatrix}, \qquad L_6=[G_{6}] := L(6),
\\
G_{7} &=
\begin{pmatrix}
2 & 0 & -1 & -1\\
0 & 2 & -1 & -1\\
-1 & -1 & 2 &-3\\
-1 & -1 & -3 & 2
\end{pmatrix} , \qquad L_7=[G_{7}] := L(7).
\end{align*}
}

The lattices  $L(1)$--$L(5)$ are maximal. The lattice $L(6)$ has a unique extension of index  $2$
generated by the vectors
\begin{equation}
\label{eq7}
\biggl\{\frac{e_1+e_2}{2}, \frac{e_1-e_2}{2}, e_3, e_4\biggr\},
\end{equation}
where $\{e_1, e_2, e_3, e_4\}$ is a basis of the lattice $L(6)$. The Gram matrix of system
\eqref{eq7} is equivalent to the matrix $\mathrm{diag}(-7,1,1,1)$, whence it follows that
the only extension in this case is the lattice $L(1)$. Similarly, for the lattice
 $L(7)$ we find a unique extension of index $2$ generated by the vectors
\begin{equation}
\label{eq8}
\biggl\{\frac{e_1+e_2}{2}, \frac{e_1-e_2}{2}, e_3, e_4\biggr\}
\end{equation}
that is isomorphic to $L(2)$; here $\{e_1, e_2, e_3, e_4\}$ is a basis of the lattice  $L(7)$.

We observe that the lattices $L(6)$ and $L(7)$ are (unique) even sublattices of index $2$
of the lattices $L(1)$ and $L(2)$, respectively.

\section{Verification of $(1{,}2)$-reflectivity and proof of Theorem~\ref{t1.2}}
\label{s6}

It remains to verify for $(1{,}2)$-reflectivity a small number of lattices.
The lattices  $L(1)$, $L(2)$, $L(6)$ and $L(7)$ are
$2$-reflective (see \cite{Vin07}), hence, are $(1{,}2)$-reflective.

\begin{proposition}
\label{p6.1}
The lattice  $L(3)=[-3] \oplus [5] \oplus [1] \oplus [1]$ is reflective, but not
$(1{,}2)$-reflective.
\end{proposition}

\begin{proof}
For the lattice  $L(3)$ we apply Vinberg's algorithm. Our program finds seven roots:
\begin{gather*}
\begin{alignedat}{4}
a_1&=(0;0,0,-1), &\quad (a_1,a_1)&=1, &\qquad a_2&=(0;0,-1,1), &\quad (a_2,a_2)&=2,
\\
a_3&=(0;-1,0,0), &\quad (a_3,a_3)&=5, &\qquad a_4&=(1;0,3,0), &\quad (a_4,a_4)&=6,
\\
a_5&=(1;1,0,0), &\quad (a_5,a_5)&=2, &\qquad a_6&=(2;1,2,2), &\quad (a_6,a_6)&=1,
\end{alignedat}
\\
a_7=(10;6,10,5), \quad (a_7,a_7)=5.
\end{gather*}
The Gram matrix of these roots has the form
$$
G(a_1,a_2,a_3,a_4,a_5,a_6,a_7) =
\begin{pmatrix}
1 & -1 & 0 & 0 & 0 & -2 & -5\\
-1 & 2 & 0 & -3 & 0 & 0 & -5\\
0 & 0 & 5 & 0 & -5 & -5 &-30\\
0 & -3 & 0 & 6 & -3& 0 & 0\\
0 & 0 &-5 & -3 & 2 & -1 & 0\\
-2 & 0 & -5 & 0 & -1 & 1 & 0\\
-5 & -5 & -30 & 0 & 0 & 0 & 5
\end{pmatrix}.
$$
The Coxeter diagram corresponding to this Gram matrix is represented in Figure~\ref{fig5}.

\begin{figure}[!htp]
\begin{center}
\includegraphics{pict-l-5-3.png}
\caption{The Coxeter diagram of the fundamental polyhedron of the lattice $L(3)$}
\label{fig5}
\end{center}
\end{figure}

It determines a three-dimensional Coxeter polyhedron of finite volume. This diagram has no
parabolic subdiagrams, so this polyhedron is bounded
(however, this is verified by CoxIter, created and published by
R.~Guglielmetti). We observe that the roots  $a_3$, $a_4$, $a_7$ determine a group generated
by ``bad'' reflections, it is infinite, since the corresponding subdiagram
(its vertices are colored in black) contains a dotted edge. Therefore, the lattice
$L(3)$ is reflective, but not $(1{,}2)$-reflective. The proposition is proved.
\end{proof}

The non-reflectivity of the lattice $L(4)$ was proved in~\cite{Mark15}
(see also the dissertation \cite{Mcl13}).

\begin{proposition}
\label{p6.2}
The lattice  $L(5)=[-55] \oplus [1] \oplus [1] \oplus [1]$ is not
$(1{,}2)$-reflective.
\end{proposition}

\begin{proof}
It turns out that for verification of the $(1{,}2)$-reflectivity there is no need to complete Vinberg's
algorithm. The program we use finds the first eight roots
\begin{alignat*}{4}
a_1&=(0, -1, -1, 0), &\quad (a_1,a_1)&=2;&\qquad a_2&=(0, 0, 1, -1), &\quad (a_2,a_2)&=2;
\\
a_3&=(0, 1, 0, 0), &\quad (a_3,a_3)&=1;&\qquad a_4&=(2, 0, 11, 11), &\quad (a_4,a_4)&=22;
\\
a_5&=(1, -4, 4, 5), &\quad (a_5,a_5)&=2;&\qquad a_6&=(1, -2, 2, 7), &\quad (a_6,a_6)&=2;
\\
a_7&=(2, -5, 10, 10), &\quad (a_7,a_7)&=5;&\qquad a_8&=(2, 0, 0, 15), &\quad
(a_8,a_8)&=5;
\end{alignat*}
the Gram matrix of which has the form
$$
G(a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8){=}\!\!\begin{pmatrix}
2 & -1 & -1 & -11 & 0 & 0 &-5 & 0\\
 -1 & 2 & 0 & 0 & -1 & -5 & 0 &-15\\
-1 & 0 & 1 & 0 & -4 &-2 &-5 & 0 \\
-11 & 0 & 0 & 22 & -11 & -11 & 0 & -55\\
0 & -1& -4 & -11 & 2 & -4 & 0 & -35 \\
 0 & -5 & -2& -11& -4 & 2 & -10 & -5\\
-5 & 0& -5 & 0 & 0 &-10 & 5 &-70\\
0 & -15 & 0 & -55 & -35 & -5 & -70 & 5
\end{pmatrix}.
$$

It suffices to consider the subgroup generated by ``bad'' reflections with respect to the mirrors
 $H_{a_7}$ and $H_{a_8}$. Since these mirrors diverge, this subgroup is infinite. The proposition is
 proved.
\end{proof}

Thus, among the seven anisotropic candidate-lattices selected in the process of solution only four
are $(1{,}2)$-reflective. These are the lattices $L(1)$, $L(2)$, $L(6)$, $L(7)$.
Theorem \ref{t1.2} is proved.

redpajama_set_name: RedPajamaArXiv


Gazing into a crystal ball for inspiration is probably not the best way to plan your business, but based on 2016, it is probably going to be as accurate as any other informed prediction when looking ahead to the new year.
(1) Rents will rise, but probably not as fast as many commentators are predicting.
The various changes to the way in which landlords are taxed, and the proposals to ban fees to tenants will undoubtedly see increased costs passed on to tenants. However, the expectation that rents will increase by inflation busting percentages may be a little overblown.
It is likely that investment levels will remain somewhat subdued, but there still seems an appetite for new acquisitions.
Recent research conducted by the NLA suggests that tenant demand is starting to slow, particularly in London, which has been the engine room for rental inflation over recent years.
(2) Red tape will cause more and more problems for landlords and agents.
This is almost inevitable given the number of tweaks to the PRS in every country of the UK. But the big issue could be tenancies now coming to an end which began post-October 2015 when the requirement to supply prescribed information to tenants at the beginning of their tenancy.
2017 could well be the year of the s21 notice rejected on the basis that no-one can remember if the 'How to Rent Guide' was provided.
(3) The argument over tenant fees will rumble on for months.
(4) Value will be the watch-word for agents.
As landlords costs go up, so will their desire to preserve their margin. And in many cases this will mean a march towards self-management.
The onus is on good agents to demonstrate exactly how much value they provide, and the pitfalls of getting it wrong when you go alone.
Some things never change, and many would argue that landlords fall into this category. However, the Government's blinkered drive for 'professionalism' and 'level playing-fields' will see landlords looking at their businesses in a different light.
Making tax digital, the growing divide between Corporation and Income Tax, the swathes of new administration and the emphasis on building to rent will see the development of an enlarged professional class of landlords and savvy investors.
Long-term plans will be (even more) essential. As will high quality management and customer service.


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Hi, a long time i was finding how to administrate sqlite with other form than console commands, last week i got a miracle named SQLite Manager!.
It is a tool you can intall in your firefox browser.
I recommend it, it easy and really fast.
I use http://sqlitebrowser.sourceforge.net/ some months ago and it works great.
But this is not a firefox plugin, I think.
Maybe you are talking about a different program? Where did you download it?


The problems with passwords will go away with a new device trying to get funding on Kickstarter. Mark Thacker, the COO and co-founder of myIDkey, has a startup that developed a USB-sized device which stores and displays secure passwords and information, according to a report on dvice.com.
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text:
Friday Evening Videos: "Thing Called Love"
For her first 18 years as a recording artist, roots rocker Bonnie Raitt was essentially a cult act. The critics loved her but few people beyond a small circle of hardcore fans had even heard of her. That all changed with the release of her tenth album, Nick of Time, her first on the Capitol Records label and also the first that she made while sober (according to Bonnie herself).
Nick of Time would hit number one on the Billboard 200 chart within months of its release, and go on to sell some five million copies. It earned Bonnie Raitt four hit singles and four Grammy Awards, but more importantly, it rescued her flagging career. Nick of Time was my introduction to Bonnie, who became one of my favorite musicians of the '90s, and its mixture of authentic, analog-style R&B, blues, pop and country pointed the way ahead for my musical interests when the pop music of the '80s morphed into something that no longer spoke to me, and grunge emerged from the shadows to kill off rock and roll.
The album also became an unlikely soundtrack for one of the most pivotal years of my life. I was 19 going on 20 in the summer of 1989. I started my first real (and in many ways still my best) job, working at that infamous multiplex movie theater I've mentioned so many times. I was making new friends, I still believed in my dreams for my future… and I was in love. The first single from Nick of Time, "Thing Called Love," was the background for these good times. And then later that year, when the weather grew cooler and the good times started to curdle, the plaintive sounds of the album's second single "Have a Heart" articulated all too keenly what I was feeling. So keenly in fact, that I had to stop listening to Nick of Time for a while.
But let's not dwell on that. To celebrate the 30th anniversary of Nick of Time — good lord, how can it be 30 years since that summer?! — let's instead listen to "Thing Called Love." This was actually a cover of a John Hiatt song, but Bonnie takes full possession of it through her sassy delivery and slinky slide-guitar playing. It was her biggest hit to that point, peaking at number 11 on the rock charts. I imagine the video added a bit of lift, thanks to an unexpected appearance by Hollywood hunk Dennis Quaid and the joyful, flirtatious energy passing between him and Bonnie. God, how I loved this song. On the strength of this one song, I went to see Bonnie Raitt live that summer, when she opened for the Steve Miller Band at an outdoor show that was cut short by a torrential rainstorm. I didn't buy the full album until after that, from what I recall.
One final observation: in the video, Quaid wears a t-shirt that sports the logo of Sun Records, which is of course the legendary Memphis recording studio that hosted Elvis Presley, Johnny Cash and Jerry Lee Lewis, among others. Quaid would star as Jerry Lee Lewis in the biopic Great Balls of Fire later in that very same summer of 1989. Everything is connected, man…
If I've piqued your interest at all with my little rambling here, I recommend this oral history that's all about how Nick of Time came to be… fascinating stuff for a true music lover!
Posted in: Friday Evening Videos
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 — железнодорожная линия японского железнодорожного оператора East Japan Railway Company, протянувшаяся от станции  расположенной в специальном районе Тиёда,Токио до станции  в городе Тёси префектуры Тиба.

Название линии происходит от названия древних провинций бывшей территории которых проходит линия: Мусаси (武蔵国), Симоса (下総国) and Кадзуса (上総国). Официальный цвет линии — жёлтый.

Западная городская часть линии носит название . Так же определённые виды обслуживания на линии носят другие названия упомянутые ниже. Название «Главная» используется касательно данной линии на участке к востоку от станции . Надписи на станциях и в вагонах и голосовые сообщения на станциях так же придерживаются этих обозначений.

Описание
 Расстояния:
  — : 120,5 км
  — : 4,3 км
  — Эттюдзима (Грузовая линия Эттюдзима): 11,7 км
  —  (Грузовая линия Синканэ): 8,9 км
 Пути
 Четырёхпутный участок: Кинситё — Тиба
 Двухпутные участки: Токио — Кинситё, Кинситё — Отяномидзу, Тиба — Сакура
 Однопутные: Сакура — Тёси, Син-Коива — Эттюдзима, Син-Коива — Канамати
 Система управления движением поездов
 АТОС: Токио — Тиба, Отяномидзу — Тиба.

История
Линия была построена честной железнодорожной компанией  . Первый участок от станции Итикава до станции Сакура был открыт 20-го июля 1984-го года. В декабре того же года линия была продленная до станции Кинситё, а к 1904 году до станции  (ныне Рёгоку) на берегу реки Сумида. До конечной нынешней конечной станции на востоке Тёси линия была доведена в 1987-м году.

В 1907 году линия была национализирована, а до конца 1908-го года был проложен второй путь на участке от станции Рёгокубаси до станции Тиба. В 1932-м году была открыта новая пассажирская линия от станции Отяномидзу до Рёгоку. В 1935-м году была завершена электрификация участка до станции Тиба, и началось сквозное сообщение с линией Тюо.

Долгое время в сельское местности префектуры Тиба использовались паровозы. Рыба и соевый соус из Тёси являлись основними грузами на данной линии. До 1958 года на линии не было скорых поездов из-за близости её к Токио.

в 1960-х годах, была начата  призванное улучшить транспортную систему. Она включала в себя планы по прокладке скорой линии от станции Токио до станции Тиба. Новая подземная ветка от Токио до станции Кинситё, а также четырёхпутный участок до станции Цуданума были открыта в 1972-м году. В 1987-м году линия перешла в ведение East Japan Railway Company (JR East). Идущая параллельно, но ближе к берегу линия Кэйё, первоначально предназначенная для грузовых перевозок, но адаптированная для позже для пассажирских, была открыта в 1990-м году. Несмотря на это в часы пик линия Собу всё ещё сильно загружена.

На востоке Тибы, линия была полностью электрифицирована в 1974-м году. В 1975-году начинает ходить экспресс «Сиосай» от станции Токио до станции Тёси. Грузовые перевозки от станции Тёси были прекращены в 1986-м году. 29 мая 1999 года на участке Отяномидзу — Тиба была введена система АТОС. 30 сентября 2000 года на участке Токио — Макухари была введена система АТОС.

Виды обслуживания
Местные составы ходят от станций  и  по линии которая носит название  или . Скорые поезда отходят от станции и идут до станции Тиба по отдельным путям по линии под названием . Поезда используют одни и те же пути вне города Токио, по достижении которого местные поезда продолжают движение по линии Тюо, а скорые поезда по линии Йокосука.

Далее описан только участок линии расположенный к востоку от станции Тиба. Остальные участки описаны в соответствующих статьях.

Limited Express
Narita Express до станции  и Аямэ, Сиосай до станции  идут по линии Собу.

Rapid and Local
 и
 Все составы идущие на запад от станции Тиба. От станции Сакура, на линию Нарита. Некоторые поезда до станции Тоганэ.

 Не идут на запад от станции Тиба.

Станции

Токио — Кинситё — Тиба
Информация в статье Линия Собу (Скорая).

Отяномидзу — Кинситё — Тиба
Информация в статье Линия Тюо-Собу.

Тиба — Тёси
 Расстояния и остановки скорых поездов указаны на диаграмме линии.
 Все станции расположены в префектуре Тиба

Примечания

Ссылки
 Сайт JR East

redpajama_set_name: RedPajamaWikipedia


Leveraging political support in the respective regions while planning visits and activities overseas and aligning with the regional and EU export strategies and the economic diplomacy support initiatives in the targeted markets.
Planning of next-year's missions: the consortium will explore the possibility of organising the missions to Singapore and Malaysia, Montreal (Canada) and South Africa.
Planning the enlargement of the partnership": the consortium will contact potential clusters in the field of medical technologies, including health-related electronics, ICT.