text:
If you saw, heard, or know something
HELP US SOLVE THIS CASE
Friends of David Buller
David Buller Memorial Scholarship
Letter from Arezu Salamzadeh 2022
History Counsels Patience
Tribute to a "Most Impactful" Painting Instructor
by Karyn Sandlos
Christopher Eamon, 2020
"My most impactful painting instructor was murdered in his office at U of T. Stabbed to death in his sixth floor office. Nobody was ever arrested for that back in 2001.
I share this painting I made on a train in his memory, David Buller."
Christopher Eamon
In early July 2020, my friend Sally Lee shared with me a Facebook post by her friend, Christopher Eamon. Chris had studied painting with my uncle, David Buller, in the Visual Studies department at the University of Toronto.
Almost 20-years after his death, this beautiful landscape painting, and Chris's words, are a welcome reminder of David's ongoing influence on the people who were lucky enough to know him.
David was a serious artist and painter, a gay man, a loyal friend, and a beloved member of our family. He was also a dedicated and passionate art professor. David was murdered in his office in the afternoon, a few hours prior to an evening class he was scheduled to teach. It was unheard of for David to miss a class, and the students waited for him. A few students went looking for David, knocking on the door of his office to see if everything was ok. The students' memories of what it felt like to wait for their professor, the lingering feeling of unease they describe, is a residue that haunts me to this day.
When David's students talk about his teaching, they inevitably mention something he would say when they were struggling with an artwork: "More paint, just use more paint on the canvas." This was David's way of encouraging students to risk more, get messier, lose themselves in the work, persist.
Today, Jan 18th, 2021, is the 20th anniversary of David's murder.
We miss David very much, and we still haven't given up trying to solve the terrible mystery of what happened to him.
Seventeen Years Later
I often take pleasure in thinking about my late uncle, David Buller. If you knew him, or if you've read about him on this website, then you know that he was a funny, irreverent, and irreplaceable person. Today is the 17th anniversary of the day David was yanked from our lives, and I'm thinking about what that means this many years later. On Jan 18th, 2001, David was the victim of a terrible crime. A brazen crime. He was murdered in his office at 1 Spadina Cr. at the University of Toronto, and the person who took David's life has still not been found.
As an artist, a teacher, a gay man, and all-around decent human being, David had a profound influence on me. He rarely gave advice, but he had the guts to follow his passions in life, and that example was plenty to work with.
The loss of David left a hole in our hearts, and in our family. This week my mom said to me, "It still feels unreal, like a dream. I can't believe this is happening to us." When you lose someone, suddenly and traumatically, the experience never moves into the past tense.
Inspired by this poster from the art collective Fierce Pussy, For the Record - Visual AIDS (2014), I could write a hundred "If you were alive today's…" for David, maybe more.
Fierce Pussy, For the Record - Visual AIDS (2014)
If You Were Alive Today (unfinished)
If you were alive today we would not have to live with the unanswered question. If you were alive today you would know that I live in Chicago now. If you were alive today you'd be cracking your great niece and nephews up with your sarcasm at family get togethers. If you were alive today we would have known each other as adults (I was 32 when you were murdered, with so much growing up to do, as it turns out). If you were alive today I would have so many things to ask you about art and teaching. If you were alive today you would hate being 67 (but of course you would still be handsome and fit). If you were alive today maybe I would tell you all the ways in which you influenced me (I hope you could see it). If you were alive today I would not be fighting tears on the El train while I write these words into my phone. If you were alive today I would not be getting emails from psychics and amateur investigators who think they can solve the case. If you were alive today we would be planning your retirement party from the Visual Art Department at the University of Toronto. If you were alive today I probably wouldn't appreciate how lucky I was to know you. If you were alive today we would not be finding ways, because you were you, to keep a little hope.
Notes on the Fifteenth Anniversary
David Bowie died last week. Of course my uncle, David Buller (1950-2001), was a Bowie fan. For an artistic, gay kid growing up in the suburbs of Toronto in the 1950s and 60s, David Bowie must have been a beacon of the possibilities for a life that lay just ahead: art school at (what was then) the Ontario College of Art (OCA), coming out, traveling across Canada on an Art in the Parks grant, and living for a short but formative time in Paris. Later still, there would be an apartment in Montreal, an MFA in the painting program at Concordia, and teaching in the Fine Arts Department at the University of Toronto.
January 18th, 2016 marks the 15th anniversary of the still unsolved murder of David Buller in his office at 1 Spadina Cr., University of Toronto. The last few years have been busy ones for David's family and friends. In 2014, there was renewed interest in this story from a TV program, "To Catch a Killer," where a team of civilians—specialists in different fields—takes a fresh look at unsolved murders.
While the episode didn't give us any magical answers, it did help us to remain in active conversation with the Toronto Homicide Cold Case Unit, a relatively new department under the leadership of Detective Sergeant Brian Borg. And the episode has also helped to keep an important, albeit difficult, conversation about what may have happened to David alive in Toronto's queer, arts and academic circles.
A few weeks ago, I found these undated photographs of David tucked inside an envelope in the back of an old family photo album. My mom (David's sister, Betty-Lou) thinks they were probably taken when David was 15 or 16 years old, around 1965-66. David didn't actually play the guitar, and as a teenager he probably wouldn't have been permitted to smoke in my grandparents' house. But he knew how to strike a pose.
Thank you, Uncle David, for showing us what it could look like.
As our uncle, David rarely handed out advice to my sister Lisa, my brother John, and me. But once in awhile he would say something that stuck in our hearts and minds. Like the time he told us, "Just do something interesting with your time on earth." This was less a piece of advice and more a provocation.
David Buller: 50 + 15
by John Sandlos
Among the many thoughts that flood inward on the fifteenth anniversary of my uncle's murder, there is this: he would have turned 65 this year. David Buller the senior citizen, distinguished grey, and elder statesman. I wonder how would he have felt about that?
Of course he would not have liked the first hints of muscle sag, the acceleration of the greying process, the extra aches, the wrinkles, the doctor telling him for the umpteenth time to give up alcohol (or become a vegetarian, according to one story he told). I think he would have studiously avoided any attraction or institution (even major art galleries) that offered a senior's discount. He would have rather moved to Kandahar than settle any where designated as a "retirement community." I don't really see him lawn bowing, golfing, or playing bridge, either. Suduko or crosswords were never really his thing.
So what would Dave be doing as he embarked on this new life stage? Certainly, I think, he would still be working. His passion for teaching undiminished, he would have found it difficult to walk away from the classroom at the standard retirement age. I have no doubt he would still be making art, fifteen years further along the evolution that had driven him to include multimedia content in his work.
I have to think that he would be content, no hint of old fartism anywhere despite the occasional gripe about aging. A sense of humour intact, a few more debts paid, excellent health and some opportunities for travel under his belt -- things would be looking up for David as the calendar rolled around to 2016. Most importantly, he could look back on a rich career as an academic and artist with quiet satisfaction.
So often I place David in the present moment, imagining what he might say or do in a particular situation. During the Toronto Blue Jays playoff run this past fall, I could practically imagine him pacing in my living room, then walking out of the room (as he did in 1992) to pour another drink because he couldn't take the suspense any more ("it's just too much," he said). Maybe he would take my own kids to a baseball game as he did with me decades ago now.
Of course, the problem with imagining those we have lost into our present is that they are never really there.
Like Orpheus we try to lead them back to our world, only to find they have slipped away. Which I suppose is a really complicated way of saying that I miss Uncle Dave.
Wedding gift to John and Yolanda from David Buller
Notes on the Thirteenth Anniversary
by Lisa Sandlos
Days away from the thirteenth anniversary of the murder of my uncle, David Buller, I have been sifting through my memories and my notes. I came across the script for a short speech I gave at the internment of Uncle Dave's ashes in 2001 in which I spoke of his presence in and influences on my life. Reflecting on this speech now makes me imagine what I might say to my uncle if I could talk to him now:
I still feel your presence deeply in my life after thirteen years. I know I am not alone in this feeling. More and more over time, I appreciate the positive and lasting imprints you have made on so many people in your roles as teacher, artist, friend, and family member. I, along with numerous others who care and remember your legacies, are doing what we have to do in our continuing process of grieving, memorializing the experiences we shared with you, and hoping for answers. Meanwhile, we miss you profoundly.
Much love, Lisa
Lives Lived column, The Globe and Mail
Wednesday, Oct 10, 2001
Visual artist, professor, much loved son, brother, uncle, and friend.
Born August 10, 1950 in Willowdale, Ontario. Murdered in Toronto, January 18, 2001, aged 50.
There was little in my uncle David's childhood to suggest the innovative painter and charismatic professor that he became in later years. He wasn't interested in school; summer camp was torture. He used to rub snow in his face when sent outside to play so he would have the requisite rosy cheeks his mother insisted upon before allowing him back inside, where he could retreat into the world of I Love Lucy.
He was captivated by drawing and painting from his early teens. A pencil drawing that now hangs on my wall, completed when Dave was only seventeen, shows a young Bob Dylan slyly winking at the viewer, the singer's last name spilling across the page in bright psychedelic colours. Uncle Dave was a 60's hipster; I remember the long wavy hair and bell bottoms he wore when I was very young, and the vastness of his canvasses from my perspective much closer to the ground. I'm taller now, but the paintings that hang in my apartment still seem immense.
David went to the Ontario College of Art from 1969 to 1973, but his career did not really take shape until a group show at the Pollock Gallery in 1979 and solo shows in 1980 and 1981. The paintings from this period are difficult and uncompromising; large, dramatic abstracts overlap regular patterns of lines or shapes with bursting swirls of colour.
A sojourn in Paris and the completion of an MFA at Concordia in 1985 inspired further growth and change in David's work. He began to experiment with three dimensional shapes in a series of oil on paper paintings, and with layering and stratification of colour and texture in his larger canvasses. Meanwhile, he pursued a parallel teaching career, taking part-time teaching appointments at McGill University, Central Technical School, and the AGO. In 1985, he was appointed to a full-time position in the studio program in the University of Toronto Department of Fine Arts studio program.
For him, teaching was inspiration, not obligation. His students gave him exceptional reviews, and it soon became difficult to get into his courses. He was instrumental in developing an innovative interdisciplinary Masters Program in Visual Studies (scheduled to open at the U of T in the Fall of 2002).
Although his teaching schedule was demanding, he remained a prolific artist until his death. Exposure to the creative energy of young artists in classes on postmodernism and the politics of visual culture led to a fertile experimental phase: He began to explore aspects of gay male sexuality, painting on photo portraits of nude males or including photo collages within smaller abstract canvasses. The introduction of phrases such as "grains of truth" or "view on monogamy" reveal a more overt political stance, matched in his personal life through his support of AIDS research and activism.
Toronto's Gay Pride Parade was, for Dave, Christmas in June. The last painting he completed includes the words "history counsels patience,' a quote from the poet W.H. Auden who denounced the Nazi persecution of homosexuals.
An intensely serious artist and teacher, David had an irreverent wit. Glass of scotch in hand, he'd recall the time at the height of his starving artist days when he "borrowed" twenty dollars from the collection plate at the baptism of a friend's child.
My uncle was found stabbed to death in his office last January. His death has profoundly shaken his family and close friends. He is much loved and sorely missed.
Cheers to you, David.


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There should be no difference in meaning or acceptability, any more than there is a difference between "oeconomics" and "economics", or "oestrogen" and "estrogen". As with (o)estrogen, the "oe" spelling is generally preferred in British English, and the "e" spelling is associated with American English. As far as I can tell, the sources you found prefer "homoeopathy" basically as a matter of branding and crankism (not really surprising that this kind of thing would be associated with a pseudoscientific worldview).
There is no actual connection between Latin "homo" and Greek "hom(o)eo-", by the way. The actual Greek root meaning "same" is just "homo-" (ὁμός) (as in homology, homotopy, homosexual), which is also not connected to the Latin word "homo" meaning "man"; both "homoeo-" and "homeo-" are variant spellings representing the same Greek root (ὅμοιος) meaning "similar". So the first source you link to is very confused from a linguistic perspective.
The Oxford English Dictionary lists a third spelling variant "homoio-", but I can't say I've ever seen the spelling "homoiopathy".
The etymological pronunciation would be /hɒˈmiːəʊ/, as in /hɒˈmɔɪəʊ/; but usage favours /ˈhɒmɪəʊ/, or in popular use /ˈhəʊmɪəʊ/; the last esp. in homoeopathy and its family (the only really popular members of the group).
While the spelling "homoeo" is more likely to suggest the pronunciation /hɒˈmiːəʊ/, this isn't really useful for "homoeopathy" because it isn't usually pronounced that way (a comparable case is ornithological, which according to the OED's view of these things "should" be pronounced as or-NIGH-thological based on the etymology, but isn't; similarly arch(a)eological, pal(a)eographic are not pronounced as ar-KEE-ological, pa-LEE-ographic despite having long/diphthongal vowels in the original Greek).
History of general variation in spelling between "ae/oe/e"
The Latin language at one point had dipthongs something like [ai̯] and [oi̯] (more or less the sounds of English "eye" and "boy"). In Classical Latin, the corresponding sounds were written "ae" and "oe"; some people think this indicates a pronunciation change that had occurred by this point whereby the off-glide was somewhat lowered, making them something like [aɪ̯] and [oɪ̯] or [ae̯] and [oe̯]. Eventually, they developed in pronunciation to monopthongs [ɛː] and [eː] respectively in Vulgar Latin. (Actually, even from early on there was a tendency, seen as rustic, for Latin "ae" to become a monophthong that merged with "long e", so sometimes "ae" corresponds to [eː] instead of [ɛː] in Vulgar Latin.) I have read somewhere about a hypothesis that "oe" monophthongized before losing rounding, passing through a stage like [øː], but I don't know if there is any particularly good evidence for this, although it seems plausible to me.
Anyway, the [eː] that developed from "oe" was pronounced the same as the [eː] that developed from long "e" or "ae", and the [ɛ] that developed from "ae" was pronounced the same as the [ɛ] that developed from "short e", so by Middle Latin the letter "e" could represent either of the sounds associated with "ae/æ" and "oe/œ". This led to a lot of variation between spellings with "æ/ae" and "e", or "œ/oe" and "e", or in some cases between "æ/ae" and "œ/oe" (e.g. the Classical Latin word "caelum" could be spelled in the Middle Ages as "cœlum").
This variation continued into a number of languages that borrowed words from Latin; in English, the most relevant thing affecting this later on was possibly Webster's preference for spellings with "e" which seems to have contributed to present-day differences in custom between American English and British English.
It has no special etymological significance.
Greek αι was conventionally Latinized as "ae"; Greek οι was conventionally Latinized as "oe", so the same variation usually applies to words we get from Greek. Occasionally, people try to use a less Latin-based Romanization of Greek, and use "ai" and "oi" instead (e.g "Koine", "ailurophobia"). This is not significant either (a variant form "aelurophobia" exists, for example).
Not the answer you're looking for? Browse other questions tagged translation prefixes greek or ask your own question.


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Itachi-Sama Luuuuver...!!
Itachi-Sama Luuuuver...!! asked in Arts & HumanitiesHistory · 1 decade ago
Catharism...11th - 13th Century?
Okay, I'm doing some research for a personal interist, and I was wondering if anybody could help me. Okay, I know that the Cathars were a religious sect during the 11-13th century, but I want REGULAR words, not fancy ones...sheesh, I went on Wikipedia, and I had to constantly go back and forth between that and Webster's online dictionary...so please, if you could help it would be appriciated...
~Itachi-sama Luuuuuver...!!~
stevhep
The Cathars were also known as the Albigensians because they were centered around the town of Albi in Southwest France. Here the historian Will Durant writes about them in his 1950 book The Age of Faith
For some time the Cathari [a variant of Albigensianism] received a broad toleration from the ecclesiastics and the secular powers of southern France. Apparently the people were allowed to choose freely between the old religion and the new. Public debates were held between Catholic and Catharist theologians; one such took place at Carcassonne in the presence of a papal legate and King Pedro II of Aragon (1204). In 1167 various branches of the Cathari held a council of their clergy, attended by representatives from several countries; it discussed and regulated Catharist doctrine, discipline, and administration, and adjourned without having been disturbed . . .
. . . the nobles, relatively poor, began to seize Church property. In 1171 Roger II, Viscount of Béziers, sacked an abbey, threw the bishop of Albi into prison, and set a heretic to guard him. When the monks of Allet chose an abbot unsatisfactory to the Viscount, he burned the monastery and jailed the abbot; when the latter died the merry Viscount installed his corpse in the pulpit and persuaded the monks to choose a pleasing substitute. Raymond Roger, Count of Foix, drove abbot and monks from the abbey of Pamiers; his horses ate oats from the altar; his soldiers used the arms and legs of the crucifixes as pestles to grind grain, and practised their marksmanship upon the image of Christ. Count Raymond VI of Toulouse destroyed several churches, persecuted the monks of Moissac, and was excommunicated (1196) . . .
Innocent III, coming to the papacy in 1198, saw in these developments a threat to both Church and state. He recognized some excuse for criticism of the Church, but he felt that he could hardly remain idle when the great ecclesiastical organization for which he had such lofty plans and hopes, and which seemed to him the chief bulwark against human violence, social chaos, and royal iniquity, was attacked in its very foundations, robbed of its possessions and dignity, and mocked with blasphemous travesties.
(The Age of Faith, New York: Simon & Schuster, 1950, 772-773)
From Kenneth Scott Latourette [Baptist]: A History of Christianity: vol. 1: Beginnings to 1500 (NY: Harper & Row, 1953, 454-455):
The Cathari were dualists, believing that there are two eternal powers, the one good and the other evil, that the visible world is the creation of the evil power, and that the spiritual world is the work of the good power . . .
Some put forth a variant of this dualism, saying that the good God had two sons, one of whom, Satanal, rebelled, and the other, Christ, became the redeemer . . .
They held that since flesh is evil, Christ could not have had a real body or have died a real death . . .
From The Oxford Dictionary of the Christian Church [Anglican] (ed. F.L. Cross, 2nd ed., ed. F.L. Cross & E.A. Livingstone, Oxford Univ. Press, 1983, "Albigenses," 31):
. . . Their doctrine in its purest form was strongly dualist, akin to the Manichaean beliefs, and they rejected the flesh and material creation as evil . . . [they taught that] Christ was an angel with a phantom body who, consequently, did not suffer or rise again, and whose redemptive work consisted only in teaching man the true (i.e., Albigensian) doctrine.
Rejecting the sacraments, the doctrines of hell, purgatory, and the resurrection of the body, and believing that all matter was bad, their moral doctrine was of extreme rigorism, condemning marriage, the use of meat, milk, eggs, and other animal produce . . .
Source(s): http://socrates58.blogspot.com/2005/03/were-albige... http://socrates58.blogspot.com/2006/02/inquisition...
Right - first things first, I don't know a great deal about the inner workings of Catharism, but I am an ameteur historian of the Inquisition, which destroyed them. But I will tell you what I know and try to put it as simply as I can.
The Cathar's core belief was that they were usually reincarnated when they died, and the only way to stop this was to attain a form of perfection. The people who tried to do this were called the "Perfecti". They were basically Cathar priests and regular Cathar followers would confess to them before they died. The Cathars were strongly persecuted by the Inquisition - who eventually destroyed them. Catharism was based prmarily in France.
As a point of interest for you, I can remember one story which stuck in my mind, which was from the point of view of the Inquisition.
The Inquisition in Orleans heard about an old woman on her death bed. One Inquisitor ordered some of the town guard to accompany him and he went to her house. He had the guards break down the door and entered the room in which the old woman was dying, ordering the guards to restrain her family members. As she was dying the old woman didn't recognise the Inquisitor, and thought he was a Perfecti - so she began to confess to him. This was enough evidence for him and the Inquisitor declared her a heretic. He then had the guards drag her into the street, still on the bed, and burned her to death.
Pretty grim, huh? I'm sorry I can't be of more help, but I certainly hope you continue with your historical interest - it's a very interesting topic and it can be very rewarding! Good luck!
Sorry I cannot add anything to the other entry but just to say that sometimes subjects are complex and you need technical words. This should not put you off, it is all part of your research. If you are interested in History you may find documents written in Latin or Old English! Hard work it may be: but worth it, education is never wasted and effort is sometimes the key! Not everything worth doing is easy and you should not expect it to be so.
Source(s): I am an Historian.
Why people were generally happier decades ago than now?
Do you think many people suffered from depression and anxiety in the 18th century and before that era? Why do people suffer it now so much?
Can we argue that no side was "good" or "bad" during World War II?
Did titanic not have any pumps to pump out all the water?
most native American's sided with the _ in the war? what's the missing Answer?
Athlete died first time drinking alcohol?
Essay Question: Responding to Jefferson Davis?
Did the allies in the WW2 European theatre have any weapons or vehicles better than the Germans?
why did the & English Spanish have to go to Africa to get slaves to work in the Americas ?
Why America was at their prime during the 1950s?

redpajama_set_name: RedPajamaCommonCrawl


Impacting 'structural soundness' means removing or altering any element of the building which contributes to the support of any other element of the building.
making an internal change that affects a supporting element of the building, for example removing a loadbearing wall.
If you are unsure of whether your change impacts structural soundness, contact Building Services.
Reminder: If you are carrying out any alterations to a verandah, refer to the Build a verandah guide.
Will your works involve removing or altering any structural elements of the building?


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Learn how to market and sell your self-published ebook
How to Format Dialogue: A Guide for Authors
Book EditingBook Writing 0
By Rosie A. Point
If you want to be a professional writer, present your stories, and have them taken seriously, you've got to understand how to format dialogue correctly. That means punctuating in the right places.
As someone who reads, you've probably picked up on how to do this in a general sense–but there are a few things you've missed. So, let's check out what it takes to format dialogue correctly from the simple (commas and dialogue tags) to the slightly more complex (dialogue interruptions with actions).
How to properly punctuate dialogue
An easy way to format interruptions
The rules of using dialogue tags
Step 1: Understand What's Dialogue and What's Not
All right, now this might seem like a pretty simple step, but you've got to understand the difference between spoken and told dialogue. Or direct and indirect dialogue.
Direct dialogue is written between inverted commas or quotes. This is someone actually speaking the words you've written down. It looks like this:
"Hello, my name is Dave," he said.
Indirect dialogue is basically you telling someone about what another person said. It looks like this:
He said hello and that his name was Dave.
Note that no quotation marks are required because it's not a direct quote — the speaker is paraphrasing.
However, most of the formatting and punctuation tips we'll be working with in this article pertain to direct dialogue.
Step 2: Place Commas, Periods, and Speech Marks Correctly
Here's a simple trick to remember when formating dialogue. If the speech is tagged with a 'she said' or 'he said,' you'll use a comma inside the quotation marks.
"I love writing books," he said.
If the speech has no dialogue tag (he said or she said) or if it's followed by an action, there's no need for a comma before the quotation marks.
"I love writing books." He scratched his nose with the end of his pen.
That's because the action isn't a dialogue tag at all.
Here's another important point: Never place a comma or punctuation mark outside of the quotation marks.
So don't do this:
"I love writing books".
"I love writing books",
Or even this:
"I love writing books"!
All the punctuation that belongs to the dialogue or the sentence between the quotation marks should be in there too. When formatting punctuation like question marks or exclamations, the dialogue tags outside of the sentence remain in lower case as follows:
"Do you love writing books?" he asked.
"I love writing books!" he shouted.
Step 3: Understand Dialogue Tags
Fortunately, dialogue tags aren't as simple as just 'he said' or 'she said.' If they were, we'd have a lot less variety to work within our stories or to read in the books we love. And they wouldn't be nearly as interesting.
Dialogue can be many things. It can be fast and choppy, between two characters who don't want to talk, or it can be long and flowing, as a character tries to explain themselves or over-explains under pressure.
Using just the simple dialogue tags might limit you, so let's explore your options.
The formatting basics for a tag are as above. The comma directly after the spoken sentence, then the quotation mark, followed by the tag and a period. The word 'he' is never capitalized but Proper Nouns and 'I' will be.
"I love writing books," Dave said.
"I love writing books," I said.
And what if you have two characters talking? A continuous 'she said' and 'he said' list of tags next to each line of dialogue will be repetitive and, in some cases, downright annoying. That's why you can leave tags out if you need to.
If you don't put a dialogue tag after the line, you close the sentence inside the speech marks, as above.
For even more variety, you can add an action next to a dialogue tag. That would be as follows:
"I love writing books," Dave said, rubbing his hands together.
As you can see here, you'd place the action after the 'Dave said' or 'he/she said' — separating it from the tag with a comma. You can also do this before the dialogue.
Rubbing his hands together, Dave said, "I love writing books."
In this case, you would use a comma after the 'said,' start your dialogue as usual and end that dialogue with a period this time.
Weird rules about dialogue tags:
Each sentence of dialogue from each different speaker must be separated out onto a new line.
They can be actions in their own right, but they are never fully written out action sentences. Example: "I like cake," he laughed. This is because the speaker is laughing as he says that he likes cake. If the speaker is laughing after he said he liked the cake. It would be, "I like cake." He laughed.
Step 4: Use Actions Outside of Quotation Marks
If you wanted to show your character performing an action while speaking, you can do that without using dialogue tags as follows:
"I love writing books." Dave rubbed his hands together.
Both sentences are separated from each other–they're equal sentences that stay on the same line because the action is helping the reader identify who spoke the words. The same rule applies here as it would for dialogue tags. If you use too many actions consecutively, it can get annoying.
"Why is that?" Kyle took a sip of water.
"Because I enjoy being creative." Dave drummed his fingers on the tabletop.
"Then you'd better get to writing." Kyle left the room.
It gets pretty irritating reading that extra information after every sentence. You can vary how you format your dialogue sentence after sentence, as long as you remember to separate out each speaker onto a separate line, always.
"Why is that?" Kyle asked.
"Because I enjoy being creative."
Kyle nodded. "Then you'd better get to writing."
Remember this: actions are always separated into another sentence. Your speech must be closed with a period and a quotation mark.
Step 5: Learn to Format Interruptions
In dialogue, as in life, there are interruptions. Unfortunately, I can't teach you how to format interruptions in your everyday activities, but I can advise you on how to format them in your writing.
If dialogue is interrupted by a tag and action…
You can format it in two ways. First of all:
"I love writing books," Dave said, rubbing his hands together, "but I don't like editing them that much."
In this first example, you write your starting dialogue, tag, and action as usual, but instead of finishing the sentence with a period, you place a comma, open a new quotation mark and continue the sentence with a conjunction. At the end of that sentence, you'd use a period and close the speech.
But you can also format that interruption by separating the spoken pieces into two separate sentences as follows:
"I love writing books," Dave said, rubbing his hands together. "But I don't like editing them that much."
Here, the sentence ends after Dave has rubbed his hands together. Because of that, when you start your new line of dialogue, you format it with a capitalized 'But' and end it with a period.
If dialogue is interrupted by just an action…
Say your speaker is being erratic, or just doing something that would interrupt his speech, like taking a sip of water or coughing uncontrollably, you wouldn't have a well-planned and inserted interruption. The text would look broken because the dialogue is being broken by the action.
You'd format that as follows:
"I love writing books"–Dave took a sip of water–"but I'm not a fan of editing them."
Note: the em dashes are outside of the dialogue for this type of formatting.
That's about it. Formatting dialogue doesn't have to be a pain in the neck–you can bookmark this article and come back to it anytime you need a reminder. I know, I'm always double-checking whether I've gotten something right, and I've been writing for a while.
Also, for further reading, The Elements of Style, Fourth Edition, by William Strunk Jr, is a book I recommend. It covers a lot of the questions you'll have about formatting dialogue, punctuation and sentence structure.
Rosie A. Point is an Amazon bestselling cozy mystery author who's been publishing for two years and writing for eight. Once an outlining aficionado and business owner, she's helped authors plot, write, and edit their novels during her career.
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Jen and Derrick recently sent Coco, a lab, to be trained with Unleashed Potential in Halifax. This is the e-mail that followed after Coco returned home.
After training with Unleashed Potential, the difference in our dog is amazing. We don't use the leash to take her out to use the bathroom and she doesn't bolt and take off like she used to. She was out to play with her doggie friends today and everyone was taken back to see her not on a leash. When I called her and she came to me, and their jaws just dropped (because they would help me chase her for 45min to an hour), lol! She didn't jumped on anyone, not even my Dad, and she would go crazy with excitement over him. No one can believe it is Coco. We can actually call her Coco now and not CooCoo, LOL! To say we are happy about the training you did with her would be an understatement. You saved her life (by her not getting hit by cars when she would run away) and our sanity. We can enjoy her now; we look forward to taking her to play fetch instead of getting prepared for an hour long chase.


text:
\section{Introduction}\label{section-introduction}

In this article we are interested in the classification of contact Fano manifolds.
We review the relevant definitions in \S\ref{section-preliminaries}.
So far the only known examples of contact Fano manifolds are obtained as follows.
For a simple Lie group $G$ consider its adjoint action on $\PP(\gotg)$,
where $\gotg$ is the Lie algebra of $G$.
This action has a unique closed orbit $X$ and this $X$ has a natural contact structure.
In this situation $X$ is called a \emph{projectivised minimal nilpotent orbit},
or the \emph{adjoint variety} of $G$.
By the duality determined by the Killing form,
equivalently we can consider the coadjoint action of $G$ on $\PP(\gotg^*)$ and $X$ is isomorphic
to the unique closed orbit in $\PP(\gotg^*)$.

In order to study the non-homogeneous contact manifolds (potentially non-existent) it is natural to assume
$\Pic X  \simeq \ZZ$ and further that $X$ is not isomorphic to a projective space.
This only exludes the adjoint varieties of types $A$ and $C$
(see \S\ref{section-preliminaries} for more details).

With this assumption,
we take a closer look at three pieces of the homogeneous structure on adjoint varieties:
the Killing form $B$ on $\gotg$, the Lie algebra grading
$\gotg = \gotg_{-2} \oplus \gotg_{-1} \oplus \gotg_{0} \oplus \gotg_{1} \oplus \gotg_{2}$
(see \cite[\S6.1]{landsbergmanivel02} and references therein)
and a part of the Lie bracket on $\gotg$.
Understanding the underlying geometry allows us to define the appropriate generalisations of these notions
on arbitrary contact Fano manifolds.

An essential building block for our constructions is the notion of a \emph{contact line}
(or simply \emph{line})
on $X$.
These contact lines were studied by Kebekus \cite{kebekus_lines1}, \cite{kebekus_lines2} and
\Wisniewski{} \cite{wisniewski}.
Also they are an instance of minimal rational curves, which are studied extensively.
The geometry of contact lines was the original motivation to study Legendrian subvarieties in projective space
(see \cite{jabu_dr} for an overview and many details).
We briefly review the subject of lines on contact Fano manifolds in \S\ref{section-review-lines}.

The key ingedient is the construction of a family of divisors $D_x$ parametrised by points $x \in X$
(see \S\ref{section-define-Dx}).
These divisors are swept by pairs of intersecting contact lines, one of which passes through $x$.
In other words, set theoretically $D_x$ is the set of points of $X$,
which can be joined with $x$ using at most $2$ intersecting contact lines.
The idea to study these loci comes from \Wisniewski{} \cite{wisniewski}
where he observed, that (under an additional minor assumption)
 these loci contain some non-trivial divisorial components
and he studied the intersection numbers of certain curves on $X$ with the divisorial components.
Here we prove all the components of $D_x$ are divisorial and draw conclusions from that observation
going into a different direction than those of \cite{wisniewski}.

\begin{thm}\label{thm-exist-Dx-and-duality}
   Let $X$ be a contact Fano manifold with $\Pic X \simeq \ZZ $
   and assume $X$ is not isomorphic to a projective space.
   Then the locus $D_x \subset X$
   swept by the pairs of intersecting contact lines, one of which passes through $x \in X$
   is a of pure codimension $1$ and thus $D_x$ determines a divisor on $X$.
   Let ${\left\langle D \right\rangle} \subset H^0(\ccO(D_x))$ be the linear system spanned by these divisors.
   Let $\phi\colon X \to \PP{\LinSyst^*}$ be the map determined by the linear system ${\left\langle D \right\rangle}$
   and let $\psi\colon X \to \PP{\left\langle D \right\rangle}$ be the map $x\mapsto D_x$.
   Then:
   \begin{enumerate}
     \item both $\phi$ and $\psi$ are regular maps.
     \item there exists a unique up to scalar non-degenerate bilinear form $B$ on ${\left\langle D \right\rangle}$,
           which determines an isomorphism $\PP{\LinSyst^*} \simeq \PP {\left\langle D \right\rangle}$
	   making the following diagram commutative:
	   \[
              \xymatrix@C1.5pc@R0.5pc{
              & & & \PP {\LinSyst^*} \ar[dd]^{\simeq} \\
              X\ar[urrr]^{\phi} \ar[drrr]^{\psi} && \\
              & & & \PP {\left\langle D \right\rangle}.}
	   \]
     \item The bilinear form $B$ is either symmetric or skew-symmetric.
     \item If $X \subset \PP(\gotg^*)$ is the adjoint variety of simple Lie group $G$,
          then ${\left\langle D \right\rangle} = \gotg$ and $B$ is the Killing form on $\gotg$.
   \end{enumerate}
\end{thm}

With the notation of the theorem,
after fixing a pair of general points $x, w \in X$
there are certain natural linear subspaces of ${\left\langle D \right\rangle}$,
which we denote ${\left\langle D \right\rangle}_{-2}$, ${\left\langle D \right\rangle}_{-1}$, ${\left\langle D \right\rangle}_{0}$, ${\left\langle D \right\rangle}_{1}$ and ${\left\langle D \right\rangle}_{2}$
(see \S\ref{section-grading} for details).
\begin{thm} \label{thm-grading-for-homogeneous}
  If $X \subset \PP(\gotg^*)$ is the adjoint variety of a simple Lie group $G$ with $\Pic X \simeq Z$
  and $X$ not isomorphic to a projective space,
  then there exists a choice of a maximal torus of $G$
  and a choice of order of roots of $\gotg$,
  such that
  ${\left\langle D \right\rangle}_{i} = \gotg_{i}$ for every $i \in \set{-2,-1,0,1,2}$,
  where $\gotg = \gotg_{-2} \oplus \gotg_{-1} \oplus \gotg_{0} \oplus \gotg_{1} \oplus \gotg_{2}$
  is the Lie algebra grading of $\gotg$.
\end{thm}

Finally, if $X$ is the adjoint variety of $G$,
then there is a rational map
\[
  [\cdot, \cdot]\colon X \times X \dashrightarrow \PP(\gotg),
\]
which is the Lie bracket on $\gotg$ (up to projectivisation).
Also there is a divisor $D \subset X \times X$,
such that for general $(x,z) \in D$
the Lie bracket $[x,z]$ is in $X$.
We recover this bracket restricted to $D$  for general contact manifolds:
\begin{thm} \label{thm-bracket}
  For $X$ and $D_x$ and in Theorem \ref{thm-exist-Dx-and-duality},
  let $D \subset X \times X$ be the divisor consisting of pairs $(x,z) \in X \times X$,
  such that $z \in D_x$.
  There exists a rational map $[\cdot , \cdot ]\colon D \dashrightarrow X$,
  such that $[x,z] = y$,
  where $y$ is an intersection point of a pair of contact lines that join $x$ and $z$.
  In particular, this intersection point $y$ and the pair of lines are unique for general pair $(x,z)$ $ \in D$.
  Moreover, if $X$ is the adjoint variety of a simple Lie group $G$, then $[ \cdot, \cdot]$ is the restricion of the Lie bracket.
\end{thm}

In \S\ref{section-preliminaries} we introduce and motivate our assumptions and notation.

In \S\ref{section-loci-swept-by-lines} we review the notion of contact lines and their properties.
We continue by studying certain types of loci swept by those lines and calculate their dimensions.
In particular we prove there Theorem~\ref{thm-Dx-is-a-divisor},
which is a part of results summarised in Theorem~\ref{thm-exist-Dx-and-duality}.
We also study the tangent bundle to $D_x$ as a subspace of $TX$.

In \S\ref{section-duality} we study the duality of maps $\phi$ and $\psi$
introduced in Theorem~\ref{thm-exist-Dx-and-duality} together with the consequences of this duality.
This section is culminated with the proof of Theorem~\ref{thm-exist-Dx-and-duality}.

In \S\ref{section-grading} we generalise the Lie algebra grading to arbitrary contact manifolds
and prove Theorem~\ref{thm-grading-for-homogeneous}.

In \S\ref{section-cointegrable} we prove that certain lines are integrable with respect to
a special distribution on $D_x$ and we apply this to prove Theorem~\ref{thm-bracket}.

\subsection*{Acknowledgements}

The author is supported by Marie Curie International Outgoing Fellowship.

The author would like to thank the following people for many enlightening
discussions:
Jun-Muk Hwang,
Stefan Kebekus,
Joseph M.~Landsberg,
\Jaroslaw{} \Wisniewski{}
and Fyodor L.~Zak.
Comments of Laurent Manivel helped to improve the presentation in the paper.

\section{Preliminaries}\label{section-preliminaries}

Throughout the paper all our projectivisations $\PP$ are naive.
This means, if $V$ is a vector space, then $\PP V = (V \setminus 0)/ \CC^*$,
and similarly for vector bundles.

A complex manifold $X$ of dimension $2n+1$ is \emph{contact}
if there exists a vector subbundle $F \subset TX$ of rank $2n$ fitting into an exact sequence:
\ContactDistributionSequence{X}
such that the derivative $\ud \theta \in H^0(\Wedge{2} F^* \otimes L)$
of the twisted form $\theta \in H^0(T^*X \otimes L)$ is nowhere degenerate.
In particular, $\ud \theta_x$ is a symplectic form on the fibre of contact distribution $F_x$.
See \cite[\S{}E.3 and Chapter~C]{jabu_dr} and references therein for an overview of the subject.

A projective manifold $X$ is Fano, if its anticanonical divisor ${K_X}^*=\Wedge{\dim X} TX$ is ample.

If $X$ is a projective contact manifold,
then by Theorem of Kebekus, Peternell, Sommese and \Wisniewski{} \cite{4authors}
combined with a result by Demailly \cite{demailly},
$X$ is either a projectivisation of a cotangent bundle
to a smooth projective manifold
or $X$ is a contact Fano manifold, with $\Pic X \simeq \ZZ$.
In the second case, since $K_X \simeq (L^*)^{\otimes(n+1)}$, by \cite{kobayashi_ochiai},
either $X \simeq \PP^{2n+1}$ or $\Pic X = \ZZ \cdot [L]$.
Here we are interested in the case $X \not \simeq \PP^{2n+1}$.
Thus our assumption spelled out below only exclude some well understood cases
(the projectivised cotangent bundles and the projective space)
and they agree with the assumptions of
Theorems~\ref{thm-exist-Dx-and-duality}, \ref{thm-grading-for-homogeneous} and~\ref{thm-bracket}.
\begin{notation}\label{notation-with-assumptions-about-contact-mflds}
   Throughout the paper $X$ denotes a contact Fano manifold with $\Pic X$ generated by the class of $L$,
   where $L= TX/F$ and $F\subset TX$ is the contact distribution on $X$.
   We also assume $\dim X = 2n +1$.
\end{notation}
From Theorem of Ye \cite{ye} it follows that $n \ge 2$.

We will also consider the homogeneous examples of contact manifolds (i.e. the adjoint varieties).
Thus we fix notation for the Lie group and its Lie algebra.

\begin{notation}\label{notation-with-assumptions-about-group}
   Throughout the paper $G$ denotes a simple complex Lie group, not of types $A$ or $C$
   (i.e.~not isomorphic to $SL_n$ nor $Sp_{2n}$ nor their discrete quotients).
   Further $\gotg$ is the Lie algebra of $G$.
   Thus $\gotg$ is one of $\gotso_n$ (types $B$ and $D$),
   or one of the exceptional Lie algebras
   $\gotg_2$, $\gotf_4$, $\gote_6$, $\gote_7$ or $\gote_8$.
\end{notation}

The contact structure on $\PP^{2n-1} = \PP(\CC^{2n})$ is determined by a symplectic form
$\omega$ on $\CC^{2n}$.
 The precise relation between the contact and symplectic structures is decribed for instance in
\cite[\S{}E.1]{jabu_dr} (see also \cite[Ex.~2.1]{lebrun}).
In particular, for all $x \in X$, the projectivisation of a fibre of the contact distribution $\PP F_x$
comes with a natural contact structure.

Let $M$ be a projective contact manifold
(in our case $M = X$ with $X$ as in Notation~\ref{notation-with-assumptions-about-contact-mflds}
 or $M = \PP^{2n-1}$).
A subvariety $Z \subset M$ is \emph{Legendrian},
 if for all smooth points $z \in Z$ the tangent space $T_z Z$
 is contained in the contact distribution of $M$ and $Z$ is of pure dimension $\half(\dim M -1)$.

Recall from \cite[Lecture~20]{harris} or \cite[III.\S3,\S4]{mumford} the notion of \emph{tangent cone}.
For a subvariety $Z \subset X$, and a point $x \in Z$ let $\tau_x Z \subset T_x X$
be the tangent cone of $Z$ at $x$.
In this article we will only need the following elementary properties of the tangent cone:
\begin{itemize}
 \item $\tau_x Z$ is an affine cone (i.e.~it is invariant under the standard action of $\CC^*$ on $T_x X$).
 \item $\dim_x Z = \dim \tau_x Z$ and thus if $Z$ is irreducible, then $\dim Z = \dim \tau_x Z$.
 \item If $x \in Z_1 \subset Z_2 \subset X$, then  $\tau_x Z_1 \subset \tau_x Z_2$.
 \item If $Z$ is smooth at $x$, then $\tau_x Z = T_x Z$.
\end{itemize}
Since $\tau_x Z$ is a cone, let $\PP \tau_x Z \subset \PP T_x$ be the corresponding projective variety.

\section{Loci swept out by lines}
\label{section-loci-swept-by-lines}

A rational curve $l \subset X$ is a \emph{contact line} (or simply a \emph{line})
if \mbox{$\deg L|_{l} =1$}.

Let $RatCurves^n(X)$ be the normalised scheme parametrising rational curves on $X$,
as in \cite[II.2.11]{kollar_book_rational_curves}.
Let $Lines(X) \subset RatCurves^n(X)$ be the subscheme parametrising lines.
Then every component of $Lines(X)$ is a minimal component of $X$ in the sense
of \cite{hwang_mok_birationality}.
We fix $\ccH \ne \emptyset$ a union of some irreducible components of $Lines(X)$.

By a slight abuse of notation, from now on we say $l$ is a \emph{(contact) line}
if and only if $l \in \ccH$.
For simplicity, the reader may choose to restrict his attention to one of the extreme cases:
either to the case $\ccH = Lines(X)$
(and thus be consistent with \cite{wisniewski} and the first sentence of this section)
or to the case where $\ccH$ is one of the irreducible components of $Lines(X)$
(and thus be consistent with \cite{kebekus_lines1, kebekus_lines2}).
In general it is expected that $Lines(X)$
(with $X$ as in Notation~\ref{notation-with-assumptions-about-contact-mflds})
is irreducible and all the cases are the same.

\subsection{Legendrian varieties swept by lines}\label{section-review-lines}

We denote by $C_x \subset X$ the locus of contact lines through $x \in X$.
Let $\mathscr{C}{}_x:= \PP\tau_x C_x \subset \PP(T X)$.
Note that with our assumptions both $C_x$ and $\mathscr{C}{}_x$ are closed subsets of
 $X$ or $\PP(T_x X)$ respectively.

The following theorem briefly summarises results of \cite{kebekus_lines2} and earlier:

\begin{thm} \label{thm-properties-of-Cx}
  With $X$ as in Notation~\ref{notation-with-assumptions-about-contact-mflds} let $x\in X$ be any point.
  Then:
  \begin{enumerate}
     \item\label{item-exist-lines}
        There exist lines through $x$, in particular $C_x$ and $\mathscr{C}{}_x$ are non-empty.
     \item\label{item-Cx-Yx-Legendrian}
        $C_x$ is Legendrian in $X$ and $\mathscr{C}{}_x \subset \PP(F_x)$ and $\mathscr{C}{}_x$
        is Legendrian in $\PP(F_x)$.
     \item\label{item-Yx}
        If in addition $x$ is a general point of $X$,
        then $\mathscr{C}{}_x$ is smooth and each irreducible component of $\mathscr{C}{}_x$
         is linearly non-degenerate in $\PP(F_x)$.
        Further $C_x$ is isomorphic to the projective cone over $\mathscr{C}{}_x \subset \PP(F_x)$,
        i.e.~$C_x \simeq \tilde{\mathscr{C}{}_x} \subset \PP(F_x \oplus \CC)$,
        in such a way that lines through $x$ are mapped bijectively onto the generators of the cone
        and restriction of $L$ to $C_x$ via this isomorphism
        is identified with the restriction of $\ccO_{\PP(F_x \oplus \CC)}(1)$ to $\tilde{\mathscr{C}{}_x}$.
        In particular all lines through $x$ are smooth and two different lines intersecting at $x$ will
        not intersect anywhere else, nor they will share a tangent direction.
\end{enumerate}
\end{thm}

\begin{prf}
   Part~\ref{item-exist-lines} is proved in \cite[\S2.3]{kebekus_lines1}.

   The proof of \ref{item-Cx-Yx-Legendrian} is essentially contained in \cite[Prop.~2.9]{4authors}.
   Explicit statements are in \cite[Prop.~4.1]{kebekus_lines1} for $C_x$
   and in \cite[Lemma~5]{wisniewski} for $\mathscr{C}{}_x$.
   Also \cite{hwang_mok_VMRTs} may claim the authorship of this observation,
   since the proof in the homogeneous case is no different than in the general case.

   Assume  $x\in X$ is a general point.
   The statements of \ref{item-Yx} are basically \cite[Thm~1.1]{kebekus_lines2},
   which however assumes (in the statement) that $\ccH$ is irreducible.
   This is never used in the proof, with the exception of the argument for the irreducibility of $C_x$
   --- see however Remark~\ref{remark-kebekus-irreducibility}.
   Thus $\mathscr{C}{}_x$ is smooth and $C_x$ is isomorphic to the cone over $\mathscr{C}{}_x$ as claimed.
   Each irreducible component $\mathscr{C}{}_x$ is non-degenerate on $\PP F_x$ by \cite[Thm~4.4]{kebekus_lines1}
   --- again the statement is only for $\mathscr{C}{}_x$, not for its components,
   however the proof stays correct in this more general setup.
   In particular, \cite[Lemma~4.3]{kebekus_lines1} implies that
   $C_x$ polarised by $L|_{C_x}$ is not isomorphic with a linear subspace
   with polarised by $\ccO(1)$.
   Thus the other results of this theorem give alternate (but more complicated)
   proof of that generalised non-degeneracy.
\end{prf}

\begin{rem}\label{remark-kebekus-irreducibility}
Note that (assuming $\ccH$ is irreducible)
Kebekus \cite{kebekus_lines2} also stated that  $C_x$ and $\mathscr{C}{}_x$
are irreducible for general $x$.
However it was observed by Kebekus himself together with the author that
there is a gap in the proof.
This gap is on page 234 in Step~2 of proof of Proposition~3.2
where Kebekus claims to construct ``a well defined family of cycles'' parametrised by a divisor $D^0$.
This is not necessarily a well defined family of cycles:
Condition (3.10.4) in \cite[\S{}I.3.10]{kollar_book_rational_curves}
is not necessarily satisfied if $D^0$ is not normal
and there seem to be no reason to expect that $D^0$ is normal.
As a consequence the map $\Phi\colon D^0 \to \operatorname{Chow}(X)$
is not necessarily regular at non-normal points of $D^0$ and it might contract some curves.
\end{rem}

Let us define:
\begin{align*}
 C^2 & \subset X \times X \\
 C^2 & := \set{ (x, y) \mid  y \in C_{x}},
\end{align*}
i.e.~this is the locus of those pairs $(x, y)$, which are both on the same contact line.
Again this locus is a closed subset of $X \times X$.

Analogously, define:
\[
   C^3 : = C^2 \times_X C^2
\]
so that:
\begin{align*}
 C^3 & \subset X \times X  \times X\\
 C^3 & := \set{ (x, y, z) \mid  y \in C_{x}, z \in C_{y}}.
\end{align*}
Finally, for $x \in X$ we also define $C^2_x$:
\begin{align*}
 C^2_x & \subset X  \times X  \simeq \set{x} \times X \times X \\
 C^2_x & := \set{ (y, z) \mid  y \in C_{x}, z \in C_{y}},
\end{align*}
with the scheme structure of the fibre of $C^3$ under the projection on the first coordinate.
Since for all $x \in X$ all irreducible components of $C_x$ are of dimension~$n$
 (see Theorem~\ref{thm-properties-of-Cx})
we conclude:

 \begin{prop}\label{prop-dimensions-of-Cs}
  All $C^2$, $C^2_x$, $C^3$ are projective subschemes, they are all of pure dimension,
  and their dimensions are:
  \begin{itemize}
   \item $\dim C^2 = 3n+1$.
   \item $\dim C^2_x = 2n$.
   \item $\dim C^3 = 4n+1$.
  \end{itemize}
\end{prop}
\noprf

\subsection{Joins and secants of Legendrian subvarieties}

For subvarieties $Y_1, Y_2 \subset \PP^N$ recall that their \emph{join} $Y_1*Y_2$
is the closure of the locus of lines between points $y_1\in Y_1 $ and $y_2 \in Y_2$.
Note that the expected dimension of $Y_1*Y_2$ is $\dim Y_1 + \dim Y_2 +1$.
We are only concerned with two special cases: either $Y_1$ and $Y_2$ are disjoint or
$Y_1=Y_2$.

\begin{lemma}\label{lemma-join-of-two-disjoint-varieties}
  If $Y_1, Y_2 \subset \PP^N$ are two disjoint subvarieties of dimensions $k-1$ and $N-k$
  respectively, then their join $Y_1 * Y_2$ fills out the ambient space,
  i.e.~this join is of expected dimension.
\end{lemma}

\begin{prf}
  Let $p\in \PP^N$ be a general point and consider the projection
  $\pi:\PP^N \dashrightarrow \PP^{N-1}$ away from $p$.
  Let $Z_i = \pi(Y_i)$ for $i =1,2$.
  Since $p$ is general, $\dim Z_i = \dim Y_i$ and thus $Z_1 \cap Z_2$ is non-empty.
  Let $q\in Z_1 \cap Z_2$ be any point.
  The preimage $\pi^{-1}(q)$ is a line in $\PP^N$ intersecting both $Y_1$ and $Y_2$ and passing through $p$.
\end{prf}

Recall, that the special case of join is when $Y=Y_1 =Y_2$
and $\sigma_2(Y) := Y*Y$ is the \emph{secant variety} of $Y$.

\begin{prop}\label{prop-secant-of-Legendrian}
     \begin{itemize}
      \item Let $Y \subset \PP^{2n-1}$ be an irreducible linearly non-degenerate Legendrian variety.
            Then $\sigma_2(Y) = \PP^{2n-1}$.
      \item Let $Y_1,Y_2 \subset \PP^{2n-1}$ be two disjoint Legendrian subvarieties.
            Then $Y_1*Y_2 = \PP^{2n-1}$.
     \end{itemize}
\end{prop}

\begin{prf}
     If $Y$ is irreducible, then this is proved in the course of proof of Prop.~17(2)
     in \cite{landsbergmanivel04}.

     If $Y_1$ and $Y_2$ are disjoint,
     then the result follows from Lemma~\ref{lemma-join-of-two-disjoint-varieties}.
\end{prf}

\subsection{Divisors swept by broken lines}\label{section-define-Dx}

Following the idea of \Wisniewski{} \cite{wisniewski}
we introduce the locus of broken lines
(or reducible conics, or chains of $2$ lines) through $x$:
\[
 {D}_x := \bigcup_{y \in C_x} C_y.
\]
Note that ${D}_x$ is a closed subset of $X$
as it can be interpreted as the image of projective variety $C^2_x \subset X \times X$ under a proper map,
which is the projection onto the last coordinate.
By analogy to the case of lines consider also:
\begin{align*}
 {D}^2 & \subset  X \times X \\
 {D}^2 & := \set{ (x, z) \mid   \exists_{y \in C_{x}} \text{ s.t. } z \in C_{y}},
\end{align*}
i.e.~${D}^2$ is the projection of $C^3$ onto first and third coordinates.
Thus again ${D}^2$ is a closed subset of the product.
Set theoretically ${D}_x$ is the fibre over $x$ of
(either of) the projection ${D}^2 \to X$ and if we consider ${D}^2$ as a reduced scheme,
then we can assign to ${D}_x$ the scheme structure of the fibre.

It follows immediately from the above discussion and Proposition \ref{prop-dimensions-of-Cs},
that every component of ${D}_x$ has dimension at most $2n$ and
every component of ${D}^2$ has dimension at most $4n+1$.
In fact the equality holds.

\begin{thm}
  \label{thm-Dx-is-a-divisor}
  Let $x \in X$ be any point.
  Then the locus ${D}_x$ is of pure codimension~$1$.
\end{thm}

\begin{prf}
   Assume first that $x\in X$ is a general point.
   Recall, that $C^2_x \subset X \times X$ has two projections:
   \[
     \xymatrix{ C^2_x  \ar@{->>}[r]^{\pi_2} \ar@{->>}[d]^{\pi_1} & D_x \\
                C_x}
   \]
   Fix $\component{D_x}$ to be an irreducible component of $D_x$.
   Then $\component{D_x}$ is dominated
   by some component $\component{C^2_x}$ of $C^2_x$.
   Dimension of $\component{C^2_x}$ is equal to $2n$ by Proposition~\ref{prop-dimensions-of-Cs}.

   For $y \in C_x$ the fiber ${\pi_1}^{-1}(y) \subset C^2_x$
   is equal to $\set{y} \times C_y$.
   In particular,
   by Theorem~\ref{thm-properties-of-Cx}\ref{item-Cx-Yx-Legendrian}
   the fibers of $\pi_1$ have constant dimension $n$.
   Thus $\component{C^2_x}$ is mapped onto an irreducible component $\component{C_x}$ of $C_x$.
   Finally, let $C'$ be an irreducible component of the preimage ${\pi_1}^{-1}(x)$
   which is contained in $\component{C^2_x}$.
   Note that $C'$ can be identified with an irreducible component of $C_x$,
   because ${\pi_1}^{-1}(x) = \set{x} \times C_x$.

   We claim that the projectivised tangent cone $\PP \tau_x \component{D_x}$ contains
   the join of two tangent cones
   \[
    (\PP\tau_x C') * (\PP\tau_x \component{C_x}) \subset \PP F_x \subset \PP T_x X.
   \]
   The proof of the claim is a baby version of \cite[Thm~3.11]{hwang_kebekus_chains}.
   There however Hwang and Kebekus assume $C_x$ is irreducible and thus their
   results do not neccessarily apply directly here.
   Let $l_0$ be a general line through $x$ contained in $C'$
   and let $l$ be a general line through $x$ contained in $\component{C_x}$.
   To prove the claim it is enough to show there exists
   a surface $S\subset D_x$ containing $l_0$ and $l$ which is smooth at $x$,
   since in such a case $T_x S \subset \tau_x D_x$ and
   $\PP T_x S$ is the line between $\PP T_x l$ and $\PP T_x l_0$.

   We obtain $S$ by varying $l_0$.
   Consider $\ccH_l \subset \ccH$ the parameter space for lines on $X$,
   which intersect $l$.
   By Theorem~\ref{thm-properties-of-Cx}\ref{item-Yx} the space $\ccH_l$ comes with a projection
   $\xi\colon \ccH_l \dashrightarrow l$,
   which maps $l'\in \ccH_l$ to the intersection point of $l$ and $l'$,
   and which is well defined on an open subset containg all lines through $x$.

   By generality of our choices, $l_0$ is a smooth point of $\ccH_l$ and $\xi$ is submersive at $l_0$.
   In the neighbourhood of $l_0$ choose a curve $A \subset \ccH_l$ smooth at $l_0$
   for which $\xi|_A$ is submersive at $l_0$.
   Then the locus in $X$ of lines which are in $A$ sweeps a surface $S \subset X$,
   which is smooth at $x$, contains $l_0$, and contains an open subset of $l$ around $x$.
   Thus the claim is proved and:
   \begin{equation}\label{equ-join-is-contained-in-tangent-to-Dx}
      (\PP\tau_x C') * (\PP\tau_x \component{C_x}) \subset \PP \tau_x \component{D_x}
   \end{equation}

   Now we claim that $F_x \subset \tau_x D_x$.
   For this purpose we separate two cases.

   In the first case $C' = \component{C_x}$.
   Then $\PP\tau_x C'$ is non-degenerate by Theorem~\ref{thm-properties-of-Cx} and thus
   \[
     (\PP\tau_x C') * (\PP\tau_x \component{C_x}) = \sigma_2(\PP\tau_x C') = \PP(F_x)
   \]
   by Proposition~\ref{prop-secant-of-Legendrian}.
   Combining with \eqref{equ-join-is-contained-in-tangent-to-Dx} we obtain the claim.

   In the second case $C'$ and $\component{C_x}$ are different components of $C_x$.
   Then by generality of $x$ and by Theorem~\ref{thm-properties-of-Cx},
   the two tangent cones $(\PP\tau_x C')$ and $(\PP\tau_x \component{C_x})$ are disjoint.
   Thus again
   \[
     (\PP\tau_x C') * (\PP\tau_x \component{C_x}) = \PP(F_x)
   \]
   by Proposition~\ref{prop-secant-of-Legendrian}.
   Combining with \eqref{equ-join-is-contained-in-tangent-to-Dx} we obtain the claim.

   Thus in any case for a general $x \in X$, every component of $D_x$ has dimension at least $2n$.
   The dimension can only jump up
   at special points when one has a fibration,
   thus also at special points every component of $D_x$ has dimension at least $2n$.
   Earlier we observed that $\dim D_x \le 2n$, thus the theorem is proved.
\end{prf}

\begin{prop}\label{prop-what-is-Dx-for-X-homogeneous}
  If $X$ is the adjoint variety of $G$, and $x\in X$, then $D_x$
  is the hyperplane section of $X \subset \PP(\gotg)$
  perpendicular to $x$ via the Killing form.
\end{prop}

\begin{prf}
  Let $X = G/P$, where $P$ is the parabolic subgroup preserving $x$.
  Notice, that $D_x$ must be reduced (because $D$ is reduced and $D_x$ is a general fibre of $D$).
  Also $D_x$ is $P$-invariant, because the set of lines is $G$ invariant
  and $D_x$ is determined by $x$ and the geometry of lines on $X$.
  We claim, there is a unique $P$-invariant reduced divisor on $X$,
  and thus it must be the hyperplane section as in the statment of proposition.

  So let $\Delta$ be a $P$-invariant divisor linearly equivalent to $L^k$ for some $k \ge 0$.
  Also let $\rho_{\Delta}$ be a section of $L^k$ which determines $\Delta$.
  The module of sections $H^0(L^k)$ is an irreducible $G$-module by Borel-Weil theorem
  (see \cite{serre_representations}), with some highest weight  $\omega$.
  Since the Lie algebra $\gotp$ of $P$ contains all positive root spaces,
  by \cite[Prop.~14.13]{fultonharris} there is a unique $1$-dimensional $\gotp$-invariant
  submodule of $H^0(L^k)$, it is the highest weight space $H^0(L^k)_{\omega}$.
  So $\rho_{\Delta} \in H^0(L^k)_{\omega}$ and $\Delta$ is unique.

  The hyperplane section of $X \subset \PP(\gotg)$
  perpendicular to $x$ via the Killing form is a divisor in $\lvert L \rvert$,  and it is $P$-invariant,
  and so are its multiples in $\lvert L^k \rvert$.
  So by the uniqueness $\Delta$ must be equal to $k$ times this hyperplane section.
  Thus $\Delta$ is reduced if and only $k=1$ and so $D_x$ is the hyperplane section.
\end{prf}

\subsection{Tangent bundles restricted to lines}

   Let $l$ be a  line through a general point $y\in X$.
   Recall from  \cite[Fact~2.3]{kebekus_lines2} that:
   \begin{align}
      T X|_{l} & \simeq {\ccO_l}(2) \oplus {\ccO_l}(1)^{n-1} \oplus {\ccO_l}^{n-1} \oplus {\ccO_l}^{2}
        \nonumber
        \label{equ-splitting-type-of-TX-on-l}\\
        \nonumber
        F|_{l} & \simeq {\ccO_l}(2) \oplus {\ccO_l}(1)^{n-1} \oplus {\ccO_l}^{n-1} \oplus {\ccO_l}(-1) \\
        \nonumber
      T l      & \simeq {\ccO_l}(2)\\
   \intertext{and for general $z \in l$:}
        \nonumber
      T C_z|_{l \setminus \set{z}} & \simeq {\ccO_l}(2) \oplus {\ccO_l}(1)^{n-1}.
   \end{align}

   If $x \in X$ is a general point and $y \in C_x$
    is a general point of any of the irreducible components of $C_x$
   and $l$ is a line through $y$,
   then we want to express $T D_x |_{l}$ in terms of those splittings.
   In a neighbourhood of $l$ the divisor $D_x$ is swept by deformations $l_t$ of $l =l_0$
   such that $l_t$ intersects $C_x$.
   By the standard deformation theory argument taking derivative of $l_t$ by $t$ at a point $z \in l$,
   we obtain that:
   \begin{equation}\label{equ-for-tangent-space-of-Dx}
     T_{z} D_x
      \supset \set{s(z) \in T_z X  \mid  \exists s \in H^0(T X|_{l}) \  \text{ s.t. } s(y) \in T_y C_x}
   \end{equation}
   Moreover, at a general point $z$ we have equality in \eqref{equ-for-tangent-space-of-Dx}.
   If we mod out $T X|_l$ by the rank $n$ positive bundle $(T X|_l)^{>0}:= {\ccO_l}(2) \oplus {\ccO_l}(1)^{n-1}$,
   then we are left with a trivial bundle ${\ccO_l}^{n+1}$.
   Thus, since by Theorem~\ref{thm-Dx-is-a-divisor} the dimension of $T_z D_x = 2n$ for general $z \in l$,
   the vector space $T_y C_x$ must be transversal to  $(TX|_l)^{>0}$ at $y$.
   In particular, if $z\ne y$,
   then dimension of the right hand side in \eqref{equ-for-tangent-space-of-Dx} is $2n$
   and thus \eqref{equ-for-tangent-space-of-Dx} is an equality for each point $z\in l$,
   such that $z$ is a smooth point of $D_x$.

   We conclude:
   \begin{prop}\label{prop-tangent-to-Dx}
   Let $x \in X$ be a general point and $y \in C_x$
   be a general point of any of the irreducible components of $C_x$
   and $l$ be any line through $y$.
   Then there exists a subbundle $\varGamma \subset TX|_l$
   such that:
   \begin{align*}
     \varGamma & = {\ccO_l}(2) \oplus {\ccO_l}(1)^{n-1} \oplus {\ccO_l}^{n},\\
     \varGamma \cap F|_{l}
                 & = {\ccO_l}(2) \oplus {\ccO_l}(1)^{n-1} \oplus {\ccO_l}^{n-1} = (F|_l)^{\ge 0}
   \end{align*}
    and if $z \in l$ is a smooth point of $D_x$, then $T_z D_x = {\varGamma}_z$.
   \end{prop}
   \noprf

\section{Duality}
\label{section-duality}

An effective divisor $\Delta$ on $X$ is an element of divisor group
(and thus a positive integral combination of codimension $1$ subvarieties of $X$)
and also a point in the projective space $\PP(H^0 \ccO_X(\Delta))$ or a hyperplane
in $\PP(H^0 \ccO_X(\Delta)^*)$.
In this section we will constantly interchange these three interpretations of $\Delta$.
In order to avoid confusion we will write:
\begin{itemize}
  \item $\divisor{\Delta}$ to mean the divisor on $X$;
  \item $\point{\Delta}$ to mean the point in $\PP(H^0 \ccO_X(\Delta))$
                                                     or in a fixed linear subsystem.
  \item $\hyperplane{\Delta}$ to mean the hyperplane in $\PP(H^0 \ccO_X(\Delta)^*)$
                                                     or in dual of the fixed subsystem.
\end{itemize}

In \S\ref{section-define-Dx} we have defined $D \subset X \times X$,
which we now view as a family of divisors on $X$ parametrised by $X$.
Since the Picard group of $X$ is discrete and $X$ is smooth and connected,
it follows that all the divisors $D_x$ are linearly equivalent.
Thus let $E \simeq L^{\otimes k}$ be the line bundle $\ccO_X(D_x)$.
Consider the following vector space ${\left\langle D \right\rangle} \subset H^0(E)$:
\[
  {\left\langle D \right\rangle} := \sspan \set{s_x : x \in X} \text { where $s_x$ is a section of $E$ vanishing on $D_x$.}
\]
Hence $\PP {\left\langle D \right\rangle}$ is the linear system spanned by all the $D_x$.

Further, consider the map
\[
  \phi: X \to \PP {\LinSyst^*}
\]
determined by the linear system ${\left\langle D \right\rangle}$,
i.e.~mapping point $x \in X$ to the hyperplane in $\PP {\left\langle D \right\rangle}$
consisting of all divisors containing $x$.
\begin{rem}\label{rem-phi-is-regular}
Note that $\phi$ is regular, since for every $x \in X$ there exists $w \in X$,
such that $x \notin D_w$
(or equivalently, $w \notin D_x$).
\end{rem}
Since $E$ is ample, it must intersect every curve in $X$ and hence
$\phi$ does not contract any curve. Therefore $\phi$ is finite to one.

\begin{prop} \label{prop-homogeneous-iff-k-eq-1}
  If $X$ is an adjoint variety, then $k=1$, i.e.~$E \simeq L$.
  If $k=1$ and the automorphism group of $X$ is reductive,
  then $X$ is isomorphic to an adjoint variety.
\end{prop}
\begin{prf}
  If $X$ is the adjoint variety of $G$, and $x\in X$, then $D_x$
  is the hyperplane section of $X \subset \PP(\gotg)$
  by Proposition~\ref{prop-what-is-Dx-for-X-homogeneous}.

  If $k=1$ and the automorphism group of $X$ is reductive,
  since $\phi$ is finite to one, we can apply
  Beauville Theorem \cite{beauvillefano}.
  Thus $X$ is isomorphic to an adjoint variety.
\end{prf}

\subsection{Dual map}\label{section-dual-map}

In algebraic geometry it is standard to consider maps
determined by linear systems (such as $\phi$ defined above).
However in our situation, we also have another map determined by
the family of divisors $D$.
Namely:
\begin{align*}
  \psi\colon X & \to \PP {\left\langle D \right\rangle} \\
             x & \mapsto \point{D_x}.
\end{align*}
So let $\ccS \subset \ccO_X \otimes {\LinSyst^*} \simeq  X  \times {\LinSyst^*}$
be the pullback under $\phi$ of the universal hyperplane bundle, i.e.
the corank 1 subbundle
such that the fibre of $\ccS$ over $x$
is $\hyperplane{D_x} \subset {\LinSyst^*}$.
We note that $\PP(\ccS)$  is both a projective space bundle on $X$
 and also it is a divisor on $X\times \PP {\LinSyst^*}$.
Also $D= (\id_X \times \phi)^* \PP(\ccS)$ as divisors.

We can also consider the line bundle dual to the cokernel of $\ccS \to \ccO_X \otimes {\LinSyst^*}$,
i.e.~the  subbundle ${\ccS}^{\perp} \subset  \ccO_X \otimes {\left\langle D \right\rangle}$.
This line subbundle determines section $X \to X\times \PP{\left\langle D \right\rangle}$,
where $x \mapsto (x , \point{D_x})$.
So $\psi$ is the composition of the section and the projection:
\[
  X \to X\times \PP{\left\langle D \right\rangle} \to \PP{\left\langle D \right\rangle}.
\]

Every map to a projective space is determined by some linear system.
We claim the $\psi$ is determined by ${\left\langle D \right\rangle}$,
precisely the system that defines $\phi$
and thus that there is a natural linear isomorphism between
$\PP {\left\langle D \right\rangle}$ and $\PP {\LinSyst^*}$.

\begin{prop}
We have $\psi^* \ccO_{\PP {\left\langle D \right\rangle}}(1) \simeq E$ and
the linear system cut out by hyperplanes
\[
 \psi^* H^0 \left( \ccO_{\PP {\left\langle D \right\rangle}}(1) \right) := \set{\psi^* s : s \in {\LinSyst^*}} \subset H^0(E)
\]
 is equal to ${\left\langle D \right\rangle}$.
\end{prop}

\begin{prf}
  For fixed $x\in X$ let $\phi(x)^{\perp} \subset \PP {\left\langle D \right\rangle}$ be the hyperplane dual to
  $\phi(x) \in \PP {\LinSyst^*}$.
  To prove the proposition it is enough to prove
  \begin{equation} \label{equ-pullback-by-psi-is-Dx}
    \psi^*(\phi(x)^{\perp}) = \divisor{D_x}.
  \end{equation}
  Since we have the following symmetry property of $D$:
  \[
     x \in D_y \iff y \in D_x,
  \]
  the set theoretic version of \eqref{equ-pullback-by-psi-is-Dx}
  follows easily:
  \[
     y \in \psi^*(\phi(x)^{\perp})
      \iff \psi(y) \in \phi(x)^{\perp}
      \iff \hyperplane{D_y} \ni \phi(x)
      \iff D_y \ni x.
  \]
  However, in order to prove the equality of divisors in \eqref{equ-pullback-by-psi-is-Dx}
  we must do a bit more of gymnastics, which translates the equivalences above into
  local equations.
  The details are below.

  The pull back of $\phi(x)^{\perp}$ by the projection $X\times \PP{\left\langle D \right\rangle} \to \PP{\left\langle D \right\rangle}$ is just
  $X\times \phi(x)^{\perp}$.
  Then the pull-back of the product by the section $X \to X\times \PP{\left\langle D \right\rangle}$
  associated to ${\ccS}^{\perp}$ is just the subscheme of $X$ defined by
  $\set{y \in X \mid ({\ccS}^{\perp})_y \subset  \phi(x)^{\perp}}$
  (locally, this
  is just a single equation: the spanning section of ${\ccS}^{\perp}$
  satisfies the defining equation of $\phi(x)^{\perp}$).
  But this is clearly
  equal to the dual equation $\set{y \mid \PP(\ccS_y) \ni \phi(x) }$.
  If we  let $\rho_x$ be the section
  \begin{align*}
    \rho_x\colon X&\to X \times X \\
    \rho_x (y) & :=  (y,x)
  \end{align*}
  then we have:
  \[
    \psi^*(\phi(x)^{\perp}) = {\rho_x}^* \circ (\id_X \times \phi)^* (\PP(\ccS)) = {\rho_x}^* (D)
    = \divisor{D_x}
  \]
  as claimed.
\end{prf}

Thus we have a canonical linear isomorphism
$f\colon \PP {\LinSyst^*} \to \PP {\left\langle D \right\rangle}$ giving rise to the following commutative diagram:
\begin{equation} \label{equ-duality-diagram}
              \xymatrix@C1.5pc@R0.5pc{
              & & & \PP {\LinSyst^*} \ar[dd]^{\simeq} \\
              X\ar[urrr]^{\phi} \ar[drrr]^{\psi} && \\
              & & & \PP {\left\langle D \right\rangle}.}
\end{equation}
We will denote the underlying vector space isomorphism ${\LinSyst^*} \to {\left\langle D \right\rangle}$
(which is unique up to scalar) with the same letter $f$.
The choice of $f$
 combined with the canonical
pairing ${\left\langle D \right\rangle} \times {\LinSyst^*} \to \CC$,
determines a non-degenerate bilinear form
$B: {\left\langle D \right\rangle} \times {\left\langle D \right\rangle} \to \CC$,
with the following property:
\begin{equation}\label{equ-defining-h}
  B(\phi(x),\phi(y)) =0 \iff (x,y) \in D \iff x \in D_y \iff y \in D_x.
\end{equation}
\begin{prop}\label{prop-Killing-form-for-homogeneous}
  If $X$ is the adjoint variety of $G$,
  then ${\left\langle D \right\rangle} = H^0(L) \simeq \gotg$ and $B$ is (up to scalar) the Killing form on $\gotg$.
\end{prop}
\begin{prf}
  Follows immediately from Proposition~\ref{prop-what-is-Dx-for-X-homogeneous}
  and Equation \ref{equ-defining-h}.
\end{prf}

\begin{cor}
          \label{cor_when_D_x_equals_D_y}
  $\phi(x) = \phi(y)$ if and only if $D_x = D_y$.
\end{cor}

\begin{prf}
  It is immediate from the definition of $\psi$
  and from Diagram \eqref{equ-duality-diagram}.
\end{prf}

\subsection{Symmetry}

Note that $B$ has the
property that for $x\in X$,
\[
  B(\phi(x),\phi(x))=0
\]
(because $x\in D_x$).

\begin{prop}
              \label{h_is_sym_or_skew}
  The bilinear form $B$ is either symmetric or skew-symmetric.
\end{prop}

\begin{prf}
  Consider two linear maps ${\left\langle D \right\rangle} \to {\LinSyst^*}$:
  \[
    \alpha(v):= B(v, \cdot) \ \text{ and } \  \beta(v):= B( \cdot,v).
  \]
  If $v= \phi(x)$ for some $x \in X$, then
  \[
    \ker\bigl(\alpha(v)\bigr)
    = \sspan\Bigl(\ker\bigl(\alpha(v)\bigr) \cap \phi(X)\Bigr)
    = \sspan\bigl(\phi(D_x)\bigr)
  \]
  and analogously $\ker(\beta(v)) = \sspan(\phi(D_x))$.
  So  $\ker(\alpha(v)) = \ker(\beta(v))$
  and hence $\alpha(v)$ and $\beta(v)$ are proportional.
  Therefore  there exists a function $\lambda: X \rightarrow \CC$ such that:
  \[
    \lambda(x) \alpha(\phi(x)) = \beta(\phi(x)).
  \]

  So for every $x,y \in X$ we have:
  \[
     B(\phi(x),\phi(y))  =
    \lambda(x) B(\phi(y),\phi(x)) =
    \lambda(x)\lambda(y) B (\phi(x),\phi(y))
  \]
  and hence:
  \[
    \forall {(x,y) \in X\times X \setminus D}  \qquad \lambda(x)\lambda(y) = 1.
  \]
  Taking three different points we see that $\lambda$ is constant and $\lambda \equiv \pm 1$.
  Therefore $\pm \alpha(\phi(x)) = \beta(\phi(x))$
  and by linearity this extends to
  $\pm \alpha = \beta$ so $B$ is either symmetric or skew-symmetric as stated in the proposition.
\end{prf}

\begin{example}
  If $X$ is one of the adjoint varieties, then $B$ is symmetric (because the Killing form is symmetric).
\end{example}

\begin{rem}
  Consider $\PP^{2n+1}$ with a contact structure arising from a symplectic form $\omega$ on $\CC^{2n+2}$.
  Recall, that this homogeneous contact Fano manifold does not satisfy our assumptions,
  namely, its Picard group is not generated by the equivalence class of $L$
  --- in this case $L \simeq \ccO_{\PP^{2n+1}}(2)$.
  However, \Wisniewski{} in \cite{wisniewski} considers also this generalised situation
  and defines $D_x$ to be the divisor swept by contact conics
  (i.e.~curves $C$ with degree of $L|_C = 2$)
  tangent to the contact distribution $F$.
  Then for the projective space $D_x$ is just the hyperplane perpendicular to $x$ with respect to $\omega$.
  And thus in this case ${\left\langle D \right\rangle} = H^0\left(\ccO_{\PP^{2n+1}}(1)\right)$
  and the bilinear form $B$ defined from such family of divisors would be proportional to $\omega$,
  hence skew-symmetric.
\end{rem}

\begin{prf}[ of Theorem~\ref{thm-exist-Dx-and-duality}]
  $D_x$ is a divisor by Theorem~\ref{thm-Dx-is-a-divisor}.
  $\phi$ is regular by Remark~\ref{rem-phi-is-regular}.
  $\psi$ is regular by \eqref{equ-duality-diagram}.
  The non-degenerate bilinear form $B$ is constructed in \S\ref{section-dual-map}.
  It is either symmetric or skew-symmetric by Proposition~\ref{h_is_sym_or_skew}.
  In the adjoint case $B$ is the Killing form by Proposition~\ref{prop-Killing-form-for-homogeneous}.
\end{prf}

\begin{cor}
If $B$ is symmetric, then $\psi(X) \subset \PP{\left\langle D \right\rangle}$ is contained in the quadric
 $B(v,v)=0$.
\end{cor}

\begin{cor}
If $x\in X$, then $\psi(C_x)$ is contained in a linear subspace of dimension
at most $\left\lfloor \frac{\dim {\left\langle D \right\rangle}}{2} \right\rfloor$.
\end{cor}

\begin{prf}
If $y,z \in C_x$, then $z\in D_y$, so $B(\psi(y), \psi(z))=0$.
Therefore $\sspan(\psi(C_x))$ is an isotropic linear subspace, which
cannot have dimension bigger than $\left\lfloor\frac{\dim {\left\langle D \right\rangle}}{2}\right\rfloor$.
\end{prf}

\section{Grading}\label{section-grading}
Suppose $X \subset \PP \gotg$ is the adjoint variety of $G$.
Assume further that a maximal torus and an order of roots in $\gotg$ has been chosen,
then $\gotg$ has a natural grading (see \cite[\S6.1]{landsbergmanivel02}):
\[
  \gotg = \gotg_{-2} \oplus \gotg_{-1} \oplus \gotg_{0} \oplus \gotg_{1} \oplus \gotg_{2}
\]
where:
\begin{enumerate}
  \item $\gotg_{0} \oplus \gotg_{1} \oplus \gotg_{2}$ is the parabolic subalgebra $\gotp$
        of $X$.
  \item $\gotg_{0}$ is the maximal reductive subalgebra of $\gotp$.
  \item for all $i \in  \set{-2,-1,0,1,2}$ the vector space $\gotg_{i}$ is a $\gotg_0$-module.
  \item $\gotg_{2}$ is the $1$-dimensional highest root space,
  \item $\gotg_{-2}$ is the $1$-dimensional lowest root space.
  \item The restriction of the Killing form to each
        $\gotg_{2} \oplus \gotg_{-2}$, $\gotg_{1} \oplus \gotg_{-1}$ and $\gotg_0$
	is non-degenerate,
	and the Killing form $B(\gotg_i, \gotg_j)$ is identically zero
	for $i \ne -j$.
  \item The Lie bracket on $\gotg$ respects the grading, $[\gotg_i, \gotg_j] \subset \gotg_{i+j}$
        (where $\gotg_k = 0$ for $k \notin \set{-2,-1,0,1,2}$).
\end{enumerate}

In fact the grading is determined by $\gotg_{-2}$ and $\gotg_{2}$ together with the geometry of $X$ only.
So let $X$ be as in Notation \ref{notation-with-assumptions-about-contact-mflds}
 and let $x$ and $w$ be two general points of $X$.
Define the following subspaces of ${\left\langle D \right\rangle}$:
\begin{itemize}
  \item ${\left\langle D \right\rangle}_{2}$ to be the $1$-dimensional subspace $\psi(x)$;
  \item ${\left\langle D \right\rangle}_{-2}$ to be the $1$-dimensional subspace $\psi(w)$;
  \item ${\left\langle D \right\rangle}_{1}$ to be the linear span of affine cone of $\psi(C_x \cap D_w)$;
  \item ${\left\langle D \right\rangle}_{-1}$ to be the linear span of affine cone of  $\psi(C_w \cap D_x)$;
  \item ${\left\langle D \right\rangle}_{0}$ to be the vector subspace of ${\left\langle D \right\rangle}$,
        whose projectivisation is:
        \[
          \bigcap_{y\in C_x \cup C_w} f(\hyperplane{D_y})
        \]
\end{itemize}

In the homogeneous case this is precisely the grading of $\gotg$.

\begin{prf}[ of Theorem~\ref{thm-grading-for-homogeneous}]
  First note that the classes of the 1-dimensional linear subspaces $\gotg_{2}$ and $\gotg_{-2}$
  are both in $X$ (as points in $\PP \gotg$).
  Moreover, they are a pair of general points, because the action of the parabolic subgroup $P < G$
  preserves $\gotg_2$ and moves freely $\gotg_{-2}$.
  This is because $\hat{T}_{[\gotg_{-2}]} X  = [\gotg_{-2} , \gotg] = [\gotg_{-2} , \gotp]$.

  So fix $x = [\gotg_2]$ and $w= [\gotg_{-2}]$.
  We claim the linear span of $C_x$ (respectively $C_w$)
  is just $\gotg_2 \oplus \gotg_1$ (respectively $\gotg_{-2} \oplus \gotg_{-1}$).
  To see that, the lines on $X$ through $x$ are
  in the intersection of $X$ and the projectivised tangent space $\PP(\hat{T}_x X) \subset \PP(\gotg)$.
  In fact this intersection is equal to $C_x$:
  if $y \ne x$ is a point of the intersection,
  then the line in $\PP \gotg$ through $x$ and $y$ intersects $X$ with multiplicity at least $3$,
  but $X$ is cut out by quadrics (see for instance \cite[\S10.6.6]{procesi_book}),
  so this line must be contained in $X$.
  Also $C_x$ is non-degenerate in $\PP(\hat{F}_x) \subset \PP(\hat{T}_x X)$.
  However $\hat{F}_x$ is a $\gotp$-invariant hyperplane in $\PP(\hat{T}_x X)$
  and the unique $\gotp$-invariant hyperplane in
  \[
    \hat{T}_x X = [ \gotg, \gotg_2] = [ \gotg_{-2}, \gotg_2] \oplus \gotg_{1} \oplus \gotg_2
  \]
  is
  \[
    \hat{F}_x = [\gotg_{-1} \oplus \gotg_0 \oplus \gotg_1 \oplus \gotg_2, \gotg_2 ] = \gotg_{1} \oplus \gotg_2.
  \]

  Further we have seen in Proposition~\ref{prop-what-is-Dx-for-X-homogeneous}
  that $D_x$ (respectively $D_w$)
  is the intersection of
   $\PP({\gotg_2}^{\perp_{B}}) = \PP(\gotg_2 \oplus \gotg_1 \oplus \gotg_0 \oplus \gotg_{-1})$ and $X$
  (respectively $\PP(\gotg_{-2} \oplus \gotg_{-1} \oplus \gotg_0 \oplus \gotg_{1})$ and $X$).
  Equivalently, $f(\hyperplane{D_x}) = \PP(\gotg_2 \oplus \gotg_1 \oplus \gotg_0 \oplus \gotg_{-1})$.
  Thus:
  \[
      C_x \cap D_w = C_x \cap f(\hyperplane{D_w})
    = C_x \cap \PP(\gotg_{-2} \oplus \gotg_{-1} \oplus \gotg_0 \oplus \gotg_{1})
    = C_x \cap \PP(\gotg_{1}).
  \]
  $C_x \cap \PP(\gotg_{1})$ is non-degenerate in $\PP(\gotg_{1})$,
  thus ${\left\langle D \right\rangle}_{1} = \gotg_1$ and analogously ${\left\langle D \right\rangle}_{-1} = \gotg_{-1}$.

  It remains to prove ${\left\langle D \right\rangle}_{0} = \gotg_0$.
  \begin{align*}
    \PP{\left\langle D \right\rangle}_{0} & = \bigcap_{y\in C_x \cup C_w} f(\hyperplane{D_z})\\
                    & = (C_x \cup C_w)^{\perp_{B}} \\
                    & = \PP(\gotg_2 \oplus \gotg_1 \oplus \gotg_{-1} \oplus \gotg_{-2})^{\perp_{B}} \\
                    & = \PP(\gotg_0).
  \end{align*}
\end{prf}

We also note the following lemma in the homogeneous case:
\begin{lemma}\label{lemma-homogeneous-then-X-cap-g1-subset-Cx}
  If $X$ is the adjoint variety of $G$,
  then
  \[
    X \cap \PP(\gotg_1) \subset C_x
  \]
  where $x$ is the point of projective space corresponding to $\gotg_2$.
\end{lemma}

\begin{prf}
  Suppose $y \in X \cap \PP \gotg_1$
  and let $l \subset \PP \gotg$ be the line through $x$ and $y$.
  Note that $l \subset \PP(\gotg_1 \oplus \gotg_2)$
  Since $\gotg_1 \oplus \gotg_2 \subset [\gotg,\gotg_2] = \hat{T}_x X$,
  hence $l \cap X$ has multiplicity at least $2$ at $x$.
  Thus $l\cap X$ has degree at least $3$ and since $X$ is cut out by quadrics,
  $l$ is contained in $X$.
\end{prf}

\section{Cointegrable subvarieties}
\label{section-cointegrable}

\begin{defin}
  A subvariety $\Delta \subset X$ is \emph{$F$-cointegrable}
  if  $T_x \Delta \cap F_x \subset F_x$ is a coisotropic subspace for general point $x$ of each
  irreducible component of $\Delta$.
\end{defin}
Note that this is equivalent to the definition given in \cite[\S{}E.4]{jabu_dr}
--- this follows from the local description of the symplectic form
    on the symplectisation of the contact manifold (see \cite[(C.15)]{jabu_dr}).

Clearly, every codimension $1$ subvariety of $X$ is $F$-cointegrable.

Assume $\Delta \subset X$ is a subvariety of pure dimension, which is $F$-cointegrable
 and let $\Delta_0$
be the locus where $T_x \Delta \cap F_x \subset F_x$ is a coisotropic subspace
of dimension $\dim \Delta-1$.
We define the \emph{$\Delta$-integrable distribution $\Delta^{\perp}$}
 to be the distribution defined over $\Delta_0$ by:
\[
  {\Delta^{\perp}}_x:=\left(T_x \Delta \cap F_x \right)^{\perp_{\ud \theta}} \subset F_x
\]
We say an irreducible subvariety $A \subset X$ is \emph{$\Delta$-integral}
if $A \subset \Delta$, $A \cap \Delta_0 \ne \emptyset$,
and $T A \subset \Delta^{\perp}$ over the smooth points of $A \cap \Delta_0$.

\begin{lemma}\label{lemma-integral-is-unique}
  Let $A_1$ and $A_2$ be two irreducible $\Delta$-integral subvarieties.
  Assume $\dim A_1 =\dim A_2 = \codim_X \Delta$.
  Then either $A_1=A_2$ or $A_1 \cap A_2 \subset \Delta \setminus \Delta_0$.
\end{lemma}
\noprf

\begin{thm}
  Consider a general point $x \in X$.
  Then:
  \begin{enumerate}
     \item \label{item-Dx-cointegrable}
        $D_x$ (as reduced, but possibly not irreducible subvariety of $X$) is $F$-cointe\-grable.
     \item \label{item-lines-are-Dx-integrable}
        For general $y$ in any of the irreducible components of $C_x$ all lines through $y$
        are $D_x$-integral.
     \item \label{item-unique-presentation-as-sum-of-2-lines}
        For general $z$ in any of the irreducible components of $D_x$
        the intersection $C_x \cap C_z$ is a unique point
        and the chain of two lines connecting $x$ to $z$ is unique.
  \end{enumerate}
\end{thm}

\begin{prf}
   Part \ref{item-Dx-cointegrable} is immediate, since $D_x$ is a divisor,
   by Theorem \ref{thm-Dx-is-a-divisor}.

   To prove part \ref{item-lines-are-Dx-integrable} let $l$ be a line through $y$.
   Then by Proposition \ref{prop-tangent-to-Dx}:
    \[
      T_z D_x \cap F_{z} = (F|_{l})^{\ge 0}
    \]
    and  for general $z \in l$ we have $(T_z D_x \cap F_{z})^{\perp_{\ud \theta_z}} \subset F_{z}$ is the $\ccO(2)$ part,
    i.e.~the part tangent to $l$. So $l$ is $D_x$-integral as claimed.

    To prove \ref{item-unique-presentation-as-sum-of-2-lines},
    let $U \subset X$ be an open dense subset of points $u \in X$ where two different lines through $u$
    do not share the tangent direction and do not meet in any other point.
    Note that since $x$ is a general point, $x \in U$
    and thus each irreducible component of $C_x$ and $D_x$ intersects $U$.
    Thus generality of $z$  implies that $z \in U$
    and thus each irreducible component of $C_z$ and $D_z$ intersects $U$.
    Also $C_x \cap C_z$ intersects $U$. So fix $y \in C_x \cap C_z \cap U$.

    By \ref{item-lines-are-Dx-integrable} and Lemma \ref{lemma-integral-is-unique}
    the line $l_z$ through $z$ which intersects $C_x$ is unique.
    In the same way let $l_x$ be the unique line through $x$ intersecting $C_z$.
    Thus
    \[
      C_x \cap C_z = l_x \cap l_z.
    \]
    In particular, $y \in  l_x \cap l_z$.
    But since $y \in U$ the intersection $l_x \cap l_z$ is just one point and therefore:
    \[
      C_x \cap C_z = \set{y}.
    \]
\end{prf}

As a consequence of part \ref{item-unique-presentation-as-sum-of-2-lines} of the theorem
the surjective map $\pi_{13}\colon C^3 \to D$ is birational.
Thus consider the inverse rational map $D \dashrightarrow C^3$ and compose it with
the projection on the middle coordinate $\pi_2: C^3 \to X$.
We define the composition to be the \emph{bracket map}:
\[
  [\cdot, \cdot]^D \colon D \dashrightarrow C^3 \stackrel{\pi_2}{\to} X.
\]
In this setting, for $(x,z) \in D$, one has $[x,z]^D = y = C_x \cap C_z$,
 whenever the intersection is just one point.

\begin{thm}
    If $X$ is the adjoint variety of $G$, then the bracket map defined above agrees with the Lie bracket on
    $\gotg$, in the following sense:
    Let  $\xi, \zeta \in \gotg$ and set $\eta := [\xi,\zeta]$ (the Lie bracket on $\gotg$).
    Denote by $x$, $y$ and  $z$ the projective classes in $\PP \gotg$
    of $\xi$, $\eta$ and $\zeta$ respectively.
    If $x \in D_z$ and $\eta \ne 0$,
    then the bracket map satisfies $[x,z]^D=y$.
\end{thm}

\begin{prf}
    It is enough to prove the statement for a general pair $(x,z) \in D$.
    Suppose further $w \in C_z$ is a general point.
    Then the pair $(x,w) \in X \times X$ is a general pair.
    Thus by Proposition~\ref{thm-grading-for-homogeneous},
    we may assume $\xi \in \gotg_2$ and $\zeta \in \gotg_{-1}$.
    The restriction of the Lie bracket to $[\xi , \gotg_{-1}]$
    determines an isomorphism  $\gotg_{-1} \to \gotg_{1}$
    of $\gotg_0$-modules.
    In particular the minimal orbit $X \cap \PP\gotg_{-1}$
    is mapped onto $X \cap \PP\gotg_{1}$ under this isomorphism.
    In particular $y \in X \cap \PP\gotg_{1} \subset C_x$
    (see Lemma~\ref{lemma-homogeneous-then-X-cap-g1-subset-Cx}).
    Analogously $y \in C_z$, so $y \in C_x \cap C_z$.
\end{prf}

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Q: Initializing Firebase Admin via Environment Variables without storing serviceAccount.json I am trying to initialize firebase-admin on my Flask API. Based off the documentation, the method initialize_app can take a dictionary as an input as seen here:
https://github.com/firebase/firebase-admin-python/blob/6d826fd15c87db62f7501f569b6e0a762d17a05e/firebase_admin/credentials.py#L85
That said, I structured my code as follows:
import firebase_admin
from firebase_admin import auth, credentials

...

firebase_admin.initialize_app({ \
    credentials.Certificate({ \
        "type": "service_account", \
        "project_id": os.environ.get('FIREBASE_PROJECT_ID'), \
        "private_key_id": os.environ.get('PRIVATE_KEY_ID'), \
        "private_key": os.environ.get('FIREBASE_PRIVATE_KEY').replace('\\n', '\n'), \
        "client_email": os.environ.get('FIREBASE_CLIENT_EMAIL'), \
        "client_id": os.environ.get('CLIENT_ID'), \
        "auth_uri": os.environ.get('AUTH_URI'), \
        "token_uri": os.environ.get('TOKEN_URI'), \
        "auth_provider_x509_cert_url": os.environ.get('AUTH_PROVIDER_X509_CERT_URL'), \
        "client_x509_cert_url": os.environ.get('CLIENT_X509_CERT_URL'), \
    }),
})

Now, I'm getting this error:

ValueError: Illegal Firebase credential provided. App must be initialized with a valid credential instance.

I would ideally want to set up the application like this as I would prefer not to store the serviceAccount.json on the cloud. Many of the examples I find are simply doing what I don't want to do.
What am I missing here?
Edit:
I am using the standard export <property_name>="..." on my Mac OS terminal which I presume is the same as Linux environment. As a result, os.environ.get(<property_name>) gets the corresponding value.
For reference:
I am trying to do the same thing as this:
https://www.benmvp.com/blog/initializing-firebase-admin-node-sdk-env-vars/
But with Python
Edit:
Looking at the source code here:
https://github.com/firebase/firebase-admin-python/blob/6d826fd15c87db62f7501f569b6e0a762d17a05e/firebase_admin/__init__.py#L209
It appears that the exception is being thrown here. However, in my Flask API, I have the following:
cert = { \
        "type": "service_account", \
        "project_id": os.environ.get('FIREBASE_PROJECT_ID'), \
        "private_key_id": os.environ.get('PRIVATE_KEY_ID'), \
        "private_key": os.environ.get('FIREBASE_PRIVATE_KEY').replace('\\n', '\n'), \
        "client_email": os.environ.get('FIREBASE_CLIENT_EMAIL'), \
        "client_id": os.environ.get('CLIENT_ID'), \
        "auth_uri": os.environ.get('AUTH_URI'), \
        "token_uri": os.environ.get('TOKEN_URI'), \
        "auth_provider_x509_cert_url": os.environ.get('AUTH_PROVIDER_X509_CERT_URL'), \
        "client_x509_cert_url": os.environ.get('CLIENT_X509_CERT_URL'), \

print(type(credentials.Certificate(cert)), isinstance(credentials.Certificate(cert), credentials.Certificate), isinstance(credentials.Certificate(cert), credentials.Base))

To which the output is:

<class 'firebase_admin.credentials.Certificate'> True True

This doesn't make sense.. Since the following block:
if not isinstance(credential, credentials.Base):

Runs if isinstance(credential, credentials.Base) is false. But I have the values as true.

A: Thanks to, smac89, for pointing me in the right direction.
The change is very minor to make it work:
firebase_admin.initialize_app({ \
    credential=credentials.Certificate({ \  # <------- This change
        "type": "service_account", \
        "project_id": os.environ.get('FIREBASE_PROJECT_ID'), \
        "private_key_id": os.environ.get('PRIVATE_KEY_ID'), \
        "private_key": os.environ.get('FIREBASE_PRIVATE_KEY').replace('\\n', '\n'), \
        "client_email": os.environ.get('FIREBASE_CLIENT_EMAIL'), \
        "client_id": os.environ.get('CLIENT_ID'), \
        "auth_uri": os.environ.get('AUTH_URI'), \
        "token_uri": os.environ.get('TOKEN_URI'), \
        "auth_provider_x509_cert_url": os.environ.get('AUTH_PROVIDER_X509_CERT_URL'), \
        "client_x509_cert_url": os.environ.get('CLIENT_X509_CERT_URL'), \
    }),
})

The method signature for initialize_app is:
def initialize_app(credential=None, options=None, name=_DEFAULT_APP_NAME):

By not specifying the input to be credential, the input was not read correctly. Guess I learnt something new about Python default values. :P

A: This answer was quite useful for me with a related problem. I was ok setting the key content as a secret environment variable. However, I wasn't sure how to set it without constructing the dictionary. In the end, I could just json.loads() the content of the secret environment key, and it worked. I thought it was worth sharing this:
import json, os
import firebase_admin
from firebase_admin.credentials import Certificate

key_dict = json.loads(
    os.environ["FIREBASE_SERVICE_ACCOUNT_CREDENTIALS_FROM_ENV"]
)

firebase_admin.initialize_app(Certificate(key_dict))

A: Here are a few things to consider when loading a service account with environment variables. Notably if you get the error Caused by: "No key could be detected."

*

*When copy/pasting the private key, you want to do it directly from the serviceAccount.json file to the file that contains your environment configuration to avoid bad formatting.

*If you are using a .yml/Jinja2 Ansible variable, you might want to try modifying the private key's \n by \\n (inspect the output of your private key)
To make sure the key is actually valid, I used pycryptodome to validate the key. While it is not necessary, it is still good to have.
Here's an example:

import os
import firebase_admin
from firebase_admin import credentials

from app.core.config import settings
from Crypto.PublicKey import RSA

def init_firebase():
    private_key_raw = os.environ.get("FIREBASE_PRIVATE_KEY")
    private_key = RSA.importKey(private_key_raw)
    info = {
        # firebase credential details here
        "private_key": private_key.exportKey().decode("utf-8"),
    }
    creds = credentials.Certificate(info)
    firebase_admin.initialize_app(creds)

Note that you might encounter some solutions advising you to do something akin to credentials.Certificate(json.loads(json.dumps(info))), telling you that Certificate does not support anything else than JSON files.
However as of version 5.2.0 (my current version), the source file credentials.py of the firebase_admin package specifies that dict type is supported:

To instantiate a credential from a certificate file, either specify
the file path or a dict representing the parsed contents of the file.

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