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Using the theme of illusion or reality, do a comparison of Jay Gatsby to Chip Carraway. How does one pursue a dream as the other pursuit of truth Dissertation Example
Using the theme of illusion or reality, do a comparison of Jay Gatsby to Chip Carraway. How does one pursue a dream as the other pursuit of truth Dissertation Example GATSBY In S. Scott Fitzgerald's American ic The Great Gatsby, two distinctive characters, good friends, can be when compared and in comparison. Nick Carraway, the story's shy, logical, unassuming narrator, is in many ways the opposite for his companion and neighbor Jay Gatsby, the brash, romantic, flashy millionaire. Each characters demonstrate a key theme inside the story, likewise: the concept of the illusion compared to reality. Our present-day research enquiry compares the particular motivations about Jay Gatsby, a character who may be searching for ideal, to Computer chip Carraway, a personality who is looking for the truth. By looking at the ways that characters happen to be defined along with illustrated throughout works associated with fiction, tips can be desired as to their similarities and even differences.
Fitzgerald employs a good supportive device in the narrative character associated with Nick
Carraway, Jay's buddy and tool, who perceives that 'life is much more properly looked at from the single display, after all' (Fitzgerald, 3). Gatsby is normally defined by means of his efforts to become less of an man or women and more associated with a representation of his idealized sort of gleaming society: he / she wants to dwell a dream, and possesses devoted his / her life to that romantic preferred. He wishes class to go along with his success, so your dog constructs a fictional past to go along with it. Compared, Nick Carraway is very straightforward about the meager experiencing as a attachment salesman, great position at West Egg cell. Though typically alone, Shape j Gatsby decisions through the fresh towards the goal to become a visible perhaps the external environment in which he could be accepted since both rich and dependable. His pursuit is indomitable due to a, unwavering beliefs in an best.
Jay's morals is in the earth to represent her dream along with hold the dog up. This individual completely
thinks that Daisy will provide your man with the graphic of his or her happiness, which will he has
potentially mistaken pertaining to happiness alone. Jay angles all of the energies with providing a faç ade
that could lure Daisy to your ex, as he or she believes in the exact utter fact that facades. Nick, as he will be courting Test Baker, is definitely ultimately even more rational in addition to realistic. Actually his outlines of Test are very another little known and overlooked fact and physical: 'Her system asserted once more with a exploratory movement involving her leg, and your lover stood up' (Fitzgerald, 12). In their marriage, what is at stake is the real truth of a potential and a life together, not the embodiment of getting some sort of dream or illusion.
Around Gatsby's case, self-discovery are at least superficially an constituent, and therefore
most likely doomed, process of manufacture in his dreams. Fitzgerald sees the last as subtractive, and ideas progress simply because additive, because evinced just by his account of motorboats against the existing. By the end, almost all his thinking has generated Gatsby's loss, and the book closes utilizing Nick musing self-consciously that humanity continues on, 'boats resistant to the current, debut back continually into the past' (Fitzgerald 189). Subtraction includes Gatsby directly into confusion, because he is an unrealistic dreamer. One can possibly see the person as Jay does: while something that ought to be hidden pertaining to external scenarios to occur. There exists
nowhere to cover: Jay is often a wealthy individual who has never construct a proper explanation
intended for his money; he has never live up to typically the responsibilities it entails. Chips, on the other hand, is more realistic.
The character of The author Gatsby, who represents wealth as a express of being that
arises from doubledealing, is contrasted against the underperforming passivity of your character connected with
Nick Carraway, who, for the reason that representative of some sort of wealthy family members, personifies a lot more
conservative together with even-keeled vision of flux. It would be an oversight to make
often the assumption, nevertheless that Y. Scott Fitzgerald is trying to state that the ones whose riches
is in order to from an started patriarchy tend to be innately more stable versus the self-made
(or self-constructed) prosperous: we see inside character with Tom Buchanan that the stagnancy of
continued wealth can make conditions wherein equally immense and negative
irresponsibility occurs.
Less very sensitive readers might not exactly pick up on the point that F. Martin Fitzgerald methods for disparage Gatsby's kind of huge selection, because the plot websites to write essays voice about Nick Carraway could be interpreted to lead someone into a talk about of an over-objectified and residual desire to justification Gatsby's flaws. This image could next be blended with the specific and a bit blasé information of the character's material success to lead to a serious sense that absolutely nothing ironic is happening at all. The writer Gatsby can be ostensibly a classic American hero, a rags-to-riches product for bootstrap-pulling and grit, nevertheless the Great Gatsby has the unique, confusing, in addition to certainly one of a kind position throughout American materials of being a deep world wide greatly focused on superficiality.
All round, in The Excellent Gatsby we see a complicated and also involving narrative
retrospective whereby Nick Carraway, the narrator, is given hindsight with which for you to
reflect self-consciously on the functions described simply by him. This leads him the realist, and even Gatsby often the dreamer. This time of watch is useful for the author because it lets the dog pass provide judgments regarding events which occurred in the previous while nonetheless providing a appearance of prompt dialogue. Nick is the perfect lone associated with a lout bourgeois patriarchy, 'making the name' in an urban vicinity. He is thorough and natural about the prospects, and features surrounded himself with as well as social aid. Gatsby, on the contrary, is the perfect self-made arcano, making her life storyline up as it goes on. He is the loner, along with a dreamer.


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Why the Nirav Modi fraud is much more than just a fraud
During the course of the last one week, the hottest news-story in India has been that of a jeweller named Nirav Modi, allegedly defrauding one of India's largest government owned banks, the Punjab National Bank (PNB).
PNB is India's second largest government owned bank (with assets of around Rs 7,203 billion ($111.7 billion, assuming $1 = Rs 64.5) as on March 31, 2017). The total amount of the fraud has been estimated to be at $1.8 billion (or around Rs 114 billion). News report suggest that Modi (no relation to the current prime minister of India Narendra Modi) fled the country in early January. His immediate family also left India, during the course of the month.
Nirav Modi is believed to be holed up in a luxury hotel in New York and was last seen in Davos, as a part of a business delegation which got a picture clicked with the prime minister Narendra Modi. Before Nirav Modi, Vijay Mallya, another businessman, who hasn't repaid loans worth Rs 90 billion ($1.4 billion) due to Indian banks, fled the country.
The latest fraud basically involves PNB guaranteeing loans issued to Nirav Modi by issuing a letter of undertaking (LOU). Every time a loan became due, Nirav Modi got PNB to open another LOU equivalent to the loan amount plus the interest that was due on it. The money from the new LOU was used to pay off the loan and the interest due on the previous LOU. In the process, Modi never repaid the loan.
Currently, it is being suggested that he was helped in the process by two employees of PNB. That such a huge Ponzi scheme could be run without the top or the middle management of the bank knowing about it, is a little difficult to believe.
Thus, Modi managed to operate a Ponzi scheme, with money from the new LOU being used to pay off the previous one. Of course, like all Ponzi schemes, Nirav Modi's scheme collapsed as well. And before the authorities came after him, he left the country, along with his family.
How does Nirav Modi's fraud look in light of the other frauds that Indian banks face? In July 2017, the ministry of finance had shared some interesting data in this context.
Between the years 2012-2013 and 2016-2017, the banks in the country had seen a total number of 22,949 frauds, with total losses to banks amounting to Rs 698 billion ($10.8 billion). The average loss on a fraud thus amounted to Rs 30.4 million ($0.47 million). The interesting thing here is that of the 78 banks on the list, PNB faced the highest losses when it came to frauds. Over the five-year period, the bank faced 942 frauds with losses of Rs 90 billion ($1.4 billion). The losses amounted to around 12.9% of the total losses faced by the Indian banks due to frauds.
In fact, the average loss for PNB due to frauds stood at Rs 95.5 million ($1.48 million), which was three times the total average of Rs 30.4 million. Also, more than that, PNB faced more frauds than the State Bank of India, the country's largest bank, with an asset base which is 4.6 times that of PNB.
What this tells us is that PNB's control systems were in bad shape and hence, the bank got defrauded significantly more than the other banks did. Having said that, the average fraud at PNB between 2012-2013 and 2016-2017 had cost the bank Rs 95.5 million. In Nirav Modi's case, the size of the fraud is around Rs 114 billion, which is much bigger than the size of the average fraud PNB has faced in the recent years.
What this tells us is that Nirav Modi's case is more than a petty bank fraud. It is basically more along the lines of a large bank loan default; which many of India's crony capitalists specialise in.
India's government owned banks have been facing a huge pressure of corporate loan defaults over the last few years. As of September 2017, the bad loans ratio of these banks stood at 13.5%. This basically means that of every Rs 100 of loans given by these banks, Rs 13.5 had been defaulted on. A bad loan is a loan which hasn't been repaid for a period of 90 days or more. The corporate default rate has been even higher.
Largely due to corporate loan defaults, the Indian banks have had to write off loans worth around Rs 2,500 billion ($38.8 billion) for the period of five years ending March 31, 2017. Nirav Modi's bank fraud will only add to this.
To keep these banks going, the government of India has to regularly keep infusing capital in them. In fact, an estimate made by The Times of India suggests that the government has infused Rs 2,600 billion ($40.3 billion) in the banks that it owns, over the last 11 years. Every rupee that goes into these banks is taken away from more important areas like agriculture, education, health, defence etc.
The reason why many Indian businessmen blatantly default on loans is because they know that given India's slow judicial system and their closeness to politicians, their chances of getting away with a loan default are very high. Nirav Modi is just a small part of this significant whole.
No wonder, former governor of the Reserve Bank of India, Raghuram Rajan, in a November 2014 speech had said that, India was a "country where we have many sick companies but no "sick" promoters".
A slightly different version of this column appeared on BBC.com on February 20, 2018.
Categories BBC, Ponzi Post navigation
For their size, Public Sector Banks Have Had Fewer Frauds Than Private Sector Ones.
The Nirav Modi Fraud Tells Us That the Business of Govt Should Not Be Business

redpajama_set_name: RedPajamaCommonCrawl


Something to look forward to: Samsung has announced it has started mass production of the world's first 32GB DDR4 memory designed for gaming laptops. Designed in the SoDIMM form factor and based on the 10nm-class process, the new modules bring a slew of benefits to portable gaming machines, including higher capacities, faster speeds, and more energy efficiency.
The modules double the capacity of Samsung's 16GB SoDIMMs based on 20nm-class 8-gigabit DDR4 that were introduced back in 2014, making them 11 percent faster and 39 percent more energy efficient.
Samsung says the 32GB SoDIMMs are made up of sixteen of its newest 16Gb DRAM chips to create a double-sided RAM module with eight chips mounted on the front and back, allowing gaming laptops to reach speeds of 2,666Mbps—an increase over the previous top SoDIMM speed of 2,400Mbps.
Power consumption—often one of the biggest issues when it comes to gaming noteboks—is also improved. Samsung claims that a 64GB laptop containing two of its 32GB modules consumes less than 4.6 watts in active mode and less than 1.4W when idle, which is a reduction in power usage compared to gaming-focused laptops equipped with 16GB modules.
Samsung didn't reveal any prices for its new SoDIMMs. It said the company is aggressively expanding its 10nm-class DRAM lineup, which includes 16Gb LPDDR4, 16Gb GDDR5, and 16Gb DDR4 for the mobile, graphics, PC, and server markets.


Fruits | Cedric Grolet | ISBN: | Kostenloser Versand für alle Bücher mit Versand und Verkauf duch Amazon. Oct 9, We tasted a lot of different fruits during our vacation in Colombia in Tsd. Abonnenten, 9 folgen, Beiträge - Sieh dir Instagram-Fotos und - Videos von true fruits (@truefruitssmoothies) an.
The ovules are fertilized in a process that starts with pollination , which involves the movement of pollen from the stamens to the stigma of flowers.
After pollination, a tube grows from the pollen through the stigma into the ovary to the ovule and two sperm are transferred from the pollen to the megagametophyte.
Within the megagametophyte one of the two sperm unites with the egg, forming a zygote , and the second sperm enters the central cell forming the endosperm mother cell, which completes the double fertilization process.
As the ovules develop into seeds, the ovary begins to ripen and the ovary wall, the pericarp , may become fleshy as in berries or drupes , or form a hard outer covering as in nuts.
In some multiseeded fruits, the extent to which the flesh develops is proportional to the number of fertilized ovules. In some fruits, especially simple fruits derived from an inferior ovary , other parts of the flower such as the floral tube, including the petals , sepals , and stamens , fuse with the ovary and ripen with it.
When such other floral parts are a significant part of the fruit, it is called an accessory fruit. Since other parts of the flower may contribute to the structure of the fruit, it is important to study flower structure to understand how a particular fruit forms.
Plant scientists have grouped fruits into three main groups, simple fruits, aggregate fruits, and composite or multiple fruits. Simple fruits can be either dry or fleshy, and result from the ripening of a simple or compound ovary in a flower with only one pistil.
Dry fruits may be either dehiscent they open to discharge seeds , or indehiscent they do not open to discharge seeds. Fruits in which part or all of the pericarp fruit wall is fleshy at maturity are simple fleshy fruits.
Types of simple, fleshy, fruits with examples include:. An aggregate fruit, or etaerio , develops from a single flower with numerous simple pistils.
The pome fruits of the family Rosaceae , including apples , pears , rosehips , and saskatoon berry are a syncarpous fleshy fruit, a simple fruit, developing from a half-inferior ovary.
Schizocarp fruits form from a syncarpous ovary and do not really dehisce , but rather split into segments with one or more seeds; they include a number of different forms from a wide range of families.
Aggregate fruits form from single flowers that have multiple carpels which are not joined together, i. Each pistil forms a fruitlet, and collectively the fruitlets are called an etaerio.
Four types of aggregate fruits include etaerios of achenes, follicles, drupelets, and berries. Ranunculaceae species, including Clematis and Ranunculus have an etaerio of achenes, Calotropis has an etaerio of follicles, and Rubus species like raspberry, have an etaerio of drupelets.
Annona have an etaerio of berries. The raspberry , whose pistils are termed drupelets because each is like a small drupe attached to the receptacle.
In some bramble fruits such as blackberry the receptacle is elongated and part of the ripe fruit, making the blackberry an aggregate-accessory fruit.
A multiple fruit is one formed from a cluster of flowers called an inflorescence. Each flower produces a fruit, but these mature into a single mass.
In the photograph on the right, stages of flowering and fruit development in the noni or Indian mulberry Morinda citrifolia can be observed on a single branch.
First an inflorescence of white flowers called a head is produced. After fertilization , each flower develops into a drupe, and as the drupes expand, they become connate merge into a multiple fleshy fruit called a syncarp.
Berries are another type of fleshy fruit; they are simple fruit created from a single ovary. Types include examples follow in the table below:.
Some or all of the edible part of accessory fruit is not generated by the ovary. Accessory fruit can be simple, aggregate, or multiple, i.
Seedlessness is an important feature of some fruits of commerce. Commercial cultivars of bananas and pineapples are examples of seedless fruits.
Some cultivars of citrus fruits especially grapefruit , mandarin oranges , navel oranges , satsumas , table grapes , and watermelons are valued for their seedlessness.
In some species, seedlessness is the result of parthenocarpy , where fruits set without fertilization. Parthenocarpic fruit set may or may not require pollination, but most seedless citrus fruits require a stimulus from pollination to produce fruit.
Seedless bananas and grapes are triploids , and seedlessness results from the abortion of the embryonic plant that is produced by fertilization, a phenomenon known as stenospermocarpy , which requires normal pollination and fertilization.
This dispersal can be achieved by animals, explosive dehiscence , water, or wind. Some fruits have coats covered with spikes or hooked burrs, either to prevent themselves from being eaten by animals , or to stick to the feathers, hairs, or legs of animals, using them as dispersal agents.
Examples include cocklebur and unicorn plant. The sweet flesh of many fruits is "deliberately" appealing to animals, so that the seeds held within are eaten and "unwittingly" carried away and deposited i.
Likewise, the nutritious, oily kernels of nuts are appealing to rodents such as squirrels , which hoard them in the soil to avoid starving during the winter, thus giving those seeds that remain uneaten the chance to germinate and grow into a new plant away from their parent.
Other fruits are elongated and flattened out naturally, and so become thin, like wings or helicopter blades, e. This is an evolutionary mechanism to increase dispersal distance away from the parent, via wind.
Other wind-dispersed fruit have tiny " parachutes ", e. Coconut fruits can float thousands of miles in the ocean to spread seeds.
Some other fruits that can disperse via water are nipa palm and screw pine. Many hundreds of fruits, including fleshy fruits like apple , kiwifruit , mango , peach , pear , and watermelon are commercially valuable as human food, eaten both fresh and as jams, marmalade and other preserves.
Fruits are also used in manufactured foods e. Many "vegetables" in culinary parlance are botanical fruits, including bell pepper , cucumber , eggplant , green bean , okra , pumpkin , squash , tomato , and zucchini.
Spices like allspice , black pepper , paprika , and vanilla are derived from berries. Fresh fruits are generally high in fiber , vitamin C , and water.
Because fruits have been such a major part of the human diet, various cultures have developed many different uses for fruits they do not depend on for food.
For food safety , the CDC recommends proper fruit handling and preparation to reduce the risk of food contamination and foodborne illness. Fresh fruits and vegetables should be carefully selected; at the store, they should not be damaged or bruised; and precut pieces should be refrigerated or surrounded by ice.
Fruits and vegetables should be kept separate from raw foods like meat, poultry, and seafood, as well as from utensils that have come in contact with raw foods.
Fruits and vegetables that are not going to be cooked should be thrown away if they have touched raw meat, poultry, seafood, or eggs.
Fruit allergies make up about 10 percent of all food related allergies. All fruits benefit from proper post harvest care, and in many fruits, the plant hormone ethylene causes ripening.
Therefore, maintaining most fruits in an efficient cold chain is optimal for post harvest storage, with the aim of extending and ensuring shelf life.
From Wikipedia, the free encyclopedia. For other uses, see Fruit disambiguation. Berry botany and Berry. The pome fruits, stone fruits and brambles are fruits of plants in Rosaceae.
Most gourds and many melons are regarded as culinary vegetables. The following are generally regarded as fruits:.
The accessory fruits , seed organs which are not botanically berries at all:. From Wikipedia, the free encyclopedia.
Blackberry , including many species and hybrids Cloudberry Loganberry Raspberry of several species Salmonberry Thimbleberry Wineberry.
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All of us at Master Plumbing and Mechanical would like to wish everyone a Happy Thanksgiving.
As a friendly reminder, now is time to winterize all outdoor faucets, hose connections, and irrigation systems. It is also important to unhook any and all hoses you may have connected. Removing hoses allows water to have someplace to go when it drains out. Although many homes have "frost free" outdoor faucets, the water still needs to be drained from them in order to not damage you pipes. The cold weather will freeze your faucets and once they thaw, you have a mess.
To drain your outdoor faucets, all you need to do is shut off their water supply and open them up. Since you've shut off the faucets' supply of water, they will just drain out any existing water that is there.
As always, please feel free to contact us if you have any questions at all! We are here to help!
This entry was posted on Tuesday, November 21st, 2017 at 4:52 pm. Both comments and pings are currently closed.


text:
\section{Introduction}
Originating in physics, contact geometry is a mathematical formulation of classical mechanics.
Contact geometry describes a geometric structure which appears in any constant energy hypersurface in the even-dimensional phase space of a mechanical system.
In mathematics, the concept of contact structure appears explicitly in the work of Sophus Lie, and implicitly perhaps much earlier.
By using a sheaf coefficient cohomology theory, Gray developed the idea and introduced a concept of {\it almost contact structure}~\cite{Gr}.
He considered the deformation of a global contact structure in the terminology of homological algebra.
Boothby and Wang studied the homogeneous manifolds associated with the contact transformation group~\cite{BW}.
Furthermore, Kobayashi introduced the complex contact structure and developed several results of complex contact geometry~\cite{K1}.
The complex contact structure is associated with a quaternionic structure with respect to the twistor correspondence \cite{I} \cite{Sa} \cite{Le}.
From the beginning of the study of contact structures, it has been known that contact structures have a deep relationship with sheaf cohomology, for example, Gray'work.

Let $M$ be a complex manifold of dimension $2n+1$.
We consider a system $\{(U_i,\gamma_i) \}$ of an open covering $\{U_i\}$ of $M$ and holomorphic 1-forms $\gamma_i$ on $U_i$
such that $\gamma_i$ is a contact 1-form, that is, $(d\gamma_i)^n \wedge \gamma_i\neq 0$ on $U_i$, and $\gamma_i=f_{ij}\gamma_j$ for a holomorphic function $f_{ij}$ on $U_i\cap U_j$.
We say that such systems $\{(U_i,\gamma_i )\}$ and $\{(U'_{i'},\gamma'_{i'} )\}$ are equivalent
if there exists a holomorphic function $g_{ii'}$ on each intersection $U_i\cap U_{i'}$ so that $\gamma_{i}=g_{ii'}\gamma'_{i'}$ on $U_i\cap U_{i'}$,
and call an equivalent class of $\{(U_i,\gamma_i) \}$ a {\it complex contact structure} on $M$.
We denote by $\gamma=\{(U_i,\gamma_i) \}$ the contact structure on $M$.
A pair $(M, \gamma)$ is called a \textit{complex contact manifold}.
The contact structure $\gamma$ induces a line bundle $L$ on $M$ by the transition function $\{f_{ij}\}$.
The canonical bundle $K_M$ is equal to $L^{-n-1}$
since $L^{n+1}\otimes \bigwedge^{2n+1}\mathbb{T}^* \cong \underline{\mathbb{C}}$ by the global section $(d\gamma_i)^{n+1}\wedge \gamma_i$.
The contact structure $\gamma$ is an $L$-valued $1$-form on $M$.
We regard $\gamma$ as a bundle map from the holomorphic tangent bundle $\mathbb{T}$ of $M$ to $L$
and denote by $\Ker \gamma$ the kernel of the map $\gamma$ :
\[
\Ker \gamma=\{v\in \mathbb{T}\mid \gamma(v)=0\}.
\]
There exists a short exact sequence of sheaves :
\[
0\to \mathcal{O}(\Ker \gamma) \to \mathcal{O}(\mathbb{T}) \stackrel{\gamma}{\to} \mathcal{O}(L) \to 0.
\]
Let $X$ be a holomorphic vector field on $M$.
The Lie derivative $L_{X}\gamma_i$ of $\gamma_i$ with respect to $X$ is a $1$-form on $U_i$.
The set $\{L_{X}\gamma_i \}$ is not a global form on $M$.
However, the restriction $L_{X}\gamma_i|_{\Ker \gamma}$ to $\Ker \gamma$ defines a global section $\{L_{X}\gamma_i|_{\Ker \gamma} \}$
of the tensor $L\otimes\Ker \gamma^*$ of $L$ and the dual of $\Ker \gamma$.
We call a vector field $X$ a {\it contact vector field} of $\gamma$
if $L_{X}\gamma_i|_{\Ker \gamma}=0$ for each $i$.
The system of equations $\{L_{X}\gamma_i|_{\Ker \gamma}=0\}$ is a global and holomorphic equation for $X$.
Such a vector field generates a contact automorphism.
We define $\mathfrak{aut}(M, \gamma)$ to be the set of contact vector fields of $\gamma$.
As an analogy of the real contact structures~\cite{K2}, Nitta and Takeuchi showed that
for any element $s$ of $\mathcal{O}(L)$,
there exists a unique contact vector field $X$ of $\gamma$ such that $\gamma(X)=s$.
Moreover, the correspondence $\mathcal{O}(L) \to \mathfrak{aut}(M, \gamma)$ is isomorphic~\cite{NT}.
It means that the holomorphic tangent sheaf $\mathcal{O}(\mathbb{T})$ splits into
$\mathcal{O}(\Ker \gamma)$ and $\mathcal{O}(L)$ as sheaves of $\mathbb{C}$-module.
LeBrun showed that $\mathcal{O}(\mathbb{T})$ does not split into the sum of $\mathcal{O}(\Ker \gamma)$ and $\mathcal{O}(L)$
as sheaves of $\mathcal{O}$-module on Fano manifolds~\cite{Le}.
Therefore, the splitting of $\mathcal{O}(\mathbb{T})$ does not directly induce that of the sheaf $\mathcal{O}(\bigwedge^k \mathbb{T})$ of $k$-vectors.

In this paper, we show the splitting of $\mathcal{O}(\bigwedge^k \mathbb{T})$ into the sum of
$\mathcal{O}(\bigwedge^{k} \Ker \gamma)$ and $\mathcal{O}(L\otimes \bigwedge^{k-1} \Ker \gamma)$ as sheaves of $\mathbb{C}$-module.
We extend the map $\gamma$ to a bundle map from $\bigwedge^k \mathbb{T}$ to $L\otimes \bigwedge^{k-1} \Ker \gamma$ by
\[
v_{1}\wedge\dots \wedge v_{k}\mapsto \sum_{j=1}^k (-1)^{j-1}\gamma(v_{j})\otimes v_{1}\wedge\dots\wedge v_{j-1}\wedge v_{j+1} \wedge\dots \wedge v_{k}
\]
and denote the map also by $\gamma$ for simplicity.
Since the kernel of the map $\gamma$ is just the space $\bigwedge^k \Ker \gamma$, we obtain the short exact sequence of the sheaves :
\[
0\to \mathcal{O}(\bigwedge^k \Ker \gamma) \to \mathcal{O}(\bigwedge^k \mathbb{T}) \stackrel{\gamma}{\to} \mathcal{O}(L\otimes \bigwedge^{k-1}\Ker \gamma) \to 0.
\]
We shall extend to the equation $L_{X}\gamma_i|_{\Ker \gamma}=0$ for a $1$-vector $X$ to a global and holomorphic equation for $k$-vectors.
Let $\nabla$ be a connection of the line bundle $L$ such that $\nabla^{0,1}=\bar{\partial}$.
For a $1$-vector $X$, the local equation $L_{X}\gamma_i|_{\Ker \gamma}=0$ is given by
the global equation $(d^{\nabla}\gamma )(X)+d^{\nabla}(\gamma (X))|_{\Ker \gamma}=0$ which is holomorphic.
The direct extension $(\otimes^k d^{\nabla}\gamma)(X)+d^{\nabla}(\otimes^{k-1} d^{\nabla}\gamma(\gamma(X)))|_{\Ker \gamma}=0$ for a $k$-vector $X$
is global but not holomorphic whenever $\nabla$ is holomorphic.
In order to find a global and holomorphic equation for holomorphic k-vector fields,
we decompose the space $\bigwedge^k \Ker \gamma$ as a sum of primitive parts with respect to the symplectic structure on $\Ker \gamma$.
Then we obtain such a equation which induces the following splitting theorem for $\mathcal{O}(\bigwedge^k \mathbb{T})$ :
\begin{thm}\label{s1t2}
Let $k$ be an integer from $1$ to $2n+1$.
The sequence
\[
0\to \mathcal{O}(\bigwedge^k \Ker \gamma) \to \mathcal{O}(\bigwedge^k \mathbb{T}) \to \mathcal{O}(L\otimes \bigwedge^{k-1} \Ker \gamma) \to 0
\]
splits as sheaves of $\mathbb{C}$-module. In particular, the sequence
\[
0\to H^i(\bigwedge^k \Ker \gamma) \to H^i(\bigwedge^k \mathbb{T}) \to H^i(L\otimes \bigwedge^{k-1} \Ker \gamma) \to 0
\]
is exact for each $i=0,\dots,2n+1$.
\end{thm}
We generalize the theorem to the splitting of $\mathcal{O}(L^{m}\otimes \bigwedge^k \mathbb{T})$
and the exact sequence of $H^i(L^{m}\otimes \bigwedge^k \mathbb{T})$ under a condition for $m$ and $k$ (see Theorem~\ref{s3t2}).
As an application, we obtain the following vanishing theorem for $H^i(L^{m}\otimes \bigwedge^k \Ker \gamma)$ on compact K\"ahler manifolds
by Kodaira-Akizuki-Nakano vanishing theorem :
\begin{thm}\label{s1t4}
If $M$ is a compact K\"ahler complex contact manifold with $c_1(M)>0$,
then $H^i(M, L^m \bigwedge^{k}  \Ker \gamma)=\{0\} $ for $k$ and $m$ satisfying one of following three conditions
\begin{equation*}
\quad
\left\{
\begin{array}{lll}
i\le 2n-k,& m \le -[\frac{k+1}{2}]-n-1,& 1\le k\le 2n+1, \\
\forall i,& -n\le m \le -k-1,& 1\le k\le n-1, \\
i\ge k+1,& m \ge -[\frac{k}{2}],&  1\le k\le 2n+1.
\end{array}
\right.
\end{equation*}
\end{thm}
We also have a similar result for vanishing of cohomology in the case of $c_1(M)<0$ (see Theorem~\ref{s4.1t1}).
Moreover, on $\mathbb{CP}^{2n+1}$, we show the vanishing theorem for $H^i(\mathcal{O}(m) \bigwedge^k \Ker \gamma)$
by Bott vanishing theorem (see Theorem~\ref{s4.2t2}).

This paper is organized as follows.
In Section 2, we prepare some propositions for complex symplectic vector spaces.
The space of $k$-vectors is decomposed into a sum of primitive parts with respect to the symplectic structure.
In Section 3, the equation $L_{X}\gamma_i|_{\Ker \gamma}=0$ for a $1$-vector $X$ is extended to a global and holomorphic equation for $k$-vectors.
By using the equation, we prove Theorem~\ref{s1t2} (see Theorem~\ref{s3t1}).
The theorem is generalized to the splitting theorem for $\mathcal{O}(L^{m}\otimes \bigwedge^k \mathbb{T})$ (Theorem~\ref{s3t2}).
On $\mathbb{CP}^{2n+1}$, $L$ is given by $\mathcal{O}(2)$ and we also obtain the splitting of $\mathcal{O}(m)\otimes \bigwedge^k \Ker \gamma$.
In the last section, we show two kinds of vanishing theorems (Theorem~\ref{s4.1t1} and Theorem~\ref{s4.2t2}).

\section{Complex symplectic vector spaces}
In this section, we prepare some propositions for complex symplectic vector spaces
in order to show our main theorems in Section 3.

\subsection{Symplectic structures}
Let $V$ be a complex vector space of dimension $2n$.
A {\it complex symplectic vector space} is a pair $(V, \omega)$ of $V$
and a non-degenerate skew-symmetric bilinear form $\omega$ on $V$.
If we regard the symplectic structure $\omega$ as the isomorphism from $V$ to the dual space $V^*$,
then the $2$-tensor $\otimes^2 \omega$ of $\omega$ is the isomorphism
\[
\otimes^2 \omega : \otimes^2 V \to \otimes^2 V^*.
\]
It induces the isomorphism $\otimes^2 \omega : \wedge^2 V \to \wedge^2 V^*$.
A $2$-vector $w_0$ is defined by
\[
\otimes^2 \omega(w_0)=\omega.
\]
If we take a basis $e_1,\dots,e_{2n}$ of $V$ such that $\omega=e_1^*\wedge e_2^*+\dots+e_{2n-1}^*\wedge e_{2n}^*$,
then $w_0$ is represented as $w_0=e_1\wedge e_2+\dots+e_{2n-1}\wedge e_{2n}$ and $\omega(w_0)=n$.

\subsection{Decomposition of the space of $k$-vectors}
Let $k$ be an integer from $1$ to $2n$.
We define an operator $\mathcal{L}: \wedge^k V \to\wedge^{k+2}V$ by $\mathcal{L}(X)=X\wedge w_0$ for $X\in \wedge^k V$.
We consider a $1$-form $\theta$ as a map from $\wedge^k V$ to $\wedge^{k-1} V$ by
\[
\theta(v_{1}\wedge\dots \wedge v_{k})=\sum_{j=1}^k (-1)^{j-1}\theta(v_{j}) v_{1}\wedge\dots\wedge v_{j-1}\wedge v_{j+1} \wedge\dots \wedge v_{k}
\]
for $v_{1},\dots, v_{k} \in V$ and denote the map also by $\theta$ for simplicity.
Let $l$ be an integer with $0\le l\le k$.
We regard an $l$-form $\theta_1\wedge \dots \wedge \theta_l$ as
a map $\theta_1\wedge \dots \wedge \theta_l : \wedge^k V \to \wedge^{k-l} V$ by
\[
\theta_1\wedge \dots \wedge \theta_l(X)= \theta_l\circ \dots \circ \theta_2\circ \theta_1(X)
\]
for $X\in \wedge^k V$.
Let $\Lambda$ be an operator $\Lambda : \wedge^k V \to \wedge^{k-2} V$ defined by $\Lambda(X)= \omega(X)$ for $X\in \wedge^k V$.
Then we obtain the formula $(\Lambda \mathcal{L}-\mathcal{L}\Lambda)(X)=(n-k)X$ for $X\in \wedge^k V$,
and inductively,
\begin{equation}
(\Lambda \mathcal{L}^r-\mathcal{L}^r\Lambda)(X)=r(n-k-r+1)\mathcal{L}^{r-1}X \label{s2eq2}
\end{equation}
where we define $\mathcal{L}^0={\rm id}$.
A $k$-vector $X$ is called {\it primitive} if $\Lambda(X)=0$.
If follows from the formula (\ref{s2eq2}) that
\begin{equation}
\Lambda^s \mathcal{L}^r X=
\left\{
\begin{array}{ll}
\frac{r!}{(r-s)!}(n-k-r+1)(n-k-r+2)\cdots(n-k-r+s)\mathcal{L}^{r-s}X, &\quad r\ge s, \\
0 , &\quad r<s
\end{array}
\right. \label{s2eq3}
\end{equation}
for a primitive $k$-vector $X$.
Let $\wedge_e^k V$ denote the space of primitive $k$-vectors :
\[
\wedge_e^k V=\{X\in \wedge^k V \mid \Lambda(X)=0 \}.
\]
Then we have the following decomposition of the space $\wedge^k V$ of $k$-vectors :
\begin{prop} \label{s2p1}
If $k\le n$, then
\[
\wedge^k V=\wedge_e^k V+\mathcal{L}\wedge_e^{k-2} V+\dots+ \mathcal{L}^{[\frac{k}{2}]}\wedge_e^{k-2[\frac{k}{2}]} V.
\]
If $k> n$, then
\[
\wedge^k V=\mathcal{L}^{k-n}\wedge_e^{2n-k} V+\mathcal{L}^{k-n+1}\wedge_e^{2n-k-2} V+\dots+ \mathcal{L}^{[\frac{k}{2}]}\wedge_e^{k-2[\frac{k}{2}]} V
\]
where $[m]$ means the Gauss symbol of $m$. $\hfill\Box$
\end{prop}

\subsection{Transformation associated with the decomposition}
We define a linear transformation $T$ on $\wedge^k V$ by
\[
T(X)=c_0X+c_1\mathcal{L}\Lambda X+c_2\mathcal{L}^2\Lambda^2 X+\dots+c_{[\frac{k}{2}]}\mathcal{L}^{[\frac{k}{2}]}\Lambda^{[\frac{k}{2}]}X=\sum_{i=0}^{[\frac{k}{2}]}c_i\mathcal{L}^i\Lambda^i X
\]
for constants $c_0, c_1, c_2,\dots, c_{[\frac{k}{2}]}$, then we obtain the following :
\begin{prop}\label{s2p2}
The transformation $T$ is isomorphic if and only if the constants $c_0, c_1,\dots, c_{[\frac{k}{2}]}$ satisfy
$\sum_{s=0}^{r} c_s \frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}\neq 0$ for any $r=0,\dots,[\frac{k}{2}]$.
\end{prop}
\begin{proof}
In the case $k\le n$ and $0\le r \le [\frac{k}{2}]$,
the equation (\ref{s2eq3}) implies that
\[
\mathcal{L}^s\Lambda^s \mathcal{L}^r X=
\left\{
\begin{array}{ll}
\frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}\mathcal{L}^r X, &\quad s\le r, \\
0 , &\quad s>r
\end{array}
\right.
\]
for $X\in \wedge_e^{k-2r}$.
It yields that
\begin{equation}
T( \mathcal{L}^r X)=\sum_{s=0}^{ [\frac{k}{2}]} c_s \mathcal{L}^s\Lambda^s \mathcal{L}^r X=\left(\sum_{s=0}^{r} c_s \frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}\right)\mathcal{L}^r X
\label{s2eq4}
\end{equation}
for $X\in \wedge_e^{k-2r}$.
In the case $k>n$ and $k-n\le r \le [\frac{k}{2}]$, we also have the same equation (\ref{s2eq4}) for $X\in \wedge_e^{k-2r}$.
Hence Proposition~\ref{s2p1} implies that $T$ is an isomorphism from $\wedge^k V$ to itself
if and only if $\sum_{s=0}^{r} c_s \frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}\neq 0$ for each $r=0,\dots,[\frac{k}{2}]$.
\end{proof}

\section{Splitting of sheaves on complex contact manifolds}
Let $(M, \gamma)$ be a complex contact manifold of dimension $2n+1$ and $L$ the line bundle associated with the contact structure $\gamma$.
We denote by $D$ the subbundle $\Ker \gamma$ of $\mathbb{T}$.
Let $\nabla$ be a connection of $L$ with $\nabla^{0,1}=\bar{\partial}$.
The covariant exterior differentiation $d^{\nabla}\gamma$ of $\gamma$ is a smooth section of $L\otimes \bigwedge^2 \mathbb{T}^*$.

\subsection{The isomorphisms associated with $d^{\nabla}\gamma$}
Let $d^{\nabla}\gamma|_D$ denote the restriction of $d^{\nabla}\gamma$ to $D$.
Then, $d^{\nabla}\gamma|_D$ is a holomorphic section of $L\otimes \bigwedge^2 D^*$ which is independent of the choice of a connection $\nabla$.
We identify $d^{\nabla}\gamma|_D$ with the holomorphic bundle map from $D$ to $L\otimes D^*$.
Then the map $d^{\nabla}\gamma|_D: D\to L\otimes D^*$ is isomorphic since $d\gamma_i$ is non-degenerate on $D$.
In general, the $k$-th tensor $\otimes^k d^{\nabla}\gamma$ of $d^{\nabla}\gamma$ is a smooth section of $L^k\otimes^{k}\bigwedge^2 \mathbb{T}^*$
which is regarded as the smooth bundle map
\begin{equation*}
\otimes^k d^{\nabla}\gamma : \bigwedge^k \mathbb{T} \to L^k\otimes\bigwedge^k \mathbb{T}^*.\label{s3eq1}
\end{equation*}
Let $e$ be a local frame of $L$ and $A$ a connection form of $\nabla$ with respect to $e$.
The contact structure $\gamma$ is given by $\gamma=e\otimes \gamma_0$ for a holomorphic $1$-form $\gamma_0$.
Then $d^{\nabla}\gamma=e\otimes (d\gamma_0 +A\wedge \gamma_0)$.
For $1$-vectors $v_1,\dots , v_k$,
the $L^k$-valued $k$-form $\otimes^k d^{\nabla}\gamma(v_1\wedge \cdots \wedge v_k)$ is written by
\begin{eqnarray*}
&&e^k\otimes \{(d\gamma_0 +A\wedge \gamma_0)(v_1)\wedge\cdots \wedge(d\gamma_0 +A\wedge \gamma_0)(v_k)\} \\
&=&e^k\otimes \{\otimes^{k} d\gamma_0(v_1 \wedge\cdots \wedge v_k)+\gamma_0\wedge (\otimes^{k-1}d\gamma_0)(A(v_1\wedge \cdots \wedge v_k))\\
&&-A\wedge (\otimes^{k-1}d\gamma_0)(\gamma_0(v_1\wedge \cdots \wedge v_k)) \}.
\end{eqnarray*}
It implies
\begin{equation}
\otimes^k d^{\nabla}\gamma(X)=e^k\otimes \{\otimes^k d\gamma_0(X)+\gamma_0
\wedge (\otimes^{k-1}d\gamma_0)(A(X))-A\wedge (\otimes^{k-1}d\gamma_0)(\gamma_0(X))\} \label{s3eq3}
\end{equation}
for any $k$-vector $X$.
We remark that the map $\otimes^k d^{\nabla}\gamma$ is not holomorphic whenever $\nabla$ is holomorphic.
However, the restriction $\otimes^k d^{\nabla}\gamma|_D$ is the holomorphic map from $\bigwedge^k D$ to $L^k\otimes\bigwedge^k D^*$.
The induced map $\otimes^k d^{\nabla}\gamma|_D: \mathcal{O}(\otimes \bigwedge^k D) \to \mathcal{O}(L^{k}\otimes\bigwedge^k D^*)$ is isomorphic on sheaves,
and it is extended to an isomorphism
\begin{equation*}
\otimes^k d^{\nabla}\gamma|_D : \mathcal{O}(L^m\otimes \bigwedge^k D) \to \mathcal{O}(L^{k+m}\otimes\bigwedge^k D^*) \label{s3eq2}
\end{equation*}
for each $m\in\mathbb{Z}$ and $k=1,\dots,2n+1$.

In the case $m=-1$ and $k=2$, we have the isomorphism $\otimes^2 d^{\nabla}\gamma|_D : \mathcal{O}(L^{-1}\otimes \bigwedge^2 D) \to \mathcal{O}(L\otimes\bigwedge^2 D^*)$.
By considering $d^{\nabla}\gamma|_D$ as a holomorphic section of $L\otimes\bigwedge^2 D^*$,
there exists a holomorphic section $w$ of $L^{-1}\otimes \wedge^2 D$ such that
\[
\otimes^2 d^{\nabla}\gamma|_D(w)=d^{\nabla}\gamma|_D
\]
as $w_0$ in Section 2.1.
The section $w$ is independent of the choice of a connection $\nabla$.

\subsection{The map $F:\bigwedge^k \mathbb{T} \to L^k\otimes \bigwedge^k \mathbb{T}^*$}
We fix an integer $k$ such that $1\le k\le 2n+1$ and define a map $F:\bigwedge^k \mathbb{T} \to L^k\otimes \bigwedge^k \mathbb{T}^*$ by
\[
F(X)=\sum_{i=0}^{[\frac{k}{2}]}\frac{(k-i)!}{k!i!}\otimes^{k-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(X))\wedge (d^{\nabla}\gamma)^i
\]
for $X\in \bigwedge^k \mathbb{T}$, where $(d^{\nabla}\gamma)^0$ is the identity map ${\rm id}$ and
$(d^{\nabla}\gamma)^i$ is the $i$-th wedge $d^{\nabla}\gamma\wedge \dots\wedge d^{\nabla}\gamma$ of $d^{\nabla}\gamma$.
We have
\begin{equation}
(d^{\nabla}\gamma)^i(X)=e^i\otimes \{(d\gamma_0)^i(X)-i(d\gamma_0)^{i-1}(A(\gamma_0(X)))\} \label{s3eq4}
\end{equation}
for each $i=0,\dots ,[\frac{k}{2}]$ since $(d^{\nabla}\gamma)^i=e^i\otimes \{(d\gamma_0)^i+i A\wedge\gamma_0 \wedge (d\gamma_0)^{i-1} \}$.
The equations (\ref{s3eq3}) and (\ref{s3eq4}) imply that
\begin{eqnarray}
&&\otimes^{k-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(X)) \notag\\
&=&e^{k-i}\otimes \{ \otimes^{k-2i} d\gamma_0((d\gamma_0)^i(X))-i\otimes^{k-2i} d\gamma_0((d\gamma_0)^{i-1}(A(\gamma_0(X)))) \notag\\
&&+\gamma_0\wedge (\otimes^{k-1-2i}d\gamma_0)((d\gamma_0)^i(A(X)))-A\wedge (\otimes^{k-1-2i}d\gamma_0)((d\gamma_0)^i(\gamma_0(X))) \} \quad \label{s3eq5}
\end{eqnarray}
where we use $A((d\gamma_0)^{i-1}(A(\gamma_0(X))))=\gamma_0((d\gamma_0)^{i-1}(A(\gamma_0(X)))=0$ in the first line.
It yields that
\begin{eqnarray}
\otimes^{k-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(X))|_D
&=&e^{k-i}\otimes \{ \otimes^{k-2i} d\gamma_0((d\gamma_0)^i(X))-i\otimes^{k-2i} d\gamma_0((d\gamma_0)^{i-1}(A(\gamma_0(X)))) \notag\\
&&-A\wedge (\otimes^{k-1-2i}d\gamma_0)((d\gamma_0)^i(\gamma_0(X))) \}. \label{s3eq6}
\end{eqnarray}
The second term in the right hand side of the equation (\ref{s3eq6}) is $0$ except for $1\le i\le [\frac{k}{2}]$ and
the third term is $0$ except for $0\le i\le [\frac{k-1}{2}]$. Hence
\begin{eqnarray}
F(X)|_D
&=&\sum_{i=0}^{[\frac{k}{2}]}\frac{(k-i)!}{k!i!}\{ \otimes^{k-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(X))\wedge (d^{\nabla}\gamma)^i \}|_D \notag\\
&=&e^{k}\otimes \Big\{ \sum_{i=0}^{[\frac{k}{2}]}\frac{(k-i)!}{k!i!}\{ \otimes^{k-2i} d\gamma_0((d\gamma_0)^i(X)) \wedge (d\gamma_0)^i \notag\\
&&-\sum_{i=1}^{[\frac{k}{2}]}\frac{(k-i)!}{k!(i-1)!}\otimes^{k-2i} d\gamma_0((d\gamma_0)^{i-1}(A(\gamma_0(X)))) \wedge (d\gamma_0)^i \notag\\
&&-\sum_{i=0}^{[\frac{k-1}{2}]}\frac{(k-i)!}{k!i!}A\wedge (\otimes^{k-1-2i}d\gamma_0)((d\gamma_0)^i(\gamma_0(X))) \wedge (d\gamma_0)^i \Big\} \Big|_D  \label{s3eq6.5}.
\end{eqnarray}
The restriction $F|_{D}:\bigwedge^k D \to L^k\bigwedge^k D^*$ satisfies $F|_{D}(X)=F(X)|_{D}$ for $X\in \bigwedge^k D$.
If $X$ is a holomorphic section of $\bigwedge^k D$, then it follows from $\gamma_0(X)=0$ that
\[
F|_D(X)=F(X)|_D = e^{k}\otimes \Big\{ \sum_{i=0}^{[\frac{k}{2}]}\frac{(k-i)!}{k!i!} \otimes^{k-2i} d\gamma_0((d\gamma_0)^i(X)) \wedge (d\gamma_0)^i\Big\}  \Big|_D
\]
is a holomorphic section of $L^k\otimes \bigwedge^k D^*$.
Hence we obtain the following :
\begin{lem}\label{s3l1}
An $L^k$-valued $k$-form $F|_{D}(X)$ restricted to $D$ is holomorphic for any holomorphic section $X$ of $\bigwedge^k D$. $\hfill\Box$
\end{lem}

The lemma implies that $F|_D$ is regarded as a map from $\mathcal{O}(\bigwedge^k D)$ to $\mathcal{O}(L^k\otimes \bigwedge^k D^*)$.
The map $F|_{D}$ is written by
\[
F|_{D}(X)=\otimes^k d^{\nabla}\gamma|_D\Big(\sum_{i=0}^{[\frac{k}{2}]}\frac{(k-i)!}{k!i!}(d^{\nabla}\gamma|_D)^i(X)\wedge (w)^i\Big)
\]
for $X\in \mathcal{O}(\bigwedge^k D)$
since $(d^{\nabla}\gamma|_D)^i=\otimes^{2i} d^{\nabla}\gamma|_D(w)^i$.
We define the transformation $f: \mathcal{O}(\bigwedge^k D)\to \mathcal{O}(\bigwedge^k D)$ by
\[
f(X)=\sum_{i=0}^{[\frac{k}{2}]}\frac{(k-i)!}{k!i!}(d^{\nabla}\gamma|_D)^i(X)\wedge (w)^i
\]
for $X\in \mathcal{O}(\bigwedge^k D)$.
Then the map $F|_{D}$ is the composition of $\otimes^k d^{\nabla}\gamma|_D$ and $f$.
Proposition~\ref{s2p2} implies that the following :
\begin{prop}\label{s3p1}
The map $F|_{D}:\mathcal{O}(\bigwedge^k D) \to \mathcal{O}(L^k\otimes \bigwedge^k D^*)$ is isomorphic.
\end{prop}
\begin{proof}
It suffices to show that $f$ is isomorphic since $\otimes^k d^{\nabla}\gamma|_D$ is isomorphic.
At each point $x\in M$, the linear map $f_x:\bigwedge^k D_x\to \bigwedge^k D_x$ is written by
\[
f_x=\sum_{i=0}^{ [\frac{k}{2}]} \frac{(k-i)!}{k!i!} \mathcal{L}^i\Lambda^i
\]
where $\mathcal{L}$ and $\Lambda$ are operators as in the previous section associated with the symplectic structure $(d^{\nabla}\gamma|_D)_x$ on $D_x$.
The map $f_x$ is isomorphic by Proposition~\ref{s2p2} since each coefficient $c_i=\frac{(k-i)!}{k!i!}$ is positive.
Hence $F|_{D}$ is also isomorphic.
\end{proof}

\subsection{The map $G:\Gamma(L\otimes\bigwedge^{k-1} D) \to \Gamma(L^k\otimes\bigwedge^k \mathbb{T}^*)$}
We define a map $G:\Gamma(L\otimes\bigwedge^{k-1} D) \to \Gamma(L^k\otimes\bigwedge^k \mathbb{T}^*)$ by
\[
G(s)=\sum_{i=0}^{[\frac{k-1}{2}]}\frac{(k-1-i)!}{k!i!}d^{\nabla}\left( \otimes^{k-1-2i}d^{\nabla}\gamma((d^{\nabla}\gamma)^i(s))\right)\wedge (d^{\nabla}\gamma)^i
\]
for $s\in \Gamma(L\otimes\bigwedge^{k-1}D)$.
Let $s$ be a section of $L\otimes\bigwedge^{k-1}D$.
The section $s$ is locally written as $s=e\otimes s_0$ for a section $s_0\in \bigwedge^{k-1}D$.
It follows from the equations (\ref{s3eq5}) and $(d^{\nabla}\gamma)^i(s)=e^{i+1}\otimes (d\gamma_0)^i(s_0)$ that
\begin{equation}
\otimes^{k-1-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(s))
=e^{k-i}\otimes \{ \otimes^{k-1-2i} d\gamma_0((d\gamma_0)^i(s_0)+\gamma_0\wedge (\otimes^{k-2-2i}d\gamma_0)((d\gamma_0)^i(A(s_0))) \}. \label{s3eq7}
\end{equation}
We remark that the second term in the right hand side of the equation (\ref{s3eq7}) is $0$ except for $0\le i\le [\frac{k}{2}]-1$.
It yields that
\begin{eqnarray*}
d^{\nabla} \otimes^{k-1-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(s))|_D
&=&e^{k-i}\otimes \{ d \otimes^{k-1-2i} d\gamma_0((d\gamma_0)^i(s_0) \\
&&+d\gamma_0\wedge (\otimes^{k-2-2i}d\gamma_0)((d\gamma_0)^i(A(s_0)))\\
&& +(k-i)A \wedge \otimes^{k-1-2i} d\gamma_0((d\gamma_0)^i(s_0)) \} |_D.
\end{eqnarray*}
Hence
\begin{eqnarray}
G(s)|_D&=&\sum_{i=0}^{[\frac{k-1}{2}]}\frac{(k-1-i)!}{k!i!} d^{\nabla} (\otimes^{k-1-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(s))|_D) \wedge (d^{\nabla}\gamma)^i|_D \notag\\
&=&e^{k}\otimes \Big\{ \sum_{i=0}^{[\frac{k-1}{2}]}\frac{(k-1-i)!}{k!i!} d (\otimes^{k-1-2i} d\gamma_0((d\gamma_0)^i(s_0)))\wedge (d\gamma_0)^i  \notag\\
&&+\sum_{i=1}^{[\frac{k}{2}]}\frac{(k-i)!}{k!(i-1)!} (\otimes^{k-2i}d\gamma_0)((d\gamma_0)^{i-1}(A(s_0)))\wedge (d\gamma_0)^i  \notag\\
&& +\sum_{i=0}^{[\frac{k-1}{2}]}\frac{(k-i)!}{k!i!}A \wedge \otimes^{k-1-2i} d\gamma_0((d\gamma_0)^i(s_0))\wedge (d\gamma_0)^i \Big\} \Big|_D. \label{s3eq8}
\end{eqnarray}
Then we have
\begin{prop}\label{s3p2}
An $L^k$-valued $k$-form $\{F(X)+G(\gamma(X))\}|_{D}$ restricted to $D$ is holomorphic for any holomorphic $k$-vector $X$.
\end{prop}
\begin{proof}
If we take $s$ as the image $\gamma(X)$ of a $k$-vector $X$, then $s_0=\gamma_0(X)$,
and it follows from the equations (\ref{s3eq6.5}) and (\ref{s3eq8}) that
\begin{eqnarray*}
\{F(X)+G(s)\}|_{D}&=&e^{k}\otimes \Big\{ \sum_{i=0}^{[\frac{k}{2}]}\frac{(k-i)!}{k!i!}\{ \otimes^{k-2i} d\gamma_0((d\gamma_0)^i(X))\wedge (d\gamma_0)^i \\
&&+\sum_{i=0}^{[\frac{k-1}{2}]}\frac{(k-1-i)!}{k!i!} d(\otimes^{k-1-2i} d\gamma_0((d\gamma_0)^i(\gamma_0(X))))\wedge (d\gamma_0)^i \Big\} \Big|_D.
\end{eqnarray*}
Hence $\{F(X)+G(\gamma(X))\}|_{D}$ is holomorphic for a holomorphic $k$-vector $X$.
\end{proof}

\subsection{The splitting of sheaves $\mathcal{O}(\bigwedge^k\mathbb{T})$ as $\mathbb{C}$-module}

We have the following theorem:
\begin{thm}\label{s3t1}
Let $k$ be an integer from $1$ to $2n+1$.
The sequence
\[
0\to \mathcal{O}(\bigwedge^k D) \to \mathcal{O}(\bigwedge^k \mathbb{T}) \to \mathcal{O}(L\otimes \bigwedge^{k-1} D) \to 0
\]
splits as sheaves of $\mathbb{C}$-module.
In particular, the sequence
\[
0\to H^i(\bigwedge^k D) \to H^i(\bigwedge^k \mathbb{T}) \to H^i(L\otimes \bigwedge^{k-1} D) \to 0
\]
is exact for each $i=0,\dots,2n+1$.
\end{thm}
\begin{proof}
Let $s$ be a holomorphic section of $L\otimes \bigwedge^{k-1} D$.
We take an open set $U$ of $M$ where the bundles $L$ and $D$ are trivial.
Then we can take a holomorphic $k$-vector $Y$ on $U$ such that $\gamma(Y)=s$ as follows.
We fix a local frame $e$ of $L$ on $U$.
The contact form $\gamma$ is given by $\gamma=e\otimes \gamma_0$ for a holomorphic 1-form $\gamma_0$ on $U$.
We can choose a local frame $\{e_1,\dots,e_{2n}\}$ of $D$ and a local section $e_{2n+1}$ of $\mathbb{T}$ on $U$
such that $\gamma (e_{2n+1})=e$.
If $s$ is written by
$s= \sum s_{i_1 \ldots i_k} e \otimes e_{i_1}\wedge \cdots\wedge e_{i_{k-1}}$
on $U$, then we take a section $Y$ by
$Y = \sum s_{i_1 \ldots i_k} e_{2n+1}\wedge e_{i_1}\wedge \cdots\wedge e_{i_{k-1}}$.

Now we consider the smooth section $F(Y)+G(s)$ of $L^{k}\otimes \bigwedge^{k} \mathbb{T}^*$.
Proposition~\ref{s3p2} implies that $\{F(Y)+G(s)\}|_{D}$ restricted to $D$ is a holomorphic section of $L^k\otimes \bigwedge^k D^*$.
Hence, there exists a holomorphic section $h$ of $\bigwedge^k D$ such that
\[
\{F(Y)+G(s)\}|_{D}=F|_{D}(h)
\]
by the isomorphism $F|_{D}$ in Proposition~\ref{s3p1}.
We define $X$ by $X=Y-h$.
Then we obtain the holomorphic $k$-vector $X$ on $U$ satisfying the following equations
\begin{eqnarray*}
{\rm (i)}&&\gamma(X)=s, \\
{\rm (ii)}&&\{F(X)+G(s)\}|_{D} =0.
\end{eqnarray*}
We take such local sections $X_1$ and $X_2$ on open sets $U_1$ and $U_2$, respectively.
The first condition $\rm (i)$ implies that the difference $X_1-X_2$ is in $\bigwedge^k D$ on $U_1\cap U_2$.
We also have the equation $F|_{D}(X_1-X_2)=F(X_1-X_2)|_{D}=0$ by the second condition $\rm (ii)$.
Then $X_1=X_2=0$ on $U_1\cap U_2$ since $F|_{D}$ is isomorphic.
Hence the correspondence of $s$ to $X$ provides a right inverse of
the map $\gamma : \mathcal{O}(\bigwedge^k \mathbb{T}) \to \mathcal{O}(L\otimes \bigwedge^{k-1} D)$ as a $\mathbb{C}$-module map.
It induces the splitting of $\mathcal{O}(\bigwedge^k \mathbb{T})$ and the exactness of $H^i(\bigwedge^k \mathbb{T})$ for each $i$.
It completes the proof.
\end{proof}

\begin{rem}\label{s3r1}
{\rm
In the proof, $h$ is independent of the connection $\nabla$
since $\{F(Y)+G(s)\}|_{D}$ and $F|_{D}$ do not depend on $\nabla$ by Proposition~\ref{s3p1} and~\ref{s3p2}.
The $k$-vector $X=Y-h$ is also independent of $\nabla$.
Hence, the splitting of the sequence in Theorem~\ref{s3t1} is independent of the choice of the connection.
}
\end{rem}

\subsection{The splitting of the sheaves $\mathcal{O}(L^{m}\otimes\bigwedge^k\mathbb{T})$ as $\mathbb{C}$-module}
In this section, we generalize Theorem~\ref{s3t1} to the splitting of $\mathcal{O}(L^{m}\otimes \bigwedge^k \mathbb{T})$
under a condition for $m$ and $k$.
Let $m$ be an integer such that $m\le -k-1$ or $m\ge -[\frac{k}{2}]$.
We define a constant $c_{m,i}$ by $c_{m,0}=1$ and
\[
c_{m,i}=\frac{1}{(k+m)(k+m-1)\cdots(k+m-i+1)\ i!}
\]
for each $i=1,\dots,[\frac{k}{2}]$.
These constants are well-defined
since $m+k<0$ and $m+k-[\frac{k}{2}]+1>0$ in the cases of $m\le -k-1$ and $m\ge -[\frac{k}{2}]$, respectively.
We define a map $F_m : L^{m}\otimes \bigwedge^k \mathbb{T} \to L^{m+k}\otimes \bigwedge^k \mathbb{T}^*$ by
\[
F_m(X)=\sum_{i=0}^{[\frac{k}{2}]}c_{m,i}\otimes^{k-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(X))\wedge (d^{\nabla}\gamma)^i
\]
for $X\in L^{m}\otimes \bigwedge^k \mathbb{T}$.
By the same argument in Lemma~\ref{s3l1}, the restriction $F_m|_{D}$ induces the map from $\mathcal{O}(L^{m}\otimes \bigwedge^k D)$ to $\mathcal{O}(L^{m+k}\otimes \bigwedge^k D^*)$.

\begin{prop}\label{s3p3}
The map $F_m|_{D}:\mathcal{O}(L^{m}\otimes \bigwedge^k D) \to \mathcal{O}(L^{m+k}\otimes \bigwedge^k D^*)$ is isomorphic
if $k$ and $m$ satisfy one of following three conditions
\[
\left\{
\begin{array}{ll}
m \le -n-[\frac{k}{2}]-2, &\quad 1\le k\le 2n+1, \\
-n-1 \le m \le -k-1, &\quad 1\le k\le n, \\
m \ge -[\frac{k}{2}], &\quad 1\le k\le 2n+1.
\end{array}
\right.
\]
\end{prop}
\begin{proof}
The map $F_m|_{D}$ is the composition of $\otimes^k d^{\nabla}\gamma|_D$ and the transformation $f_m$ of $\mathcal{O}(L^{m}\otimes \bigwedge^k D)$
defined by $f_m(X)=\sum_{i=0}^{[\frac{k}{2}]}c_{m,i}(d^{\nabla}\gamma|_D)^i(X)\wedge w^i$ for $X\in \mathcal{O}(L^{m}\otimes \bigwedge^k D)$.
At each point $x\in M$, the linear map $(f_m)_x:L^{m}_x\otimes \bigwedge^k D_x\to L^{m}_x\otimes \bigwedge^k D_x$ is written
by $(f_m)_x=\sum_{i=0}^{ [\frac{k}{2}]} c_{m,i} \mathcal{L}^i\Lambda^i$.
Proposition~\ref{s2p2} implies that $(f_m)_x$ is isomorphic if $\sum_{s=0}^{r} c_{m,s} \frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}$ is not zero for each $r$.
If $k\ge 2$, then
\[
\sum_{s=0}^{r} c_{m,s} \frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}=\frac{(m+n+r+1)\cdots(m+n+2)}{(m+k)\cdots(m+k-r+1)}
\]
for any $r=1,\dots,[\frac{k}{2}]$.
It yields that $\sum_{s=0}^{r} c_{m,s} \frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}\neq 0$ for each $r$
if $m\le -n-[\frac{k}{2}]-2$ or $m \ge -n-1$.
By the assumption $m\le -k-1$ or $m\ge -[\frac{k}{2}]$, the linear map $(f_m)_x$ is isomorphic
if
\[
\left\{
\begin{array}{ll}
m \le -n-[\frac{k}{2}]-2, &\quad \forall k, \\
-n-1 \le m \le -k-1, &\quad 1\le k\le n, \\
m \ge -[\frac{k}{2}], &\quad \forall k.
\end{array}
\right.
\]
Hence, we finish the proof.
\end{proof}

We define a constant $c'_{m,i}$ by
\[
c'_{m,i}=\frac{1}{(k+m)(k+m-1)\cdots(k+m-i)\ i!}
\]
for each $i=0,1,\dots,[\frac{k-1}{2}]$.
We remark that these constants are well-defined since $m+k<0$ and $m+k-[\frac{k-1}{2}]>0$ in the cases of $m\le -k-1$ and $m\ge -[\frac{k}{2}]$, respectively.
We define a map $G_m:\Gamma(L^{m+1}\otimes\bigwedge^{k-1} D) \to \Gamma(L^{m+k}\otimes\bigwedge^k \mathbb{T}^*)$ by
\[
G_m(s)=\sum_{i=0}^{[\frac{k-1}{2}]}c'_{m,i}d^{\nabla}\left( \otimes^{k-1-2i}d^{\nabla}\gamma((d^{\nabla}\gamma)^i(s))\right)\wedge (d^{\nabla}\gamma)^i
\]
for $s\in \Gamma(L^{m+1}\otimes\bigwedge^{k-1}D)$.
Similarly to Proposition~\ref{s3p2}, we have
\begin{prop}\label{s3p4}
An $L^{k+m}$-valued $k$-form $\{F_m(X)+G_m(\gamma(X))\}|_{D}$ restricted to $D$ is holomorphic for any holomorphic $L^m$-valued $k$-vector $X$.
$\hfill\Box$
\end{prop}
By repeating the proof of Theorem~\ref{s3t1} with $F_m$ and $G_m$ instead of $F$ and $G$, then we obtain
\begin{thm}\label{s3t2}
The sequence
\[
0\to \mathcal{O}(L^{m}\otimes \bigwedge^k D) \to \mathcal{O}(L^{m}\otimes \bigwedge^k \mathbb{T}) \to \mathcal{O}(L^{m+1}\otimes \bigwedge^{k-1} D) \to 0
\]
splits as sheaves of $\mathbb{C}$-module, and the sequence
\[
0\to H^i(L^{m}\otimes \bigwedge^k D) \to H^i(L^{m}\otimes \bigwedge^k \mathbb{T}) \to H^i(L^{m+1}\otimes \bigwedge^{k-1} D) \to 0
\]
is exact for each $i=0,\dots,2n+1$
if $k$ and $m$ satisfy one of following three conditions
\[
\left\{
\begin{array}{ll}
m \le -n-[\frac{k}{2}]-2, &\quad 1\le k\le 2n+1, \\
-n-1 \le m \le -k-1, &\quad 1\le k\le n, \\
m \ge -[\frac{k}{2}], &\quad 1\le k\le 2n+1. \\
\end{array}
\right.
\]
$\hfill\Box$
\end{thm}

\subsection{The splitting of the sheaves $\mathcal{O}(l^{m}\otimes\bigwedge^k\mathbb{T})$ on $\mathbb{CP}^{2n+1}$}
On the odd dimensional projective space $\mathbb{CP}^{2n+1}$,
there exists a standard contact structure $\gamma$ written by
$\gamma =z^0 dz^1 - z^1dz^0 + \cdots + z^{2n}dz^{2n+1} - z^{2n+1}dz^{2n}$ in the homogeneous coordinate.
Let $l$ denote the hyperplane bundle $\mathcal{O}(1)$ on $\mathbb{CP}^{2n+1}$.
The associated bundle $L$ is given by $l^2=\mathcal{O}(2)$ and the contact structure $\gamma$ is regarded as a section of $l^2\otimes \mathbb{T}^*$.

We consider the short exact sequence
\begin{equation*}
0\to l^m \otimes \bigwedge^k D \to l^m \otimes \bigwedge^k \mathbb{T} \stackrel{\gamma}{\to} l^{m+2} \otimes \bigwedge^{k-1}D \to 0
\end{equation*}
for $m\in \mathbb{Z}$ and $1\le k \le 2n+1$.
If $m$ is even, then the splitting of $\mathcal{O}(l^{m}\otimes\bigwedge^k\mathbb{T})$ is induced by Theorem~\ref{s3t2} since $L=l^2$.
From now on, we assume that $m$ is odd.
We define a constant $\tilde{c}_{m,i}$ by $\tilde{c}_{m,0}=1$ and
\[
\tilde{c}_{m,i}=\frac{1}{(k+\frac{m}{2})(k+\frac{m}{2}-1)\cdots(k+\frac{m}{2}-i+1)\ i!}
\]
for each $i=1,\dots,[\frac{k}{2}]$.
These constants are well-defined since $k+\frac{m}{2}-i+1\neq 0$ for any $i$.
We fix a connection $\nabla$ of $l$ and define a map $\widetilde{F}_m : l^{m}\otimes \bigwedge^k \mathbb{T} \to l^{m+2k}\otimes \bigwedge^k \mathbb{T}^*$ by
\[
\widetilde{F}_m(X)=\sum_{i=0}^{[\frac{k}{2}]}\tilde{c}_{m,i}\otimes^{k-2i} d^{\nabla}\gamma((d^{\nabla}\gamma)^i(X))\wedge (d^{\nabla}\gamma)^i
\]
for $X\in l^{m}\otimes \bigwedge^k \mathbb{T}$.
The restriction $\widetilde{F}_m|_{D}$ induces the map from $\mathcal{O}(l^{m}\otimes \bigwedge^k D)$ to $\mathcal{O}(l^{m+2k}\otimes \bigwedge^k D^*)$.
\begin{prop}\label{s4.2p1}
The map $\widetilde{F}_m|_{D}:\mathcal{O}(l^{m}\otimes \bigwedge^k D) \to \mathcal{O}(l^{m+2k}\otimes \bigwedge^k D^*)$ is isomorphic
if $m$ satisfies $m \le -2n-2[\frac{k}{2}]-3$ or $-2n-3 \le m$.
\end{prop}
\begin{proof}
If $k\ge 2$, then
\[
\sum_{s=0}^{r} \tilde{c}_{m,s} \frac{r!}{(r-s)!}\frac{(n-k+r+s)!}{(n-k+r)!}=\frac{(\frac{m}{2}+n+r+1)\cdots(\frac{m}{2}+n+2)}{(\frac{m}{2}+k)\cdots(\frac{m}{2}+k-r+1)}
\]
for any $r=1,\dots,[\frac{k}{2}]$.
By the same argument in Proposition~\ref{s3p3}, the map $\widetilde{F}_m|_{D}$ is isomorphic
if $m \le -2n-2[\frac{k}{2}]-3$ or $-2n-3 \le m$.
\end{proof}

We define a constant $\tilde{c}'_{m,i}$ by
\[
\tilde{c}'_{m,i}=\frac{1}{(k+\frac{m}{2})(k+\frac{m}{2}-1)\cdots(k+\frac{m}{2}-i)\ i!}
\]
for each $i=0,1,\dots,[\frac{k-1}{2}]$.
These constants are well-defined since $\frac{m}{2}+k-[\frac{k-1}{2}]\neq 0$.
We define a map $\widetilde{G}_m:\Gamma(l^{m+1}\otimes\bigwedge^{k-1} D) \to \Gamma(l^{m+2k}\otimes\bigwedge^k \mathbb{T}^*)$ by
\[
\widetilde{G}_m(s)=\sum_{i=0}^{[\frac{k-1}{2}]}\tilde{c}'_{m,i}d^{\nabla}\left( \otimes^{k-1-2i}d^{\nabla}\gamma((d^{\nabla}\gamma)^i(s))\right)\wedge (d^{\nabla}\gamma)^i
\]
for $s\in \Gamma(l^{m+2}\otimes\bigwedge^{k-1}D)$.
Then we have
\begin{prop}\label{s4.2p2}
An $l^{2k+m}$-valued $k$-form $\{\widetilde{F}_m(X)+\widetilde{G}_m(\gamma(X))\}|_{D}$ restricted to $D$ is holomorphic for any holomorphic $l^m$-valued $k$-vector $X$.
\end{prop}
\begin{proof}
We fix a frame $\tilde{e}$ of $l$ on $U$. Then $\gamma$ is given by $\gamma=\tilde{e}^2\otimes \gamma_0$ for a holomorphic $1$-form $\gamma_0$ on $U$.
Let $X$ be a holomorphic $l^m$-valued $k$-vector $X$ on $M$.
By the same argument in the proof of Proposition~\ref{s3p2}, we have
\begin{eqnarray*}
\{\widetilde{F}_m(X)+\widetilde{G}_m(s)\}|_{D}&=&\tilde{e}^{2k+m}\otimes \Big\{ \sum_{i=0}^{[\frac{k}{2}]}\tilde{c}_{m,i}\{ \otimes^{k-2i} d\gamma_0((d\gamma_0)^i(X))\wedge (d\gamma_0)^i \\
&&+\sum_{i=0}^{[\frac{k-1}{2}]}\tilde{c}'_{m,i} d (\otimes^{k-1-2i} d\gamma_0((d\gamma_0)^i(\gamma_0(X))))\wedge (d\gamma_0)^i \Big\} \Big|_D.
\end{eqnarray*}
Hence $\{\widetilde{F}_m(X)+\widetilde{G}_m(\gamma(X))\}|_{D}$ is holomorphic.
\end{proof}
By repeating the proof of Theorem~\ref{s3t1} with $\widetilde{F}_m$ and $\widetilde{G}_m$ instead of $F$ and $G$, then we obtain
\begin{thm}\label{s4.2t1}
The sequence
\[
0\to \mathcal{O}(l^{m}\otimes \bigwedge^k D) \to \mathcal{O}(l^{m}\otimes \bigwedge^k \mathbb{T}) \to \mathcal{O}(l^{m+2}\otimes \bigwedge^{k-1} D) \to 0
\]
splits as sheaves of $\mathbb{C}$-module, and the sequence
\[
0\to H^i(l^{m}\otimes \bigwedge^k D) \to H^i(l^{m}\otimes \bigwedge^k \mathbb{T}) \to H^i(l^{m+2}\otimes \bigwedge^{k-1} D) \to 0
\]
is exact for each $i=0,\dots,2n+1$ if $k$ and $m$ satisfy one of following conditions
\[
\left\{
\begin{array}{lll}
m \le -2n-2[\frac{k}{2}]-4, &\quad 1\le k\le 2n+1, &\quad m : \text{ even}, \\
-2n-2 \le m \le -2k-2, &\quad 1\le k\le n, &\quad m : \text{ even}, \\
m \ge -2[\frac{k}{2}], &\quad 1\le k\le 2n+1, &\quad m : \text{ even}, \\
m \le -2n-2[\frac{k}{2}]-3, &\quad 1\le k\le 2n+1,&\quad m : \text{ odd}, \\
m\ge -2n-3, &\quad 1\le k\le 2n+1,&\quad m : \text{ odd}. \\
\end{array}
\right.
\]
$\hfill\Box$
\end{thm}

\section{Vanishing theorems for cohomology of $\bigwedge^k D$}
In this section, we apply the splitting theorems to the cohomology of $\bigwedge^k D$ and obtain the vanishing theorems.
From now on, we denote by $E \bigwedge^k D$ the tensor $E\otimes \bigwedge^k D$ of vector bundles $E$ and $\bigwedge^k D$ for simplicity.

\subsection{Vanishing of the cohomology on compact K\"ahler complex contact manifolds}
We have the following vanishing theorem of the cohomology on compact K\"ahler manifolds :
\begin{thm}\label{s4.1t1}
If $M$ is a compact K\"ahler complex contact manifold with $c_1(M)>0$, then
\begin{equation*}
H^i(M, L^m \bigwedge^{k} D)=\{0\} \quad
\left\{
\begin{array}{lll}
i\le 2n-k,& m \le -[\frac{k+1}{2}]-n-1,& 1\le k\le 2n+1, \\
\forall i,& -n\le m \le -k-1,& 1\le k\le n-1, \\
i\ge k+1,& m \ge -[\frac{k}{2}],&  1\le k\le 2n+1.
\end{array}
\right.
\end{equation*}
If $c_1(M)<0$, then
\begin{equation*}
H^i(M, L^m \bigwedge^{k} D)=\{0\} \quad
\left\{
\begin{array}{lll}
i\ge k+1,& m\le -n-[\frac{k}{2}]-2,&  1\le k\le 2n+1, \\
i\ge k+2,& m=-n-1,& 1\le k\le n, \\
i\le 2n-k-1,& m=-k,& 1\le k\le n, \\
i\le 2n-k,& m\ge -[\frac{k}{2}],&  1\le k\le 2n+1.
\end{array}
\right.
\end{equation*}
\end{thm}
\begin{proof}
By Theorem~\ref{s3t2}, the sequence
\begin{equation}
0\to H^i(M, L^m \bigwedge^k D) \to H^i(M, L^m \bigwedge^k \mathbb{T}) \to H^i(M, L^{m+1} \bigwedge^{k-1} D)\to 0 \label{s4.1eq1}
\end{equation}
is exact for each $i$ if $m$ and $k$ satisfy one of following three conditions
\begin{equation}
\left\{
\begin{array}{ll}
m \le -n-[\frac{k}{2}]-2, &1\le k\le 2n+1, \\
-n-1 \le m \le -k-1, & 1\le k\le n, \ \\
m \ge -[\frac{k}{2}], &1\le k\le 2n+1.
\end{array}
\right.\label{s4.1eq1.5}
\end{equation}
It follows from Serre's duality that
\begin{equation}
H^i(M, L^m \bigwedge^k \mathbb{T})^* \cong H^{2n+1-i}(M, \Omega^k L^{-m}K_{M})\cong H^{2n+1-i}(M, \Omega^k L^{-m-n-1}). \label{s4.1eq2}
\end{equation}

In the case of $c_1(M)>0$,
the first Chern class $c_1(L^{-m-n-1})$ of the line bundle $L^{-m-n-1}$ is negative
if $m>-n-1$ since $c_1(L)=-\frac{1}{n+1}c_1(K_M)>0$.
By applying the Kodaira-Akizuki-Nakano vanishing theorem~\cite{AN} to the last cohomology in (\ref{s4.1eq2}), we have
$H^i(M, L^m \bigwedge^k \mathbb{T})=\{0\}$ for $k+1\le i$ if $m>-n-1$.
Hence, the sequence (\ref{s4.1eq1}) implies
\begin{eqnarray}
&&H^i(M, L^m \bigwedge^{k} D)=\{0\}, \label{s4.1eq3} \\
&&H^i(M, L^{m+1}\bigwedge^{k-1} D)=\{0\} \label{s4.1eq4}
\end{eqnarray}
for $k+1\le i\le 2n+1$, $m>-n-1$ and $1\le k\le 2n+1$.
The condition (\ref{s4.1eq4}) is written by $H^i(M, L^{m'}\bigwedge^{k'} D)=\{0\}$ for $k'+2\le i\le 2n+1$, $m'>-n$ and $0\le k'\le 2n$.
We only consider the vanishing for $k'\ge 1$
since the case $k'=0$ that $H^i(M, L^{m'})=\{0\}$ for $2\le i$ and $m'>-n$ is induced by the Kodaira-Akizuki-Nakano vanishing theorem.
The first condition (\ref{s4.1eq3}) induces the second one (\ref{s4.1eq4}) for $k\ge 1$.
Hence, (\ref{s4.1eq1.5}) and (\ref{s4.1eq3}) imply that $H^i(M, L^m \bigwedge^{k}D)=\{0\}$ if $m$ satisfies one of two conditions
\begin{equation}
\left\{
\begin{array}{lll}
i\ge k+1,& -n\le m \le -k-1,& 1\le k \le n-1, \\
i\ge k+1,& m \ge -[\frac{k}{2}],& 1\le k\le 2n+1.
\end{array}
\right.\label{s4.1eq5.1}
\end{equation}
The Serre's duality implies that
\begin{equation}
H^i(M, L^m \bigwedge^{k} D)^* \cong H^{2n+1-i}(M, K_{M} L^{-m} \bigwedge^{k} D^*)\cong H^{2n+1-i}(M, L^{-k-m-n-1}\bigwedge^{k}D)\label{s4.1eq6}
\end{equation}
since $D^*=L^{-1}D$.
We apply the vanishing in (\ref{s4.1eq5.1}) to the last cohomology in (\ref{s4.1eq6}),
and obtain $H^i(M, L^m \bigwedge^{k}D)=\{0\}$ if $m$ satisfies one of two conditions
\begin{equation}
\left\{
\begin{array}{lll}
i \le 2n-k,& m \le -[\frac{k+1}{2}]-n-1,&  1\le k\le 2n+1, \\
i \le 2n-k,& -n\le m \le -k-1,& 1\le k\le n-1.
\end{array}
\right.\label{s4.1eq8.1}
\end{equation}
We remark that the condition $k\le -n-1$ implies that $2n-k\ge k+1$.
It follows from (\ref{s4.1eq5.1}) and (\ref{s4.1eq8.1}) that $H^i(M, L^m \bigwedge^{k} D)=\{0\}$ for
\begin{equation*}
\left\{
\begin{array}{lll}
i\le 2n-k,& m \le -[\frac{k+1}{2}]-n-1,& 1\le k\le 2n+1, \\
\forall i,& -n\le m \le -k-1,& 1\le k\le n-1, \\
i\ge k+1,& m \ge -[\frac{k}{2}],&  1\le k\le 2n+1.
\end{array}
\right.
\end{equation*}

In the case of $c_1(M)<0$, the first Chern class $c_1(L^{-m-n-1})$ is negative if $m<-n-1$.
Then the Kodaira-Akizuki-Nakano vanishing theorem implies
$H^i(M, L^m \bigwedge^{k} \mathbb{T})=\{0\}$ for $k+1\le i$ if $m<-n-1$.
It follows from (\ref{s4.1eq1}) that
\begin{equation}
H^i(M, L^m \bigwedge^{k} D)=\{0\} \quad
\left\{
\begin{array}{lll}
i\ge k+1,& m \le -n-[\frac{k}{2}]-2,& 1\le k\le 2n+1, \\
i\ge k+2,& m=-n-1,& 1\le k\le n.
\end{array}
\right.\label{s4.1eq13.5}
\end{equation}
Serre's duality implies
\begin{equation}
H^i(M, L^m \bigwedge^{k} D)=\{0\} \quad
\left\{
\begin{array}{lll}
i\le 2n-k,& m \ge -[\frac{k}{2}],& 1\le k\le 2n+1, \\
i\le 2n-k-1,& m=-k,& 1\le k\le n.
\end{array}
\right.\label{s4.1eq15}
\end{equation}
Hence it completes the proof.
\end{proof}

\begin{rem}
{\rm
Salamon proved that any $(p,q)$-cohomology $H^{p,q}(M)$ vanishes for $p\neq q$
if $M$ is the twistor space of a quaternion manifold with a positive scalar curvature~\cite{Sa}.
In the proof, he also obtained the vanishing of the cohomology of $\bigwedge^k D$ on the twistor space (Equation (6.4) in \cite{Sa}).
He used the notation of $L$ and $E$ as $L^{\frac{1}{2}}$ and $L^{-\frac{1}{2}}D$ in our notation, respectively.
These results are improved to the case of compact K\"ahler complex contact manifolds with $c_1(M)>0$ by the same argument,
and the vanishing is translated into
\begin{equation}
H^i(M, L^m \bigwedge^{k} D)=\{0\} \quad
\left\{
\begin{array}{lll}
\forall i,& -n\le m \le -k-1,& 1\le k\le n-1, \\
i\neq k,& m=-k,& 1\le k\le n.
\end{array}
\right.\label{s4.1eq16}
\end{equation}
The first condition in (\ref{s4.1eq16}) is equal to the second condition in Theorem~\ref{s4.1t1}.
The second one in (\ref{s4.1eq16}) is independent of our theorem.
However, we remark that the first and third conditions in Theorem~\ref{s4.1t1} are not induced by the vanishing (\ref{s4.1eq16}).
}
\end{rem}

\subsection{The vanishing of the cohomology on $\mathbb{CP}^{2n+1}$}
In this section, we show the vanishing theorem for
$H^i(l^{m}\wedge^k D)=H^i(\mathcal{O}(m)\wedge^k D)$ on $\mathbb{CP}^{2n+1}$ by using Bott's vanishing formula~\cite{Bo}.
We have the short exact sequence
\begin{equation*}
0\to D \to \mathbb{T} \stackrel{\gamma}{\to} l^2 \to 0. \label{s4.2eq1}
\end{equation*}
It induces the exact sequence
\begin{equation}
0\to l^{m} \bigwedge^k D \to l^{m} \bigwedge^k \mathbb{T}\stackrel{\gamma}{\to} l^{m+2} \bigwedge^{k-1} D \to 0 \label{s4.2eq2}
\end{equation}
for $m\in \mathbb{Z}$.
Then we have a vanishing of the cohomology as follows
\begin{thm}\label{s4.2t2}
$H^i(l^{m} \bigwedge^k D)=\{0\}$ if, in the case $m$ is even
\begin{equation*}
\left\{
\begin{array}{lll}
i\neq 2n+1,&  m\le -2n-2-2[\frac{k+1}{2}], & 1\le k\le 2n+1, \\
i\neq 2n+1-k,&  m= -2n-2, & 1\le k\le n, \\
\forall i,&  -2n\le m\le -2k-2, & 1\le k\le n, \\
i\neq k,&  m= -2k, & 1\le k\le n, \\
i\neq 0,&  m\ge -2[\frac{k}{2}], & 1\le k\le 2n+1
\end{array}
\right.
\end{equation*}
and, in the case $m$ is odd
\begin{equation*}
\left\{
\begin{array}{lll}
i\neq 2n+1,&  m\le -2n-3-k, & 1\le k\le 2n+1, \\
\forall i,& -2n-2-k\le m\le -2k+1, & 1\le k\le 2n+1, \\
\forall i,& -2n-3\le m\le -k, & 1\le k\le 2n+1, \\
i\neq 0,&  m\le -k+1, & 1\le k\le 2n+1.
\end{array}
\right.
\end{equation*}
\end{thm}
\begin{proof}
By applying Theorem~\ref{s4.2t1} to the sequence (\ref{s4.2eq2}),
we obtain the short exact sequence
\begin{equation}
0\to H^i(l^{m} \bigwedge^k D) \to H^i(l^{m} \bigwedge^k \mathbb{T}) \to H^i(l^{m+2} \bigwedge^{k-1} D)\to 0 \label{s4.2eq3}
\end{equation}
if
\begin{equation}
\left\{
\begin{array}{lll}
m \le -2n-2[\tfrac{k}{2}]-4, & 1\le k\le 2n+1, &\quad m : \text{ even}, \\
-2n-2 \le m \le -2k-2, & 1\le k\le n, &\quad m : \text{ even}, \\
m \ge -2[\tfrac{k}{2}], & 1\le k\le 2n+1, &\quad m : \text{ even}, \\
m \le -2n-2[\tfrac{k}{2}]-3, & 1\le k\le 2n+1,&\quad m : \text{ odd}, \\
m \ge -2n-3, & 1\le k\le 2n+1,&\quad m : \text{ odd}.
\end{array}
\right. \label{s4.2eq3.3}
\end{equation}
By applying Bott's vanishing formula to Serre's duality $H^i(l^{m} \bigwedge ^k \mathbb{T})^* \cong H^{2n+1-i}(\Omega^k (l^{-m-2n-2}))$,
$H^i(l^{m} \bigwedge ^k \mathbb{T})=\{0\}$ holds except for the following cases
\begin{equation}
\left\{
\begin{array}{ll}
i=2n+1-k, & m=-2n-2, \\
i=2n+1, & m< -2n-2-k, \\
i=0, & m> -k-1.
\end{array}
\right. \label{s4.2eq5}
\end{equation}
It follows from (\ref{s4.2eq3}) and (\ref{s4.2eq5})
that $H^i(l^{m} \bigwedge^k D)=H^i(l^{m+2} \bigwedge^{k-1} D)=\{0\}$
if
\begin{equation}
\left\{
\begin{array}{ll}
i\neq 2n+1-k, & m=-2n-2, \\
i\neq 2n+1, & m\le -2n-3-k, \\
\forall i, & -2n-2-k \le m\le -k-1,\ (m\neq -2n-2), \\
i\neq 0, & m\ge -k.
\end{array}
\right. \label{s4.2eq6}
\end{equation}
In the case that $m$ is even, (\ref{s4.2eq3.3}) and (\ref{s4.2eq6}) imply that
$H^i(l^{m} \bigwedge^k D)=\{0\}$ and $H^i(l^{m+2} \bigwedge^{k-1} D)=\{0\}$ if
\begin{equation}
\left\{
\begin{array}{lll}
i\neq 2n+1-k,&  m= -2n-2, & 1\le k\le n, \\
i\neq 2n+1,&  m\le -2n-4-2[\frac{k}{2}], & 1\le k\le 2n+1, \\
\forall i,&  -2n-1\le m\le -2k-2, & 1\le k\le n, \\
i\neq 0,&  m\ge -2[\frac{k}{2}], & 1\le k\le 2n+1.
\end{array}
\right. \label{s4.2eq9}
\end{equation}
We replace the vanishing $H^i(l^{m+2} \bigwedge^{k-1} D)=\{0\}$ for (\ref{s4.2eq9}) by $H^i(l^{m} \bigwedge^k D)=\{0\}$
for
\begin{equation}
\left\{
\begin{array}{lll}
i\neq 2n-k,&  m= -2n, & 0\le k\le n-1, \\
i\neq 2n+1,&  m\le -2n-2-2[\frac{k+1}{2}], & 0\le k\le 2n, \\
\forall i,&  -2n+1\le m\le -2k-2, & 0\le k\le n-1, \\
i\neq 0,&  m\ge -2[\frac{k+1}{2}]+2, & 0\le k\le 2n.
\end{array}
\right. \label{s4.2eq10}
\end{equation}
We remark that the case $k=0$ of (\ref{s4.2eq10}) is contained in Bott's vanishing formula.
We only consider the vanishing of $H^i(l^{m} \bigwedge^k D)$ for $k\ge 1$.
We summarize (\ref{s4.2eq9}) and (\ref{s4.2eq10}) as $H^i(l^{m} \bigwedge^k D)=\{0\}$ for even $m$ and
\begin{equation}
\left\{
\begin{array}{lll}
i\neq 2n+1,&  m\le -2n-2-2[\frac{k+1}{2}], & 1\le k\le 2n+1, \\
i\neq 2n+1-k,&  m= -2n-2, & 1\le k\le n, \\
\forall i,&  -2n\le m\le -2k-2, & 1\le k\le n, \\
i\neq 0,&  m\ge -2[\frac{k}{2}], & 1\le k\le 2n+1.
\end{array}
\right. \label{s4.2eq11}
\end{equation}
In the case that $m$ is odd, by repeating the above argument the conditions (\ref{s4.2eq3.3}) and (\ref{s4.2eq6}) imply that
$H^i(l^{m} \bigwedge^k D)=\{0\}$ for
\begin{equation}
\left\{
\begin{array}{lll}
i\neq 2n+1,&  m\le -2n-3-k, & 1\le k\le 2n+1, \\
\forall i,&  m= -2n-2-k, & 1\le k\le 2n+1, \\
\forall i,&  m= -2n-1-k, & 1\le k\le 2n, \\
\forall i,&  -2n-3\le m\le -k-1, & 1\le k\le 2n+1, \\
\forall i,&  m=-k, & 1\le k\le 2n-1, \\
i\neq 0,&  m= -k, & k= 2n+1, \\
i\neq 0,&  m\ge -k+1, & 1\le k\le 2n+1.
\end{array}
\right. \label{s4.2eq18}
\end{equation}
Applying (\ref{s4.2eq11}) and (\ref{s4.2eq18}) to Serre's duality
$H^i(l^{m} \bigwedge^k D)^*\cong H^{2n+1-i}(l^{-m-2n-2-2k} \bigwedge^k D)$,
we obtain $H^i(l^{m} \bigwedge^k D)=\{0\}$ if, in the case $m$ is even
\begin{equation}
\left\{
\begin{array}{lll}
i\neq 0,&  m\ge -2[\frac{k}{2}], & 1\le k\le 2n+1, \\
i\neq k,&  m= -2k, & 1\le k\le n, \\
\forall i,&  -2n\le m\le -2k-2, & 1\le k\le n, \\
i\neq 2n+1,&  m\le -2n-2-2[\frac{k+1}{2}], & 1\le k\le 2n+1
\end{array}
\right. \label{s4.2eq20}
\end{equation}
and, in the case $m$ is odd
\begin{equation}
\left\{
\begin{array}{lll}
i\neq 0,&  m\ge -k+1, & 1\le k\le 2n+1, \\
\forall i,&  m= -k, & 1\le k\le 2n+1, \\
\forall i,&  m= -k-1, & 1\le k\le 2n, \\
\forall i,&  -2n-1-k\le m\le -2k+1, & 1\le k\le 2n+1, \\
\forall i,&  m=-2n-2-k, & 1\le k\le 2n-1, \\
i\neq 2n+1,&  m= -2n-2-k, & k= 2n+1, \\
i\neq 2n+1,&  m\le -2n-3-k, & 1\le k\le 2n+1.
\end{array}
\right. \label{s4.2eq21}
\end{equation}
The conditions (\ref{s4.2eq11}) and (\ref{s4.2eq20}) in the case of even $m$,
and (\ref{s4.2eq18}) and (\ref{s4.2eq21}) in the case of odd $m$ induce the conditions in theorem.
Hence it completes the proof.
\end{proof}
\begin{rem}
{\rm
A contact structure on $\mathbb{CP}^{2n+1}$ is unique, up to automorphisms~\cite{NT}.
Hence Theorem~\ref{s4.2t1} and~\ref{s4.2t2} hold for any contact structure on $\mathbb{CP}^{2n+1}$.
}
\end{rem}

\begin{rem}
{\rm
In algebraic geometry, the {\it null correlation bundle} $N$ on $\mathbb{CP}^{2n+1}$ is defined by the short exact sequence
\[
0\to N \to \mathbb{T}(-1)\to \mathcal{O}(1) \to 0
\]
where $\mathbb{T}(-1)=\mathcal{O}(-1)\otimes \mathbb{T}$ (\cite{OSS}).
It induces to the following :
\[
0\to N(1)\to \mathbb{T} \to \mathcal{O}(2) \to 0.
\]
The bundle $N(1)=\mathcal{O}(1)\otimes N$ is the kernel of a transformation $A: \mathbb{T} \to \mathcal{O}(2)$
which is given by $a=\sum a_{ij}z_idz_j$ in the homogeneous coordinate.
If $a_{ij} = -a_{ji}$ and $(a_{ij})$ is non-degenerate, then $A$ induces a contact structure on $\mathbb{CP}^{2n+1}$.
By applying Theorem~\ref{s4.2t2} to $D= N(1)$,
we obtain the vanishing formula for the cohomology $H^i(\wedge^k N(m+k))$.
}
\end{rem}

\noindent
\textbf{Acknowledgements}.
The authors would like to thank Professor S. Nayatani
for his useful comments and advice.
The first named author is supported by Grant-in-Aid for Young Scientists (B) $\sharp$17K14187 from JSPS.

redpajama_set_name: RedPajamaArXiv


This time last year Met Eireann released three weather warnings, as eight counties faced the threat of flooding. With the April Showers approaching, it seems like a good time to assess the geography of flood risk in Ireland. Using our risk mapping platform Perilfinder, using a detailed 3D flood model from our partners in JBA, we can get an overview of where the greatest risk is. As output, we have calculated the number of residential buildings, from the Eircode database, that are at significant risk of flooding in each county. So where does the greatest risk lie?
As a percentage of total housing stock, the two outstanding counties with the highest risk score are Co Clare and Co Louth. Clare has just over 10% of its housing stock currently classified in areas deemed of significant risk in the flood model, and Louth's risk is just under this at 9%. The two counties are scoring highest because they are at risk of both coastal and river flooding and this brings more homes into the reckoning. At the other end of the scale, Longford is the place you are likely to remain driest in your abode, with only 1% of the housing stock under significant threat of flooding. Cavan and Westmeath also have a relatively low percentage of residences in high flood risk zones, at around 1% and 2% respectively.
In absolute terms, the county with the highest count of residential buildings at risk is as expected Dublin, with 12,347 residences at risk, followed by Cork with 6,722 and then Clare with 5,273.
At a national level, just under 60,000 residences are at significant risk from flooding, or just over 3% of the residential housing stock.
As well as an overall score, flood risk is further categorised in the model by the source of risk, covering river, coastal and surface and by likely water depth, so that damage impact can be assessed. For further details contact us at info@gamma.ie.
Gamma's Perilfinder risk mapping platform is a global cloud-hosted underwriting solution used by insurers to assess environmental risk. Perilfinder is a complete risk assessment and reporting solution for underwriters, incorporating address matching, rooftop-level geocoding, risk and accumulation scoring, and spatial catastrophe models. Technically advanced yet user-friendly, Perilfinder helps its users control their risk portfolios by better managing policy accumulations and through identifying geographical perils in real time. Perilfinder is a world-class, no-fuss, intuitive and feature-rich location intelligence system.


Having a strict "black and white" rain check policy isn't something we're comfortable with here at Beachwood. As we know, Mother Nature's mood is quite mercurial at times.
"If you tee off in the rain, then no rain check will be issued." That doesn't seem fair to us.
"Tee off on hole #1; a nine hole rain check will be issued, tee off hole #10; NO rain check will be given." That's pretty harsh too.
Moral compasses and common sense should prevail in these matters. We promise to be fair and honest with you; and in return, we ask that you be fair and honest with us.
Beachwood's dress code is quite lenient, comparatively, although we do have a few rules to maintain some sense of decorum. Men are not allowed to play wearing tank tops or sleeveless t-shirts (muscle shirts). Cut-off jean shorts are also not permissible for men or women. Blue jeans and t-shirts are allowed provided that they are clean and presentable. Women can wear just about anything, including tank tops. In recent history we've only had to say something to one woman, and that was when she was spotted playing in a bikini top.
Walking is NOT permitted at Beachwood at any time during the year. Carts are always required and therefore, all rates on this website and all rates quoted over the phone include cart fee.
Beachwood permits golfers to bring personal coolers to the course, provided that they fit easily into the basket on the golf cart. Don't worry if you forget, though. We have a grill, snack bar and a fully stocked beverage cart out on the course everyday!
When visiting the North Myrtle Beach area, you'll find that policies regarding junior golfers vary greatly from golf course to golf course. We recommend calling ahead when playing other courses (although we don't know why you would play anywhere but Beachwood) to get each individual course's junior policy. Our policy follows, along with some scenarios that we see every year to help better explain and avoid confusion at check-in.
Junior golfers age 16 and under play for just $9, one per each paid adult, from Memorial Day weekend through Labor Day weekend and after 10:00am only. No further discounts are available.
Scenario #1 – Father comes in with two juniors, ages 16 and 12. The sixteen year old has a valid driver's license, so Dad pays for himself and one of the juniors; remember one junior plays for $9 per each paid adult, and they're good to go.
Scenario #2 – Dad comes in with two juniors, ages 13 and 10. When we tell Dad that you must have a valid driver's permit or license to drive the golf cart he says, "No problem. My kids will take turns walking, or we'll all just ride in one cart." One problem… well, two problems – we don't allow walking and only two players are allowed to ride on a cart. Dad's only options at this point are to get paired with a single golfer old enough to drive a cart, or call Mom, Grandma, uncle Charlie, etc. and have them come along to drive the other cart. Beachwood charges $20 for non-playing riders, or in this case drivers, if an extra cart is not required.


text:
Q: Using the State design pattern in Objective-C I'm new to Objective-C and I'm learning OOP design patterns, but I'm an experienced functional programmer. I believe I need to use the State design pattern to implement my button, but I'm not sure what the best way to do it is.
Presently, the button has three states: ACTIVE, INACTIVE, and FINISHED. Depending on which state the button is in, it will be drawn in a different way. Instead of using a switch or subclassing the button for each state, my plan was to use three different categories, each overwriting the drawRect: method in MyButton. However, after looking around, I've discovered that this is highly discouraged.
What is the best way to tackle this? Should I subclass the button, use a category or a delegate, or is there another solution that I'm not thinking of?
EDIT: I should add that while I'm not building a game, I am basing my State pattern off of the explanation here.

A: The answer depends on how you are trying to change the appearance of the button.  If you can get what you want by setting the image or other standard attributes for the button, then I would leave the button class alone and just configure it from an external view model class.  Let the button be responsible for managing the UIControlState and configure it from your view controller or from a separate view model class.
If you need to drastically change the appearance of the button, then subclassing and providing your own drawRect: is the proper approach, although even there I would consider defining the class to have the minimal code needed to handle you different rendering needs and keep the state machine outside of the button.  Conceptually the button should be responsible for drawing itself and detecting user events, more complicated application logic should be handled by a delegate class.  Keep your UI components simple and reusable.

A: Consider delegate (by protocol) as a first option. Because delegate provides you a loosely coupled, flexible way to customize behaviors. (delegate = observer pattern) Anyway this is often feels too verbose.
Consider subclassing-and-overriding only when you have to. Subclassing/overriding is a powerful construct, and it 's very fragile like other too powerful construct. Because you need to properly handle calling of superclass method not to break the feature. Then it should be used very carefully. (recent compiler may warn on some cases) Unfortunately many Cocoa features are based on subclassing, then subclassing often becomes the only option. In this case, you're using UIView class, and UIView class does not support customizing drawing behavior by delegate, then you have to use subclassing with proper state value.
Never consider category to override a method. The feature is not intended for overriding, and actually implicit and silent overriding of existing method is one of the issue of category. (now you will be warned by the compiler in some cases)
You have another way called Objective-C blocks instead of delegate pattern which is an Objective-C version of closures. Blocks can replace delegate and target-action pattern in many cases, but  sometimes discouraged because its implicit capturing behavior may cause memory management issue.
Optimization
Though drawRect: is enough to provide custom look of a button, drawing is very expensive operation, and you need to avoid it as much as possible. If your button has a static look, then it would be better to use a bitmap image. UIButton already have methods to support this.
Not only for UIButton, many UIKit classes provides built-in look customization features, and I recommend you to consider using of the built-in features primarily.
And then, how to run a code for a specific state? UIButton already have a feature for that. Its superclass UIControl has a built-in method called addTarget:action:forControlEvents: and you can use this to register an observer method. This is an analogue to the delegate pattern, then it's more robust. Also, this is designed, and intended way to handle button behavior. Many stuffs in Cocoa are predefined like this, and you might need to spend some time to look for existing design intentions.
Consistency
These features are baked into the button class, and less flexible then providing your own drawing implementation by implementing your own state pattern/machine. But strongly recommended because this is the only way to provide semantically/practically exact behavior of typical system-default iOS app behavior.

A: If you want to overwrite the drawRect: in the button state then subclass will be the best option.

A: You should subclass a UIButton and create a new method to handle this. For example, you might call your subclass MYButton and put a new method called:

-(void) setButtonState: (NSInteger) state;

In it, you would want to do something like:

-(void) setButtonState: (NSInteger) state {

switch (state) {

case ACTIVE:

//Do ACTIVE CHANGES and so on like [self setImage: forState:] if you want to change the image, etc.

}

redpajama_set_name: RedPajamaStackExchange


text:
recreation time!
Launched in 1997, Renault Kangoo, originally designed as a van, won over an array of private customers looking for a vehicle to suit their family and leisure pursuits. Renault had set out to design a compact, spacious vehicle that didn't look like a utilitarian. In its passenger version, Renault Kangoo came with a liftgate and became part of a brand new trend on the automotive market, the "ludospace" or leisure activity vehicle (LAV).
THE DOOR THAT
MADE THE DIFFERENCE
Renault Kangoo quickly became a benchmark, offering a sliding rear side door to put it one step ahead of its French rivals, who lacked this feature on their launch. However, the other brands took note and eventually added one to their cars. That's what you call pioneering!
GET LOADING!
While Renault Kangoo replaced a commercial vehicle (Renault Express), its loading volume was greater than that of its predecessor (a plus point for a family vehicle!). The additional volume came from its two-box bodywork with a roof line running uninterrupted from front to rear, and the rear swing doors on the commercial version (while the LAV had a tailgate). The sliding side doors also made all the difference!
The 1990s were the golden age for summer hits. Renault Kangoo echoes the trend – what better way to combine family and leisure with a holiday at the wheel of this LAV, with the Macarena on the stereo?
THE END OF THE
STEPPED LINE
Renault Kangoo launched in 1997, taking over from the Renault Express, a small van derived from Renault Supercinq from which it inherited the front passenger space and rear cubic body constituting the loading area. This design was common on the small van market, and usually featured a "stepped" roofline between the front and rear sections. The pioneering Renault Kangoo did away with this step, giving it a more harmonious profile. And to forget its "utility" origins, Renault gave it cheerful-looking front face.
LPG AND NGV
from 1997 to 2007 in Maubeuge, France
from 1999 to 2018 in Argentina 2,300,000 units
transverse engine, petrol and diesel
from 999 cm3 to 1,870 cm3
55-95 hp depending on version electric, hybrid, LPG and NGV versions
135-170 kph depending on version
Height 1,86 m
two-box body with uninterrupted roofline
from front to back front-wheel drive + 4x4 version available
swing doors to the rear for the small van version,
tailgate on the leisure activity vehicle.
RENAULT PASSION NOW
HAS ITS OWN STORE!
A Renault fan or just a classic car enthusiast? There's something here for you!
Alongside its commercial versions (Renault Kangoo Express), Renault Kangoo was also available as an electric vehicle (Renault Kangoo Z.E.) used by various municipal services and authorities (like the French postal service) and company fleets. In 2008, the new Renault Kangoo Be-Bop stood out with its 3-door configuration, two-tone bodywork, sliding sunroof to the rear and rear tailgate window that slid down into the door. In Japan, Renault Kangoo gained cult status and features at big meets known as the "Kangoo Jamboree" near Mount Fuji.


change has a way of burning through your life - just when you get comfortable. choose. hunker down and hope you don't get burned or stand up and let the fire make you better. stand up and affirm that change isn't the enemy. comfort is the enemy.
I had such high hopes after averaging 18mph on a chunk of my ride two weeks ago. I envisioned climbing hills with ease, hauling a$$ down the bike lane, commanding Rabbonni (my husband's bike, f.k.a. Teacher, renamed in Aramaic. Because, well, seminary) with such skill and confidence that I look like someone who actually, I don't know, rides. So. Yeah. That so didn't happen.
There are a number of factors that contributed to the crazy slowness of this ride. It was a route I'd not ridden before. The roads were crazy rough. There were some hills. My legs were sore from yesterday's intervals and swim. But, mostly, it was just that I'm still such a newb.
I almost fell over at the first light - started to lean to the left, where I was still clipped in, and had to frantically unclip before I toppled over at the intersection in front of God and everybody.
Things went alright for a few miles after that, until I dropped a chain at another light - it was quite comical, really. I'd been waiting for what seemed like forever for the light to change. Low traffic means not very many cars to set off the light sensor. I was beginning to consider walking over to the crosswalk and pressing the button when a car came up behind me. Finally! I clip in and realize that there is no tension - the chain had dropped from the front ring. I then had to scurry over to the sidewalk - only on the left side of the road, of course - and fix my chain. As I was doing this, of course, the light changed. grrr. So, I used the crosswalk - walking the bike across the road in case I lost the chain in the intersection (terrifying thought!!) and continued.
About three miles from the end of my ride, it happened again. This time, I fixed it and it dropped again when I was about to take off. Three strikes and I was out. I called TriHubby and asked him to come and get me. Defeat.
As I was waiting for him, I looked at my ride stats on Strava. 11.9 mph Average?!?!? Then, I felt a little better when I saw I was awarded QOM for one segment. Cool! Until, I saw that I was the only one to attempt that hill this year. Looking at all time women - I was DFL. Insert deflating sound effect here. Crap.
So, I have a much longer way to go than I thought. At least as far as today's ride is concerned. You win some, you lose some, and sometimes, you fall over.
I'm a brand new pastor at St. Philip's UCC in South County St. Louis, a mom, a wife, a daughter, sister, friend. And, in my spare time, I like to swim, bike and run - with varying degrees of success. I love a challenge, I love new chapters, I love life in all its crazy turning.


LaFleur wanted to bring Arthur with him to Green Bay, per Kuharsky.
if we had actually conducted a search and he was the best candidate then i wouldn't care, it's just absurd that we didn't even bother and something that it seems like this organization likes to do on the regular, with typically horrible results.
How do you know we didn't bother interviewing anyone else?
We didn't fail to examine the coaching landscape. I don't think we just blindly stuck with him.
I think we looked around at what was possible and then looked in house.
I think we looked at the pros and cons of bringing in a guy outside to be oc.
For example, I don't think Josh McDaniels is the best OC in football. But he works best with that team.
We might not have the best OC in football ( yet to see) but we might have the best for our team. Someone who knows every player on that team. Who was around when each one of them was drafted. He has seen their worst and likely has an idea what their best is.
With this being said, I think that says he was on track for this promotion and more.
Kuharsky reported we looked in to 3 outside candidates as well, but ultimately went with Smith. So not sure "didn't conduct a search" is accurate based off that report.
I'm glad to hear the positive vibes from the players - seems like more than the usual lip service.
The big question to me - is he good at implementing other people's systems or does he have ideas of his own?
Love the continuity with Marcus and should have a good idea what Marcus likes.
Not crazy about the search as it seems he was the only interview, but I trust JRob and Vrabel.
Hopefully he is like what I read about some other OCs including McVay.
They had their foundation but was not against "stealing' something that works somewhere else. They would add it to their playbook and then adapt it to their scheme.
define "looked into." if we're not bringing people in for interviews it can't be reasonably said that we conducted a thorough search that placed the candidates on even ground.
I see this offense as a mix of whiz passing game. Mularkey's run game and protection. Lafluer's formation and set up.
With how loved Lafleur was and how not loved this hire is, maybe we'll get the same inverse effect on the field.