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Carrie Underwood Is Pregnant! [PHOTO]
Thomas Chau
Thomas Chau Published: September 1, 2014
Neilson Barnard, Getty Images
Congratulations to Carrie Underwood, who has announced today, Sept. 1, that she is officially pregnant!
Underwood announced via an Instagram post that she is expecting her first child with husband, NHL player Mike Fisher.
"In honor of 'Labor' Day, Ace & Penny would like to make an announcement," she wrote in the caption of the photo of her two dogs. "Their parents couldn't be happier..."
The photo features both Ace and Penny wearing "I'm going to be a big brother" and "I'm going to be a big sister" shirts.
Rumors of Underwood's pregnancy began around the middle of the summer, but the 31-year-old country star and former 'American Idol' winner kept quiet about her upcoming bundle of joy until now.
Underwood and Fisher began dating in 2008 and later became engaged in 2009. The two were then married on July 10, 2010 in Georgia.
Congratulations to Carrie and Mike! We can't wait to see photos of the little boy or girl!
Next: Carrie Underwood Rumored for 2015 Super Bowl Show?
Filed Under: Carrie Underwood
Categories: Celebrity News, Music News, News, Pictures, Pregnant Celebrities


"After a catastrophic fire destroyed parts of the Notre Dame cathedral, thousands of Parisians and tourists prayed on the streets for the future of the twelfth century landmark, not just building is the visiting hardest Harrison sing it like been heart on the medical table. It's it's ho- breaking French president Emmanuel Macron addressed the nation and said he wants to see the cathedral rebuilt within five years. We hear them through an interpreter. What we have to do. Now. Is a great deal, and we will and we will succeed if long live the Republic and long live France CBS's, Elaine Cobb reports some firefighters fought the flames others acted quickly to save priceless artifacts. Firefighters police stuff at the casino formed a human chain to ferry out as much as they possibly could anything they could get their hands on. And in fact, the chaplain for the fire fighters was among those who was going in getting chalices. Of course, the blessed sacrament taken out immediately. And they got out a lot of the artworks the religious artifacts. The French news agency says six hundred sixty five million dollars already been pledged to help rebuild the cathedral, much more from Paris tonight. CBS evening news, two days until attorney general William ball releases a redacted version of special counsel, Robert Muller's report. Correspondent Paula Reed says the president's legal team. Isn't waiting."


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Tag Archives: University Choir
Alumni, College of Liberal Arts, EVENTS, Featured, SCHOOLS AND COLLEGES, Uncategorized
TSU Remembers Founders During 2019 Homecoming
October 15, 2019 Michael McLendon
NASHVILLE, Tenn. (TSU News Service) –Tennessee State University held it's Founder's Day Convocation on Oct. 15 in Kean Hall.
TSU President Glenda Glover, accompanied by keynote speaker Nashville General Chief Executive Officer, Dr. Joseph Webb, led a procession of faculty, student leaders and administrators to mark the university's 107th birthday.
President Glenda Glover presents 2018 Founders' Day speaker Dr. Joseph Webb with a plaque at the ceremony in Kean Hall. (Photo by Emanuel O. Roland II)
The University Wind Ensemble, led by Dr. Reginald McDonald, offered selections to a cheering audience, following presentation of colors by the Air Force ROTC Color Guard.
"We honor our founders, and we celebrate our excellence," Glover said, as she reflected on the significance of the university's contributions. "It is my esteemed pleasure to be here on this founders day."
Miss TSU, Jada Crisp, and Mr. TSU, Damyr Moore, shared a brief history of TSU, followed by a musical selection from the University Choir.
In his keynote address, Webb reminisced about his days as a student at TSU.
Miss TSU Jada Crisp, joined by Mr. TSU Damyr Moore, gives the university history at the Founders' Day program. (Photo by Michael McLendon, TSU Media Relations)
"For me this is truly a homecoming because I am back in what we used to call 'Kean's Little Garden,'" Webb said. "They call it Kean Hall now, but we had a lot of experiences here as an athlete. As a basketball player, I had two goals in mind: one was on this end, and one was on that end."
He reminded the students, faculty and alumni that as members of the TSU family, they are part of a proud legacy.
"TSU, founded in 1912, has a rich history of producing public service and great leaders. It is still that beacon of hope where many aspire to gain a better life with the motto of 'Think.Work.Serve,' and a charge of 'Enter to learn, and go forth to serve.'" he said.
The University Wind Ensemble, led by Dr. Reginald McDonald, offer a selection at Founder's Day Convocation. (Photo by Michael McLendon, TSU Media Relations)
Webb, who joined Nashville General Hospital in 2015, has more than 25 years of experience managing for–profit and not-for-profit healthcare organizations. He obtained his doctorate of science in health services administration from the University of Alabama at Birmingham and holds a bachelor of science and master of science in health and physical education from TSU.
Founded in 1912, Tennessee State University is Nashville's only public university, and is a premier, historically black university and land-grant institution offering 38 bachelor's degree programs, 24 master's degree programs, and seven doctoral degrees. TSU is a comprehensive research intensive institution with a R-2 Carnegie designation, and has a graduate school on its downtown Avon Williams Campus, along with the Otis Floyd Nursery Research Center in McMinnville, Tennessee. With a commitment to excellence, Tennessee State University provides students with a quality education in a nurturing and innovative environment that prepares them as alumni to be global leaders in every facet of society. Visit the University online at tnstate.edu.
Damyr MooreKean HallMichael McLendonTSU President Glenda GloverUniversity Choir

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Pechlivanoglou (Eunice Wind): Emobility without "green" charging is an empty letter
By aspamet In Uncategorized
Along with the generation of clean energy and the incorporation of innovative technologies into the energy system, e-mobility and green charging should go hand in hand, as EUNICE WIND Executive Director and Wind Energy Committee Chairman ASME Dr. Giorgos Pechlivanoglou pointed out, inter alia, speaking at the Economist conference with the title "Third Sustainability Summit for South-East Europe and the Mediterranean" held on October 16 & 17 at Goulandris Museum of Natural History in Athens.
He characteristically said that "e-mobility is inextricably linked to the production of renewable energy as well as the EV charging with green energy". Moreover, he added that charging an electric car makes no sense unless it is done with clean energy.
In any other case any premium and promotion of e-mobility will not necessarily imply an immediate reduction in CO2 emissions when the providers' energy mix is not 100%.
Finally, he emphasized that electromobility and charging of electric cars must be both technically and economically viable solutions.
Source: Energypress
Following are the key points of Mr. Giorgos Pechlivanoglou's speech: "Third Sustainability Summit for South-East Europe and the Mediterranean" October 16-17, Goulandris Museum of Natural History, Kifissia, Athens, Greece
The follow-up to Tilos' innovative project, which practically created the first "green island" in Greece, is being presented today at this conference and is the starting as well as reference point for the following solutions presented at our last meeting in the same conference, which concern the development of "smart grids" and the integration of all innovative technologies, such as electric drive, into the solutions of the new energy system.
What essentially follows the "smart islands" has to do with autonomous energy production, independent energy management, based on smart meters and energy management devices and is completed with e-mobility that practically converts the conventional fossil fuel automation into a form embedded in the grid, and launches a new practice in transportation.
E-mobility and green charging
E-mobility is being promoted as the new order of things in humanity's struggle to reduce CO2 emissions and greenhouse gases, which is directly linked to both electricity generation and transportation sector.
Therefore, electricity is inextricably linked to the production of energy from renewable sources, as well as charging cars with green energy.
Charging an electric car makes no sense unless it comes from green energy. This is extremely important considering that at the moment the energy mix of the largest providers in our country and abroad is not 100% green.
There are only a few providers that have a 100% green energy mix. So, if we charge a car with non-RES energy and at the same time ask the states to subsidize or boost the introduction of electric cars into the market then we would have made a very big, logical, maybe even legal, jump towards the goal of subsidy. But maybe we would also be committing an offense. We have virtually subsidized a technology, without ensuring that it would lead to an immediate reduction in CO2 emissions. It is therefore extremely important to link the promotion of electric cars to green charging.
Sustainability and economy in e-mobility
Electric drive and electric car charging must be both technically and economically viable solutions. While there is a lot of talk about accelerating the new charger technology and the "supercars" that are coming into the market with huge power and very large batteries, it is important to think about what sustainable electricity really means. The promotion of fast-charging networks, which require huge amounts of grid power and ultimately are not economically viable, is apparently wrong. There is a need to rationalize the grid with chargers that follow consumer habits rather than standard practice.
Fast-charging and the best sustainable electric car charging solution
A vivid example is gas stations. The gas stations that according to the old model were powerful provided a fast-charging technical solution for a car. In such a case, within a minute a quantity of energy capable of charging our car for about 500-700 km was received. This electric drive is not feasible, at least in terms of economy and sustainability, since in an attempt to gain autonomy for about 1000 km, an amount of energy should be drawn from the grid equal to the instant consumption of a small town.
The requirements on infrastructures, networks, transformers and other network components are immense and so this is not a viable way to charge my car. What is important is to realize that electric drive, unlike gasoline and petrol, is available in many places as the electricity grid reaches everywhere.
The sustainable way to set up an electric car charging network is one that is customer-centric, follows the car and its driver, and "finds" the car where it is parked.
Let's not forget that cars are parked for about 80% of their life. If we find where the car is parked then we will be able to set up a charging network that is sustainable and can offer a lower-cost solution. If we simply tried to make an electrical replica of a gasoline system then we would have failed as the result would be a system with very high costs and at the same time extremely low efficiency.


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Q: Maximum outdegree of internal vertex in a multi-source single sink directed acyclic graph I'm looking to try and get a bound on the maximum outdegree of an internal vertex in a DAG w/ multiple-source vertexes and a single sink vertex given the following constraints:

*

*$\text{deg}^-(\text{sink}) = 2$

*For any internal vertex, $1 \le \text{deg}^-(v) \le 2$

*Source vertexes can have any outdegree.

Using the degree sum formula, my gut instinct says that the maximum outdegree of an internal vertex is 2 but I'm having difficulty formally proving it. However, I'm new to graph theory so I'm not sure if I'm even on the right track or if my intuition is leading me astray. Any help would is much appreciated!

A: Consider the following graph
            -> v_1 ->
            -> v_2 -> v_5 ->
            -> v_3 ->
Source -> v -> v_4 -> v_6 -> sink

i.e. the graph $G=(V,E)$, with $V=\{source,v,v_1,v_2,...,v_6,sink\}$ and $E=\{(source,v),(v,v_1),(v,v_2),(v,v_3),(v,v_4),(v_1,v_5),(v_2,v_5),(v_3,v_6),(v_4,v_6),(v_5,sink),(v_6,sink)\}$. Source and sink here are two vertices and thus sink has in-degree exactly 2.
You can easily add more "layers" in the graph and thus increase $\deg^+(v)$. If you do not bound the total number of vertices in your graph, vertex $v$ can have arbitrarily high out-degree (something around $\deg^+(v)=\frac{n-1}{2}$, where $n=|V|$). So the answer would be that there can only be an upper bound that is depending on the number of vertices. Note however, that this upper bound might even be higher than the one in my example.
(Excuse me for that ugly scetch but I only have my phone with me and I am not allowed to upload pictures.)

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20 - Sep
Cibo Restaurant presents The Rob Lopes Project
(Nearby Event: Food & Dining)
Where:Cibo Ristorante Italiano, 301 Alvarado St., Monterey, California, United States, 93940
map-Cibo Restaurant presents The Rob Lopes Project Get Directions Cibo Restaurant presents The Rob Lopes Project Add to my Calendar
Cibo Restaurant invites you to join us for a night of live R&B, funk, rock, and pop music featuring The Rob Lopes Project, 9pm-12:30am.
Rob Lopes is a producer, engineer, and musician who has worked with many artists around the globe over the years. From his early days in the 70s as a lead guitarist in groups such as Rain, Avalon, and Coaltrain (a Doobie Bros sideband) and subsequently in the early 80s as a producer and electronic engineer who was one of the pioneers of computerized music for television and radio. More recently, Rob has worked and performed with friends around the Bay Area and Europe including Polo Jones, Gregg Kofi Brown, Alan Glass, and Bridget Marie.


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Fantasy Football For Dummies Cheat Sheet
Your Blog » Fantasy Football For Dummies Cheat Sheet
Previous: All The Single Ladies Meet The Football Players Who Can Dance! Higher School Group Spoofs
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He is also a figurehead for the young players coming by means of. You only had to see how all the substitutes celebrated with him when he scored against Chelsea. Wilshere is showing them the way. Jose Mourinho's side have won all three of their home Premier League games kicking off at 3pm on a Saturday this season, scoring seven objectives and conceding none.Allardyce didn't mention regardless of whether any players had been doubts for the game on Sunday so I'm not confident where that leaves us, but we'll assume that Walcott and Rooney are doubts given that they missed 1-1 draw with Southampton. A GRANDDADS' army of 1980s football yobs are reportedly heading to Russia on a bloody World Cup revenge mission.In a grassroots setting I worked in previously we had a rule - if the children turned up then we would play. It was the option of the group and their parents if we played. This meant that at times we played for just click the following internet site 30 minutes as an alternative of the complete hour, at times the session program went out of the window and we just played small games and often we led sessions with only 3 or 4 children.Understand and take benefit of position runs. Normally, there will be a waterfall effect when a player from a specific position is chosen (for instance — when the first team goes soon after a tight end, there will at some point be a mass-scramble from the rest of the league to acquire high-talent tight ends).We apply the worlds most effective fantasy football algorithm to your squad, giving you transfer ideas to increase your FPL performance. Swansea have won their last two Premier League games against 'big six' sides, albeit in house games (vs Liverpool and Arsenal) this following a run of just click the following internet site 1 win in 20 league games against these clubs (W1 D4 L15). Nonetheless, players such as Loftus-Cheek and Trent Alexander-Arnold have not been incorporated basically for the sake of displaying them the ropes - these youngsters have repeatedly impressed this season and deserve their place on merit. Southgate undoubtedly implies company, and with a lot of hardened heads among the squad - Cahill, Young, Henderson, and so on - this fascinating fusion of youth and encounter makes for a curious prospect.Mkhitaryan needs to be nurtured and cared for. When I watched United lately, there have been two players who were clearly not thriving: Luke Shaw, and Mkhitaryan. The significance of a skill depends to an extent on the player's position on the field. All round, football capabilities can be divided into 3 major locations, namely outfield technical, physical, mental and goalkeeping technical skills.The away targets rule was very first employed in the UEFA Cup Winners' Cup in 1970. Because then, the rule has spread to other competitions and leagues, even though it can be really confusing to know to which competitions the away goals rule applies. No one particular cares about style, Wembley only remembers winners. Our football will only get us credit this season since it got us promoted. If we'd lost here, no one would have talked about us.Competitors with Rugby League, Rugby Union and Soccer for players and spectators is constant, specifically at the grassroots levels with underage and regional competitions. Geographically, Aussie Rules dominates the southern regions of Australia, and is growing into the northern centres exactly where the rugby codes have traditionally held sway.Indirect free-kicks can throw a defence into confusion when taken quickly, or in unexpected directions. The third aim of Brazil's 1970 Globe Cup victory over Italy illustrates the approach. Brazil's midfield common Gerson Nunes approached quickly and lofted the dead ball in a high arc virtually from the midfield line. It identified Pele completely positioned near the Italian aim. The Brazilian headed down softly and accurately, straight into the path of the onrushing winger Jairzinho, who virtually walked the ball into the net. If you loved this article and you would certainly like to receive even more information pertaining to just click the following internet site kindly see our own web-site. Italy's catenaccio defence was caught out of position right here, not only failing to neutralise Pele, but offering little cover to stop Jairzinho. A speedy, daring set-piece kick will typically accomplish this unbalancing of defences.Manchester City will obtain about £3.9million from FIFA, the highest sum of any club contributing players to the ongoing Globe Cup in Russia. New players only. 18+, UK & Ire only. Promo code 'G30' required. Min first bet £10. Need to be placed within 14 days of account reg. £30 credited as £3 x £10 free of charge bets. Not valid just click the following internet site with CashOut. Totally free bet valid for 4 days. T&Cs apply.Down South, SEC football is a way of life. Beckham joins MLS side Los Angeles Galaxy from Real Madrid in July. One of the clauses in his $250million deal is the choice to obtain an MLS expansion franchise in any marketplace except for New York City at the fixed value of $25m anytime he stopped playing in the league. The league's owners had by no means provided this allowance to a player prior to.


Veteran Joe Troupe had a gift for woodwork and motivation.
Joseph Randall Troupe, better known as Joe, was a veteran, a skilled carpenter and a powerful motivator. Troupe died Friday after battling lymphoma. He was 67. Troupe's wife, Mary Troupe, is the executive director of the Mississippi Coalition for Citizens with Disabilities. She credits him with helping her live a normal life in spite of being wheelchair bound.
Joe Troupe was born in Columbus, Miss., in 1942. His family later moved to West Point, and Troupe graduated from West Point High School. After graduation, he joined the Army and served as a Morse code operator in Okinawa, Japan, before returning to serve with the Army National Guard. He pursued his love of Morse code back home, as well, joining the Jackson Amateur Radio Club.
Troupe retired from the Army in 1992 and switched gears to pursue another great passion: woodworking. He formed a company, J.R. Troupe Construction, and did custom cabinetry and woodwork for private builders.
"One of his pride and joys was having the wife in a new home come in with a picture from Southern Living and say, 'I want this mantle,'" Mary Troupe said today. "He'd build it, and they'd say, 'I didn't know it was going to be just like it!'"
Mary Troupe met Joe at a trying time. In 1972, after returning from rehabilitation after an injury that left her unable to walk, Mary was in the process of divorcing her then-husband. Her divorce attorney's secretary, one of Mary's friends, was dating Joe at the time.
"That was when a lot of people—and of course some still don't—didn't think that people with disabilities can have a life," Mary said. "He never ever thought that. He was right there with me."
When Mary got her first set of hand controls for her car, she thought she would never master the apparatus and drive again. Joe drove the two of them to the reservoir and told her: "Slide over. You're fixing to drive."
He was right: Mary learned to drive again.
Joe retired from his construction company in January 2009 and was diagnosed with lymphoma in February. Mary praises the Veterans Administration for taking excellent care of her husband.
Troupe is survived by his wife, their three children and six grandchildren. A memorial service will be held this afternoon at 2 p.m. at the Mississippi Agriculture Museum Chapel on Lakeland Drive.
Do you know a Jacksonian or Mississippian who should be the JFP Person of the Day? Briefly tell us his or her story, sending it to [e-mail missing]. Please include contact information and a picture, if possible.
ON THE ISSUES: Special Needs: Left Behind?


His nine career professional major championships tie him (with Gary Player) for fourth all-time, trailing only Jack Nicklaus (18), Tiger Woods (14) and Walter Hagen (11). and after his 1949 auto accident, Hogan struggled to manage more than 18 holes a day.
His career has not been without controversy, then he quit school and began working as a mechanic. and is eclipsed only by Tiger Woods who has been nominated seven times.
FIC Woods, S Dcd D.C. dead : [a Stone Barrington novel] / Stuart Woods. Rock paper tiger [electronic resource] / Lisa Brackmann. Brackmann, Lisa. College and career ready standards and assessments [electronic resource].


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\section{Introduction}

The first systematic treatment
of the orbit structure of a complex flag manifold $X=G/P$
under the action of a real form $G_0 \subset G$
is due to J.~Wolf \cite{W}.
Forty years after his paper,
these real group orbits and their cycle spaces
are still an object of intensive research.
We present here some results in this area, together with
other related results
on transitive and locally transitive actions
of Lie groups on complex manifolds.

\medskip
\noindent
The paper is organized as follows.
In Section 2 we prove the celebrated finiteness theorem for $G_0$-orbits
on $X$ (Theorem 2.3).
We also state a theorem
characterizing open $G_0$-orbits on $X$ (Theorem 2.4).
All results of Section 2 are taken from \cite{W}.
In Section 3 we recall for future use
a theorem, due to B.~Weisfeiler \cite{We}
and A.~Borel and J.~Tits \cite{BT}. Namely, let
$H$ be an algebraic subgroup
of a connected reductive group $G$. Theorem 3.1 shows that
one can find
a parabolic subgroup
$P \subset G$ containing $H$, such that the unipotent radical of
$H$ is contained in the unipotent radical of $P$.
In Section 4 we consider the factorizations
of reductive groups. The results of this
section are due to A.L.~Onishchik \cite{On1, On2}.
We take for granted his list
of factorizations $G = H_1\cdot H_2$,
where $G$ is a simple algebraic group over ${\Bbb C}$
and $H_1, H_2 \subset G$ are reductive complex subgroups
(Theorem 4.1), and deduce from it
his theorem on real forms. Namely, a
real form $G_0$ acting locally transitively
on an affine homogeneous space $G/H$
is either $SO_{1,7}$ or $SO_{3,5}$.
Moreover, in that case $G/H =
SO_8/Spin_7$ and the action of $G_0$ is in fact transitive
(Corollary 4.7).
This very special homogeneous space
of a complex group $G$ has on open orbit of
a real form $G_0$,
the situation being typical for flag manifolds.
One can ask what homogeneous spaces
share this
property. It turns out that if a real form of inner type
$G_0 \subset G$ has an open orbit on a
homogeneous space $G/H$ with $H$ algebraic, then $H$
is in fact parabolic and so $G/H$ is a flag manifold.
We prove this in Section 5 (see Corollary 5.2)
and then retrieve the result of F.M.~Malyshev
of the same type
in which the isotropy subgroup is not necessarily algebraic
(Theorem 5.4). It should be noted that,
the other way around, the statement
for algebraic homogeneous spaces can be deduced from
his theorem. Our proof of both results is new.

\medskip
\noindent
Let $K$ be the complexification
of a maximal compact subgroup $K_0 \subset G_0$.
In Section 6 we briefly recall the Matsuki correspondence
between $G_0$- and $K$-orbits on a flag manifold.
In Section 7 we define,
following the paper of S.G.~Gindikin and the author \cite{AG}, the
crown $\Xi $ of $G_0/K_0$ in $G/K$. We also introduce the cycle
space of an open $G_0$-orbit on $X=G/P$, first
considered by
R.~Wells and J.~Wolf \cite{WeW},
and state a theorem describing the cycle spaces
in terms of the crown (Theorem 7.1). In fact,
with some exceptions which are well-understood,
the cycle space of an open $G_0$-orbit on $X$
agrees with $\Xi $
and, therefore, is independent of the flag manifold.
In Sections 8 and 9,
we give an outline of the original proof
due to G.~Fels, A.~Huckleberry and J.~Wolf \cite{FHW},
using the methods of complex analysis.
One ingredient of the proof is a theorem of
G.~Fels and A.~Huckleberry \cite{FH},
saying that $\Xi $ is a maximal $G_0$-invariant, Stein and
Kobayashi hyperbolic domain in $G/K$ (Theorem 8.4).
Another ingredient is the construction of the Schubert domain,
due to A.~Huckleberry and J.~Wolf \cite{HW} and
explained in Section 9.
Finally, in Section 10
we discuss complex geometric properties of flag domains.
Namely, let $q$ be the dimension
of the compact $K$-orbit in an open $G_0$-orbit.
We consider measurable open $G_0$-orbits and
state the theorem of W.~Schmid and J.~Wolf \cite{SW}
on the $q$-completeness of such flag domains.

\medskip
\noindent
Given
a $K$-orbit $O$
and the corresponding $G_0$-orbit
$O^\prime $ on $X$,
S.G.~Gindikin and T.~Matsuki suggested
to consider the subset $C(O) \subset G$ of all $g\in G$, such that
$gO\cap O^\prime \ne \emptyset$ and
$ gO \cap O^\prime $ is compact, see \cite{GM}.
If $O$ is compact then $O^\prime $ is open and $O \subset O^\prime $.
Furthermore,
in this case $C(O) = \{g\in G \ \vert \ gO \subset O^\prime\}$
is precisely the set whose connected component $C(O)^\circ$ at
$e \in G$ is the cycle space of $O^\prime $ lifted to $G$.
This gives a natural way of generalizing the notion
of a cycle space to lower-dimensional $G_0$-orbits.
Recently, using this generalization, T.~Matsuki
carried over Theorem 7.1 to
arbitrary $G_0$-orbits on flag manifolds, see \cite{M4} and Theorem 7.2.
His proof is beyond the scope of our survey.

\medskip
\noindent

\section{Finiteness theorem}

Let $G$ be a connected
complex semisimple Lie group,
$\goth g$ the Lie algebra of $G$,
and $\goth g_0$ a real form of
$\goth g$.
The complex conjugation of $\goth g$ over $\goth g_0$ is denoted
by $\tau $.
Let $G_0$ be the connected real Lie subgroup
of $G$ with Lie algebra $\goth g_0$.
We are interested
in $G_0$-orbits on flag manifolds of $G$.
By definition, these manifolds are the quotients of the form $G/P$,
where $P \subset G$ is a parabolic subgroup.
It is known that the intersection of two parabolic
subgroups in $G$ contains a maximal torus of $G$. Equivalently,
the intersection of two parabolic subalgebras
in $\goth g$ contains a Cartan subalgebra of $\goth g$.
We want to prove a stronger statement in the case when
the parabolic subalgebras are $\tau $-conjugate.
We will use the notion of a Cartan subalgebra for an arbitrary
(and not just semisimple) Lie algebra $\goth l$
over any field $k$.
Recall that a Lie subalgebra
$\goth j \subset \goth l$
is called a Cartan subalgebra if $\goth j$ is nilpotent
and equal to its own normalizer.
Given a field extension $k \subset k^\prime $,
it follows from that definition that
$\goth j$ is a Cartan subalgebra in $\goth l$
if and only if $\goth j\otimes _k k^\prime $ is a Cartan
subalgebra in $\goth l\otimes _k k^\prime $.
We start with a simple general observation.

\medskip \noindent {\bf Lemma 2.1.}\ {\it
Let $\goth g$ be any complex Lie algebra,
${\goth g }_0 $ a real form of $\goth g$, and
$\tau : \goth g \to \goth g$ the complex conjugation of $\goth g$
over $\goth g_0$.
Let $\goth h \subset \goth g$ be a complex Lie
subalgebra. Then $\goth h \cap \goth g _0$ is a real form of
$\goth h \cap \tau(\goth h)$.}

\medskip \noindent
{\it Proof.}\ For any $A \in \goth h \cap \tau (\goth h)$ one has
$2A = (A+\tau (A))+ (A - \tau (A))$, where the first summand is contained in
$\goth h \cap \goth g_0$ and the second one gets into that subspace after
multiplication by $i$.
\hfill{$\square $}

\medskip
\noindent
The
following corollary will be useful.

\medskip
\noindent
{\bf Corollary 2.2.}\ {\it If $\goth p$ is a parabolic subalgebra
of a semisimple algebra $\goth g$, then $\goth p \cap \tau (\goth p)$
contains a $\tau $-stable Cartan subalgebra $\goth t $ of $\goth g$.}

\medskip
\noindent
{\it Proof.}\ Choose a Cartan subalgebra
$\goth j$ of
$\goth p \cap \goth g _0$. Its complexification $\goth t $
is a Cartan subalgebra
of $\goth p \cap \tau (\goth p)$, which is $\tau $-stable.
Now, $\goth p \cap \tau (\goth p)$ contains a Cartan subalgebra
$\goth t ^\prime $ of $\goth g$.
Since $\goth t$ and $\goth t^\prime $ are conjugate
as Cartan subalgebras of $\goth p \cap \tau (\goth p)$,
it follows that
$\goth t$ is itself
a Cartan subalgebra of $\goth g$.
\hfill {$\square $}

\medskip
\noindent
The number of conjugacy classes of Cartan subalgebras of a real semisimple
Lie algebra is finite. This was proved independently by A.~Borel and B.~Kostant
in the fifties of the previous century, see \cite{Ko}. Somewhat later,
M.~Sugiura determined explicitly the number of conjugacy classes and
found their representatives for each simple Lie algebra, see \cite {Su}.
Let $\{\goth j_1, \ldots, \goth j_m\}$ be a complete system of representatives
of Cartan subalgebras of $\goth g_0$. For each $k, \ k=1,\ldots, m,$
the complexification $\goth t_ k $ of $\goth j_k$
is a Cartan subalgebra of $\goth g$.

\medskip
\noindent
{\bf Theorem 2.3 ({\rm J.~Wolf \cite {W}, Thm.\,2.6}).}\
{\it For any parabolic subgroup $P \subset G$
the number of $G_0$-orbits on $X=G/P$ is finite.}

\medskip
\noindent{\it Proof.}\ Define
a map $\iota : X \to \{1,\ldots, m\}$ as follows.
For any point $x \in X$ let $\goth p_x$ be the isotropy
subalgebra of $x$ in $\goth g$. By Corollary 2.2, we can choose a Cartan
subalgebra $\goth j _x$ of $\goth g _0$ in $\goth p _x$.
Take $g\in G_0$ so that ${\rm Ad} g \cdot \goth j_x = \goth j_k$
for some $k, k=1, \ldots , m$. Since $\goth j_k $ and $\goth j_l$
are not conjugate for $k \ne l$, the number $k$ does not depend on
$g$.
Let $k =\iota (x)$. Then $\iota (x)$ is constant along the orbit $G_0(x)$.
Now, for $\iota (x) $ fixed
there exists $g\in G_0$ such that $\goth p _{gx} $
contains $\goth t_k$ with fixed $k$. Recall that
a point of $X $ is uniquely determined by its isotropy subgroup.
Since there are only finitely many parabolic subgroups containing
a given maximal torus,
the fiber of $\iota $ has finitely many $G_0$-orbits.
\hfill {$\square $}

\medskip \noindent
As a consequence of Theorem
2.3, we see that at least one $G_0$-orbit is open in
$X$. We will need a description
of open orbits in terms of isotropy subalgebras of their points.
Fix a Cartan
subalgebra $\goth t \subset \goth g$. Let $\Sigma = \Sigma (\goth g, \goth t)$
be the root system,
$\goth g_\alpha \subset \goth g,\ \alpha \in \Sigma$,
the root subspaces, $\Sigma ^+ = \Sigma^+(\goth g, \goth t)
\subset \Sigma $ a positive subsystem,
and $\Pi $ the set of simple roots
corresponding to $\Sigma ^+$.
Every $\alpha \in \Sigma $ has a unique expression $\alpha = \sum
_{\pi \in \Pi}\ n_{\pi}(\alpha )\cdot \pi $, where $n_{\pi}(\alpha )$
are integers, all non-negative for $\alpha \in \Sigma ^+  $
and all non-positive for $\alpha \in \Sigma ^- = -  \Sigma ^+$.
For an arbitrary subset $\Phi \subset \Pi $
we will use the notation
$$\Phi ^r =\{ \alpha \in \Sigma \ \vert \ n_{\pi }(\alpha ) = 0
\ {\rm whenever}\ \pi \not \in \Phi \},\ \ \Phi ^u
=\{ \alpha \in \Sigma ^+ \ \vert \ \alpha \not \in \Phi ^r\}.$$
Then the standard parabolic subalgebra $\goth p _\Phi\subset \goth g$
is defined by
$$\goth p _{\Phi } = \goth p _{\Phi }^r + \goth p_{\Phi }^u,$$
where
$$\goth p _{\Phi }^r = \goth t + \sum _{\alpha \in \Phi ^r}\ \goth g_{\alpha }
\ \
{\rm and}
\ \
\goth p _{\Phi }^u = \sum _{\alpha \in \Phi ^u}\ \goth g _{\alpha }$$
are the standard reductive Levi subalgebra and, respectively, the
unipotent radical of $\goth p_{\Phi }$. In the sequel,
we will also use the notation
$$\goth p_{\Phi}^{-u} = \sum _{- \alpha \in \Phi ^u} \goth g_{\alpha }.$$
Now, let
$\goth k_0$ be a maximal compact subalgebra of $\goth g_0$.
Then we have the Cartan involution $\theta : \goth g_0 \to \goth g_0$
and the Cartan decomposition
$\goth g_0 = \goth k_0 + \goth m_0$,
where $\goth k_0$ and $\goth m_0$ are the eigenspaces of $\theta $
with eigenvalues 1 and, respectively, -1.
A $\theta $-stable Cartan subalgebra $\goth j \subset\goth g_0$
is called fundamental (or maximally compact) if $\goth j \cap \goth k_0$
is a Cartan subalgebra of $\goth k_0$. More generally,
a Cartan subalgebra $\goth j \subset \goth g_0$
is called fundamental if $\goth j$ is conjugate to a
$\theta $-stable fundamental Cartan subalgebra.
It is known that any two fundamental Cartan subalgebras of $\goth g_0$
are conjugate under an inner automorphism of $\goth g_0$.
We will assume that a Cartan subalgebra $\goth t \subset \goth g$
is $\tau $-stable. In other words, $\goth t = \goth j ^{\Bbb C}$,
where $\goth j$ is a Cartan subalgebra in $\goth g_0$. Then $\tau $
acts on $\Sigma $ by
$\tau (\alpha ) (A) = \overline {\alpha (\tau\cdot A)}$,
where $\alpha \in \Sigma ,\, A \in \goth t$.

\medskip
\noindent
{\bf Theorem 2.4 ({\rm J.~Wolf \cite{W}, Thm.\,4.5}).}\
{\it Let $X = G/P$ be a flag manifold. Then the $G_0$-orbit
of $x_0 = e\cdot P$ is open in $X$ if and only if $\goth p = \goth p_{\Phi}$,
where

{\rm (i)} $\goth p \cap \goth g_0 $ contains a fundamental Cartan subalgebra
$\goth j \subset \goth g_0$;

{\rm (ii)} $\Phi $ is a subset of simple roots for
$\Sigma ^+ (\goth g, \goth t),
\ \goth t = \goth j ^{\Bbb C}$,
such that $\tau \Sigma^+ = \Sigma ^-$.

}
\medskip \noindent
The proof can be also found in \cite {FHW}, Sect. 4.2.

\section{Embedding a subgroup into a parabolic one }

Let $G$ be a group. The normalizer of a subgroup $H \subset G$
is denoted by $N_G(H)$.
For an algebraic group $H$ the unipotent radical is denoted
by $R_u(H)$.

\medskip
\noindent
Let $U$ be an algebraic unipotent subgroup of a complex semisimple group $G$.
Set $N_1 = N_G(U)$, $U_1 = R_u(N_1) $, and continue inductively:

$$N_k = N_G(U_{k-1}),\ \ U_k = R_u(N_k),\ \ k \ge 2.$$
Then
$U \subset U_1$ and
$ U_{k-1} \subset U_k ,\ \ N_{k-1}\subset N_k$
for all $k \ge 2$. Therefore there exists an integer $l$, such that
$U_l = U_{l+1}$. This means that $U_l$ coincides with
the unipotent radical of its normalizer. We now recall the
following general theorem of fundamental importance.

\medskip \noindent
{\bf Theorem 3.1 ({\rm B.~Weisfeiler \cite{We}, A.~Borel, J.~Tits \cite{BT},
Cor. 3.2}).}\ {\it Let $k$ be an arbitrary field, $G$ a connected
reductive
algebraic group defined over $k$, and $U$ a
unipotent algebraic subgroup of $G$.
If the unipotent radical of the normalizer
$N_G(U)$ coincides with $U$, then $N_G(U)$
is a parabolic subgroup of $G$.}

\medskip \noindent
For $k ={\Bbb C}$, which is the only case we need,
the result goes back to V.V.~Morozov, see \cite{BT}, Remarque 3.4.
In the above form, the theorem was conjectured by I.I.~Piatetski-Shapiro, see
\cite{We}. For future references, we state the following corollary
of Theorem 3.1.

\medskip \noindent
{\bf Corollary 3.2.}\ {\it Let $k = {\Bbb C}$ and
let $G$ be as above. The normalizer $N_G(U)$
of a unipotent algebraic subgroup $U \subset G$
embeds into a parabolic subgroup $P \subset G$ in such a way that
$U \subset R_u(P)$. For any algebraic subgroup $H \subset G$
there exists an embedding into a parabolic subgroup $P$,
such that $R_u(H) \subset R_u(P)$.}

\medskip \noindent
{\it Proof.} Put $P = N_G(U_l)$ in the above construction.
Then $U \subset U_l = R_u(P)$.
This proves the first assertion. To prove the second one,
it suffices to take $U = R_u(H)$.
\hfill$\square$

\section{Factorizations of reductive groups}

The results of this section are due to A.L.~Onishchik.
Let $G$ be a group, $H_1, H_2 \subset G$ two subgroups.
A triple $(G;H_1,H_2)$ is called a factorization of $G$
if for any $g\in G$ there exist
$h_1 \in H_1$
and $h_2 \in H_2$, such that $g = h_1 \cdot h_2$.
In the Lie group case a factorization
$(G;H_1,H_2)$ gives rise to the factorization $(\goth g; \goth h_1, \goth h_2)$
of
the Lie algebra $\goth g$. By definition, this means that $\goth g =
\goth h_1 + \goth h_2$. Conversely, if $(\goth g; \goth h_1, \goth h_2)$
is a factorization of $\goth g$ then the product $H_1\cdot H_2$
is an open subset in $G$ containing the neutral element. In general,
this open set
does not coincide with $G$, and so
a factorization $(\goth g; \goth h_1, \goth h_2)$ is sometimes called a local
factorization of $G$. But, if $G, H_1$ and $H_2$ are
connected reductive (complex or real) Lie groups then
every local factorization is (induced by) a global one,
see \cite {On2}. We will give a simple proof of this fact below,
see Prop. 4.3 and Prop. 4.4.

\medskip
\noindent
All factorizations of connected compact Lie groups are classified
in \cite {On1}, see also \cite {On3}, \S\,14.
If $G, H_1$ and $H_2$ are connected reductive
(complex or real) Lie groups, then the same problem is solved in \cite{On2}.
The core of the classification is the complete list of factorizations
for simple compact Lie groups. We prefer
to state the result for simple algebraic groups over $\Bbb C$.
If both subgroups $H_1,H_2$
are reductive algebraic, then the list
is the same as in the compact case.

\medskip \noindent
{\bf Theorem 4.1 ({\rm A.L.~Onishchik \cite {On1, On2}}).}
{\it If $G$ is a simple algebraic group over $k ={\Bbb C}$
and $H_1, H_2$ are proper reductive algebraic subgroups of $G$, then,
up to a local isomorphism and renumbering of factors, the factorization
$(G; H_1, H_2)$ is one of the following:

{\rm (1)}\ $(SL_{2n}; Sp_{2n},
SL_{2n-1}), \ n\ge 2 $;

{\rm (2)}\ $(SL_{2n}; Sp_{2n},
S(GL_1\times GL_{2n-1}),\  n\ge 2$;

{\rm (3)}\ $(SO_7; G_2, SO_6)$;

{\rm (4)}\ $(SO_7; G_2, SO_5)$;

{\rm (5)}\ $SO_7; G_2, SO_3\times SO_2)$;

{\rm (6)}\ $(SO_{2n}; SO_{2n-1}, SL_n),\ n\ge 4 $

{\rm (7)}\ $(SO_{2n}; SO_{2n-1}, GL_n),\ n\ge 4$;

{\rm (8)}\ $(SO_{4n}; SO_{4n-1}, Sp_{2n}), \ n\ge 2$;

{\rm (9)}\ $(SO_{4n}; SO_{4n-1}, Sp_{2n}\cdot Sp_2),\ n\ge 2$;

{\rm (10)}\ $(SO_{4n}; SO_{4n-1}, Sp_{2n}\cdot k^*), \ n \ge 2$;

{\rm (11)}\ $(SO_{16}; SO_{15}, Spin _9)$;

{\rm (12)}\ $(SO_8; SO_7, Spin _7)$.}

\medskip
\noindent
Although this result is algebraic by its nature, the only known proof
uses topological methods. We want to show how Theorem 4.1 applies
to factorizations of complex Lie algebras involving their real forms.

\medskip
\noindent
{\bf Lemma 4.2.}\ {\it
Let $\tau : \goth g \to \goth g$
be the complex
conjugation of a complex Lie algebra over its real form $\goth g_0$.
Let $\goth h \subset \goth g$ be a complex Lie subalgebra.
Then $\goth g = \goth g_0 + \goth h$ if and only if $\goth g = \goth h +
\tau (\goth h)$.}

\medskip
\noindent
{\it Proof.}\ Let $\goth g = \goth g_0 + \goth h$.
For any $X \in {\goth g}_0$ one has $iX = Y + Z$,
where $Y \in \goth g_0$ and $Z \in \goth h$. This implies
$$2X=-iZ - \tau (iZ) \in \goth h +  \tau (\goth h).$$
Conversely, if $\goth g = \goth h + \tau (\goth h)$,
then for any $X \in \goth g$ there exist $Z_1, Z_2 \in \goth h$, such that
$$X  = Z_1 + \tau (Z_2)
=(Z_1 - Z_2) + (Z_1 + \tau (Z_2)),$$
hence $X \in \goth h + \goth g_0$.
\hfill $\square $

\medskip
\noindent
{\bf Proposition 4.3.}\ {\it Let $G$ be a connected
reductive algebraic group over
${\Bbb C}$ and let $H_1, H_2 \subset G$ be two reductive algebraic subgroups.
Then $\goth g = \goth h_1 + \goth h_2$ if and only if $G= H_1\cdot H_2$.}

\medskip
\noindent
{\it Proof.}\ It suffices to prove that the local factorization
implies the global one. Let $X = G/H_2$ and let $n ={\rm dim}(X)$.
If $L $
is a maximal compact subgroup of $H_2$
and $K$ is a maximal compact subgroup of $G$, such that $L \subset K$,
then $X$ is diffeomorphic to a real vector bundle over $K/L$.
Therefore $X$ is homotopically equivalent to a compact manifold
of (real) dimension $n$.
On the other hand,
$H_1$ has an open orbit on $X$. Since $X$ is an affine variety,
closed $H_1$-orbits are separated by $H_1$-invariant regular functions.
But such functions are constant, so there is only one closed orbit.
Assume now that
$H_1$ is not transitive on $X$, so that the closed $H_1$-orbit
has dimension $m < n $. A well-known corollary of Luna's Slice Theorem
displays $X$ as a vector bundle over the closed orbit, see \cite{Lu}.
Thus $X$ is homotopically equivalent to that orbit and,
by the same argument as above, to a compact manifold
of (real) dimension $m$.
Now, for a compact connected manifold $M$ of dimension $n$ one has
$H_i(M, {\Bbb Z}_2) = 0$ if $i > n$ and $H_n(M, {\Bbb Z}_2)
\cong {\Bbb Z}_2 $,
see e.g. \cite{D}, Prop. 3.3 and Cor. 3.4.
Therefore two compact manifolds of dimensions $m$
and $n, m \ne n,$
are not homotopically equivalent,
and we get a contradiction.
\hfill $\square $

\medskip \noindent
As a corollary, we have a similar proposition for real groups.

\medskip \noindent
{\bf Proposition 4.4.}\ {\it Let $G, H_1$ and $H_2$ be real forms
of complex reductive algebraic
groups $G^{\Bbb C}, H_1^{\Bbb C}$ and $H_2^{\Bbb C}$.
For $G$ connected one has
$\goth g = \goth h_1 + \goth h_2$ if and only if $G=H_1\cdot H_2$.}

\medskip \noindent
{\it Proof.}
If $\goth g = \goth h_1 + \goth h_2$ then
$\goth g^{\Bbb C} = {\goth h_1}^{\Bbb C} + {\goth h_2}^{\Bbb C}$. Thus
$G^{\Bbb C} = H_1^{\Bbb C}\cdot H_2^{\Bbb C}$
by Proposition 4.3. The action of $H_1^{\Bbb C}
\times H_2^{\Bbb C}$ on $G^{\Bbb C}$, defined by
$$g \mapsto h_1gh_2^{-1},\ \ g \in G^{\Bbb C}, h_i \in H_i^{\Bbb C},$$
is transitive. For $g\in G \subset G^{\Bbb C}$ we have
the following estimate of the dimension of $(H_1 \times H_2)$-orbit
through $g$:
$${\rm dim}\, H_1gH_2 = {\rm dim}\, H_1 + {\rm dim}\, H_2 - {\rm dim}
\, (H_1 \cap gH_2g^{-1}) \ge $$
$$ {\rm dim}_{\Bbb C}\, H_1^{\Bbb C}
+ {\rm dim}_{\Bbb C}\, H_2 ^{\Bbb C}
-{\rm dim} _{\Bbb C}\, H_1 ^{\Bbb C} \cap gH_2^{\Bbb C} g^{-1}
= {\rm dim}_{\Bbb C}\, G^{\Bbb C} = {\rm dim}\, G.$$
But $G$ is connected and each coset $H_1gH_2$ is open,
hence $G = H_1\cdot H_2$.
\hfill $\square $

\medskip
\noindent
We will use the notion of an algebraic subalgebra
of a complex Lie algebra $\goth g$, which corresponds
to an algebraic group $G$. A subalgebra $\goth h \subset \goth g$
is said to be algebraic, if the associated connected subgroup $H \subset G$
is algebraic. In general, this notion depends
on the choice of $G$. However, if $\goth g$ is semisimple,
which will be our case, then $\goth h$ is algebraic for some
$G$ if and only if $\goth h$ is algebraic for any other $G$.
An algebraic subalgebra of $\goth g$ is said to be reductive,
if $H$ is a reductive algebraic subgroup of $G$. Again,
for $\goth g$ semisimple the choice of $G$ does not matter.

\medskip
\noindent
{\bf Theorem 4.5 ({\rm cf. \cite{On2}, Thm. 4.2}).}
{\it Let $\goth g$ be a simple complex Lie algebra, $\goth h \subset \goth g$,
$\goth h \ne \goth g$,
a reductive algebraic subalgebra, and $\goth g _0$ a real form of $\goth g$.
If $\goth g = \goth g_0 + \goth h$ then $\goth g$
is of type ${\rm D}_4$, $\goth h$ is of type ${\rm B}_3$,
embedded as the spinor subalgebra,
and $\goth g_0$ is either $\goth {so}_{1,7}$
or $\goth {so}_{3,5}$.}

\medskip
\noindent
{\it Proof.}\ In the notation of Lemma 4.2 we have $\goth g = \goth h +
\goth \tau (\goth h)$.
Note that $\tau (\goth h)$ is a reductive
algebraic subalgebra
of $\goth g$.
Choose $G$ simply connected. Then $\tau $ lifts
to an antiholomorphic involution of $G$, which we again
denote by $\tau $. Let $H_1$ and $H_2$ be the connected
reductive algebraic subgroups of $G$ with Lie algebras $\goth h$ and,
respectively, $\tau (\goth h)$. By Proposition 4.3 we have
the global decomposition $G = H_1 \cdot H_2$. Since $H_1$ and $H_2$
are isomorphic, it follows from Theorem 4.1 that
the factorization $(G; H_1, H_2)$ is obtained from factorization (12).
More precisely,
$G$ is isomorphic to
$Spin _8$, the universal covering group of $SO _8$, and $H_1, H_2$
are two copies of $Spin _7$ in $Spin _8$. We assume that $H_1$
is the image of the spinor representation $Spin _7 \to SO_8$
and $H_2$ comes from the embedding
$SO_7 \to SO_8 $.
The conjugation $\tau $ interchanges $H_1$ and $H_2$.
We want to replace $\tau $ by a holomorphic involutive automorphism
of $G$ with the same behaviour with respect to $H_1$ and $H_2$.
For this we need the following lemma.

\medskip
\noindent
{\bf Lemma 4.6.}\ {\it Let $G$ be a connected reductive algebraic group
over ${\Bbb C}$. Take a maximal
compact subgroup in $G$ which is invariant under $\tau $.
Let
$\theta : G \to G$ be the corresponding Cartan involution
and let $\sigma = \theta \tau (= \tau \theta)$.
For a reductive algebraic subgroup $H \subset G$, the factorization
$(G; H, \tau (H))$ implies the factorization $(G; H, \sigma (H))$,
and vice versa.}

\medskip
\noindent
{\it Proof.}\
First of all, if $(G; H_1, H_2)$ is a factorization
of a group
then one also has the factorization
$(G; \tilde H_1, \tilde H_2)$, where
$\tilde H_1 = g_1 H_1 g_1 ^{-1}, \tilde H_2
= g_2 H_2 g_2 ^{-1}$
are conjugate subgroups.
In the setting of the lemma,
choose $\tilde H = gHg^{-1}$
so that a maximal compact subgroup of $\tilde H$ is contained in the
chosen maximal compact subgroup of $G$.
Then $\theta (\tilde H) = \tilde H$
and, consequently,
$$\tau (H) \simeq \tau (\tilde H) = \sigma (\tilde H) \simeq \sigma (H),$$
where $\simeq $ denotes conjugation by an inner automorphism.
By the above remark, one of the
two factorizations
$(G;H , \tau (H)),\,(G; H, \sigma (H))$ implies the other.
\hfill $\square $

\medskip
\noindent
{\it End of proof of Theorem 4.5.}\ We can replace $H_2$
by a conjugate subgroup so that $H_1 $ and $H_2  $ are
interchanged by $\sigma $. The factorization is in fact defined for $SO_8$,
in which case the subgroups are only locally isomorphic. For
this reason, $\sigma $ is an outer automorphism. It follows that
the restriction of $\sigma $ to the real form,
i.e., the Cartan involution of the latter, is also an outer automorphism.
There are precisely two real forms of ${\rm D}_4$ with this property,
namely, ${\goth s \goth o}_{1,7}$ and ${\goth s \goth o}_{3,5}$.
The remaining non-compact real forms ${\goth s \goth o}_{2,6},
\ {\goth s \goth o}_{4,4}$, and ${\goth s \goth o}^* _8$
are of inner type, see Sect 5.
We still have to show that ${\goth s \goth o}_{1,7}$, as well as
${\goth s \goth o}_{3,5}$,
together with the complex spinor subalgebra,
gives a factorization of
$\goth g = {\goth s \goth o}_8$. So let $\tau $
be the complex conjugation of $\goth g$
over ${\goth s \goth o}_{1,7}$ or ${\goth s \goth o}_{3,5}$.
Define $\sigma $ as in the lemma
and denote again by $\sigma $ the corresponding automorphism of $\goth g$.
The fixed point subalgebra of $\sigma $ has rank 3,
whereas $\goth g$ has rank 4. Thus $\sigma $
is an outer automorphism of $\goth g$.
There are three
conjugacy classes
of subalgebras of type ${\rm B}_3$
in $\goth g$. Let $\Upsilon $ be the set of these conjugacy classes.
The group of outer isomorphisms
of $\goth g$ acts on $\Upsilon $ as the group
of all permutations of $\Upsilon $, isomorphic
to the symmetric group $S_3$. Choose ${\cal C }\in \Upsilon $ so that
$\sigma (\cal C) \ne \cal C$ and let $\goth h \in \cal C$.
Applying an outer automorphism of
$\goth g$, we can arrange that
$\goth h$ corresponds to $Spin _7$ and $\sigma (\goth h) = {\goth s \goth o}_7$.
Therefore $\goth g = \goth h + \sigma (\goth h)$ by Theorem 4.1
and $\goth g = \goth g_0 + \goth h$ by Lemma 4.6 and Lemma 4.2.
\hfill $\square $

\medskip
\noindent
{\bf Corollary 4.7.}\ {\it Let $G$ be a simple algebraic group over ${\Bbb C}$,
$G_0$ a real form of $G$, and $H \subset G$ a proper
reductive algebraic subgroup.
Then the following three conditions are equivalent:

{\rm (i)} $G_0$ is locally transitive on $G/H$;

{\rm (ii)} $G_0$ is transitive on $G/H$;

{\rm (iii)} up to a local isomorphism, $G = SO_8$, $H= Spin _7$,
${G_0} = SO_{1,7}$ or $SO_{3,5}$.}

\medskip
\noindent
{\it Proof.} Theorem 4.5 says that (i) and (iii) are equivalent.
Proposition 4.4 shows that (i) implies (ii).
\hfill $\square $

\section{Real forms of inner type}

Let $\goth g_0$ be a real semisimple Lie algebra
of non-compact type. Let
$\goth g_0 = \goth k_0 + \goth m_0$ be a Cartan
decomposition with the
corresponding Cartan involution $\theta $. It is known
that $\theta $ is an inner automorphism of $\goth g_0$ if and
only if $\goth k_0$ contains a Cartan subalgebra of $\goth g_0$.
If this is the case, we will say that the Lie algebra $\goth g_0$
and the corresponding Lie group $G_0$ is of inner type.
Clearly, $\goth g_0  $
is of inner type if and only if all simple
ideals of $\goth g_0$ are of inner type. The Cartan classification
yields the following list of simple
Lie algebras
of inner type:
$${\goth s \goth l}_2(\Bbb R),
{\goth s \goth u}_{p,q},
{\goth s \goth o}_{p,q}\ (p\ {\rm or}\ q \ {\rm even}),
{\goth s \goth o}^*_{\, 2n},
{\goth s \goth p}_{\, 2n}({\Bbb R}),
{\goth s \goth p}_{p,q},$$
$${\rm EII, EIII, EV, EVI, EVII, EVIII, EIX, FI, FII, G}.$$
As we have seen in Sect.1,
a conjugacy class of parabolic subalgebras
has a representative $\goth p$, such that $\goth g = \goth g_0 + \goth p$.
In other words, for any parabolic subgroup
$P \subset G$ the real form $G_0$ has an open orbit on $G/P$.
For real forms of inner type the converse is also true.

\medskip \noindent
{\bf Theorem 5.1.}\
{\it Let $\goth g$ be a complex semisimple Lie algebra, $\goth g_0$
a real form of $\goth g$ of inner type, and
$\goth j$ a compact Cartan subalgebra of $\goth g_0$.
If $\goth h$ is an
algebraic Lie subalgebra of $\goth g$ satisfying
$\goth g = \goth g_0 + \goth h$
then $\goth h$ is parabolic. Moreover,
there exists an inner automorphism ${\rm Ad}(g),\ g\in G_0$,
such that $\goth h = {\rm Ad}(g)\cdot \goth p_{\Phi} $, where
$\Phi $ is a subset of simple roots for
some ordering of
$\Sigma (\goth g, \goth j\, ^{\Bbb C})$. Conversely,
any such $\goth h$ satisfies $\goth g = \goth g_0 + \goth h$.}

\medskip \noindent
{\bf Corollary 5.2.}\ {\it Let $G$ be a complex semisimple
Lie group, $G_0\subset G$ a real form of inner type,
and $H\subset G$ a complex algebraic subgroup.
If $G_0$ has an open orbit on $G/H$ then $H$ is parabolic.}

\medskip
\noindent For an algebraic Lie algebra $\goth h$
we denote by $R_u(\goth h)$ the unipotent radical
and by $L(\goth h)$
a reductive Levi subalgebra.
For the proof of Theorem 5.1 we will need a lemma which rules
out certain factorizations with semisimple factors.

\medskip
\noindent
{\bf Lemma 5.3.}\ {\it Let $\goth g$ be a simple complex Lie algebra
and let
$\goth h_1, \goth h_2 \subset \goth g$ be two semisimple real Lie subalgebras,
such that ${\goth h}_1 \cap {\goth h}_2 = \{0\}$.
Then $\goth g \ne {\goth h}_1 + {\goth h}_2$.}

\medskip
\noindent
{\it Proof.}\ Assume ${\goth h}_1 +{\goth h}_2 = \goth g$.
Let $G$ be a simply connected Lie group with Lie algebra $\goth g$
and let $H_1, H_2$ be connected subgroups of $G$ with Lie algebras $\goth h_1$,
$\goth h_2$.
Then $G = H_1 \cdot H_2$ by Prop. 4.4. Therefore one can write
$G$ as a homogeneous space $G = (H_1\times H_2)/(H_1\cap H_2)$,
where $H_1 \cap H_2$ embeds diagonally into the product.
Because $G$ is simply connected, we see that the
intersection $H_1 \cap H_2$ is in fact trivial.
But $H^3(G, {\Bbb R}) \cong {\Bbb R}$,
whereas  ${\rm dim}\, H^3(H_i, {\Bbb R}) \ge 1$,
see e.g. \cite{On3}, Ch.\,3, \S\,9, and
so the decomposition $G = H_1 \times H_2$
yields a contradiction.
\hfill $\square $

\medskip
\noindent
{\it Proof of Theorem 5.1.}\
Write $\goth g_0$ as the sum
of simple ideals
$\goth g_{k,0},\, k=1,\ldots , m$.
Each of them is stable under the Cartan involution $\theta $,
because $\theta $ is an inner automorphism. Furthermore,
each $\goth g_{k,0}$ is again of inner type. Thus the
complexification $\goth g _k= (\goth g_{k,0})^{\Bbb C}$ is a simple
ideal of $\goth g$,
and $\goth g = \goth g _1 \oplus \ldots
\oplus \goth g _m$. Let $\pi _k  : \goth g \to \goth g _k$
be the projection map.

Assume that $\goth h $ is reductive. We want to show that
then $\goth h = \goth g$.
For each $k$ we have $\goth g _k = \goth g _{k,0} + \pi _k (\goth h)$.
If $\pi _k (\goth h) \ne \goth g _k$, then
$\goth g_{k,0}$ is isomorphic to ${\goth s \goth 0}_{1,7}$ or
${\goth s \goth o}_{3,5}$ by Corollary 4.7. Since $\goth g_{k,0}$ is
of inner type, this can not happen. Hence $\pi _k (\goth h)
= \goth g _k$ for all $k$. In particular, $\goth h$ is semisimple, and
so we write
$\goth h$ as the sum of simple ideals $\goth h = \goth h_1
\oplus \ldots \oplus \goth h_n$. Since $\pi _k({\goth h}_l)$ is
an ideal in $\goth g_k$, there
are only two possibilities: $\pi _k ({\goth h}_l) = \goth g_k$
or $\pi _k ({\goth h}_l) =\{0\}$. If, for $k$ fixed, we have
$\pi _k (\goth h_l) \ne \{0\}$ and $\pi _k(\goth h _s) \ne \{0\}$
then in fact $ s= l$, because otherwise
$$ \goth g_k =[\goth g_k, \goth g_k] = [\pi _k(\goth h_l), \pi _k (\goth h_s)]
= \pi _k ([\goth h_l, \goth h_s]) =\{0\}.$$
We want to make sure that $m = n$. In that case, renumbering the
simple ideals of $\goth g$, we get
$\goth h_l \subset \goth g_l$ for all $l$. This implies $\goth h_l =
\goth g_l$ for all $l$ and $\goth h = \goth g$.
Now, if $n < m $ then one and only one $\goth h_l$
projects isomorphically onto $\goth g_k$ and $\goth g _p$ for
$p \ne k$. Let $\omega _k =(\pi _k \vert \goth h_l)^{-1}$
and $\omega _p =(\pi _p \vert \goth h_l)^{-1}$. Then
$$\goth g_k \oplus \goth g _p = (\pi _k \oplus \pi _p)(\goth h_l) +
(\goth g_{k,0} \oplus \goth g_{p,0}), $$
hence
$$\goth h_l = \omega _k(\goth g_{k,0}) + \omega _p(\goth g_{p,0}),$$
and so a simple complex Lie algebra $\goth h_l$
is written as the sum of two real forms.
This contradicts Lemma 5.3.

Assume from now on that $R_u(\goth h) \ne \{0\}$
and take an embedding of $\goth h$ into a parabolic subalgebra $\goth p$,
such that $R_u(\goth h) \subset R_u(\goth p)$, see Cor. 3.2.
Then $\goth g = {\goth g}_0 + \goth p$, i.e., the $G_0$-orbit
of the base point is open in $G/P$. By Theorem 2.4 $\goth p$
is a standard parabolic subalgebra, $\goth p = \goth p_{\Phi}$, where:

(i) $\goth p \cap {\goth g}_0 $ contains a fundamental Cartan subalgebra
$\goth j \subset {\goth g}_0$, which is now compact (recall that
${\goth g}_0 $ is of inner type);

(ii) $\Phi $ is a subset of simple roots for some ordering of
$\Sigma (\goth g, \goth t),\ \goth t = \goth j ^{\Bbb C}$ (since
$\goth j$ is compact, $\tau (\alpha) =
- \alpha $ for all $\alpha \in \Sigma (\goth g, \goth t)$
and $\tau \Sigma ^+ = \Sigma ^-$ for any choice of $\Sigma ^+$).

\smallskip \noindent By our construction,
$R_u(\goth h) \subset {\goth p}_{\Phi }^u$.
Applying an inner automorphism of $\goth p _{\Phi}$,
assume that
$L(\goth h) \subset \goth p_{\Phi}^r$.
Since $\tau (\goth g_\alpha ) = \goth g _{-\alpha }$
for all root spaces, we have
$$\tau (\goth p_{\Phi}^r) = \goth p_{\Phi} ^r \ \ {\rm and}\ \
\tau (\goth p_{\Phi}^u) = \goth p_{\Phi}^{-u}.$$
Observe that $\tau (\goth h) + \goth h = \goth g$ by Lemma 4.2.
Therefore
$$\tau (R_u(\goth h)) + R_u(\goth h) = \tau (\goth p_{\Phi }^u) +
\goth p_{\Phi }^u = \goth p _{\Phi }^{-u} + \goth p _{\Phi}^u,$$
and so we obtain
$$R_u(\goth h) = \goth p_{\Phi }^u.$$
We also have
$$\tau (L(\goth h)) + L(\goth h) = \goth p_{\Phi}^r,$$
hence, again by Lemma 4.2,
$$\goth p_{\Phi }^r = (\goth p _{\Phi}^r)_0 + L(\goth h),
\ \ {\rm where}
\ \ (\goth p _{\Phi}^r)_0 = \goth p_{\Phi }^r \cap \goth g_0.$$
Write $\goth p_{\Phi }^r = \goth s + \goth z$, where $\goth s$
is the semisimple part and $\goth z$ the center of $\goth p_{\Phi }^r$,
denote by $\pi _s, \pi _z$ the corresponding projections, and put
$\goth s _0 = \goth s \cap \goth g _0,\ \ \goth z_0 =
\goth z \cap \goth g _0$. Then
$\goth s = \goth s _0 + \pi _s(L(\goth h))$.
Since $\goth s _0$ is a semisimple algebra of inner type and
$\pi _s(L(\goth h))$ is reductive, we get $\pi _s(L(\goth h)) = \goth s$
by what we have already proved. Therefore $L(\goth h) = \goth s +\goth z_*$,
where $\goth z_* $ is an algebraic subalgebra in $\goth z$.
On the other hand, $\goth z = \goth z_0 + \pi _z(L(\goth h))
= \goth z_0 + \goth z_*$.
But $\goth z_0$ is compact, so $\goth z_* = \goth z$
and $L(\goth h) = \goth p_{\Phi }^r$. Together with the
equality $R_u(\goth h )=
\goth p _{\Phi }^u$, this gives $\goth h = \goth p_{\Phi}$.
To finish the proof, recall that any two compact
Cartan subalgebras of $\goth g_0$ are conjugate by an inner automorphism.
For the converse statement of the theorem,
note that, $\goth j$ being
compact,
(ii) in Theorem 2.4 is fulfilled for any ordering of $\Sigma (\goth g,
\goth t)$.
\hfill $\square $

\medskip
\noindent
We now recover a theorem of F.M.~Malyshev,
in which $\goth h$ is not necessarily algebraic.
Of course, our Theorem 5.1 is a special case of his result.
We want to show
that
the general case can be obtained from that special one.
We adopt the notation introduced in the above proof.
Namely,
$\goth s = \goth s_{\Phi }$ and $\goth z = \goth z_{\Phi}$ are the semisimple
part and, respectively,
the center of the reductive algebra $\goth p_{\Phi}^r$.

\medskip
\noindent
{\bf Theorem 5.4\ ({\rm F.M.~Malyshev \cite {Ma1}}).}\
{\it Let $\goth g$, $\goth g_0$
and
$\goth j$ be as in Theorem 5.1.
If $\goth h$ is a complex
Lie subalgebra of $\goth g$ satisfying
$\goth g = \goth g_0 + \goth h$
then
there exists an inner automorphism ${\rm Ad}(g),\ g\in G_0$,
such that $\goth h =
{\rm Ad}(g) (\goth a + \goth s _{\Phi}+ \goth p_{\Phi}^u)$, where
$\Phi $ is a subset of simple roots for
some ordering of
$\Sigma (\goth g, \goth j\, ^{\Bbb C})$ and
$\goth a$ is a complex subspace of $\goth z _{\Phi}$
which projects onto the real form $(\goth z_{\Phi})_0$.
Conversely,
any such $\goth h$ satisfies $\goth g = \goth g_0 + \goth h$.}

\medskip \noindent
{\it Proof.}\ Let $\goth h_{\rm alg}$ be the algebraic closure of $\goth h$,
i.e., the smallest algebraic subalgebra of $\goth g$ containing $\goth h$.
According to a theorem of C.~Chevalley, the commutator algebras
of $\goth h$ and $\goth h_{\rm alg}$ are the same, see \cite{C}. Chap.\,II,
Th\'eor\`eme 13. Applying
an inner automorphism ${\rm Ad}(g),\ g\in G_0$,
we get
$$\goth h_{\rm alg} = \goth p_{\Phi}=
\goth z_{\Phi} + \goth s_{\Phi} + \goth p_{\Phi}^u$$
by Theorem 5.1. Since $\goth h$
contains
$[\goth h_{\rm alg}, \goth h_{\rm alg}] = \goth s_{\Phi} + \goth p_{\Phi }^u$,
it follows that
$$\goth h = \goth a + \goth s_{\Phi} + \goth p_{\Phi}^u,$$
where $\goth a \subset \goth z_{\Phi}$ is a complex subspace.
Observe that
$$\tau (\goth h) = \tau (\goth a) + \goth s_{\Phi} + \goth p_{\Phi}^{-u}.$$
By Lemma 4.2 we have
$\goth g = \goth h + \tau (\goth h)$.
Clearly, $\goth z_{\Phi}$ is $\tau $-stable.
The above expression for $\tau (\goth h)$ shows that
$\goth z_{\Phi} = \goth a + \tau (\goth a)$.
Again by Lemma 4.2,
this implies $\goth z_{\Phi} = (\goth z_{\Phi})_0 + \goth a$ or,
equivalently,
$\goth z_{\Phi} = i\cdot (\goth z_{\Phi})_0 + \goth a$.
Thus $\goth a$ projects onto $(\goth z_{\Phi})_0$.
Since $\tau \goth p_{\Phi}^u = \goth p_{\Phi}^{-u}$,
the converse statement is obvious.
\hfill $\square $

\medskip \noindent
If $\goth g_0$ is a real form of outer type (= not
of inner type), then a Lie subalgebra
$\goth h \subset \goth g$, satisfying $\goth g = \goth g_0 + \goth h$,
is in general very far from being parabolic. Some classification of such
$\goth h$ is known for type ${\rm D}_n$, see \cite{Ma2}. Here is
a typical example of what can happen for other Lie algebras.

\medskip \noindent
{\bf Example 5.5.}\ Let $\goth g =\goth s \goth l _{2n}(\Bbb C),\ n> 1,$ and
let $\tau (A) = \bar A$ for $A \in \goth g$, so that $\goth g_0 =
\goth s \goth l _{2n}(\Bbb R)$.
Then there is a fundamental Cartan subalgebra $\goth j \subset \goth g_0$
and an ordering of the root system $\Sigma (\goth g, \goth t),\ \goth t =
\goth j \, ^{\Bbb C},$ such that the set of simple roots
$\Pi$ is of the form $\Pi = \Phi \sqcup \Psi \sqcup \{\gamma \}$,
where $\Phi $ and $\Psi $ are orthogonal,
$\tau(\Phi ) = -\Psi$ and $\tau (\gamma ) = -\gamma $.
The standard Levi subalgebra of $\goth p_{\Phi \sqcup \Psi}$
can be written as
$$\goth p_{\Phi \sqcup \Psi}^r = \goth s_1+ \goth s_2 + \goth z,$$
where $\goth s_1 $ and $\goth s_2$ are isomorphic simple algebras of type
${\rm A}_{n-1}$
interchanged by $\tau $ and $\goth z$ is a $\tau$-stable one-dimensional torus.
Set
$$\goth h = \goth s_1 + \goth z + \goth p_{\Phi \sqcup \Psi}^u,$$
then
$$\tau (\goth h) = \goth s_2 + \goth z + \goth p_{\Phi \sqcup \Psi} ^{-u}.$$
Therefore $\goth h + \tau (\goth h) = \goth g$,
showing that $\goth g = \goth g_0 + \goth h$. Note that $\goth h$ is an
ideal in the parabolic subalgebra $\goth p = {\goth p}_{\Phi \sqcup \Psi}$,
such that $\goth p/\goth h$ is a simple algebra.

The construction of $\goth j$ and the ordering in $\Sigma (\goth g, \goth t)$
goes as follows. Take the Cartan decomposition
$\goth g_0 = \goth k_0 + \goth m_0$, where $\goth k_0 =
\goth s \goth 0_{2n}(\Bbb R)$. Define $\goth j$ as
the space of block matrices

$$        \left(
           \matrix{ a_1  & b_1  & &  & & & \cr
                  -b_1  & a_1  & &  & & & \cr
              & & a_2 & b_2 & & & \cr
              & & -b_2 & a_2 & &  &\cr
              &     & & &  \ddots & &\cr
              &     & & & & a_n & b_n \cr
              &     & & & & -b_n & a_n \cr
              } \right)$$

\noindent with real entries and
$\Sigma \,a_i = 0 $. Then $\goth j$ is a fundamental Cartan subalgebra and
$\goth j = \goth j \cap \goth k_0 + \goth j \cap \goth m_0$. Consider
$a_i$ and $b_i$ as linear functions on $\goth j$ and $\goth t =
\goth j\, ^{\Bbb C}$. Then it is easy
to determine the root system $\Sigma (\goth g, \goth t)$.
We list the roots that we declare positive:
$$i(b_p - b_q) \pm (a_p - a_q),\ i(b_p + b_q) \pm (a_p - a_q)\ \ (p< q),
\ {\rm and } \ 2ib_p\ \ (p,q=1,\ldots,n).$$
Let
$\Phi = \{\alpha_1, \ldots , \alpha_{n-1}\},
\Psi = \{\beta _1, \ldots , \beta _{n-1}\}$, where
$$\alpha _p = i(b_p - b_{p+1}) + a_p - a_{p+1},\
  \beta _p = i(b_p - b_{p+1}) - a_p + a_{p+1}\ (p=1,\ldots, n-1),$$
and let
   $\gamma = 2ib_n$.
Then the set of simple roots $\Pi$ is the union $\Pi = \Phi \sqcup \Psi
\sqcup \{\gamma \}$, $\Phi $ and $\Psi $ are orthogonal,
$\tau (\alpha _p) = - \beta _p$ for all $p$ and $\tau (\gamma) = -\gamma$.

\section{Matsuki correspondence}
Recall that $G_0$ is a real form of
a complex semisimple group $G$ and both $G_0$ and $G$ are connected.
Let $\goth g_0= \goth k_0 + \goth m_0$ be a Cartan decomposition,
$\goth k$ the complexification in $\goth g$ of $\goth k_0$, and $K$
the corresponding connected reductive subgroup of $G$.

\bigskip
\noindent
{\bf Theorem 6.1 {(\rm T.~Matsuki \cite{M2})}.}
{\it
Let $O$ be a $K$-orbit and let $O^\prime $ be a $G_0$-orbit on $G/P$,
where $P \subset G$ is a parabolic subgroup.
The relation
$$O \leftrightarrow O^\prime \ \ \iff \
\ O\cap O^\prime \ne \emptyset {\rm \ \ and } \ O\cap O^\prime
{\rm \ is\ compact} $$
defines a bijection between $K\setminus G\,/P$ and $G_0\setminus G\,/P$.}

\medskip
\noindent
A geometric proof of this result, using the moment map technique,
is found in \cite{MUV}, \cite{BL}. Note that $K$ is a spherical subgroup
of $G$,
i.e., a Borel subgroup of $G$ has an open orbit on $G/K$.
It that case $B$ has finitely many orbits on $G/K$,
see \cite{Br}, \cite{Vin}.
Thus the set $K\setminus G\,/P$ is finite,
and so $G_0\setminus G\,/P$ is also finite
(another proof of Theorem 2.3).

\medskip
\noindent
It can happen that both $K\setminus G\,/P$ and $G_0\setminus G\,/P$
are one-point sets. For $G$ simple, there are only two types
of such actions.

\medskip
\noindent
{\bf Theorem 6.2 {(\rm A.L.~Onishchik \cite{On2}, Thm. 6.1)}.}
{\it If $G$ is simple and $G_0$ or, equivalently,
$K$ is transitive on $X=G/P$ then,
up to a local isomorphism,

{\rm (1)}\ $G= SL_{2n}(\Bbb C),\ K= Sp_{2n}(\Bbb C),\ G_0 = SU^*_{2n},
\ X = {\Bbb P}^{2n-1}(\Bbb C)$, or

{\rm (2)}\ $G = SO_{2n} (\Bbb C),\ K = SO_{2n-1}(\Bbb C),\ G_0 = SO^o_{2n-1,1},
\ X= SO_{2n}(\Bbb R)/U_n.$                            }

\medskip
\noindent
There are two important cases of the correspondence
$O \leftrightarrow O^\prime $, namely, when one of the two orbits is open
or when it is compact. The first of the following two propositions is evident,
and the second one is due to T.~Matsuki \cite{M1}.

\medskip
\noindent
{\bf Proposition 6.3.}\ {\it If $O$ is open then $O^\prime $ is compact
and $O^\prime \subset O$. If $O^\prime $ is open then $O$ is compact
and $O \subset O^\prime $.}
\hfill $\square $

\medskip \noindent
{\bf Proposition 6.4.}\ {\it If $O$ is compact then $O^\prime $ is open
and $ O \subset O^\prime $. If $O^\prime $ is compact then $O$ is open
and $O^\prime \subset O$.}

\medskip
\noindent
{\it Proof.}\ We prove the second statement. The proof of the
first one is similar. Take a base point $x_0 \in O \cap O^\prime $
and let $P$ be the isotropy subgroup of $x_0$.
Note that $G_0\cap P$
has only finitely many connected components,
since it is an open subgroup of a real algebraic group. By a
theorem of D.~Montgomery \cite{Mo}, $K_0$
is transitive on the compact homogeneous space $G_0/(G_0\cap P)$, hence
$$\goth g_0 = \goth k_0 + \goth g_0 \cap \goth p \subset \goth k + \goth p.$$
On the other hand, $\goth k_0 + i\goth m_0$
is the Lie algebra of a maximal compact subgroup of $G$,
which is transitive on $G/P$. Therefore
$$\goth g = \goth k_0+i\goth m_0 + \goth p$$
or, equivalently
$$\goth g = i\goth k_0 + \goth m_0 + \goth p,$$
and it follows that
$\goth g \subset \goth k + \goth g_0 + \goth p \subset \goth k + \goth p$,
i.e., $\goth g = \goth k+ \goth p$.
This means that $O = K(x_0)$ is open in $G/P$,
and the inclusion $O^\prime \subset O$ follows from Proposition 6.3.
\hfill $\square $

\section{Cycle spaces}

First, we recall the definition of the complex crown
of a real symmetric space $G_0/K_0$, see \cite {AG}.
Let $\goth a \subset \goth m_0$ be a maximal abelian subspace and let
$\goth a^+ \subset \goth a$ be the subset given by the inequalities $\vert
\alpha (Y)
\vert < {\pi \over 2}$, where $Y \in \goth a$ and $\alpha $ runs
over all restricted roots,
i. e., the roots of $\goth g_0$ with respect to $\goth a$. Then the
crown is the set
$$\Xi = G_0 ({\rm exp} \, i\goth a^+) \, o \subset G/K,$$
where $o = e\cdot K \in G/K $ is the base point.
The set $\Xi $ is open and the $G_0$-action on $\Xi $
is proper, see \cite {AG}. We discuss some
properties of the complex manifold $\Xi $ in the next section.
Because
all maximal abelian subspaces in $\goth m_0$ are $K_0$-conjugate,
it follows that
$\Xi $ is independent of the choice of $\goth a$ and is therefore
determined by $G_0/K_0$ itself. Some authors call $\Xi$
the universal domain, see \cite{FHW}. We reserve this term
for the lift of $\Xi $ to $G$ and define the universal domain by
$$\Omega = G_0 ({\rm exp} \, i\goth a^+) K \subset G, $$
due to the properties of $\Omega $ which will soon
become clear. Of course, $\Omega $ is invariant under the right $K$-action
and $\Omega /K = \Xi $.

\medskip
\noindent
Next, we define the (linear) cycle space for an open $G_0$-orbit on $X=G/P$,
see \cite{WeW}.
Since full cycle spaces (in the sense of D.Barlet) are not discussed here,
we will omit the adjective ``linear".
Let $D$ be such
an orbit and let $C_0$ be the corresponding
$K$-orbit, so that if $O\leftrightarrow O^\prime $
for $O = C_0$ and $O^\prime = D$. The orbit $C_0$ is a compact complex
manifold contained in $D$.
Consider the open set $$G\{D\} = \{g\in G \ \vert \ gC_0 \subset D\}
\subset G$$
and denote by $G\{D\}^\circ $ its connected component containing $e \in G$.
Observe that $G\{D\}$ is invariant under the right
multiplication by $L = \{ g\in G \ \vert \ gC_0 = C_0\}$
and left multiplication by $G_0$.
Since $L$ is
a closed complex subgroup of $G$,
we have a natural complex structure on $G/L$.
By definition, the cycle space
${\cal M}_D$ of $D$ is the connected
component of $C_0(=e\cdot L)$ in
$G\{D\}/L$
with the inherited $G_0$-invariant complex structure.

\medskip
\noindent
In what follows, we assume $\goth g$ simple.
We will say that $G_0$ is
of Hermitian type if the symmetric space $G_0/K_0$
is Hermitian.
If this is the case then $\goth g$
has three irreducible components as $({\rm ad}\, {\goth k})$-module,
namely,
$\goth g = \goth s^- + \goth k + \goth s^+$,
where $\goth s ^+, \goth s^-$
are abelian subalgebras. The subalgebras $\goth k + \goth s^+$
and $\goth k + \goth s^-$ are
parabolic. The corresponding parabolic subgroups are denoted $P^+$
and $P^-$.
We have two flag manifolds
$X^+ = G/P^+ ,\ X^-=G/P^-$ with base points $x^+ = e\cdot P^+,\  x^-=
e\cdot P^-$
and two $G_0$-invariant complex structures on $G_0/K_0$
defined by the equivariant embeddings
$g\cdot K_0 \mapsto gx^\pm \in X^\pm $.
Each of the two orbits ${\cal B} = G_0(x^+)$ and $\bar {\cal B} = G_0(x^-)$
is biholomorphically isomorphic
to the bounded symmetric domain associated to $G_0$.
The Lie algebra $\goth l$ of $L$ contains $\goth k$.
If $G_0$ is of Hermitian type
and $\goth l$ coincides with $\goth p^+$
or $\goth p^-$ then we say that $D$ and, also,
the corresponding compact $K$-orbit $C_0$ is of (Hermitian)
holomorphic type.
If $G_0$ is of non-Hermitian type
then $\goth k$
is a maximal proper subalgebra of $\goth g$. Thus,
if $\goth l \ne \goth g$ then
$\goth l = \goth k$.
For $G_0$ of Hermitian type, each
flag manifold has exactly two $K$-orbits of holomorphic type.
All other $K$-orbits for $G_0$ of Hermitian type
and all $K$-orbits for $G_0$ of non-Hermitian type
are said to be of non-holomorphic type.
In the following theorem, we exclude the actions
listed in
Theorem 6.2. The symbol $\simeq $
means a $G_0$-equivariant biholomorphic isomorphism.

\medskip
\noindent
{\bf Theorem 7.1 {(\rm G.~Fels, A.~Huckleberry, J.A.~Wolf \cite{FHW},
Thm. 11.3.7)}.}
{\it \ Assume $G$ simple and suppose $G_0$
is not transitive on $X =G/P$.
Let $D$ be an open $G_0$-orbit on $X$.
If $D$ is of holomorphic type
then ${\cal M}_D \simeq {\cal B}$ or ${\cal M}_D \simeq \bar
{\cal B}$.
In all other cases,
$G\{D\}^\circ$ coincides with
the universal domain $\Omega \subset G$. Moreover,
$\pi: G/K \to G/L$ is a finite covering map, which induces
a $G_0$-equivarint biholomorphic map $\pi\vert _{\Xi}: \Xi \to {\cal M}_D$.}

\medskip
\noindent
If $G_0$ is of Hermitian type then $\Xi \simeq {\cal B}\times \bar
{\cal B}$, see \cite{BHH}, Sect.3, \cite{GM},
Prop. 2.2, or \cite{FHW}, Prop. 6.1.9. The cycle space in that case
was first described by J.~Wolf and R.~Zierau \cite{WZ1,WZ2}.
Namely,
in accordance with the above theorem,
${\cal M}_D \simeq {\cal B}\times \bar {\cal B}$
if $D$ is of non-holomorphic type and ${\cal M}_D \simeq {\cal B}$
or ${\cal M}_D \simeq \bar {\cal B}$ if $D$ is of holomorphic type.
For $G_0$ of non-Hermitian type,
the crucial equality $G\{D\}^\circ = \Omega $
is proved by G.~Fels and A.T.~Huckleberry in \cite{FH},
see Thm. 4.2.5,
using Kobayashi hyperbolicity of certain $G_0$-invariant
domains in $G/K$.
In the next section, we consider
some properties of the crown $\Xi$,
which are important for that proof and are interesting in themselves.
After that,
we explain the strategy of their proof, without going into the details.

\medskip
\noindent
Meanwhile,
the notion of the cycle space has been generalized to lower-dimensional
orbits and it turned out that its description in terms of
the universal domain holds in this greater generality. Namely,
given any $K$-orbit $O$ on $X=G/P$,
S.G.~Gindikin and T.~Matsuki \cite{GM}
defined a subset of $G$ by
$$C(O) = \{g \in G \ \vert \ gO \cap O^\prime \ne \emptyset {\rm \ and \ }
gO \cap O^\prime {\rm \ is \ compact}\},$$
where $O^\prime $ is the corresponding $G_0$-orbit, i.e.,
$O \leftrightarrow O^\prime $.
Let $C(O)^\circ $ be the connected component of $C(O)$ containing $e \in G$.
Of course, if $D = O^\prime $
is open, then $C(O) = G\{D\}$ is
the open set considered above.
The following theorem was stated as a conjecture
in \cite{GM}, see Conjecture 1.6.

\medskip
\noindent
{\bf Theorem 7.2 {(\rm T.~Matsuki \cite{M4})}.}\ {\it Let $G$, $G_0$ and $X$
be as above.
Then $C(O)^\circ = \Omega $
for all $K$-orbits on $X$ of non-holomorphic
type.}

\medskip
\noindent
{\it Remark.} The proof in \cite{M4}
uses combinatorial description of the
inclusion relations between the closures of $K$-orbits
on the flag manifolds of $G$. As a corollary, we get that $C(O)^\circ $
is an open set, which is not clear a priori.
If this is known,
then Theorem 7.2 follows from \cite{HN} or
from Theorem 12.1.3 in \cite{FHW}. The latter
asserts that the connected component of
the interior of $C(O)$, containing the neutral element $e\in G$,
coincides with $\Omega $.

\section{Complex geometric properties of the crown }

The following theorem proves the conjecture
stated in \cite{AG}.

\medskip
\noindent
{\bf Theorem 8.1 {(\rm D.~Burns, S.~Halverscheid, R.~Hind \cite{BHH})}.}
\ {\it The crown $\Xi $ is a Stein manifold.
}

\medskip
\noindent
The crucial ingredient of the proof is the
construction of a smooth strictly plurisubharmonic function on $\Xi $
that is $G_0$-invariant and gives an exhaustion of the orbit space
$G_0\setminus \Xi$. We call such a function a BHH-function.
Let $\Gamma \subset
G_0$ be a discrete cocompact subgroup acting freely on $G_0/K_0$.
Then
$\Gamma $ acts properly and freely on $\Xi $ and
any BHH-function induces a plurisubharmonic exhaustion of $\Gamma
\setminus \Xi$. Thus $\Gamma \setminus \Xi$ is a Stein manifold
and its covering $\Xi $
is also Stein.

\medskip
\noindent
We now want to give another application of BHH-functions.
Let $G_0 = K_0A_0N_0$ be an Iwasawa decomposition and
let $B$ be a Borel subgroup of $G$ containing
the solvable subgroup $A_0N_0$.
Then $B$ is called an Iwasawa-Borel
subgroup, the orbit $B(o) \subset G/K$
is Zariski
open and its complement, to be denoted
by ${\cal H}$, is a hypersurface.
The set
$$\Psi = \bigcap _{g\in G_0} \, gB(o) = \bigcap _{k\in K_0}\, kB(o)$$
is open as the intersection of a compact family of open sets.
Let $\Xi_I $
be the connected component of $\Psi $ containing $o$.
L.~Barchini \cite{Ba} showed that
$\Xi_I \subset \Xi$. The reverse inclusion was checked
in many special cases including all classical groups and all real forms
of Hermitian type, see
\cite {GM,KS}. The proof in the general case
is due to A.~Huckleberry, see \cite{Ha, FH} and \cite{FHW},
Remark 7.2.5. His argument is as follows.
It is enough to prove that ${\cal H}\cap \Xi = \emptyset$.
Assuming the contrary,
observe that ${\cal H}\cap \Xi$ is $A_0N_0$-invariant
and so $G_0\cdot ({\cal H}\cap \Xi) $ is closed in $\Xi$.
Pick a BHH-function, restrict it
to ${\cal H }\cap \Xi$ and take a minimum point $x_* \in {\cal H}
\cap \Xi $ of the restriction. Then all points of
the orbit $A_0N_0(x_*)$ are
minimum points. Therefore $A_0N_0(x_*)$ is
a totally real submanifold of dimension equal to
${\rm dim}\, G_0/K_0 = {\rm dim}_{\Bbb C}\, G/K$ that is contained
in $\cal H$, contrary
to the fact that ${\cal H}$ is a proper analytic subset.
From these considerations we get the following
description of $\Xi $, see Theorem 8.2.

\medskip
\noindent
{\it Remark.}
For a proof of the inclusion $\Xi \subset \Xi_I$
in a more general setting see \cite{M3}.
Namely, the result is true for a connected real semisimple
Lie group with two commuting involutions whose product
is a Cartan involution.
The corresponding fixed point subgroups generalize $G_0$ and $K$.
The universal domain is defined similarly.
The proof is based on a detailed study of
double coset decompositions. Complex analytic techniques and,
in particular, the existence of BHH-functions are not used.

\medskip
\noindent
{\bf Theorem 8.2.}\ $\Xi = \Xi_I$.

\medskip
\noindent
Since $\Xi_I$ is a connected component of the open
set $\Psi $, which is obtained
by removing a family of
hypersurfaces from the affine variety $G/K$, we see again that $\Xi $ is Stein.
Since $\Psi $ is the set of all points
for which the $kBk^{-1}$-orbit is open for every $k \in K_0$,
we have
$$\Psi = \{x\in G/K\ \vert \ {\goth g}_x + ({\rm Ad}k)\cdot {\goth b} = \goth g
\ \ {\rm for\ all}\ \ k\in K_0\}.$$
Let $N$ be the normalizer of $K$. Then $\Gamma = N/K$
is a finite group with a free action $x\to x^\gamma $ on $G/K$.
From the last description of $\Psi $, it follows
that $\Psi ^{\gamma }=
\Psi $ for all $\gamma \in \Gamma $.
Thus $\Gamma $ interchanges the connected components of $\Psi $.
It follows from the definition
that $\Xi $
is contractible, so a non-trivial
finite group cannot act freely on $\Xi$. Hence
$\Gamma $ interchanges simply transitively the open sets
$ \Xi ^{\gamma}$.
Moreover,
for any group $\tilde K \subset G$ with connected component $K$
the covering map $G/K \to G/\tilde K$ induces
a biholomorphic map of $\Xi $ onto its image, cf. \cite{FHW}, Cor. 11.3.6.

\medskip
\noindent
{\bf Theorem 8.3\ {\rm (A.~Huckleberry \cite{Ha}}).}\
{\it $\Xi $ is Kobayashi hyperbolic.}

\medskip
\noindent
{\it Proof.}\ By Frobenius reciprocity,
there
exist a $G$-module $V$ and a vector $v_0 \in V$ such that
$K \subset G_{v_0} \ne G$.
If $G_0$ is of non-Hermitian type then
$K$ is a maximal connected subgroup of $G$.
If $G_0$ is of Hermitian type then there are precisely two intermediate
subgroups between $K$ and $G$, both of them being parabolic.
In any case,
the connected component of
the stabilizer of the line $[v_0]$ equals $K$
and the natural maps $G/K \to Gv_0 \to G[v_0]$ are finite coverings.
Let ${\Bbb C}[V]_d \subset {\Bbb C}[V]$
be the subspace of homogeneous polynomials of degree $d$,
let $I_d$ be the intersection of ${\Bbb C}[V]_d$
with the ideal of (the closure of) $G[v_0]$
and let $M_d$ be a $G$-stable complement to $I_d$ in ${\Bbb C}[V]_d$.
The space of all polynomials in $M_d$ vanishing on
$G[v_0]\setminus B[v_0]$
is $B$-stable and non-trivial for some $d$, so $B$ has
an eigenvector $\varphi $ in that space.
The zero set of $\varphi $ on the orbit $G[v_0]$
is exactly
the complement to the open $B$-orbit $B[v_0]$.
Replacing $V$ by its symmetric power $S^kV$ and $v_0$ by $v_0^k \in S^kV$,
we obtain a linear form $\varphi $ with the same property.
Now, let $V_0$ be the intersection
of all hypersurfaces $g^* \varphi = 0,\ g\in G$.
Then $V_0$ is a $G$-stable linear subspace of $V$
and we have the $G$-equivariant linear projection map $\pi : V \to W= V/V_0$.
Let $w_0 = \pi (v_0)$ and let $\psi \in W^*$
be the linear form defined by $\pi ^*\psi = \varphi$. Then $K \subset G_{w_0}$
and $G_{w_0} \ne G$,
because $\varphi $ is non-constant on the orbit $Gv_0$. Therefore
$\pi $ gives rise to the finite coverings
$Gv_0 \to Gw_0$ and $G[v_0] \to G[w_0]$. By construction, the orbit
$G\psi =\{g^*\psi \ \vert \ g\in G\}$
generates
$W^*$ and the same is true for $G_0\psi$.
By \cite{Ha}, Cor. 2.13, there exist
hyperplanes ${\cal H}_i = \{g_i^*\psi = 0\} \subset {\Bbb P}(W),\
g_i \in G_0,\  i=1, \ldots, 2m + 1,m ={\rm dim}\
{\Bbb P}(W)$, satisfying the normal crossing
conditions. It is then known that
${\Bbb P}(W) \setminus \bigcup_{_i}\, {\cal H}_i$
is Kobayashi hyperbolic, see \cite{K}, Cor. 3.10.9.
The intersection of this set with the orbit $G[w_0]$ equals
$\bigcap _i \, g_iB[w_0]$ and is likewise hyperbolic.
Recall that we have an equivariant fibering $G/K \to G/\tilde K = G[w_0]$.
As we have seen before stating the theorem,
$\Xi $ is mapped biholomorpically onto its image.
The latter is contained in
the connected component of
$\bigcap _i \, g_iB[w_0] $
at $[w_0]$ and is therefore hyperbolic.
\hfill $\square $

\medskip
\noindent
{\bf Theorem 8.4\ {\rm (G.~Fels, A.~Huckleberry \cite{FH}}).}
{\it If $\Xi ^\prime $ is
a $G_0$-invariant, Stein,
and Kobayashi hyperbolic domain
in $G/K$ that contains $\Xi $ then
$\Xi ^\prime = \Xi$.}

\medskip
\noindent
The proof requires analysis of the boundary ${\rm bd}(\Xi )$.
First,
one considers
the special case of $G_0 = SL_2({\Bbb R})$
and proves Theorem 8.4 for the
crown $\Xi _{{\goth sl}_2}$
of $SL_2({\Bbb R})/SO_2({\Bbb R})$.
Note that $G = SL_2({\Bbb C})$
has precisely two
non-isomorphic affine homogeneous surfaces.
Namely, if $T \simeq {\Bbb C}^* $
is a maximal torus in $SL_2({\Bbb C})$
and $N \subset SL_2({\Bbb C})$
is the normalizer of $T$, then these surfaces are
of the form
$Q_1 = {SL_2}({\Bbb C})/T
\simeq ({\Bbb P}^1({\Bbb C})\times
{\Bbb P}^1({\Bbb C}) ) \setminus \Delta $ and
$Q_2 = {SL_2}({\Bbb C})/N
 \simeq {\Bbb P}^2({\Bbb C}) \setminus C $,
where $\Delta $ is the diagonal and $C$ is a non-degenerate
curve of degree 2.
The crown $\Xi _{\goth sl_2}$
can be viewed as a domain in $Q_1$ or in $Q_2$.
In the general case one constructs a $G_0$-stable
open
dense subset ${\rm bd}^{\rm gen}(\Xi ) \subset
{\rm bd}(\Xi)$, such that
for $z\in {\rm bd}^{\rm gen}(\Xi )$
there exists a simple 3-dimensional subalgebra
$\goth s _0 \subset \goth g_0$
with the following properties:

\smallskip
(i) the orbit of the corresponding complex group
$S = {\rm exp}(\goth s _0 ^{\Bbb C}) \subset G$ through $z$
is an affine surface, i.e., $Sz \simeq Q_1$ or $Sz \simeq Q_2$;

(ii) under this isomorphism
$Sz \cap \Xi $ is mapped biholomorphically onto $\Xi _{{\goth sl}_2}$.

\smallskip
\noindent
Now, if $ \Xi^\prime \setminus
\Xi \ne \emptyset$ then one can find a point $z$
as above in $\Xi^\prime \cap {\rm bd}
(\Xi )$. Then $Sz \cap \Xi ^\prime $
properly contains $Sz \cap \Xi$,
contrary to the fact that $\Xi _{\goth sl_2}$ is a maximal
$SL_2({\Bbb R})$-invariant, Stein
and Kobayashi hyperbolic domain in $Q_1$ or in $Q_2$.
The details are found in \cite{FHW}, see Thm. 10.6.9.

\medskip
\noindent
{\it Remark.}\ In fact, $\Xi $ is the unique maximal
$G_0$-invariant, Stein, and Kobayashi hyperbolic domain in $G/K$
that contains the base point $o$, see \cite{FHW}, Thm. 11.3.1.

\medskip
\noindent
{\it Remark.}\ We refer the reader to \cite{GK} for the
definition of the Shylov-type boundary of the crown
and to \cite{KO} for its simple description and
applications to
the estimates of automorphic forms.

\section{The Schubert domain}

We assume here that $G_0$ is of non-Hermitian type. Then the map
$G/K \to G/L$
is a finite covering.
We have an open $G_0$-orbit $D \subset X = G/P$
and the corresponding compact $K$-orbit $C_0 \subset D$.
Let $q$ denote the complex dimension of $C_0$.
Translations $gC_0, \ g\in G$, are called cycles and
are regarded as points of ${\cal M}_X := G/L$.
The cycle space $\cal{M}_D$
is a domain in ${\cal M}_X$ and the crown $\Xi $
is mapped biholomorphically
onto a domain $\tilde \Xi \subset {\cal M}_X$. We want to prove
the statement of Theorem 7.1, namely, that
$G\{D\}^\circ $ agrees with $\Omega $. Equivalently,
we will prove that $\cal{M}_D $ agrees with $\tilde  \Xi  $.
A.~Huckleberry and J.~Wolf \cite{HW} defined the Schubert domain ${\cal  S}_D$
in ${\cal M}_X$ as follows. Let $B$ be an Iwasawa-Borel subgroup of $G$.
The closures of $B$-orbits on $X$ are called Schubert varieties
(with respect to $B$).
The group $B$ has an open orbit
on any such variety $S$.
Since the open orbit is affine,
its complement $S^\prime $ is a hypersurface in $S$.
For topological reasons,
the (finite) set ${\cal S}_{C_0}$ of Schubert varieties
of codimension $q$ intersecting
$C_0$ is non-empty.
One shows that $S^\prime \cap D = \emptyset$
for any $S \in {\cal S}_{C_0}$. Thus the
incidence variety
$${\cal H}(S):=I(S^\prime )= \{
gC_0 \in {\cal M}_X\ \ \vert \ \
gC_0 \cap S^\prime \ne \emptyset \}$$
is contained in ${\cal M}_X \setminus {\cal M}_D$.
Clearly,
${\cal H}(S)$ is $B$-invariant.
Furthermore, one can show
that ${\cal H}(S)$ is an analytic hypersurface in ${\cal M}_X$, see
\cite{FHW}, Prop.7.4.11.
For any $k \in K_0$ we have
${\cal M}_D \subset {\cal M}_X \setminus k{\cal H}(S)$.
The set
$$\bigcup _{S \in {\cal S}_{C_0}}\bigl \{
\bigcup _{k\in K_0}\, k{\cal H}(S) \bigr \}$$
is closed in ${\cal M}_X$. Its complement is denoted by ${\cal S}_D$
and is called the Schubert domain. By construction,
${\cal S}_D$ is a $G_0$-invariant Stein domain and
$${\cal M}_D \subset {\cal S}_D. \eqno{(*)}$$
On the other hand,
for any boundary point $z \in {\rm bd}(D)$
there exist
an Iwasawa decomposition $G_0 = K_0A_0N_0$,
an Iwasawa-Borel subgroup
$B$ containing $A_0N_0$ and a $B$-invariant variety $S_z$ of
codimension $q + 1 $, such that
$z \in S_z $ and $D \cap S_z = \emptyset$
(a supporting Schubert variety at $z$), see \cite {FHW}, Prop. 9.1.2.
Take a boundary point of
${\cal M}_D$ and consider the
corresponding cycle. It has a point $z \in {\rm bd}(D)$,
hence is contained in the incidence variety
$$I(S_z) := \{gC_0\ \vert \  gC_0 \cap S_z \ne \emptyset\}.$$
Obviously, $I(S_z)$ is $B$-invariant and $I(S_z)
\subset {\cal M}_X \setminus {\cal M}_D$,
in particular, $I(S_z) \ne {\cal M}_X$.
But $\tilde \Xi$ is contained
in the open $B$-orbit by Theorem 8.2. Thus
a point of ${\tilde \Xi}$
cannot be a boundary point of ${\cal M}_D$, and it follows that
$$\tilde \Xi \subset {\cal M}_D. \eqno{(**)}$$
Finally, one can modify the proof of Theorem 8.3 to show that
${\cal S}_D$ is hyperbolic. Namely, take the linear
bundle ${\Bbb L}$ over $G/L$ defined by the hypersurface ${\cal H}(S)$,
which appears in the definition of ${\cal S}_D$.
Then some power ${\Bbb L}^k$ admits a $G$-linearization. Thus
we obtain a non-degenerate equivariant map $G/L
\to
{\Bbb P}(W)$, where a $G$-module $W$ is generated
by a weight vector of $B$. The map is in fact a finite
covering over the image, which is a $G$-orbit in ${\Bbb P}(W)$
containing the image of ${\cal H}(S)$ as a hyperplane section.
Since $W$ is irreducible,
the same argument as in the proof of Theorem 8.3
shows that ${\cal  S}_D$ is hyperbolic.
The inclusions ${(*)}$ and ${(**)}$, together with
Theorem 8.4, imply
$$\tilde {\Xi} = {\cal M}_D = {\cal S}_D.$$

\section{Complex geometric properties of flag domains}

An open $G_0$-orbit in a complex flag manifold $X = G/P$
is called a flag domain.
One classical example of a flag domain
is a bounded symmetric domain in the dual compact
Hermitian symmetric space.
In this example, a flag domain is a Stein manifold.
However,
this is not the case for an arbitrary flag domain $D$, because
$D$ may contain compact complex submanifolds of
positive dimension.
As we have seen,
the cycle space of $D$ is always Stein.
Here, we consider the properties of $D$ itself.

\medskip
\noindent
An open orbit $D =G_0(x_0) \subset X$ is said to be measurable if $D$ carries
a $G_0$-invariant volume element.
We retain the notation of Section 2.
In particular, $x_0 = e\cdot P,\,\goth p = \goth p_{\Phi }$,
where $\goth p$ and $\Phi $ satisfy (i), (ii) of Theorem 2.4.

\medskip
\noindent
{\bf Theorem 10.1\ {(\rm J.~Wolf \cite{W}, Thm. 6.3)}.}
{\it The open orbit
$G_0(x_0)$ is measurable if and only if
$\tau \Phi ^r = \Phi ^r $ and $\tau \Phi ^u = - \Phi ^u$.
Equivalently, $G_0(x_0)$
is measurable if and only if $\goth p \cap \tau\goth p $
is reductive.}

\medskip
\noindent
Since two fundamental Cartan subalgebras in $\goth g_0$
are conjugate by an inner automorphism of $G_0$,
it follows from the above condition and from
Theorem 2.4 that all open $G_0$-orbits on $X$
are measurable or non-measurable simultaneously.
The proof of Theorem 10.1 can be also found in \cite{FHW}, Sect. 4.5.

\medskip
\noindent
{\bf Example 10.2.}\ Let $\goth g_0$ be a real form of inner type.
Since the Cartan subalgebra $\goth t \subset \goth g$ contains
a compact Cartan subalgebra $\goth j \subset \goth g_0$, it follows that
$\tau (\alpha ) = -\alpha $ for any root $\alpha $. Thus
the open orbit $G_0(x_0)$
is measurable.

\medskip
\noindent
{\bf Example 10.3.}\ If $P = B$ is a Borel subgroup of $G$,
then $\Phi = \emptyset $,
$\Phi ^u = \Sigma ^+$ and $\tau \Phi ^u = - \Phi ^u$. Therefore an open
$G_0$-orbit in $G/B$ is measurable.

\medskip
\noindent
A complex manifold $M$ is said to be $q$-complete
if there is a smooth non-negative
exhaustion function $\varrho : M \to {\Bbb R}$,
whose Levi form has at least $n - q$ positive eigenvalues
at every point of $M$. A fundamental theorem
of A.~Andreotti and H.~Grauert says
that for any coherent sheaf ${\cal F}$ on a $q$-complete
manifold and for all $k > q$
one has
$H^k(M,{\cal F})= 0$, see \cite{AnG}.
Note that in
the older literature including \cite{AnG}
the manifolds that we call $q$-complete
were called
$(q+1)$-complete.

\medskip
\noindent
{\bf Theorem 10.4 \ {(\rm W.~Schmid, J.~Wolf \cite{SW})}.}
{\it If $D$ is a measurable open $G_0$-orbit in a flag manifold of $G$ and
$q$ is the dimension of the compact $K$-orbit in $D$ then
$D$ is $q$-complete. In particular, $H^k(D, {\cal F}) = 0$
for all coherent sheaves on $D$ and for all $k > q$.}

\medskip
\noindent
The authors of \cite{SW} do not say that $D$ is measurable, but they
use the equivalent condition that the isotropy group of $D$
is the centralizer of a torus.
The proof of Theorem 10.4 can be also found in \cite{FHW},
see Thm. 4.7.8.

\medskip
\noindent
{\bf Example 10.5.}\ Let $X = {\Bbb P}^n({\Bbb C}),\
G =  SL_{n+1}({\Bbb C})$, and $G_0 =  SL_{n+1}({\Bbb R})$.
Let $\{e_1, e_2, \ldots, e_{n+1}\}$ be a basis of ${\Bbb R}^{n+1}$.
If $n > 1$ then $G_0$ has two orbits on $X$, the open one
and the closed one, with representatives
$x_0 = [e_1+ie_2]$ and $[e_1]$, respectively.
The isotropy subgroup $(G_0)_{x_0}$ is not reductive. Its
unipotent radical consists of all $g\in G_0$, such that
$$g(e_i) = e_i\ (i=1,2),\  g(e_j) \equiv  e_j\ {\rm mod }\,({\Bbb R}e_1 +
{\Bbb R}e_2)\ (j\ge 3).$$
Hence the open orbit $D=G_0(x_0)={\Bbb P}^n({\Bbb C}) \setminus
{\Bbb P}^n({\Bbb R})$
is not measurable. Note that $K =  SO_{n+1}(\Bbb C)$.
Thus the
compact $K$-orbit $C_0 \subset D$ is the projective
quadric $z_1^2 + z_2^2 + \ldots +
z_{n+1}^2 = 0$ and its dimension equals $n-1$.
In this case, we have $n - q = 1$ and we show how to construct
a smooth non-negative
exhaustion function $\varrho: {\Bbb P}^n({\Bbb C})\setminus
{\Bbb P}^n({\Bbb R})
\to {\Bbb R}$, whose Levi form has at least one positive eigenvalue
at every point.
For $z=x+iy \in {\Bbb C}^{n+1}$ put
$$\varrho _1(z) = \sum\, x_k^2 + \sum\, y_k^2,\ \ \varrho_2(z) =
\sqrt {\sum\, (x_ky_l -x_ly_k)^2} $$
and
notice that
$$\varrho _1(\zeta z) = \vert \zeta \vert^2 \varrho _1(z),\ \
\varrho _2(\zeta z) = \vert \zeta \vert ^2 \varrho _2 (z) \ \ {\rm for\ any}
\ \zeta \in {\Bbb C}^*.$$
Thus
$$\varrho ([z]) = {\varrho _1(z) \over \varrho _2(z)}$$
is well-defined for all $[z] \in {\Bbb P}^n({\Bbb C})
\setminus {\Bbb P}^n({\Bbb R})$.
Obviously, $\varrho $ is a smooth exhaustion function for ${\Bbb P}^n({\Bbb C})
\setminus {\Bbb P}^n{(\Bbb R})$.
Given a point $[z] = [x+iy] \in {\Bbb P}^n({\Bbb C})\setminus
{\Bbb P}^n({\Bbb R})$,
take the line $L$ in ${\Bbb P}^n({\Bbb C})$,
connecting $[z]$ with $[x] \in {\Bbb P}^n({\Bbb R})$,
and restrict $\varrho $ to that line.
Clearly, $L$ is the projective image of the affine line
$$\lambda = \alpha + i\beta \mapsto w = x+i\lambda y = x - \beta y
+ i\alpha y$$
and the restriction $\varrho \vert _L$ equals
$$\varphi (\lambda) := \varrho ([w]) = {\vert \alpha \vert \over D}\sum y_k^2
\ + {1\over \vert \alpha \vert D}\sum (x_k-\beta y_k)^2,$$
where $D = \varrho _2(x+iy)$.
Computing the Laplacian
$$\Delta \varphi = {\partial ^2 \varphi \over \partial \alpha ^2} +
{\partial ^2 \varphi \over \partial \beta ^2}$$
for $\alpha \ne 0 $, we get
$$\Delta \varphi = {2\over D\vert \alpha \vert ^3}
\sum \, (x_k-\beta y_k)^2
+ {2\over D\vert\alpha \vert} \sum \, y_k^2 \, > 0,$$
showing that $\varphi $ is strictly subharmonic.
Hence the Levi form of $\varrho $
has at least one positive eigenvalue at $[z] = [w]\vert_{\alpha =1, \beta =0}$.

\medskip
\noindent
{\it Concluding remark.} The open orbit in the last
example is not measurable. As a matter of fact, the conclusion
of Theorem 10.4 holds true in this
case. In general, the author does not know whether one can drop the
measurability assumption in Theorem 10.4.

redpajama_set_name: RedPajamaArXiv


text:
Re-Stocking the roster: Linebackers
By Keith ArnoldFeb 22, 2016, 11:33 AM EST
Notre Dame's spring roster at linebacker is one of the most interesting position groups on the roster. Jaylon Smith is gone, the junior All-American taking his talents—and healing knee—to the NFL. Joe Schmidt is no longer in the middle of the defense, the two-year starter and team captain no longer a coach-on-the-field. Jarrett Grace is gone as well, a player who'll be missed by more than the 115 snaps he played in 2015.
A new generation awaits, nearly all of them recruited under Brian VanGorder. James Onwualu remains at Sam linebacker, a potential three-year starter who has never been a truly full-time player. Nyles Morgan's wait is over, the starting middle linebacker job is his to lose. While injuries and youth will impact how the Irish decide to fill Smith's shoes, there are some intriguing young athletes ready to see if they're capable of stepping forward.
No group has more to do this spring than Mike Elston's crew. So before spring practice begins, let's take a look at the state of the linebacking corps.
Jaylon Smith, Jr. (114 tackles, 9 TFLs)
Joe Schmidt, Grad Student (78 tackles, 4 TFLs)
Jarrett Grace, Grad Student (26 tackles, 2.5 TFLs)
2015-16 ADDITIONS
Josh Barajas*
Asmar Bilal*
Te'von Coney
Daelin Hayes
Jamir Jones
*Fifth year of eligibility available
PRE-SPRING DEPTH CHART
James Onwualu, OLB
Nyles Morgan, MLB
Te'von Coney, OLB
Greer Martini
Josh Barajas
Asmar Bilal
Where'd all the linebackers go? That's the first thing that jumps out, just how thins the numbers seem to be. It'll be very interesting to see how spring practice goes, especially considering the injuries that have wreaked havoc on this group. Coney is expected to be out for spring, healing from a shoulder injury that happened just plays after Jaylon Smith went down. Greer Martini also needed work done to fix an injury that all but kept him out against Ohio State, how that impacts his spring remains to be seen as well. Daelin Hayes has everybody excited, but he's coming off a late-November shoulder surgery, so spring practice isn't necessarily the best bet for him to be unleashed.
It's a very big spring for two young redshirts, with Asmar Bilal and Josh Barajas looking at nothing but opportunity in front of them. This defense badly needs playmakers and both guys were recruited because of their ability to make an impact. But Barajas was never healthy last season after getting hurt in fall camp, and he also added heft to his frame that the staff didn't necessarily think he needed. Bilal is a great-looking athlete, though probably could use some of the extra weight Barajas was lugging around to protect him in the trenches.
On paper, it's easy to see some weakness at the position, especially after attrition took guys like Michael Deeb, Kolin Hill and Bo Wallace out of the program. And while some of that will be shored up come summer when Jonathan Jones and Jamir Jones hit campus, this position may also be impacted by how well the secondary's rebuild goes. A season after not being able to play a nickel or dime package, those may be preferred looks in 2016.
The biggest question that faces this group is knowledge base. Replacing two multiyear starters is difficult. Now add in the challenges of learning position fits and scheme under Brian VanGorder and it's no wonder some Irish fans are calling for a dumbing down of the playbook.
But before things get too remedial, it's worth pointing out that this is Morgan's third year learning under VanGorder and the only defense he's known at the college level. He should be ready. And whoever slides into Smith's shoes, they've been in the program for at least a full season. The key to all of this is Morgan. If he's able to take his instincts and athleticism and pair that with a solid grasp of the system, there's a big year in store. Throw in Onwualu, some intriguing athletes and ascending talent and while it might take some time to learn new jersey numbers, there's plenty of promise on the horizon as the next wave of linebackers step into battle.