The Committee is composed of elected ELAC representatives from each of TRUSD schools with 21 or more English Language Learners.
All meetings will take place at the Twin Rivers Unified School district office building located at 5115 Dudley Blvd., McClellan, CA 95652.
Interpreters will be available in Spanish & Hmong. Intérpretes disponibles en Español & Hmong. Cov neeg txhais lus yuav muaj rau lus Mev & Hmoob.


text:
Q: Why does Mathematica not give the simple solution of the given equations? I want to solve this system of two equations for $x$
$$\{  \sin (2\pi x)+\sin (\frac{2\pi}y)=0   \quad ,  \quad \cos (2\pi x)-\cos (\frac{2\pi}y)=0   \}   \qquad (1)$$
I use Solve:
Solve[  {  Sin[ 2 π  x] + Sin[(2 π )/y ] == 0, Cos[ 2 π  x] - Cos[(2 π )/y ] == 0  }, x ] // Simplify

(*{{x -> ConditionalExpression[ArcTan[Cos[(2 π )/y], -Sin[(2 π )/y]]/(2 π ) + C[1], C[1] \[Element] Integers]}}*)

The result is in the form of a conditional expression:
$$x=\frac{1}{2\pi}\text{ArcTan}[\cos (\frac{2\pi}y),-\sin (\frac{2\pi}y)]+c_1, \qquad c_1\in\mathbb{Z}$$
Substituting $x=\dfrac{1}{2\pi}\arctan\left(\frac{-\sin (\frac{2\pi}y)}{\cos (\frac{2\pi}y)}\right)+m$ into $(1)$, I get
Sin[ 2 π  x] + Sin[(2 π )/y] /. x -> ((ArcTan[-(Sin[(2 π )/y]/Cos[(2 π )/y])]/(2 π )) + m) // FullSimplify[#,  Assumptions ->  y > 0 && y \[Element] Reals && m \[Element] Integers] &
Cos[ 2 π  x] - Cos[(2 π )/y] /. x -> ((ArcTan[-(Sin[(2 π )/y]/Cos[(2 π )/y])]/(2 π )) +  m) // FullSimplify[#,  Assumptions ->  y > 0 && y \[Element] Reals && m \[Element] Integers] &

(* (-1 + Sign[Sec[(2 π )/y]]) Sin[(2 π )/y]*)
(* (-1 + Sign[Sec[(2 π )/y]]) Cos[(2 π )/y] *)

which means that the equations vanish simultaneously if $\text{sgn} (\frac{2\pi}y)=+1$, i.e., only for those $y$ for which $\text{sgn} (\frac{2\pi}y)=+1$, we can simplify $x=\dfrac{1}{2\pi}\arctan\left(\frac{-\sin (\frac{2\pi}y)}{\cos (\frac{2\pi}y)}\right)+m$ as $x=m-\frac1y$

On the other hand, it is evident that both equations $(1)$ vanish at $x=m-\frac1y$ for all values of $y$ (while some of them may not satisfy $\text{sgn} (\frac{2\pi}y)=+1$).

 Sin[ 2 π x] + Sin[(2 π)/y] /. x -> m - 1/y //  FullSimplify[#, Assumptions ->  y > 0 && y \[Element] Reals && m \[Element] Integers] &

Cos[ 2 π x] - Cos[(2 π)/y] /. x -> m - 1/y // FullSimplify[#,  Assumptions ->  y > 0 && y \[Element] Reals && m \[Element] Integers] &

(*0*)
(*0*)

A: The reason that the second code snippet in the OP doesn't evaluate to zero is because ArcTan[x, y] is not the same as ArcTan[y/x] for every choice of x and y.  This is explained in the documentation for ArcTan: it matters which quadrant the points $(x,y)$ is in. To see this, notice:
Plot[{
   ArcTan[Cos[(2 π)/y], -Sin[(2 π)/y]]/(2 π),
   ArcTan[-Sin[(2 π)/y]/Cos[(2 π)/y]]/(2 π)},
 {y, 1, 6},
 PlotRange -> {{0, 6}, All}, PlotPoints -> 100]

results in

Indeed, if we plug the solution that Solve actually spits out, we get zero:
FullSimplify[
  Sin[2 π x] + Sin[(2 π)/y] /. x -> ArcTan[Cos[(2 π)/y], -Sin[(2 π)/y]]/(2 π) + m,
  m ∈ Integers]
(* 0 *)

The second issue in the OP is that Mathematica doesn't simplify to the simple solution m - 1/y, which is a solution (with caveats), which we can see from
Simplify[ArcTan[Cos[(2 π)/y], -Sin[(2 π)/y]]/(2 π) + m, {y > 4}]

However, the issue is that the function actually has an infinite number discontinuities as we approach $y=0$. On each continuous piece of the function, it is equal to m - 1/y for some integer m, but it seems like Mathematica is unable to come up with a general expression for the location of the discontinuities, and hence it doesn't simplify the expression to m - 1/y until it's sure that there will be no discontinuities.
To see this, consider the plot:
Plot[ArcTan[Cos[(2 π)/y], -Sin[(2 π)/y]]/(2 π), {y, 0.2, 6}, PlotRange -> {{0, 6}, All}, PlotPoints -> 300]

Now, interestingly enough, we can paste the solutions that Mathematica spits out together (i.e., the set of solutions for every m), and it turns out that it maps out the set of curves m - 1/y!
GraphicsRow[{
  Plot[Table[ArcTan[Cos[(2 π)/y], -Sin[(2 π)/y]]/(2 π) + m, {m, -5, 5}] // Evaluate, {y, 0.2, 6}, PlotRange -> {{0, 6}, {-6, 6}}],
  Plot[Table[m - 1/y, {m, -5, 5}] // Evaluate, {y, 0.2, 6}, PlotRange -> {{0, 6}, {-6, 6}}]
  }]

I suspect that because Mathematica tries to find general solutions first, it writes things in terms of ArcTan[x, y], and once it's there, it gets stuck, because it needs to care about an infinite number of discontinuities, and it's hard for a computer algebra system to dealt with and recognize that m - 1/y is a general solution to the problem.

Finally, note that Mathematica does at least know that this family of curves solves the original set of equations:
Sin[2 π x] + Sin[(2 π)/y] /. x -> m - 1/y /. Sin[a_] :> Sin@Expand@a // TrigExpand // Simplify
Cos[2 π x] - Cos[(2 π)/y] /. x -> m - 1/y /. Cos[a_] :> Cos@Expand@a // TrigExpand // Simplify
(* 2 Cos[π (m - 2/y)] Sin[m π] *)
(* -2 Sin[m π] Sin[π (m - 2/y)] *)

From which we can see that both of these are clearly zero when m is an integer.

A: I think your assumption is that $y\in \mathbb{R}$ is real, so you can put it as the condition (otherwise it is assumed to be complex):
SolveValues[
Sin[2π x]+Sin[2π/y]==0&&
Cos[2π x]-Cos[2π/y]==0&&
y∈Reals
,x]//FullSimplify

So after FullSimplify you get 2 answers that you are looking for:
$\begin{array}{l}
 \fbox{$-\frac{1}{y}+2 c_1\text{ if }y\in \mathbb{R}\land c_1\in \mathbb{Z}$} \\
 \fbox{$-\frac{1}{y}+1+2 c_1\text{ if }y\in \mathbb{R}\land c_1\in \mathbb{Z}$} \\
\end{array}$
If you are looking for a similar compact answer in the complex domain then this should do:
SolveValues[
Sin[2π x]+Sin[2π/y]==0&&
Cos[2π x]-Cos[2π/y]==0
,x]//FullSimplify

$\fbox{$c_1-\frac{i \log \left(e^{-\frac{2 i \pi }{y}}\right)}{2 \pi }\text{ if }c_1\in
   \mathbb{Z}$}$

redpajama_set_name: RedPajamaStackExchange


Putting our Agreement in Place In this agreement we are wideshoe4u.co.uk the online shop of Ian W Elsby Footfitters Ltd., 95, St. Johns Road, Waterloo, Merseyside, L22 9QD and you are the person detailed on the order form. Our agreement is made on the date we deduct payment from your credit card the details of which you provided on your order form. Our acceptance of your order by delivery of the goods constitutes a legally binding contract between us on these terms and conditions.
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Category: Radio Turn-key Solutions. Tag: TURNKEY SOLUTION.
The goal is to provide broadcasters with a complete plug and play solution ready to go On Air.
The Turnkey Radio is a comprehensive On Air Studio designed by Eletec Broadcast to integrate all the essential equipment for a Radio Station to have the best ratio between quality, functionality and economical investment.


The streets of Kuala Lumpur were taken over by bicycles between the 17th and 19th of January. The OCBC Cycle Malaysia events were back in town for the third time.
This year there were seven events on Saturday, including a tricycle ride for 2 to 5 year olds, kids rides for various age groups, an Ultimate Foldies Challenge, and a Criterium.
3,000 of us rode in one of the two events on Sunday. The 48 km Challenge ride. Four loops on closed roads in the center of KL. Click on the photograph below for a larger image. Zoom in and you'll see the PETRONAS Twin Towers just outside the loop on the right.
The start was scheduled for 6.30am, in front of the iconic Sultan Abdul Samad Building.
The guest of honor and a participant in the Challenge Ride was the Minister of Youth and Sports, Yang Berhormat Encik Khairy Jamaluddin (number V9028).
It was nice to have his support, but I wish he had been on time. The poor drummers had to work overtime to keep the riders entertained while we all waited.
We got underway at a few minutes past 7.00am. Once the speed demons had squeezed their way through to the front we all settled down to a relaxed spin through the city. I struck up a conversation with a rider next to me on the second lap. Alan and I kept each other company for the rest of the ride.
While I was waiting after the finish a reader of this blog came up to me and said "hello." What a treat that was! He came all the way from Kuala Terengganu to ride in The Challenge.
Azlan, Shahfiq and I met up here. I'm not sure how anyone found their friends in crowds like this before we all had mobile phones.
The rest of the Racun and Flipsiders groups met up on the other side of the finishing area to take a photograph with their medals.
Azlan, Shahfiq and I cycled back onto the ride route to get to the Twin Towers area, where they had parked. As we passed Kedai Makanan Yut Kee on Jalan Dang Wangi we noticed some familiar faces at one of the tables.
Jason, Cedric, Keat and others were tucking into breakfast. The three of us stopped and had soft-boiled eggs, toast with kaya and coffee too. A great end to a fun morning on my bicycle with good friends.
Posted in Cycling in Malaysia and tagged Kedai Makanan Yuk Kee, Kuala Lumpur, OCBC Cycle Malaysia 2014, Sultan Abdul Samad Building. Bookmark the permalink.


Hostal Boquete i Boquete, Panama: Reservér nu!
Cancellations made later than 48 hours before the date of arrival are non-refundable.
We pride ourselves in giving the very best of customer service and our bi-lingual staff is always available to help you in anyway possible.
Without question, you will find our charming rooms and unique riverside community areas as tranquil as the mountain breezes that make this area so special. We also provide free WIFI and computer access to our guests.
We strive to have each one of our rooms be uniquely decorated and spotlessly clean. All rooms are private rooms, featuring private bathrooms, and hot water. Our helpful and knowledgeable staff are always available and are happy to assist you during your stay with us.
In addition to our gorgeous riverside community sitting area, Hostal Boquete also features a guest-only kitchen and dining area.
Please note that our office hours are 7am- 10pm. If you plan to arrive later than 10pm, please contact us before your arrival so that we can make other arrangements to welcome you.


text:
Hotel Mumbai: Anupam Kher gives us a glimpse of the official trailer
Actor Anupam Kher treated fans with a part of the official trailer of Hotel Mumbai on Thursday.
Anupam Kher has been enjoying a fair bit of attention over the last couple of weeks, thanks to his upcoming film The Accidental Prime Minister. The film releases on Friday but there's no rest for the man. He's all ready to promote his next. The actor will be featured alongside Dev Patel in Hotel Mumbai, the teaser of which was dropped on Thursday. The short promotional video takes us back to the horrifying 26/11 attacks of Mumbai. The film has an interesting line up of casts including Dev Patel, Nazanin Boniadi, Armie Hammer, Tilda Cobham-Hervey, among others and is helmed by Anthony Maras.
The film focuses on the devastating Mumbai terror attacks that took place at the Taj Mahal Hotel in Mumbai on November 26, 2008. The short view of the trailer showcases Kher, who plays a renowned chef, and Dev, who plays a waiter, risk their lives to ensure the safety of their guests during the attack. "Remember always, here at the Taj, your guest is God," instructs Kher to the group of waiters at the start of the video. The video also shows us how the group of terrorists enter the city and open fire at the opulent hotel. A portion of the video will make you relive the shock of the explosion - the part where Kher asks everyone to run for their lives but just at the moment they begin whooshing, the big blast takes place. We also get a glimpse of a warm family moment shared by Dev with his wife and children towards the end.
Ladies & Gentlemen!! Presenting a part of OFFICIAL TRAILER of #HotelMumbaiFilm starring #DevPatel, @ArmieHammer, @NazaninBoniadi @therealjasonisaacs aur aapka apna Anupam. 😍 The gripping true story of 2008 siege on Mumbai's Taj Hotel arrives in theaters this March. 🙏
A post shared by Anupam Kher (@anupampkher) on Jan 9, 2019 at 12:29pm PST
The Indian film industry has made many attempts to make films on the subject, but have failed to make their mark. In the year 2013, Ram Gopal Varma helmed The Attacks of 26/11 which featured Sanjay Jaiswal, Nana Patekar and more. In 2015, Nicolas Saada wrote and directed Taj Mahal, a French-Belgian thriller drama which is based on the story of a teen French girl who was alone on the hotel room when the terrorists attacked the hotel. In 2009, Terror Mumbai featured exclusive audio tapes of the intercepted phone calls between the gunmen and their controllers in Pakistan. And, Mumbai Siege: 4 Days of Terror made in 2018 essays the situation of some foreigners inside Taj during the attack.
Will Hotel Mumbai live up to expectations? Let us know what you think.
26/11 AttacksAnupam KherBollywoodDev PatelEntertainmentHotel MumbaiTaj Hotel
nextWhy Cheat India: Emraan Hashmi's Cheat India gets a new title after CBFC objection

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After sending to bat by MS Dhoni, RR got the drizzling start as their openers Ajinkya Rahane and Jos Buttler were again looking good. But again their misfortune continues and did not manage to score big. There was a collapse after them. Sanju Samson also did not click today. Steve Smith again struggle as he scored only 15 runs off 22 balls. Rahul Tripathi also got out cheaply on 10. Ben Stokes scored 28 runs off 26 balls, Riyan Parag also made an early exit on 16 runs. Jofra Archer and Shreyas Gopal added some valuable runs in the last over for Rajasthan Royals.
Shreyas Gopal scored 19 runs off 7 balls and Jofra Archer scored quick 13 in the end guided their team to a respectable total of 151 runs in 20 overs.
Rajasthan Royals Score : 151-7 in 20 overs.
In reply of Rajasthan Royals 151 runs, Chennai Super Kings were never looking comfortable out there. Shane Watson out on 0, Faf Du Plessis on 7, Suresh Raina on 4 and Kedar Jadhav on 1 run. At one stage they were down and out as a scorecard looking very pathetic with 24 runs on the board and 4 wickets were down in 6 overs. After that, it was Ambati Rayudu and MS Dhoni who took the responsibility and added 95 runs in 68 balls. But still, it was looking way too far from the reach of Chennai Super Kings.
They however took the match to the last over with 18 runs to get. Contention directly toward the end as Dhoni is out on the ground from the uncovered and he is completely angry with the umpires. The square leg umpire didn't call no-ball however the other umpire had give a full hurl from Stokes as a no-ball. Dhoni is telling his inclination as he out in the center.
Chennai Super Kings have beaten Rajasthan Royals by four wickets as Mitchell Santner's last-ball six wins for the meeting group in Jaipur. Prior, RR was limited by CSK bowlers for 151/7 of every 20 overs. CSK lost four early wickets in the pursuit however association among Dhoni and Rayudu got them close. Stirs surrendered 21 keeps running from the last over as CSK won their third match on the jog.
Result: Chennai Super Kings beat Rajasthan Royals by 4 wickets.


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Measuring diabetes self-care: a psychometric analysis of the Self-Care Inventory-Revised with adults.
The relationship of depressive symptoms to symptom reporting, self-care and glucose control in diabetes.
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The impact of barriers and self-efficacy on self-care behaviors in type 2 diabetes.
Deviation from developmentally appropriate self-care autonomy. Association with diabetes outcomes.


text:
"Gandhi is a Babe!" - Why Every Hindu Should Read Sankaran Nair's Verdict on Mohandas Gandhi
Excerpts from Sri C. Sankaran Nair's searing assessment of Mohandas Gandhi's disastrous role in the Indian freedom struggle
ENTIRELY CONSISTENT WITH the despotic and nepotistic methods of the Congress party after India attained a questionable Independence, the Nehruvian Congress bulldozed every attempt that told the truth about the party, its "great leaders" and their role in the freedom struggle. Some of these despotic methods included but were not limited to banning books, jailing people, destroying careers, and creating amnesia among the public about the real stalwarts who fought and sacrificed so much only to see India free.
One such stalwart was Chettur Sankaran Nair, a fearless, blunt and acerbic truth-teller, especially about the fake saint, Mohandas Gandhi and his serial blunders as a public figure who usurped the freedom struggle from Lokamanya Tilak.
Ever since the said usurpation, Sankaran Nair quickly saw through and noted both with alarm and anger how Gandhi systematically began to hoodwink the Congress and the gullible Indian masses with his loin-cloth trickery. Nair's fury peaked after the Moplah genocide of Malabar Hindus.
The outcome was the nuclear book titled Gandhi and Anarchy, still a classic. Needless, the book was promptly made to disappear after the British left India in a hurry. It is an early record chronicling Gandhi's list of sins spread over seven years and is still a primary source to understand the real Gandhian side of the Indian freedom movement.
The following are some excerpts from its brilliant preface, which every Hindu needs to read for reasons that will become evident once you finish reading it.
Gandhi and Anarchy: Preface
"The struggle for Indian Home Rule which was started with the inauguration of the Indian National Congress has many difficulties to encounter, has strong and powerful opponents and has received many checks. But its strongest opponent is Mr. Gandhi and perhaps the most severe check it has received is the adoption by the National Congress at his instance in Calcutta and Nagpur of the so-called-Non-violent Non-co-operation.
"Non-co-operation as advocated by Mr. Gandhi may be a weapon to be used when constitutional methods have failed to achieve our purpose. Non-violence and passive suffering will lead to bloodshed or be unfruitful of any satisfactory results...It was rightly realised by many..."that the path of the progress of the Gandhi movement fused with the Khilafat element will be bloody.
The Misguided Mystic's Misadventures: R.C. Majumdar's Assessment of Mohandas Gandhi's Ideals and Methods
The claim for Indian Home Rule rests upon very different grounds. The Hindus have nothing to do with the Khilafat agitation...While the claim itself rested on such slender grounds, the means first adopted to enforce the claim were grotesque.
The methods advocated by Mr. Gandhi and the Congress are directed against Western civilization; against the class which fought for and won the reforms; and the Montague reforms scheme of constitutional progress. They have failed miserably and as was natural more violent methods leading to direct conflict with the forces of Government have been advocated which would in all probability have been carried out but for the arrest and imprisonment of Mr. Gandh
Mr. Gandhi belongs to a class of thought which has attracted some of the noblest minds in this world, but in applying his the gospel of life to politics, he has shown himself a babe and his interference has been generally mischievous.
In South Africa he is responsible for creating a situation which makes a peaceful and satisfactory solution practically impossible. His factious policy in India stands in the way of further reforms. The opposition to Gandhi was however not strenuous. The so-called Moderates only whispered their protests against his policy so as not to be heard beyond a few feet.
They are loud however, in their denunciation of Government action to check the illegal activities of Mr. Gandhi and his followers. It can hardly be doubted that their cautious attitude has contributed to the growth of the Gandhi movement.
Mohandas Gandhi Orders the Indian National Congress to Succumb to Moplah Jihadis in the Infamous Resolution No. 3
There is scarcely any item in the Gandhi programme which is not a complete violation of everything preached by the foremost sons of India till 1919; which has not been strongly even vehemently denounced by those old respected members of the Congress who now follow Mr. Gandhi: Pandit Malaviya, Messrs. Vijayaragavachari, Lajapat Rai, Natarajam, S. Kasturiranga Iyengar, the Editor of the 'Hindu."
Mr. Gandhi's emotional outbursts, fastings, penances, Sanyasi waist cloth, may carry away the emotional masses, women and students. But whether this wave of emotionalism submerged the men above named I would not care to guess.
No one of course has any right to find fault with his genuine followers like Mr. Prakasam, Editor, 'Swaraj' whose motives, however much we might differ from his politics, no one will question. He is one of those genuine patriots who believes in the efficacy of Mr. Gandhi's methods to obtain Home Rule. By far the great majority however, follow him for other reasons.
Mohandas Gandhi's Dogmatic Ahimsa and Adharmic Experiments with Truth: An Analysis
The severe simplicity and austerity of Mr. Gandhi's life combined with his appeal to the principle of 'Ahimsa' non-injury inherited from Buddists and now ingrained in Hindu life, has secured him the support of the Hindu masses and particularly vegetarians.
His indiscriminating support of the extreme Khilafat demands has ensured the Mahomedan support. Some politicians who naturally desire to use him and the influence he has acquired for putting pressure on the Government to concede further reform, also have joined him. But I am satisfied he is using them all to further his own ends. An attempt in which he is bound to fail.
His success to obtain what they call Dominion status or Home Rule will only lead to bloodshed and anarchy...I have attempted in the following pages to give my reasons for these conclusions. I have also given a list of riots or disturbances. These give a fair idea of the activities of Mr. Gandhi."
Do read Sankaran Nair's work in full.
Khilafat Movement
Moplah Genocide of Malabar Hindus
Sankaran Nair
Mohandas Gandhi
History of the Indian National Congress
Indian freedom movement
Home Rule Movement
Role of Muslims in the Freedom Struggle
Gandhi and Anarchy


That ShopBop Sale is On Now!
The ShopBop sale is back! Can I get a hell yeah?!
One of my favorite places to shop would be this site because, well, A) you get Amazon Prime shipping (perfection) and B) they always have really stylish stuff for deep discounts if you hit the sales right!
And this week is one of those weeks, friends. The universe has thrown us a gift in the middle of a tough week and it's a good one.
Not going to lie, I've been kinda hanging in the background when it comes to fashion lately as I try to get the rest of my life together but I always keep my eye out for good deals and what I might be wanting to add to my wardrobe at the change of the seasons.
I recently scooped this suede moto jacket!
Click to access to Shopbop's biggest sale of the season! With code EOTS17 you can get 20% off your whole purchase, even sale items, or 25% off if you purchase $500 or more (some brand exclusions)! Since it's now officially fall–enjoy all the jewel-toned extras that are sure to make your wardrobe pop! Have a great Pre-Friday!


text:
Omicron COVID Variant Symptoms: What We Know
On Nov. 26, the World Health Organization (WHO) classified a new COVID-19 variant of concern: Omicron or B.1.1.529. There's much scientists need to learn about the variant in order to fully gauge its transmissibility and how effective current vaccines are against it, as well as what the Omicron variant symptoms are. The CDC has not identified any instances of the new variant, reported first to WHO by South Africa, in the United States as of now, and President Joe Biden recently told Americans it was not a cause for panic.
What Are Omicron COVID Variant Symptoms?
More research is needed to clearly identify Omicron variant symptoms. WHO states that it's not yet clear whether people infected by the Omicron COVID-19 variant have worse cases or differing symptoms compared to other variants such as the highly contagious Delta variant, which accounts for the heavy majority of cases in the US currently, according to the CDC's national genomic surveillance system.
WHO noted in an update about Omicron that initial reports of COVID-19 cases caused by the new variant were in university students — and younger people tend to present with more mild cases. Dr. Angelique Coetzee, who is reported as being one of the first known doctors treating patients with the Omicron COVID-19 variant, told BBC that consistent symptoms she saw in patients were as follows:
Body aches and pain
Scratchy throat
These symptoms, in her opinion, don't "fit in with" the Delta variant. Dr. Coetzee, also the chairwoman of the South African Medical Association, told CNN that patients at this point "are not severely ill," don't need to be hospitalized, and can be treated at home. Reuters also spoke to the doctor, who noted that her patients are not reporting loss of smell or taste and there are no major drops in oxygen levels.
WHO does state that preliminary data suggest there are higher rates of hospitalization in South Africa, where Omicron is spreading, "but this may be due to increasing overall numbers of people becoming infected, rather than a result of specific infection with Omicron."
Understanding the severity of the new variant, WHO said, will take up to several weeks, further stating, "All variants of COVID-19, including the Delta variant that is dominant worldwide, can cause severe disease or death, in particular for the most vulnerable people, and thus prevention is always key."
Omicron has large number of mutations, some of which WHO said are cause for concern; however, Dr. Noc on TikTok, a scientist with a PhD in pharmaceutical sciences and immunology, said in a Nov. 29 video he wanted to emphasize that "simply having a large number of mutations does not necessarily make the virus much, much worse." He also said it's hard to predict right now what that combination of mutations will do and, echoing WHO, that pinpointing actual severity of the disease caused by the Omicron variant could take months.
POPSUGAR aims to give you the most accurate and up-to-date information about the coronavirus, but details and recommendations about this pandemic may have changed since publication. For the latest information on COVID-19, please check out resources from the WHO, the CDC, and local public health departments.
Sam Thompson in hysterics after thieves ditch girlfriend Zara's stolen suitcase because items inside 'suck'
Low-income families and single parents will be 'devastated' by rising energy bills


text:
\section{Introduction}

Anomaly flow is a geometric flow of  Hermitian metrics studied in \cite{AF1,AF9,AF2,AF8,AF3,AF4,AF5,AF6,AF7,UgartePujia}. The flow was originally considered in \cite{AF8} on complex threefolds to study the Strominger system \cite{Strominger} and involves a real parameter $\alpha'$. The flow was subsequently generalized to any complex dimension $n \geq 3$ for $\alpha'=0$ in \cite{AF7}.  The latter evolves an initial Hermitian metric $\omega_0$ on a compact complex manifolds $M$
of complex dimension $n\geq 3$ with $c_1(M)=0$ by
\begin{equation}\label{phongflow}
\tfrac{\partial}{\partial t}(|\Omega|_{\omega}\omega^{n-1})=i\partial\bar\partial \omega^{n-2}\,,\quad \omega(0)=\omega_0\,,
\end{equation}
where $\Omega$ is a fixed complex volume form and $|\cdot|_{\omega}$ is the pointwise norm with respect to $\omega$. By \lq\lq complex volume form\rq\rq $\,$ we just mean a nowhere vanishing $(n,0)$-form, indeed for the purpose of the present paper we do not need to assume $\Omega$ to be holomorphic.

The well-posedness of the flow is proved in \cite[Theorem 1]{AF7}  under the assumption on $\omega_0$ to be conformally balanced (in such a case the components of $\omega(t)$ satisfy a parabolic system \cite[Theorem 4]{AF7}). Moreover, when $\omega_0$ is conformally balanced, \eqref{phongflow} is conformally equivalent to the Hermitian curvature flow introduced by Ustinovskiy in \cite{YU1} (see \cite{AF9}).

The flow \eqref{phongflow} can only converge when $M$ is K\"ahler. The research of the present paper is motivated by the following theorem about the long time existence and convergence of the flow when  $|\Omega|_{\omega_0}^{1/(n-1)}\omega_0$ is a K\"ahler metric:

\begin{theorem}[Phong, Picard and  Zhang  {\cite[Theorem 2]{AF7}}]\label{phong}
Let $(M, \chi)$ be a compact K\"ahler manifold with vanishing first Chern class and let $\Omega$ be a complex volume form on  $M$ with constant norm with respect to $\chi$. Let $\omega_0$ be a Hermitian metric on $M$ such that
$$
|\Omega|_{\omega_0}\omega^{n-1}_0=\chi^{n-1}\,,
$$
then \eqref{phongflow} starting from $\omega_0$ has a long-time solution which converges in $C^{\infty}$--topology to the unique Ricci-flat  K\"ahler metric $\omega_{\infty}\in [\chi] \in H^{1,1}(M)$ .
\end{theorem}
The theorem gives an alternative proof of the Calabi-Yau theorem \cite{yau}. In \cite{AF7} it is raised the problem of studying the convergence of the flow in K\"ahler manifolds when $|\Omega|_{\omega_0}^{1/(n-1)}\omega_0$ is not K\"ahler, for instance when $|\Omega|_{\omega_0}\omega^{n-1}_0$ is just closed and $[|\Omega|_{\omega_0}\omega^{n-1}_0]=[\chi^{n-1}]$, with $\chi$ K\"ahler.
Here we prove that \eqref{phongflow} is stable around Calabi-Yau metrics $\chi$ with $|\Omega|_{\chi}$ constant. In particular we have the convergence of \eqref{phongflow}, when $|\Omega|_{\omega_0}\omega^{n-1}_0$ is just close to the $(n-1)$-th power of a K\"ahler metric.

\begin{theorem}\label{main}
Let $(M, \chi)$ be a compact K\"ahler manifold with vanishing first Chern class and let $\Omega$ be a complex volume form on  $M$ with constant norm with respect to $\chi$. For every $\epsilon>0$ there exists $\delta>0$ such that if $\omega_0$ is a Hermitian metric on $M$ satisfying
\begin{equation}\label{vicino}
\|\,|\Omega|_{\omega_0}\omega^{n-1}_0-\chi^{n-1}\|_{C^{\infty}}<\delta\,,
\end{equation}
then flow \eqref{phongflow} has a long-time solution $\omega(t)\in C^{\infty}(M\times [0,\infty),\Lambda_{+}^{1,1})$ such that
$$
\|\,|\Omega|_{\omega(t)}\omega(t)^{n-1}-\chi^{n-1}\|_{C^{\infty}}<\epsilon\,,\mbox{ for every }t\in [0,\infty)
$$
and $|\Omega|_{\omega(t)}\omega(t)^{n-1}$ converges in $C^{\infty}$--topology to a positive $(n\!-\!1,n\!-\!1)$-form $\omega_{\infty}^{n-1}$ with $\omega_{\infty}$ astheno-K\"ahler.

If further $\omega_0$ satisifies the conformally balanced condition $d(|\Omega|_{\omega_0}\omega_{0}^{n-1})=0$, then $\omega_\infty$ is K\"ahler Ricci flat.
\end{theorem}

We prove Theorem \ref{main} as follows:

\smallskip
after the change of variable $|\Omega|_{\omega}\omega^{n-1} = \tilde\omega^{n-1}$, equation \eqref{phongflow} rewrites as
\begin{equation}
\label{ourflow}
\tfrac{\partial}{\partial t} \tilde\omega^{n-1}=i\partial\bar\partial (|\Omega|_{ \tilde\omega}^{-2} \tilde\omega^{n-2})\,,
\end{equation}
(see Lemma \ref{lemma31}).

If we set $E(\tilde\omega^{n-1})=i\partial\bar\partial (|\Omega|_{ \tilde\omega}^{-2} \tilde\omega^{n-2})$, then the linearization $DE(\alpha^{n-1})$ of $E$ at any positive $(n-1,n-1)$-real form $\alpha^{n-1}$ satisfies
$$
DE(\alpha^{n-1})(\psi)=-\frac{1}{n-1}|\Omega|_{\alpha}\square \psi+{\rm l.o.t.}
$$
for any {\em closed} $\psi\in C^{\infty}(M,\Lambda^{n-1,n-1}_\R)$ (see Lemma \ref{lin_E}) and the flow \eqref{ourflow} fits in Hamilton's framework \cite[Section 5 and 6]{positive} with integrability condition $L=d$ (this is analogous to the argument used in \cite{AF7}). That in particular implies the well-posedness of the flow \eqref{phongflow} also when $\omega_0$ is not conformally balanced.

Moreover if $\chi$ is K\"ahler, then
$$
DE(\chi^{n-1})(\psi)=-\frac{1}{n-1}|\Omega|_{\omega}\square \psi
$$
for any closed $\psi\in C^{\infty}(M,\Lambda^{n-1,n-1}_\R)$. This allows us to apply a general result about the stability of second order geometric flows with an integrability condition. We state this theorem in section \ref{description} and we prove it in the last section.

\bigskip
\noindent{\em Notation.} Given a vector bundle $F$ on a manifold $M$, we denote by $C^{\infty}(M,F)$ the space of smooth sections of $F$. If further $I\subset \R$ is an interval we denote by $C^{\infty}(M\times I,F)$
the space of smooth time depending sections of $F$. When we write $\|f\|_{C^{\infty}}<\delta$, we mean that $\|f\|_{C^{k}}<\delta$ for every $k\in \mathbb N$.

\bigskip
\noindent{\em Acknowledgments.} The authors would like to thank the anonymous referee for pointing out several inaccuracies and helping to considerably improve the presentation of the paper.

\section{A stability result for  second order geometric flows with an integrability condition} \label{description}
In \cite{positive} Hamilton proved the following general result about the short-time existence of second order geometric flows on compact manifolds.

\medskip
Let $M$ be an oriented compact manifold, $F$ a vector bundle over $M$, $U$ an open subbundle of $F$ and
$$
E\colon C^{\infty}(M,U)\to  C^{\infty}(M,F)
$$
a second order differential operator. Consider the geometric flow
\begin{equation}\label{flow_Ham}
\frac{\partial f}{\partial t}=E(f)\,,\quad f(0)=f_0\,,
\end{equation}
where $f_0$ belongs to $C^{\infty}(M,U)$.
For $f\in C^{\infty}(M,U)$, we denote by $D E(f)\colon C^{\infty}(M,F)\to C^{\infty}(M,F)$ the linearization of $E$ at $f$ and by
$\sigma D E(f)$ the principal symbol of
$D E(f)$. Following Hamilton's paper we assume that there exists a first order linear differential operator
$$
L\colon C^{\infty}(M,F)\to C^{\infty}(M,G)\,,
$$
with values  in another vector bundle $G$ over $M$, such that
\begin{enumerate}
\item[1.] $L(E(f))=0$ for all $f\in C^{\infty}(M,U)$;

\vspace{0.1cm}
\item[2.] for every $f\in C^{\infty}(M,U)$ and for every $(x,\xi) \in T^*M$ all the eigenvalues of $\sigma DE(f)(x,\xi)$ restricted to $\ker \sigma L(x, \xi)$
 have strictly positive real part\,.
 \end{enumerate}
Because of the following result $L$ is called an {\em integrability condition for $E$}.

\begin{theorem}[Hamilton {\cite[Theorem 5.1]{positive}} ]\label{Ham_int}
Under the above  assumptions the geometric flow \eqref{flow_Ham} has a unique short-time solution.
\end{theorem}

\begin{rem} {\em Theorem 5.1 in \cite{positive} is in fact more general since the integrability condition $L$ is allowed to smoothly depend on $f \in C^\infty(M,U)$. This generality is needed to prove the short time existence of the Ricci flow.}
\end{rem}

Using Theorem \ref{Ham_int} we will be able to prove the following stability theorem for geometric flows with an integrability condition $L$.

\begin{theorem}\label{stabilityham}
Assume that $E$ and $L$ are as above.
Let $\bar f\in C^{\infty}(M,U)$ be such that $E(\bar f)=0$. Let $\bar h$ be a fixed metric along the fibers of $F$. Assume
\begin{enumerate}[(i)]
\item \label{semidef} $D E(\bar f)\colon \ker L \to \ker L$ is symmetric and negative semidefinite with respect to $\bar h$;
\item \label{miracolo} $E(f)$ is $L^2$-orthogonal to $\ker D E(\bar f)$ for every $f \in C^\infty(M,U)$;
\item $DE(\bar f): C^\infty(M,F) \to C^\infty(M,F)$ is elliptic.
\end{enumerate}
Then for every $\epsilon>0$ there exists $\delta>0$ such that if
$f_0\in C^{\infty}(M,U)$ satisfies
$$
\|f_0-\bar f\|_{C^\infty}<\delta,
$$
then \eqref{flow_Ham} has a long-time solution $f\in C^{\infty}(M\times [0,\infty),U)$ such that
$$
\|f(t)-\bar f\|_{C^	\infty}<\epsilon, \mbox{ for every $t\in[0,\infty)$}.
$$
Moreover, $f(t)$ converges to $f_{\infty}\in C^{\infty}(M,U)$ in $C^{\infty}$--topology which satisfies $E(f_{\infty})=0$.
\end{theorem}

\section{Proof of Theorem \ref{main}}
Let $(M,\omega_0)$ be a compact  $n$-dimensional  Hermitian manifold with vanishing first Chern class and  let $\Omega$ be a fixed complex volume form.
\begin{lemma}
\label{lemma31}
Let $\omega(t)$ be a solution to the geometric flow \eqref{phongflow} on $M$;  then $\tilde{\omega}=|\Omega|_{\omega}^{1/(n-1)}\omega$ satisfies
\begin{equation}\label{phongflow2}
\tfrac{\partial}{\partial t}\tilde \omega^{n-1}=i\partial\bar\partial (|\Omega|_{\tilde \omega}^{-2}\tilde \omega^{n-2})\,.
\end{equation}
\end{lemma}
\begin{proof}
Since
$$
\omega=|\Omega|_{\omega}^{-1/(n-1)}\tilde{\omega}\,,
$$
we have
$$
\tfrac{\partial}{\partial t}(\tilde{\omega}^{n-1})=i\partial\bar\partial(|\Omega|_{\omega}^{-(n-2)/(n-1)} \tilde{\omega}^{n-2})\,.
$$
Now in general  for any conformal factor $f\in C^{\infty}(M,\R_{+})$ one has
$$
|\Omega|_{f\tilde{\omega}}=f^{-n/2} |\Omega|_{\tilde{\omega}}
$$
and thus
$$
 |\Omega|_{\omega}=(|\Omega|_{\omega}^{-1/(n-1)})^{-n/2} |\Omega|_{\tilde{\omega}}=
 |\Omega|_{\omega}^{n/(2n-2)} |\Omega|_{\tilde{\omega}}
$$
from which we deduce
$$
|\Omega|_{\omega}= |\Omega|_{\tilde{\omega}}^{(2n-2)/(n-2)}\,,
$$
and the claim follows.
\end{proof}

Now we focus on the geometric flow \eqref{phongflow2} and we show that it fits in the set-up of Theorem \ref{stabilityham}. The flow is governed by the operator
$$
E\colon C^{\infty}(M,\Lambda^{n-1,n-1}_{+})\to C^{\infty}(M,\Lambda^{n-1,n-1}_{\R})
$$
defined by
\begin{equation}\label{EEE}
E(\omega^{n-1})=i\partial \bar \partial (|\Omega|_{\omega}^{-2}\,\omega^{n-2})
\end{equation}
where $\Lambda^{n-1,n-1}_{+}$ is the bundle of positive real $(n\!-\!1,n\!-\!1)$-forms on $M$ and $\Lambda^{n-1,n-1}_{\R}$ is the bundle of real $(n\!-\!1,n\!-\!1)$-forms.

\medskip
In order to study the linearization of $E$, we describe the principal part of the operator $\square$ in terms of the components of $(n\!-\!1,n\!-\!1)$-real forms on $M$.

Let $\omega$ be any Hermitian metric on $M$ and $\psi\in C^{\infty}(M,\Lambda_{\R}^{n-1,n-1})$. Then $\psi$ writes in a unique way as
\begin{equation}\label{psi}
\psi=\frac{1}{(n-1)!}h_0\omega^{n-1}-\frac{1}{(n-2)!}h_2\wedge\omega^{n-2}\,,
\end{equation}
where $h_0$ is a smooth function and $h_2\in C^{\infty}(M,\Lambda^{1,1}_{\R})$ satisfies
$$
h_2	\wedge \omega^{n-1}=0\,.
$$
Since
$$
*h_2=-\frac{1}{(n-2)!}h_2\wedge\omega^{n-2}\,,
$$
the form $\psi$ can be alternatively written as
$$
\psi =*(h_0\omega +h_2)\,.
$$

\begin{lemma}\label{lap}
If $\psi$ is closed, then
$$
\square\,\psi=\partial \partial^*\psi=-\frac{2i}{(n-2)!}\partial \bar\partial h_0\wedge \omega^{n-2}+\frac{i}{(n-3)!}\partial\bar\partial h_2\wedge\omega^{n-3}+{\rm l.o.t.}
$$
where {\rm \lq\lq l.o.t.\rq\rq} stands for \lq\lq lower order terms\rq\rq\, in  $\psi$. Moreover if $\omega$ is K\"ahler we have
$$
\square\,\psi=\partial \partial^*\psi=-\frac{2i}{(n-2)!}\partial \bar\partial h_0\wedge \omega^{n-2}+\frac{i}{(n-3)!}\partial\bar\partial h_2\wedge\omega^{n-3}\,.
$$
\end{lemma}
\begin{proof}
Since $\psi$ is closed we have
$$
\square\,\psi=\partial \partial^*\psi =-\partial *\bar\partial *\psi =
-\partial *(\bar\partial h_0\wedge \omega)-\partial *\bar\partial h_2 +{\rm l.o.t.}
$$
On the other hand, from the closure of $\psi$, we deduce
$$
\bar \partial h_2\wedge\omega^{n-2}=\frac{1}{n-1}\bar \partial h_0\wedge \omega^{n-1}+{\rm l.o.t}\,.
$$
Now we use the well-known splitting of $(1,2)$-forms as
$$\gamma = \gamma_+ + \gamma_-$$
where $\gamma_+$ is of the form $\alpha \wedge \omega$ with $\alpha$ a $(0,1)$-form and $\gamma_-$ is such that $\gamma_- \wedge \omega^{n-2}=0$. \\
The fact we use is that $* \gamma_+ = \frac{i}{(n-2)!} \gamma_+ \wedge \omega^{n-3}$ and $*\gamma_-=- \frac{i}{(n-3)!} \gamma_- \wedge \omega^{n-3}$.
Thus in our case, taking into account that $(\bar\partial h_2)_+=\frac{1}{n-1} \bar\partial h_0 \wedge \omega +{\rm l.o.t.}$,
we have
$$
\begin{aligned}
*\bar\partial h_2=&\frac{i}{(n-1)!}\bar \partial h_0\wedge \omega^{n-2}-\frac{i}{(n-3)!}\left(\bar\partial h_2-\frac{1}{n-1}\bar \partial h_0\wedge \omega\right)\wedge\omega^{n-3}+{\rm l.o.t.}\\
&=\frac{i}{(n-2)!}\bar \partial h_0\wedge \omega^{n-2}-\frac{i}{(n-3)!}\bar\partial h_2\wedge\omega^{n-3}+{\rm l.o.t.}
\end{aligned}
$$
Therefore
$$
\square\,\psi=-\frac{2i}{(n-2)!}\partial \bar\partial h_0\wedge \omega^{n-2}+\frac{i}{(n-3)!}\partial\bar\partial h_2\wedge\omega^{n-3}+{\rm l.o.t.},
$$
where these lower order terms vanish if $\omega$ is closed since they all come from $d\omega$. \end{proof}

\begin{lemma}
\label{lin_E}
Let $\omega$ be a Hermitian metric on $M$ and let $\psi \in C^{\infty}(M,\Lambda^{n-1,n-1}_{\R})$ be closed; then
\begin{equation}\label{DEnonK}
DE(\omega^{n-1})(\psi)=-\frac{1}{n-1}|\Omega|_{\omega}\square \psi+{\rm l.o.t.}
\end{equation}
Moreover if we assume that $\omega$ is K\"ahler then we have
$$
DE(\omega^{n-1})(\psi)=-\frac{1}{n-1}|\Omega|_{\omega}\square \psi\,.
$$
\end{lemma}

\begin{proof}
Let $\omega(t)$, $t\in (-\epsilon,\epsilon)$ be a smooth curve of Hermitian metrics with $\omega(0)=\omega$ on $M$ and we assume that
$$
\psi:=\tfrac{\partial}{\partial t}_{|t=0}\omega^{n-1}(t)
$$
is closed. In order to simplify the notation  we set
$$
r(t)=|\Omega|_{\omega(t)}^{-2}\,,\quad r=r(0)\,,\quad \dot r=\tfrac{\partial}{\partial t}_{|t=0}r(t)\,,\quad \dot \omega=\tfrac{\partial}{\partial t}_{|t=0}\omega(t)\,.
$$

We directly compute
$$
\begin{aligned}
\tfrac{\partial}{\partial t}_{|t=0}E(\omega(t)^{n-1})=&\,i\partial \bar \partial (\dot{r}\,\omega^{n-2}+(n-2) r\, \dot{\omega}\wedge\omega^{n-3})\\
=&i\partial \bar \partial \dot{r}\,\wedge \omega^{n-2}+(n-2)i r\, \partial \bar \partial\dot{\omega}\wedge \omega^{n-3}+{\rm l.o.t.}
\end{aligned}
$$
We decompose $\psi$ according to \eqref{psi}.  Then \cite[Lemma 2.5]{Crelle} implies
$$
\dot \omega=\frac{h_0}{(n-1)(n-1)!}\omega-\frac{1}{(n-1)!}h_2
$$
and using
$$
\dot r=\frac{nh_0}{(n-1)(n-1)!} r
$$
we obtain
\begin{multline*}
\tfrac{d}{dt}_{|t=0}E(\omega(t)^{n-1})=\frac{n}{(n-1)(n-1)!}i r\, \partial \bar \partial h_0\wedge \omega^{n-2}\\+\frac{n-2}{(n-1)(n-1)!}i r\, \partial \bar \partial h_0\wedge \omega^{n-2}-\frac{n-2}{(n-1)!}i r \,\partial \bar \partial h_2\wedge \omega^{n-3}+{\rm l.o.t.}\,,
\end{multline*}
i.e.
\begin{equation}\label{pre}
\tfrac{d}{dt}_{|t=0}E(\omega(t)^{n-1})=\frac{2}{(n-1)!}i r \,\partial \bar \partial h_0\wedge \omega^{n-2}-\frac{n-2}{(n-1)!}i r \, \partial \bar \partial h_2\wedge \omega^{n-3}+{\rm l.o.t.}
\end{equation}
and Lemma \ref{lap} implies \eqref{DEnonK}. Since the lower order terms in \eqref{pre} vanish if $\omega$  is closed, the claim follows.
\end{proof}

\begin{proof}[Proof of Theorem $\ref{main}$]
Let $(M,\chi)$ be a compact K\"ahler manifold of complex dimension $n$. Let $\Omega$ be a complex volume form on $M$ with constant norm with respect to $\chi$ and let
$E\colon C^{\infty}(M,\Lambda^{n-1,n-1}_{+})\to C^{\infty}(M,\Lambda^{n-1,n-1}_{\R})$ be as in \eqref{EEE}. Then we have
$$
E(\chi^{n-1})=0\,.
$$
Now Lemma \ref{lin_E} and Hodge theory imply that all the assumptions of Theorem \ref{stabilityham} are satisfied when we consider
$$
\begin{array}{llll}
& F=\Lambda^{n-1,n-1}_\R\,, & U= \Lambda^{n-1,n-1}_+\,, & L=d\colon C^{\infty}(M,\Lambda^{n-1,n-1}_\R)\to C^{\infty}(M,\Lambda^{2n-1}_\C)\,.
\end{array}
$$
Hence for every $\epsilon>0$ there exists $\delta>0$ such that if $\tilde \omega_0$ is a Hermitian metric on $M$ satisfying
\begin{equation}\label{vicinotilde}
\|\tilde \omega_0^{n-1}-\chi^{n-1}\|_{C^{\infty}}<\delta\,,
\end{equation}
then there exists  a smooth family of Hermitian metrics $\tilde \omega(t)$, $t\in [0,\infty)$, such that
\begin{equation}\label{vicinotilde1}
\|\tilde \omega(t)^{n-1}-\chi\|_{C^{\infty}}<\epsilon\,,\quad \tfrac{\partial}{\partial t}\tilde \omega=i\partial \bar\partial(|\Omega|_{\tilde \omega}^{-2}\tilde\omega^{n-2})\,,\quad \tilde \omega(0)=\tilde \omega_0
\end{equation}
and $\tilde \omega(t)$ converges in $C^{\infty}$--topology to a Hermitian metric $\tilde\omega_{\infty}$ such that
\begin{equation}\label{vicinotilde2}
\partial \bar\partial (|\Omega|_{\tilde \omega_{\infty}}^{-2}\,\tilde \omega_{\infty}^{n-2})=0\,.
\end{equation}

Now let $\omega_{0}$ be a Hermitian metric on $M$ satisfying \eqref{vicino}; then $\tilde\omega_0:=|\Omega|_{\omega_0}^{1/(n-1)}\omega_0$ satisfies \eqref{vicinotilde}. Thus there exists $\tilde \omega \in C^{\infty}(M\times[0,\infty),\Lambda^{1,1}_{\R})$ satisfying \eqref{vicinotilde1} and  converging in $C^{\infty}$-topology to a $\tilde \omega_\infty$ for which \eqref{vicinotilde2} holds. Therefore
$\omega=|\Omega|_{\tilde \omega}^{-2/(n-2)}\tilde{\omega}$ is a solution to the anomaly flow \eqref{phongflow} satisfying
$$
\||\Omega|_{\omega(t)} \omega(t)^{n-1}-\chi\|_{C^{\infty}}<\epsilon\,,
$$
and $|\Omega|_{\omega(t)}\omega(t)^{n-1}$ converges in $C^{\infty}$-topology to
$$
\omega_{\infty}^{n-1}=|\Omega|_{\tilde\omega_{\infty}}^{-2(n-1)/(n-2)}\tilde{\omega}_{\infty}^{n-1}\,.
$$
By a change of variable we obtain that $\omega_{\infty}$ is astheno-K\"ahler, i.e.
$$
\partial 	\bar\partial (\omega_{\infty}^{n-2})=0\,,
$$
and the first part of the claim follows.

\medskip
About the second part of the claim, we use that if $\omega_0$ satisfies the conformally balanced condition
$d(|\Omega|_{\omega_0}\omega^{n-1}_0)=0$, then $d(|\Omega|_{\omega(t)}\omega(t)^{n-1})=0$ for every $t$
and hence $\omega_{\infty}$ satisfies
$$
d(|\Omega|_{\omega_\infty}\omega_\infty^{n-1})=0	\,,\quad \partial 	\bar\partial (\omega_{\infty}^{n-2})=0\,.
$$
In view of \cite[Lemma 1]{AF7} (see also \cite{Matsuo}), $\omega_\infty$ is K\"ahler-Einstein.
\end{proof}

\section{Proof of Theorem \ref{stabilityham}}

In this section we prove the general result about the stability of geometric flows in Hamilton's set-up described in section \ref{description}.
\begin{proof}
We adapt the proof of the main theorem in \cite{Advances}.  \\
Fix a metric connection $\nabla$ on $(F,\bar h)$ and a volume form on $M$. The space $C^{\infty}(M,F)$  has the natural structure of tame Fr\'echet space given by the Sobolev norms $\|.\|_{H^n}$ induced by $\bar h$, $\nabla$ and the volume form of $M$.

\medskip
On $C^{\infty}(M\times [a,b], F)$ we consider the grading
$$
\|f\|_{n,[a,b]}=\sum_{2j\leq n}\int_{a}^{b}\|\partial_t^{j} f(t) \|_{H^{n-2j}}\, dt.
$$

Hamilton in \cite[sections 5 and 6]{positive} proved that, with respect to this grading, for any $T>0$ the map
$$
\mathcal F\colon C^{\infty}(M\times[0,T],U)\to C^{\infty}(M\times[0,T],F)\times C^{\infty}(M,F)
$$
$$
\mathcal F(f)=(\partial f/\partial t-E(f), f(0))
$$
satisfies the assumptions of the Nash-Moser theorem, i.e. $\mathcal F$ is smooth tame, $D\mathcal F(f)$ is bijective for every
$f\in C^{\infty}(M\times[0,T],U)$ and the family of the inverses
$$
C^{\infty}(M\times[0,T],U)\times  C^{\infty}(M\times[0,T],F)\times C^{\infty}(M,F)\to C^{\infty}(M\times[0,T],F)
$$
$$
(f,(g,k))\mapsto D \mathcal F (f)^{-1}(g,k)
$$
is a smooth tame map. Hence the Nash-Moser theorem can be applied and $\mathcal F$ is locally invertible with smooth tame inverse. As a direct consequence, arguing as in Proposition 5.3 in \cite{Advances} we have the following

\medskip
{\bf Claim 1.}
{\em For all $\epsilon,T>0$, there exists $\delta'>0$ such that if $f_0\in C^{\infty}(M,U)$ satisfies
$$
\|f_0-\bar f\|_{C^\infty}<\delta'\,,
$$
then \eqref{flow_Ham} has a solution $f\in C^{\infty}(M\times[0,T],U)$ and
$$
\|f-\bar f\|_{n,[0,T]}<\epsilon
$$
for all $n \in \mathbb N$.}

\medskip
Then we show the following

\medskip
{\bf Claim 2.}
{\em For $\delta'$ small enough the $L^2$-norm of $E(f)$ with respect to $\bar h$ has an exponential decay.}

\medskip
Fix a small time $\tau > 0$ arbitrary. By Claim 1 we have that there exists $\delta' >0$ such
that if $\|f_0-\bar f\|_{C^\infty}<\delta'$, then problem \eqref{flow_Ham} has a solution $f \in C^\infty(M \times [0, T + 2\tau], U)$ with $\|f - \bar f\|_{l, [0, T+2\tau]}$
bounded for every $l$.
Now since
$$
\frac{\partial}{\partial t} E(f) = DE(f)(E(f))\,,
$$
we have
$$
\frac{d}{dt} \|E(f)\|^2_{L^2}=2 \langle \frac{\partial}{\partial t}E(f),E(f)\rangle_{L^2}=2 \langle DE(f)(E(f)),E(f)\rangle_{L^2}\,.
$$

Moreover a general result for families of symmetric operators on Hilbert spaces combined with the Sobolev Embedding theorem and elliptic regularity of $DE(\bar f)$ (see \cite{Advances} Corollary 5.6)
imply that given $a>0$ we can choose $\delta'$ so small that we have
$$
\langle D E(f)(E(f)),E(f) \rangle_{L^2}\leq (1-a)\langle D E(\bar f)(E(f)),E(f) \rangle_{L^2}+a\|E(f)\|_{L^2}^2
$$
for every time in the interval $[0,T+\tau]$.
Now let $\lambda$ be half the smallest positive eigenvalue of $-DE(\bar f)_{| \ker L}$.
Take $a=\frac{\lambda}{2\lambda+1}$ so that
$$
\langle D E(f)(E(f)),E(f) \rangle_{L^2}\leq \frac{\lambda+1}{2\lambda+1}\langle D E(\bar f)(E(f)),E(f) \rangle_{L^2}+\frac{\lambda}{2\lambda+1}\|E(f)\|_{L^2}^2\,.
$$
By assumption \eqref{semidef} and \eqref{miracolo} we have $\langle D E(\bar f)(E(f)),E(f) \rangle_{L^2}\leq -2\lambda \|E(f)\|^2_{L^2} $ thus
$$
\langle D E(f)(E(f)),E(f) \rangle_{L^2}\leq- \lambda \|E(f)\|^2_{L^2}
$$
which implies
$$
\frac{d}{dt}\|E(f)\|_{L^2}^2 \leq -2\lambda \|E(f)\|_{L^2}^2\,.
$$
Using Gronwall's lemma we get
$$
\|E(f(t))\|_{L^2}^2\leq e^{-2\lambda t} \|E(f_0)\|_{L^2}^2\,,\quad\mbox{ for all $t\in [0,T+\tau)$}\,,
$$
and the Claim 2 follows.

\medskip
By integrating the last formula we get
\begin{equation}
\label{exp2}
\|E(f)\|^2_{{0},[t,T+\tau]} = \int_{t}^{T+\tau}\|E(f(s))\|_{L^2}^2\,ds\leq \|E(f_0)\|^2_{L^2}\frac{e^{-2\lambda t}}{2\lambda}\,.
\end{equation}
Since using the parabolic Sobolev embedding theorem (see \cite{Advances} Corollary 5.8)
there exist  $m$ and $C>0$ such that for every $t\in [0,T]$
$$
\|E(f(t))\|_{H^{n}} \leq C \|E(f)\|_{{m},[t,T+\tau]}\,,
$$
we will need the following estimate in order to prove $H^n$-exponential decay.

\medskip
{\bf Claim 3.} {\em Let $\tau_0\in (0,T)$. For $\delta'$ small enough we have that for every
 $m\in \mathbb N$ there exists $C>0$  such that
$$
\|E(f)\|_{{m},[t,T+\tau]} \leq C\|E(f)\|_{{0},[t-\tau_0,T+\tau]}\,,
$$
for every $t\in [\tau_0,T]\,.$}

\medskip
We prove by induction on $m$ that for every $\tau_0 \in (0,T)$ we can choose $\delta'$ small enough such that there exists  a positive $C$ (depending on  $m$,
$\tau_0$ and an upper bound on $\delta'$) such that for every $g\in C^{\infty}(M\times [0,T+\tau], F)$ solving
$$
\frac{\partial}{\partial t}g=DE(f)(g)\,,
$$
the following estimate holds
\begin{equation}
\label{est1}
\|g\|_{{m},[t,T+\tau]} \leq C\|g\|_{{0},[t-\tau_0,T+\tau]}\,,
\end{equation}
for every $t\in [\tau_0,T]$.
Then we deduce the claim by setting $g=E(f)$. \\
For $m=0$ the estimate \eqref{est1} is trivial. We assume the above statement true up to $m=N$. For a smooth family of linear second order differential operator $P$ we set
$$
|[P]|_{N} =\sum_{2j\leq N}[\tfrac{\partial^j}{\partial t^j}P]_{N-2j}
$$
where $[P]_N$ is the supremum of the norm of $P$ and its space covariant derivatives up to degree $N$. By [2, Lemma 6.10]
for $\delta'$ small enough there exists $C >0$,
depending on
$T$ and an upper bound on $\delta'$, such that for $t\in  [0, T+\tau)$ and $g\in C^{\infty}(M\times [0,T+\tau], F)$ we have
\begin{multline}
\|g\|_{N+2,[t,T+\tau]} \leq C\left(\|\tfrac{\partial}{\partial t}g-DE(f)(g)\|_{N,[t,T+\tau]}+\|g(t)\|_{H^{N+1}}\right)\\
+C|[DE(f)]|_{N}\left(\|\tfrac{\partial}{\partial t}g-DE(f)(g)\|_{0,[t,T+\tau]}+\|g(t)\|_{H^{1}}\right)\,,
\end{multline}
which implies
$$
\|g\|_{N+2,[t,T+\tau]} \leq C(1+|[DE(f)]|_{N}) \left(\|\tfrac{\partial}{\partial t}g-DE(f)(g)\|_{N,[t,T+\tau]}+\|g(t)\|_{H^{N+1}}\right)\,.
$$
Up to shrink $\delta'$ we may assume
$$
 |[DE(f)]|_{N}\leq  1+ |[DE(\bar f)]|_N
$$
in order to rewrite the last estimate as
\begin{equation}
\label{est2}
\|g\|_{N+2,[t,T+\tau]} \leq C(2+|[DE(\bar f)]|_{N}) \left(\|\tfrac{\partial}{\partial t}g-DE(f)(g)\|_{N,[t,T+\tau]}+\|g(t)\|_{H^{N+1}}\right)\,.
\end{equation}
Now assume that $g$ satisfies
$$
\frac{\partial}{\partial t}g=DE(f)(g)
$$
and let $\chi\colon \R\to [0,1]$ be smooth and such that
$$
\begin{cases}
\chi(s)=0	\quad \mbox{ for } s \leq t-\frac{\tau_0}{2}\\
\chi(s)=1    \quad \mbox{ for } s \geq t\,.
\end{cases}
$$
Then $\tilde g=\chi g$ satsfies
$$
\frac{\partial}{\partial t}\tilde g=DE(f)(\tilde g)+\dot\chi g
$$
and from \eqref{est2} we deduce
$$
\|g\|_{N+2,[t,T+\tau]}\leq \|\tilde g\|_{N+2,[t-\frac{\tau_{0}}{2},T+\tau]} \leq C(2+|[DE(\bar f)]|_{N}) \|\dot\chi g\|_{N,[t-\frac{\tau_0}{2},T+\tau]}
\leq C' \|g\|_{N,[t-\frac{\tau_0}{2},T+\tau]}
$$
and the claim follows using the inductive hypothesis.

\medskip
Finally putting these together with \eqref{exp2} we have that for $\delta'$ small enough we have
\begin{equation}
\label{Hndecay}
\|E(f(t))\|_{H^{n}} \leq C  \|E(f_0)\|_{L^2}{\rm e}^{{-\lambda} t}\,
\end{equation}
for $t \in [\tau_0,T]$. The constant $C$ may depend on $T,\tau_0$ and an upperbound on $\delta'$, but not on
$t$ and $f_0$.

\medskip
Now we choose $\delta \leq \delta'$ such that  if $\|f_0-\bar f\|_{C^{\infty}}\leq \delta$ then
\begin{equation}\label{condizione}
C\|E(f_0)\|_{L^2}\frac{{\rm e}^{-\lambda \tau_0}}{\lambda}\sum_{j=0}^{\infty}{\rm e}^{-\lambda j(T-\tau_0)}+\|f({\tau_0})-\bar f\|_{H^{n}}\leq \delta'\,.
\end{equation}

Using \eqref{Hndecay} and \eqref{condizione} and working as in \cite{Advances} we have that for  any $t\in [NT-(N-1)\tau_0,(N+1)T-N\tau_0]$ with $N \in \mathbb N$
$$
\|f-\bar f\|_{H^{n}}\leq C\|E(f_0)\|_{L^2}\frac{{\rm e}^{-\lambda \tau_0}}{\lambda}\sum_{j=0}^{N}{\rm e}^{-\lambda j(T-\tau_0)}+\| f({\tau_0})-\bar f\|_{H^{n}}\leq \delta'.
$$
This allows us to conclude that the solution $f$ is defined in $M\times [0,\infty)$.

Now let $f_\infty=f_0+\int_{0}^\infty E(f)ds\in  C^{\infty}(M,F)$;
since
$$
\lim_{t\to \infty}\|f(t)-f_{\infty}\|_{H^n}\leq \lim_{t\to \infty}C\|E(f_0)\|_{L^2}{\rm e}^{-\lambda t}=0\,, \mbox{ for $n$ large enough}
$$
$f(t)$ converges to $f_\infty$ in $C^\infty$-topology. Possibly shrinking $\delta$, we will have $f_\infty\in C^{\infty}(M,U)$.
Using again \cite[Proposition 5.7]{Advances}, we have that up to shrink $\delta$,
$$
\|f(t)-\bar f\|_{C^{\infty}}<\epsilon
$$
for every $t\in [0,\infty)$.

Finally
$$
E(f_\infty)=\lim_{t\to 0} E(f(t))=0
$$
and the claim follows.
\end{proof}

redpajama_set_name: RedPajamaArXiv