text:
Yes, It's Real. Here's How it Happened
What re-enacting the unlikeliest accident on Chesapeake Bay tells us about boating safety.
By Jeff Moag
The first thing everyone wanted to know is if the photo was real.
It seemed too perfect not to be staged: A charter fishing boat just sitting atop a sailboat as if placed there by a giant toddler. Only this was no bubble bath. It was the Chesapeake Bay near Annapolis, Maryland and the photo was very real. A firefighter snapped it from the deck of the Ann Arundel County fireboat on what may have been the strangest call of his career: a 34-foot deadrise boat, Hunter teetering on top of Levitation, a 35-foot J/105 sailboat.
When this image went viral in the hours after the accident, boaters of all stripes wanted to know if it was real, and then how it could have happened. Maryland Natural Resources Police
A person on the sailboat is talking on the phone, casual as can be. A guy on the charter boat is leaning against a stanchion, as if waiting for a bus. The effect is both terrifying and a little bit funny, and as soon as folks realized nobody had been seriously hurt, they started sharing like crazy on social media. The photo went viral, sparking worldwide news coverage and heated debates on online sailing and powerboating forums.
Nobody could say for sure how it happened, but that didn't stop people from speculating. Even when the only information about the accident was that picture, everybody seemed to have an opinion about who was to blame—and those opinions varied a great deal depending on whether they came from sailors or powerboaters.
One sailor opened a new account on a popular power boating forum just so he could heckle the motor boaters there. Didn't they know the sailboats always have right of way? (It's not quite that simple, as we'll soon see.) Some power boaters contended the charter boat had priority because he was fishing. (He wasn't, but the point is moot. That privilege only applies to commercial fishing vessels dragging big nets or longlines.)
John Martino says all that finger pointing misses the point. The founder of the Annapolis School of Seamanship was teaching a course on the Rules of the Road when the accident happened a few minutes before noon on August 17, 2018. Before the class was over, a half-dozen people had texted the image to him.
"When I first saw that picture, my first thought is that neither captain saw the other boat," Martino says. If either captain had seen the other boat and followed the Rules of the Road the accident should never have happened, Martino said when I asked him to analyze the accident for Lessons For Life, a safe boating campaign administered by the U.S. Coast Guard. Martino is a master mariner with deep experience in both power and sail, and he knows the Chesapeake as well as anyone. Still, the more we talked about the accident, the less sense it made.
Damage to Levitation is visible on her starboard side as she's towed from the accident scene. The sailboat was heeled to starboard when Hunter struck her, likely reducing the severity of the impact. Maryland Natural Resources Police
Seven people were on the charter boat. How could all of them, including a licensed captain, fail to see a 35-foot sailboat with a 50-foot mast, in broad daylight? And if the sailor saw the motorboat coming for several minutes, why didn't he steer out of the way when he realized the other boater didn't see him?
To answer those questions, Martino agreed to re-create the moments leading up to the collision, reproducing the sightlines from both boats. He obtained the Coast Guard's accident report, which included the GPS tracks from both vessels. Then he pulled together a dream team of experienced captains to reenact the moments leading up to the collision, using a deadrise boat similar to the one involved in the accident and a J/105 sailboat identical to the one it T-boned.
Conditions for the reenactment were almost identical to the day of the accident: Clear and bright with light winds. Capt. Matt Benhoff, an instructor at the Annapolis School of Seamanship, was in the deadrise. John Stefancik, a lifelong boater and publisher of Chesapeake Bay Magazine, was on the sailboat with Capt. Karl Richter, also of the Annapolis School. Torrey Pocock and his Riggo Productions crew filmed from both boats and a drone. The result was some very dramatic video, and important insights into the Rules of the Road, the maritime traffic rules designed to avoid collisions at sea.
Before the reenactment, Martino and his team thought the view from one or both of the vessels would be obstructed. Deadrise boats have a tendency to squat down in the stern, and it's sometimes difficult to see over the bow. And a sailing vessel's sails can sometimes block the helmsman's view. At the time of the accident, Levitation's jib was on the starboard side—the same direction from which Hunter was approaching.
In the very first pass of the reenactment, however, it became obvious that neither captain's view had been blocked. Stefancik could easily see around the J/105's sails, and Benhoff's view of the sailboat was unnervingly clear. In fact, the hardest part of the whole exercise was fighting the urge to swerve away, he says.
As the distance to the sailboat kept decreasing, its position as seen from the motorboat didn't change. It was a textbook example of "constant bearing, decreasing range," which is how mariners are taught to know they're on a collision course. Just watching the reenactment video is a little uncomfortable for any experienced boater. For Benhoff, it was excruciating.
"All of my training and experience is telling me to turn," he says. "It felt really unnatural."
After the accident, people speculated online that the charter captain had been running on autopilot or steering from the deadrise's aft helm. According to the Coast Guard investigation, that was not the case. He also tested clean for alcohol and drugs.
In the very first pass of the reenactment, it became obvious that neither captain's view had been obstructed. Stefancik could easily see around the J/105's sails, and Benhoff's view of the sailboat was unnervingly clear.
The reenactment suggests a simpler conclusion: The charter captain simply didn't notice the sailboat in front of him. "It just screams Rule Five, which is lookout," Martino says. "It's one sentence long, but it says a ton. It says we're supposed to look out by sight and hearing and all available means at all times."
If you're looking for one root cause of the accident, that's it. But if you want to understand how both captains failed to avoid the collision, you need to go a little deeper.
Remember the sailor claiming he had "right of way?" That's not quite accurate. In this scenario, as in most other crossing situations involving a vessel under power and another under sail, the sailboat is the stand on vessel. Its job is to maintain course and speed, or "stand on" as Popeye would say. That's not a right, though. It's a responsibility. And as Martino explains, that responsibility evolves with the situation.
"Rule Five is one sentence long, but it says a ton. It says we're supposed to look out by sight and hearing and all available means at all times."
Capt. John Martiono, Annapolis School of Seamanship
The motorboat is the give way vessel. Its job is to avoid the sailboat, by making an obvious course change that makes its intentions clear. That's important, because if the powerboat doesn't take action to avoid a collision, the sailboat's options and responsibilities change.
"The rule is very interesting, because when it becomes apparent that the give way vessel is not taking appropriate action, the stand on vessel may take action to avoid a collision. You're released from your obligation to maintain course and speed," Martino says. Based on the reenactment, this would have been the best time for Levitation to alter course and avoid the motorboat.
The 34-foot charter boat Hunter was brand new on the day of the accident. It was repaired and put back in service, but the J/105 Levitation was a total loss. Maryland Natural Resources Police
"In the final stage the stand on vessel is directed that it shall take action when action by the give way vessel alone is not going to be sufficient to avoid a collision," Martino says. By that time, though, it might not be so easy to get out of the way. Stefancik, who was in the J/105 during the reenactment, estimates he could easily have tacked or jibed out of the motorboat's path until about 20 seconds before impact. After that, he was out of good options.
Re-enacting one of the Chesapeake Bay's unlikeliest collisions led to a few surprises, but mostly it confirms what we already know about safe boating. Keep a good lookout. Know and follow the Rules of the Road. But more simply, don't forget about courtesy and common sense.
"You just have to keep your head on a swivel, and then make decisions and take action to avoid a collision," Martino says. "In fact, that's even one of the rules."
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text:
Surawahadi: Promotion has to be accompanied by positive change
Home BALINEWS Surawahadi: Promotion has to be accompanied by positive change
JBN.CO.ID ■ Promotion in essence is a form of recognition, trust and appreciation from the nation and state, for their dedication and achievement in carrying out the duties and responsibilities of their assigned positions.
This was conveyed by Pangdam XIV Hasanuddin Mayjen TNI Surawahadi while presiding over the report on the corps of the Hasanuddin Kodam XIV commander for the period of April 1, 2019, taking place at the Hasanuddin Meeting Hall, Jln Urip Sumohardjo Makassar, on Monday (01/04/2019).
Surawahadi said that promotion must be grateful as a gift accompanied by positive changes to thinking that is further ahead, more creative, innovative and productive in working to realize better unit performance.
"The higher the rank a person achieves, the heavier the burden and responsibility on his shoulders," said the Hasanuddin Commander.
The atmosphere of excitement and happiness and happiness radiated to the look on the faces of the soldiers and their families who were on the rise, when the Hasanuddin Military Commander along with other key Kodam officials, gave greetings while shaking hands to the rank-and-file personnel.
As for the total number of ranks of Hasanuddin Kodam XIV who rose in the period of April 1, 2019, there were 922 people, including the rank of 82 officers, Bintara 587 people, enlisted 178 people and civil servants totaling 75 people.
■ Rasyid
Tags BALINEWS
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BRITAIN'S most decorated Winter Olympian Lizzy Yarnold has announced her retirement from skeleton, saying she is "ready for a new chapter" in her life.
"It has been an amazing 10 years of my life, and I have been lucky enough to learn from amazing coaches and be part of this team," Yarnold told the BBC.
Super Bowl 2019: Who is playing the halftime show in Atlanta?


Hi my name is Steve Garrity; my passion for photography began 40 years ago when my dad bought me my first SLR camera. I had no idea how much photography would play in my life. I was totally hooked; I read up on photography and was influenced by the best pro photographers of that time people like Norman Parkinson, Patrick Litchfield, David Bailey, Terry O'Neil, Duffy and later Rankin. At first I would just take photos for my own pleasure I had a passion for portrait and fashion photography. Very soon I was asked to photograph, friends and family, and then friends of friends wanted their photos taken too. Finally I had found something that I enjoyed and was good at. A good friend suggested and encouraged me to setup up as a professional photographer.....the idea terrified me, but with a lot of support I eventually created Garrity Photography.
I try to minimise the amount of manipulation I do, but I believe manipulation is part of the photographer's art as much as darkroom work. As long as it starts with a photograph it is okay by me.
Please comment upon my photos, I am always looking for ways to improve my shots and would welcome any tips. All comments are welcome, especially critiques from which I can learn and hopefully improve.
I try to leave constructive critique on any photo I view (if possible). If you don't agree with what I say you can always ignore it.


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text:
Q: Missing right parenthesis at line 1 So i'm very new to oracle just doing it for my assignment. i created a table before and it was created successfully. Now i'm creating an other table and i want to give eid the reference of my first table but it is giving me the error of missing right parenthesis. I searched it but couldn't solve it. Please help me for the sake of my assignment.
Create Table Tab2 (eid number reference tab1 (eid), address1 varchar2 (25), address2 varchar2 (15));

A: As mentioned in comment, use REFERENCES
CREATE TABLE Tab2
(eid number REFERENCES tab1 (eid),
address1 varchar2 (25),
address2 varchar2 (15)
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A: Create Table Tab2 (eid number references tab1 (eid), address1 varchar2 (25), address2 varchar2 (15));

A: Welcome to Oracle!
What are you using to work with the database? SQL*Plus? SQL Developer?
If you're using SQL Developer, it can help you answer these questions.

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Who am I it doesn't really matter. What is important is what I'm trying to do: escape the rat race and look for a tropical paradise. Is it going to be the way I see it in my dreams? Stay tuned to find out!
If you wonder how was my first day of being jobless, it was good, thanks for asking. Felt like Saturday. I also got a lot done.
The first time I thought about taking a long journey was few years ago, right after I came back from Peru. I was quite surprised how easy I could communicate and find my way around. I was thinking that will be easy to travel through the whole Central and South America, since they mostly speak Spanish (except for Brazil). At that time, it seemed to be just another daydream.
Very long time ago I made a habit to read the news every morning, right after I read my emails. I like to know what's going on in the world. One day I came across an article that caught my attention. It was about a British company that offers overland expeditions, London to Sydney, in 3 months, at a reasonable price. Pretty cool, huh? So I searched more and found another company that was doing the same thing, just a little bit more expensive, but in 8 months. I definitely think it was worth it. One problem was that I wasn't interested in a big part of the route, London to Pakistan. The other problem was the time. How the hell could I get 8 months vacation? It seemed impossible. Yeah, right, like this is ever gonna happen. But the desire was there. It will come back now and then, and the more I ignored it, the stronger it resurfaced. It would just not go away.
A couple of months ago, being very unhappy with the salary I was getting, I looked for another job. I didn't take long at all to get at interview for a job that way paying 80% (you read correctly, eighty percent) more. But when I had to take the job I freaked out. Should I have taken the job I would just kiss goodbye my dream of traveling around the world. Basically, I had to chose between time and money. So I look down into my soul and I made the decision.(Now, let me state here, just for the record, that a straight up alpha male type cool dude like me never uses this kind of expression - looking down into the soul - but at the moment I lack the inspiration).
It wasn't until I bought the plane ticket when I realized that I am actually going to do it. I felt butterflies into my stomach. The level of excitement was so high I could barely sleep for 3 nights in a row. I believe this marked the point of no return. As for the reasons I am doing this, I could make a really long list, but the bottom line is I want it, I can do it, and there is no better time than now.
Oh, by the way, I have created another blog here.


text:
\section{Introduction and Preliminaries}\label{intro}
\setcounter{equation}{0}

Let $h=(h_1,\dots,h_k)$ be a ${\mathbb C}^k$-valued Hermitian form on ${\mathbb C}^n$, with $n,k\ge 1$. The form $h$ is called {\it non-degenerate}\, if the following two conditions are satisfied:
\vspace{-0.3cm}\\

\noindent (i) the scalar Hermitian forms $h_1,\dots,h_k$ are
linearly independent\linebreak over ${\mathbb R}$;
\vspace{-0.3cm}\\

\noindent (ii) $h(z,z')=0$ for all $z'\in{\mathbb C}^n$ implies $z=0$.
\vspace{-0.3cm}\\

\noindent For a non-degenerate $h$ one has $k\le n^2$. Note that many authors define a non-degenerate Hermitian form as a form satisfying condition (ii) alone.

To any ${\mathbb C}^k$-valued Hermitian form $h$ on ${\mathbb C}^n$ one associates the quadric $Q_h\subset{\mathbb C}^{n+k}$ of CR-dimension $n$ and CR-codimension
$k$ as follows:
$$
Q_h:=\{(z,w)\in{\mathbb C}^{n+k}:\mathop{\rm Im}\nolimits w=h(z,z)\},
$$
where $z=(z_1,\dots,z_n)$ is a point in ${\mathbb C}^n$, and $w=(w_1,\dots,w_k)$ is a point in ${\mathbb C}^k$. The CR-manifold $Q_h$ is called the {\it quadric associated to $h$}.

If $h$ is non-degenerate, then any $C^1$-smooth CR-isomorphism between domains in $Q_h$ extends to a birational map of ${\mathbb C}^{n+k}$ (see the classical papers \cite{Po}, \cite{Tan1}, \cite{A} for $k=1$ and the papers \cite{KT}, \cite{F}, \cite{Tum}, \cite{Ka1}, \cite{Su}, \cite{B1}, \cite{B2} for $1<k\le n^2$). These birational maps form a group (this is not obvious at all and requires a justification -- see Remark \ref{propstar}). We denote this group by $\mathop{\rm Bir}\nolimits(Q_h)$ and call the {\it group of birational transformations of $Q_h$}. For $k=1$ every element of $\mathop{\rm Bir}\nolimits(Q_h)$ is a linear fractional transformation induced by an automorphism of ${\mathbb C}{\mathbb P}^{n+1}$ (see \cite{Po}, \cite{Tan1}, \cite{A}). For some Hermitian forms $h$ with $1<k\le n^2$ formulas for elements of certain subgroups of $\mathop{\rm Bir}\nolimits(Q_h)$ were given in \cite{ES2}, \cite{ES3}. It was shown in \cite{Tum} that the group $\mathop{\rm Bir}\nolimits(Q_h)$ can be endowed with the structure of a Lie group (possibly with uncountably many connected components) with Lie algebra isomorphic to the Lie algebra of all infinitesimal CR-automorphisms of $Q_h$, where a smooth vector field on $Q_h$ is called an infinitesimal CR-automorphism if in a neighborhood of every point of $Q_h$ translations along the integral curves of the vector field form a local group of local CR-automorphisms. Every infinitesimal CR-automorphism of $Q_h$ is known to be polynomial. We will see below that $\mathop{\rm Bir}\nolimits(Q_h)$ can be embedded in a natural way into the complex group $\mathop{\rm PGL}\nolimits_N({\mathbb C})$ as a closed real subgroup (see Corollary \ref{generalcor} and Remark \ref{quadr}).

We are interested in regularizing the elements of the group $\mathop{\rm Bir}\nolimits(Q_h)$ as stated in Definition \ref{regulariz} below. This definition applies to more general CR-submanifolds $M$ of a finite-dimensional complex vector space $E$ than quadrics, and we will first introduce $\mathop{\rm Bir}\nolimits(M)$, the {\it group of birational transformation of $M$}. Throughout the paper $M$ is assumed to be connected, locally closed, real-analytic and generic in $E$.

For a rational map $g$ of $E$, we denote by $\mathop{\rm reg}\nolimits(g)$ the subset of all regular points of $g$. Let $\mathop{\rm Bir}\nolimits(E)$ be the group of all birational transformations of $E$. The restriction of $g\in\mathop{\rm Bir}\nolimits(E)$ to the Zariski open subset $\mathop{\rm reg}\nolimits(g)$ defines a biholomorphic map $\mathop{\rm reg}\nolimits(g)\rightarrow\mathop{\rm reg}\nolimits(g^{-1})$. Denote by $\mathop{\rm BR}\nolimits(M)$ the collection of all elements $g\in\mathop{\rm Bir}\nolimits(E)$ with the following property: there exists a non-empty domain $V\subset M$ with $V\subset\mathop{\rm reg}\nolimits(g)$ and $g(V)\subset M$. Note that in general elements of $\mathop{\rm BR}\nolimits(M)$ are not defined on all of $M$: they may have poles and points of indeterminacy on $M$. It is clear that $(\mathop{\rm BR}\nolimits(M))^{-1}=\mathop{\rm BR}\nolimits(M)$. In general, however, $\mathop{\rm BR}\nolimits(M)$ is a proper subset of $\mathop{\rm BR}\nolimits(M)\cdot\mathop{\rm BR}\nolimits(M)$. We define $\mathop{\rm Bir}\nolimits(M)$ to be the subgroup of $\mathop{\rm Bir}\nolimits(E)$ generated by $\mathop{\rm BR}\nolimits(M)$.

One can give a sufficient condition that guarantees that $\mathop{\rm Bir}\nolimits(M)=\mathop{\rm BR}\nolimits(M)$. Recall, first of all, that $M$ is called {\it minimal at a point}\, $a\in M$, if there does not exist a CR-submanifold $M_0\subset M$, with $\dim M_0<\dim M$ and\linebreak CR-dim$M_0$=CR-dim$M$, passing through $a$. The manifold $M$ is called {\it minimal}\, if it is minimal at its every point.

Let $M$ be a connected real-analytic generic CR-submanifold of $E$. For such $M$ we introduce the following
$$
\begin{array}{l}
\hbox{{\bf Condition \boldmath{$(*)$}:}}\\
\vspace{-0.1cm}\\
\hbox{(a) $M$ is minimal,}\\
\vspace{-0.1cm}\\
\hbox{(b) $M_1\subset M$ holds for every connected real-analytic submanifold}\\
\vspace{-0.3cm}\\
\hspace{0.7cm}\hbox{$M_1\subset E$ such that $W\cap M=W\cap M_1\ne\emptyset$ for some domain $W$ in $E$.}
\end{array}
$$
In Proposition \ref{cond8} in Section \ref{birational} we show that if $M$ satisfies Condition $(*)$ then $\mathop{\rm Bir}\nolimits(M)$ coincides with $\mathop{\rm BR}\nolimits(M)$. This condition is satisfied, for example, if $M$ is minimal and closed in $E$. In particular, Condition $(*)$ is satisfied for any quadric $Q_h$ (note that part (i) of the definition of the non-degeneracy of an Hermitian form $h$ given at the beginning of Section \ref{intro} is equivalent to $Q_h$ being minimal). There are also a large number of examples of non-closed everywhere Levi degenerate CR-submanifolds satisfying Condition $(*)$. An interesting family of such CR-submanifolds is presented in Example \ref{exx2} in Section \ref{propp}.

\begin{remark}\label{propstar}\rm Proposition \ref{cond8} plays a key role in understanding the group $\mathop{\rm Bir}\nolimits(Q_h)$, but it appears to have been overlooked in the literature on quadrics so far. Indeed, many authors seem to assume without proof that the set of maps $\mathop{\rm BR}\nolimits(Q_h)$ is a group.
\end{remark}

We will now give an exact definition of what we mean by regularization. For a complex manifold $Y$ we denote by $\mathop{\rm Aut}\nolimits(Y)$ the group of all biholomorphic automorphisms of $Y$.

\begin{definition}\label{regulariz}\sl Let $M$ be a connected real-analytic generic CR-submanifold $M$ of a finite-dimensional complex vector space $E$. A subgroup $G\subset\mathop{\rm Bir}\nolimits(M)$ is said to be
\vspace{0.3cm}

\noindent {\rm (i)} {\it regularizable}\, on a complex manifold $Y$, if there exists an open holomorphic embedding $\varphi:E\rightarrow Y$ and a group homomorphism $\tau: G\rightarrow\mathop{\rm Aut}\nolimits(Y)$, such that for every $g\in G$ one has $\varphi\circ g=\tau(g)\circ\varphi$ on $\mathop{\rm reg}\nolimits(g)$;
\vspace{0.3cm}

\noindent {\rm (ii)} {\it projectively regularizable} if for a suitable integer $N$ there exists an irreducible complex algebraic subvariety $X\subset{\mathbb C}{\mathbb P}^N$, a group homomorphism $\tau: G\rightarrow\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$, and an algebraic isomorphism $\varphi: E\mapsto X_0$, where $X_0$ is a Zariski open subset of $X$, such that $\varphi\circ g=\tau(g)\circ\varphi$ on $\mathop{\rm reg}\nolimits(g)$ for every $g\in G$.
\vspace{0.3cm}

\noindent The map $\varphi$ is called a {\it regularization map}.

\end{definition}

\noindent Clearly, if $G$ is projectively regularizable, it is regularizable on the connected Zariski open subset
\begin{equation}
\hat E:=\bigcup_{g\in G}\tau(g)\varphi(E),\label{setZ}
\end{equation}
of the non-singular part $X_{\mathop{\rm reg}\nolimits}$ of $X$. The set $\hat E$ is the smallest $\tau(G)$-invariant domain in $X$ that contains $\varphi(E)$. Note also that one can assume that $X$ is not contained in any projective hyperplane in ${\mathbb C}{\mathbb P}^N$.

Regularization results for certain groups of birational transformations can be found in \cite{HZ}, \cite{Z1}. If $Q_h$ is a hyperquadric (i.e. $k=1$), the group $\mathop{\rm Bir}\nolimits(Q_h)$ is known to be projectively regularizable with $N=n+1$ due to the classical work \cite{Po}, \cite{Tan1}, \cite{A}. Further, it was shown in \cite{ES1} (see also \cite{B2}, \cite{Mi}) that for $2\le k\le n^2-1$, excluding the situation $k=n=2$, a quadric in general position has only affine automorphisms, in which case $\mathop{\rm Bir}\nolimits(Q_h)$ is projectively regularizable with $N=n+k$ for trivial reasons. In fact, we show in Section \ref{generalproof} that $\mathop{\rm Bir}\nolimits(Q_h)$ is projectively regularizable for any non-degenerate form $h$. This is a consequence of our main theorem, which applies to much more general CR-manifolds than quadrics. In order to state the theorem we need to introduce some notation and give necessary definitions.

Let $M$ be a real-analytic generic CR-submanifold of a complex manifold $Z$. In what follows all local CR-automorphisms and infinitesimal CR-automorphism of $M$ are assumed to be {\it real-analytic} (note that a $C^1$-smooth CR-isomorphism between Levi non-degenerate real-analytic CR-manifolds is in fact real-analytic -- see Theorem 3.1 in \cite{BJT}). We denote by ${\mathfrak{hol}}(M)$ the real Lie algebra of all real-analytic infinitesimal CR-automorphisms of $M$. A vector field $\xi$ on $M$ lies in ${\mathfrak{hol}}(M)$ if and only if $\xi$ extends to a holomorphic vector field on a neighborhood $U$ of $M$ in $Z$. [We think of holomorphic vector fields on $U$ as holomorphic sections over $U$ of the tangent bundle $TU$. In particular if $Z=E$, a holomorphic vector field $f(z)\dd z$ is just given by a holomorphic map $f: U\rightarrow E$.]

For $a\in M$ we denote by ${\mathfrak{hol}}(M,a)$ the real Lie algebra of all germs at $a$ of vector fields in ${\mathfrak{hol}}(V)$, with $V$ running over all open neighborhoods of $a$ in $M$. Clearly, ${\mathfrak{hol}}(M,a)$ is a real Lie subalgebra of the complex Lie algebra ${\mathfrak{hol}}(Z,a)$. By Proposition 12.5.1 of \cite{BER} the finite-dimensionality of ${\mathfrak{hol}}(M,a)$ implies that $M$ is {\it holomorphically non-degenerate at}\, $a$, i.e. the Lie algebra ${\mathfrak{hol}}(M,a)$ is totally real in ${\mathfrak{hol}}(Z,a)$ for all $a\in M$. Indeed, if $\xi$ lies in ${\mathfrak{hol}}(M,a)\cap i{\mathfrak{hol}}(M,a)$, then $\psi\cdot\xi\in{\mathfrak{hol}}(M,a)$ for any germ $\psi$ of a holomorphic function near $a$. Thus the formal complexification of ${\mathfrak{hol}}(M,a)$ is isomorphic to ${\mathfrak{hol}}(M,a)+i{\mathfrak{hol}}(M,a)\subset{\mathfrak{hol}}(Z,a)$ if $\hbox{dim}\,{\mathfrak{hol}}(M,a)<\infty$.

Let $M$ be a real-analytic generic CR-submanifold of a finite-dimensional complex vector space $E$. For such $M$ and a point $a\in M$  we introduce the following
$$
\begin{array}{l}
\hbox{{\bf Property (P) at $a$:}}\\
\vspace{-0.1cm}\\
\hbox{(a) the Lie algebra ${\mathfrak{hol}}(M,a)$ is finite-dimensional,}\\
\vspace{-0.1cm}\\
\hbox{(b) the complex Lie algebra ${\mathfrak{hol}}(M)+i{\mathfrak{hol}}(M)$ contains the complex solv-}\\
\vspace{-0.3cm}\\
\hspace{0.7cm}\hbox{able Lie algebra}
\end{array}
$$
\begin{equation}
{\mathfrak s}:=\left\{(\alpha+cz)\dd z:\alpha\in E,c\in{\mathbb C}\right\}.\label{s}
\end{equation}
Further, we say that $M$ has {\bf Property (P)} if it has Property (P) at its every point. In Section \ref{propp} we give sufficient conditions for $M$ to have Property (P) (see Proposition \ref{suffcondpropp}), and discuss several examples. In particular, every non-degenerate quadric $Q_h$ has Property (P).

We now state our main result that provides projective regularization of $\mathop{\rm Bir}\nolimits(M)$ for a large class of CR-submanifolds.

\begin{theorem}\label{general}\sl Let $M$ be a connected real-analytic generic CR-sub\-man\-i\-fold of a finite-dimensional complex vector space $E$. Assume further that $M$ has Property {\rm (P)}. Then the following holds:
\vspace{-0.3cm}\\

\noindent{\rm (I)}  for every $a\in M$ the Lie algebra ${\mathfrak{hol}}(M,a)$ consists of polynomial vector fields, hence ${\mathfrak{hol}}(M,a)={\mathfrak{hol}}(M)$;
\vspace{-0.3cm}\\

\noindent{\rm (II)} every real-analytic CR-isomorphism $g$ between non-empty domains in $M$ extends to a map lying  in $\mathop{\rm Bir}\nolimits(M)$ of the form $q(z)^{-1}p(z)$, where $p:E\rightarrow E$, $q:E\rightarrow\mathop{\rm End}\nolimits(E)$ are polynomial maps, and $\mathop{\rm reg}\nolimits(g)=\mathop{\rm reg}\nolimits(q^{-1})=\{z\in E:\det q(z)\ne 0\}$;
\vspace{-0.3cm}\\

\noindent{\rm (III)} $\mathop{\rm Bir}\nolimits(M)$ is projectively regularizable.
\end{theorem}

Our next theorem provides information on the extension of $\varphi(M)$ into ${\mathbb C}{\mathbb P}^N$. Recall that a real-analytic CR-manifold $M$ is called {\it locally homogeneous at a point}\, $a\in M$ if the evaluation map ${\mathfrak{hol}}(M,a)\rightarrow T_aM$, $\xi\mapsto\xi_a$, is surjective, and $M$ is called {\it locally homogeneous}\, if $M$ is locally homogeneous at every point (see \cite{Z2} for equivalent definitions of local homogeneity). In the theorem to follow we assume that $M$ has Property (P) at some point, satisfies part (b) of Condition $(*)$, and is locally homogeneous. Observe that these assumptions imply that $M$ has Property (P) and satisfies Condition $(*)$. Indeed, local homogeneity implies that $M$ has Property (P). Further, by Proposition 4.2 of \cite{Z2} the finite-dimensionality of ${\mathfrak{hol}}(M,a)$ and local homogeneity at $a$ for all points $a\in M$ yield that $M$ is minimal. Hence $M$ satisfies Condition $(*)$. For such a manifold $M$, we denote by $\hat M$ the unique $\mathop{\rm Bir}\nolimits(M)$-orbit in ${\mathbb C}{\mathbb P}^N$ containing $\varphi(M)$. Clearly, $\hat M$ is a connected immersed generic CR-submanifold of $\hat E$ (see (\ref{setZ})), and we denote by $\mathop{\rm Aut}\nolimits(\hat M)$ the group of all real-analytic CR-automorphisms of $\hat M$.

We now state our next result.

\begin{theorem}\label{general2}\sl Let $M$ be a connected real-analytic generic CR-sub\-man\-i\-fold of a finite-dimensional complex vector space $E$. Assume that $M$ has Property {\rm (P)} at some point, satisfies part (b) of Condition $(*)$, and is locally homogeneous. Then for the regularization map $\varphi$ and homomorphism $\tau$ arising in Theorem \ref{general} the set $\varphi(M)$ is open and dense in $\hat M$, and $\mathop{\rm Aut}\nolimits(\hat M)=\tau(\mathop{\rm Bir}\nolimits(M))$. Furthermore, if $\overline{M}\setminus M$ does not contain a CR-submanifold of $E$ locally CR-equivalent to $M$, then $\tau(\mathop{\rm Bir}\nolimits(M))$ is closed in $\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$, and the Lie algebra of $\tau(\mathop{\rm Bir}\nolimits(M))$ is canonically isomorphic to ${\mathfrak{hol}}(M)$.
\end{theorem}

For $M$ satisfying the assumptions of Theorem \ref{general2} we now introduce a Lie group structure on $\mathop{\rm Bir}\nolimits(M)$ by pulling back the Lie group structure from $\tau(\mathop{\rm Bir}\nolimits(M))$ by means of $\tau$. In this Lie group topology $\mathop{\rm Bir}\nolimits(M)$ has at most countably many connected components. In Section \ref{liegrstr} we give another sufficient condition for the existence of a Lie group structure on $\mathop{\rm Bir}\nolimits(M)$ with this property (see Theorem \ref{liegr}). It comes from the natural faithful representation of $\mathop{\rm Bir}\nolimits(M)$ on ${\mathfrak{hol}}(M)$.

Applying Theorems \ref{general}, \ref{general2}, \ref{liegr} to any quadric $Q_h$ we obtain the following corollary.

\begin{corollary}\label{generalcor}\sl If $h$ is non-degenerate, then $\mathop{\rm Bir}\nolimits(Q_h)$ is projectively regularizable, and for the regularization map $\varphi$ the set $\varphi(Q_h)$ is open and dense in a $\mathop{\rm Bir}\nolimits(Q_h)$-orbit in ${\mathbb C}{\mathbb P}^N$. The corresponding homomorphism $\tau$ maps $\mathop{\rm Bir}\nolimits(Q_h)$ onto a closed real subgroup of\, $\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$, and $\mathop{\rm Bir}\nolimits(Q_h)$ admits the structure of a Lie group with at most countably many connected components and Lie algebra isomorphic to ${\mathfrak{hol}}(Q_h)$.
\end{corollary}

For the case when $Q_h$ is the \v Silov boundary of a Siegel domain, the regularization statement of Corollary \ref{generalcor} is essentially contained in Theorem 9 of \cite{KMO}.

\begin{remark}\label{quadr}\rm For quadrics the degrees of the polynomial maps $p$ and $q$ arising in statement (II) of Theorem \ref{general} do not exceed 2. The rationality property for local automorphisms of quadrics can be derived already from the results of \cite{Ka1} (see Satz 2, p. 134). This property was also obtained in \cite{Tum}, but our arguments are simpler even for more general CR-manifolds. In addition, a Lie group structure on $\mathop{\rm Bir}\nolimits(Q_h)$ with Lie algebra ${\mathfrak{hol}}(Q_h)$ has been constructed in \cite{Tum} by means of considering the natural faithful representation $\rho$ of $\mathop{\rm Bir}\nolimits(Q_h)$ on ${\mathfrak{hol}}(Q_h)$ that maps every $g\in\mathop{\rm Bir}\nolimits(Q_h)$ into the corresponding push-forward transformation $g_*$ of vector fields in ${\mathfrak{hol}}(Q_h)$. By a general theorem due to Palais (see \cite{Pa}, Theorem VII, p. 103) the image $\rho(\mathop{\rm Bir}\nolimits(Q_h))\subset\mathop{\rm GL}\nolimits({\mathfrak{hol}}(Q_h))$ has the structure of a Lie group with Lie algebra ${\mathfrak{hol}}(Q_h)$, but this Lie group may a priori have uncountably many connected components, if $\rho(\mathop{\rm Bir}\nolimits(Q_h))$ is not closed in $\mathop{\rm GL}\nolimits({\mathfrak{hol}}(Q_h))$. No proof of closedness was given in \cite{Tum}. Our construction of a Lie group structure on $\mathop{\rm Bir}\nolimits(M)$ in Theorem \ref{general2} relies on the algebraic regularization map $\varphi:E\rightarrow{\mathbb C}{\mathbb P}^N$, while the Lie group structure arising in Theorem \ref{liegr} comes from the natural representation $\rho$ of $\mathop{\rm Bir}\nolimits(M)$ on ${\mathfrak{hol}}(M)$. In Theorem \ref{general2} we show that $\mathop{\rm Bir}\nolimits(M)$ embeds as a closed subgroup into $\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$, whereas in Theorem \ref{liegr} we prove that $\rho(\mathop{\rm Bir}\nolimits(M))$ is closed in $\mathop{\rm GL}\nolimits({\mathfrak{hol}}(M))$. The Lie group structures on $\mathop{\rm Bir}\nolimits(Q_h)$ arising from Theorems \ref{general2} and \ref{liegr} for $M=Q_h$ are identical. We also note that since the extension $\hat Q_h$ of $Q_h$ is Levi non-degenerate and has pairwise equivalent Levi forms at all points, the existence of the structure of a Lie group on $\mathop{\rm Aut}\nolimits(\hat Q_h)$ (and hence on $\mathop{\rm Bir}\nolimits(Q_h)$) with Lie algebra ${\mathfrak{hol}}(Q_h)$ in a certain topology follows from the results of \cite{Tan2}. We refer the reader to \cite{BRWZ}, \cite{LMZ} and references therein for results on the existence of Lie group structures on the groups of CR-automorphisms of more general CR-manifolds.
\end{remark}

If one does not insist on finding a projective regularization, the group $\mathop{\rm Bir}\nolimits(Q_h)$ (in fact, the group $\mathop{\rm Bir}\nolimits(M)$ for much more general $M$) can be regularized on some complex manifold in the sense of part (i) of Definition \ref{regulariz}, as follows. Consider the complexification ${\mathfrak l}$ of ${\mathfrak{hol}}(Q_h)$. The complex Lie algebra ${\mathfrak l}$ consists of polynomial vector fields of degree not exceeding 2 and has a natural grading ${\mathfrak l}={\mathfrak l}^{-1}\oplus{\mathfrak l}^0\oplus{\mathfrak l}^1$, where the Lie subalgebra ${\mathfrak l}^{-1}$ consists of all constant vector fields on $E$, and all vector fields in ${\mathfrak l}_0:={\mathfrak l}^0\oplus{\mathfrak l}^1$ vanish at the origin (see e.g. Section \ref{generalproof}). Since $[\xi,{\mathfrak l}^0]$ is not contained in ${\mathfrak l}^0$ for every non-zero $\xi\in{\mathfrak l}^{-1}$, the normalizer of ${\mathfrak l}_0$ in ${\mathfrak l}$ coincides with ${\mathfrak l}_0$. Let ${\mathfrak L}$ be the connected simply-connected group with Lie algebra ${\mathfrak l}$. The stabilizer ${\mathfrak L}_0$ of ${\mathfrak l}_0$ under the adjoint representation of ${\mathfrak L}$ is a closed complex subgroup of ${\mathfrak L}$. Since the normalizer of ${\mathfrak l}_0$ in ${\mathfrak l}$ coincides with ${\mathfrak l}_0$, the Lie algebra of ${\mathfrak L}_0$ coincides with ${\mathfrak l}_0$. Thus ${\mathfrak L}_0^{\circ}$ is a closed complex connected subgroup of ${\mathfrak L}$ with Lie algebra ${\mathfrak l}_0$, and we consider the simply-connected complex homogeneous manifold $Y_h:={\mathfrak L}/{\mathfrak L}_0^{\circ}$. One can show that the vector group $E^{+}:=(E,+)$ naturally lies in ${\mathfrak L}$, and therefore $E$ embeds into $Y_h$ as an an open (and dense) subset. Let $\mathop{\rm Bir}\nolimits(Q_h)^{\circ}$ denote the connected component of the identity of $\mathop{\rm Bir}\nolimits(Q_h)$ with respect to the Lie group topology on $\mathop{\rm Bir}\nolimits(Q_h)$ provided, say, by the results of \cite{Tum}. It can be easily shown that $\mathop{\rm Bir}\nolimits(Q_h)^{\circ}$ is regularizable on the manifold $Y_h$.

Further, let $\mathop{\rm Bir}\nolimits_0(Q_h):=\{g\in\mathop{\rm Bir}\nolimits(Q_h):\, \hbox{$0\in\mathop{\rm reg}\nolimits(g)$ and $g(0)=0$}\}$. The full group $\mathop{\rm Bir}\nolimits(Q_h)$ is generated by $\mathop{\rm Bir}\nolimits(Q_h)^{\circ}$ and $\mathop{\rm Bir}\nolimits_0(Q_h)$. For an element $g\in\mathop{\rm Bir}\nolimits_0(Q_h)$ the corresponding push-forward map $g_*$ is a Lie algebra automorphism of ${\mathfrak l}$ leaving ${\mathfrak l}_0$ invariant. This automorphism induces an automorphism of ${\mathfrak L}$ leaving ${\mathfrak L}_0^{\circ}$ invariant, and therefore gives rise to an element of $\mathop{\rm Aut}\nolimits(Y_h)$. Hence the full group $\mathop{\rm Bir}\nolimits(Q_h)$ is regularizable on $Y_h$.

While the approach that we have just outlined solves the regularization problem for $\mathop{\rm Bir}\nolimits(Q_h)$ in principle (in the sense of part (i) of Definition \ref{regulariz}), our Theorem \ref{general} contains a much stronger result. It provides an algebraic solution to this problem and applies to a large class of CR-manifolds.

We would like to thank Michael Eastwood for many helpful discussions. This research is supported by the Australian Research Council and was initiated while the second author was visiting the Australian National University.

\section{Birational Transformations of a Vector Space}\label{birational}
\setcounter{equation}{0}

In this section we state two general propositions on birational maps of a finite-dimensional complex vector space $E$.

The first proposition will be used in the proofs of Theorems \ref{general}, \ref{general2} but is also of independent interest (cf. \cite{Ko1}, \cite{Ko2}, \cite{Ka2}). For every $\alpha\in E$ we consider the constant holomorphic vector field $\alpha\dd z$, and denote by $\eta$ the {\it Euler vector field}\, $z\dd z$.

\begin{proposition}\label{birat}\sl Let $D_1,D_2\subset E$ be non-empty domains and $g:D_1\to D_2$ a biholomorphic map with induced Lie algebra isomorphism $g_*:{\mathfrak{hol}}(D_1)\rightarrow{\mathfrak{hol}}(D_2)$. With $g^*:=g_*^{-1}$ define the holomorphic maps
$$
p_g:D_1\rightarrow E\quad \hbox{and}\quad  q_g:D_1\rightarrow\mathop{\rm End}\nolimits(E)
$$
by
\begin{equation}
g^*(\eta)=p_g(z)\dd z,\quad g^*(\alpha\dd z)=\big(q_g(z)\alpha\big)\dd z\label{pandq}
\end{equation}
for all $\alpha\in E$. Then $q_g( D_1)\subset\mathop{\rm GL}\nolimits(E)$ and \begin{equation}
\hbox{$g(z)=q_g(z)^{-1}p_g(z)$, with $g'(z)=q_g(z)^{-1}$, for all $z\in D_1$.}\label{explform}
\end{equation}
\end{proposition}

\noindent {\bf Proof:} For every $h(z)\dd z\in{\mathfrak{hol}}(D_2)$ we have by definition
$$
g^*(h(z)\dd z)=\Bigl(g'(z)^{-1}h(g(z))\Bigr)\dd z\in{\mathfrak{hol}}(D_1),
$$
where $g'(z)\in\mathop{\rm GL}\nolimits(E)$ for $z\in D_1$, is the derivative of $g$ at $z$. For $h(z)\equiv \alpha$, with $\alpha\in E$, this implies $g'(z)^{-1}\alpha=q_g(z)\alpha$, and for $h(z)\equiv z$ we get $g'(z)^{-1}g(z)=p_g(z)$. Formula (\ref{explform}) follows from these two relations. {\hfill $\Box$}
\vspace{0.3cm}\\

Recall that ${\mathfrak s}$ is the complex solvable Lie subalgebra of ${\mathfrak{hol}}(E)$ spanned by all constant vector fields $\alpha\dd z$ and the Euler vector field $\eta$ (see (\ref{s})). Proposition \ref{birat} yields the following corollary.

\begin{proposition}\label{corprop1}\sl Suppose that for the biholomorphic map $g:D_1\rightarrow D_2$ from Proposition \ref{birat} all vector fields in both $g^*({\mathfrak s})$ and $g_*({\mathfrak s})$ extend to rational vector fields on $E$. Then $g$ extends to an element of $\mathop{\rm Bir}\nolimits(E)$, with $\mathop{\rm reg}\nolimits(g)=\mathop{\rm reg}\nolimits(g')=\mathop{\rm reg}\nolimits(q^{-1})$.
\end{proposition}

\noindent{\bf Proof:} We only need to show that $\mathop{\rm reg}\nolimits(g)=\mathop{\rm reg}\nolimits(g')$. Clearly, we have $\mathop{\rm reg}\nolimits(g)\subset\mathop{\rm reg}\nolimits(g')$. To obtain the opposite inclusion, we suppose that $\mathop{\rm reg}\nolimits(g')\setminus\mathop{\rm reg}\nolimits(g)$ is non-empty. We let $n:=\dim E$, identify $E$ with ${\mathbb C}^n$, and write $g$ as $g=(g_1,\dots,g_n)$. Then there exists $j$ such that $A:=\mathop{\rm reg}\nolimits(g')\setminus\mathop{\rm reg}\nolimits(g_j)$ is non-empty. It then follows that one can find a point $a\in A$, which is not an indeterminacy point of $g_j$, that is, $g_j=r_j/s_j$, where $r_j$ and $s_j$ are polynomials with $r_j(a)\ne 0$, $s_j(a)=0$. Hence for some $k$ the order of vanishing of $s_j \partial r_j/\partial z_k- r_j \partial s_j/\partial z_k$ at $a$ is finite and strictly less than that of $s_j^2$. Therefore, $a$ is not a regular point of $\partial g_j/\partial z_k$, which contradicts our choice of $a$. {\hfill $\Box$}
\vspace{0.3cm}\\

\begin{remark}\label{regset}\rm We will use Proposition \ref{corprop1} in Section \ref{generalproof} in the case when all vector fields in $g^*({\mathfrak s})$ and $g_*({\mathfrak s})$ extend to polynomial vector fields on $E$. In this situation $\mathop{\rm reg}\nolimits(g)=\mathop{\rm reg}\nolimits(q_g^{-1})=\{z\in E:\det q_g(z)\ne 0\}$. In fact, $\det q_g$ is a denominator of the rational map $g$, that is, $(\det q_g) g$ is a polynomial map. As the following example shows, $\det q_g$ need not be an exact denominator (a denominator of minimal degree) of $g$.
\end{remark}

\begin{example}\label{exrational}\rm Let $E:={\mathbb C}^{n\times m}$, $b\in{\mathbb C}^{m\times n}$ a fixed
matrix, and
$$
g(z):=({1\kern-3.8pt 1}-zb)^{-1}z,
$$
where ${1\kern-3.8pt 1}$ is the $n\times n$ identity matrix. Then $g\in\mathop{\rm Bir}\nolimits(E)$ (indeed $g^{-1}(w)=({1\kern-3.8pt 1}+wb)^{-1}w$). Differentiation yields
$$
g'(z)\alpha=({1\kern-3.8pt 1}-zb)^{-1}\alpha({1\kern-3.8pt 1}-bz)^{-1}
$$
for all $\alpha\in E$. In particular, for the functions $p,q$ from Proposition \ref{birat} we have
$$
\hbox{$q_g(z)\alpha=({1\kern-3.8pt 1}-zb)\alpha({1\kern-3.8pt 1}-bz)$ and $p_g(z)=z-zbz$}
$$
for all $\alpha\in E$. Thus $\det q_g$ is not an exact denominator of $g$. Further, a moment's thought gives $\det q_g(z)=\det({1\kern-3.8pt 1}-zb)^m\det({1\kern-3.8pt 1}-bz)^n$, hence $\mathop{\rm reg}\nolimits(g)=\{z\in E:\det({1\kern-3.8pt 1}-zb)\ne 0\}$.
\end{example}
\vspace{0.3cm}

In the next proposition we relate the group $\mathop{\rm Bir}\nolimits(M)$ of birational transformations of a CR-submanifold $M\subset E$ to the subset $\mathop{\rm BR}\nolimits(M)\subset\mathop{\rm Bir}\nolimits(E)$ by means of Condition $(*)$, as stated in Section \ref{intro}.

\begin{proposition}\label{cond8}\sl Let $M$ be a connected real-analytic generic CR-sub\-man\-i\-fold of $E$. If Condition $(*)$ is satisfied for $M$, then $\mathop{\rm Bir}\nolimits(M)=\mathop{\rm BR}\nolimits(M)$. Moreover, for every $g\in\mathop{\rm BR}\nolimits(M)$ we have $g(M\cap\mathop{\rm reg}\nolimits(g))=M\cap\mathop{\rm reg}\nolimits(g^{-1})$.
\end{proposition}

\noindent{\bf Proof:} Fix $g\in\mathop{\rm BR}\nolimits(M)$, and let $V\subset M$ be a non-empty domain such that $V\subset\mathop{\rm reg}\nolimits(g)$ and $g(V)\subset M$. By Lemma 2.2 of \cite{FK2} the non-empty set $M\cap\mathop{\rm reg}\nolimits(g)$ is connected, and therefore $M_1:=g(M\cap\mathop{\rm reg}\nolimits(g))$ is a real-analytic connected submanifold of $E$. Since $W:=g(V)$ is a non-empty domain in $M$ such that $W\cap M_1=W$, Condition $(*)$ implies that $M_1\subset M\cap\mathop{\rm reg}\nolimits(g^{-1})$. Interchanging the roles of $g$ and $g^{-1}$ gives $g(M\cap\mathop{\rm reg}\nolimits(g))=M\cap\mathop{\rm reg}\nolimits(g^{-1})$.

Now for any $g_1,g_2\in\mathop{\rm BR}\nolimits(M)$ we choose a non-empty domain $V\subset M$ with $V\subset\mathop{\rm reg}\nolimits(g_1)$ and $g_1(V)\subset\mathop{\rm reg}\nolimits(g_2)$. Then $g_2\circ g_1\in\mathop{\rm BR}\nolimits(M)$. Therefore, $\mathop{\rm BR}\nolimits(M)=\mathop{\rm Bir}\nolimits(M)$, as required.{\hfill $\Box$}
\vspace{0.3cm}

We stress the importance of Proposition \ref{cond8} for the correct understanding of $\mathop{\rm BR}\nolimits(M)$ and $\mathop{\rm Bir}\nolimits(M)$. In particular, if $M$ does not satisfy the assumptions of Proposition \ref{cond8}, then the set $\mathop{\rm BR}\nolimits(M)$ may not be a group.

As we stated in Section \ref{intro}, a connected real-analytic generic submanifold $M\subset E$ satisfies Condition $(*)$ if $M$ is minimal and closed. There is, however, a large class of examples of non-closed CR-submanifolds satisfying Condition $(*)$. An interesting family of such manifolds is given in Example \ref{exx2} in Section \ref{propp}.

\section{Proof of Theorems \ref{general} and \ref{general2}}\label{generalproof}
\setcounter{equation}{0}

We will first prove Theorem \ref{general}.

Without loss of generality we assume that $M$ contains the origin, and let ${\mathfrak l}$ be the complexification of ${\mathfrak{hol}}(M,0)$. Arguing as in the proof of Proposition 4.2 of \cite{FK1}, we obtain that ${\mathfrak l}$ admits a ${\mathbb Z}$-grading
\begin{equation}
{\mathfrak l}=\bigoplus_{m\in{\mathbb Z}}{\mathfrak l}^m,\quad[{\mathfrak l}^m,{\mathfrak l}^{\ell}]\subset{\mathfrak l}^{m+\ell},\label{grad}
\end{equation}
where ${\mathfrak l}^m$ is the $m$-eigenspace of $\mathop{\rm ad}\nolimits\eta $ in ${\mathfrak l}$, and ${\mathfrak l}^m=0$ for $m<-1$, as well as for $m$ large enough. Every ${\mathfrak l}^m$ consists of polynomial vector fields homogeneous of degree $m+1$, with
$$
{\mathfrak l}^{-1}=\{\alpha\dd z:\alpha\in E\}
$$
being the Lie algebra of all constant vector fields on $E$. Thus every vector field in ${\mathfrak{hol}}(M,0)$ is polynomial. Arguing in this way for every $a\in M$ we see that all Lie algebras ${\mathfrak{hol}}(M,a)$ are polynomial and hence coincide with ${\mathfrak{hol}}(M)$. Thus we have obtained statement (I).

For a non-empty domain $D\subset E$ we identify ${\mathfrak l}$ with a Lie subalgebra of ${\mathfrak{hol}}(D)$ by restriction. Let $V_1$, $V_2$ be non-empty domains in $M$, and $g:V_1\to V_2$ a real-analytic CR-isomorphism. Then there exist domains $D_1,D_2\subset E$ and a biholomorphic extension $g:D_1\to D_2$ with $g_*({\mathfrak l})={\mathfrak l}$. Since all vector fields in ${\mathfrak l}$ are polynomial, Proposition \ref{corprop1} yields that $g$ extends to an element of $\mathop{\rm Bir}\nolimits(M)$ of the form $q^{-1}p$, where $p:E\rightarrow E$ and $q:E\rightarrow\mathop{\rm End}\nolimits(E)$ are polynomial maps (see (\ref{pandq})). By Remark \ref{regset} we have $\mathop{\rm reg}\nolimits(g)=\mathop{\rm reg}\nolimits(q^{-1})=\{z\in E:\det q(z)\ne 0\}$. Thus we have obtained statement (II).

Further, for every $a\in E$ the isotropy Lie subalgebra
$$
{\mathfrak l}_a:=\{\xi\in{\mathfrak l}:\xi_a=0\}
$$
has codimension $n:=\dim E$ in ${\mathfrak l}$, and ${\mathfrak l}$ is the direct sum of subspaces ${\mathfrak l}={\mathfrak l}^{-1}\oplus\,{\mathfrak l}_a$, with ${\mathfrak l}_a\ne{\mathfrak l}_b$ for all $a,b\in E$, $a\ne b$. Let ${\mathbb G}$ be the Grassmannian of all complex linear subspaces $\Lambda\subset{\mathfrak l}$ of codimension $n$. Then ${\mathbb G}$ is a rational projective algebraic complex manifold on which the complex linear group $\mathop{\rm GL}\nolimits({\mathfrak l})$ acts transitively and algebraically by means of the canonical projection $\mathop{\rm GL}\nolimits({\mathfrak l})\rightarrow\mathop{\rm PGL}\nolimits({\mathfrak l})\subset\mathop{\rm Aut}\nolimits({\mathbb G})$.

The subset
$$
U:=\{\Lambda\in{\mathbb G}:{\mathfrak l}={\mathfrak l}^{-1}\oplus\,\Lambda\}
$$
is Zariski open in ${\mathbb G}$ and is algebraically equivalent to the complex vector space of all linear operators $\lambda:{\mathfrak l}_0\to{\mathfrak l}^{-1}$ (just identify every $\lambda$ with its graph $\{\xi+\lambda(\xi):\xi\in{\mathfrak l}_0\}\in{\mathbb G}$). In this coordinate chart every automorphism of ${\mathbb G}$ arising from the action of $\mathop{\rm GL}\nolimits({\mathfrak l})$ can be written as a matrix linear fractional transformation.

Consider the injective holomorphic map
$$
\varphi:E\to{\mathbb G}\,,\quad a\mapsto{\mathfrak l}_a.
$$
Then $\varphi(E)\subset U$, and since all vector fields in ${\mathfrak l}$ are polynomial, the map $\varphi$ is an algebraic morphism. As a consequence, the set $\varphi(E)$ is constructible. Let $X$ be the Zariski closure of $\varphi(E)$. Clearly, $X$ is an irreducible algebraic subvariety in ${\mathbb G}$ and $\varphi(E)$ contains a Zariski open (and dense) subset of $X$, hence the topological closure of $\varphi(E)$ in ${\mathbb G}$ coincides with $X$.

Define
$$
\mathop{\rm Bir}\nolimits(E,{\mathfrak l}):=\left\{g\in\mathop{\rm Bir}\nolimits(E):g_*({\mathfrak l})={\mathfrak l}\right\}.
$$
Observe that $\mathop{\rm Bir}\nolimits(E,{\mathfrak l})$ contains the set $\mathop{\rm BR}\nolimits(M)$. Since every element of $\mathop{\rm Bir}\nolimits(M)$ is the composition of a finite number of elements of $\mathop{\rm BR}\nolimits(M)$, it follows that $\mathop{\rm Bir}\nolimits(M)\subset\mathop{\rm Bir}\nolimits(E,{\mathfrak l})$.

For any $g\in\mathop{\rm Bir}\nolimits(E,{\mathfrak l})$, we regard the push-forward map $g_*$ as an element of $\mathop{\rm Aut}\nolimits({\mathfrak l})\subset\mathop{\rm GL}\nolimits({\mathfrak l})$, where $\mathop{\rm Aut}\nolimits({\mathfrak l})$ is the complex algebraic subgroup of $\mathop{\rm GL}\nolimits({\mathfrak l})$ that consists of all Lie algebra automorphisms of ${\mathfrak l}$. Define $\nu$ to be the homomorphism
\begin{equation}
\nu:\,\,\mathop{\rm Bir}\nolimits(E,{\mathfrak l})\rightarrow\mathop{\rm Aut}\nolimits({\mathfrak l}),\quad g\mapsto g_*.\label{mapPhi}
\end{equation}
By formula (\ref{explform}) the homomorphism $\nu$ is injective. Note that the canonical homomorphism $\pi:\mathop{\rm Aut}\nolimits({\mathfrak l})\rightarrow\mathop{\rm PGL}\nolimits({\mathfrak l})$ is injective as well.

Since for $g\in\mathop{\rm Bir}\nolimits(E,{\mathfrak l})$ we have $g_*({\mathfrak l}_a)={\mathfrak l}_{g(a)}$ for all $a\in\mathop{\rm reg}\nolimits(g)$, the map $\pi(g_*)$ preserves $X$ and the following holds:
\begin{equation}
\varphi\circ g=\sigma(g)\circ\varphi\quad\hbox{on $\mathop{\rm reg}\nolimits(g)$,}\label{commutative}
\end{equation}
where $\sigma:=\pi\circ\nu$. Formula (\ref{commutative}) applies, in particular, to every translation $g(z)=z+\beta$, $\beta\in E$ (note that every translation is an element of $\mathop{\rm Bir}\nolimits(E,{\mathfrak l})$). It is straightforward to see that the action of the complex vector group $E^{+}:=(E,+)$ on ${\mathbb G}$ through the homomorphism $\sigma$ is algebraic, and formula (\ref{commutative}) implies that $\varphi(E)$ is an orbit of this action. It then follows that $\varphi(E)$ is a Zariski open subset of $X$ lying in the non-singular part $X_{\mathop{\rm reg}\nolimits}$ of $X$. By Zariski's Main Theorem (see \cite{Mu}, III.9.1), $\varphi:E\to\varphi(E)$ is an algebraic isomorphism.

We now embed ${\mathbb G}$ into ${\mathbb C}{\mathbb P}^N$ for a sufficiently large integer $N$ by the Pl\"ucker map, and regard $X$ as an algebraic subvariety of ${\mathbb C}{\mathbb P}^N$, and $\mathop{\rm PGL}\nolimits({\mathfrak l})$ as a subgroup of $\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$. Formula (\ref{commutative}) then yields that $\mathop{\rm Bir}\nolimits(M)$ is projectively regularizable with $\tau:=\sigma|_{\mathop{\rm Bir}\nolimits(M)}$. This proves statement (III).

The proof of Theorem \ref{general} is complete.{\hfill $\Box$}
\vspace{0.5cm}

We will now prove Theorem \ref{general2}.

Recall that $\hat M$ is a connected generic immersed submanifold of the Zariski open subset $\hat E$ of the non-singular part $X_{\mathop{\rm reg}\nolimits}$ of $X$ (see (\ref{setZ})). The minimality of $M$ implies that $\hat M$ is minimal as well. Therefore, it follows from Lemma 2.2 of \cite{FK2} that $\hat M\cap\varphi(E)$ is connected. Since $\hat M$ contains $\varphi(M)$ as an open subset, part (b) of Condition $(*)$ yields that $\hat M\cap\varphi(E)=\varphi(M)$ and that $\varphi(M)$ is dense in $\hat M$.

To show that $\mathop{\rm Aut}\nolimits(\hat M)=\tau(\mathop{\rm Bir}\nolimits(M))$, observe that for every $g\in \mathop{\rm Aut}\nolimits(\hat M)$ there exist domains $V_1, V_2\subset\varphi(M)$ such that $g(V_1)=V_2$. Then the composition $\varphi^{-1}\circ g\circ\varphi$ is a real-analytic CR-diffeomorphism between the domains $\varphi^{-1}(V_1),\varphi^{-1}(V_2)$ in $M$. By statement (II) of Theorem \ref{general}, the map $\varphi^{-1}\circ g\circ\varphi$ extends to an element $g_0$ of $\mathop{\rm Bir}\nolimits(M)$, hence $g=\tau(g_0)$. Thus $\mathop{\rm Aut}\nolimits(\hat M)=\tau(\mathop{\rm Bir}\nolimits(M))$.

Assume now that $\overline{M}\setminus M$ does not contain a CR-submanifold of $E$ locally CR-equivalent to $M$. Let $g_n$ be a sequence in $\tau(\mathop{\rm Bir}\nolimits(M))$ converging to an element $g\in\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$. We claim that $g(\varphi(M))\cap\varphi(E)\ne\emptyset$. Indeed, otherwise $g(\varphi(M))$ lies in a Zariski closed subset of $X_{\mathop{\rm reg}\nolimits}$, which is impossible since $g(\varphi(M))$ is generic in $X_{\mathop{\rm reg}\nolimits}$. Thus for some domain $V\subset\varphi(M)$ we have $g(V)\subset \varphi(E)$. Clearly, $g(V)$ is a CR-submanifold of $\varphi(E)$ locally equivalent to $\varphi(M)$ and contained in $\overline{\varphi(M)}$. It then follows that there exists $x_0\in V$ for which $g(x_0)\in \varphi(M)$. Since $M$ is locally closed in $E$, for some neighborhood $V'\subset V$ of $x_0$ in $\varphi(M)$ we have $g(V')\subset\varphi(M)$. Hence $g\in\tau(\mathop{\rm Bir}\nolimits(M))$, and therefore $\tau(\mathop{\rm Bir}\nolimits(M))$ is closed in $\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$.

Let ${\mathfrak a}$ be the Lie subalgebra of the Lie algebra of $\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$ corresponding to the closed subgroup $\tau(\mathop{\rm Bir}\nolimits(M))\subset \mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$. Every element $v\in{\mathfrak a}$ is a holomorphic vector field on ${\mathbb C}{\mathbb P}^N$ tangent to $\hat M$ and gives rise to a holomorphic vector field on $E$ tangent to $M$.

Conversely, consider a vector field $\xi\in{\mathfrak{hol}}(M)$ and fix $a\in M$. Near $a$ the vector field $\xi$ can be integrated to a local 1-parameter group $t\mapsto g_t$, with $|t|<\varepsilon$, of local real-analytic CR-isomorphisms of $M$. By statement (II) of Theorem \ref{general} every $g_t$ extends to an element of $\mathop{\rm Bir}\nolimits(M)$. Further, one can define by composition a map $g_t\in\mathop{\rm Bir}\nolimits(M)$ for every $t\in{\mathbb R}$ and obtain a 1-parameter subgroup of $\mathop{\rm Bir}\nolimits(M)$. Then $t\mapsto \tau(g_t)$ is a continuous 1-parameter subgroup of $\tau(\mathop{\rm Bir}\nolimits(M))$ and hence $g_t=\exp(tv)$ for some $v\in{\mathfrak a}$.

Thus we have established an isomorphism between ${\mathfrak a}$ and ${\mathfrak{hol}}(M)$, and the proof of Theorem \ref{general2} is complete.{\hfill $\Box$}
\vspace{0.3cm}

For the remainder of the article we set ${\mathfrak g}:={\mathfrak{hol}}(M)$. Let $M$ satisfy the assumptions of Theorem \ref{general}, and set $\rho:=\nu|_{\mathop{\rm Bir}\nolimits(M)}$, with $\nu$ defined in (\ref{mapPhi}). The homomorphism $\rho$ is injective, and the image $\rho(\mathop{\rm Bir}\nolimits(M))$ lies in the group $\mathop{\rm Aut}\nolimits({\mathfrak g})$ of all Lie algebra automorphisms of ${\mathfrak g}$, which is a real algebraic subgroup of each of $\mathop{\rm GL}\nolimits({\mathfrak g})$ and $\mathop{\rm Aut}\nolimits({\mathfrak l})$. Thus $\mathop{\rm Bir}\nolimits(M)$ can be endowed with the structure of a Lie group (possibly with uncountably many connected components). An argument identical to that at the end of the proof of Theorem \ref{general2} shows that the Lie algebra of $\mathop{\rm Bir}\nolimits(M)$ with respect to this Lie group structure is isomorphic to ${\mathfrak g}$. For $\mathop{\rm Bir}\nolimits(M)$ with this Lie group structure the map
$$
\rho:\mathop{\rm Bir}\nolimits(M)\rightarrow\mathop{\rm Aut}\nolimits({\mathfrak g})
$$
is just the adjoint representation. In the next section we will show that under additional assumptions  the image $\rho(\mathop{\rm Bir}\nolimits(M))$ is closed in $\mathop{\rm Aut}\nolimits({\mathfrak g})$.

\section{Closed Embedding of $\mathop{\rm Bir}\nolimits(M)$ into $\mathop{\rm Aut}\nolimits({\mathfrak g})$}\label{liegrstr}
\setcounter{equation}{0}

Throughout this section we suppose that $M$ has Property (P) and satisfies Condition $(*)$. In particular, by Proposition \ref{cond8} we have $\mathop{\rm Bir}\nolimits(M)=\mathop{\rm BR}\nolimits(M)$. Under these assumptions, which are weaker than those of Theorem \ref{general2}, we obtain the existence of a Lie group structure on $\mathop{\rm Bir}\nolimits(M)$ with at most countably many connected components and Lie algebra ${\mathfrak g}$. Instead of investigating the closedness of $\tau(\mathop{\rm Bir}\nolimits(M))$ in $\mathop{\rm PGL}\nolimits_{N+1}({\mathbb C})$, as we did in the proof of Theorem \ref{general2}, we investigate the closedness of $\rho(\mathop{\rm Bir}\nolimits(M))$ in $\mathop{\rm Aut}\nolimits({\mathfrak g})$. Note that a simple example shows that there is always a closed subgroup of $\mathop{\rm GL}\nolimits({\mathfrak l})$ whose image in $\mathop{\rm PGL}\nolimits({\mathfrak l})$ under the canonical projection $\mathop{\rm GL}\nolimits({\mathfrak l})\rightarrow \mathop{\rm PGL}\nolimits({\mathfrak l})$ is not closed.

The result of this section is the following theorem.

\begin{theorem}\label{liegr}\sl Let $M$ be a connected real-analytic generic CR-sub\-man\-i\-fold of a finite-dimensional complex vector space $E$. Assume that $M$ has Property {\rm (P)} and satisfies Condition $(*)$. Then $\rho(\mathop{\rm Bir}\nolimits(M))$ is closed in $\mathop{\rm Aut}\nolimits({\mathfrak g})$.
\end{theorem}

\noindent {\bf Proof:} Let $g_n$ be a sequence in $\mathop{\rm Bir}\nolimits(M)$ such that the sequence $f_n:=\rho(g_n)$ converges in $\mathop{\rm Aut}\nolimits({\mathfrak g})$ to an element $f$. By Proposition \ref{birat} every map $g_n$ can be written as $g_n=q_n^{-1}p_n$, where $p_n:=p_{g_n}$ and $q_n:=q_{g_n}$ are polynomial maps on $E$ found from formulas (\ref{pandq}). Since the maps $f_n^{-1}$ converge in $\mathop{\rm Aut}\nolimits({\mathfrak g})$ to $f^{-1}$, the sequences $p_n$, $q_n$ converge (uniformly on compact subsets of $E$) to polynomial maps $p:E\rightarrow E$ and $q:E\rightarrow\mathop{\rm End}\nolimits(E)$, respectively, and we have $f^{-1}(\alpha\dd z)=\big(q(z)\alpha\big)\dd z$ for all $\alpha\in E$.

Similarly, every map $g_n^{-1}$ can be written as $g_n^{-1}=\tilde q_n^{-1} \tilde p_n$, where $\tilde p_n:=p_{g_n^{-1}}$ and $\tilde q_n:= q_{g_n^{-1}}$ are polynomial maps. The sequences $\tilde p_n$ and $\tilde q_n$ converge to polynomial maps $\tilde p:E\rightarrow E$ and $\tilde q:E\rightarrow\mathop{\rm End}\nolimits(E)$, respectively, and we have $f(\alpha\dd z)=\big(\tilde q(z)\alpha\big)\dd z$ for all $\alpha\in E$.

Since $f_k^{-1} f\rightarrow\hbox{id}\in\mathop{\rm Aut}\nolimits({\mathfrak g})$, for every neighborhood ${\mathcal V}$ of the identity in $\mathop{\rm Aut}\nolimits({\mathfrak g})$ one can find an element $\hat f\in\rho(\mathop{\rm Bir}\nolimits(M))$ with $\hat f f\in{\mathcal V}$. Choosing ${\mathcal V}$ such that ${\mathcal V}={\mathcal V}^{-1}$ one can also assume that $f^{-1} \hat f^{-1}\in{\mathcal V}$. Hence by replacing $g_n$ by $\rho^{-1}(\hat f)g_n$ and $f$ by $\hat f f$ we can assume without loss of generality that $\det q\not\equiv 0$ and $\det\tilde q\not\equiv 0$. We then define the rational maps $g:= q^{-1}p$ and $\tilde g:=\tilde q^{-1}\tilde p$.

Let
$$
A:=\{z\in E: \det q(z)=0\},\quad B:=\{z\in E:\det\tilde q(z)=0\}.
$$
Since $\det q_n\rightarrow \det q$ and $\mathop{\rm reg}\nolimits(g_n)=\{z\in E:\det q_n(z)\ne 0\}$ (see Proposition \ref{corprop1}), it follows that $g_n\rightarrow g$ uniformly on compact subsets of $E\setminus A$. Similarly, $g_n^{-1}\rightarrow\tilde g$ uniformly on compact subsets of $E\setminus B$. For every $\xi\in{\mathfrak l}$,\linebreak $\xi=h(z)\dd z$, we have
$$
f(\xi)(z)=\tilde q(z)h(\tilde g(z))\dd z,\quad f^{-1}(\xi)(z)=q(z)h(g(z))\dd z.
$$
Applying the identity
$$
\begin{array}{l}
\xi=f(f^{-1}(\xi))=\tilde q(z)q(\tilde g(z))h(g(\tilde g(z)))\dd z
\end{array}
$$
to constant vector fields $\xi=\alpha\dd z$, we see that $\tilde q(z) q(\tilde g(z))\equiv \hbox{id}$. Applying this identity to the Euler vector field $\eta$ and interchanging the roles of $f$ and $f^{-1}$ we obtain $\tilde g=g^{-1}$, thus $g\in\mathop{\rm Bir}\nolimits(E)$. By Proposition \ref{cond8} we have $g_n\in\mathop{\rm BR}\nolimits(M)$ and $g_n(M\cap\mathop{\rm reg}\nolimits(g_n))=M\cap\mathop{\rm reg}\nolimits(g_n^{-1})$, which yields that $g\in\mathop{\rm Bir}\nolimits(M)$. Finally, since $g_n\rightarrow g$ on $E\setminus A$ and $g_n^{-1}\rightarrow g^{-1}$ on $E\setminus B$, it follows that $(g_n)_*\rightarrow g_*$. Hence $f=\rho(g)$.

The proof of the theorem is complete. {\hfill $\Box$}

\section{Property (P)}\label{propp}
\setcounter{equation}{0}

In this section we give sufficient conditions for a CR-submanifold to have Property (P) and discuss examples of manifolds with this property (for the statement of Property (P) see Section \ref{intro}).

We call a CR-submanifold $M$ of a complex manifold $Z$ {\it semi-homogeneous at a point}\, $a\in M$, if the span over ${\mathbb C}$ of the values $\xi_a$ of elements $\xi\in{\mathfrak{hol}}(M,a)$ at $a$ contains $T_a(M)$. Also, we call $M$ {\it semi-homogeneous}\, if $M$ is semi-homogeneous at every point. Clearly, local homogeneity implies semi-homogeneity for a real-analytic CR-submanifold.

Recall that ${\mathfrak g}={\mathfrak{hol}}(M)$. With this notation we state the following proposition.

\begin{proposition}\label{suffcondpropp}\sl Let $M$ be a connected real-analytic generic CR-sub\-man\-i\-fold of a finite-dimensional complex vector space $E$. Let $a$ be a point in $M$. Assume that:
$$
\begin{array}{ll}
(1) & \hbox{$M$ is holomorphically non-degenerate at $a$;}\\
\vspace{-0.4cm}\\

(2) & \hbox{$M$ is minimal at $a$;}\\
\vspace{-0.4cm}\\

(3) & \hbox{$M$ is semi-homogeneous at $a$;}\\
\vspace{-0.4cm}\\

(4) & \hbox{the complex Lie algebra ${\mathfrak g}+i{\mathfrak g}$ contains the vector}\\
&\hbox{field $(z-a)\dd z$.}
\end{array}
$$

\noindent Then $M$ has property {\rm (P)}.
\end{proposition}

\noindent{\bf Proof:} By Theorem 12.5.3 in \cite{BER}, assumptions (1) and (2) imply that ${\mathfrak{hol}}(M,b)$ is finite-dimensional for all $b\in M$.

Without loss of generality we assume that $a=0$. Then using assumption (4) we obtain, as at the beginning of Section \ref{generalproof}, that ${\mathfrak l}:={\mathfrak{hol}}(M,0)+i{\mathfrak{hol}}(M,0)$ admits grading (\ref{grad}) where ${\mathfrak l}^m$ is the $m$-eigenspace of $\mathop{\rm ad}\nolimits \eta$ in ${\mathfrak l}$, and ${\mathfrak l}^m=0$ for $m<-1$, as well as for $m$ large enough. Every ${\mathfrak l}^m$ consists of polynomial vector fields homogeneous of degree $m+1$. Since all vector fields in ${\mathfrak l}^m$ for $m>-1$ vanish at the origin, assumption (3) implies that ${\mathfrak l}^{-1}$ is the space of all constant vector fields on $E$. The proof is complete.{\hfill $\Box$}
\vspace{0.5cm}

We now give examples of CR-submanifolds that have Property (P).

\begin{example}\label{exx3}\rm Every quadric $Q_h\subset{\mathbb C}^{n+k}$ associated to a non-degenerate Hermitian form $h$ has Property (P). Indeed, $Q_h$ is homogeneous under the action of the group of maps
$$
\begin{array}{lll}
z&\mapsto& z+\alpha,\\
\vspace{-0.3cm}\\
w&\mapsto&w+2ih(z,\alpha)+\beta,\\
\end{array}
$$
where $(\alpha,\beta)\in Q_h$. Thus all Lie algebras ${\mathfrak{hol}}(Q_h,a)$ coincide. In fact, they coincide with the Lie algebra ${\mathfrak{hol}}(Q_h)$, which is finite-dimensional (see \cite{B1}, \cite{B2}, \cite{Tum}). Furthermore, ${\mathfrak{hol}}(Q_h)$ clearly contains the vector fields $s\dd w$, $r\dd z+2ih(z,r)\dd z$, $z\dd z+2w\dd w$, $iz\dd z$, with $s\in{\mathbb R}^k$, $r\in{\mathbb C}^n$. Therefore, the complexification of ${\mathfrak{hol}}(Q_h)$ contains all constant vector fields and the Euler vector field $\eta$. Hence the quadric $Q_h$ has Property (P).
\end{example}

\begin{example}\label{exx1}\rm Let $F\subset{\mathbb R}^{n}$ be an arbitrary connected real-analytic submanifold and
$$
M:=F+i{\mathbb R}^n\subset {\mathbb C}^n
$$
the corresponding {\it tube submanifold with base}\, $F$. Then $M$ is a generic semi-homogeneous CR-submanifold of $E$, and ${\mathfrak g}+i{\mathfrak g}$ contains all constant holomorphic vector fields on $E$. Furthermore (see Lemma 4.1 of \cite{FK2}), the tube manifold $M$ is minimal at a point if and only if
\begin{equation}
\hbox{\sl $F$ is not contained in any affine hyperplane of ${\mathbb R}^n$,}\label{conda}
\end{equation}
hence a tube manifold is minimal if it is minimal at one point. Next (see Proposition 4.3 of \cite{FK2}), $M$ is holomorphically non-degenerate at a point if and only if
\begin{equation}
\hbox{\sl the only constant vector field $\xi=\alpha\dd x$ tangent to $F$ is $\xi=0$,}\label{condb}
\end{equation}
hence a tube manifold is holomorphically non-degenerate if it is holomorphically non-degenerate at one point.

To meet conditions (1), (2), (4) of Proposition \ref{suffcondpropp} it is therefore sufficient to require besides (\ref{conda}), (\ref{condb}) that $F$ is a cone, that is, $tF=F$ for every real $t>0$. For every cone $F$ the Levi form of $M$ is degenerate at every point (condition (ii) stated at the beginning of Section \ref{intro} does not hold).
\end{example}

From the large class of tube manifolds that have Property (P) we single out the following special one.

\begin{example}\label{exx2}\rm Fix integers $p\ge q\ge1$ with $n=p+q\ge3$. Then
\begin{equation}
H_{p,q}:=\{x\in{\mathbb R}^{n}:x^{2}_{1}+\dots
x^{2}_{p}=x^{2}_{p+1}+\dots+x^{2}_{n}\}\label{BA}
\end{equation}
is a real hyperquadric with 0 as the only singularity. Let $F$ be a connected component of the non-singular part of $H_{p,q}$ and $M:=F+i{\mathbb R}^{n}$ the corresponding tube submanifold. Then $M$ is a homogeneous CR-submanifold of ${\mathbb C}^{n}$ that has Property (P) and satisfies Condition $(*)$. The Levi form of $M$ is degenerate at every point. For $q=1$ the non-singular part of $H_{p,q}$ has two connected components (given by $x_n>0$ and $x_n<0$), the future light cone and the past light cone. In this case the group $\mathop{\rm Bir}\nolimits(M)$ can be canonically identified with an open subgroup (having two connected components) of $O(n,2)$. For $q>1$ the non-singular part of $H_{p,q}$ is connected, and $\mathop{\rm Bir}\nolimits(M)$ is a real algebraic group. For every $a\in M$ the Lie algebra ${\mathfrak{hol}}(M,a)$ is isomorphic to ${\mathfrak{so}}(p{+}1,q{+}1)$ (cf. \cite{FK1}, p. 21).
\end{example}

\begin{example}\label{exx4}\rm Let $D\subset E$ be an irreducible bounded symmetric domain of rank $r$ given in its Harish-Chandra realization, and $Z$ its compact dual containing $E$ as a Zariski open subset. Then $D$ is convex and invariant under the circle group $\exp(i{\mathbb R}\eta)\subset\mathop{\rm GL}\nolimits(E)$. The complex manifold $Z$ is a compact rational algebraic variety on which the simple complex Lie group $L:=\mathop{\rm Aut}\nolimits(Z)$ acts transitively. All transformations in $G:=\mathop{\rm Aut}\nolimits(D)$ extend to elements of $L$ and, in this way, $G$ is a real form of $L$ and also acts on $Z$. On $Z$ the group $G$ has exactly ${r+2}\choose{2}$ orbits. Among these there are exactly $r+1$ open orbits (including $D$) and a unique closed orbit, the \v Silov boundary of $D$, which coincides with the extremal boundary $\partial_{e}D$ of the convex set $\overline{D}$. Every $G$-orbit is a generic homogeneous CR-submanifold of $Z$ invariant under the circle group $\exp(i{\mathbb R}\eta)\subset G$. Furthermore, for every orbit $G(b)$, $b\in Z$, the intersection $M:=G(b)\cap E$ is a connected CR-submanifold that has Property (P), and for every $a\in M$ the Lie algebra ${\mathfrak{hol}}(M,a)$ is isomorphic to the Lie algebra of $G$, provided neither $G(b)$ is open in $Z$ nor $G(b)=\partial_{e}D$ in the case when $D$ is of tube type (in this last case $\partial_{e}D$ is totally real). The \v Silov boundary (except when $D$ is of tube type) can be locally realized as a standard quadric in $E$ and, in particular, has non-degenerate Levi form. Every $G$-orbit that is neither open nor closed in $Z$ is Levi degenerate (more precisely 2-nondegenerate) and hence cannot be locally realized as a standard quadric in $E$. The group $\mathop{\rm Bir}\nolimits(M)$ is regularizable on the simply-connected complex manifold $Z$, and this is the only possibility up to isomorphism.
\end{example}

\begin{example}\label{exx5}\rm We specialize Example \ref{exx4} to the case
$$
E=\{z\in{\mathbb C}^{2\times2}:z'=z\}\quad\hbox{and}\quad D=\{z\in E:zz^{*}<{1\kern-3.8pt 1}\},
$$
where $z'$ is the transpose and $z^{*}$ is the transpose conjugate of a matrix $z$. Then $D$ is irreducible symmetric of rank 2, and $Z$ can be identified with a complex projective quadric in ${\mathbb C}{\mathbb P}^4$. Also, $G$ is isomorphic to an open subgroup of $O(2,3)$ of index 2. The boundary $\partial D$ of $D$ decomposes into two $G$-orbits: the totally real \v Silov boundary $\partial_{e}D\simeq{\mathbb R}{\mathbb P}^3$ and the smooth part $M$ of $\partial D$ that has Property (P). The manifold $M$ is locally CR-equivalent to the tube submanifold over the future the light cone (see (\ref{BA}) for $p=2$, $q=1$). Here $Z$ is simply-connected while the homogeneous manifold $M$ has fundamental group ${\mathbb Z}_{2}$.
\end{example}

redpajama_set_name: RedPajamaArXiv


Kendall Coyne Schofield is a pretty big deal in the world of women's hockey. The forward won five gold medals as part of the U.S. national title at the IIHF World Women's Championships as well as the gold medal at the 2018 Winter Olympics. She made headlines over the weekend as the first woman to ever compete in the NHL All-Stars Skills Competition. And following her performance, in which she finished seventh overall in the fastest skater event, she's become a pretty big deal in the world of hockey, period.
NBC is well aware of this and has hired her to be an analyst for the Wednesday Night Hockey game between Tampa Bay Lightning and Pittsburgh Penguins. The game will be televised on NBCSN at 8:00 p.m. EST.
Kendall is no stranger to the world of media. She was a communications major at Northeastern and worked as a sideline reporter while there. As for what she'll be doing during this broadcast, CoyneSchofield will appear on the pregame show and Inside the Glass with Pierre McGuire. She will also make appearances during intermissions.
This won't be the first time NBC has featured a women's hockey player during an NHL broadcast. Former U.S. Olympian A.J. Mleczko previously worked as an analyst for the network. She is currently working as an Islanders studio analyst for Madison Square Garden network.


text:
Charlotte Gainsbourg + Beck = an album I will buy
Beautiful French actress/musician Charlotte Gainsbourg is probably most familiar to American audiences by way of Michel Gondry's film "The Science of Sleep," in which she played the female lead. Muisc nerds, film buffs, and all of France, however, knows her as the daughter of Serge Gainsbourg and Jane Birkin. So far, she's only released two albums in her entire career and they've each contained the pop goodness that her father always brought to the table. Having previously worked with Jarvis Cocker and Nigel Godrich, Gainsbourg isn't done collaborating with qualified musicians as it's been confirmed that Beck wrote all of the new record's music, some of the lyrics, and produced and mixed the tracks. Hm, Charlotte's voice and Beck's musicianship together? Sign me up.
Daughter of legendary French songwriter Serge Gainsbourg, Charlotte Gainsbourg worked with Air, Jarvis Cocker and The Divine Comedy in the making of her debut album 5:55, and she hasn't scaled down the contributor notoriety for her forthcoming follow-up LP. On KCRW's Morning Becomes Eclectic, Beck revealed that he's currently working on Gainsbourg's second album in Los Angeles and Paris.
"I was supposed to work on her last album, but I couldn't get over there," Beck said. "Now I'm working on her follow-up record."
The album, entitled IRM (French for MRI), is set for a January release.
Posted in: News, Videos
Tags: 2009 Beck, Beck IRM, Charlotte Gainsbourg, Charlotte Gainsbourg and Beck, Charlotte Gainsbourg and Beck IRM, Charlotte Gainsbourg IRM, Charlotte Gainsbourg new album


Your team will face a task of constructing a conveyor belt that will deliver the product safely through all stages of production and all departments to the display shelves of the shop. All skills and talents will need to come together - marketing tricks, engineering ideas, precision and determination of hands-on manual labour as well as soft skills will all prove equally important. Each team's section of the pipeline must interface with those before and after it. Prepare yourself for working on a small budget, with limited supplies and with strict safety standards.
Perfect finale to any team building programme of your choice. When run on the grounds of Killary Adventure Centre it can be combined with a variety of land and water based adventure activities.


text:
Pentagram's Yuri Suzuki has designed a machine that lets you cut your own vinyl
The Easy Record Maker comes with blank discs for people to record their own musical creations, and has in-built speakers to play it back.
By Henry Wong March 27, 2020 3:53 pm April 2, 2020 11:50 am
Pentagram partner and sound designer Yuri Suzuki has designed the Easy Record Maker in collaboration with Gakken, a Japanese publishing and toy company.
The machine comes with ten blank five-inch discs. You can plug in the audio source from any device, such as a computer or phone and then "engrave sound directly from the recording stylus," Suzuki tells Design Week. You can then instantly playback sound using the tone arm and in-built speaker.
While record players and records have had a resurgence in recent years, the experience of making your own record in this way will be a new experience to most people.
"I wanted to create a machine that makes it easy and cheap to create your own bespoke record without pressing a whole batch," the designer says.
"Vinyl has more value than other media, in my opinion"
The machine taps into a "value" people attach to vinyl records. "Vinyl has more value than other media, in my opinion," Suzuki says. "Recording your voice message or your music onto bespoke vinyl and sending it to someone feels very special and is more valuable (and long-lasting) than just sending a voice message on WhatsApp."
It might also seem even more valuable during the coronavirus lockdown, where "internet-based telecommunication has become the primary way of talking to other people".
"This could be an alternative way to communicate with friends and families," he adds. The machine is portable which "opens up a lot of possibilities" according to Suzuki. He describes the sound as "lo-fi".
"This device is my teenage dream machine"
For Suzuki, the project also has a personal appeal.
"When I was a high school student, I was in a punk band – my friends and I always dreamed of pressing a record, but we knew how expensive it was," he explains. "Personally this device is my teenage dream machine!"
You can also design your own label and sleeves, which is another "platform to express yourself", Suzuki adds.
"If we lost electricity in the future, we could still playback the sound"
Suzuki says that he has been working on the record and vinyl as a material for a while, and the analogue elements of the medium has a special appeal.
He says: "Vinyl records are the only physical recorded media. If we lost electricity in the future, we could still playback the sound mechanically.
"With music media like CDs, cassette tapes, open reel, mp3, WAV and Flac, it's possible to make copies by yourself at home, but there's no easy way to create a bespoke record."
The Easy Record Maker can be bought in Japan now, and will be made available in the UK and US later this year.
"Vinyl has definitely come back"
The product launch is a coming-together of trends in music-consumption. "Although people mainly listen and distribute music through streaming, we can see that the demand for record pressing has rocketed in the past decade and the fascination with records and vinyl has definitely come back," Suzuki says.
Last year, he talked to Design Week about sound design's big moment, in conjunction with a career retrospective at the Design Museum.
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The "loss-making" Conran Shop has been sold to new owners
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Poser – A Review
On February 17, 2019 By tietzeIn Review
In a world where the comics mainstream remains largely superhero fare, interspersed with lighter content drawn largely from "funnies" strips and cartoons, it's pleasing that independent comic creators continue to stretch the possibilities of the form.
It's into this context that Matt Miner and Clay McCormack's Poser drops like a dose of acid gob. At once a throwback to a quainter, cruder art style and a progressive exploration of the contemporary LA punk scene, this four-part mini dares us to stare at its garish excess.
Miner's narrative doesn't feel as though it belongs to this milieu, yet this is where he scores the book's first points. We open on Redondo Beach in 1982, a fireside gathering of drunk punks crashed by "that weirdo from the pier", a strait-laced figure in bomber jacket and glasses. Things quickly get violently out of hand, and a local serial killer legend is born.
Rush forward to "TODAY", and we meet Ash, a member of the local punk community who still lives at home with her dad and works at his record store. Ash used to be in a punk band, and she and her friends – who we also get to know – are thick in the local punk community. Inevitably it isn't long before the "Poser" killer, so called because he targets poseurs, strikes again. Or is it a copycat?
The Poser
This is ultimately what Miner gives us: an authentic snapshot of a contemporary punk community overlaid with a serial killer mystery. The narrative trappings are familiar: successive victims, a core group of protagonists being picked on (and off), an impotent police force doing little to quell the situation, and decades-old grudges being settled in the worst way possible. There's an aimlessness, too, the protagonists trying desperately to survive the threat without doing much to counter it – or even investigate it.
But it's compelling all the same. The first three issues are absolute page-turners, daring the reader to guess the identity of the next victim, to play detective and piece the clues together.
In truth, the fourth act is a letdown, as is almost inevitable when the first three are so thick with mystery. The blurb on issue #4 dares us to "prepare for a twist as shocking and unexpected as the edge of a knife". It's hard to agree with the hyperbole, but the twist is undeniably welcome. There are enough clues to figure out the identity of the killer long before the final act, so the presence of a twist to push events and revelations a little further at least gives the attentive reader a modicum of surprise by way of a reward.
Nevertheless, the twist comes across precisely as that: an artificial and unconvincing turn of the screw by a writer who's realised his big reveal may not be as impressive as he'd originally intended. The subterfuge is a handy reminder, at least, that there is a precision and expertise inherent in writing a flawless whodunnit.
Where Poser really shines is in the details. Miner fashions an impressive world in his punk-infused Redondo. It is no accident that 'Redondo Beach' is a seminal Patti Smith proto-punk track (much) later covered by Morrissey. It lends the book a thematic resonance apparent from the opening caption. This is an American punk aesthetic so seminal that it has adapted into post-punk universality. It's a crucial hint: Poser isn't populated by one punk aesthetic, it's an organic multitude of youthful responses to conservative culture.
Ash, her friends and the background players are visualised by McCormack with meticulous individuality. Authenticity is further heaped on by some impressive artistic ploys. Lesser known punk acts have their names splashed across t-shirts, some of them providing crucial subtext. Is Whitney's destiny to be a Mercy Kill when her singlet is thus adorned? Is the guy in a Minor Threat tee exactly that? When Whitney wears "I Can't Breathe" – (a nod to the 2014 death of Eric Garner?) – is it because she's falling apart emotionally?
Ash and her friends
The characters are likewise written expertly for the most part. There's a sense of genuine depth, friendship and playfulness among them. The only false step is the relationship Ash has with her father. Their occasional familial eruptions are regularly diffused by a perverse "I love you" exchange that never sits quite comfortably. There's a good reason for it, but the reader might come away from the experience wondering if there was an equally idiosyncratic but ultimately less awkward way of depicting the dynamic.
Ash's odd relationship with her father
Violence and virtue are stunningly rendered throughout. McCormack's art – and especially Doug Garbark's colouring – sets and maintains the ideal tone. The art style is appropriately garish throughout, especially McCormack's four covers. There's a clear "fuck you" punk aesthetic at its heart, and the colouring continually screams "retro four-colour pastel" without ever stooping so low on a technical level.
The violence is incredibly brutal. There's a first-wave punk aggression to the killings that will leave readers gasping. Murders are full-page splashes amid a sea of rectangular panels, bloodied like frame blowups from a Lucio Fulci giallo. Both 1982 and present-day punk aesthetics nail the periods effortlessly. The 1982 prologue is a Sex Pistols/Billy Idol afterthought of genuine punk posers, while the "TODAY" representation is extremely diverse, reflecting a genuine community while serving as a love letter to the creators' favourite indie acts.
A murder splash page
While undeniably flawed, particularly in the conclusion of its whodunnit framing narrative, Poser is a delightfully direct treat for fans of 1980s slashers and punk music of all eras. It's also a reminder of the potency and potential of independent visual storytelling, unshackled from commercial concerns and self-censorship. Precisely what punk is all about.
Though review copies are sadly missing the feature, issue #1 is accompanied by an impressive gimmick: a 7" EP featuring songs actually "played" in the comic. A genuine soundtrack album!
Lyrics corresponding to the EP
Poser is available at all comic stores not worth stomping to the ground while The Saints' debut album plays on scratchy vinyl loop in the background, or direct from Waxwork Records.
7"billy idolclaymccormackdoug garbarkeperic garnerfulcigiallographic novelhorrorindiematt minermercy killmercykillminor threatpatti smithpunkredondorevengereviewsex pistolssoundtrackthe saintsvinylwaxwork records
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4 thoughts on "Poser – A Review"
Ty Sparks
excellent review, I followed Matt Miner on twitter
tietze
Thanks for the kind words, Ty.
Matt is definitely worth a follow.
My Bookish Blog - Ayushi
Very well written review 🙂 I don't read comics usually but I found this very interesting.


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Can Intel's 'boy marvel' pull a Steve Jobs?
By lalaine	 Last updated Jan 13, 2021
Pat Gelsinger left Intel Corp. more than a decade ago when it became clear that after 30 years with the company, he would not become a CEO.
He finally got his dream job on Wednesday but is faced with quite the task of bringing the chip giant back to the glory days he was a part of.
Gelsinger is leaving his position as managing director of VMWare Inc. VMW, -6.79% to take over his former employer at what is perhaps the most critical point in Intel's long and storied history. An active investor who was once an industry leader demands change and is the target of numerous shareholder lawsuits.
More from Therese: How did Intel lose its Silicon Valley crown?
Gelsinger started his career at Intel at the age of 18. During his 30-year tenure, he said in a statement to employees that he was "honored to be cared for at the feet of Intel co-founders Bob Noyce, Gordon Moore and Andy," Hain. In 2000 he eventually became the company's first chief technology officer and held other senior positions. In 2009 he moved to EMC Corp., now part of Dell Technologies Inc. DELL, -7.19%.
"It's the biggest return on a prodigal son since Steve Jobs returned to Apple AAPL, + 1.62%,
Nathan Brookwood, an analyst at Insight64, told MarketWatch.
He was also the architect of the company's 80486 processor, also known as the 486, for personal computers. He has both technical qualifications and, more recently, experience leading an enterprise software company in the cloud computing market. This market – more than Gelsinger's past PC market – is currently Intel's biggest growth area and the place where the next big battles will begin, as graphics chip maker Nvidia Corp. The processor designer ARM Holdings of Softbank Group Corp. has made NVDA clear with its purchase plans with + 0.35%. 9984, + 2.82%.
"He's a 'perfect storm' CEO for Intel," said Robert Enderle, principal analyst at Enderle Group, to MarketWatch. "Because of his experience at EMC / VMware, he has a good understanding of Intel and what makes it successful and what Intel currently needs."
As Intel has fallen behind in its manufacturing process technology, the company has considered transferring some of its manufacturing to leading contract manufacturer Taiwan Semiconductor Manufacturing Corp. TSM to outsource or not, -3.07%..
Change is expected to be slow, but generally speaking, Gelsinger is seen as a strong hope to bring the company back to its leadership position.
"We couldn't imagine a better candidate as Intel's CEO," said Raymond James analyst Chris Caso in a statement to customers.
That sentiment was widespread, causing Intel stocks to surge nearly 8% on Wednesday.
As Gelsinger Intel in 2009 for EMC Corp. left, Intel had made a number of leadership changes as part of its well-planned CEO succession, and analysts said at the time it was clear that Gelsinger was ultimately not in the running to be named CEO.
Brian Krzanich was named CEO in 2013, but his tenure was far from stellar and, under his control, there were significant delays in manufacturing the company. He left in a storm of controversy over a previous relationship with another Intel employee and was eventually replaced by then Chief Financial Officer Bob Swan.
As of 2019: With the election of Bob Swan, Intel has blamed the bean counters for the most famous technology company in Silicon Valley
At the time, Gelsinger's name emerged as a perfect candidate for returning to Intel after being outside the company for a while, also combining insider knowledge from the past. However, he said he was not interested in the job.
Gelsinger is now arriving at the right time. Intel needed Swan to clean up the mess Krzanich had made, and now it takes Gelsinger to break new ground.
"[Krzanich] Intel has broken so much that Intel has lagged behind a two-year technical lead to three years, "said Enderle. He said the situation was similar to the US when he moved on to President-elect Biden, a former vice president and senator "who understands the country's plumbing".
"Gelsinger is like a top racing driver and Bob Swan was like the mechanic," he said. "They needed Swan to fix the plumbing, get the company back up and running, and hire a top driver to fix the rest."
See also: Biden inherits a technical cold war with China after Trump ends the battle
He added that Krzanich also killed Intel's Developer Forum (IDF), a developer conference that Gelsinger started there in his day and that was useful for both entering information and for establishing relationships with industry developers. Nvidia used this concept to create support for its CUDA platform.
While at Intel, Gelsinger was often referred to as "the boy's wonder" because he rose up the ranks at a young age, impressing tough and famously paranoid former Intel CEO Andy Grove. Gelsinger's ties to Grove and the earnest recognition of his leadership should help raise morale at Intel, which struggled under Swan, its first non-Intel CEO.
The big question now is what Gelsinger will do about Intel's manufacturing situation. It's likely that, as the old school Intel executive, Gelsinger is tending to expand on his current hybrid approach of keeping most of Intel manufacturing in-house but potentially more outsourcing. But he will also have new ideas for change after disappearing from the giant for so many years.
Caso put it bluntly. "Gelsinger will not bring a miracle drug on February 15," he wrote. "We see Intel as a 'burning platform', analogous to Nokia a decade ago. If Intel chooses to fully outsource, they will never again have transistor dominance, making it difficult to maintain their dominant stake and margin structure. "
Read: Big Tech Regulation Will Be Difficult, and California is proving it
Insight64's Brookwood said another important role for Gelsginer will be to stop the drain of engineering talent from the company. "If Gelsinger can stop the loss of tech talent and restore people's confidence that Intel can return to technology leadership, I think that's all for the good," he said.
Regardless of what Gelsinger does, he's likely to get more board and Intel support than Swan.
"It is the greatest honor of my career to return to Intel at such a critical time for innovation as the digitization of everything is accelerating," said Gelsinger in his email.
With a long history at Intel and technical qualifications, employees will know Gelsinger is serious. After his previous love-hate relationship with the chip giant, Gelsinger appears to be the leader who can lead Intel out of the chaos it currently finds itself in.
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