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Q: Picasso image loading using own Target very slow when I use my own Target to put the Bitmap loaded by Picasso into my ImageView like this (I do that because I want to compress the bitmap to a byte array to store it in sqlite):
private ImageView imageView;
.
.
.

Picasso.with(getContext()).load(uri).into(target);

private Target target=new Target()
{
    @Override
    public void onBitmapLoaded(Bitmap bitmap, Picasso.LoadedFrom from)
    {
        imageView.setImageDrawable(new BitmapDrawable(getContext().getResources(), bitmap));
    }

    @Override
    public void onBitmapFailed(Drawable errorDrawable)
    {
    }

    @Override
    public void onPrepareLoad(Drawable placeHolderDrawable)
    {
    }
};

I do that in the getView method of my ArrayAdapter, which serves a GridView.
it is very slow (without compressing yet), showing the pics in the final gridview only bit by bit, in contrast to putting it directly into the imageview. why is that? how can I combine fast Picasso loading with retrieving the bytes of every loaded pic?

A: use this code may be its help you
Picasso.with(this).load("http://api.androidhive.info/json/movies/1.jpg").placeholder(getResources().getDrawable(R.drawable.ic_launcher)).into(imageViewCenter);

redpajama_set_name: RedPajamaStackExchange


Cupping is a type of traditional therapy that uses suction to bring blood and lymph that is considered stagnant or not flowing well to the surface. There are many forms of cupping. We use pneumatic pumps. Sometimes the cups are glass and fire is used to cause the vacuum effect. We do not use this type because of the potential for injury to both the patient and provider.
Wet cupping, where there are small incisions placed on the skin to pull blood out is another form of the therapy. We do not find it necessary and there is too much potential for infection and scar tissue development.


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Home » Boston » Mattapan
Court dismisses Mattapan murder conviction; says judge should have followed her initial instinct and tossed the case for lack of evidence before it went to the jury
By adamg on Wed, 10/16/2019 - 1:31pm
The Supreme Judicial Court today overturned Abdullah Yasin's conviction for second-degree murder for Chaz Burton's 2016 death outside a Mattapan party, saying the trial judge in the case should not have allowed the case to go to a jury after she concluded Suffolk County prosecutors had failed to produce evidence that showed Yassin was responsible for Burton's death.
In its decision, the state's highest court said Suffolk Superior Court Judge Janet Sanders managed to make decisions that harmed both Burton's case and the prosecution's case, but that because Burton's constitutional right to due process was harmed, it was ruling for him and dismissing his case.
A Suffolk Superior Court jury had convicted Yasin of second-degree murder for Burton's stomping and shooting murder early on March, 2016 outside 227 River St. A co-defendant was convicted of assault and battery with a dangerous weapon. Both men had been charged with first-degree murder for Burton's death in a fight involving two groups.
After prosecutors concluded making their case to the jury in March, 2018, Yasin's attorney filed a motion asking the judge to issue a "required finding of not guilty." In a hearing on the motion, the SJC writes in its summary of the case, Sanders agreed prosecutors had not proven beyond a reasonable doubt that Yasin caused Burton's death:
[T]he judge observed that murder premised on a theory of extreme atrocity or cruelty was "totally unsupported" by the evidence. In addition, she noted that there was no evidence as to the identity of the shooter or the circumstances under which the shooter had acted. The judge summarized the Commonwealth's case as, "some unknown person came down in the middle of [a] melee and shot [the victim] as he was apparently waving a knife" at others, after having stabbed "at least" two people. Based on the evidence of the victim's conduct, the judge said that "there [was] certainly evidence raised of self-defense, [and] defense of others, enough so that the Commonwealth then [had] to prove beyond a reasonable doubt that the perpetrator was not acting in self-defense" or in defense of others. The judge also said that the Commonwealth had failed to present any such evidence and thus that the Commonwealth had not proved that the killing was unlawful. She noted, as well, that the Commonwealth had presented no evidence that the defendant had aided or abetted the shooter. For all practical purposes, the judge thus deemed the evidence insufficient to convict the defendant of murder.
But despite her misgivings, Sanders then decided to take no action on the case, saying she reserved that right until after the jury made its decision. She similarly "reserved" any action after Yasin's lawyer repeated the request after the close of Yasin's case - in which Yasin was not called, because his attorney felt the case was so weak. But the jury found Yasin guilty of second-degree murder, which carries a sentence of life, with the possibility of parole after at least 15 years.
Yasin's attorney then filed a similar motion to dismiss the case, and this time Sanders agreed, issuing a nunc pro tunc ruling (from the Latin for "now for then," essentially, to overturn an earlier ruling) overturning the murder charge (Yasin was also convicted of assault and battery with a dangerous weapon; he did not appeal his 2 1/2-year sentence for that).
In a hearing on the post-verdict motion, Sanders told both sides that she figured the jury would conclude, as she had, that prosecutors had failed to make their case and that jurors would find Yasin not guilty, because her experience had taught her that "juries are able to recognize when a case is founded on speculation or conjecture, and they will return a of not guilty, thus making it unnecessary for me to intervene." Also, "no rational jury following the instructions I gave them could have concluded that there was an unlawful killing."
She added that had she granted the motion to dismiss before the jury ruled, that would legally let the DA appeal her decision to appellate courts and "then the time and expense invested in selecting a jury and presenting evidence at trial would be wasted."
She then granted Yasin's motion, in part because "the verdict was based on emotion rather than reason and therefore was not consonant with justice."
The DA's office then appealed.
In its ruling, the justices concluded that the way Sanders did not act on Yasin's motion until after the trial was a violation of his rights, but that the judge's decision to essentially overturn the verdict was also wrong:
When the judge reserved decision on the defendant's motion for a directed verdict at the close of the Commonwealth's case, she deprived the defendant of his right to insist that the Commonwealth prove each element of murder beyond a reasonable doubt before he decided whether to rest or to present a defense. Such prejudice is manifest where, as here, the judge indicates at the time of the reservation that she strongly favors allowing the motion. Immediately prior to reserving a decision, the judge observed that the Commonwealth had presented no evidence to identify the killer or the circumstances under which the shooter acted, or to show that the killing was unlawful, given evidence that the shooter may have acted in self-defense or in defense of others. In addition, the judge noted that murder predicated on extreme atrocity or cruelty was "totally unsupported" by the evidence. In effect, the judge told the parties that the Commonwealth had presented insufficient evidence to convict the defendant of murder. After the judge made these statements, however, the trial proceeded, and the defendant was put to the choice of deciding whether to rest or to present a defense.
Writing for the United States Supreme Court in Smith v. Massachusetts, 543 U.S. 462, 471-472 (2005), Justice Antonin Scalia explained that "when, as here, the trial has proceeded to the defendant's presentation of his case, the possibility of prejudice arises." A "false assurance of acquittal on one count may induce the defendant to present defenses to the remaining counts that are inadvisable." Id. at 472. "The seeming dismissal" of an indictment "may induce a defendant to present a defense to the undismissed charge[] when he would be better advised to stand silent." Id. See Cote, 15 Mass. App. Ct. at 240.
The justices added:
Because the defendant's motion under rule 25 (a) [of state trial rules] was filed at the close of the Commonwealth's case, the judge was required to rule on it "at that time."
And that hurt prosecution, because nunc pro tunc rulings are supposed to apply just to such things as clerical errors in the court record, not to fundamental issues of guilt or innocence.
But, given the seriousness of the violation of Yasin's rights, the court added, it was ruling for him and reversing the verdict. Prosecutors, it continued, had no right to appeal the judge's decision, but just in this case, the court said.
Free tagging:
Chaz Burton
Complete ruling 151.03 KB
Was this judge appointed
By StillFromDorchester on Wed, 10/16/2019 - 3:03pm.
Because of political donations she made to someone? Or for her judicial abilities?
I bet I know.
Voting closed 15
Pretty easy to look up if you were really interested
By WalkingTheDog on Wed, 10/16/2019 - 5:55pm.
Stanford U., Harvard Law. Federal clerkships, 6 years civil litigation, 5 years criminal defense. 13 years various judicial appointments.
But don't keep us in suspense! Why do you think she was appointed?
Voting closed 0
By GrampaNurgle on Wed, 10/16/2019 - 6:11pm.
A home in some extremely safe, extremely expensive, extremely white suburb where she doesn't have to face the consequences of letting murderous gangbangers out of jail.
She didn't
She didn't let anyone out of jail. The SJC dismissed the case but said the judge should have dismissed the case due to lack of evidence.
I'm curious. Did you actually read the article before posting your comment?
Let me guess
By SwirlyGrrl on Wed, 10/16/2019 - 7:46pm.
You have never been arrested and tried for something you didn't do.
You also failed civics.
"Innocent until PROVEN guilty" means something. It means that if you can't PROVE guilt, you don't take a case to trial.
You just want kangaroo courts and summary executions for your feels.
Always bleed from a safe distance, where the chances of running into one of those "theoretically innocent" fine fellas are about as high as running into a great white down in Key West.
But please do tell about that cousin's cousin of yours who lived in perfectly safe Mattaroxchester, where every single violent criminal let out by an idiot activist judge immediately turns into a model citizen and opens up a kitten sanctuary.
It can be helpful
By adamg on Wed, 10/16/2019 - 8:30pm.
To have an understanding of the Constitutions rights to due process before one makes a fool of oneself.
The Supreme Judicial Court ruled that it agreed with the judge that the prosecution failed to make its case and that it was not going to let somebody who might be innocent rot in prison for possibly the rest of his life. If you object to the way the judicial system handled the case, please explain why you feel the Fifth and Fifteenth Amendments are no longer in force in Massachusetts. Feel free to explain in a footnote or something why you think assistant district attorneys under Tough Guy Dan Conley were unable to make their case.
For what it's worth, five of the seven justices on the SJC were appointed by the current Republican governor. All seven signed off on today's decision.
http://archive.suffolkdistrictattorney.com/guilty-verdict-in-2016-beatin... - he sure sounds innocent.
As it says in the ruling, no
By *yawn* on Thu, 10/17/2019 - 7:25am.
As it says in the ruling, no one knows who had the gun and thus fired the fatal shot.
The man you think is out on the street is not appealing is 2.5 year sentence for assault and battery.
So... he will in fact spend time in prison.
But sure, we should throw people in prison for life when there is not sufficient evidence to support the charge.
You have supposedly been here for 4 hours and you know exactly ... what, again?
You clearly do not know my absolute redneck bonafides - or anything else about any sort of "safe distance" from anything. Hint: I can probably out-poverty your childhood on just about any measure. I have also seen family hauled off because the cops wanted to "solve" a crime and grabbed someone who was innocent.
Even if you are released, being put through the court mess is at the very least harassment.
Sanders has a reputation in the legal community.
By anon on Wed, 10/16/2019 - 11:19pm.
She's ill tempered, arrogant and not nearly as smart as she likes to think she is. This rebuke from the SJC is an embarrassment and one can only hope she will learn from it. She has an important job and the people who appear in front of her every day deserve reasoned and thorough jurisprudence. Not someone who does exactly what the SCOTUS prohibits.
Character assassination
By perruptor on Thu, 10/17/2019 - 6:51am.
Not one of the better features of the site.
I bet you don't know
By anon on Wed, 10/16/2019 - 5:55pm.
And you're shooting off your mouth with imagined accusations, and not being a good American citizen in that regard.
So who was responsible?
And the NRA, of course, who else?
The DA's office
Didn't make their case, or had the wrong guy. Whichever, it's on them that a murderer is still on the street.
This should be terrifying to anyone who might ever be wrongly
accused, or someone they know wrongly accused. Which is basically everyone.
3 Called Strikes
By Oscar Worthy on Wed, 10/16/2019 - 6:44pm.
Did nothing on 1st motion for directed verdict, did nothing on 2nd motion for directed verdict, did nothing on motion for judgment notwithstanding verdict.
Bat never left her shoulders . .
The jury system can be silly sometimes...
By Pete Nice on Thu, 10/17/2019 - 7:53am.
And in the long run, rulings like this help everyone and make it a fairer system.
Real interesting stuff listening to District Attorneys talk about juries and who they want on them in which cases. I haven't been to a lot of sexual assault or rape trials, but DA's always talk about how they don't want any women on those juries because they put themselves in the vicitm's shoes and try to justify how the assault wasn't an actual assault because of their own personal history.
Anyway, the jury system isn't perfect, and rulings like this make it a little better for everyone.
Wrong name
By tachometer on Thu, 10/17/2019 - 10:38am.
"Suffolk Superior Court Judge Janet Sanders managed to make decisions that harmed both Burton's case and the prosecution's case"
I don't think the victim had a case, should be Yasin instead of Burton twice in that paragraph.
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In other Mattapan news
Person shot in the head on Morton Street in Mattapan
Somebody thought he found a good place to stash a loaded gun on Morton Street but he was wrong
American Legion Highway between Franklin Hill and Canterbury remains a tire-destroying menace
Woman found on fire on Edgewater Drive in Mattapan
Citizen complaint of the day: American Legion Highway has become a tire-eating monster
Who can help tell us where the annoying reverberating nightly noise is coming from, affecting Hyde Park & Milton?
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redpajama_set_name: RedPajamaCommonCrawl


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\begin{document}
\begin{center}
{\Large{\bf Rigidity of action of compact quantum groups III: the general case}}\\
{\large  \bf Debashish Goswami\footnote{Partially supported by Swarnajayanti Fellowship from D.S.T. (Govt. of India)}}\\
Indian Statistical Institute\\
203, B. T. Road, Kolkata 700108\\
Email: goswamid@isical.ac.in\\
\end{center}
\begin{abstract}

If a compact quantum group
 acts faithfully and smoothly (in the sense of \cite{goswami}) on a smooth, compact, oriented, connected Riemannian manifold such that the action induces a natural bimodule morphism on the module of sections of the co-tangent bundle, then it is proved that the
 quantum group is necessarily commutative as a $C^{*}$ algebra i.e. isomorphic with
$ C(G)$ for some compact group $G$. From this, we deduce that the quantum isometry group of such a manifold $M$ coincides with $C(ISO(M))$ where $ISO(M)$ is the group of  (classical) isometries, i.e. there is no genuine quantum isometry of such a manifold.
\end{abstract}
\section{Set-up and statement of the main result}
In this note, we complete the programme taken up in \cite{rigid2} to extend the main result of \cite{rigid2} to  an arbitrary compact connected oriented Riemannian manifold and thereby proving Conjecture II of that paper which states that there is no genuine quantum isometry of such a manifold. We give only sketches of arguments here, and plan to give full details in an article (or a couple of them) combining \cite{rigid2} with the present one.  We will  freely use all the notations and terminologies of \cite{rigid2}.  In particular, we will use the Frechet algebras, (Hilbert) modules over them and topological tensor product (denoted by $\hat{\otimes}$) between such algebras or modules, discussed in details in \cite{rigid2} and references therein.

Let $M$ be a smooth, compact, connected, oriented manifold of dimension $n$ and assume that a compact quantum group (CQG) ${\cal Q}$ has a faithful, smooth (in the sense of \cite{rigid2}) action $\alpha: C(M) \rightarrow C(M) \hat{\otimes} {\cal Q}$ which satisfies condition (1) of Theorem 3.4 of \cite{rigid2}.
Our goal is to prove the following:
\begin{thm}
\label{maingeneral}
${\cal Q}$ is commutative as $C^*$ algebra, i.e. of the form $C(G)$ for some compact group $G$.
\end{thm}
From this, we get the following
\begin{crlre}
The quantum isomery group of any compact conencted oriented Riamannian manifold $M$ coincides with $C(ISO(M))$ where $ISO(M)$ denotes the group of Riemannian isometries. In other words, there is no genuine quantum isometry of such manifolds.
\end{crlre}
Let us explain our strategy.
  Using the techniques of \cite{rigid2}, we can equip $M$ with a Riemannian structure $\langle\cdot, \cdot \rangle$ for which $\alpha$ is isometric. Then, we emulate the classical technique of lifting an isometric group action to a smooth action on the total space of  the orthonormal frame bundle of $M$, which is always a parallelizable manifold (see, e.g. \cite{bsp_crt}. We verify that the lifted action satisfies the hypothesis of Theorem 6.6 of \cite{rigid2}, hence    the conclusion of Theorem \ref{maingeneral} follows.
  \section{Sketch of proof}
  We recall from \cite{rigid2} the equivariant unitary representation $d \alpha$ on the Hilbert module of smooth forms. For $x \in M$ let us denote by ${\cal Q}_x$ the unital $C^*$-subalgebra of ${\cal Q}$ generated by elements of the forms $ \alpha(f)(x), (\phi \otimes {\rm id})(\alpha(g))(x)$, where $f,g \in C^\infty(M)$ and $\phi$ is any smooth vector field.  By assumption, $\alpha(f)(x)$ commute among themselves  for different $f$'s and also commute with $(\phi \otimes {\rm id})(\alpha(g))(x)$'s. We now claim that actually $(\phi \otimes {\rm id})(\alpha(g))(x)$'s commute among themselves too, for different choices of $\phi$ and $g$. In other words:
  \begin{lmma}
  \label{commute}
   ${\cal Q}_x$ is commutative.
   \end{lmma}
{\it Proof:}\\
The statement of the lemma is clearly equivalent to proving $\tilde{d}\alpha(f)$ and $\tilde{d}\alpha(g)$ commute for $f,g \in C^\infty(M)$, where $\tilde{d}=d \hat{\otimes} {\rm id}$ as in \cite{rigid2}. For $x \in M$, choose smooth one-forms $\{ \omega_1,\ldots, \omega_n  \}$ such that they form a basis of $T^*M$ at $x$. Let $F_i(x), G_i(x),i=1,\ldots, n$ be elements of ${\cal Q}$ (actually in ${\cal Q}_x$) such that  $\tilde{d} \alpha(f)(x)=\sum_i \omega_i F_i(x), $ $\tilde{d}\alpha(g)=\sum_i \omega_i G_i(x)$. Now, in the notation of Lemma 4.5 of \cite{rigid2}, $d\alpha_{(2)} $ leaves invariant the submodules of symmetric and antisymmetric tensor product of $\Lambda^1(M)$, thus in particular, $C^s_{ij}=C^s_{ji}, $ $C^a_{ij}=-C^a_{ji}$ for all $i,j$, where $C^s_{ij}$ and $C^a_{ij}$  denote the ${\cal Q}$-valued coefficient of $w_i \otimes w_j$ in the expression of $d\alpha_{(2)}(df \otimes dg+dg \otimes df)$ and $d\alpha_{(2)}(df \otimes dg - dg \otimes df)$ respectively. By a simple calculation using these relations, we get the commutativity of $F_i(x), G_j(x)$ for all $i,j$.
 \mbox{}\hfill $\sqare$\vspace{1ex}\vspace{2mm}\\
Let $O_M$ be the bundle of orthonormal frames of $M$. It is convenient to identify it as a subbundle of $E={\rm Hom}(M \times \mathbb{R}^n, \Lambda_{\mathbb{R}}^1(M))$ (where $\Lambda^1_{\mathbb{R}}(M)$ denotes the real vector bundle of smooth one-forms on $M$) with fibres at $x$ isomorphic with linear maps from $\mathbb{R}^n$ to $T^*_xM$. Then, $O_M$ can be viewed as the subset of $E$ consisting of those linear maps $e$ which are orthonormal w.r.t. the canonical Euclidean inner product on $\mathbb{R}^n$ and the Riemannian inner product on $T^*_xM$ where $x=\pi(e)$, $\pi: O_M \rightarrow M$ be the projection map from the total space onto the base space $M$.  Let $\{ b_1,\ldots, b_n\}$ be the canonical basis of $\mathbb{R}^n$. We will also use them as the canonical basis of ${I\! \! \!\! C}^n$. Then any element $e \in O_M$ is determined by the orthonormal basis $e(b_1),\ldots,e(b_n)$ of $T^*_{\pi(e)}M$. Moreover, we can define local coordinates of $O_M$ as follows. Consider any open subset $U$ of $M$ such that $\Lambda^1(U)$ is trivial, diffeomorphic to $U \times \mathbb{R}^n$. Choose $\omega_1,\ldots, \omega_n$ in $\Lambda^1(M)$ such that for each $x \in U$, $\{ \omega_1(x),\ldots, \omega_n(x)\}$ is an orthonormal basis w.r.t. the Riemannian inner product $\langle \cdot, \cdot \rangle_x$. Define  $t_{ij}(e):=<\omega_j(x), e(b_i)>_x$, $i,j=1,\ldots, n$, $e \in \pi^{-1}(U) \cong U \times O(n)$.  It is clear that $(( t_{ij}(e) ))$ is in $O(n)$ for all $e \in \pi^{-1}(U)$. Thus, choosing any $n(2n-1)$'s of $t_{ij}$'s, along with a choice of $n$ local coordinates for $U \subseteq M$, we get a choice of local coordinates for $O_M$. Let us call $t_{ij}(e)$ the `matrix coefficients' of $e$ corresponding to the basis $\{ \omega_1,\ldots, \omega_n\}$.

Let ${\cal B}=C(O_M)$, ${\cal B}^\infty=C^\infty(O_M)$,  ${\cal B}_U:=C_c(\pi^{-1}(U)) \subset C(O_M)$, ${\cal C}_U^\infty=C_c^\infty(\pi^{-1}(U))$, where $C_c(V) $ (respectively $C_c^\infty(V)$) denotes continuous (respectively smooth) compactly supported functions on a manifold $V$. Consider the $\ast$-algebra  ${\cal P}_U$ consisting of functions of the form below:
$$ \sum_{l=1}^N \hat{f_l} P_l(t_{ij};i,j=1,\ldots n),$$ where $N \geq 1$, $\hat{f_l}\equiv f_l \circ \pi \in {\cal B}^\infty_U$ for some $f_l \in C_c^\infty(U)$, and $P_l$ denotes a polynomial with complex coefficients of $n^2$ variables. It is easy to see that ${\cal P}_U$ is dense in ${\cal B}_U$ and ${\cal B}^\infty_U$ in the norm and Frechet topologies respectively.  We shall also use the notation $\hat{F}$ for $F \in C(M) \hat{\otimes} {\cal Q} $ to denote the element $F \circ \pi$ of $C(O_M) \hat{\otimes }{\cal Q}$.

We want to construct a smooth action $\eta$ of ${\cal Q}$ on $C^\infty(O_M)$.
 For this, we need the concept of symmetric algebra ${\cal S}({\cal E})$ of a module ${\cal E}$ over a unital commutative algebra say ${\cal M}$. This is by the definition the algebraic direct sum of symmetric tensor products (over ${\cal M}$) of copies of ${\cal E}$, i.e.${\cal S}({\cal E}):=\bigoplus_{n \geq 0} {\cal E}^{(n)}_{\rm sym}, $ where ${\cal E}^{(0)}_{\rm sym}:={\cal M}$ and for $n \geq 1$, ${\cal E}^{(n)}_{\rm sym}$ is the symmetric tensor product of $n$ copies (which is possible to define as ${\cal M}$ is commutative). This has a natural algebra structure coming from  tensor multiplication. Moreover, it has the following well-known universal property:
 \begin{ppsn}
  Given any  unital commutative algebra ${\cal F}$ with an algebra inlusion $i$ (unital), which naturally gives ${\cal F}$ an ${\cal M}\equiv i({\cal M})$-module structure, and an ${\cal M}$-module map $\theta: {\cal E} \rightarrow {\cal F}$, there is a unique lift $\hat{\theta}: {\cal S}({\cal E})\rightarrow {\cal F}$ which is an algebra homomorphism.
  \end{ppsn}
Now, recall the vector bundle $E={\rm Hom}(M \times {\mathbb{R}}^n, \Lambda_{\mathbb{R}}^1(M))$ and consider its complexification $E^{{I\! \! \!\! C}}={\rm Hom }(M \times {{I\! \! \!\! C}}^n, \Lambda^1(M))$, where $\Lambda^1(M)$ stands for the complexification of $\Lambda^1_{\mathbb{R}}(M)$. Denote by ${\cal E}$ the $C^\infty(M)$-module of smooth sections of $E_{{I\! \! \!\! C}}$, viewed as a Hilbert Frechet module as in \cite{rigid2}. There is a natural $C^\infty(M)$-valued inner product $\langle \langle \cdot, \cdot \rangle \rangle^{\cal E}$ on ${\cal E}$ using matrix traces in fibres: $\langle \langle \xi, \eta \rangle  \rangle(x):={\rm Tr}(\xi^*(x)\eta(x)).$ Define $\theta: {\cal E} \rightarrow C^\infty(O_M)$ by $$ \theta(\xi)(e):={\rm Tr}( e^*, \xi (\pi(e)))=\langle \langle \zeta, \xi \rangle \rangle^{\cal E} (\pi(e)),$$ for any section $\zeta \in {\cal E}$ such that $\zeta(\pi(e))=e$. It is easy to check that it is a module map where we view $C^\infty(M)$ as a subalgbera of $C^\infty(O_M)$ in the obvious was, by sending $f$ to $\hat{f}$. So, we get algebra homomorphism $\hat{\theta}$ from ${\cal S}({\cal E})$ to $C^\infty(O_M)$, and it is easy to verify that it is in fact a $\ast$-homomorphism and also its range is dense in $C^\infty(O_M)$ in the Frechet topology.

 On the other hand, identifying the Frechet module ${\cal E} \hat{\otimes} {\cal Q}$ with $${\rm Hom}_{C^\infty(M)}
\left( C^\infty(M) \otimes {I\! \! \!\! C}^n, \Lambda^1(M) \hat{\otimes} {\cal Q} \right),$$  we can define an equivariant representation $\beta: {\cal E} \rightarrow {\cal E} \hat{\otimes} {\cal Q}$ by $$ \beta(\xi)(v):=d \alpha (\xi(v)),$$ for $\xi \in {\cal E},$ $v \in C^\infty(M) \otimes {I\! \! \!\! C}^n$, say $v=(v_1,\ldots, v_n),$ $v_i \in C^\infty(M)$. Now, define $$ \Gamma(\xi)(e):=\langle \langle  \zeta \otimes 1_{\cal Q}, \beta(\xi) \rangle \rangle^{{\cal E} \hat{\otimes} {\cal Q}} (\pi(e)),$$ where $e \in O_M,$ $\xi \in {\cal E}$, $\zeta \in {\cal E}$ such that $\zeta(\pi(e))=e$ and $\langle \langle \cdot, \cdot, \rangle \rangle^{{\cal E} \hat{\otimes} {\cal Q}}$ denotes the natural $C^\infty(M) \hat{\otimes} {\cal Q}$ valued inner product. In fact, one can easily verify that $\Gamma(\xi) \in {\cal A} \subset C^\infty(O_M) \hat{\otimes} {\cal Q}$,  and consider the inclusion $i : C^\infty(M) \rightarrow {\cal A}$ given by $i(f)=\alpha(f)$.  We verify that $\Gamma$ is a module map, hence lifts to a homomorphism $\hat{\Gamma} : {\cal S}({\cal E}) \rightarrow {\cal A}$.

 As in \cite{rigid2}, we  get a Frechet-dense submodule ${\cal E}_0$ on which the representation $\beta$ is algebraic, so that in particular $\beta({\cal E}_0) \subseteq {\cal E}_0 \otimes {\cal Q}_0$, where ${\cal Q}_0$ is the canonical Hopf $\ast$ algebra of ${\cal Q}$ and $\otimes$ (as in \cite{rigid2}) denotes the algebraic tensor product, It is easily seen that $\hat{\theta}({\cal E}_0)$ is dense in $C^\infty(O_M)$. From the discussion of subsection 2.4 on Hilbert modules over topological algebras,as well as using the commutativity of the algebra ${\cal Q}_x$ for all $x$, we can conclude that the equivariant representation $\beta$  lifts to an equivariant representation say $\beta^{(n)}_{\rm sym}$ of ${\cal Q}$ on each ${\cal E}^{(n)}_{\rm sym}$, hence we get  $\hat{\beta} : {\cal S}({\cal E}_0) \rightarrow {\cal S}({\cal E}_0) \otimes {\cal Q}_0$,  which is a representation and also  an algebra homomorphism. We also have  on ${\cal S}({\cal E}_0)$: \begin{equation} \label{22} (\hat{\theta} \otimes {\rm id}) \circ \hat{\beta}=\hat{\Gamma}. \end{equation}

We claim that there is a well-defined continuous  map which sends $\hat{\theta}(s)$ to $\hat{\Gamma}(s)$ for $s \in {\cal E}({\cal S})$, which will define the desired action of ${\cal Q}$.
To this end, consider the $\ast$-subalgebra ${\cal A}$ of $C(O_M) \hat{\otimes} {\cal Q} \equiv C(O_M, {\cal Q})$ consisting of those functions $F$ for which $F(e) \in {\cal Q}_{\pi(e)}$ for all $e \in O_M$. By Lemma \ref{commute}, this is a commutative algebra. Fix the open set $U$ as before, with $\omega_1,\ldots,\omega_n$ which give a local section for  orthonormal frame bundle on $U$, and  define $T_{ij} \in C^\infty(O_M) \hat{\otimes} {\cal Q}$  by $$T_{ij}(e):=\langle \langle d\alpha(\omega_j), e(b_i) \otimes 1_{\cal Q} \rangle \rangle(\pi(e)), $$ where $\langle \langle\cdot, \cdot \rangle \rangle$ denotes the $C^\infty(M) \hat{\otimes} {\cal Q}$-valued inner product of $\Lambda^1(M) \hat{\otimes} {\cal Q}$.
 If  $\omega^\prime_1,\ldots, \omega^\prime_n$ in $\Lambda^1(M)$ are chosen such that $\{ \omega^\prime_1(\pi(e)), \ldots, \omega^\prime_n(\pi(e))\}$ is an orthonormal basis for $T^*_{\pi(e)}M$, and $d\alpha(\omega_j)(\pi(e))=\sum_k \omega_k^\prime(\pi(e)) F_{jk}(\pi(e)), $ where $F_{jk}(\pi(e)) \in {\cal Q}_{\pi(e)}$, we have $$ T_{ij}(e)=\sum_k t^\prime_{ik}(e) F_{jk}(\pi(e)),$$ where $t^\prime_{ik}(e)$ are the matrix coefficients of $e$ corresponding to $\{ \omega^\prime_1,\ldots, \omega^\prime_n\}$. It follows that $T_{ij}$'s belong to ${\cal A}$. Moreover, we have
 \begin{lmma}
 \label{orthogonality}
 For any smooth real-valued function $\chi$ supported in $U$, we have $$\widehat{\alpha(\chi)} \sum_k T_{ik}T_{jk}=
 \widehat{\alpha(\chi)} \sum_k T_{ki}T_{kj}=\widehat{\alpha(\chi)}\delta_{ij},$$ for all $i,j=1,\ldots,n$, where $\delta_{ij}$ are the Kronecker's delta.
\end{lmma}
{\it Proof:}\\ By a simple calculation using the definition of $F_{mn}'s$ as well as the fact that $d\alpha$ is inner-product preserving,  we obtain  \begin{eqnarray*} \lefteqn{ \widehat{\alpha(\chi)}^2 \sum_k T_{ki}T_{kj}}\\
&=& \langle \langle d\alpha(\chi \omega^\prime_i), d\alpha(\chi \omega^\prime_j)\rangle \rangle =\delta_{ij}\widehat{\alpha(\chi)}^2.\end{eqnarray*}
Now, as the both sides of the above equality belong to the commutative $C^*$ algebra $\overline{{\cal A}}$, we get the same equality replacing $\widehat{\alpha(\chi)}^2$ by $\widehat{\alpha(\chi)}$, and moreover, the other equality in the statement of the lemma. \mbox{}\hfill $\sqare$\vspace{1ex}\\

Recall the algebra ${\cal P}_U$. Denote by ${\cal E}_U$ the smooth sections of $E_{I\! \! \!\! C}$ supported in $U$ and let ${\cal S}^0_U$ be the subslgabra of ${\cal S}({\cal E}_U) \subseteq {\cal S}({\cal E})$ consisting of elements of the form $\sum_{l=1}^N \hat{f_l} P_l(\xi_{ij};i,j=1,\ldots n),$  where $P_l$, $f_l$ as in the definition of ${\cal P}_U$, and $\xi_{ij}$ denotes the section $|\omega_j><b_i|$, i.e. $\xi_{ij}(x)$ sends $b \in {I\! \! \!\! C}^n$ to $<b_i,b>\omega_j$.  Clearly, $\hat{\theta}({\cal S}^0_U)={\cal P}_U$.  It is easily seen that
$$\hat{\theta}(\sum_{l=1}^N \hat{f_l} P_l(\xi_{ij};i,j=1,\ldots n)):=
\sum_{l=1}^N \widehat{\alpha(f_l)} P_l(t_{ij};i,j=1,\ldots n),$$
$$\hat{\Gamma}(\sum_{l=1}^N \hat{f_l} P_l(\xi_{ij};i,j=1,\ldots n)):=
\sum_{l=1}^N \widehat{\alpha(f_l)} P_l(T_{ij};i,j=1,\ldots n).$$
\begin{lmma}
\label{normest}
 $\| \hat{\Gamma}(g)\| \leq \| \hat{\theta}(g) \|$ for all $g \in {\cal S}^0_U$.
\end{lmma}
Indeed, as $T_{ij}$'s are self-adjoint elements of the commutative $C^*$ algebra $\overline{{\cal A}}$,
  Note the natural identification of the $C^*$-algebra ${\cal B}_U$ with $C(O(n), C_c(U)) \subset C(O(n), C(M))$, so that we have $\| \psi \|=\sup_{v \in O(n)} \| \psi^v\|$, for $\psi \in C(O(n) \times M) \equiv C(O(n), C(M)), $ where $\psi^v(x)=\psi(v,x)$ for $v \in O(n))$. A similar fact holds if we replace $C(M)$ by $C(M) \hat{\otimes} {\cal Q}$. We also denote by $\alpha^\sharp$ the $C^*$-homomorphism sending $\Psi \in C(O(n) \times M)$ to $\alpha^\sharp(\psi)$ in $C(O(n),C(M)\hat{\otimes} {\cal Q})$ given by, $\alpha^\sharp(\psi)(v)=\alpha(\psi^v)$, $v \in O(n)$. Clearly, $\| \alpha^\sharp(\psi)\| \leq \| \psi \|$.

  Now, consider $g\in {\cal P}_U$ of the form $\sum_{l=1}^N \hat{f_l} P_l(t_{ij};i,j=1,\ldots n)$, and assume that $K$ is a compact subset of $U$ in which supports of all $f_l$'s are contained. Let $\chi \in C^\infty(U)$ be such that $0 \leq \chi \leq 1$ and  $\chi|_K \equiv 1$. Fix $e \in O_M$ and let $\theta$ be a character (multiplicative functional) of the commutative $C^*$ algebra ${\cal Q}_{\pi(e)}$. By Lemma \ref{orthogonality}, we see that if $\theta(\widehat{\alpha(\chi)}(e))$ is nonzero, then $\left( \tau_{ij}:=\theta(T_{ij}(e)) \right)$ is an orthogonal matrix, i.e. in $O(n)$. On the other hand, we have (using the fact that $f_l=\chi f_l$ for each $l$): \begin{eqnarray*} \lefteqn{\theta(\eta_U(g)(e))}\\
  &=& \theta(\widehat{\alpha(\chi)}(e)) \theta(\eta_U(g)(e))\\
  &=&
  \sum_{l=1}^N \theta(\alpha(f_l)(\pi(e))) P_l(\tau_{ij};i,j=1,\ldots n)\\
  &=& \theta\left( \sum_{l=1}^N \alpha(f_l)(\pi(e)) P_l(\tau_{ij};i,j=1,\ldots n) \right).\end{eqnarray*} Thus, if $\theta(\widehat{\alpha(\chi)}(e))$ is zero, $\theta(\eta_U(g)(e))=0 $, and when it is  nonzero, we obtain $|\theta(\eta_U(g)(e))| \leq \sup_{v=((v_{ij}))  \in O(n)}  \| \left( \sum_{l=1}^N \alpha(f_l)(\pi(e)) P_l(v_{ij};i,j=1,\ldots n)\right)\|$, which is equal to $\|\alpha^\sharp(g) \| \leq \| g\|$.  \mbox{}\hfill $\sqare$\vspace{1ex}\\

  This allows us to get a $C^*$-homomorphism $\eta_U$ on ${\cal B}_U$ satisfying $\eta_U(\hat{\theta}(g))=\hat{\Gamma}(g)$. We can now `patch' $\eta_U$'s by $C^\infty$-partition of unity to get  $C^*$ homomorphism $\eta$ from $C(O_M)$ to $C(O_M) \hat{\otimes} {\cal Q}$ satisfying $\eta (\hat{\theta}(g))=\hat{\Gamma}(g)$ for all $g \in {\cal S}({\cal E})$. Indeed, as  $T_{ij}$ are in $C^\infty(O_M) \hat{\otimes}{\cal Q}$, we do get $\eta(C^\infty(O_M)) \subseteq C^\infty(O_M) \hat{\otimes} {\cal Q}$, and also that $\eta$ is Frechet continuous.
Moreover,  we have $$ (\hat{\theta} \otimes {\rm id}) \circ \hat{\beta}=\hat{\Gamma} =\eta \circ \hat{\theta}$$ on ${\cal S}({\cal E}_0)$. From this, and the Frechet density of $\hat{\Gamma}({\cal S}({\cal E}_0) \otimes {\cal Q}_0)$ in $C^\infty(O_M) \hat{\otimes} {\cal Q}$, we get the Frechet density of $\eta({\cal S}({\cal E}_0))(1 \otimes {\cal Q}_0)$ in $C^\infty(O_M) \hat{\otimes} {\cal Q}$.
  It follows that  $\eta$ is  a smooth faithful  action of ${\cal Q}$ in the sense of \cite{rigid2}.

   Finally, the condition (1) of Theorem 3.4 of \cite{rigid2} follows by verifying it on ${\cal P}_U$ for every $U$, with the help of the commutativity of ${\cal A}$ and the expression of $\eta_U$ in terms of $F_{ij}$'s and the coordinate functions $t^\prime_{ij}$'s. This completes the proof.

redpajama_set_name: RedPajamaArXiv


text:
Home News 1960's female rockers, Ace Of Cups, will release debut studio album on...
1960's female rockers, Ace Of Cups, will release debut studio album on 11.9
Ace Of Cups, the beloved all-female rock group from the 1960s San Francisco psychedelic scene, will release their self-titled debut studio album on November 9th via High Moon Records. Produced by Dan Shea (Jennifer Lopez, Mariah Carey, Santana), the record's twenty-one tracks span fifty years of brilliant songwriting.
From 1967 to 1972, Ace Of Cups – Mary Gannon (Bass), Marla Hunt (Organ, Piano), Denise Kaufman (Guitar, Harmonica), Mary Ellen Simpson (Lead Guitar), and Diane Vitalich (Drums) – were at the epicenter of the '60s cultural and social revolution. From the Acid Tests to the protests, from the free concerts in Golden Gate Park to the ballrooms of San Francisco, they shared stages with everyone from The Band to the Grateful Dead. Michael Bloomfield, Jerry Garcia, and Buddy Miles were their fans and the Ace were chosen to open for Jimi Hendrix the week after his groundbreaking performance at The Monterey Pop Festival. When asked by Melody Maker magazine to name his favorite musical discoveries on his U.S. trip, the Ace Of Cups was the first thing he mentioned.
"I heard some groovy sounds last time in the States, like this girl group, Ace Of Cups, who write their own songs and the lead guitarist is hell, really great." -Jimi Hendrix, 1967
Despite eliciting some music industry interest for their exceptional songs, sublime harmonies and exuberant live performances, the Ace Of Cups never got the chance to make a record of their own.
As the decades passed, the band members pursued other personal and creative endeavors, playing music both individually and collectively when opportunities arose. While performing at Wavy Gravy's 75th birthday party and SEVA Foundation benefit, the Ace met with High Moon Records' founder George Baer Wallace, who was there to talk to the band about releasing archival concert recordings. After hearing them play live, Wallace was so moved by their spirit and spark that he hatched a new plan on the spot, offering the Ace Of Cups the opportunity to record their first ever studio album.
With the enthusiastic support of their new record label, and the guidance of celebrated producer Dan Shea, four of the band's original members began exploring their back catalog and writing new material. Right from the start the Ace understood the art of the song, and the band has kept that spirit true and close to everything they do. Ace Of Cups blends pure rock, folk, blues and gospel influences with a pop sensibility and a garage band rush; all tinted by an intoxicating psychedelic sheen. As the news spread that the Cups were recording, old friends began to come by the studio to offer support and musical contributions. In total 36 songs were recorded during those sessions, and what started out as a chance to set the record straight turned into a history-making second act.
Album guests include legendary players – and long-time friends of the band – Jorma Kaukonen, Jack Casady, David Grisman, Pete Sears (Jefferson Airplane/Starship, Moonalice), Steve Kimock (Zero/RatDog), and Charlie Musselwhite, as well as lead vocal turns by Bob Weir, Taj Mahal, Buffy Sainte-Marie and Peter Coyote. Even with this illustrious line-up, this record belongs wholeheartedly to the Ace Of Cups, who can convey the joy of humanity's greatest art form as simply and as eloquently as anyone ever has:
"Music – everything will be alright . . ." – Ace of Cups
This is the first release from the Ace of Cups project. Volume Two – coming in 2019 – includes 16 more songs from the band, as well as contributions from Jackson Browne, Wavy Gravy, Sheila E and the Escovedo family, Bakithi Kumalo (Paul Simon), and more.
Visit http://www.aceofcups.com for more information about the band and to pre-order Ace of Cups on CD/LP/Digital.
Listen to the first single "Feel Good": https://www.billboard.com/articles/columns/rock/8469266/the-ace-of-cups
Learn more about the Ace Of Cups in this award-winning KQED piece, which has garnered 5 million views on social media: http://bit.ly/TheAceOfCups
Ace Of Cups Tracklist:
Introduction: There's a Record Being Made
Fantasy 1&4
We Can't Go Back Again
The Well (feat. Bob Weir)
Taste of One
Mama's Love
Interlude: Transistor
Interlude: Baby from the Forest of Knolls
Life in Your Hands (feat. Taj Mahal)
Macushla/Thelina
As the Rain (feat. Peter Coyote)
Daydreamin' (feat. Taj Mahal)
Pepper in the Pot (feat. Buffy Sainte-Marie)
Interlude: Breath
Grandma's Hands
Medley (The Hermit / The Flame Still Burns / Gold & Green / Living In The Country)
Outroduction: It's Always Safe…
High Moon Records is a boutique record label with a catalog including Love, Gene Clark, Terry Dolan, Marvin Gardens, and Ryan Martin. Upcoming releases include Sopwith Camel, The Matrix and Avalon Live archival series, and Lotti Golden's cult-classic Motor-Cycle. All releases feature deluxe packaging, extensive liner notes by iconic writers, deluxe packaging and artwork, never-before-seen photos, and ultra-rare bonus tracks.
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Xavier Haller has defended his thesis on 23rd October 2015 in Montpellier.
The irradiated uranium dioxide (UO2), which is the nuclear fuel of pressurized water reactors, contains two populations of cavities saturated by fission gaz : i. intergranular cavities almost lenticular in shape whose size ranges between few tens to several hundred nanometers, ii. intragranular cavities, almost spherical in shape whose size is of the order of the nanometer. Recent studies have shown the existence of a surface effect at the scale of nanometric cavities, which influences the effective elastic behavior of the nuclear fuel. In this work, an analytical micromechanical model, which is able to take into account this heterogeneous microstructure and the surface effect at the nanometric scale, is proposed to describe the macroscopic behavior of the irradiated UO2. The approach is based on a multiscale modeling and homogenization techniques in mechanics of materials. The irradiated UO2 is described as a porous media, which contains pressurized spherical nanocavities (intragranular cavities) and randomly oriented pressurized spheroidal cavities (intergranular cavities). The surface effect is taken into account with imperfect coherent interfaces between the matrix and the cavities. A novel model based on the morphologically representative pattern approach has been developed to describe the effective elastic behavior of this heterogeneous medium. The proposed model relies on assumptions whose relevance is evaluated with finite element simulations which require a specific formulation to take into account the imperfect coherent interfaces.


text:
Q: Showing boundedness of $f(x,y)$ How can I show that the set $k=\{(x,y)\in \mathbb{R}^2:x^2+y^2+e^{xy}=5\}$ is bounded? I've simply got no idea how to go on with $2$-dimensional functions..

A: $e^{xy} > 0$ for any real $x$, $y$. Therefore $x^2+y^2<5$, which means that all points $x,y$ lie within a circle of radius $\sqrt{5}$.

A: Note that $|x|>3$ implies $x^2+y^2+e^{xy}> 9+0+0>5$, for example.


text:
« Born Noble
Restless »
December 9, 2019 by d2k
We associate relaxation, and vacations, and the beach with being in the south. It seems as if things are easier there. And our dream images can reflect this, too. Here we have an image of heading into the cold, up a mountain, where it's harsher. Yet it's also a place where fewer people have made the effort to journey. This is the story of a spiritual life, where we must go, inside, to a place where few people have consciously chosen to aspire, but where the experience is magnificent. (At the end of this post there are instructions and a link to download this recording to your computer.)
https://dream2know.files.wordpress.com/2019/12/northward.mp3
Jeane: In the next dream, I seem to be joining a group of people who are going on some kind of a trek and this involves going up into the mountain, and there's snow and ice, and two or three of us finally make it up to like this ice cave.
There's some people that have gone in ahead of us, and there's the guide, and when we make it up kind of crawl up into this ice cave. And you almost feel like there's no going back; you know, we're going to have to continue the trek. But there may be people that don't make it that far, too.
I feel like I'm excited about going on this trek, but that's really all I remember of the dream. It's almost like there's no going back at one point. And I'm glad we've made it to the takeoff point where we are.
John: It's an interesting way of trying to catch up with a quality of a vibration. In other words, it's kind of like kind of like being introduced to a type of letting go, and, at the same time going in the direction of a kind of spiritual inner North inside of yourself, where it's cold and the conditions are harsh.
And yet there is something there, even though it may seem more barren. Barren on one level, but relieving on another. So what you're doing is you're trying to find what that feels like. In other words, you're trying to come close to that quality, to soak up that quality, in kind of an inner vibration.
In the South, things are comfortable, and your senses are quickened to all kinds of things. When you go to the North, there's actually more going on, in one regard, especially if you looked at it in nature. But it's at a depth within.
This is hard to describe, other than to maybe compare the traits of North and South, as they are in nature. In a southern slope, for example, which gets the sun, that condition is never as fertile, it's a little more barren; always is a little bit more barren. It's almost as if something isn't quite protected, or fully properly embraced, or something. It's like something is a little too harsh, transcendent, or something.
And that the importance of the evening, to the South, is where you get your dew, and something really sweet that can descend. Harder, of course, when the conditions of something of a fire energy so prevalently prevail.
Now, the experience, again in nature, of the North side is quite different and profound, in that the trees grow much, much better there. They're more protected. They that's where the water source is hidden, it's latent. Everyone likes the feel of the wide open spaces of the South. Yet, in the North, it's a greater closeness.
Those who are in the business, we'll say, of husbandry, which is growing trees, timber business, know the value of northern slopes: that's where the trees are the best. They also know, that from the perspective of man selecting things for himself, the South is what is chosen. You have your better views, seemingly, there. But the nurturing, the uplifting, comes from the North.
And so even in nature you have this whole imagery of going to the spiritual North. North is going to the spiritual side, and South is going more out into the outer of things. And so when you're going into the ice caves, and you're trekking, it's an interesting compacted effect, in which the quality of the sun, and its overall transcendence, it's not an overriding condition. And it's really neat, too, in that there's such a stillness in it. Even in the full daytime of things you can find a stillness, in nature, on a North slope – much more so than on a South slope.
Somehow or another, what you're doing is you're seemingly needing to bring to your attention something really, really deep last night that was operating on a level that was not apparent, in terms of dreams, that was kind of a training, or a teaching, a kind of adab kind of teaching where one is meant to come closer to the subtler levels, as if there is a system, so to speak, on an inner level, to the letting go.
And so you are being shown this quality, only you're being shown it in terms of ice, a way of going and looking at that. In other words, there's an impingement in there that is for you to see. It's the going to kind of a North side of yourself, but still doing it with an aspect of putting yourself into things. And yet, somehow you're meant to see the difference. You have to see it that way.
In your case, how do you experience the essence of the ice cave? You're kind of doing an interesting tandem: the ice cave and the spiritual North simultaneously. It's a tiny, subtle alignment that's required.
To download this file, Right Click (for PCs) or Control Click (for Macs) and Save: Northward
Posted in Audio Posts, Fundamentals, Inner / Outer | Tagged heading north in a dream, ice cave in a dream, north and south in dreams, on a trek in a dream, spiritual north | Leave a Comment


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Silver & Red Forest Premium Holiday Appreciation Card
25 @ $62.50 ($2.50 each) 30 @ $75.00 ($2.50 each) 35 @ $87.50 ($2.50 each) 40 @ $100.00 ($2.50 each) 45 @ $112.50 ($2.50 each) 50 @ $125.00 ($2.50 each) 55 @ $137.50 ($2.50 each) 60 @ $150.00 ($2.50 each) 65 @ $162.50 ($2.50 each) 70 @ $175.00 ($2.50 each) 75 @ $187.50 ($2.50 each) 80 @ $200.00 ($2.50 each) 85 @ $212.50 ($2.50 each) 90 @ $225.00 ($2.50 each) 95 @ $237.50 ($2.50 each) 100 @ $239.00 ($2.39 each) 105 @ $250.95 ($2.39 each) 110 @ $262.90 ($2.39 each) 115 @ $274.85 ($2.39 each) 120 @ $286.80 ($2.39 each) 125 @ $298.75 ($2.39 each) 130 @ $310.70 ($2.39 each) 135 @ $322.65 ($2.39 each) 140 @ $334.60 ($2.39 each) 145 @ $346.55 ($2.39 each) 150 @ $358.50 ($2.39 each) 155 @ $370.45 ($2.39 each) 160 @ $382.40 ($2.39 each) 165 @ $394.35 ($2.39 each) 170 @ $406.30 ($2.39 each) 175 @ $418.25 ($2.39 each) 180 @ $430.20 ($2.39 each) 185 @ $442.15 ($2.39 each) 190 @ $454.10 ($2.39 each) 195 @ $466.05 ($2.39 each) 200 @ $478.00 ($2.39 each) 205 @ $489.95 ($2.39 each) 210 @ $501.90 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Wishing you joy at the holidays and a new year filled with promise and possibility.
Wishing you a wonderful holiday break and thanks for your hard work and dedication.
May your holidays be filled with joy and peace.
Merry Christmas and Happy New Year! Looking forward to another great year of working together.
Wishing you a holiday season and new year filled with hope, health and happiness.
Warm wishes to you and yours this holiday season and all year long.
Best wishes for all the season's gifts and a wonderful new year.
As the year comes to an end, it's a great time to tell you how much we appreciate you. Thank you for all that you do.
We hope you know how much you are appreciated.
Sending warm wishes for happiness at the holidays.
Thank you for placing your trust in us another year. Happy Holidays and best wishes for a bright and happy new year!
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Peel & Stick Envelopes White, Silver Foil Lining
Bring a peaceful winter scene to their mailbox during the busy holiday season. Red berries and a pair of cardinals add festive contrast to the silver trees, and selective emboss adds dimension to the red ribbon design. Warm appreciation on the cover and foil that continues to the inside left all combine for a highly memorable, highly displayable holiday card. Features double-thick cover. Printed on premium paper with a superfine texture. Made in USA.
Standard, Double-Thick Cover (6.93 x 4.81)
Foil, Premium Paper
Designed Message, Logo, Personal Message, Recipient Addressing, Sender Addressing
1-Day: 1/31/2023


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DC Restaurant Closings
J. Paul's Plans to Close, Marking the End of an Era in Georgetown
The tavern classed up the bar scene when it opened in 1983
by Tierney Plumb Nov 19, 2018, 4:50pm EST
Share All sharing options for: J. Paul's Plans to Close, Marking the End of an Era in Georgetown
J. Paul's will serve its last burger before the year is up.
J. Paul's/Facebook
In the latest sign that Georgetown's dining scene is entering a new phase, J. Paul's will close its doors before the end of the year.
The saloon debuted at 3218 M Street NW in 1983, kicking off a mini restaurant empire in Georgetown for Capital Restaurant Concepts. The group owned Paolo's, which just flipped into High Street Cafe, and Old Glory, which reinvented itself as America Eats Tavern this summer.
"There is a life span to independent restaurant groups — they are graciously retiring after a long successful career. It's a changing-of-the-guard moment," Georgetown restaurant broker Bill Miller, principal at Miller Walker, tells Eater.
He adds that, with J. Paul's, Capital Restaurant Concepts can be credited with changing the "strictly bar culture" of Georgetown into more of a "cafe culture" and "a more polished bar and grill" scene.
The wood-lined restaurant features 5,000 square feet across two levels, and Miller says the space is currently entertaining potential restaurants to take its place. Eater reached out to the building's owner Richard Levy, a prolific Georgetown landlord, but did not immediately hear back.
"While I'm sad to see eras end, Georgetown is due for another chapter," Miller says.
A manager at J. Paul's declined to discuss the closure, but says it'll stay open until at least past the end of the month.
The restaurant announced the news about its final few weeks on Twitter on Sunday, November 18:
We will truly miss you all! So please come by and say your goodbyes! @ J.Paul's Georgetown https://t.co/NsTAc63Bc6
— J.PAUL'S (@jpauls_gtown) November 19, 2018
High Street Cafe Gives Georgetown a Brasserie With a Hispanic Twist [EDC]
J. Paul's
3218 M St NW, Washington, D.C. 20007 (202) 333-3450 Visit Website
All of 2018's D.C. Restaurant Closures
Penn Quarter's Upscale 701 Restaurant Will Close Next Week
CF Folks Shuts Down Following Alleged Bathroom Battle With Landlord [Updated]
15 Standout Thai Restaurants Around D.C. Where to eat pad see ew, larb, and more


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Q: API redirect with ProxyPassMatch I'm trying to redirect OpenTSDB API with apache2. Here is my configuration(Apache doc)
ProxyPassMatch /api/* http://127.0.0.1:4343/api/$1

The server has apache2 and TSDB. I'm able to access <ip>:4343/api/version which is showing version info.
But If I try to access <ip>/api/version, showing Error Code:404 "Endpoint not found". So, what's happening here?

A: The problem is the regular expression; the apache is forwarding the incorrect API URL to TSDB.
After playing with that apache config, I'm able to redirect the request with below config.
ProxyPassMatch "/api/*" "http://127.0.0.1:4343"


We represent all those who consider themselves 'connoisseurs of life' and represent the properties they so desire. In today's market, local, national and international exposure is now paramount: it is what attracts the most qualified buyers, and the highest price for your home.
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Our listings get maximum exposure to a global audience of international buyers and real estate investors. We offer a sophisticated, immersive approach to marketing your home, offering the highest calibre of service and discretion, regardless of your home's size, neighbourhood or price point.


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All profits over January go to Red Cross Bushfire Relief
A London Love Story
Ms. Kensingtons Musings
We like to take it slowly
I had to be as hands on as possible in the creation of the label – teaching myself fashion illustration to design each garment and surface design to create the 'London' print. This took a lot of time, but I feel very strongly that a luxury brand has to be unique and special – something rare that you can't find anywhere else.
Lingerie that sparks the imagination
Having spent years working in advertising agencies in London and Sydney as an art director. I wanted the brand to spark the imagination and for every garment to tell a story that would inspire the wearer. For modern romantics who love to travel - each collection has its own story revealing the inspiration behind it and the moment it is seeking to capture. A Londoner, born and bred – I was inspired by the London IT girls – Kate, Naomi, Alexa, Rita, Cara etc and wanted to capture their rebellious, sexy and daring attitude in the debut 'London' Collection.
We want women to look in the mirror and think YES!
Over a year ago, I was coming to the end of parental leave with my second child Zoe (now two). Through many conversations with friends in their 30s it struck me just how many of us found it hard to focus on ourselves because of our busy demanding lifestyles. Specifically, in the areas of intimacy, sexuality and relationships.
© 2020 mskensington.
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Simple Technique, Stunning Results: Messing Around in the Studio [feedly]
Simple Technique, Stunning Results: Messing Around in the Studio
// 500px ISO
"The entire shoot took around 2 hours," photographer Justin Lim told me when I emailed him about the photo above. "But this was actually just at the end—the last 10 mins or so—when we were messing around afterwards!"
The portrait above, titled "I bet that you look good on the dancefloor", is one of my favorite simple shots on 500px. It was obvious to me that it was a quick and fun shot to capture, but something about it makes it stick.
So I asked Justin for the details, and between the exif data and Justin's revelations, we can break down this shot:
On the technical side: the image was captured using a Nikon D610 and 50mm f/1.4G lens. The settings were 1/320s shutter, f/1.8, and ISO 1600. Continuous light was provided by a Manfrotto Spectra900F LED panel, which was set up behind the model (Justin's friend Gabi). Beyond that there was no ambient light or other source… the studio was completely dark.
"We didn't rig any background," explained Justin. "[The shot] was just achieved by Gabi sitting on the floor of my studio, The Kandid, and throwing her hair back and forth."
"I guess you could say it's just one of those moments where you get a bit lucky," he continued. "Some of the shots looked like the hair was a bit limp, but in this one it comes alive!"
As for post-processing, that was pretty straight forward as well. Since Gabi has, in Justin's words, "flawless skin," no touchups were needed there. It was all basic adjustments in Lightroom:
[I added] some sharpness and made some levels adjustments […] The original shot was underexposed then lifted in Lightroom. I wouldn't say this was a particularly complex edit, maybe 10 mins in post at most! A little bit of grain was also introduced, I seem to recall.
That's it. A simple shot captured by simply "messing around" in the studio. Never forget to have fun with your shoots… some of your favorite work might come out of it:
To see a lot more from Justin, follow him on 500px, visit his website, or show him some love on Facebook and Instagram. And be sure to keep an eye out for his selections in the Editors' Choice section, he's a 500px Guest Editor this month!
Shared via my feedly reader
Robert Gallagher
Www.studiol7.com


The Eastern Pike Regional Police Department will be providing free medical IDs to senior citizens from 9 a.m.-2 p.m. Oct. 8 at the Matamoras Borough Hall, 10 Ave. I, Matamoras, Pa.
Medical IDs are custom-made cards that outline a person's medical conditions, medications and emergency contacts. Carrying this card can help ensure that patients receive appropriate treatment and that the right people are notified in an emergent situation.
"Emergency responders look for a medical ID," said Chad Stewart, regional chief of police. "We want to provide these to prevent emergencies from escalating or any medical error; many symptoms can be misdiagnosed without a medical ID."
Healthcare companies estimate that the average American age 65 or older takes about 15 prescribed medications, and many suffer from chronic health conditions.
The Milford Lions Club needs help to continue an annual holiday tradition.
Since 1986, the Milford Lions Club has put lights in the trees along Broad and Harford Streets for the holiday season.
"For the last several years, we have asked the community to assist with this project, with only limited success," said Bill Harkness, club secretary. "However, due to declining membership and the reduced success of our fundraising efforts, without significant contributions of both physical assistance and financial contributions from the community, we will not be able to continue this very popular project this year."
If there are any organizations or individuals in the community that would be willing to partner with the Lions Club, they can contact Harkness to discuss the details by calling 570-296-934 or emailing him at bghark@ptd.net.
Belle Reve Senior Living will host "Medicare & You," a seminar by Elizabeth Snyder of Main Street Financial Partners, at 1:30 p.m. Oct. 4.
Snyder will provide information about co-pays, accessing services, and what benefits are available.
This event is free and open to the public, but seating is limited. RSVP by Oct. 2 to 570-409-9191.
Belle Reve Senior Living is at 404 E. Harford St., Milford, Pa.
The March of Dimes will hold its annual March for Babies Oct. 6 at Firefly Field in Matamoras, Pa.
March for Babies is the largest annual fundraising event for the March of Dimes, which will celebrate its 75th anniversary in 2013.
More than 4 million babies are born in the United States each year. The March of Dimes supports research, education, vaccines and breakthroughs that help babies to be born healthy.
Registration for the walk will begin at 10 a.m. The walk will start at 11 a.m.
The finish line will feature lunch, music, activities for kids, goodie bags and more. To register for this event, visit marchforbabies.org.
Volunteers are needed for the walk. To volunteer or for more information, call 570-829-1019.


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Non-View of the Lake from Underground | Torreya Cummings | April 2019
Non-View of the Lake from Underground
by Torreya Cummings
5 April – 4 May 2019
In 1873, Collis P. Huntington, partner in the First Transcontinental Railroad, commissioned a painting of Donner Lake. He wanted to commemorate the tracks crossing the Donner summit in 1867, a feat of both engineering and sheer human misery. The resulting work by Albert Bierstadt succeeds in romanticising Manifest Destiny and the Sierra Nevada but nonetheless failed to please Huntington, who hoped for a more detailed view that highlighted the railroad itself.
Aggregate Space Gallery and Torreya Cummings present Non-View of the Lake from Underground, a collection of landscape leftovers, geologic scraps, photographs, partial environments, and a performance set. These works elevate immersion and complication over a totalizing "view", using Bierstadt's View of Donner Lake as a departure point, and creating a new moment that foregrounds the foreground.
5 April, 6-10pm
Second Saturday Artist Talk
13 April, 11am
First Friday Event
3 May, 5-8pm
Thursdays, Fridays, and Saturdays 1-5pm
Entrance on West Street
Aggregate Space is a warehouse and has a similar temperature to the outside; please dress accordingly.
Aggregate Space Gallery programs are generously supported by: National Endowment for the Arts, Andy Warhol Foundation for the Visual Arts, Neda Nobari Foundation, Creative Work Fund, Kenneth Rainin Foundation, Zellerbach Family Foundation, Clorox Foundation, Awesome Foundation, and ArtNow International.
Image courtesy of Torreya Cummings
An Incredible 2018!


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Enjoy Korea's diverse summer festivals
By Hallie Bradley
There's just something about the summer that screams "take me outside and find a festival."
Luckily, there are plenty of festivals around the Korean Peninsula, from beaches to music, during the hot humid summer months to keep those summer loving vibes alive and well and those weekends full of fun.
Here are six you should get on the calendar now and start planning for ASAP.
Boryeong Mud Festival
One of the most popular festivals of summer, no list would be complete without at least a mention.
Located on Daecheon Beach in Boryeong, Chungcheongnam-do, this festival is all about having fun while getting dirty.
Well, it's not so much dirt as it is nutrient-rich mud known for its health benefits for the skin.
Play games, soak in a giant mud bath or get a mud massage after running through the mud maze this July 13 to 22.
Pohang International Fireworks Festival
Held on Yeongildae Beach in Pohang, Korea, the city is known as the "City of Light and Fire" and during this fest it truly lives up to that name.
From a fireworks competition to street performances and a parade of lights, there is plenty of light and fire to enjoy at this beach fest.
Head to Pohang from July 25 to 29 to partake in the numerous hands on activities and experience programs during the day to await the spectacular show at night.
Incheon Pentaport Rock Festival
Pentaport has been a staple on the music festival scene since 2006.
Rock and electronic music fans can be assured of fun in the sun while listening to the likes of Nine Inch Nails, Daybreak, Hoobastank, Starsailor, Walk The Moon, Mike Shinoda of Linkin Park, The Koxx and plenty more.
Bands will take the stage from Aug. 10 to 12 out in Pentaport Park in Songdo, Incheon. Don't miss the fun.
Busan Sea Festival
Taking place from Aug. 1 to 5 this year, this festival encompasses events at Haeundae Beach, Gwangalli Beach, Songdo Beach, Dadaepo Beach and Songjeong Beach in Busan.
From cultural events to hands-on experiences, there are pool parties, beach parties, fireworks shows and a ton more to keep any visitor actively involved while digging their feet into the sand.
Busan International Rock Festival
If you're down south, don't miss out on the Busan International Rock Festival. This fest was first held in 1999 and each year thousands of fans head to Samnak Ecology Park in Sasang-gu, Busan, to dance away the weekend.
This fest is just one part of the Busan Sea Festival which rocks 30 festivals around the city at once! See Busan's own homegrown musicians as well as rock and metal acts from across the country and abroad from Aug. 10 to 12.


Two years ago I authored a blogpost, Everyone's Worst Nightmare , about a family's experience with genetic variant interpretation, communication (or lack of) by healthcare providers with families, and an outcome that couldn't be more tragic – the death of a child. Here I am providing an update on the legal status of the lawsuit brought on behalf of the child by his mother. In my original posting, I did not identify the child's syndrome or the specific court case. However, since then, the story has been picked up by a variety of media outlets (Turna Ray at GenomeWeb has done the best reporting on the details) and it now being a matter of public record, I have included some particulars here .
Briefly, the story began about 12 years ago when the child was experiencing multiple, intractable seizures and had a clinical picture consistent with Dravet syndrome. Unbeknownst to the parents, genetic testing was ordered and the child was found to carry a mutation in SCN1A, the gene linked to Dravet syndrome. The mutation was interpreted as a variant of uncertain significance, though at the time there was reason to believe that it could be pathogenic. Based on the genetic test result, it was felt that the child did not have Dravet syndrome and was kept on a sodium channel blocker, which unfortunately is contra-indicated for patients with this syndrome. Not long afterwards the child died of seizures at the age of two.
The patient did not find out that genetic testing had been performed until about 7 years after the test was ordered. Shortly after the mother learned of the test result and inquired into its meaning, the lab reclassified the variant as pathogenic.
In February 2016 the mother initiated a lawsuit on behalf of her deceased child in the fifth judicial circuit court in Richland County, South Carolina. The defendant's lawyers requested that the case be dismissed on the grounds of restrictions imposed by the state's statute of repose, i.e., a law that states legal action must be initiated within a certain period from the time the alleged offense occurred (it is similar to but slightly different from a statute of limitations). The defendants presented the argument that a genetic testing laboratory is a licensed health care provider and South Carolina has a 3 year statute of repose for lawsuits brought against licensed health care providers. Since the events took place a decade ago, the defendants asserted that the case should be dismissed. The plaintiff countered that, under South Carolina state law, a genetic testing lab that is separate from a hospital or a clinic cannot be considered a licensed health care provider and therefore the statute of repose did not apply. The plaintiff contended that this is a case of ordinary negligence, not medical malpractice, since the lab should not be considered a licensed health care provider and therefore the suit should be allowed to proceed.
The South Carolina Supreme Court heard the case on February 14th, 2018. For those of you who have never witnessed a state Supreme Court hearing, I recommend that you watch the ~40 minute video of the session. The court's decision, which will be about whether the suit can proceed rather than determining liability, will depend on how it interprets whether a laboratory is a "similar category" to the health professionals listed in the state code. I thought that the five justices were insightful and asked thoughtful questions. As a side note, at about the 29-30 minute mark of the hearing, Justice Few gives a shout out to our genetics colleagues at the Greenwood Genetics Center.
The court does not have a set date on when they will issue an opinion; as the Supremes, they call that shot (the South Carolina Supreme Court's motto is Nil ultra, which roughly translates as "Nothing is above us"). Typically, though, the time frame on a ruling is in weeks or months. If the court decides that the lab is not a healthcare provider, then the plaintiff's suit will be allowed to continue, though I got the sense that the court felt that even if the ruling were in favor of the defendants that the plaintiff may still have alternate legal pathways to pursue a case. I will keep the good readers of The DNA Exchange posted on important developments in this case, which I suspect will continue to drag on for some time after the Supreme Court's ruling.
With the rapid expansion of genetic testing in the clinical and consumer spaces, and the growing involvement of non-genetics professionals in ordering genetic testing, bad clinical outcome scenarios are likely to become more common. Critical questions about variant interpretation and legal liability aside, from a genetic counseling standpoint, this case highlights the importance of clear and ongoing communication with patients and their families about the limits and clinical interpretation of genetic testing. This can be extraordinarily difficult when a family is trying to cope with caring for a child with a life-threatening disease, but genetic counselors are trained to work precisely in those situations. Genetic testing may be simple to order and widely available but it benefits no one without good clinical care and counseling.
Would knowing about the presence of this mutation modify the treatment given by clinicians?
Marc – a pathogenic SCN1A mutation is diagnostic for Dravet syndrome. The seizure med the child was on would have been discontinued immediately because the med is specifically contraindicated in this syndrome and a different med would have been chosen. Instead the med was continued since it was assumed the child did not have Dravet syndrome based on the genetic test result.


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SuperSummary, a modern alternative to SparkNotes and CliffsNotes, offers high-quality study guides that feature detailed chapter summaries and analysis of major themes, characters, quotes, and essay topics. This one-page guide includes a plot summary and brief analysis of 61 Hours by Lee Child.
The fourteenth book in the Jack Reacher series, British author Lee Child's crime thriller novel, 61 Hours (2010), concerns a snowy small town in South Dakota that is host to a large prison and a dangerous criminal element. In a positive review in The New York Times, Janet Maslin writes, "What heats 61 Hours to the boiling point is Mr. Child's decision to defy his own conventions."
Born in 1960, Jack Reacher is a retired military vet who drifts from town to town investigating suspicious activity. Six-foot-five, he is described as having "hands the size of supermarket chickens." Outside the town of Bolton, South Dakota, Jack rides on a bus carrying a senior citizen tour, "like a hitchhiker, but not quite." In avoiding a crash with a reckless motorist, the bus spins off the icy road into a snowbank. Jack helps save and protect the elderly victims while they wait in the cold for local authorities to arrive. With the wind chill below 50 degrees Fahrenheit, the famously light-packing Jack thinks maybe he should have brought warmer clothes.
A few hours later, Jack arrives in the nearby town of Bolton, South Dakota. Snowy and remote, Bolton is home to one of the largest correctional facilities in the country. As a result, the town's industry—along with its law enforcement officers—are beholden to the demands of the prison and its administrators. All of the motels are full because friends and families travel there to visit loved ones in the prison. Jack also learns that the leader of a biker gang has been arrested on drug charges. As the gang members settle ominously on the outskirts of town, their leader awaits trial. The prosecution's key witness is Janet Salter. Fearing that the gang plans to murder Janet, Jack decides to help the local police in their efforts to protect her until the trial. A bookish old biddy, Janet is startlingly direct with Jack, criticizing his extremely ascetic lifestyle.
Jack learns that the mastermind behind the gang and its drug activities is a Mexican drug lord and criminal mastermind named Plato. Plato sends a merciless, nameless assassin to Bolton, instructing him to murder Janet and anyone who else who knows anything or gets in his way. Feeling a bit in over his head, Jack contacts Major Susan Turner, the military police officer who took over command of his old unit, the 110th Special Investigations Unit. In a meeting in Virginia, Susan tells Jack there is a suspicious abandoned military facility in Bolton that was supposedly converted into an orphanage.
Doubtful that the facility was ever really an orphanage, Jack returns to Bolton and searches every square inch of the compound with the help of the local police. While there, he discovers a group of pseudo-hippies squatting in the compound. On the verge of giving up, Jack stumbles upon a secret passageway to a vast underground section of the facility. Though designed as a shelter for children, it is now used by the U.S. Department of Defense to store massive amounts of methamphetamine. While the use of methamphetamines by Nazi Germany and even Adolf Hitler himself is well-documented, both the British and American armies also administered vast quantities of amphetamines, such as Benzedrine, to their soldiers. According to a PBS documentary on the subject, U.S. General and future president Dwight D. Eisenhower ordered over a half-million amphetamine tablets to North Africa for use by Allied troops.
Jack infers that Plato plans to steal the military cache of methamphetamine. To do so, he needs an inside man; Jack quickly discovers that Major Holland, the local police chief, is Plato's accomplice. After killing Holland, Jack poses as the dead man in a phone call with Plato, luring him to Bolton. When Plato arrives in the underground facility for the meeting, Jack kills him. To avenge their boss's death, Plato's men flood the underground facility with airplane fuel while Jack is still inside. While escaping, Jack also discovers a cache of diamonds that the Department of Defense hid in the structure.
Worth Dying For, a novel about Jack's exploits in rural Nebraska, follows 61 Hours in the Jack Reacher series.
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Five Discovery Gateways named among country's best green spaces
Five Valleys Regional Park Discovery Gateways have achieved Green Flag Award status – the international mark of a quality park or green space.
Flags will be flying at Bryngarw Country Park, Cwmcarn Forest, Cyfarthfa Park, Parc Bryn Bach and Ynysangharad Park in recognition of their excellent visitor facilities, high environmental standards, and commitment to delivering great quality green space.
More than just parks, the Discovery Gateways are 'stepping stones' into the wider landscape of the Valleys, offering visitors the chance to learn about the rich natural and cultural heritage, to experience the iconic landscape with its seasonal changes and to reconnect with the nature on our doorsteps, with all of the benefit that brings.
Phil Lewis, who is leading on the Valleys Regional Park, said:
"It's fantastic to see our Discovery Gateways being recognised for excellence once again in this year's Green Flag Awards. We have been shown more clearly than ever before the vital importance of such places for our health and well-being and we're incredibly lucky in the Valleys to have these amazing natural landscapes on our doorsteps, and the outstanding Gateways that help connect us to them. The award once again illustrates the hard work of everyone involved and the dedication to delivering such important and high quality visitor experiences. Well done to all."
This year, 224 parks and green spaces across the country have received the prestigious Green Flag Award and Green Flag Community Award – from country parks and formal gardens, to allotments, woodlands and churchyards.
The Green Flag Award programme is delivered in Wales by environmental charity Keep Wales Tidy, with support from Welsh Government. Independent green space experts volunteered their time in early autumn to judge applicant sites against eight strict criteria, including biodiversity, cleanliness, environmental management, and community involvement.
Lucy Prisk, Green Flag Coordinator at Keep Wales Tidy said:
"The pandemic has shown just how important high-quality parks and green spaces are to our communities. For many of us, they have been a haven on our doorstep, benefitting our health and well-being.
"The five flags flying in the Valleys Regional Park this year are a testament to the hard work of staff and volunteers who have maintained excellent standards under the most challenging circumstances. I'd like to congratulate and thank them all for their outstanding commitment."
A full list of award winners can be found on the Keep Wales Tidy website
www.keepwalestidy.cymru
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VRP is bringing together a network of twelve urban and country parks and nature reserves — also known as 'Discovery Gateways' — as part of a project to establish a permanent regional park for the South Wales Valleys. Regional parks can be found both in the UK and internationally, providing recreational and wellbeing opportunities for […]
We're so proud of how much our Valleys Regional Park Guardians Scheme — delivered by Groundwork Wales — has achieved since May 2021. From environmental training and workshops to land improvement across our Discovery Gateways and working in partnership with the Big Bocs Bwyd programme — the impact on the community and local environment has […]
New Education and Wellbeing Centre — now open at Bryngarw Country Park
On Thursday 02 December, a brand-new education and wellbeing centre was officially unveiled by Deputy Minister for Arts and Sport, Dawn Bowden MS at Bryngarw Country Park.
Guardians Contract Awarded to Groundwork Wales
Valleys Regional Park are pleased to have contracted Groundwork Wales to continue delivering the VRP Guardians Scheme.


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The 29-year-old actor, who played Justin Russo on the Disney Channel's 'Wizards of Waverly Place', announced that he and his wife, Maria Cahill, will welcome their first child.
Congratulations are in order for former Disney Channel star David Henrie. The Wizards of Waverly Place alum announced this week that and he his wife, Maria Cahill, are expecting their first child.
David and Maria have been married since April 2017 and shared a video of what looked to be an adorably exciting gender reveal gathering.
In the video, David popped a balloon to reveal that they were having a girl and joyous celebrations rang out in the room.
Judging by the caption he wrote on Instagram, David is VERY excited to be a first time dad. "IT'S A GIRL IT'S A GIRL IT'S A GIRL. I'm a Poppa," he wrote. David also joked that he couldn't wait to sit his daughter down and tell her "a 9 season story of How I Met Your Mother".
David and Maria were immediately showered with well wishes in response to the video. His Waverly Place TV mom, Maria Canals Barrera, congratulated him, writing, "CONGRATULATIONS!!! What a blessed baby girl she is to have you both as parents!"
Some fans referenced his Waverly Place character being all grown up, which he responded to with a sense of humour.
David Henrie, played Justin Russo on Wizards of Waverly Place, which ran as a Disney Channel series from 2007 to 2012.
The 29-year-old actor and dircetor seems fully ready for fatherhood and has already grabbed the necessary dad accessories needed for the job.


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2015 • Comedy/Crime/Drama • Rated R • 105 Minutes
In Theaters February 27, 2015
On Blu-ray June 2, 2015
On DVD June 2, 2015
Robert Taylor
Stephanie Honore
Brennan Brown
Trailer + 1 More
Will Smith stars as Nicky, a seasoned master of misdirection who becomes romantically involved with novice con artist Jess (Margot Robbie). As he's teaching her the tricks of the trade, she gets too close for comfort and he abruptly breaks it off. Three years later, the former flame-now an accomplished femme fatale-shows up in Buenos Aires in the middle of the high stakes racecar circuit. In the midst of Nicky's latest, very dangerous scheme, she throws his plans for a loop.... and the consummate con man off his game.
Rated R for language, some sexual content and brief violence.
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Lincoln Park's Adopt a Park Program.
The Lincoln Park Neighborhood Watch Program (NWP) is proud to support the city of Lincoln Park's Adopt a Park Program.
We have chosen to adopt Jaycee Park, located at the corner of Harrison and Ferris. This is a small park, but has a playscape for younger children, a shelter, and a barbecue grill, as well as a basketball court.
As a part of the program, we will be responsible for visiting the park regularly to do minor clean up, such as picking up trash, pulling large weeds. Any major issues noted, such as graffiti, damaged equipment, should be reported to the program supervisor. In addition, we are required to hold at least one large public event at the park during the summer season.
We are reaching out to our members and asking you to help us support this effort to make our city a cleaner, safer place to live, work, and play.
Our first cleanup effort will be held on Saturday May 16th from 10:00 am until about 12:00 noon. We expect to be picking up trash, pulling weeds, and possibly painting the playscape, so wear work clothes, and bring your own gloves. Other supplies will be provided. Specific dates and times for subsequent organized group cleanups are undetermined at this time; we will notify you as soon as we have that information. In the meantime, please feel free to stop by Jaycee Park, or any of the parks in our city, whenever you have a chance.
For more information or to assist, please contact Mary Ellen Mitchell (Volunteer Coordinator) mare48146@yahoo.com or Michael Parr (Program Coordinator) gmparr714@gmail.com or by calling 313-381-3217.