Preheat oven to 230 ºC. Line an oven tray with baking paper.
Prep & cook pumpkin sabzi. Toss pumpkin, carrot, cauliflower, onion and sabzi spice with oil on prepared tray. Season and bake for 18-20 minutes, until tender and lightly golden.
Marinate chicken. Add chicken, tandoori spice, first measure of yoghurt, ginger and garlic to a medium bowl. Toss to coat and leave to marinate for 5 minutes.
Make yoghurt. In a small bowl, mix together yoghurt and saffron and season. Set aside.
Cook chicken. Heat oil in a large fry-pan on medium-high heat. Season chicken and cook for 2-3 minutes each side (depending on thickness), or until cooked through. Set aside, covered, to rest. Slice thickly.
Finish pumpkin sabzi. Toss kale, peanuts and lime juice through cooked veggies on tray and season.
To serve, spoon yoghurt onto each plate. Top with pumpkin sabzi and chicken. Squeeze over lime wedges and sprinkle over coriander.


The paper – dubbed the 'Pausebuster' – claimed that contrary to what scientists had been saying for several years, there was no 'pause' or 'slowdown' in the rate of global warming in the early 21st Century, and that in fact it had been taking place even faster than before.
The 'pause' had been seized on by climate sceptics, because throughout the period, carbon dioxide emissions continued to rise.
Last week Peter Stott, head of climate monitoring at the UK Met Office, admitted that notwithstanding the Pausebuster, it was clear 'the slowdown hasn't gone away'.


Everything you need to know about visiting the Chicago Theatre (175 N State St, Chicago, IL 60601).
The Chicago Theatre is best known for its iconic marquee, a remnant of its past as an opulent French baroque-style movie palace. When it opened 1921, the 3,600-seat theater screened films and hosted live jazz performances, dubbed by its owners the "Wonder Theatre of the World." The theater (which has been renovated several times over) is equally magnificent today, hosting performances by bands like the Avett Brothers and Sigur Rós as well as comedians such as Chris Rock and John Mulaney. in one of the most ornate venues in the city.
Even if you've seen a few concerts at the Chicago Theatre, you haven't experienced everything the historic facility has to offer. Daily Marquee Tours allow visitors to stand on the venue's stage and take a peek at the autographed walls in the backstage dressing rooms (see if you can spot signatures left by Frank Sinatra and Dolly Parton).
Some of the biggest names in music, comedy and entertainment stop by the Chicago Theatre, but you'll need to keep your eyes peeled if you want to score tickets. Surprise sets from Louis C.K. and performances by contemporary acts like Harry Styles and Father John Misty tend to sell out quickly. You'll need to pay close attention to the Chicago Theatre concert calendar if you want to catch your favorite acts in one of the city's most beautiful venues.
Our Chicago editorial team is constantly updating and reviewing the best attractions, activities and venues across the city, so that you're always in the know, with the best of Chicago at your fingertips. Chicago Theatre was most recently updated with new tips on June 30, 2017.
El stop: Brown, Green, Orange, Pink, Purple (rush hrs) to State/Lake; Red to Lake.


Do you know how to play with kids? In this episode of Mom Docs, Dr. Dehra Harris shares a few tips to keep in mind when playing with children.
"It's all about connection," explains Dr. Harris. When you play with your kids, ask questions and let them know you enjoy the activity. Let them take the lead! Whether you're cooking or just taking a walk, there are always opportunities for connection when playing with your child.


text:
Google Employees Across The Country Walk Out: Demand Better Handling Of Sexual Misconduct
Max Moeller on November 5, 2018 / 0 Comments
Last week, we reported on Google paying off Andy Rubin, the "father of Android" after sexual misconduct allegations. This story was broken by The New York Times. It revealed that the company had given the former employee a $90 million exit package. It was also revealed that Rubin isn't the first time this happened.
Karma Always Hits
On Thursday, November 1st, that story came back to bite. As reported by The Verge, hundreds of Google employees stormed a New York City park to protest the handling of these situations. This was all part of a worldwide gathering. It that took place in areas like Dublin, London, Berlin, Tokyo, and other cities. Workers are demanding a clearer process for handling sexual misconduct and generate a better company culture overall.
Google Employees Walk out, Photo by the Verge
One organizer, Claire Stapleton, gave a speech to the crowd of disgruntled employees:
"This walkout is the culmination of a fast, furious week and the work of more than 1,000 people. I don't know what it will take to change the system, but I do know that we are a crazy force to be reckoned with."
"This is an emergent movement," says another organizer who wishes to be kept anonymous. "But, it's building on a lot of work that's already been done this year and over the years by a lot of people pushing for structural change," they continued. However, the speaker went on to say they have faith in the infrastructure of Google.
One clever protester held up a sign playing off of Google's coding criteria, which is "C++ Style, no exceptions". The sign said "Code of conduct, no exceptions."
As time went on, even more offices joined in. Google's main campus, Mountain View, held their own as well. However, journalists were asked to step away there. They were told the situation was a "private company event."
Addressing The Situation
One participant at the Mountain View office space, Max Timkovich, revealed that they just "want them to take harassment claims more seriously." He details stories about HR doing nothing about reported allegations. Another anonymous member was more firm – she says Google has a "broken system of reporting sexual harassment" alongside a male-dominated executive board. "Anywhere decisions need to be made, there need to be more women," she says.
Soon after the report from last week, Google CEO Sundar Pichai sent an email to employees apologizing about the situation. The email read that the company was "deeply sorry for the past actions and the pain they have caused employees." He continued, divulging that Google had fired 48 people – including at least 13 members in senior positions – over sexual harassment and they had received no exit packages.
Pichai sent out a public statement during the walkout, in which he partly reinforced the email statements:
"Earlier this week, we let Googlers know that we are aware of the activities planned for today and that employees will have the support they need if they wish to participate. Employees have raised constructive ideas for how we can improve our policies and our processes going forward. We are taking in all their feedback so we can turn these ideas into action."
Demanding Change
On Twitter and as revealed by The Cut, the protestors have five demands:
"An end to forced arbitration in cases of harassment and discrimination". Additionally, Google workers could bring a co-worker, representative, or supporter when meeting with Human Resources.
"A commitment to end pay and opportunity inequity, for example making sure there are women of color at all levels of the organization, and accountability for not meeting this commitment." Google would release internal reports on any salary or professional advancement gaps across employees of different races, genders, and ethnicities.
"A publicly disclosed sexual harassment transparency report". This would include the number of harassment claims and the division where they were made, the types of claims submitted, how many of the victims and accused have left Google, and the value of any exit packages — like the alleged payout for Rubin.
"A clear, uniform, globally inclusive process for reporting sexual misconduct safely and anonymously". The new process would need to make Google's HR department more independent from its senior management.
"Elevate the Chief Diversity Officer to answer directly to the CEO and make recommendations directly to the board of directors. In addition, appoint an employee representative to the board."
According to Yana Calou, an employee at the worker nonprofit website, Coworker.org, the first claim should be no problem for Google. Should these demands make any changes from within, however, we'll have to wait and see.
Posted in:News, Technology
Tagged in:Android, Andy Rubin, Google, Misconduct
Apple Reveals New MacBook Air and iPad Pro During New Showcase
Tech Startup Tarform Introduces Electric Motorcyles To U.S. Market

redpajama_set_name: RedPajamaCommonCrawl


RE: IGAD, a meddler or a mediator.
In December 2013, IGAD offered itself as a mediator after deadly conflict erupted in Juba. Many South Sudanese were happy and full of hope because IGAD was part of the process that led to the resolution of the conflict between the then Southern Sudan and the Sudan government.
We were convinced that IGAD has the niceties to navigate the difficult route in the current South Sudan conflict and, is therefore, better equipped with lessons learnt, political maturity and delicate diplomacy. We reasoned that the regional bloc was therefore, equipped with expertise to exert pressure, provide incentives, and offer desirable alternatives to all the parties to the conflict.
Sadly, our hopes and aspirations are dashed as we begin to see an incoherent, inconsistent, sometimes inconsiderate and biased muscular style of mediation by IGAD. We have painfully witnessed IGAD moving gradually away from helping the parties to overcome their deep-seated incompatible goals, to entrenching mistrusts and fears.
IGAD's poor mediation formulation since 2013, has gradually pushed the parties further apart from de-escalation, concession and compromise.
We are not blaming IGAD for the lack of peace in South Sudan. Any good mediator dealing with many belligerents in South Sudan who are blood-thirsty, greed-based, power hungry and devoid of well articulate creed, is bound to face enormous challenges.
Therefore, it will be irresponsible to blame IGAD for the failure to achieve a sustainable peace in South Sudan. But as a regional bloc shouldering the responsibility of mediating peace in South Sudan, IGAD should be held accountable for showing competing and conflicting interests in South Sudan. IGAD's actions shows that it has now moved away from being a mediator to becoming a meddler. This new role means that IGAD cannot succeed to satisfy the interests of all its members without side-lining other warring parties to the conflict.
IGAD's selfish approach is not helping South Sudanese, but rather blocking the road to peace in South Sudan. The result of IGAD's lack of impartiality and its exclusionary approach to peace making in South Sudan is deadly. For instance, IGAD poor rationale for isolating the SPLA (IO) through coercion actions has not help to de-escalate conflict. The isolation of the main actors has backfired and led to counter-escalation. This and other missteps of IGAD have created a dire humanitarian conditions for millions of South Sudanese.
Due to IGAD's rag-tagged approach to mediation, the excluded parties have resorted to bargains by bullets. These groups have banked their hopes on violence and other beyond-the-table tactics, to get IGAD's attention, hence maximizing their bargaining positions to weaken opponents, to prevent being side-lined, or to sabotage the talks altogether.
The civilians are paying huge costs, directly and indirectly. IGAD's continuous involvement in peace making in South Sudan has produced more harm than good.
It is against this backdrop, that we, the undersigned, are firmly convinced that sustainable peace in South Sudan and the region will not come through IGAD. We appeal to the African Union, the United Nations and Troika to appoint a neutral and credible mediator who work to bring an end to the conflict South Sudan.
We thank IGAD for their efforts so far! We hope that the views and voices of the petitioners will find audience.
Dr. Remember Miamingi needs your help with "African Union: Neutral, credible and effective mediation of the conflict in South Sudan". Join Dr. Remember and 562 supporters today.


text:
Player's club faked death to avoid match
By PAA
A Spanish soccer player whose former club in Ireland faked his death in an apparent bid to avoid a match has said he learned of his demise from work colleagues.
Fernando LaFuente said he had no role in the ruse attempted by the club in Dublin, Ballybrack FC.
The club has apologised for a "gross error of judgment" after falsely reporting that Mr LaFuente had died in a traffic accident on the way home from training last week.
Ballybrack's game against Arklow Town on Saturday was postponed, while other teams in the Leinster Senior League held a minute's silence in honour of the amateur player.
The league also posted a notice of his death in a newspaper before it was discovered that Mr LaFuente was actually alive and well.
Mr LaFuente moved to Galway a number of months ago. He said the club had told him they might claim that he had had an accident as a way of avoiding playing the game against Arklow.
He said he only learned that his death had been faked on Tuesday night.
"I was aware there was going to be some story on me but I thought it was going to be me breaking a leg," he told Ireland's RTE news.
"I was home yesterday after my work finished. I was playing some video games. Suddenly I got a call from work and they said what's happened. They told me, 'you're a celebrity'.
"They started sending me all these news articles and mass media. And that's how I found out I was dead."
Mr LaFuente said the club has since apologised to him.
"As soon as I heard the news, I wrote to them," he said. "They got straight back to me and told me what was going on and they apologised.
"It's serious on their part but I'm finding it a little bit funny. Because basically, I'm not dead. And no one has actually been harmed here."
Mr LaFuente said he would have had no issue with the club falsely claiming he had broken a limb.
But he said the reports on his death had been a "bit sad".
"A man in his 20s, with his whole life in front of him, dying in a car accident. I don't know how I feel about that. It's funny for me because I've been watching my own death," he said.
The software engineer suggested his ex-club might have been keen to get Saturday's game called off due to a lack of players.
In a statement, Ballybrack FC said: "At this stage we can only offer our sincere apologies to the Leinster Senior League, our opponents Arklow Town FC and the host of clubs and football people who made contact with us or offered messages of support in recent days."
The club added that "the person in question has been relieved of all footballing duties" following an emergency meeting.
Leinster Senior League chairman David Moran told RTE that the story "started to unravel" after they had enquired about offering support to the family.
The league issued a statement saying that it had acted in "good faith" after being informed of the death.
©PAA2018
Jose Mourinho Champions League celebration has football greats calling full-time
EFL: Derby County's Bradley Johnson under fire for appearing to bite Stoke City's Joe Allen


When it's time to say 'au revoir' to Lyon, end your trip the easy way with this departure transfer in a climate-controlled minivan to Lyon-Saint Exupery. Transfer service is available 24 hours a day, 7 days a week. It's a great option for groups of up to 7 people.
You will be met at your Lyon city hotel by your driver at the hotel reception. Please advise your preferred pickup time in the Special Requirements field on booking.
Enjoy a stress-free exit from Lyon with when you book this transfer to Lyon-Saint Exupery airport. Just enter your flight and hotel details when you book, pay in advance, and show your voucher to your driver. Please call the number on your voucher to confirm pickup details 24-48 hours before your departure.
After pickup at your Lyon hotel in a comfortable minivan with room for up to 7 passengers, relax on the journey to the airport. Arrive with plenty of time to catch your flight. This one-way transfer service is available 24 hours a day, 7 days a week.
Price is per person, based on 7 adults per car/vehicle.
Driver very helpful, arrived ahead of time as we had been advised to get to the airport early because of the alterations. The vehicle was clean and comfortable.
On time. Beautiful van. Worked like clockwork.


The International Society for Minimal Intervention in Spinal Surgery (ISMISS) was conceived in occasion of the GIEDA-BRUXELLES Meeting in 1988 and structurally organized as a non-profit organization in 1989 under the leadership of Professor Parviz Kambin, Philadelphia, first President of ISMISS, in collaboration with his European counterparts, Professor Adam Schreiber (Zurich), and the Asian pioneer Professor Sadahisa Hijikata (Tokyo). Further they were supported by european pioneers as Professor Mario Brock, Berlin, PD Dr. Hansjoerg Leu, Zurich and Dr. Jan Sheppard, Hastings. Thus the term, "minimally invasive spinal surgery" was coined. ISMISS has held regular biannual ISMISS symposiums in Philadelphia (USA) and annual meetings in Zurich (Switzerland) already running since the mid-eighties. In its annual session of Janaury 2015 in Zurich, the ISMISS board has decided to held from now annual meetings in triennial cycles in Europe/Asia/America. So the last 34th meeting was held in Chong-Quing/China and the next upcoming 35th will be in Chicago/USA (see also under "meetings").
The membership and participants have witnessed the evolution of percutaneous nucleotomy to endoscopic fragmentectomy, foraminoplasty and segmental stabilization. In recent years ISMISS auspiced several periodical meetings and courses worldwide.
"Guidelines for Endoscopic Spinal Surgery"
thatwere updated on a regular base under the lead of PD Dr. Christof Birkenmiyer, Munich.


I am going to be a married woman in two weeks. The thought astounds me. It is the craziest feeling that my entire life is going to change in so many ways. Only a few more days being Michelle Miller. I am so very happy to be marrying the most incredible man. God has overwhelmingly blessed me by putting Chance in my life so many years ago.
I love you Chance Marble!


text:
Google introduces Android L Developer Preview
Google today at I/O developer conference, announced a next big android update i.e the developer preview version of Android L, the upcoming android operating system.
"That comes out tomorrow," Android user-interface engineer, Chet Haase Haase said. "Please get started developing today." Taking elements from windows Phone 8 and iOS 7, L features new colors, new notifications and totally new UI design. Google will use the design across Chrome and all of Google's services. "We challenged ourselves to create a design that was not just for Android phones and tablets," said Matias Duarte, head of Android user experience, today.
The most amazing feature of Android L is a new design language called Material Design, which includes shadows, subtle textures and bright colors to Android. Also the design will work on a smartphone, tablet, smartwatch or TV. With android L the android apps will have better performance and improved battery life and new battery enhancements. To improve the battery performance, Google introduces Project Volta
Google said in a blog post – "Bringing material design to Android is a big part of the L-Release of Android, the version we previewed today. We've added the new Material theme, which you can apply to your apps for a new style: it lets you easily infuse your own color palette into your app, and offers new system widgets, screen transitions and animated touch feedback. We've also added the ability to specify a view's elevation, allowing you to raise UI elements and cast dynamic, real-time shadows in your apps."
Android L also introduces a new Android Runtime "ART" to replace Dalvik, and integration with its Android Wear platform. "ART" will allow android apps to run faster which will be compatible with every kind of processor, including ARM, x86, and MIPS chips.
More often there are a total of 5,000 new APIs in the new L version.
Android L preview also introduces the new look and feel for notifications, "In L, you hear your phone buzz, you take it out out, turn it on, and that's it," Sandler said. "It's completely glanceable.". Android L will also be able to detect some network problems so that apps can adjust accordingly.
"It allows a graceful handoff from one link to another — if you're about to lose Wi-Fi when it's losing its range, you can rebuild [a video] stream on cellular [networks] with no interruption to the user," Sandler said. The final name of "L" is not yet revealed by Google.
Tags:Android L, Google, Google I/O


text:
\section{Introduction}
\label{sec:intro}

Throughout, $G$ is a connected reductive linear algebraic group
defined over an algebraically closed field
of characteristic $p > 0$ and
$H$ is a closed subgroup of $G$.
Following  Serre \cite{serre2}, we say that
$H$ is \emph{$G$-completely reducible} ($G$-cr for short)
provided that whenever $H$ is contained in a parabolic subgroup $P$ of $G$,
it is contained in a Levi subgroup of $P$;
for an overview of this concept see for instance \cite{serre1} and
\cite{serre2}.
Note that in case $G = \GL(V)$ a subgroup $H$ is $G$-cr exactly when
$V$ is a semisimple $H$-module.
Recall that if $H$ is $G$-cr,
then the identity component $H^\circ$ of $H$ is reductive,  \cite[Property 4]{serre1}.

Let $V$ denote a rational $G$-module
and let $\rho : G \to \GL(V)$ be the representation of $G$
afforded by $V$.
Following Serre \cite{serre1},
we call $V$ \emph{non-degenerate} provided
$({\rm ker} \rho)^\circ$ is a torus.

First we consider two important and deep theorems in this context,
\cite[Thm.\ 5.4]{serre2} and \cite[Thm.\ 4.4]{serre2},
which provide necessary and sufficient conditions for a subgroup
$H$ of $G$ to be $G$-cr provided $p$ is sufficiently large.

\begin{thm}{\cite[Thm.\ 5.4]{serre2}}
\label{thm:nv}
Suppose that $p > n(V)$ for some rational $G$-module $V$.
\begin{itemize}
\item[(i)]
If $H$ is $G$-completely reducible, then $V$ is a semisimple $H$-module.
\item[(ii)]
Suppose that $V$ is non-degenerate.
If $V$ is semisimple as an $H$-module, then
$H$ is $G$-completely reducible.
\end{itemize}
\end{thm}

Here the invariant $n(V)$ is
defined as follows: let $T$ be a maximal torus of $G$ and let
$\lambda$ be a $T$-weight of $V$. Define
$n(\lambda) = \sum_{\alpha>0}\inprod{\lambda,\alpha\check}$,
where the sum is taken over all positive roots of $G$
with respect to $T$. Then define
$n(V) = \sup \{n(\lambda)\}$, where the
supremum is taken over all $T$-weights $\lambda$ of $V$,
 \cite[\S 5.2]{serre2}.
The proof of Theorem \ref{thm:nv} is elaborate and complicated;
it depends on
the full force of the following result.

\begin{thm}{\cite[Thm.\ 4.4]{serre2}}
\label{thm:ag}
Suppose that $p \ge a(G)$ and that
$(H:H^\circ)$ is prime to $p$.
Then $H^\circ$ is reductive  if and only if
$H$ is $G$-completely reducible.
\end{thm}

Here the invariant $a(G)$ is defined as follows:
for $G$ simple, set $a(G) = \rk(G) +1$,
where $\rk(G)$ is the rank of $G$.
For $G$ reductive, let $a(G) = \sup(1, a(G_1), \ldots, a(G_r))$,
where $G_1, \ldots, G_r$ are the simple components of $G$,
cf.\ \cite[\S 5.2]{serre2}.

We emphasize that Theorem \ref{thm:ag} is a consequence of a
number of deep theorems
due to Jantzen \cite{jantzen0} (Theorem \ref{thm:jantzen})
and McNinch  \cite{mcninch2}
in case $G$ is classical
and Liebeck and Seitz \cite{liebeckseitz0}
in case $G$ is of exceptional type, where
the latter involves complicated and long case-by-case analyses.
Given that the proofs of both these theorems are intricate,
it is desirable to have uniform arguments for them
even under additional restrictions on $p$.
We present new concise and uniform proofs of these two results in
Theorems \ref{thm:ss-vs-cr} and \ref{thm:red-vs-cr}
with different bounds on $p$.
Here we are particularly interested in obtaining
short proofs for sufficient conditions for $G$-complete reducibility.
Unfortunately, though not unexpectedly,
the brevity and uniformity
do come at the expense of less favourable bounds on $p$.

In \cite{serre1}  and \cite{serre2}, Serre gave an alternative proof
of his tensor product theorem
\cite[Thm.\ 1]{serre0} via the concept of $G$-cr subgroups.
Theorems \ref{thm:nv} and \ref{thm:ag} are both part of this approach.
In Section \ref{sec:variations} we essentially argue the other way round:
based on a special case of
the aforementioned
tensor product theorem
(Theorem \ref{thm:serre}), we first
derive a short proof of Theorem \ref{thm:nv} (in Theorem \ref{thm:ss-vs-cr})
and in turn use part of that result
to obtain a concise and uniform
proof of Theorem \ref{thm:ag} (in Theorem \ref{thm:red-vs-cr}),
with a worse bound on $p$.

\medskip

Our next result  generalizes
Jantzen's semisimplicity Theorem \ref{thm:jantzen} to $G$-cr  subgroups
of $G$.

\begin{thm}
\label{thm:jantzen-cr-subgps}
If  $p \ge \dim V$ and $H$ is $G$-completely reducible,
then $V$ is a semisimple $H$-module.
\end{thm}

Theorem \ref{thm:jantzen-cr-subgps}  is also of interest, as the  bound $p > n(V)$ in Theorem \ref{thm:nv}(i) does not apply in case
$V$ admits a non-restricted composition factor,
cf.\ Remark \ref{rem:nv-vs-dim}.
The proof of  Theorem \ref{thm:jantzen-cr-subgps} requires
the force of Theorem \ref{thm:nv}(i) and
\cite[Thm.\ 1]{serre0}.

\medskip

Serre's notion of saturation in $\GL(V)$
(see Definition \ref{def:saturation}) is an important tool in the
theory of complete reducibility, see \cite{serre0} and \cite{serre1}.
As an application of Theorem \ref{thm:jantzen-cr-subgps} we show in Corollary \ref{cor:Hsat}
that if $p \ge \dim V$ and $H$ is a $G$-cr
subgroup of $G \le \GL(V)$, then
the saturation of $H$ in  $\GL(V)$
is completely reducible in the saturation of $G$ in $\GL(V)$.

\medskip

Our final result is similar in spirit to Theorem \ref{thm:nv}(ii)
giving a sufficient semisimplicity condition for $H$ to be $G$-cr,
but strikingly it does not require any restriction on $p$.

\begin{thm}
\label{thm:sufficient-cr}
If $H$ acts semisimply on $V \otimes V^*$
for some non-degenerate $G$-module $V$,
then $H$ is $G$-completely reducible and $\rho(H)$ is separable in $\rho(G)$.
\end{thm}

Recall that a subgroup $H$ of $G$ is said to be \emph{separable in $G$} if
its scheme-theoretic centralizer is smooth, i.e.,\ if its
global and infinitesimal centralizers have the same dimension,
cf.\ \cite[Def.\ 3.27]{BMR}.
In \cite{BMRT}, we study the interaction
between this notion of separability and the concept of $G$-complete
reducibility.
Several general theorems concerning $G$-complete
reducibility require some separability hypothesis,
e.g.,\ see \cite[Thm.\ 3.35]{BMR}, \cite[Thm.\ 3.46]{BMR}.
In \cite[Thm.\ 1.2]{BMRT}, we show that any subgroup of $G$ is
separable in $G$ provided $p$ is very good for $G$.
Note that the special case of Theorem \ref{thm:sufficient-cr}
when $G = \GL(V)$  follows from  \cite[Thm.\ 3.46]{BMR},
since $\Lie(\GL(V)) \cong V \otimes V^*$.
The proof of Theorem \ref{thm:sufficient-cr} is based on a variant of
Richardson's tangent space argument
(cf.\ the separability statement of \cite[Thm.\ 1]{slodowy}),
see Lemma \ref{lem:descend}.

\section{Preliminaries}
\label{sec:prelims}

We maintain the notation from the Introduction.
In particular, $G$ is a connected reductive linear algebraic group
defined over an algebraically closed field
of characteristic $p > 0$
and $H$ is a closed subgroup of $G$.
Moreover, $V$ is a rational $G$-module and
$\rho : G \to \GL(V)$ is the representation of $G$
afforded by $V$.

First we recall Jantzen's
fundamental semisimplicity result, \cite[Prop.\ 3.2]{jantzen0}.

\begin{thm}
\label{thm:jantzen}
If $p \ge \dim V$, then $V$ is semisimple.
\end{thm}

We continue with the special case
for connected reductive groups
of Serre's seminal tensor product
theorem, \cite[Prop.\ 8]{serre0}.

\begin{thm}
\label{thm:serre}
Suppose $p > 2\dim V - 2$.
If $V$ is semisimple, then so is $V\otimes V^*$.
\end{thm}

Both Theorems \ref{thm:jantzen} and \ref{thm:serre}
have conceptional and uniform proofs
and both bounds are sharp (cf.\ \cite[Rem.(2), p.\ 260]{jantzen0} and \cite[\S 1.3]{serre0}).

\medskip

Let $H$ be a closed
subgroup of $G$ such that $H^\circ$ is reductive.
We say that $(G,H)$ is a \emph{reductive pair} if
the Lie algebra $\Lie H$
of $H$ is an $H$-module direct
summand of the Lie algebra $\Lie G$
of $G$, cf.\ \cite{rich2}.
Our next result is \cite[Cor.\ 3.36]{BMR}.

\begin{prop}
\label{prop:red-pair-gl-cr}
Suppose that $(\GL(V),G)$ is a reductive pair.
If  $V$ is a semisimple $H$-module, then
$H$ is $G$-completely reducible.
\end{prop}

Next we recall \cite[Cor.\ 2.13]{BMRT}.

\begin{prop}
\label{prop:red-pair-gl-sep}
If $(\GL(V),G)$ is a reductive pair, then every subgroup of $G$ is separable in $G$.
\end{prop}

Suppose that $p \ge \dim V$, so that
every non-trivial unipotent element in $\GL(V)$ has order $p$.
We recall Serre's notion of \emph{saturation} in this instance,
cf.\ \cite{serre1}.
Let $u \in \GL(V)$ be unipotent. Then there is a nilpotent element $\epsilon \in \End(V)$
with $\epsilon^p = 0$ such that $ u = 1 + \epsilon$.
For $t \in \mathbb G_a$ we can define $u^t$ by
$u^t = (1 + \epsilon)^t = 1 + t\epsilon + \binom{t}{2} \epsilon^2 + \cdots + \binom{t}{p-1} \epsilon^{p-1}$.
Then $\{u^t \mid t \in  {\mathbb G}_a\}$ is a closed connected subgroup  of $\GL(V)$
isomorphic to ${\mathbb G}_a$.

\begin{defn}
\label{def:saturation}
Suppose $p \ge \dim V$.
Let $H$ be a closed subgroup of $\GL(V)$.
We say that $H$ is \emph{saturated} if
for each unipotent $u \in H$ we have
$u^t \in H$ for all $t \in  \mathbb G_a$.
The \emph{saturated closure} $H^{\sat}$ of $H$ in $\GL(V)$ is the
smallest saturated subgroup of $\GL(V)$ containing~$H$.
\end{defn}

There is a notion of saturation for any connected reductive group $G$,
but this is considerably
more subtle, see \cite{serre1} for details.

The next result is the special case when $G = \GL(V)$ in  \cite[Thm.\ 5.3]{serre2}.
It follows since parabolic and Levi subgroups of $\GL(V)$ are saturated.

\begin{lem}
\label{lem:saturation}
Suppose that $p \ge \dim V$.
Let $H$ be a closed subgroup of $\GL(V)$.
Then $V$ is semisimple as an $H$-module if and only if it is semisimple as an $H^{\sat}$-module.
\end{lem}

The following is one of the key properties of saturated subgroups,
\cite[Property 3]{serre1}.

\begin{lem}
\label{lem:saturation-index}
Suppose $H$ is a saturated subgroup of $\GL(V)$.
Then
$(H: H^\circ)$ is prime to $p$.
\end{lem}

\section{Variations on Theorems  \ref{thm:nv} and  \ref{thm:ag}}
\label{sec:variations}
We begin by showing that
Theorems \ref{thm:jantzen} and \ref{thm:serre} and the
bound on $p$ in the latter guarantee that
$(\GL(V),\rho(G))$ is a reductive pair,
which is crucial for some of our subsequent arguments.

\begin{thm}
\label{thm:red-pair}
Suppose that $p > 2\dim V-2$.
Then $(\GL(V),\rho(G))$ is a reductive pair.
\end{thm}

\begin{proof}
Since $p > 2\dim V-2$, we also have $p \ge \dim V$.
Thus $V$ is semisimple, by
Theorem \ref{thm:jantzen}.
Thanks to Theorem \ref{thm:serre},
$V \otimes V^* \cong \Lie(\GL(V))$ is also semisimple.
Consequently, $\Lie \rho(G)$ is a direct $\rho(G)$-module
summand of  $\Lie(\GL(V))$.
\end{proof}

\begin{rem}
\label{rem:sharp-bound}
The bound in Theorem \ref{thm:red-pair} is sharp.
For, let $p = 2$, let $G = \SL_2$, and let $V$ be the
natural module for $G$.
Since $G$ is not separable in itself, Proposition \ref{prop:red-pair-gl-sep}
implies that $(\GL(V), G)$ is not a reductive pair.
In fact, although Theorem \ref{thm:red-pair} asserts that generically every
representation $V$  of $G$ gives rise to a reductive pair $(\GL(V),\rho(G))$,
this is \emph{never} the case if $p$ is bad for $G$ and $V$ is non-degenerate,
cf.\ Remark \ref{rem:separable}.
\end{rem}

The following is our variant of Theorem \ref{thm:nv}:

\begin{thm}
\label{thm:ss-vs-cr}
Let $H$ be a closed subgroup of $G$ and let $V$ be a $G$-module.
\begin{itemize}
\item[(i)]
Suppose that $p \ge \dim V$ and that $(H : H^\circ)$ is prime to $p$.
If $H$ is $G$-completely reducible, then $V$ is a semisimple $H$-module.
\item[(ii)]
Suppose that $V$ is non-degenerate and $p > 2\dim V-2$.
If $V$ is semisimple as an $H$-module, then
$H$ is $G$-completely reducible.
\end{itemize}
\end{thm}

\begin{proof}
First suppose as in (i).
Since $H$ is $G$-cr, $H^\circ$ is reductive, by \cite[Property 4]{serre1}.
Since $p \ge \dim V$ and $(H : H^\circ)$ is prime to $p$,
it follows from Theorem \ref{thm:jantzen}  (applied to $H^\circ$) and
\cite[Lem.\ 3.1]{jantzen0} that $V$ is a semisimple $H$-module.

Next suppose as in (ii).
Since $(\GL(V),\rho(G))$ is a reductive pair,
by Theorem \ref{thm:red-pair}, and $V$ is a semisimple $H$-module,
it follows from
Proposition \ref{prop:red-pair-gl-cr} that $\rho(H)$ is $\rho(G)$-cr.
Since $V$ is non-degenerate, \cite[Lem.\ 2.12(ii)(b)]{BMR}
implies  that $H$ is $G$-cr.
\end{proof}

\begin{rem}
\label{rem:LieG}
Note that Theorem \ref{thm:nv}(ii) (and thus Theorem \ref{thm:ss-vs-cr}(ii))
holds  in particular cases for considerably weaker bounds.
For instance, in \cite[Thm.\ 1.7]{BMRT},
we showed that if $p$ is very good for $G$ and
$H$ acts semisimply on the adjoint module $\Lie G$, then $H$ is $G$-cr.
The reverse implication fails under this bound,
cf.\ \cite[Rem.\ 3.43(iii)]{BMR}.
Here
$n(\Lie G) = 2 h - 2$, where $h$ is the Coxeter number of $G$,
cf.\ \cite[Cor.\ 5.5]{serre2}.
\end{rem}

Here is our variation of Theorem \ref{thm:ag}:

\begin{thm}
\label{thm:red-vs-cr}
Suppose that $p > 2\dim V-2$ for a non-degenerate $G$-module $V$ and that
$(H:H^\circ)$ is prime to $p$.
Then $H^\circ$ is reductive  if and only if
$H$ is $G$-completely reducible.
\end{thm}

\begin{proof}
First suppose that $H^\circ$ is reductive.
Since $p > 2\dim V-2$, we also have $p \ge \dim V$.
Thus $V$ is a semisimple $H^\circ$-module, by
Theorem \ref{thm:jantzen} (applied to $H^\circ$).
Moreover, since $(H:H^\circ)$ is prime to $p$, it follows from
\cite[Lem.\ 3.1]{jantzen0}
that $V$ is a semisimple $H$-module.
The result now follows from Theorem \ref{thm:ss-vs-cr}(ii).

The reverse implication is immediate, by
\cite[Property 4]{serre1}.
\end{proof}

Clearly, our bound in Theorem \ref{thm:red-vs-cr}
is much worse than the bound $a(G)$
in Theorem~\ref{thm:ag}.

\begin{rem}
\label{rem:separable}
It follows from Theorem \ref{thm:red-pair}
and Proposition \ref{prop:red-pair-gl-sep}
that only separable subgroups of
$G$ are captured in Theorems \ref{thm:ss-vs-cr}(ii) and  \ref{thm:red-vs-cr}.
In \cite[Ex.\ 4.2]{herpel}, the second author showed that
if $p$ is bad for $G$, then there always exists
a non-separable subgroup of $G$
(likewise if $G$ is simple and $p$ is not very good for $G$).
Consequently, in this case $(\GL(V),\rho(G))$ can't be a reductive pair
for \emph{any} non-degenerate rational $G$-module $V$,
by Proposition \ref{prop:red-pair-gl-sep}, cf.\ \cite[Rem.\ 4.3]{herpel}.
Each of the non-separable subgroups constructed in \cite[Ex.\ 4.2]{herpel}
is a regular reductive subgroup of $G$ and hence is $G$-cr,
thanks to \cite[Prop.\ 3.20]{BMR}.
\end{rem}

\section{Proof of Theorem \ref{thm:jantzen-cr-subgps}}

We now come to our generalization of
Jantzen's semisimplicity  Theorem \ref{thm:jantzen} to $G$-cr subgroups
of $G$; Theorem \ref{thm:jantzen}
is the special  case of Theorem \ref{thm:jantzen-cr-subgps} when $H = G$.

\begin{proof}[Proof of Theorem \ref{thm:jantzen-cr-subgps}]
Suppose that $p\geq \dim V$ and $H$ is $G$-cr.
By Theorem \ref{thm:jantzen}, $V$ is a semisimple $G$-module.
So to show that $V$
is also semisimple as an $H$-module, we may assume that $V = L(\lambda)$ is simple
of highest weight $\lambda = \lambda_0 + p \lambda_1 + \dots + p^r \lambda_r$, with
restricted weights $\lambda_i$.
Set $L_i := L(p^i\lambda_i) \cong L(\lambda_i)^{[i]}$,
the $i$th \emph{Frobenius twist} of $L(\lambda_i)$.
Then
$V = L_0 \otimes L_1 \otimes \cdots \otimes L_r$,
by Steinberg's tensor product theorem.
Since $p\geq \dim V$, we also have $p\geq \dim L_i = \dim L(\lambda_i)$ for each $i$, so that
$p > n(\lambda_i) = n(L(\lambda_i))$ for each $i$,
according to \cite[Lem.\ 1.2]{jantzen0}.
By Theorem \ref{thm:nv}(i),
each $L(\lambda_i)$ is a  semisimple $H$-module and hence
so is each Frobenius twist $L_i$.
Moreover, $p\geq \dim V$ also implies that $p > \sum_i (\dim L_i -1)$.
We therefore may apply Serre's tensor product theorem
\cite[Thm.\ 1]{serre0} to deduce
that $V$ is a semisimple $H$-module.
\end{proof}

Note that Theorem \ref{thm:jantzen-cr-subgps}
shows that Theorem \ref{thm:ss-vs-cr}(i) is valid even without the restriction on
the index of $H^\circ$ in $H$.
However, the proof of Theorem \ref{thm:jantzen-cr-subgps}  requires the full force of Theorem  \ref{thm:nv}(i), whereas our proof of Theorem \ref{thm:ss-vs-cr}(i) does not, so this is of independent interest.

\begin{rem}
Proposition \ref{prop:red-pair-gl-cr} asserts that
under the assumption that $(\GL(V),G)$ is a reductive pair,
$H$ is $G$-cr provided $V$ is a semisimple $H$-module.
In Theorem \ref{thm:jantzen-cr-subgps}
we prove the reverse implication under the assumption that $p \ge \dim V$.

Even under the seemingly stronger condition
that $(\GL(V),G)$ is a reductive pair and $V$ is a semisimple $G$-module, the statement of
Theorem \ref{thm:jantzen-cr-subgps} is false without the restriction on $p$.
Such an example is already known
thanks to a construction from unpublished work of Serre,
cf.\ \cite[Ex.~4.7]{BMRT}.
We now give a different example:
Let $p=3$, $q=9$ and let $G= \SL_2$.  Set $H = G(q)$.
Clearly, $H$ is $G$-cr.
The simple $G$-module $V = L(1+q+q^2)$
is isomorphic to $L(1)\otimes L(1)^{[1]}\otimes L(1)^{[2]}$ ,
by Steinberg's tensor product theorem,
where the superscripts denote $q$-twists.
Then $\dim V = 8 > p$.
One readily checks that $V \otimes V^*$ is a semisimple $G$-module, so that
$(\GL(V),G)$ is a reductive pair.
However, as a $G(q)$-module, $V$ is isomorphic to the $G$-module
$L(1)\otimes L(1) \otimes L(1)$ which
admits the non-simple indecomposable Weyl module of highest weight $3$
as a constituent, and the latter is not semisimple for $G(q)$,
e.g., see \cite[(2D)]{Wong}.
Consequently, $V$ is not semisimple as a  $G(q)$-module.
\end{rem}

It follows from \cite[Lem.\ 2.12(ii)(a)]{BMR} that if $H$ is $G$-cr, then $\rho(H)$ is $\rho(G)$-cr.
The following result shows that the same holds
for the saturation
 $\rho(G)^{\sat}$ of the image of $G$ in $\GL(V)$.

\begin{cor}
\label{cor:Hsat}
Suppose that $p \ge \dim V$.
Then $\rho(G)^{\sat}$ is connected and reductive.
If $H$ is $G$-completely reducible, then
$\rho(H)^{\sat}$ is $\rho(G)^{\sat}$-completely reducible.
\end{cor}

\begin{proof}
By Theorem \ref{thm:jantzen}, $V$ is a semisimple $G$-module.
Lemma \ref{lem:saturation}
then shows that
$\rho(G)^{\sat}$ is $\GL(V)$-cr and thus
$(\rho(G)^{\sat})^\circ$ is reductive,
by \cite[Property 4]{serre1}.
Consider the subgroup $M$ of $\GL(V)$ generated by
$\rho(G)$ and the closed connected subgroups
$\{\rho(u)^t \mid t \in  {\mathbb G}_a\} \cong {\mathbb G}_a$  of $\GL(V)$
for each unipotent element $u \in G$.
By definition, $M \le \rho(G)^{\sat}$.
If $M \ne \rho(G)^{\sat}$, then
by  repeating this process with $M$ (possibly several times), we eventually generate all of
$\rho(G)^{\sat}$ by $\rho(G)$ and closed connected subgroups of $\GL(V)$
isomorphic to ${\mathbb G}_a$. It thus
follows from \cite[Cor.\ 2.2.7]{spr2} that $\rho(G)^{\sat}$ is connected.

Now suppose that $H$ is $G$-cr.
Theorem \ref{thm:jantzen-cr-subgps} then implies
that $V$ is a semisimple $H$-module and thus, by
Lemma \ref{lem:saturation},
$V$ is also semisimple as a $\rho(H)^{\sat}$-module.
Thanks to \cite[Cor.\ 1]{serre1}, we have
$n_{\rho(G)^{\sat}}(V) \le n_{\GL(V)}(V) = \dim V -1 < p$.
It thus follows from Theorem \ref{thm:nv}(ii)
that $\rho(H)^{\sat}$ is $\rho(G)^{\sat}$-cr, as desired.
\end{proof}

We note that a variation of  Corollary \ref{cor:Hsat} is valid for
the general notion of saturation
replacing $\GL(V)$ with an arbitrary connected reductive group $G$.
We will  return to this in a future publication.

\begin{rem}
\label{rem:Hsat-converse}
If $p > 2 \dim V - 2$ and $V$ is non-degenerate,
then also the converse of the final assertion of
Corollary \ref{cor:Hsat} holds.
For, if $p > 2 \dim V - 2$, then $p \ge \dim V$.
Thus $\rho(G)^{\sat}$ is connected and reductive, by the first part of
Corollary \ref{cor:Hsat}.
Now suppose that $\rho(H)^{\sat}$ is $\rho(G)^{\sat}$-cr.
It then follows from Lemma \ref{lem:saturation-index} and
Theorem \ref{thm:ss-vs-cr}(i) that $V$ is a semisimple
$\rho(H)^{\sat}$-module, and thus $V$ is a semisimple $H$-module,
by Lemma \ref{lem:saturation}.
The result now follows from Theorem \ref{thm:ss-vs-cr}(ii).
\end{rem}

\begin{rem}
\label{rem:nv-vs-dim}
We compare the bounds $p > n(V)$ from  Theorem \ref{thm:nv}(i) and
$p\geq \dim V$
from Theorem \ref{thm:jantzen-cr-subgps}.
Let $L(\mu_1),\dots,L(\mu_m)$ be the non-isomorphic simple factors
of a composition series of $V$.

First, suppose that all $\mu_j$ are restricted.
Then $p>n(L(\mu_j))$
for each $j$, by \cite[Lem.\ 1.2]{jantzen0}.
In particular, $p> \sup\{n(L(\mu_j))\} = n(V)$, by
\cite[\S 5.2]{serre2}, so that Serre's bound applies.
In general,  $\dim V$
is considerably larger  than $n(V)$ in this
situation. For instance, let $G$ be simple of type $E_6$ and let
$V = L(\omega_1)$ be the  simple $G$-module of
highest weight $\omega_1$, the first fundamental dominant weight.
Here we have $n(V) = 16$, while
$ \dim V = 27$.

Next assume that one of the $\mu_j$ is not restricted, i.e.,\  say
$\mu_j = \lambda_0 + p \lambda_1 + \dots + p^r \lambda_r$, with
restricted weights $\lambda_i$ and at least one $\lambda_i \neq 0$, ($i>0$).
According to \cite[\S 5.2]{serre2}, we find that
$n(V) \ge n(\mu_j) = n(\lambda_0) + pn(\lambda_1)+\dots+p^rn(\lambda_r) \ge p$,
so the bound $p > n(V)$ does not apply.

Now suppose  in addition that $p \ge \dim V$ and $H$ is $G$-cr.
Then Theorem \ref{thm:jantzen-cr-subgps}
shows that $V$ is a semisimple $H$-module.
We can also argue as in the proof of Corollary \ref{cor:Hsat}:
Since  $p \ge \dim V$, we can saturate the image of $G$
in $\GL(V)$. Then,
because $H$ is $G$-cr, it follows from  Corollary \ref{cor:Hsat} that
$\rho(H)^{\sat}$ is $\rho(G)^{\sat}$-cr.
Thanks to Lemma \ref{lem:saturation-index} and
Theorem \ref{thm:ss-vs-cr}(i), we see  that $V$ is a semisimple
$\rho(H)^{\sat}$-module, and thus $V$ is a semisimple $H$-module,
by Lemma \ref{lem:saturation}.
In place of Lemma \ref{lem:saturation-index} and
Theorem \ref{thm:ss-vs-cr}(i), we can use
Theorem \ref{thm:nv}(i) directly:
\cite[Cor.\ 1]{serre1} implies that
$n_{\rho(G)^{\sat}}(V) \le n_{\GL(V)}(V) = \dim V -1 < p$,
so that Serre's condition is satisfied for $\rho(G)^{\sat}$
even though it is not satisfied for $G$ itself.
\end{rem}

\section{Proof of Theorem \ref{thm:sufficient-cr}}
\label{sec:proof13}
Let $H \le K$ be closed subgroups of $G$.
The normalizer of $K$ in $G$ is denoted by $N_G(K)$.
By $C_G(H) = \{g \in G \mid gxg\inverse = x \ \forall x \in H\}$
and $C_K(H) = C_G(H) \cap K$ we denote the centralizer of $H$ in $G$
and the centralizer of $H$ in $K$, respectively.
Analogously,
we denote the  centralizer
of $H$ in $\gg = \Lie G$ by
$\cc_\gg(H) =\{y \in \gg \mid \Ad(x)y = y \ \forall x \in H\}$
and the centralizer of $H$ in $\kk = \Lie K$ by
$\cc_\kk(H) = \cc_\gg(H) \cap \kk$, respectively.

Given $n\in \mathbb{N}$, we let $G$ act diagonally on $G^n$ by simultaneous conjugation:
\[
g\cdot (g_1,g_2,\ldots, g_n) = (gg_1g^{-1},gg_2g^{-1},\ldots,
gg_ng^{-1}).
\]

We require the notion of a generic tuple, \cite[Def.\ 5.4]{GIT}.
Let $G\hookrightarrow\GL_m$
be an embedding of algebraic groups.
Then $\tuple{h} = (h_1, \ldots, h_n) \in H^n$ is called a
\emph{generic tuple of $H$ for the embedding $G\hookrightarrow\GL_m$}
if the $h_i$
generate the associative subalgebra of $\Mat_m = \mathfrak{gl}_m = \Lie \GL_m$
spanned by $H$. We call $\tuple{h}\in H^n$ a \emph{generic tuple of $H$}
if it is a generic tuple of $H$ for some embedding $G\hookrightarrow\GL_m$.
Generic tuples exist for any embedding $G\hookrightarrow\GL_m$ if
$n$ is sufficiently large.
The relevance to $G$-cr subgroups of this notion is as follows:
For $\tuple{h}\in H^n$ a generic tuple of $H$,
\cite[Thm.\ 5.8]{GIT} asserts that the orbit $G \cdot \tuple{h}$ is closed in
$G^n$ if and only if $H$ is $G$-cr.

Our first  result in this section generalizes  \cite[Rem.\ 3.31]{BMR}.

\begin{lem}
\label{lem:sep}
Let $H \le K \le G$ be closed subgroups of $G$.
Let $\tuple{h}$ be a generic tuple for $H$.
Then the orbit map $K \rightarrow K \cdot \tuple{h}$ is separable if and only if $H$
is separable in $K$.
\end{lem}

\begin{proof}
According to \cite[Lem.\ 5.5(i)]{GIT}, a generic tuple $\tuple{h}$ for $H$
satisfies the identity $C_K(H) = C_K(\tuple{h})$.
By the same argument, we obtain
$\cc_\kk(H) = \cc_\kk(\tuple{h})$.

Let $\pi: K \rightarrow K \cdot \tuple{h}$ be the orbit map.
Then $\pi$ is separable if and
only if $d_e \pi: \kk \rightarrow T_{\tuple{h}}(K \cdot \tuple{h})$
is surjective. Using
$\dim T_{\tuple{h}}(K \cdot \tuple{h}) = \dim K\cdot \tuple{h} = \dim K - \dim C_K(\tuple{h})$
and $\dim \im(d_e \pi) = \dim \kk - \dim \cc_\kk(\tuple{h}) = \dim K - \dim
\cc_\kk(\tuple{h})$, we find that the surjectivity of $d_e \pi$ is equivalent to the
equality $\dim C_K(\tuple{h}) = \dim \cc_\kk(\tuple{h})$. By the first paragraph of
the proof, this is equivalent to the separability of $H$ in $K$.
\end{proof}

The following generalizes part of \cite[Thm.\ 1.3]{BMRT}
(which is the special case of Lemma \ref{lem:descend}
when $(G,K)$ is a reductive pair and the $h_i$ lie in $K$).

\begin{lem}
\label{lem:descend}
Let $K \le G$ be a closed subgroup.
Let $\tuple{h}=(h_1,\dots, h_n) \in N_G(K)^n$.
Suppose that there is a decomposition $\gg = \kk \oplus \mm$ that is
$\Ad_G(h_i)$-stable for $i=1,\dots,n$,
and that the orbit map $\pi' : G \rightarrow G\cdot \tuple{h} \subseteq G^n$
is separable.
Then the orbit map $\pi: K \rightarrow K\cdot \tuple{h}$ is separable.
\end{lem}

\begin{proof}
Since $\pi':G \rightarrow G\cdot  \tuple{h}$ is separable, the
 differential $d_e\pi':\gg \rightarrow T_{\tuple{h}}(G \cdot \tuple{h})$ is surjective.
Since $T_{\tuple{h}}(K \cdot \tuple{h}) \subseteq T_{ \tuple{h}}(G \cdot \tuple{h})$,
 any $y \in T_{\tuple{h}}(K \cdot \tuple{h})$ has a preimage $z \in \gg$ such that
 $d_e\pi'(z) = y$.
Let $z=z_1 + z_2$ be a decomposition of $z$ in $\gg = \kk \oplus \mm$.
Let $\mu:G^n \rightarrow G^n$ be the automorphism of varieties that sends a tuple
 $(g_1,\dots,g_n)$ to $(g_1h_1^{-1},\dots,g_nh_n^{-1})$.
Applying the differential of $\mu$ to $y$, we get
$d_{\tuple{h}}\mu (y) = d_{\tuple{h}}\mu \circ d_e \pi' (z) = d_e
 (\mu \circ \pi') (z_1) + d_e(\mu \circ \pi')(z_2) \in \gg^n$.
Due to our assumption on the $h_i$, the map $\mu$ sends
$K \cdot \tuple{h}$ to $K^n$, so that
$d_{\tuple{h}}\mu (y) \in \kk^n$.
Likewise, $d_e
 (\mu \circ \pi') (z_1)\in \kk^n$.
However,
$d_e(\mu \circ \pi')(z_2) = ((1 -\Ad h_1)(z_2),\dots,(1-\Ad h_n)(z_2)) \in \mm^n$,
according to the stability of the decomposition $\gg = \kk \oplus \mm$.
We deduce that $d_e(\mu \circ \pi')(z_2)=0$
and hence $d_e(\mu \circ \pi')(z_1)=d_{\tuple{h}}\mu(y)$.
Since $\mu$ is an automorphism, this implies that $y = d_e\pi'(z_1) = d_e \pi(z_1)$.
We have thus shown that $d_e\pi$ is surjective, i.e., that $\pi$ is separable.
\end{proof}

Our next result generalizes \cite[Thm.\ 1.4]{BMRT}
(which is the special case of Corollary \ref{cor:descend}
when $(G,K)$ is a reductive pair).

\begin{cor}
\label{cor:descend}
Let $H \le K \le G$ be closed subgroups.
Suppose that there is a decomposition $\gg = \kk \oplus \mm$ as an $H$-module
and that $H$ is separable in $G$.
Then $H$ is separable in $K$.
\end{cor}

\begin{proof}
Let $\tuple{h} \in G^n$ be a generic tuple of $H$.
Since $H$ is separable in $G$, the orbit map
$G \rightarrow G\cdot \tuple{h}$ is separable,
thanks to Lemma \ref{lem:sep}.
It then follows from Lemma \ref{lem:descend} that
$K \rightarrow K\cdot \tuple{h}$ is separable and thus that
$H$ is separable in $K$, again by Lemma \ref{lem:sep}.
\end{proof}

Next we give an immediate consequence of Corollary \ref{cor:descend}
and \cite[Thm.\ 1.2]{BMRT}.

\begin{cor}
\label{cor:descend2}
Suppose that $p$ is very good for $G$.
Let $H \le K \le G$ be closed subgroups.
Suppose that there is a decomposition $\gg = \kk \oplus \mm$ as an $H$-module.
Then $H$ is separable in~$K$.
\end{cor}

We are now in a position to prove  Theorem \ref{thm:sufficient-cr}.

\begin{proof}[Proof of Theorem \ref{thm:sufficient-cr}]
Thanks to \cite[Lem.\ 2.12(ii)(b)]{BMR},
we may assume that $V$ is a faithful $G$-module so that $G \le \GL(V)$.
By assumption, $H$ acts semisimply on
$\Lie(\GL(V)) \cong V \otimes V^*$, and it is automatically
 separable in $\GL(V)$ (cf.\ \cite[Ex.\ 3.28]{BMR}).
Since the $H$-submodule $\gg$ must have a complement
in $\Lie(\GL(V))$, we can use Corollary \ref{cor:descend} (with ``$G=\GL(V)$''
and ``$K=G$'') to deduce that $H$ is separable in $G$. Moreover, as an $H$-submodule of $\Lie(\GL(V))$,
$\gg$ is also semisimple.
Finally, \cite[Thm.\ 3.46]{BMR} implies that $H$ is $G$-cr.
\end{proof}

We discuss some consequences of Theorem \ref{thm:sufficient-cr}.

\begin{rem}
\label{rem:linred}
It follows readily from Theorem \ref{thm:sufficient-cr} that
a linearly reductive subgroup of $G$ is $G$-cr and separable in $G$,
\cite[Lem.\ 2.6]{BMR} and \cite[Lem.\ 4.1]{rich1}.
In particular, Theorem \ref{thm:sufficient-cr} gives an alternative
proof for the first fact without any cohomology considerations,
cf.~\cite[\S 6]{rich1}.
\end{rem}

\begin{rem}
\label{rem:2}
Assume as in Theorem \ref{thm:sufficient-cr} that
$H$ is a closed subgroup of $G$ acting semisimply on $V \otimes V^*$
for some faithful $G$-module $V$.
Since  $H$ is separable in $\GL(V)$ (cf.\ \cite[Ex.\ 3.28]{BMR}),
and semisimple on $\Lie(\GL(V)) \cong V \otimes V^*$,
it follows from \cite[Thm.\ 3.46]{BMR}
that $H$ is $\GL(V)$-cr, i.e.,
$V$ is a semisimple $H$-module
(cf.\ \cite[(2.2.2), Prop.\ 3.2]{serre0.5}).

Moreover,
by  Theorem \ref{thm:sufficient-cr},
$H$ is $G$-cr and thus $H^\circ$ is reductive and the proof of
Theorem \ref{thm:sufficient-cr} shows that both
$(\GL(V),H)$ and $(G,H)$ are reductive pairs.
Note that in general $(\GL(V),G)$ need not be a reductive pair.
\end{rem}

\begin{exmp}
Let $p = 2$, $G=\SL_2$ and let $T$ be a maximal torus of $G$.
Let $H = N_G(T)$.
If $\phi\colon G\ra G'$ is a non-degenerate epimorphism
(i.e., $(\ker \phi)^\circ$ is a torus),
then $G'= \SL_2$ or $G'= \PGL_2$ and it is easily checked that
$\phi(H)= N_{G'}(T')$, where $T':= \phi(T)$ is a maximal torus of $G'$.
Hence $\phi(H)$ is not separable in $G'$.
It follows from Theorem \ref{thm:sufficient-cr}
that $H$ does not act semisimply on $V\otimes V^*$
for \emph{any} non-degenerate $G$-module $V$.
\end{exmp}

\begin{rem}
 The converse of Theorem \ref{thm:sufficient-cr} is false.
For instance, let $p = 2 $, let $H= G= \GL_2$
and let $V$ be the natural module for $G$.
Then clearly $H$ is $G$-cr and separable in $G$.
But $V\otimes V^*$ is not $H$-semisimple.
\end{rem}

\bigskip

{\bf Acknowledgments}:
The authors acknowledge the financial support of
the DFG-priority program SPP 1388 ``Representation Theory''.
Part of the research for this note was carried out while the second and fourth
authors were visiting the Department of Mathematics at the University of York.
We would like to
thank the members of the Department of Mathematics at the University of York
for their hospitality.
We are grateful to J-P.\ Serre for helpful comments on the material of this note.

redpajama_set_name: RedPajamaArXiv


text:
Ford ›
Ford Mustang Shelby GT500 2010-2014 ›
New Mustang to go four-wheel drive with hybrid V8
Currently reading: New Mustang to go four-wheel drive with hybrid V8
New Porsche Taycan Sport Turismo on sale from £73,560
Electrified V8 will spearhead the 2022 arrival of the next-generation Ford Mustang
How Autocar imagines the new Ford Mustang hybrid could look
Current Mustang's basic character will be retained
Mach-E is tipped to give style pointers to the new Mustang
by Lawrence Allan
Ford's muscle car icon, the Mustang, will enter its next generation in 2022 – and Autocar understands it will be brought into a new era with its first hybrid powertrain.
An electrified variant of the Mustang will join the electric Mustang Mach-E, which is due to be delivered to customers from this autumn, as Ford's electrification roll-out gathers momentum. Ford plans to launch 18 mild-hybrid and full-hybrid models by 2022, and the next Mustang is thought to be one of those.
The firm is hoping for the new model to build on the success of the current, S550-generation car, which has been on sale since 2015 and was the first Mustang to be marketed and sold globally. This meant the first-ever right-hand-drive Mustang sales were finally made possible. Ford has shifted 55,000 examples of the current car across Europe – more than ever before in the Mustang's history.
The cars that did a million miles - on one engine
The Mustang at 55: a look back to its '60s heyday
Promoted: The Ford Mustang story
Ford Mustang Shelby GT500 2010-2014 road test review
The next-generation Mustang, codenamed S650, will follow that template by making a return to mainland Europe in left-hand drive and the UK in right-hand drive, Autocar understands.
The new Mustang would have arrived far sooner were it not for a reversal of product plans. In 2017, Ford announced it would bring a hybrid Mustang to market in 2020. But that same year, Jim Hackett stepped up to the role of Ford CEO and decided instead to prioritise getting a Mustang-inspired EV, the Mach-E, to market as its long-awaited first mass-production electric car.
Now that the Mach-E has been developed and its launch is around the corner, attention has turned back to the standard coupé and convertible Mustang models, which are still expected to dominate sales of the nameplate for some time yet.
Ford is hoping to appease its loyal following of enthusiasts by retaining the V8 option, but this time it will be mated to a fuel-saving hybrid system.
A US patent filed by Ford in 2017, when the original plan was in place, refers to technical drawings of a "twin-motor drive system for hybrid vehicle". These show an eight-cylinder engine in a vee formation sending power to the rear wheels – a classic Mustang trait. But the filing also details battery assistance in the form of two electric motors "mounted directly to opposing sides of the engine".
Ford Mustang Shelby GT500 2010-2014
Pricier but thoroughly updated Ford Mustang Shelby GT500 packs the power to justify the Shelby tag, but also offers accessibility and a dynamic edge
This means such a model would have four-wheel drive, with electric-only power heading to the front axle and, in performance driving, stepping into play when the car's rear wheels lose traction. The system could offer torque vectoring, too.
In more eco-minded modes, that same set-up would allow the engine to be shut off in situations with low power demand and high battery charge, boosting efficiency. A similar powertrain concept is used in the recently launched Ford Explorer Hybrid in the US.
A likely factor in the decision to go hybrid is Ford's intention to base the next Mustang on the same CD6 platform as the Explorer SUV and its sibling, the Lincoln Aviator. In 2018, the company announced it was streamlining its platform line-up with five highly modular underpinnings, including one to cover all rear-wheel-drive/ four-wheel-drive unibody-construction cars (such as the new Explorer and next Mustang, among others).
However, it remains to be seen whether Ford will offer alongside the hybrid a cheaper, non-hybrid V8 option and a turbocharged four-cylinder engine, as are available today.
Ford recently dropped automatic versions of the four-pot Ecoboost Mustang in the UK due to their low efficiency. It's also unlikely that a traditional V8 without some form of electrification would continue in Europe, given the need to meet stringent fleet average emissions targets.
As for the new Mustang's design, we're expecting it to stray not too far from the formula so well loved by its customer base. The muscle car's shape has gently evolved over the decades and, given the Mach-E takes inspiration from today's Mustang and will be on sale alongside the new car, it should be recognisable, albeit with a more modernist approach to the detailing. However, we've yet to see any disguised test mules to give us a clear indication of its shape.
It's likely that a convertible version will again appear, despite some makers moving away from that market, because the drop-top still makes up a significant proportion of Mustang sales.
Inside, expect the Mustang to be more driver-focused than the tech-heavy and space-maximised Mach-E.
The infotainment and driver assistance systems on that car are expected to be moderated for their introduction in the Mustang.
The four-seat layout should continue largely as it is today, with engineering developments ensuring the hybrid system has a minimal effect on interior space.
Ford Mustang Mach-E: UK prices and specs confirmed
First ride: 2020 Ford Mustang Mach-E prototype
Ford GT vs Ford Fiesta ST: in search of the blue bloodline
FRI2 20 April 2020
Hey Rollocks, did you ever
Hey Rollocks, did you ever see Mustangs from 1975 - 2005? Ford did you a HUGE favor by not bringing it to your shores....they were econoboxes powered by elastic-bands.
Rollocks 20 April 2020
Tears of shame
I'm still furious that Ford thought so little of British consumers that it refused to sell us one of the world's greatest cars for half a century - half a freakin' century! - and decided instead to fob us of with the Crapi. And I'm appalled that so many of my countrymen fell for the deceit.
eseaton 20 April 2020
Putting the words 'Mustang'
Putting the words 'Mustang' and 'torque vectoring' in the same sentence makes me feel ill.


This was a message shared with me by a classmate of mine who is currently doing her internship. I know she is not the only one to encounter this, as I have myself. I'd like this post to serve as a place for you to share any advice/widsom with us if you have any.
I am part way through marking a grade 10 assignment (began with measuring a pop can, ended with finding the cost of producing a certain amount of aluminum) that my class had to do that incorporated everything from the last unit they learned, and the lack of effort put into this is blowing my mind.
Almost every student in my class should have been able to complete this assignment 85-100% correctly, had they even used the class time I had given them. The language was simple, it was divided into small steps, and was EXTREMELY straightforward (almost ridiculously so, for a grade 10 class).
I am finding that one of the things I am most frustrated with is many of my students just DON'T CARE.
Are any of you facing this challenge? If yes, how are you dealing with this? (I'm not looking for a discussion on grading and grades as a non-motivator, because frankly, the fact that this assignment was one of the few graded this unit still wasn't motivation for my students.) So, how are you motivating your students? Please share.
How do you deal with students who seem to just not care? Any words of wisdom?
In middle years you can "motivate them" by taking away the hands on classes. I hate taking away things like art, PAA, or PE, but on an infrequent basis it has impact. I'm not sure how this would translate to high school. It's also a different matter if it's the entire class instead of a couple students. I'm having the same issue so if anyone has any advice I'm all ears as well!
Have you tried having a conversation with the students? Explain what you need and why? They have to demonstrate a certain level of competency to pass, earn credits etc.
Then ask them what their issues are and have them explain.
Finally you and students have to find a way to have both your needs met.
Back to redo and take new mark. Hopefully process will get through to a few. Could see having to do this whole thing again with a few more assignments.
Those that don't buy in will see low marks on report cards. Good luck.
You have posed a question that all teachers have asked themselves over the years. Having taught for 23 years does not qualify me for having the answers. In the past three years I have been flabbergasted about the number of students who lack self-motivation. Perhaps it's the result of an over-indulged generation? That's talking from the viewpoint of a teacher and a parent. I have no answers for this. I'm afraid that some students will carry on with that innate ability to challenge themselves. I've taken to having students draw up contracts based on unit reflections and goals. I find myself relying more and more on support teachers and contacting parents on a consistent basis. Making sure that my students have self-created timelines has helped as has allowing them say into how they will present their learning and how they will be assessed.
Sorry to hear your students are not putting in their best effort. You mentioned that it was a relatively easy assignment and perhaps that has something to do with it? We all enjoy a good challenge. It needs to be just beyond their reach but still attainable in order to engage. If they are feeling like you have lowered they bar they may respond in kind. Don't give up. Have an honest conversation with them about your feelings. Co construct a rubric that outlines the expectations and see if that engages them. Whatever you do, don't give up!
Evening! Saw your post via Twitter and a few things came to mind. First, I'm wondering what evidence does she (you) have they don't care? In other words, why "don't care" versus "already know the content"? What lead you that conclusion rather than another? Second, I'm wondering if appearance of not caring stems from adults telling them why it's important, rather than them uncovering the significance of what they're learning. Finally, I think a challenge is often we apply adult thinking to teenage brains. We see things they can't. They feel things we don't. You CLEARLY see why it's important, what they may see is "Why do I need to know this and what possible difference does it make on my life?" To that end, I'm wondering about engaging your students in the design of your culminating project. If you go back to your learning targets or objectives, what did you want students to demonstrate their understanding of? So rather than working with a pop can, could they select a product from an area of interest (i.e. CDs, video games, a specific car part) and research the cost of that and submit their findings as evidence of their learning? You could provide grading criteria to ensure they all include the same features so you'd be grading common tasks, just with different objects and values.
First, thanks for your insight – it's amazing that within moments I can have so much input toward something I've been struggling with as a new teacher.
I suppose I came to the conclusion that my students "didn't care" because of previous patterns I'd seen with them, and because the work done appeared like they had rushed through it. At this point I feel like I've tried everything, but I know that's not the case – I've only been doing this for 2 1/2 months. I know I've got a lot to learn, and appreciate all of the feedback I've been getting.
I have also wrestled with the thought that maybe this was TOO easy, TOO direct, and in turn, not engaging enough. I had intended it to be completed progressively, as we made our way through new skills and concepts, which I probably could have enforced more. I also have no doubt that with some work this could be redesigned, taking into consideration your last suggestion.
Will I, as a teacher, ever be able to make students see the things I do? Will this come with more experience? I feel like the more experience I gain as a teacher, the farther I will actually be from the teenaged mind.
class)". Perhaps one of the problems from the learners' POV is that there isn't a lot of room to problem solve–to sink their proverbial teeth into the assignment as you have done a lot of the thinking ahead of time. Having done this as well, I found that when I scale back and build in enough ambiguity, that results are often better.
2. Might there be a range of reasons for students' lackluster responses? Can you begin to sort the responses and categorize them so that you have a clearer picture as to why students' work failed to show effort? After sorting, you may then be able to better assist students.
3. Just this evening, my husband had said how disappointed he was with several of his middle school student' essays as these students had not revised their texts based on earlier comments he had made but elected to hand in the same paper again. Why I mention this now is he said this: "I am extending their time and will revisit one-to-one with each as they are not allowed to fail this. They need to learn how to revise. I need to figure out with each what went wrong." Talking one-to-one can often yield important insights. Also, if the learning was important, have you considered what next you will do in order for the students to learn?
4. Choice matters. Had you considered offering choice to students as to how they demonstrated the 'content' of this assignment? Perhas through alternatives that they design (you certainly will learn what they know) they may be a bit more motivated.
I wish you much luck as you navigate this. I have been teacher/admin for the last 25+ years. Working with teenagers is thrilling and it is often frustrating. I think it is fabulous that you are reaching out to others. We are a sharing profession.
This is a huge issue. You simply can't blame all the apathy on your lesson design. You can't get blood from a stone. I've been trying to solve this for decades frankly.
I am now convinced that much of the lack of motivation is a culture full of many depressed children. Our kids also have much more anxiety than they once did and I think lethargy is a symptom.
Solutions are complex. I try to deal w one student at a time because it gets just too overwhelming otherwise.
Knowing how to motivate and engage students is one of the great mysteries of teaching; what works with one group may be completely unworkable with another. Therein lies the art!
From the phrasing above, it seems to me that too much work had been done for the students – there was no thinking left for them to do. An open-ended problem-solving approach where students have to develop their own methods for measuring and determining cost may have been more effective. I would suggest that the teacher in question ask the students directly – deconstruct the lesson with them. Find out from them where the break down occurred and then ask them to design a new task for meeting the learning objectives.
For me, assignments are only one part of class. Continue to build relationships with students and help them find a way to share their learning but in different ways. The more you get to know them, the better you will also know their strengths and interests. It is hard in a practicum because you are so limited in time but there is no magical solution or else someone would have shared it with the world. Continue to build relationships with them and you will see them come around. It does take time.
You are absolutely right about it being tough in practicum. I feel like I am still getting to know my students (even the ones I've been teaching since the beginning of September, never mind the ones I began teaching last week!). It is something I will definitely continue to work at, as I can see how it is so important. Thank you for the words of encouragement!
Have you learned about your students' interests? One of the first things we did at school this year was to hold an Identity Day (thank you, George Couros) to learn what our students loved doing. What I love the most about this is that I can talk with my students about the things that interest them and then use those areas to help them learn almost everything. One of my students loves making stop-motion animation videos. We can learn more about writing, math, other cultures, music, art- and he can learn through his love of making videos.
I'm curious as to the goal of the exercise you wanted them to do. Could you have asked them how they might demonstrate their understanding of the topic and allow them the option of making the choice? For those who were unsure, you could have shown them your example with the aluminum, but then asked them to devise their own example. My students have a lot of choice when it comes to demonstrating their learning, and that has made all the difference to me.
Please do not withhold hands-on activities from them as "motivation," because that really amounts to punishment. It also makes both students and parents view the more hands-on learning as fluff and fun… which translates to "less academic and less learning." That couldn't be further from the truth! Kids enjoy art, music, and P.E. because they are learning in ways that are more natural. They have more autonomy and opportunities for creativity. Why should ANY learning opportunity be less than that?
There are many great comments here and they all should be carefully read. The only thing I would add is make sure you care about what you're teaching. I mean really care. If you're teaching something just because it's in the curriculum they non-compliant student will sniff that out in a heartbeat. Even if you're doing a good job, teaching by the book, your students need to no he and why it matters. That doesn't mean you have to engage them every day in that question but you have to show them with your actions and heart that Math matters.
That might be a tough question to answer. I can't imagine teaching in 2011 when you only barely care about what you're doing. I'm not saying that's the case here but your students need to see it matters to you. Modeling is a powerful teaching method. Maybe the most powerful.
I'm assuming you're talking about maths teaching in high school in particular. If you follow #mathchat on Twitter you'll find that even experienced teachers grapple with motivating students to learn maths. Know that there are 'wins' and quite often, as a result of a combination of things including some of the things previously suggested: passion for your subject, knowing your students, etc.
Somewhere between primary and high school, students start to disengage – even dislike – maths. Know this first. Then, do something about it. And you won't always win but the wins you do have will be enough to keep you going.
My approach? "How can I make this interesting for me?" Yes, me. If I can keep myself interested, then the passion and enthusiasm comes through. Granted, I do have a playful mindset. It also means that I have a varied approach to my teaching. I have students tell me, "we like you miss, we just don't like maths. sorry". Imagine that, apologizing for something they're not interested in!
There are topics in high school maths that are just not interesting for many people and that should be fine because everyone has at least one thing they're not interested in (e.g. I'm not interested in learning about options-trading even though many say it's lucrative). What's not fine is to indulge such disinterest if we can help it…..SO, we try to motivate our students….and this is not a one-off thing.
I've got a few ideas for motivating high school maths students in my blog under the motivation tag (funnily enough). Check out the learning strategies and reflection tags as well.
Final advice: be kind to yourself.


text:
Rafael Nadal bounces back against Alexander Zverev to reach semi-final in Rome
Rafael Nadal gained revenge over Alexander Zverev to stay on track for a 10th title at the Italian Open in Rome.
The Spaniard needed nearly three-and-a-half hours to defeat Denis Shapovalov in the third round on Thursday, saving two match points, and backed it up brilliantly with a 6-3, 6-4 victory over Zverev.
The German had won his last three meetings with Nadal, including in the quarter-finals of the Madrid Open last week, but saw his opportunities consistently snuffed out.
In a reversal of his fortunes against Shapovalov, Nadal surged into a 4-0 lead before Zverev gained a foothold in the contest and made things tight.
Nadal doing Nadal things 💪@RafaelNadal is into the semi-finals in Rome after beating Zverev 6-3, 6-4. #IBI21 pic.twitter.com/OuTMESg4ue
— ATP Tour (@atptour) May 14, 2021
There was a worrying moment for Nadal when he caught his foot on the service line sliding to the ball and fell heavily, but he recovered quickly to serve out the set.
The second set was one of the best of the tournament, with the pair engaging in lung-busting rallies, but a single break of serve ultimately proved the difference as Nadal saved all eight break points he faced.
"Today I played much better than yesterday," said the second seed. "I did a lot of things well, playing much more solid. I think I played one of the more solid matches on clay this year against a very tough opponent. It's a great victory for me.
"Under pressure, under tough moments that sometimes I lost important points the last couple of weeks, today, yesterday, before yesterday too, I was able to manage it well. To hold well the pressure, to put one more ball in.
Alexander Zverev was unable to take his opportunities (Gregorio Borgia/AP)
"I think my movements have been improving. Today was a good day in that. I'm super happy because, after the battle of yesterday, to be able to be on court against a player like Sascha – he had a tough match, too, but he's more than 10 years younger than me.
"I remember well the feeling on the body when you are younger comparing to today and the feeling is completely different. And I was able to be fit, to be ready, and that's something that gives me confidence."
Nadal next faces a surprise opponent in seven-foot American Reilly Opelka, who is through to his first Masters semi-final after a 7-5 7-6 (2) victory over Federico Delbonis.
Opelka said: "I'm really pleased. We have been working hard. I made some adjustments on my game. It's nice to see them come into play.
"Rafa's brutal. But it will be a good test. I've never played Rafa."
Alexander Zverev Rafael Nadal
John McEnroe questions Iga Swiatek's emotional state after Australian Open exit
John McEnroe suggested world No 1 Iga Swiatek needs to rediscover a stable mindset to get back to winning ways.
Russian fans display pro-war symbols and banned flags at Australian Open
Russian fans have been questioned by police over flags and symbols displayed at the Australian Open.
John McEnroe on whether Novak Djokovic gets the respect he is due
John McEnroe has given his verdict on the lack of respect being shown to Novak Djokovic.
Novak Djokovic fuelled by 2022 tribulation as title march continues
The Serbian brushed aside Andrey Rublev to set up a clash with Tommy Paul.
Wimbledon 2023: Men's doubles matches to be abbreviated
The change puts Wimbledon in line with the other grand slams.
Australian Open day 10 wrap: Novak Djokovic rolls towards 10th Melbourne title
The story of the 10th day of action from Melbourne Park.
Few sports can match tennis in how great rivals can also have huge respect for one another, like how Novak Djokovic hailed Roger Federer.
Novak Djokovic reaches Australian Open semi-finals after outclassing Andrey Rublev
The nine-time champion Novak Djokovic set up a clash with American Tommy Paul for a place in the final.
Tommy Paul gets better of Ben Shelton to reach Australian Open semi-final
Tommy Paul claimed a 7-6 (6), 6-3, 5-7, 6-4 victory over Ben Shelton on Rod Laver Arena.
Stefanos Tsitsipas hails Mark Philippoussis' influence on Australian Open credentials
There is a growing belief Stefanos Tsitsipas will break his Grand Slam duck in Melbourne this week, with a local favourite in his camp.


text:
using System;
using System.IO;
using BrightData;

namespace BrightWire.ExecutionGraph.GradientDescent
{
    /// <summary>
    /// Adam gradient descent
    /// https://en.wikipedia.org/wiki/Stochastic_gradient_descent#Adam
    /// </summary>
    internal class Adam : RmsProp
    {
        float _decayRate2;
        IFloatMatrix _cache2;

        public Adam(float decay, float decay2, IFloatMatrix cache, IFloatMatrix cache2, IGradientDescentOptimisation updater) : base(decay, cache, updater)
        {
            _decayRate2 = decay2;
            _cache2 = cache2;
        }

        public override void Dispose()
        {
            _cache2.Dispose();
            base.Dispose();
        }

        public override void Update(IFloatMatrix source, IFloatMatrix delta, ILearningContext context)
        {
            var t = context.CurrentEpoch;

            using var deltaSquared = delta.PointwiseMultiply(delta);
            _cache.AddInPlace(delta, _decayRate, 1 - _decayRate);
            _cache2.AddInPlace(deltaSquared, _decayRate2, 1 - _decayRate2);

            using var mb = _cache.Clone();
            using var vb = _cache2.Clone();
            mb.Multiply(1f / (1f - Convert.ToSingle(Math.Pow(_decayRate, t))));
            vb.Multiply(1f / (1f - Convert.ToSingle(Math.Pow(_decayRate2, t))));
            using var vbSqrt = vb.Sqrt();
            using var delta2 = mb.PointwiseDivide(vbSqrt);
            _updater.Update(source, delta2, context);
        }

        public override void ReadFrom(GraphFactory factory, BinaryReader reader)
        {
            base.ReadFrom(factory, reader);

            _decayRate2 = (uint)reader.ReadSingle();
            var rows = (uint)reader.ReadInt32();
            var columns = (uint)reader.ReadInt32();
            _cache2 = factory.LinearAlgebraProvider.CreateZeroMatrix(rows, columns);
        }

        public override void WriteTo(BinaryWriter writer)
        {
            base.WriteTo(writer);
            writer.Write(_decayRate2);
            writer.Write(_cache2.RowCount);
            writer.Write(_cache2.ColumnCount);
        }
    }
}

redpajama_set_name: RedPajamaGithub


Today I am at the Canadian Post Secondary Web Editors Conference at Brock University in St. Catharines.
Great keynote to start the day. The takeaway for me was pay attention to mobile devices.
Current students are a great resource for a social media strategy because prospective students want to hear about the experiences of current students.
Discussion around how York University has been using social media. Good discussion, but too policy heavy.
This session showcased a webcasting tool. 20+ minute videos accompanied by ability to discuss underneath. Good concept but YouTube does it better… for free.
Very interesting session about how Ivey is using social media to target a very specific group of prospecitve students. The takeaway here was to find where your audience is online and interact with them.
Best session of the day for me so far. Stewart Foss from http://www.edustyle.net/ gave a walk through of numerous Canadian sites and did a very entertaining job of it.
Great first day and looking forward to more tomorrow.


What is an advanced health directive?
When does a advanced health directive come into effect?
A direction contained in an advanced health directive only comes into effect when you do not have the capacity to communicate a decision already outlined in an advanced health directive. For example, if you are in an irreversible coma, a directive to donate organs or withdraw life sustaining treatment comes into effect.
Why should I make an advanced health directive?
The purpose of an advanced health directive is to decide in advance about health treatment. If this is an issue of concern to you, you might consider whether you wish to make an advanced health directive.
How do I make an advanced health directive?
Making an advanced health directive is similar to making an enduring power of attorney. The form is a standard document that your solicitor can prepare for you or you can obtain a form from the newsagent or GoPrint. The requirements for completing the form are similar to an enduring power of attorney. In addition, a valid advanced health directive must include a certificate from your doctor that states that, at the time the directive was made, you appeared to have the necessary capacity to make it.


text:
Smart Union
SMART History
We Are SMART
SMART MAP
The Members' Journal
SMART Transportation Newspaper
Products and Svcs
A message from President Ferguson on the INVEST in America Act
Last updated on September 14th, 2020
All Members — SMART Transportation Division
Dear Brothers and Sisters:
Today, House Democrats released the text of a $500 billion five-year funding authorization bill that defines their vision for the future of transportation in America, as well as outlines their plans to refresh and renew the infrastructure of the nation's surface transportation network.
The Investing in a New Vision for the Environment and Surface Transportation in America (INVEST in America) Act is the result of countless hours of work by this Union on the Hill and in the halls of Congress. The INVEST in America Act reauthorizes funding set to expire Sept. 30, but more so, sets standards for safety, training, and transportation reform that have long been sought by the members of SMART Transportation Division including:
Two-Person Crews;
Operator Assault;
Yardmaster Hours of Service
a "Cross Border" fix.
Additionally, Amtrak would see its funding triple to $29 billion over the five-year period of the bill, allowing for expansion of national, state and regional routes and facility modernization. Funding for the Consolidated Rail Infrastructure and Safety Improvements (CRISI) grant program also would be increased to $7 billion to fund passenger and freight rail projects. Provisions for/or against the transportation of liquid natural gas (LNG) via rail tank cars, blocked railroad crossings, and excessive freight train length, among others, also have been included.
Our National Legislative Office has been hard at work in Washington, D.C., to convey our issues to both sides of the aisle in the U.S. House and Senate, and the provisions within this bill are the fruits of that labor.
Undoubtedly, House Democrats have heard our cries and have answered the call. By including our issues within the context of this bill, they have let America know that the only safe operation of a Class I freight train is with a two-person crew; that our bus drivers and operators have the right to a safe work environment; and that the public should be shielded from the risks that rail carriers will take in the name of greed.
But make no mistake, this bill still has a long road to travel and a lot of heavy-handed opposition standing before it in the Republican-controlled Senate. We will need all hands on deck to protect the provisions we have all fought so hard for to survive that journey.
I am asking you to please watch this bill as it moves through the legislative process and see who and what hurdles it faces. I'm asking you to please pay attention to the party affiliations of the individuals as the yeas and nays are registered when the bill is voted upon. And I am asking you to listen to the rhetoric and testimony that will affect its final appearance. Once the dust has settled, I will call on you to please support those who support you and your family's well-being, and I firmly believe that picture will be crystal clear.
There are only two parties at the table. The Democrats wrote it into the bill, only the Republicans will take it out.
Fraternally yours,
Jeremy Ferguson
President — Transportation Division
ABOUT THE INVEST IN AMERICA ACT
Release from Transportation Committee Chair U.S. Rep. Peter DeFazio
Full text of the INVEST in America Act
INVEST in America Act fact sheet
LMRDA Rights
© 2013 Smart Union


text:
Los Ingobernables de Tokusatsu Ep 88 – United as One Pt II: Anniversary to Anniversary Edition
June 29, 2019 July 10, 2019 Hero Club
Welcome to an all-new episode of Hero Club's first original podcast, "Los Ingobernables de Tokusatsu!" Today's episode practically features the "Ninja Talks Live" roster hosting the show since Demonwolf aka T-CosmicLatte is making a guest appearance. Now, the bulk of the episode is Super Sentai! First off, we try out a new segment called "On the Road," where we discuss a few vehicles in tokusatsu. Some versions of the segment maybe one single vehicle, or a few depending on the uses. Then for the rest of the episode we continue on United as One with Gaoranger to Boukenger. However, we attempted to run the gambit all the way to Gokaiger from the start…
Next time on an all-new episode of Los Ingobernables de Tokusatsu, the show's plot is up in the air as we are heading into a small but necessary hiatus.
Richard Monty aka Projekt
Twitter: @Projektrdm
Instagram: @Projekt1
Mucha, Janitor Joker aka The GREEN VAPER
Twitter: @JanitorJoker
Toku Warriors: YouTube & FaceBook
Soundout12
Twitter: @Soundout12
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