Our team is always happy to serve the residents of Lebanon with our quick garage door repairs. The town is located in Boone County and approximately 29 miles away from Indianapolis making it a suburb of the area. The city is famous for the Historic Cragun House that was built in 1893.
We want to give you a warm welcome into town if you're new to the area. You may want to know what fun things you can do, if so we have you covered.
For more things to do, go to the Lebanon City Page. Another great resource for information about the city is to visit the Chamber of Commerce.


text:
#if !defined(FUSION_REPLACE_08182005_0830)
#define FUSION_REPLACE_08182005_0830

#include <boost/fusion/view/transform_view/transform_view.hpp>
#include <boost/fusion/algorithm/transformation/detail/replace.hpp>
#include <boost/fusion/support/is_sequence.hpp>
#include <boost/utility/enable_if.hpp>

namespace pdalboost {} namespace boost = pdalboost; namespace pdalboost { namespace fusion
{
    namespace result_of
    {
        template <typename Sequence, typename T>
        struct replace
        {
            typedef transform_view<Sequence, detail::replacer<T> > type;
        };
    }

    template <typename Sequence, typename T>
    inline
    typename
        enable_if<
            traits::is_sequence<Sequence>
          , typename result_of::replace<Sequence const, T>::type
        >::type
    replace(Sequence const& seq, T const& old_value, T const& new_value)
    {
        typedef typename result_of::replace<Sequence const, T>::type result;
        detail::replacer<T> f(old_value, new_value);
        return result(seq, f);
    }
}}

#endif

redpajama_set_name: RedPajamaGithub


Snoring is a noise produced when someone breathes during sleep which in turn causes vibration of the soft palate and uvula (that thing that dangles down the back of the throat).
All snorers have an incomplete obstruction (a block) of the upper airway. Many regular snorers have complete episodes of upper airway obstruction where the airway is completely blocked for a short time – usually a few seconds. This silence is then usually followed by snorts and gasps as they fight to take a breath. When someone snores so loudly that it disturbs others, obstructive sleep apneoa is commonly present.
Primary Snoring, also known as simple snoring, or snoring without sleep apnoea, is characterised by loud upper airway breathing sounds during sleep. Usually in these cases the complaint of snoring is by the partner or an observer rather than the patient and there is usually no evidence of insomnia or excessive sleepiness due to the snoring. In the case of simple, or primary snoring, simple devices and sprays can be tried as advertised here.
Obstructive Sleep Apneoa Syndrome is characterized by repeated episodes of upper airway obstruction during sleep, often associated with a reduction in blood oxygen saturation. The upper airway can be obstructed by blocked nasal (nose) passages, large tonsils or a large tongue and usually includes the airway muscles relaxing and collapsing when asleep. Sometimes the structure of the jaw can be a factor.
The symptoms include excessive daytime sleepiness and reports of obstructed breathing during sleep. Other symptoms suggestive of sleep apnoea include very loud snoring, a dry mouth upon waking, morning headaches, daytime tiredness, unrefreshing sleep, irritability, change in personality, high blood pressure, being overweight, difficulty concentrating, snorting, gasping or choking in the night and restless sleep. If there is any doubt, you should be assessed by your doctor or a specialist such as Mr Frosh for sleep apnoea, because sleep apnoea is a potentially life-threatening condition that requires urgent medical attention.
Undiagnosed and untreated obstructive sleep apneoa can lead to include heart attacks, strokes, impotence, irregular heartbeat, high blood pressure and heart disease. Obstructive sleep apnea also causes daytime sleepiness that can result in accidents and problems at work and in relationships. The severity of the symptoms may be mild, moderate or severe.


text:
McSweeney's Collection
The San Francisco based not-for-profit publishing house, McSweeney's Publishing, was established by Dave Eggers in 1998. It began as a literary journal, Timothy McSweeney's Quarterly Concern, of which Eggers was the editor, publishing quality writing that was rejected by other magazines. Today the company not only produces the Quarterly but also operates a daily humour website, publishes The Believer, a bi-monthly arts and culture magazine, and publishes an increasing number of books under various imprints. It has received dozens of awards for its contributions to literature and publishing, and was recently named America's seventh most innovative media company for proving the value of print publishing through its distinctive writing and design.
Dave Eggers was born in Boston, Massachusetts, in 1970. His parents, attorney John K. Eggers and school teacher Heidi McSweeney Eggers, sent him to the public high school in Lakes Forest, where he befriended classmate and actor, Vince Vaughn. He developed an interest in journalism, enrolling in the course at the University of Illinois, but his studies were cut short by the tragic death of both of his parents from cancer in 1991 and 1992. Eggers was just 21 years of age at the time; his youngest brother, Christopher, just 8. With his two eldest siblings committed to full-time work and law studies, Eggers took on the responsibility of raising Christopher. He left University, moved to Berkeley, California, began freelance design work for a local newspaper and, with money left to him by his parents, sent his brother to a small private school. Eggers' first book, A heartbreaking work of staggering genius, released in 2000, recounted his personal struggle to raise Christopher after their parents' death. It met with widespread acclaim for its originality, quickly becoming a bestseller as well as a finalist for the Pulitzer Prize.
In the late 1990s, Eggers, together with David Moodie and Marny Requa, founded Might magazine. Might evolved out of the small San Francisco-based independent paper Cups, its humorous articles and essays on issues and personalities of the time earning it a loyal following. The magazine failed to make a profit, however, and by 1997 had ceased publication altogether. It was around this time that Eggers founded McSweeney's Publishing, producing in 1998 the first issue of Timothy McSweeney's Quarterly Concern, to some extent a successor to Might, albeit with a more literary focus. The journal, noted for its inconsistency in format and publishing style, was named after the mysterious "Timothy McSweeney" who had sent unusual letters and notes, including odd pamphlets, train schedules and drawings, to Eggers and his mother. Intrigued by the letters as a child, and curious to know whether their author was related, Eggers kept the letters, eventually naming his journal after the man who had become somewhat of an enigma. As chance would have it, an intern by the name of Ross McSweeney came to work for the publishing company in 2000. He was in fact the nephew of the real Timothy McSweeney, who had been a talented art teacher at Rutgers University and who, overcome by mental illness, had sent the letters to the Eggers family whilst in the care of a mental health institution. With the McSweeney family's permission, Eggers kept the name for the journal, and all issues since 2000 have been implicitly dedicated to the real Timothy. The full "Timothy McSweeney" story can be read here.
Renowned for its short fictional stories, the Quarterly, has helped to launch the careers of numerous aspiring authors. It has also published works by famous novelists such as Denis Johnson, Joyce Carol Oates, Michael Chabon, T.C. Boyle, Steven Millhauser and Stephen King. It has been described as the first bona fide literary movement in decades and the flagship literary quarterly, and in 2007 McSweeney's received the National Magazine Award for three stories published in 2006 – T.C. Boyle's Wild Child, Susan Steinberg's To sit, unmoving and Rajesh Parameswaran's The strange career of Dr. Raju Gopalarajan. Authors Anthony Doerr, Wells Tower and Kevin Moffett also won National Magazine Awards for their stories Memory Wall, Raw Water and Further interpretations of real-life events, published in 2010.
In addition to the Quarterly, McSweeney's also publishes The Believer, a monthly journal edited by Eggers' wife Vendela Vida, and co-edited by Heidi Julavitis. A five-time finalist itself for the National Magazine Award, it features delightfully bizarre essays, book reviews, interviews with authors, artists, musicians and directors, and letters from readers. McSweeney's also publishes books under The Believer imprint; some of its most notable titles include Nick Hornby's The polysyllabic spree (2004) and Tom Bissell's Magic hours (2012).
McSweeney's also operates a daily humour website, McSweeney's Internet Tendency, which its editor, Christopher Monks, describes as a "cool place to send funny stuff, and then some other people read it". Many of the Tendency's stories do well, generating millions of page-views and drawing visitors' attention to the fact that McSweeney's is a magazine and book publisher too. In fact, the company publishes approximately 30 book titles a year, including fiction, non-fiction, poetry, humour and children's books. Recent titles include Information doesn't want to be free (Cory Doctorow, 2014), The end of war (John Horgan, 2012) and Some recollections of a busy life (T.S. Hawkins with an introduction by Dave Eggers, 2016).
The Library's collection of McSweeney's books and journals was donated to Rare Books & Special Collections in 2016 by former University Librarian, Ray Choate. It contains approximately 60 titles, including 46 of the 48 Quarterly issues (no.'s 10 and 40 lacking). Other books in the collection include McSweeney's mammoth treasury of thrilling tales (Michael Chabon ed., 2002) and Dave Eggers' own The wild things (2009), adapted from Maurice Sendak's Where the wild things are.
The McSweeney collection of books and Quarterly issues can be accessed in the Rare Books & Special Collections reading room.
Last updated: 13 Feb 2017

redpajama_set_name: RedPajamaCommonCrawl


Formulated for white or grey hair, this shampoo cleanses hair and reduces any yellowing effect while adding subtle silver highlights, thanks to cyancentaurein, a vegetal pigment with blue hues. Hair becomes soft, easy to style and reveals all its beauty and radiance.
Centaury is traditionally used to add silver highlights to white or grey hair. We selected Centaurea cyanus L. because it is rich in cyanocentaureine. Cultivated in the Southwest of France according to certified organic farming methods, the flowers are dried away from the sunlight to preserve the quality of its pigments.
WATER (AQUA). SODIUM LAURETH SULFATE. MALTODEXTRIN. SODIUM COCOAMPHOACETATE, GLYCOL DISTEARATE. BENZOIC ACID. CENTAUREA CYANUS FLOWER EXTRACT. CETRIMONIUM CHLORIDE. CITRIC ACID. COCAMIDE MEA. FRAGRANCE (PARFUM). HEXYLENE GLYCOL. LAURETH-10. PHENOXYETHANOL. SODIUM CHLORIDE. SODIUM COCOYLAPPLE AMINO ACIDS. TETRASODIUM EDTA.


text:
\section{Introduction}
Let $k$ be a field, $\gg = \mathfrak{sl}_2(k)$ and $U=U(\gg)$ the universal
enveloping algebra of $\gg$. In this note we calculate $\operatorname{HH}^*(U)$ when $k$ has characteristic $p>2$.

$U$ is a Hopf algebra with antipode
$\eta(x)=-x$ and comultiplication $\Delta(x)=x\otimes 1 + 1 \otimes x$
for $x \in \gg$. Let $U^e = U \otimes_k
U^{\textrm{op}}$ be the enveloping algebra of $U$, so there is an
algebra homomorphism $(1\otimes \eta) \circ \Delta : U \to U^e$ making
$U^e$ into a free $U$-module (on the generators $\lambda \otimes 1$ for
$\lambda$ in a PBW basis of $U$).  The induced module $k \uparrow _U
^{U^e}$ is isomorphic to $U$ and the restricted module $U|^{U^e}_U$ is by definition
$U^{\textrm{ad}}$, so Eckmann-Shapiro gives
\begin{equation*}
\operatorname{HH}^*(U) = \operatorname{Ext}^*_{U^e}(U,U) = \operatorname{Ext}^*_{U^e}(k \uparrow
^{U^e}_U,U) \cong \operatorname{Ext}^*_U(k, U^{\textrm{ad}}).
\end{equation*}

If $k$ has characteristic zero, the structure of $\operatorname{HH}^*(U)$ follows
immediately:  in that case $\operatorname{Ext}^*_U(k,L)=0$ whenever $L$ is a
nontrivial simple $U$-module so $\operatorname{HH}^*(U) \cong \operatorname{Ext}^*_U(k,k)
\otimes_k Z(U)$.  The centre $Z(U)$ is generated by the
Casimir element of $U$ in the characteristic zero case, and
$\operatorname{Ext}^*_U(k,k)$ is an exterior algebra with one generator of degree 3
arising from the Killing form.

Let $S$ be the symmetric algebra on the adjoint $\gg$-module.  To
calculate the Hochschild cohomology in the case when the characteristic
of $k$ is greater than two we use the isomorphisms
\begin{equation*}
	\operatorname{HH}^*(U) \cong \operatorname{Ext}^*_U(k,U^{\textrm{ad}}) \cong \operatorname{Ext}^*_U(k, S)
\end{equation*}
where the first follows as above and the
second from the isomorphism of $U$-modules
$U^{\textrm{ad}}\cong S$ of \cite{FP}: this result shows
$\operatorname{HH}^*(U)$ agrees with the Poisson cohomology of $S$ equipped with the
Poisson bracket induced by the Lie bracket of $\gg$.

Under this isomorphism $\operatorname{HH}^0(U)$ corresponds to $S^\gg$, and each
$\operatorname{HH}^i(U)$ becomes a $S^\gg$-module. We describe this module structure
by generators and relations, and give the Hilbert series
(arising from the polynomial grading on $S$) explicitly.  The following
theorem summarises some of the results to be proved in sections
\ref{hh1_sec}, \ref{hh2_sec} and \ref{hh3_sec}.
\begin{thm}
	As $S^\gg$-modules,  $\operatorname{HH}^1(U)$ is generated  by three elements
	of degree $p-1$ and four of degree $p$, $\operatorname{HH}^2(U)$ is generated
	by three elements of degree $p-1$ and one of degree $p$ and
	$\operatorname{HH}^3(U)$ is generated by one element of degree one and one of
	degree $p-2$.
\end{thm}

\subsection{Notation}
Let $k$ be a field of characteristic $p>2$ and $e,f,h$ denote the usual
basis of $\gg=\mathfrak{sl}_2(k)$.  Let $S^n$ be the $n$th symmetric power of
the adjoint $\gg$-module and $S = \bigoplus_{n\geq 0}S^n$ the symmetric
algebra on the adjoint $\gg$-module. Let $c = h^2 + 4ef \in S^\gg$, let
$Z$ be the subalgebra of $S$ generated by $c, e^p, h^p, f^p$ so that
$Z=S^\gg$ and let $Z_0$ be the subalgebra of $Z$ generated by $e^p, f^p$
and $h^p$.  Multiplication by an element of $Z$ is a $\gg$-endomorphism
of $S$, so each Hochschild cohomology group is a $Z$-module.

Let $L(r)$ denote the simple highest weight $\gg$-module with dimension
$r+1$.  The natural module $L(1)$ has a basis $x,y$ with $h\cdot x = x,
h\cdot y = -y$, and we write $S^{n}(L(1))$ for the $n$th symmetric power
of $L(1)$.

There is a standard resolution of the trivial $U$-module
\begin{equation}\label{ce_res}
	0 \to U\otimes_k \wedge^3  \gg \stackrel {\delta_3}\to U\otimes_k
	\wedge ^2 \gg \stackrel {\delta_2} \to U
	\otimes_k \gg \stackrel{\delta_1} \to U \to  k \to 0
\end{equation}
in which the differentials are given by
\begin{multline*}
	\delta_n (1\otimes x_1 \wedge \cdots \wedge x_n)=\sum_i
	(-1)^{i+1} x_i \otimes x_1\wedge \cdots \wedge \widehat{x_i}
	\wedge \cdots \wedge x_n  \\ + \sum_{i<j} (-1)^{i+j}\otimes
	[x_i,x_j] \wedge x_1 \wedge \cdots \wedge \widehat{x_i} \wedge
	\cdots \wedge \widehat{x_j} \wedge \cdots \wedge x_n
\end{multline*}
for $n \geq 1$ and $\delta_1(1\otimes x)=x$ for $x, x_i \in \gg$. In
particular the Hochschild cohomology groups $\operatorname{HH}^i(U)$ are zero for $i
\geq 4$.

Both $L(1)$ and $\gg$ admit weight gradings, with $e,h,f$ in degrees
$2,0,-2$ and $x,y$ in degrees $1,-1$ respectively, hence so do their
symmetric and exterior powers.  The differentials in this
resolution respect the weight gradings, so $\operatorname{Ext}_U^n(k,S)$ has two
gradings, one from the weight grading on $S$ and one from the polynomial
grading $S=\bigoplus_{n\geq 0}S^n$.

\section{Symmetric powers of the natural module}
\begin{lem}
For each $n\geq 2$ there is a short exact sequence of $\gg$-modules
\begin{equation}\label{ses}
	0
	\to
	S^{n-2}
	\stackrel{\times c} \to
	S^n
	\stackrel{\phi}{\to}
	S^{2n}(L(1))
	\to
	0.
\end{equation}
\end{lem}
\begin{proof}
	The images of the elements $e^i h^{n-i}, h^n, f^i h^{n-i}$ for
	$1 \leq i \leq n$ in $S^n/cS^{n-2}$ form a basis, and defining
	$\phi:S^n \to S^{2n}(L(1))$ by $\phi(cS^{n-2})=0$ and
		\begin{align}
		\phi(e^i h^{n-i} ) &= (-2)^{-i}x^{n+i} y^{n-i}
		\label{map}\\
		\nonumber \phi(h^n ) & = x^n y^n \\
		\nonumber \phi(f^i h^{n-i} ) &= 2^{-i} x^{n-i} y^{n+i}
		\end{align}
	induces an isomorphism $S^{n}/cS^{n-2}\to S^{2n}(L(1))$.
\end{proof}
If $2n=qp+r$ with
$0\leq r<p-1$ then the submodule structure
of $S^{2n}(L(1))$ is
\[ \xymatrix{
		& L(r')\ar@{-}[dl] \ar@{-}[dr] &        & \,\,\,\,\,\cdots & L(r')\ar@{-}[dr] &   \\
L(r) &                                 & L(r)\ar@{-}[ur] &        &                     & L(r)
} \]
where $r'=p-2-r$ and there are $q+1$ copies of $L(r)$ and $q$ of $L(r')$.  If $2n=qp
+ (p-1)$ then $S^{2n}(L(1))$ is a direct sum of copies of $L(p-1)$.
\begin{prop}\label{ext1}
	Let $2n=qp+r$ with $0 \leq r \leq p-1$. Then
	\[\dim \operatorname{Ext} ^1 _U (k, S^{2n}(L(1))) = \begin{cases}
		q+2 & r =p-2 \\
		2q &  r = 0  \\
		0 & \text{otherwise.}
	\end{cases}\]
\end{prop}
\begin{proof}
	We identify 1-cocycles for the resolution (\ref{ce_res})
	with linear maps $\alpha : \gg \to S^{2n}(L(1))$ such that
	\begin{align}
		\label{1cocyc}	e\cdot \alpha(f)-f\cdot \alpha(e)&=\alpha(h) \\
		\nonumber	(h-2)\cdot \alpha(e)&=e\cdot \alpha(h) \\
		\nonumber (h+2)\cdot \alpha(f)&=f\cdot \alpha(h)
	\end{align}
	so that the coboundaries are the maps $\delta_z: \gg \to S^{2n}(L(1))$ given
	by $\delta_z(r)=r\cdot z$ for $z\in S^{2n}(L(1))$.  We may assume $\alpha$ is homogeneous with
	respect to the weight grading, and if its weight degree $j$ is
	not zero mod $p$ then $\alpha$ equals the coboundary
	$\delta_{\alpha(h)/j}$. Weight homogeneous cocycles $\alpha$ whose weight degree is
	divisible by $p$ take the form
	\begin{equation*}
		\alpha(h)=\lambda_hx^i y^{2n-i}\,\, \alpha(e) =
		\lambda_e x^{i+1}y^{2n-i-1}\,\, \alpha(f)=\lambda_f
		x^{i-1} y^{2n-i+1}
	\end{equation*}
	where $n\equiv i\mod p$, $-1 \leq i \leq 2n+1$ and $\lambda_e,\lambda_f,\lambda_h\in k$, with
	the convention that these coefficients are zero if the
	corresponding monomial would have a negative exponent.
	The cocycle condition becomes
	\begin{align*}
		(n+1)(\lambda_f-\lambda_e)&=\lambda_h \\
		n\lambda_h&=0
	\end{align*}
All coboundaries with the same weight degree are scalar multiples of the
map $\delta_{x^i y^{2n-i}}$ which sends $h$ to $0$, $e$ to
$ix^{i+1}y^{2n-i-1}$ and $f$ to $ix^{i-1}y^{2n-i+1}$.

Suppose first $n \not \equiv 0$, so that $\lambda_h =0$.  If $n \not
\equiv -1$ then we must have $\lambda_f=\lambda_e$, so $\alpha$ is a
coboundary. This shows that the dimension of the Ext group is as claimed
for $r \neq 0, p-2$.  If
$n \equiv -1$, so $r=p-2$, we may choose $\lambda_f$ and $\lambda_e$ freely
but $\alpha$ is a coboundary if $\lambda_e=\lambda_f$.  Each choice for
$i$ contributes one to the dimension of the Ext group, which
is $q+2$.

Now let $n \equiv 0$, so all coboundaries with this weight degree are
zero.  We have $r=0$ and $2n=qp$, and each possibility for $i$ contributes two to the
dimension of the Ext group since $\lambda_e$ and $\lambda_f$ may be
chosen freely, except for the extreme values which contribute one: one of
$\lambda_e$ or $\lambda_f$ is forced to be zero as the
corresponding monomial has a negative exponent.  This gives a total of
$2q$.
\end{proof}
\begin{prop}\label{ext2}
Let $2n=qp + r$ with $0 \leq r \leq p-1$.  Then
\begin{equation*}
	\dim \operatorname{Ext}^2 _U (k, S^{2n}(L(1))) = \begin{cases}
		2q+2 &r=p-2\\
		q-1 &  r=0\\
		0 & \text{otherwise.}
	\end{cases}
\end{equation*}
\end{prop}
\begin{proof}
	We interpret 2-cocycles as maps $\alpha: \wedge^2 \gg \to
	S^{2n}(L(1))$ such that
	\begin{equation}\label{2cocyc}
		e\cdot \alpha(h\wedge f)+f\cdot \alpha(e\wedge h)=
		h\cdot \alpha(e\wedge
		f).
	\end{equation} Coboundaries are spanned by maps of the form
	\begin{equation}\label{2bdry}
		\begin{array}{lll}
		r_z(h\wedge f)=0& r_z(e\wedge f)=-f\cdot z& r_z(e\wedge
		h)=(2-h)\cdot z \\
		s_z(h \wedge f)=(2+h)\cdot z& s_z(e\wedge f)=e\cdot z& s_z(e\wedge
		h)=0 \\
		t_z(h\wedge f) = -f\cdot z& t_z(e\wedge f)=-z& t_z(e\wedge
		h)=e\cdot z
	\end{array}
\end{equation}
	for $z \in S^{2n}(L(1))$. If $\alpha$ is weight homogeneous of
	degree $j \not \equiv 0$ then it equals $(1/j)(s_{\alpha(h\wedge
		f)}-r_{\alpha(e\wedge h)})$ so is a coboundary. Therefore
	we assume \[\alpha(h\wedge f)=\lambda_f
	x^{i-1}y^{2n-i+1},\alpha(e\wedge
	f)=\lambda_hx^i y^{2n-i},  \alpha(e\wedge h)=\lambda_e
	x^{i+1}y^{2n-i-1}\] where $n \equiv i \mod p$, $-1 \leq i \leq
2n+1$ and  $\lambda_e, \lambda_h,
	\lambda_f \in k$, with the convention that these are zero if the
	corresponding monomial would have a negative exponent.  The cocycle condition becomes
	\begin{equation*}
		(n+1)(\lambda_e+\lambda_f)=0.
	\end{equation*}

	Suppose first that $n+1 \equiv 0 \mod p$ so this condition is
	empty.  By subtracting a coboundary of the form
	$t_z$ we can assume $\lambda_h=0$, and such a cocycle is not a
	boundary unless it is zero since if $r_z$ or $s_z$ have weight
	degree congruent to zero then they kill $e\wedge h$ and $h\wedge
	f$.
It follows that each possible
	choice of $i$ contributes two to the dimension of the Ext group,
	except the extreme values $i=-1, 2n+1$ where one of $\lambda_e,\lambda_f$ is forced to be zero
	because the corresponding monomial would have a negative
	exponent, so the dimension is
	$2q$. This takes care of the case $r=p-2$.

Now suppose $n+1 \not \equiv 0$, so $\lambda_e=-\lambda_f$ and by
subtracting a suitable $t_z$ we may assume $\lambda_h=0$. If $n \not
\equiv 0$ then $\alpha= -(\lambda_f/n)
t_{x^iy^{2n-i}}-(1/n(n+1))s_{x^{i-1}y^{2n-i+1}}$  is a
coboundary.
Otherwise as before $\alpha$ is not a coboundary unless it
is zero, and each possible choice
for $i$ contributes one to the dimension of the Ext group so the
dimension is as claimed.
\end{proof}
\begin{prop}\label{ext3}
Let $2n= qp + r$ with $0 \leq r \leq p-1$. Then \begin{equation*}
	\dim \operatorname{Ext}^3 _U (k, S^{2n}(L(1))) = \begin{cases}
		1 &  n=0 \\
		q &r=p-2\\
		0 & \text{otherwise.}
	\end{cases}
\end{equation*}
\end{prop}
\begin{proof}
	The resolution (\ref{ce_res}) shows $\operatorname{Ext}^3_U(k, S^{2n}(L(1)))
	\cong S^{2n}(L(1)) / \gg \cdot S^{2n}(L(1))$.
	We have
	\begin{align*}
		h\cdot x^ay^b& =(a-b)x^ay^b\\ e\cdot
		x^{a-1}y^{b+1}&=(b+1)x^ay^b\\ f\cdot
		x^{a+1}y^{b-1} &=(a+1)x^ay^b
	\end{align*}
	so $x^ay^b \in S^{2n}(L(1))$ is in the image of the
	action of $\gg$ if and only if $a \equiv b \equiv -1 \mod p$ or
	$a=b=0$. Thus the images of
	the monomials $1$ and  $x^{q_xp +
	(p-1)}y^{q_y p + (p-1)}$ for $q_x,q_y \geq 0$ in $S(L(1)) /
\gg \cdot S(L(1))$ are a basis. \end{proof}
\begin{prop}
The first connecting homomorphism in the long exact sequence
obtained by applying $\hom_U(k,-)$ to (\ref{ses}) is zero.
\end{prop}
\begin{proof}
We need that
\begin{equation*}
	\phi^*:	\hom_U(k,S^n) \to \hom_U(k, S^{2n}(L(1)))
\end{equation*}
is onto, or equivalently that every trivial submodule of $S^{2n}(L(1))$
is the image of a trivial submodule of $S^n$ under $\phi: S^n \to
S^{2n}(L(1))$.  The trivial submodules of $S^{2n}(L(1))$ are spanned by
monomials $x^{ap}y^{bp}$ for $a,b\geq 0$. If $a>b$ then
$x^{ap}y^{bp}=\phi(e^{p(a-b)/2}h^{bp})$, and $\gg$ acts trivially on
$e^{p(a-b)/2}h^{bp}$. The cases where $a \leq b$ are similar.
\end{proof}
\section{$\operatorname{HH}^1(U)$}\label{hh1_sec}
\begin{lem}
	\label{1cyc}
	The following are cocycles with values in $S$:
	\begin{equation*}
		\begin{array}{l|lll}
			\alpha & \alpha(e) & \alpha(h) & \alpha(f)\\
			\hline
			\delta_e & f^{p-1} & 0 & 0 \\
			\delta_f & 0 & 0 & e^{p-1} \\
			\delta & \frac{h^p-hc^{(p-1)/2}}{4f} &c^{(p-1)/2}
			& \frac{h^p-hc^{(p-1)/2}}{4e} \\
			E &0 &2e^p &-he^{p-1} \\
			F &hf^{p-1} &-2f^p &0 \\
			H &-2eh^{p-1} &0 &2fh^{p-1} \\
			C &\frac{c^{(p+1)/2}-h^{p+1}}{4f}  &h^p &\frac{c^{(p+1)/2}-h^{p+1}}{4e}
		\end{array}
	\end{equation*}
\end{lem}
\begin{proof}
	Checking the cocycle condition (\ref{1cocyc}) is straightforward
	using the fact that the action of $\gg$ on $S$ is by derivations.
\end{proof}
Given a cocycle $A$ we write $\bar{A}$ for
its cohomology class.
\begin{lem}\label{hh1rels}
\begin{enumerate}
	\item $e^p \bar F + f^p \bar E = (1/2)h^p \bar H$
	\item $2e^p \bar \delta - c^{(p-1)/2} \bar E -h^p \bar
		\delta_f=0$.
	\item $2f^p \bar \delta + c^{(p-1)/2}\bar F-h^p \bar \delta_e=0$.
	\item $c^{(p-1)/2}\bar H +2e^p \bar \delta_e -2f^p \bar
		\delta_f=0$.
	\item $c^{(p-1)/2}C-h^p\delta -e^p \delta_e-f^p\delta_f=0$.
	\item $c^{(p+1)/2}\delta - h^p C+e^p F-f^p E=0$.
	\item $2f^p \bar C + h^p \bar F -c^{(p+1)/2} \bar \delta_e-f^p
		\bar H = 0$.
	\item $2e^p \bar C - h^p \bar E - c^{(p+1)/2} \bar \delta_f +
		e^p \bar H=0$.
\end{enumerate}
\end{lem}
\begin{proof}
\begin{enumerate}
	\item $s=(1/p)(c^p-4(ef)^p -h^{2p})$ makes sense as an element of
		$S$ if we expand $c^p$ and perform the division in an appropriate
		$\mathbb{Z}$-form. Then  $e^p F+f^p E -(1/2)h^p H$ is equal to $-1/4$ times the
		coboundary of $1\mapsto s$.
	\item From now on we write $(a,b,c)$ for the linear map
$\gg \to S$ sending $e$ to $a$, $h$ to $b$ and $f$ to $c$. Then
$2e^p \delta - c^{(p-1)/2}E -h^p \delta_f$ is $e^p$ times
		the cocycle \[\left( \frac{h^p-hc^{(p-1)/2}}{2f}, 0 ,
				-\frac{h^p-hc^{(p-1)/2}}{2e}\right).\]
		This is the coboundary
		of $1\mapsto -\sum_{r=0}^{(p-1)/3}
		h^{p-2r-1}c^r/(2r+1)$.

	\item Similar to (2).
	\item $c^{(p-1)/2}H-2e^p\delta_e+2f^p\delta_f$ is the cocycle

	\begin{multline*}
	-h^p\left(\frac{h^p-hc^{(p-1)/2}}{2f},0,
	-\frac{h^p-hc^{(p-1)/2}}{2e} \right) \\ +
	c^{(p+1)/2} \left( \frac{c^{(p-1)/2}-h^{p-1}}{2f}
	,0,-\frac{c^{(p-1)/2}-h^{p-1}}{2e}\right)
	\end{multline*}
	The first term is a coboundary as in (2), and the
	second bracketed term is the coboundary of $1 \mapsto
\sum_{b=0}^{(p-3)/2} c^b h^{p-2(b+1)}/(2(b+1))$.
\item Follows immediately from the definitions.
\item Follows immediately from the definitions.
\item $2f^p C + h^p F -c^{(p+1)/2} \delta_e -f^pH$ is the cocycle
\begin{equation*}
-f^pc\left( \frac{c^{(p-1)/2}-h^{p-1}}{2f} ,0,
-\frac{c^{(p-1)/2}-h^{p-1}}{2e} \right)
\end{equation*}
which is a coboundary as in (4).
\item Similar to (7).\qedhere
\end{enumerate}
\end{proof}
\begin{prop}
The second connecting homomorphism in the long exact sequence obtained
by applying $\hom_U(k,-)$ to (\ref{ses}) is zero.
\end{prop}
\begin{proof}
We show
\begin{equation*}
\phi^*: \operatorname{Ext}^1_U(k,S^n) {\to} \operatorname{Ext}^1_U(k, S^{2n}(L(1)))
\end{equation*}
is onto by finding preimages under the map induced by
$\phi$ of representing cocycles for
the basis of
the right hand side provided by Proposition \ref{ext1}.

Firstly let $n \equiv -1 \mod p$ and write $2n=qp+(p-2)$. For any choice
of scalars $\lambda_{i,e}\neq \lambda_{i,f}$ where $-1 \leq i \leq
2n+1$ and $i\equiv n \mod p$, the cocycles \begin{equation*}
	A_i=( \lambda_{i,e} x^{i+1}y^{2n-i-1},0, \lambda_{i,f} x^{i-1}y^{2n-i+1})\end{equation*}
represent a basis of $\operatorname{Ext}^1_U(k,S^{2n}(L(1)))$.

Suppose $i=n$, and choose $\lambda_{i,e}=1, \lambda_{i,f}=-1$.
Since $\phi(-2eh^{n-1})=A_n(e)$ and $\phi(-2fh^{n-1})=A_n(f)$, a cocycle represents a preimage of
$A_n$ if it is equal modulo $c$ to
\begin{equation*}
	-(2eh^{n-1},0,2fh^{n-1}).
\end{equation*}
So $2h^{p(q-1)/2} \delta$ is a suitable preimage: using $4ef
\equiv -h^2 \mod c$ we get $2\delta(e)=(h^p-hc^{(p-1)/2})/(4f) \equiv -2eh^{p-2}
\mod c$, and similarly for $2\delta(f)$.

Suppose $i<n$, and choose $\lambda_{i,e}=1, \lambda_{i,f}=0$.  Since $\phi(2^{n-i-1}h^{i+1}f^{n-i-1})=A_i(e)$, a
cocycle represents a
preimage of $A_i$ if it equals
\begin{equation*}
	(2^{n-i-1}h^{i+1}f^{n-i-1},0,0 ) \mod c.
\end{equation*}
Since $n+1 \equiv 0 \mod p$ a
scalar multiple of a power of $h^p$ and of $f^p$ times $\delta_e$ works.
The case $i>n$ is similar.

Now let $n \equiv 0 \mod p$, so cocycles representing elements of
$\operatorname{Ext}^1_U(k, S^{2n}(L(1)))$ have the form
\begin{equation*}
	B_i=(\lambda_{i,e}x^{i+1}y^{2n-i-1},
	\lambda_{i,h}x^i y^{2n-i},
	(\lambda_{i,e}+\lambda_{i,h})x^{i-1}y^{2n-i+1})
\end{equation*}
for some $\lambda_{i,e},\lambda_{i,h}$ and some $i\equiv n\mod p$.

If $i=n$ and $\lambda_h=0$
then as $\phi(h^{i-1}e) = (-2)^{-1}x^{i+1}y^{i-1}$ and $\phi(h^{i-1}f)
=2^{-1}x^{i-1}y^{i+1}$ a suitable power of $h^p$ times $H$ is a preimage
of $B_n$.

If $i=n$ and $\lambda_{i,h}=1, \lambda_{i,e}=-1/2$ then a preimage of
$B_n$ has the form
\begin{equation*}
	(eh^{n-1}, h^n, fh^{n-1}) \mod c
\end{equation*}
so a power of $h^p$ times $C$ is a preimage, for $C(e) \equiv eh^{p-1}$ and
$C(f) \equiv fh^{p-1}$ mod $c$.

Suppose $n<i<2n$. If $\lambda_{i,h}=0, \lambda_{i,e}=1$ then
$\phi((-2)^{i+1-n}h^{2n-i-1}e^{i+1-n})=B_i(e)$ and $\phi(
(-2)^{i-1-n}e^{i-1-n}h^{2n-i+1})=B_i(h)$, so  $B_i$ has a preimage which
is a scalar multiple of some cocycle $e^{ap}h^{bp}H$.
If $\lambda_{i,e}=-1/2,
\lambda_{i,h}=1$ then as before a scalar multiple of some $e^{ab}h^{bp}C$ is
a preimage for $B_i$.

Suppose $i=2n$, so that $\lambda_{i,e}=0$, and write $2n=qp$. Then a scalar multiple of
$e^{(q/2-1)p}E$ is a preimage for $B_{2n}$ since $\phi(
(-2)^{n}e^{n})=x^{2n}$ and $\phi( (-2)^{n-1} e^{n-1}h)=x^{2n-1}y$.  The
cases when $i<n$ are similar.
\end{proof}

\begin{prop}\label{hh1genZ}
	$\operatorname{HH}^1(U)$ is generated as a $Z$-module by the cohomology
	classes of the cocycles listed in Lemma \ref{1cyc}.
\end{prop}
\begin{proof}
Vanishing of the first two connecting homomorphisms gives an exact
sequence
\begin{equation}\label{ext_ses}
	0 \to \operatorname{Ext}^1_U(k, S^{n-2}) \stackrel{\times c}{\to} \operatorname{Ext}^1_U(k,
	S^n) \stackrel{\phi^*}{\to} \operatorname{Ext}^1_U (k, S^{2n}(L(1))) \to 0
\end{equation}
for each $n$. Let $\mathcal{H}$ be the $Z$-submodule of $\operatorname{Ext}^1_U(k, S)$ generated
by the cohomology classes of the cocycles of Lemma \ref{1cyc}.
The proof of the previous proposition showed that every element of $\operatorname{Ext}^1_U(k,
S^{2n}(L(1)))$ has a preimage under $\phi^*$ in $\mathcal{H}$, so the
result follows by induction on $n$.
\end{proof}

\begin{prop}
	As a $Z_0$-module $\operatorname{HH}^1(U)$ is the free module on $c^i\bar
	\delta, c^i \bar\delta_e, c^i\bar \delta_f$ for $0 \leq i \leq
	(p-1)/2$ and $c^i \bar E, c^i\bar  F, c^i\bar  H, c^i \bar C$
	for $0 \leq i \leq (p-3)/2$ modulo relations $c^i R$ where $R$
	is the first relation from Lemma \ref{hh1rels} and $0 \leq i
	\leq (p-3)/2$.
\end{prop}
\begin{proof}
	Lemma \ref{hh1rels} and Proposition \ref{hh1genZ} imply that these
	elements generate $\operatorname{HH}^1(U)$ over $Z_0$ and the $Z_0$-module
	with these generators and relations surjects onto $\operatorname{HH}^1(U)$. We
	need only check the dimensions of their graded pieces agree.
	Since $Z_0$ is polynomial on three generators of degree $p$, its
	degree $n$ part has dimension $d_n={n/p +2 \choose 2}$ if $p$
	divides $n$ and zero otherwise.  Thus the degree $n$ part of the
	$Z_0$-module with these generators and relations has dimension
	\begin{equation}\label{hilb}
c_n=		3\sum_{i=0}^{(p-1)/2}d_{n-(p-1) - 2i} +
		4\sum_{i=0}^{(p-1)/2}d_{n-p-2i} - \sum_{i=0}^{(p-3)/2}
		d_{n-2p-2i}.
	\end{equation}

	Let $e_n = \dim \operatorname{Ext}^1_U(k, S^n)$.  Then (\ref{ext_ses}) and
	Proposition \ref{ext1} tell us that $e_0=e_1=0$ and if $n=qp+r$
	for $0 \leq r <p$,
	\[ e_n-e_{n-2}= \left\{
	\begin{array}{ll}  4q & r=0   \\
		2q+3 &r=p-1  \\
		0 & \textrm{otherwise.} \end{array}
	\right. \]
It is straightforward to check that $c_n$ obeys the same initial
conditions and recurrence relation.
\end{proof}
\begin{cor}
	Let $n=qp+r$ with $0 \leq r \leq p-1$. Then

	\[\dim \operatorname{Ext}^1_U(k, S^n) = \begin{cases}
	3{q+2\choose 2}+4{q+1\choose 2} & r=p-1 \\
	4{q+1\choose 2}-{q\choose 2} &  r \neq p-1 \text{ even } \\
	3{q+1 \choose 2} & r \text{ odd.}
	\end{cases}\]
\end{cor}
\begin{proof}
	This follows from (\ref{hilb}).
\end{proof}

\section{$\operatorname{HH}^2(U)$}\label{hh2_sec}
\begin{lem}
The following are cocycles with values in $S$:
\begin{equation*}
	\begin{array}{l|lll}
		\alpha & \alpha(h\wedge f) & \alpha( e\wedge f) &
		\alpha(e \wedge h) \\ \hline
		R_e &0 &0 &f^{p-1} \\
		R_h & fh^{p-2} &0 &eh^{p-2} \\
		R_f &e^{p-1} &0 &0 \\
		T &fh^{p-1} &0  & eh^{p-1}
	\end{array}
\end{equation*}
\end{lem}
\begin{proof}
	Check the cocycle condition (\ref{2cocyc}).
\end{proof}

\begin{lem}\label{hh2rel}
	$
	e^p \bar R_e + f^p \bar R_f - h^p \bar R_h = c^{(p-1)/2} \bar T
	$
\end{lem}
\begin{proof}
	$e^pR_e+f^pR_f-c^{(p-1)/2}T$
	equals
	\begin{equation*}
		(f( (ef)^{p-1}-c^{(p-1)/2}h^{p-1}),0,e(
		(ef)^{p-1}-c^{(p-1)/2}h^{p-1}))
	\end{equation*}
	and since
	$(ef)^{p-1}=(c-h^2)^{p-1}=\sum_{i=0}^{p-1}c^ih^{2p-2-2i}$

	this is
	\begin{equation*}
		\left(\sum _{i=0, i \neq (p-1)/2}^{p-1} fc^ih^{2p-2-2i},
		0,\sum _{i=0, i \neq (p-1)/2}^{p-1}
		ec^ih^{2p-2-2i}\right) .
	\end{equation*}
	The coboundary $r_{c^ih^{2p-1-2i}}$ equals
	\begin{equation*}
		(2(2i+1)fc^ih^{2p-2-2i}, -c^ih^{2p-1-2i},
		2(2i+1)ec^ih^{2p-2-2i})
	\end{equation*}
	so $e^pR_e+f^pR_f-c^{(p-1)/2}T$ differs by a coboundary from
	\begin{equation*}
		\left(0, \sum_{i=0, i \neq (p-1)/2}^{p-1}
		\frac{c^ih^{2p-1-2i}}{2(2i+1)},0\right).
	\end{equation*}
	Each $(0, c^i h^{2p-1-2i},0)$ with $0<i\leq p-1$ is the coboundary
	$s_{x_i}$ where $x_i=(1/i) \sum _{j=0}^{i-1} f c^i h^{2p-2-2i}$,
	so up to coboundaries $e^pR_e+f^pR_f-c^{(p-1)/2}T$ equals
	\begin{equation*}
		(0,h^{2p-1}/2,0).
	\end{equation*}
	Since \begin{equation*}
		t_{h^{p-1}}=(2fh^{p-2},-h^{p-1},2eh^{p-2}) \end{equation*}
	$h^pR_h$ differs by $(0,h^{2p-1}/2,0)$ from a boundary, and the result follows.
\end{proof}

\begin{prop}\label{3rd_con_hm}
The third connecting homomorphism in the long exact sequence
obtained by applying $\hom_U(k,-)$ to (\ref{ses}) is zero
except in the case $n=p-1$ when
it has image of dimension one.
\end{prop}
\begin{proof}
We must show
\begin{equation*}
	\phi^* : \operatorname{Ext}^2_U(k, S^n) \to \operatorname{Ext}^2_U(k, S^{2n}(L(1)))
\end{equation*}
is onto unless $n \neq p-1$ when its image has codimension one.
To do this, as before, we find preimages of cocycles
representing a basis of the right hand side.

Suppose $n \equiv 0 \mod p$.  The cocycles \begin{equation*}
	A_i = (x^{i-1}y^{2n-i+1}, 0, -x^{i+1}y^{2n-i-1} )
\end{equation*}
for $0  < i <2n$ and $i \equiv 0 \mod p$ represent a basis of
$\operatorname{Ext}^2_U(k,S^{2n}(L(1)))$, where we adopt the notation $(a,b,c)$ for the linear map on $\wedge ^2
\gg$ such that $h \wedge f\mapsto a, e\wedge f\mapsto b, e\wedge
h\mapsto c$.

If $i =n$ then $2h^{n-p}T$ is a preimage.
If $i>n$ then a scalar multiple of $e^{i-n}h^{2n-i-p}T$ is a preimage,
and if $i<n$ we get a similar result.

Now suppose $n \equiv -1 \mod p$ and $n>p-1$, so that cocycles have the
form
\begin{equation*}
	(\lambda_f x^{i-1}y^{2n-i+1}, 0, \lambda_e x^{i+1}y^{2n-i-1})
\end{equation*}
for $-1 \leq i \leq 2n+1$ and $i \equiv n \mod p$.

When $i=2n+1$, we need only find a preimage for
\begin{equation*}
	(x^{2n}, 0, 0)
\end{equation*}
and  $e^{n+1-p}R_f$ works.

When $n<i<2n+1-p$ we have to find
preimages for two linearly independent cocycles.  Taking $\lambda_f=1,
\lambda_e=0$ we get the cocycle
\begin{equation*}
	(x^{i-1}y^{2n-i+1},0,0)
\end{equation*}
which has a scalar multiple of $h^{2n-i+1}e^{i-n-p} R_f$ as a preimage.
Also $h^{2n-i+1-p}e^{i-n}R_h$ is congruent modulo $c$ to
\begin{equation*}
	(e^{i-n-1}h^{2n-i+1}/4, 0, e^{i-n+1}h^{2n-i-1})
\end{equation*}
so it is a preimage for the cocycle with
$\lambda_e=\lambda_f=(-2)^{i-n-1}$. The cases where $i < n$ are similar.

When $i=n$, $2f^ph^{n+1-2p} R_f$ is a preimage of the cocycle with
$\lambda_f=1, \lambda_e=0$ since $(fe)^{p-1} \equiv h^{2p-2} \mod c$. A
preimage for the cocycle with $\lambda_f=0, \lambda_e=1$ can be found
similarly.

Finally suppose $n=p-1$. Preimages when $i=-1$ or $2n+1$ work exactly as
before, so we look only at cocycles with weight degree zero. By
subtracting an appropriate coboundary we may assume these send $e\wedge
f$ to zero, so they take the form
\begin{equation*}
	\left(f\sum_{i=0}^{(p-3)/2} \phi_i c^i h^{p-2-i}, 0,
	e\sum_{i=0}^{(p-3)/2} \epsilon_i c^i h^{p-2-i}\right)
\end{equation*}
for some scalars $\phi_i, \epsilon_i$. Such a map is a
a cocycle if and only if $\phi_{(p-3)/2}=\epsilon_{(p-3)/2}$ and
\begin{equation*}
	\phi_{i-1}-\phi_i=\epsilon_{i-1}-\epsilon_i
\end{equation*}
for $1 \leq i \leq (p-3)/2$. This has the general solution
$\phi_i=\epsilon_i$ for $0 \leq i \leq (p-3)/2$. On the other hand,
(\ref{2bdry}) shows that if $z \in S$ has weight $-2$ then
\begin{equation*}
	(-f\cdot e \cdot z, 0, e\cdot e\cdot z)
\end{equation*}
is a coboundary. Taking
$z=fc^i h^{p-2-2i}$
we get
\begin{multline*}
	-4i(2i+1)(fc^ih^{p-2(i+1)}, 0,
	ec^ih^{p-2(i+1)}) \\+ 4(i+1)(2i+3)(fc^{i+1}h^{p-2(i+2)},0,ec^{i+1}h^{p-2(i+2)})
\end{multline*}
and so each
\begin{equation*}
	(fc^ih^{p-2(i+1)},0,ec^ih^{p-2(i+1)})
\end{equation*}
for $0<i\leq (p-3)/2$ is a coboundary. Therefore the weight zero ($i=n$)
part of $\operatorname{Ext}^1_U(k,S)$ has dimension at most one, and since
$(fh^{p-2},0,eh^{p-2})=R_h$ is a preimage
of the nonbounding cocycle
\begin{equation*}
	(x^{p-2}y^p/2, 0, -x^py^{p-2}/2)
\end{equation*}
it is exactly one.
\end{proof}

\begin{prop}
	$\operatorname{HH}^2(U)$ is generated as a $Z$-module by
	$\bar R_e,\bar  R_f, \bar R_h$ and
	$\bar T$.
	As a $Z$-module it is isomorphic to the free module
	on those generators
	modulo the relation of Lemma \ref{hh2rel}.
\end{prop}
\begin{proof}
	That these elements generate $\operatorname{HH}^2(U)$ as a $Z$-module follows
	from the preimages computed in the proof of Proposition \ref{3rd_con_hm}, as for
	$\operatorname{HH}^1(U)$. We only need show that the graded pieces of the free
	module with these generators and relations have the same
	dimensions as those of $\operatorname{HH}^2(U)$.

Proposition \ref{3rd_con_hm} implies
\begin{equation*}
	0 \to \operatorname{Ext}^2_U(k, S^{n-2}) \stackrel{\times c}{\to} \operatorname{Ext}^2_U(k,
	S^n) \stackrel{\phi^*}{\to} \operatorname{Ext}^2_U(k, S^{2n}(L(1))) \to 0
\end{equation*}
is exact except when $n=p-1$ when $\phi^*$ has image of dimension three,
which together with Proposition \ref{ext2} gives a recurrence relation
for $e_n = \dim \operatorname{Ext}^2_U(k, S^{n-2})$:
\begin{equation*}
	e_n-e_{n-2} = \begin{cases}
		3 &n=p-1\\
		4q+4 &  n=qp + p-1 \text{ for } q>0 \\
		2q-1 & n=qp \\
		0 & \text{otherwise.}
	\end{cases}
\end{equation*}

Since $Z$ is free as a $Z_0$-module on $c^i$ for $0\leq i <p$,
the dimension $f_n$ of the polynomial degree $n$ part of $Z$
satisfies
\begin{equation}\label{fn}
f_n=	\begin{cases}
		{q+2 \choose 2}& n=qp+r, r \text{ even}\\
		{q+1 \choose 2}& n=qp+r, r \text{ odd}
	\end{cases}
\end{equation}
so the free $Z$-module with our generators and
relations has degree $n$ part of dimension $3f_{n-(p-1)}+f_{n-p}-f_{n-(2p-1)}$
which equals
\begin{equation*}
	\begin{cases}
		3{q\choose 2}+{q+1 \choose 2}-{q-1\choose 2} &n=qp+r, r
		\neq p-1 \text{ even}\\
		3{q+2 \choose 2} & n=qp + p-1 \\
		3{q+1\choose 2} & n=qp + r, r \text{ odd}.
	\end{cases}
\end{equation*}
This satisfies the same recurrence
relation and initial conditions as $e_n$.
\end{proof}

\section{$\operatorname{HH}^3(U)$}\label{hh3_sec}
\begin{prop}\label{hh3_dims} Let $n=qp+r$ for $0 \leq r <p$. Then
	\begin{equation*}\dim \operatorname{Ext}^3_U(k, S^n) = \begin{cases}
		1 &n \leq p-3 \text{ even}\\
		{q+2 \choose 2} & r=p-1 \\
		{q+1 \choose 2} & r \text{ odd} \\
		{q\choose 2} & r<p-1 \text{ even}.
	\end{cases}\end{equation*}
\end{prop}
\begin{proof}
Proposition \ref{3rd_con_hm} implies that
\begin{equation}\label{end_les}
	0 \to \operatorname{Ext}^3_U(k, S^{n-2})\stackrel{\times c}{\to} \operatorname{Ext}^3_U(k,
	S^n)
	\stackrel{\phi^*}{\to} \operatorname{Ext}^3_U(k, S^{2n}(L(1))) \to 0
\end{equation}
is exact, except when $n=p-1$ in which case the map $\operatorname{Ext}^3_U(k,
S^{n-2})\stackrel{\times c}{\to} \operatorname{Ext}^3_U(k, S^n)$ has kernel of dimension
one.  Using Proposition \ref{ext3} this gives the following recurrence relation for $e_n=\dim
\operatorname{Ext}^3_U(k, S^n)$:
\begin{equation*}
	e_n-e_{n-2}=\begin{cases}
		1 &n=0 \\
		2q+1 & n=qp + p-1, q>0 \\
		0 & \text{otherwise}.
	\end{cases}
\end{equation*}
The claimed dimensions obey this recurrence
relation and agree with $e_n$ for $n=-1,0$ so the result follows.
\end{proof}

Let $I$ and $J$ be the cohomology classes of the 3-cocycles sending
$f\wedge h \wedge e$ to $1$ and $h^{p-1}$ respectively.
\begin{prop}
	As a $Z$-module, $\operatorname{Ext}^3_U(k,S)$ is
	generated by $I$ and $J$ subject to the relations
	$e^pI=f^pI=h^pI = c^{(p-1)/2}I=0$.
\end{prop}
\begin{proof}
	The free $Z$-module with these generators and relations has
	degree $n$ part of dimension 1 if $n \leq (p-3)/2$ is even and
	$f_{n-(p-1)}$ otherwise, where $f_n$ is as in (\ref{fn}). This
	agrees with $\dim \operatorname{Ext}_U^3(k,S^n)$ by
	the previous proposition, so we only need show that the given
	relations hold in $\operatorname{Ext}^3_U(k,S)$ and that $I$ and $J$ generate
	it as a $Z$-module.

That $e^pI=f^pI=h^pI=0$ is because
$\dim \operatorname{Ext}^3_U(k,S^p)=0$, and $c^{(p-1)/2}I=0$ because of the
$n=p-1$ case of (\ref{end_les}).
 That $I$ and $J$
generate $\operatorname{Ext}^3_U(k,S)$ as a $Z$-module follows, as for the lower
degree Hochschild cohomology groups, by induction on $n$ using
(\ref{end_les}): the right hand term has a basis consisting of the
element represented by the
cocycle $f\wedge h \wedge e \to 1$, which has $I$ as a preimage under
$\phi^*$, and elements represented by cocycles
\begin{equation*}
	f\wedge h \wedge e \mapsto x^{ap+p-1}y^{bp+p-1}
\end{equation*}
for $a,b \geq 0, a+b$ even by Proposition \ref{ext3}. If $a>b$ these have
$e^{(a-b)p/2}h^{bp} J$ as a preimage under $\phi^*$, if $a<b$
$f^{(b-a)p/2}h^{ap}J$ is a preimage and if $a=b$ then $h^{ap}J$ is a
preimage.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
'''
conf.py: common support for configurable objects
'''
import string
from core.api import coreapi
from functools import reduce

class ConfigurableManager(object):
    ''' A generic class for managing Configurables. This class can register
        with a session to receive Config Messages for setting some parameters
        for itself or for the Configurables that it manages.
    '''
    # name corresponds to configuration object field
    _name = ""
    # type corresponds with register message types
    _type = None

    def __init__(self, session=None):
        self.session = session
        self.session.addconfobj(self._name, self._type, self.configure)
        # Configurable key=values, indexed by node number
        self.configs = {}

    def configure(self, session, msg):
        ''' Handle configure messages. The configuration message sent to a
            ConfigurableManager usually is used to:
            1. Request a list of Configurables (request flag)
            2. Reset manager and clear configs (reset flag)
            3. Send values that configure the manager or one of its
               Configurables

            Returns any reply messages.
        '''
        objname = msg.gettlv(coreapi.CORE_TLV_CONF_OBJ)
        conftype = msg.gettlv(coreapi.CORE_TLV_CONF_TYPE)
        if conftype == coreapi.CONF_TYPE_FLAGS_REQUEST:
            return self.configure_request(msg)
        elif conftype == coreapi.CONF_TYPE_FLAGS_RESET:
            if objname == "all" or objname == self._name:
                return self.configure_reset(msg)
        else:
            return self.configure_values(msg,
                                    msg.gettlv(coreapi.CORE_TLV_CONF_VALUES))

    def configure_request(self, msg):
        ''' Request configuration data.
        '''
        return None

    def configure_reset(self, msg):
        ''' By default, resets this manager to clear configs.
        '''
        return self.reset()

    def configure_values(self, msg, values):
        ''' Values have been sent to this manager.
        '''
        return None

    def configure_values_keyvalues(self, msg, values, target, keys):
        ''' Helper that can be used for configure_values for parsing in
            'key=value' strings from a values field. The key name must be
            in the keys list, and target.key=value is set.
        '''
        if values is None:
            return None
        kvs = values.split('|')
        for kv in kvs:
            try:
                # key=value
                (key, value) = kv.split('=', 1)
            except ValueError:
                # value only
                key = keys[kvs.index(kv)]
                value = kv
            if key not in keys:
                raise ValueError("invalid key: %s" % key)
            setattr(target, key, value)
        return None

    def reset(self):
        return None

    def setconfig(self, nodenum, conftype, values):
        ''' add configuration values for a node to a dictionary; values are
            usually received from a Configuration Message, and may refer to a
            node for which no object exists yet
        '''
        conflist = []
        if nodenum in self.configs:
            oldlist = self.configs[nodenum]
            found = False
            for (t, v) in oldlist:
                if (t == conftype):
                    # replace existing config
                    found = True
                    conflist.append((conftype, values))
                else:
                    conflist.append((t, v))
            if not found:
                conflist.append((conftype, values))
        else:
            conflist.append((conftype, values))
        self.configs[nodenum] = conflist

    def getconfig(self, nodenum, conftype, defaultvalues):
        ''' get configuration values for a node; if the values don't exist in
            our dictionary then return the default values supplied
        '''
        if nodenum in self.configs:
            # return configured values
            conflist = self.configs[nodenum]
            for (t, v) in conflist:
                if (conftype is None) or (t == conftype):
                    return (t, v)
        # return default values provided (may be None)
        return (conftype, defaultvalues)

    def getallconfigs(self, use_clsmap=True):
        ''' Return (nodenum, conftype, values) tuples for all stored configs.
        Used when reconnecting to a session.
        '''
        r = []
        for nodenum in self.configs:
            for (t, v) in self.configs[nodenum]:
                if use_clsmap:
                    t = self._modelclsmap[t]
                r.append( (nodenum, t, v) )
        return r

    def clearconfig(self, nodenum):
        ''' remove configuration values for the specified node;
            when nodenum is None, remove all configuration values
        '''
        if nodenum is None:
            self.configs = {}
            return
        if nodenum in self.configs:
            self.configs.pop(nodenum)

    def setconfig_keyvalues(self, nodenum, conftype, keyvalues):
        ''' keyvalues list of tuples
        '''
        if conftype not in self._modelclsmap:
            self.warn("Unknown model type '%s'" % (conftype))
            return
        model = self._modelclsmap[conftype]
        keys = model.getnames()
        # defaults are merged with supplied values here
        values = list(model.getdefaultvalues())
        for key, value in keyvalues:
            if key not in keys:
                self.warn("Skipping unknown configuration key for %s: '%s'" % \
                          (conftype, key))
                continue
            i = keys.index(key)
            values[i] = value
        self.setconfig(nodenum, conftype, values)

    def getmodels(self, n):
        ''' Return a list of model classes and values for a net if one has been
        configured. This is invoked when exporting a session to XML.
        This assumes self.configs contains an iterable of (model-names, values)
        and a self._modelclsmapdict exists.
        '''
        r = []
        if n.objid in self.configs:
            v = self.configs[n.objid]
            for model in v:
                cls = self._modelclsmap[model[0]]
                vals = model[1]
                r.append((cls, vals))
        return r

    def info(self, msg):
        self.session.info(msg)

    def warn(self, msg):
        self.session.warn(msg)

class Configurable(object):
    ''' A generic class for managing configuration parameters.
        Parameters are sent via Configuration Messages, which allow the GUI
        to build dynamic dialogs depending on what is being configured.
    '''
    _name = ""
    # Configuration items:
    #   ('name', 'type', 'default', 'possible-value-list', 'caption')
    _confmatrix = []
    _confgroups = None
    _bitmap = None

    def __init__(self, session=None, objid=None):
        self.session = session
        self.objid = objid

    def reset(self):
        pass

    def register(self):
        pass

    @classmethod
    def getdefaultvalues(cls):
        return tuple( [x[2] for x in cls._confmatrix] )

    @classmethod
    def getnames(cls):
        return tuple( [x[0] for x in cls._confmatrix] )

    @classmethod
    def configure(cls, mgr, msg):
        ''' Handle configuration messages for this object.
        '''
        reply = None
        nodenum = msg.gettlv(coreapi.CORE_TLV_CONF_NODE)
        objname = msg.gettlv(coreapi.CORE_TLV_CONF_OBJ)
        conftype = msg.gettlv(coreapi.CORE_TLV_CONF_TYPE)

        if mgr.verbose:
            mgr.info("received configure message for %s" % cls._name)
        if conftype == coreapi.CONF_TYPE_FLAGS_REQUEST:
            if mgr.verbose:
                mgr.info("replying to configure request for %s model" %
                           cls._name)
            # when object name is "all", the reply to this request may be None
            # if this node has not been configured for this model; otherwise we
            # reply with the defaults for this model
            if objname == "all":
                defaults = None
                typeflags = coreapi.CONF_TYPE_FLAGS_UPDATE
            else:
                defaults = cls.getdefaultvalues()
                typeflags = coreapi.CONF_TYPE_FLAGS_NONE
            values = mgr.getconfig(nodenum, cls._name, defaults)[1]
            if values is None:
                # node has no active config for this model (don't send defaults)
                return None
            # reply with config options
            reply = cls.toconfmsg(0, nodenum, typeflags, values)
        elif conftype == coreapi.CONF_TYPE_FLAGS_RESET:
            if objname == "all":
                mgr.clearconfig(nodenum)
        #elif conftype == coreapi.CONF_TYPE_FLAGS_UPDATE:
        else:
            # store the configuration values for later use, when the node
            # object has been created
            if objname is None:
                mgr.info("no configuration object for node %s" % nodenum)
                return None
            values_str = msg.gettlv(coreapi.CORE_TLV_CONF_VALUES)
            defaults = cls.getdefaultvalues()
            if values_str is None:
                # use default or preconfigured values
                values = mgr.getconfig(nodenum, cls._name, defaults)[1]
            else:
                # use new values supplied from the conf message
                values = values_str.split('|')
                # determine new or old style config
                new = cls.haskeyvalues(values)
                if new:
                    new_values = list(defaults)
                    keys = cls.getnames()
                    for v in values:
                        key, value = v.split('=', 1)
                        try:
                            new_values[keys.index(key)] = value
                        except ValueError:
                            mgr.info("warning: ignoring invalid key '%s'" % key)
                    values = new_values
            mgr.setconfig(nodenum, objname, values)
        return reply

    @classmethod
    def toconfmsg(cls, flags, nodenum, typeflags, values):
        ''' Convert this class to a Config API message. Some TLVs are defined
            by the class, but node number, conf type flags, and values must
            be passed in.
        '''
        keys = cls.getnames()
        keyvalues = list(map(lambda a,b: "%s=%s" % (a,b), keys, values))
        values_str = '|'.join(keyvalues)
        tlvdata = b""
        if nodenum is not None:
            tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_NODE,
                                                nodenum)
        tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_OBJ,
                                            cls._name)
        tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_TYPE,
                                            typeflags)
        datatypes = tuple( [x[1] for x in cls._confmatrix] )
        tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_DATA_TYPES,
                                            datatypes)
        tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_VALUES,
                                            values_str)
        captions = reduce( lambda a,b: a + '|' + b, \
                           [x[4] for x in cls._confmatrix])
        tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_CAPTIONS,
                                            captions)
        possiblevals = reduce( lambda a,b: a + '|' + b, \
                           [x[3] for x in cls._confmatrix])
        tlvdata += coreapi.CoreConfTlv.pack(
            coreapi.CORE_TLV_CONF_POSSIBLE_VALUES, possiblevals)
        if cls._bitmap is not None:
            tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_BITMAP,
                                                cls._bitmap)
        if cls._confgroups is not None:
            tlvdata += coreapi.CoreConfTlv.pack(coreapi.CORE_TLV_CONF_GROUPS,
                                                cls._confgroups)
        msg = coreapi.CoreConfMessage.pack(flags, tlvdata)
        return msg

    @staticmethod
    def booltooffon(value):
        ''' Convenience helper turns bool into on (True) or off (False) string.
        '''
        if value == "1" or value == "true" or value == "on":
            return "on"
        else:
            return "off"

    @staticmethod
    def offontobool(value):
        if type(value) == str:
            if value.lower() == "on":
                return 1
            elif value.lower() == "off":
                return 0
        return value

    def valueof(self, name,  values):
        ''' Helper to return a value by the name defined in confmatrix.
            Checks if it is boolean'''
        i = self.getnames().index(name)
        if self._confmatrix[i][1] == coreapi.CONF_DATA_TYPE_BOOL and \
           values[i] != "":
            return self.booltooffon( values[i] )
        else:
            return values[i]

    @staticmethod
    def haskeyvalues(values):
        ''' Helper to check for list of key=value pairs versus a plain old
            list of values. Returns True if all elements are "key=value".
        '''
        if len(values) == 0:
            return False
        for v in values:
            if "=" not in v:
                return False
        return True

    def getkeyvaluelist(self):
        ''' Helper to return a list of (key, value) tuples. Keys come from
        self._confmatrix and values are instance attributes.
        '''
        r = []
        for k in self.getnames():
            if hasattr(self, k):
                r.append((k, getattr(self, k)))
        return r


text:
Q: Gtk object, name, type and so I Created a window and vbox with 3 buttons in Glade. All buttons have connected "Clicked" event to same handler. Handler looks like this:
CLICKED_btn(GtkObject *object, gpointer user_data)
{
    g_print("CLICKED\n");
}

CLICKED appears on terminal for any click to any button.
Is here a way, through object or other to know which button invokes event Clicked in case where all buttons uses same handler?

A: The object parameter refers to the object that generates the event, in your case the button. Then you can use gtk_widget_get_name() or any other GtkObject/GtkWidget/GtkButton function to make a difference.
UPDATE:
As it seems, newer versions of GTK/Glade do not set the name of the widgets to their id, so it is left to the default, that is the name of the type. In order to get the id of the object you can use the function gtk_buildable_get_name() that works with any buildable object.
With that you'll get button1, button2 or whatever name you put to these buttons.
Please, do not use the label to make a difference between the buttons. Yes, it works, but it is a bad habit: hard to maintain, bad with internationalization, and defeats the main purpose of Glade: to have the interface and the code separated.

A: First use the documentation to have the right prototype for the "clicked" signal of a GtkButton.
You then know that your callback's prototype should look like:
void on_button_cliked (GtkButton *button, gpointer user_data)
The button parameter is the object which received the signal, ie. the button on which you clicked.

redpajama_set_name: RedPajamaStackExchange


Even though administrators and moderators from RSVCD-Forum try to delete or edit all the unwanted posts, it is impossible for them to check all the posts. All the posts express the opinion of the author, the owners of RSVCD-Forum and WoltLab GmbH (the developers of Burning Board) cannot be held responsible for the content of those posts.
Enjoy your stay at RSVCD-Forum!


Each day our campers experience a long list of creative and exciting activities all led by highly qualified specialists. These activities are designed to tap into each camper's sense of adventure and their creative and artistic abilities. Our newly updated and expanded programs give campers an opportunity to participate in the latest in camp experiences and technology along with the camp traditions you expect from Park Shore, a leader in the day camp industry.


text:
Q: Spring Integration Images for use in Visio Does anyone know where some Spring style integration images can be found for use in Visio - just a stencil or where one could possible download the images used in the Spring Integration designer GUI as seen here: http://blog.springsource.com/wp-content/uploads/2010/10/si-sts-editor.png
FYI for anyone else interested there are these EIP stencils for Visio http://eaipatterns.com/downloads.html .
However I was hoping to use the ones that are used in the Spring Integration GUI to keep things consistent.
Cheers!

A: When You download and install SpringSource Tool Suite, it has a file: $STS_HOME/plugins/com.springsource.sts.config.ui_<version>.jar.
When You unpack it with standard unzip tool, inside You'll find a folder icons/integration. That's where all images from Spring Integration GUI editor are.

A: git clone https://github.com/spring-projects/spring-ide
# any icons
git ls-files "*/icons/*"
# integration icons:
git ls-files plugins/org.springframework.ide.eclipse.config.ui/icons/integration

valid for: df2c295 > 20 hours ago (HEAD, origin/master, origin/HEAD, master)

P.S. beware licensing

A: The images of Spring integration are part of STS bundle. You can extract them from the bundle using simple bash commands:
BASE=https://github.com/spring-projects/spring-ide/tree/master/plugins/org.springframework.ide.eclipse.config.ui/icons/integration/

RAW=https://raw.githubusercontent.com/spring-projects/spring-ide/master/plugins/org.springframework.ide.eclipse.config.ui/icons/integration/

curl "$BASE" | cheerio "a.js-directory-link" | sed "s;^;$RAW;" | xargs -I {} curl -O {}

Without the need to clone the original repository. My previous answer works on the local cloned repo.
Script uses cheerio CLI, you can donwload and install from npm. https://www.npmjs.com/package/cheerio-cli


Sacked Immigration Head of Infosys Ltd, Erin Green, has filed a lawsuit before the U.S. District Court in Texas, alleging discrimination and retaliation by the senior management of the global software company.
The civil suit, filed on June 19 in the Eastern District of Texas (Sherman Division), seeks a trial by jury on the alleged discrimination against non-Asian employees at the company's offices in the U.S., resulting in Green's dismissal nearly a year ago.
Green, who joined the IT major in October 2011 at Plano in Texas, questioned his dismissal on June 28, 2016, while he was tasked to analyze and rectify all H-1B, L-1 and I-9 and other U.S. immigration compliance-related issues.
An Infosys spokesperson, however, declined to respond to the charges in the lawsuit, saying "we cannot comment on an ongoing litigation".


text:
Spurs versus Huddersfield Town: (match preview)
Posted on September 29, 2017 by superhotspur
The sixth of September 1972 was the last time that Spurs played Huddersfield, the match was a league cup second round tie which was played in front of just over 20,000 spectators at White Hart Lane. That particular game ended in a 2-1 victory for Spurs courtesy of goals from Alan Gilzean and Martin Chivers. Now, more than 40 years later and after a fairytale promotion to the premier league, Spurs travel to Yorkshire for tomorrows game to take on the newly promoted club. Huddersfield who are so far unbeaten at home (the John Smiths stadium) this season, produced a fairy tale return to the top flight of English football last season, through the brilliance of both their players and manager David Wagner. Tomorrows league game is our last match before a two week international break, so it's imperative that Pochettino's side get the three points against a well structured Huddersfield. We've won our last three games in all competitions, with our away form being nothing short of sensational. With players such as Christian Eriksen and Dele Alli returning to the team, I'm hoping that our attacking prowess will prove too much for the terriers. Wagner has the Yorkshire club playing some fine attacking football, and one of the players at the heart of that is Huddersfield's Australian midfielder and creative force within the side, Aaron Mooy. I'm optimistic for Saturdays big game, though Pochettino's lads face a tough test, I'm hopeful that they'll achieve their objective. COYS
Absentees for Huddersfield: Mounié, Quaner, Williams, Palmer, Stankovic.
Absentees for Spurs: Lamela, Rose, Wanyama, Dembele.
My predicted lineup: (3-4-2-1) Lloris(c), Alderweireld, Sanchez, Vertonghen, Davies, Dier, Sissoko, Walker-Peters, Dele, Eriksen, Kane.
My score prediction: 3-0 to Spurs courtesy of a Harry Kane brace and a goal from Sissoko.
One to watch: Huddersfield's influential and creative midfielder, Aaron Mooy.
Posted in Match previews, Tottenham Hotspur senior sideLeave a comment
Spurs under 18's versus Reading: (match preview)
Posted on September 28, 2017 September 28, 2017 by superhotspur
Scott Parker's under 18 side travel to Reading's Hogwood park training ground this Saturday, for a league encounter with the royals, kick off: 11:30am. The game will be our sides last match before a two week international break. I will be traveling to Finchampstead for Saturdays game to report on what I'm hoping will be a positive game for the young lillywhites. Parker's side have had a mixed start to the season with three wins and three loses from their first six competitive fixtures. Meanwhile, Reading who done the double over us last season have started this campaign with six straight defeats, something which will only encourage our lads. This is after all a Spurs squad consisting of mostly first year scholars and their results and performances this season have been impressive. Despite Reading's current winless streak they boast a top academy, one which saw them excel last season, qualifying for tier one of their merit group. I suspect this game will still come too soon for the injured Rayan Clarke as too is the case for 18 year old goalkeeper Charlie Freeman. I will be providing an in depth match report for Saturdays game, which will be up on Saturday evening. COYS
My predicted lineup: De Bie(c), Lock, Dinzeyi, Lyons-Foster, Hinds, P.Maghoma, Bowden, Markanday, A.Shashoua, P.Patterson, Griffiths.
My score prediction: 2-1 to Spurs.
Posted in Match previews, U18'sLeave a comment
A dream debut for Spurs youngster Anthony Georgiou:
"It feels amazing to make my debut and to do it in the Champions League as well is a dream come true really. It's also nice to do it in Cyprus as I have a lot of family from here. It's unbelievable! Everybody dreams of playing in the Champions League, it's the biggest competition in club football, so it's really a dream come true. It's a proud moment for me and family." (Anthony Georgiou)
A couple of weeks ago Spurs youngster Anthony Georgiou was playing in front of a couple of hundred enthusiastic spectators on a cold night in Stevenage, as Spurs' development squad drew 2-2 with Swansea. Last night was a completely different matter for the promising 20 year old winger from Lewisham. Georgiou, who has represented Cyprus at youth level, came on as an 84th minute substitute to make his Spurs debut in the UEFA champions league, in a country which he has strong links to. You couldn't have scripted it any better, Georgiou making his Spurs debut in front of hundreds of thousands of football fans across the planet. The 20 year old who missed the majority of last season with a knee injury, has fought back over pre season, highly impressing Spurs boss Mauricio Pochettino who went on to include him in the first team squad that travelled to the USA in July. From there Georgiou has seized his opportunity, clearly impressing Pochettino with his performances on the training ground, so much so that he seems to be exclusively training with the senior squad since the beginning of the season. Only featuring in a handful of under 23 games, the left winger who joined Spurs' academy as a 13 year old. Is one of the hardest workers on a football pitch that you'll find. From my many years of seeing Georgiou play for Spurs across a variety of youth levels, what has struck me time and time again, is the endless amounts of energy he puts into games.
A creative left winger who can also slot in at wing back, Georgiou possesses blistering pace, (something that our squad lacks) a brilliant delivery and a work rate of the highest quality. He is the kind of impact player who as I've said before, could become a very useful option for Pochettino over the course of the season. Georgiou's a confident player who isn't afraid to take defenders on down the flank, something that impressed Spurs fans during the friendly with Roma in pre season. When, up against a quality right back in Bruno Peres, Georgiou impressed and seemed unfazed by the quality of the opposition. He becomes the ninth academy player to make their Spurs debut under Pochettino, a number which will only increase as the season progresses. My congratulations go to Anthony, on his first team debut for Spurs, the first of many more to come. COYS
Dream come true last night making my debut in the champions league 🙌🏼 also to win and do it in Cyprus🇨🇾was very special ⚽amazing experience pic.twitter.com/sZGIVKTrhC
— Anthony Georgiou (@_Ant_georgiou) September 27, 2017
Posted in Tottenham Hotspur senior sideLeave a comment
Spurs 3-0 APOEL Nicosia: Some thoughts on the game
There was little time for Mauricio Pochettino to prepare his side for our important champions league group stage game with APOEL Nicosia on Tuesday night. A pulsating derby day victory over West Ham on Saturday showed little signs of sapping our players energy for our trip to Cyprus. It was on a warm September evening in the Cypriot capital Nicosia where Spurs would continue their fine start to this seasons champions league campaign. Following on from last Saturdays hard fought victory over West Ham which was made much harder by Serge Aurier's second half seconding off, Pochettino had just the seven first team absentees for last night's game. However, he got his tactics spot on last night against a fine APOEL side. A side for which Pochettino highlighted pre match, the importance of not underestimating APOEL, a side who have a rich history in European competition. The much rotated Spurs side which saw players such as Moussa Sissoko and Son come into the team. Battled out a closely knit first half, which saw few clear cut chances come the Lilywhites way. In fact, it was the home side who came the closer of the two sides to finding the lead, Belgian Igor De Camargo's fierce strike rattling Lloris' crossbar on 19 minutes as Spurs escaped an early scare. The home team tested Spurs on a couple more occasions as they continued to impress going forward.
However, a slightly concerning lack of chances for Spurs was to be resolved on the stroke of half time. When man of the moment Harry Kane latched on to Alderweireld's superb long pass, before calmly slotting the ball past Waterman, to eradicate those early Spurs worries. It was a much smoother second half for Tottenham barring a crucial reflex save from Lloris to stop the hosts getting an equaliser shortly after the interval. Kane got his and Spurs' second of the game shortly after the hour mark to reassure the Spurs faithful. Kane clinically converting from Sissoko's cross to double our advantage. We confirmed our victory soon afterwards when that man Harry Kane scored the perfect hat trick to head home Trippier's pin point cross to cap off a remarkable team move, to ensure a vital three points. Pochettino's side had dug deep and got that all important victory in fine fashion last night, it's even more impressive when you consider the numerous key players that were all absent. Regardless of our key absentees, I thought we done a fine job at the GSP stadium, the players worked well as a team to create the goals we craved, they were focused and got the job done.
There were quite a few performances that caught my eye last night. But none more so than that of Mr.Harry Edward Kane who once again proved why he is one of the world's very best. Kane was brilliant against APOEL, he was driven, determined and incredibly clinical in front of goal. Without waxing lyrical, he was just perfect last night he carried the team. Other performances which impressed me were those of Harry Winks who dominated the midfield with his guile and neat passing. I also thought Moussa Sissoko done a fine job, he attacked with real intent, was creative and held the ball up well. An all round brilliant night for Spurs, who face Huddersfield away on Saturday. Will that be a spectacular 7th consecutive away win? COYS
32 – Harry Kane has now scored 32 goals in 30 competitive appearances for Tottenham Hotspur in 2017. Deadly. pic.twitter.com/nlIE8Y74BZ
— OptaJoe (@OptaJoe) September 26, 2017
Spurs under 23's 2-4 Liverpool: (match report)
"At 4-2 down we never stopped competing, never stopped running and never stopped working. We had three or four decent chances, with about 10 minutes to go we had an excellent chance, could have made it 4-3 and then anything can happened but the players kept going which we expect, although we just fell a little bit short today with our decision-making with and without the ball." (Wayne Burnett)
Our under 23's played out a highly entertaining contest with PL2 table toppers Liverpool on Friday night, as Wayne Burnett's side fell to a 4-2 defeat against the Reds at Anfield. Though it was a disappointing result for the team, I left Anfield proud of our teams performance and work rate. Neil Critchley's Liverpool oozed class, they were like watching Jurgen Klopp's senior team. They had an array of attacking talent which pushed us to the limits, they were incisive in their passing and devastating on the counter attack, it was certainly no disgrace to lose to a side who've won their first six league games of the season. Burnett's team were resilient as they fought back from a two goal deficit, to make it 2-2 against a possession based Liverpool side, who controlled the second half. Proceedings got off in the worst possible way for Burnett's side, when an attacking reds side opened the scoring inside just 30 seconds. Spurs were desperately unlucky to fall behind so early on after Maghoma's header unfortunately came to Wilson inside the box, the Welshman sending a looping header at goal which dipped over Austin, 0-1. Burnett's side fell under the cosh as Liverpool's attacking exploits tested the youngsters resolve, two important interceptions by Tanganga inside the first 5 minutes, stopping Liverpool from extending their lead. A dangerous free kick from Pritchard followed at the other end, as his effort flew straight into the palms of Kelleher, as Spurs struggled to get a grasp on a Liverpool side who oozed attacking quality and panache.
Welsh international Ben Woodburn flashed an effort across the face of Austin's goal, as the reds went in search of goal number two. Shortly afterwards, potent striker Rhian Brewster went close after being played through on goal. The centre forwards first attempt was thwarted by Austin, with the young Brewster's follow up somehow slicing wide of the goal, as Spurs escaped a two goal deficit. Critchley's rampant Liverpool side were not lowering their tempo, and on 16 minutes they were rewarded for their early efforts. Spurs were dealt a further blow as Wilson bagged his second of the game. After Markovic embarked on a powerful surging run, the Serbian held the ball up inside the penalty area before teeing off Wilson, who slotted clinically inside the far corner of the net, Burnett's side were totally under the cosh. In a positive turn of events, Spurs' first meaningful attack gifted them a way back into the game, after Kazaiah Sterling was brought down inside the box by Jones, as the referee pointed to the spot. Edwards confidently stood up to take, as his penalty curled into the top corner, a great penalty, taken with such composure. This was becoming a really enthralling game, as Spurs carved their way back into the flow of it. And we were unlucky not to have been given a second penalty soon after our first, after Edwards who received a pass from Duncan inside the area was brought down, my first instincts were penalty, but the referee wasn't to follow up as he waved play on. Despite our resurgence, Liverpool remained in control of the game. Brewster embarking on a weaving run down the left side of the box as he managed to get past Maghoma before squaring to Wilson. Whose effort was deflected wide for a Liverpool corner. Marko Grujic's powerful effort at goal was vitally blocked by Tanganga, as the reds looked to restore the two goal cushion.
Brewster almost made it three for the home side after he got on the end of a cross inside the area, only to smash the ball off of Austin's crossbar. A slight scuffle between the two sets of players after George Marsh was fouled shortly after his introduction, disrupted the game for a couple of minutes, as Tanganga was booked for his part. Then in a most remarkable turn of events, a resurgent Spurs side drew level. After prominent winger Keanan Bennetts worked his way into the box, the ball eventually came to Maghoma who showed the composure of a top centre forward, to superbly curl it into the top corner of Kelleher's net, the centre halves second goal of the season. Spurs were bombarding the home side, as Spurs looked to end a dramatic half on a high. Maghoma forced a close range stop out of Kellleher, then seconds later Bennetts played through Sterling down the left side of the area, his powerful effort being well saved by Kelleher. Spurs were playing brilliantly, as they pressurised Liverpool into making errors, though Liverpool were still dangerous on the break. Woodburn's long drive failed to test Austin minutes earlier, but it was Critchley's side who would retain their lead at the end of a pulsating first half. Liverpool attacked us with real ferocity, as the brilliant Harry Wilson claimed his hat trick. The Welshman receiving Woodburn's lay off on the edge of the area, Wilson cut inside before beating Austin with a low curling effort, 2-3.
The second half started well for the reds who almost extended their lead on 48 minutes, when Woodburn's powerful low effort was only denied, because of an outstanding reflex save from Austin to thwart the 17 year olds seemingly certain goal. However, Liverpool were to succeed when Rhian Brewster made it 4-2 to the reds on 50 minutes. Lloyd Jones' brilliant long ball was met by Brewster inside the area, as the youngster sent a delicious lob over Austin to cap off an admirable piece of play from the home side. Even though Liverpool were dominating possession Spurs were maintaining their high pressing game as they looked to get back into the game. They went close after Edwards surging forward run, ended in the forward slipping in Sterling inside the box. However, the centre forwards effort from a tight angle was lashed over Kelleher's crossbar. Liverpool were playing wonderful attacking football which continued to see them go close. Woodburn's long range effort going narrowly wide on the hour mark, as Burnett's lads struggled to contain a rampant and relentless Liverpool side, who just wanted to play fast attacking football. Spurs did respond however, Edwards going on a brilliant run down the right, which ended with a low shot from Marsh being comfortably gathered by Kelleher.
Then, on 70 minutes Pritchard lashed over from long range as Spurs struggled to threaten an in control Liverpool side. Burnett's team were then given a scare, after Austin rushed out to clear the ball. Austin's clearance came to hat trick hero Wilson who almost caught him out with a dipping effort. Spurs continued to look for a route back into the game, Edwards slotting through Sterling inside the Liverpool box, as the youngsters powerful effort forced another fine save out of Kelleher. Burnett was frantically urging his team to attack as the game neared an end, the rain lashing down. And Spurs ended the game solid at the back, as a late flurry of chances were halted by Spurs' defence. Dhanda's dangerous free kick was easily held by Austin minutes before, Ejaria's promising cross was flicked inches over the crossbar by Brewster. An important late block by Christian Maghoma to stop Wilsons late effort meant the game ended in a deserved 4-2 victory for Critchley's side. However, Burnett's battling side can hold their heads high after putting in a commendable performance against a very good Liverpool side. COYS
– Brandon Austin: Couldn't of done anything about goals, made some fine stops including a stunning reflex save in the second half.
– TJ Eyoma: Had a decent game at right back, was solid defensively.
– Christian Maghoma: I thought Maghoma and Tanganga had really positive games last night, yes four goals may have been conceded. But none of which were as a result of the centre halves. Maghoma was brilliant in the air, but he was also very efficient in his tackling at the back. Took his goal as well as Harry Kane would have.
– Japhet Tanganga: Had a very good game, impressed me yet again. Solid and composed in his play, made some really important blocks, and his positioning was also very good.
– Jaden Brown: Efficient and hardworking at left back, good shift.
– Luke Amos: Effective and tidy in the middle of the park.
– Dylan Duncan: Did a fine job, but was forced off early on through injury.
– Marcus Edwards: Did very well on occasions down the right, went on some brilliant runs and created numerous chances for Sterling. Despatched his penalty with real class.
– Joe Pritchard: Played deeper against Liverpool, the skipper had a good game. Won a lot of balls in the midfield.
– George Marsh: Added steel to the game, did his defensive duty very well.
– Keanan Bennetts: A player who has come on leaps and bounds over the past year. Bennetts put in a great performance on the left wing. He looked dangerous every time he had the ball at his feet, had the Liverpool defenders sweating, with his creative forays into the penalty area, had a big influence in Maghoma's goal. Very influential!
– Kazaiah Sterling: Can be very proud of his performance at Anfield, he may not have got on the scoresheet. But Sterling put in a fine performance. Made some really good runs, pressed Kelleher really high throughout the game. And won us a penalty, I rate Sterling very highly.
– Shilow Tracey: Pressed high, had a positive effect on the game from the bench.
– Samuel Shashoua: Despite it being a short cameo, Shashoua showed his creative talent, with good movement on and off the ball. He also embarked on numerous take ons.
My man of the match: Keanan Bennetts. Even though he was substituted on the hour mark, I felt Bennetts was our most influential player during the game. A player who has come on leaps on bounds over the past year. Bennetts put in a great performance on the left wing. He looked dangerous every time he had the ball at his feet, had the Liverpool defenders sweating, with his creative forays into the penalty area, had a big influence in Maghoma's goal. Very influential!
Liverpool: Kelleher, Virtue, Whelan (c), Phillips, Jones, Grujic, Wilson, Woodburn (Dhanda 79), Brewster, Ejaria, Markovic (Brannagan 63). Substitutes (not used): Masterson, Johnston, Atherton (GK).
Spurs: Austin, Eyoma, Brown, Amos, C Maghoma, Tanganga, Edwards, Duncan (Marsh 28), Sterling, Pritchard (c, S Shashoua 74), Bennetts (Tracey 61). Substitutes (not used): Loft, Whiteman (GK).
Goals: Liverpool – Wilson 1, 16, 44, Brewster 50; Spurs – Edwards 20 (pen), Maghoma 35.
Yellow cards: Liverpool – Brewster 34, Phillips 54, Wilson 81; Spurs – Tanganga 33, Amos 52.
Referee: Daniel Middleton.
Posted in Match reports, U23'sLeave a comment
Spurs under 18's versus Aston Villa: (match preview)
Scott Parker's under 18 side return to league action tomorrow afternoon, when they host Aston Villa at Hotspur way. Looking to make it three wins on the trot, Parker's side will be looking to carry on their momentum in to tomorrows encounter against the villains. Unfortunately I won't be able to report on tomorrows game, due to being up in Liverpool for our under 23's match. Though there should be a match report up on the official site on Saturday. It will be interesting to see what the team will look like for the game, taking into consideration our under 19's trip to Cyprus for the UEFA youth league in midweek. A couple of pieces of team news: I noticed that winger Rayan Clarke who is currently injured, was involved in over half an hours light training before last Saturdays win at Leicester, I doubt he'll return for tomorrow. In addition, I believe that 17 year old goalkeeper Charlie Freeman is also injured. COYS
My predicted lineup: De Bie(c), Lock, Lyons-Foster, Dinzeyi, Reynolds, P.Maghoma, Bowden, P.Patterson, A.Shashoua, Markanday, Griffiths.
Spurs versus West Ham United: (match preview)
After three home games, Spurs are back on the road again on Saturday, when Pochettino's side travel to east London for an important London Derby against rivals West Ham United. After finding out we'll be facing the Hammers again, in the fourth round of the Carabao cup, Pochettino will be hoping his side set the tone with a league victory at the weekend. This fixture last season, saw Spurs' feint title hopes dashed when the Slavan Bilic's side beat Spurs 1-0, in what was one of the most disappointing days of an other wise brilliant campaign. The rivalry between these two clubs has grown somewhat in recent seasons, seeing some of the most dramatic Spurs' games of recent times. Paul Stalteri's late winner in a 4-3 victory at the old Boleyn ground, and a late stunner from Bale back in 2013, just two that come to mind. Results and performances have been mostly impressive from our lads of recent, superb displays against Dortmund followed by uninspiring stalemates with Swansea though, still suggesting that Pochettino's new look side hasn't yet clicked. Of course all fans know that victory over the hammers is imperative, but with a champions league tie away to Apoel Nicosia in midweek, it may prove a tad tricky for Pochettino to play the same team for both encounters. Despite, West Ham's poor start to the season they've shown some consistency from recent games, with a draw and two victories from their last three games.
Bilic is a top manager who has his side very well structured, it will certainly be a tough task to break them down tomorrow. However, its their potent and dynamic front three consisting of Antonio, Hernandez and Carroll which will provide their main goal threat. I'm anticipating a tough and closely knit game tomorrow afternoon, but as always I'm feeling confident, that Pochettino's side can record their first win at the London stadium. COYS
Absentees: Victor Wanyama, Erik Lamela, Danny Rose.
My predicted lineup: (3-4-2-1) Lloris(c), Alderweireld, Sanchez, Vertonghen, Aurier, Dier, Dembele, Davies, Eriksen, Alli, Kane.
My score prediction: 3-1 to Spurs, courtesy of a brace from Harry Kane and and a goal from Sanchez.
One to watch: West Ham's powerful winger Michael Antonio, who is a real threat inside the penalty area.
Spurs versus Liverpool under 23's: (match preview)
Date: Friday 22 September, kick off- 19:00pm
Venue: Anfield
Head2Heads: 3-2 Liverpool, 6-2 Liverpool.
Wayne Burnett's development side have little time to ponder what might have been after their 2-2 draw with Swansea on Monday. His side will be making their second trip up to Merseyside in less then a month, as our under 23's face table toppers Liverpool at their world renowned Anfield stadium on Friday evening. This PL2 game is arguably Spurs' toughest test of the season yet, the reds have won their first five league matches. Neil Critchley's side who boast talents such as English striker Rhian Brewster, highly skilful winger Harry Wilson and midfielder Cameron Branagan. Have arguably the better under 23 side. However, Burnett's side have already proved they can compete against very good sides on quite a few occasions already this season, and I'm sure our encounter with Liverpool will be no different. This game comes on the back of a highly competitive 2-2 draw with Swansea early in the week, whilst just under two weeks ago we were beaten 4-1 by fellow Liverpool club, Everton. Last seasons meeting which was also played at Anfield, was a most intriguing game. Spurs took a two goal lead with in the first half, before being pegged back by a very strong Liverpool side and eventually losing 3-2, this will be my fourth trip to Liverpool in less than a year to see Spurs, the last three have all ended in defeat.
As far as injuries are concerned, I'm none the wiser on whether Shayon Harrison is nearing a return. Playing under the lights at Anfield is a special occasion for any player, but all the more so for a youth player, this will provide an excellent learning curve for the young Spurs lads. As always, I will be writing an in depth match report. Which will be published on Friday night. A quick note, Liverpool TV don't seem to be showing this game, due to the fact that the reds under 18's are playing Everton at the same time. COYS
My predicted lineup: (4-2-3-1) Austin, Eyoma, C.Maghoma, Tanganga, Brown, Skipp, Duncan, S.Shashoua, Pritchard(c), Georgiou, Sterling.
My score prediction: 3-3.
One to watch: Liverpool's tricky Welsh winger Harry Wilson, the Welsh international has a great scoring record against Spurs over recent seasons.
Spurs 1-0 Barnsley: some thoughts on the game…
Spurs booked their place in round four of the Carabao cup last night, with a 1-0 victory over Barnsley at Wembley. It may not have been a memorable occasion for those at the game, but it doesn't matter in the slightest. As Pochettino's second string side have achieved the win that we needed, and for that I am grateful. Though Spurs had the majority of possession throughout the game, they lacked creativity in the final third, and Barnsley's tenacious defending made it harder for us to find the back of the net. Pochettino's side passed the ball around well, a golden chance for Llorente to head home unmarked from a resulting corner kick, saw the new signing rise too early as his header flew over the crossbar on 26 minutes. A chance for Barnsley's Ike Ugbo forced Vorm into making an important save at the beginning of the second half, the game was far from an easy win, as the Tykes gave us a couple of scares in front of goal. As Pochettino's sides frustration grew, so did the anxiety of Spurs' fans on the terraces. However, the razor sharp instincts of Dele Alli were to ease those worries.
When the England international scored what would prove to be the winner on 65 minutes, after some good work down the right side of the penalty area by Trippier. The fullback provided a pullback inside the area, which the aware Alli raced to, to calmly slot home. It was a pretty uneventful end to the game, as both sides went in search of goals at either end. But it wasn't to matter how flowing or entertaining our football was, as Pochettino's side had achieved their aim of getting into round four, where they'll find out their opposition after Man United's game with Burton this evening. Next up for Pochettino's lads is a visit to the London stadium on Saturday, as Spurs face a tough and important London derby against West Ham. I was paticulary impressed with the performances of Juan Foyth and Walker-Peters last night. Foyth on his debut, put in a confident and composed display at centre half, and looked very solid and conformable. Whilst Walker-Peters done a superbly efficent job at left back. Special mention goes to young Tashan Oakley-Boothe (17) who became the first millenial to feature for Spurs, last night! COYS
Job done ✅ #COYS pic.twitter.com/Ve69zPa1Cl
— Mousa Dembélé (@mousadembele) September 19, 2017
Spurs versus Barnsley:(match preview)
Date: Tuesday 19 September, kick off: 8:00 pm.
Venue: Wembley stadium
Referee: Tim Robinson
Previous meeting: Spurs 1-0 Barnsley (1999)
I want our lads to go out and try and win this seasons Carabao cup which starts for us this evening with the visit of championship club Barnsley to the national stadium. Regardless of how strong a side Pochettino fields for the early stages of the competition. I vividly remember the disappointment I felt last October, sitting in the Anfield away end, as a youthful and inexperienced Spurs side crashed out of the cup to the Reds in a 2-1 defeat. Pochettino clearly expressed his intentions about the competition at yesterday's press conference at Hotspur way. You can't argue with his prioritisation of the premier and Champions league, but my view is that the supporters as well as the club as a whole, are hungry for silverware, and this competition provides the most realistic place where we can achieve that. Enough of my views, this evenings opponents Barnsley who are managed by Paul Heckingbottom, are expected to bring 3,000 fans down to London for the match. The tykes are a club who we haven't faced since the late 1990's when a memorable David Ginola goal stole the show. The Yorkshire club, who play their football in the championship, have achieved a lot on a small budget since their return to the second tier. Barnsley have a talented young side who have some great up and coming prospects on their books. Two of them are loanees Harvey Barnes (Leicester) and Ike Ugbo (Chelsea) who incidentally both caused our youth teams great bother last season. Barnes is a tricky winger, whilst Ugbo is one of England finest striking prospects. Though, the tykes most potent player is probably Welsh centre forward Tom Bradshaw who holds a good scoring record for them.
I'm looking forward to seeing some of our youngsters that I have watched for years, get their debut tonight. But I'm also mindful of the fact that we must respect our opposition who beat Morecambe and Derby on route to this tie. I've a good feeling about this game. COYS
Absentees: Lamela, Wanyama, Rose.
My predicted lineup: (4-2-3-1) Vorm(c),Walker-Peters, Dier, Foyth, Bennetts, Winks, Oakley-Boothe, Sissoko, Alli, Nkoudou, Llorente.
Youngsters on the bench (my prediction): Dylan Duncan.
My predicted score line: 4-0 to Spurs courtesy of a brace from Llorente, a goal from Alli and one from Walker-Peters!
One to watch: Barnsley's on loan Chelsea striker Ike Ugbo, a player that our youth
teams will know very, very well.


I saw a picture of a quilt on this post over a year ago and haven't been able to get it out of my mind, it is the one down the bottom of the post. I did consider buying it for a time but I really wanted to make it for myself.
So I searched online and in my two blocks books, Encyclopaedia of Pieced Quilt Patterns and 5500 Quilt Block Designs. No luck.
After a few comment exchanges I got a better idea of how it was constructed. I spent another few night staring at the little picture and trying to work out where the seams were in the original and what formed the blocks.
In the end I got some graph paper, a pencil, a ruler and a compass and drew it out as a finished block. I was pretty happy with the first version and only made a few changes in scale. I cut out the drawn block, traced it onto template plastic and added seam allowance until I had a template set for the block.
And then I got sewing. It isn't an impossible block, but it does require some technical skill - a 'Y' seam and a curved seam.
I ended up deciding to make this quilt for the Quilter's Guild of New South Wales Under 35's Quilt Competition. This year the theme is "How Does Your Garden Grow?"
Because to the size restrictions in the competition I stopped at 16 blocks - finishing at 95cm (or just over a yard) square but I would love to make a 6x6 layout version as a centre medallion. I think this would show off the crosses better and make the name make a little more sense.
I posted it off today and will not see it again until sometime in December.
If you happen to be interested in seeing the show (and maybe sending me a picture of my little quilt hanging in the show) all the entries will be on display in the Newington College Gallery, Stanmore Road, Stanmore, Sydney from the 21-23 September from 10am - 4pm daily.


... structure, which can be either cobalt blue or stainless steel, is robust and has an appealing design. Due to this, the entire packaging line is qualified.
This machine is perfect for all kind of tasks in the food packing branch but of course also for all requirements of other branches; equipped by a two-piece belt, "V" groove for the transport of round and oval products and the complying guides.