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text:
Australian Academics Criticize Political Meddling in Basic Research
In an open letter, 63 Australian Research Council laureate fellows complain vigorously to the minister and to the chief executive of the ARC about a recent instance of political interference in the funding of basic research.
Lamar Smith's National Interest Bill Passes House
Legislation that requires that future grants made by the National Science Foundation meet a test for being in the 'national interest' passed the House of Representatives on Wednesday.
The National Interest and the NSF
Current legislation calls for federally funded science to be in the 'national interest.' What does that even mean, and why do scientists fear this Republican-led effort?
Bill That Seeks 'National Interest' Justifications for NSF Grants Advances
A bill that would require the National Science Foundation to justify, in writing, that every grant it makes is in the national interest and "worthy of federal funding" passed the science committee of the U.S. House of Representatives this morning.
Bill: Make 'National Interest' Explicit in NSF Grants
In February officials with the National Science Foundation, the National Institute of Standards and Technology, and the National Science Board […]

redpajama_set_name: RedPajamaCommonCrawl


These two consolidated appeals involved two lawsuits, a derivative claim and a direct shareholder action, both arising from a series of stock transactions in two family corporations owned primarily by eight siblings. After the death of one of the sisters, the corporations attempted to repurchase her stock pursuant to the terms of a stock purchase agreement. The sister's estate refused. The corporations filed a declaratory judgment action, seeking enforcement of the agreement. Meanwhile, two siblings, aggrieved by an earlier stock transaction, filed a derivative action, alleging self-dealing and breach of fiduciary duty. The circuit court (1) granted summary judgment in favor of the corporations on the derivative action after deferring to the judgment of a special litigation committee (SLC); and (2) granted summary judgment to the corporation in the declaratory judgment proceeding. The Supreme Court (1) reversed the circuit court's judgment in the derivative action, holding that the court made an inadequate inquiry into the SLC's independence and the reasonableness of its procedures; and (2) reversed in part the circuit court's grant of summary judgment in the declaratory judgment action, holding that the circuit court erroneously applied res judicata to the issue.
The Maryland Real Estate Commission revoked the real estate licenses of Joel Pautsch pursuant to Md. Code Ann. Bus. Occ. & Prof. 17-322(b)(24)(i) based on Pautsch's convictions for child abuse. The circuit court affirmed after finding there was competent, material and substantial evidence to support the Commission's decision. The court of special appeals affirmed. The Court of Appeals affirmed, holding (1) there was substantial evidence upon which the Commission relief to support its finding that there was a nexus between Pautsch's convictions and his professional activities; and (2) the sanction was neither arbitrary nor capricious because Pautsch's crimes undermined his trustworthiness in dealing with the public during the course of providing real estate brokerage services and negatively impacted his character and reputation.


text:
gs had outshot the Sharks 19-6 by the time they had taken that 3-0 l
Future Of Harderstyles » Q-Dance » Defqon. 1 » gs had outshot the Sharks 19-6 by the time they had taken that 3-0 l
#1 | gs had outshot the Sharks 19-6 by the time they had taken that 3-0 l 19.09.2019 05:53
miaowang123
Chelsea manager Antonio Conte has apologised after Cesc Fabregas red card during Wednesdays friendly clash with Liverpool. Gary Cahills early header saw the Blues come away with a 1-0 victory during the International Champions Cup tie in California, despite going down to 10 men in the 70th minute. Cesc Fabregas was shown the red card for this dangerous lunge on Liverpool centre-back Ragnar Klavan The Spain international flew into a dangerous tackle on new Liverpool defender Ragnar Klavan before receiving a straight sending off. Six other players were booked during the feisty encounter. Speaking on the incident, Conte said: Im sorry for this and the player is the same. To the player, I said no excuses and Im sorry for him and Im sorry for the Liverpool player. Cesc Fabregas was sent off during the clash in California I think that in this period, the fatigue sometimes can [cause a player] to go too late in the situation. I know Fabregas is a good man and hes not a player to have this type of situation. Im sorry for him and Im sorry for the Liverpool player. Hazard keen to make amends Midfielder determined to recapture best form Former Italy boss Conte has made two signings since arriving at Stamford Bridge, bringing in NGolo Kante and Michy Batshuayi, and has reportedly shown interest in striker Carlos Tevez.The pair worked together at Juventus during the 2013/14 season and earlier this week, Tevez claimed Conte wanted to bring him to Stamford Bridge - but revealed he wants to finish his career with current club Boca Juniors.When asked about Chelseas reported interest in the 32-year-old, Conte only had good wishes for his former player. Antonio Conte comments on Chelseas reported interest in Carlos Tevez Carlos was my player when I was coach at Juventus and I had a fantastic season with him and I appreciated Carlos as a player and also as a man, the manager said.But now I know he is a player of Boca Juniors and he did a lot back in Argentina last season and I wish the best for him and his family.Watch the International Champions Cup live on Sky Sports as well as 126 Premier League games throughout the season, including Chelsea v West Ham on August 15. Also See: Fabregas off in feisty friendly Tevez: Chelsea want me Chelsea bid £38m for Koulibaly Chelsea must make title bid Shoes NZ 2020 . In a pregame tribute commemorating his final contest at Coors Field on Wednesday night, Helton caught the ceremonial first pitch from his daughter with his wife, younger daughter and good friend Peyton Manning watching from the field. Shoes NZ Online . Williams withdrawal came less than 24 hours after losing in the final Auckland WTA tournament final. "I dont believe she has an injury," Hobart tournament director Mark Handley said. https://www.shoesnzonline.com/ . A top pitching prospect, one who the ball club is pinning some of its future hopes, takes the spot of a veteran who once was viewed as a future ace but who, to this point, hasnt realized his potential and may never. Wholesale Shoes NZ . Louis Cardinals continued their offensive tear with a 9-5 victory over the Los Angeles Dodgers in the opener of a four-game series. Cheap Shoes NZ .Airport spokeswoman Heather Lissner said that all departing flights were back on schedule. However, arriving flights were still trying to catch up.Super Bowl Sunday got off to an inconvenient start for hundreds of passengers trying to get to Phoenix for the game.The Bruins bury the Red Wings, Pittsburgh controls play against Columbus, Colorados amazing comeback and the Kings stay alive; Scott Cullen has notes on Krug, Hamilton, Nyquist, Crosby, MacKinnon, Quick and more. BRUINS FINISH RED WINGS The Boston Bruins calmly dispatched the Detroit Red Wings, with a 4-2 win Saturday afternoon, wrapping up their first-round series in five games. While the series was short, it was relatively competitive but, in the end, the Bruins simply had too much. In Game Five, the Red Wings had a goal and an assist each from Pavel Datsyuk and Henrik Zetterberg -- and contributions from their superstars may have been their best chance for the upset -- but it wasnt enough. Boston got a goal and an assist from LW Milan Lucic and two assist from D Torey Krug on their way to the second round. The tough part for the Red Wings is that they couldnt get any production out of the young forwards that had played such vital roles this season. LW Tomas Tatar, C Riley Sheahan and RW Gustav Nyquist were held off the scoresheet in all five games and the tendency might be to think that these inexperienced forwards were overmatched in the moment, but they were Detroits best puck possession trio in the series, all three on for better than 58% of shot attempts at 5-on-5. Yet, there was ultimately some dissatisfaction from the Wings, who needed goals, not possession. In the series-deciding game, which Detroit trailed for 42 minutes, Nyquist played only 11:55, a threshold he surpassed in 44 of the previous 45 games. Detroits uphill battle was made steeper by the fact that G Jimmy Howard missed the last two games with a suspected concussion. Jonas Gustavsson actually played well, stopping 66 of 72 shots (.917), but its tough to upend the Bruins with a second-string goaltender. Datsyuk, playing on a bad knee, finished with five points in five games and was the only Red Wing with more than two points in the series. The Bruins emerged from the series, relatively easily, even though David Krejcis line struggled in possession terms. It certainly didnt hurt to have G Tuukka Rask at the top of his game, stopping 146 of 152 shots (.961) in the series. C Patrice Bergeron and Krug tied for the Bruins team lead with five points in the series. Lucic, RW Jarome Iginla and D Dougie Hamilton each had four points. That the Bruins young defencemen were able to contribute offensively proved to be a real advantage. The series win sets up Boston for a matchup with their rivals from Montreal. The top-seeded Bruins will be favoured, but the head-to-head matchups with the Canadiens have been tumultuous, dramatic affairs (well, forever, but also) in recent seasons, so it should be a fascinating battle. PENGUINS TAKE SERIES LEAD Perhaps feeling a little threatened, with the series tied 2-2, the Pittsburgh Penguins came out firing in Game Five and dominated play on their way to a 3-1 win over the Columbus Blue Jackets. The margin could have been much if not for the play of Blue Jackets G Sergei Bobrovsky, who stopped 48 of the 50 shots he faced. As the Penguins made a concerted effort to play Sidney Crosby and Evgeni Malkin together on the top line, it turned out that Pittsburghs best possession numbers came from the second and third lines. Lee Stempniak, Brandon Sutter, Jussi Jokinen and Beau Bennett were each on the ice for more than 70% of the 5-on-5 shot attempts. Pittsburghs depth was also boosted by the presence of C Marcel Goc, who played 10:34 in his first game since suffering an ankle injury March 27. The possession domination for Pittsburgh, naturally, had a counter in Columbus, where Jack Skille, R.J. Umberger, Nick Foligno and Artem Anisimov were each below 30%. Penguins D Brooks Orpik missed the game with an undisclosed injury, giving Robert Bortuzzo an opportunity to play his first NHL playoff game and Bortuzzo didnt hhurt the Penguins in 13:36 of ice time.dddddddddddd While there has been criticism of Sidney Crosby for his play in this series, it needs to be said that his play is probably better than fine. After 11 individual shot attempts, and one assist, in Game Five, Crosby has five points and a in five games and has been on the ice for 60.1% of the shot attempts during 5-on-5 play. His LW, Chris Kunitz, had a goal and an assist, with seven shots on goal and six hits, in Game Five. This hasnt been an easy series for the Penguins, by any stretch, and theyre still trying to do it with the second-best goaltender in the series, but their skill advantage has been enough to give them the lead in the series. ROCKY MOUNTAIN WAY Trailing 2-1 into the third period of Game Five, the Minnesota Wild scored twice in the first 6:25 of the third period to take the lead and appeared on their way to taking the game, before the Avalanche scored with 1:14 remaining to tie it, then finished matters 3:27 into overtime. In the end, the 4-3 overtime win gave the Avalanche a 3-2 series lead. The winning goal was authored by Avalanche rookie Nathan MacKinnon, who finished the game with a goal and two assists, giving him a playoff-leading 10 points thus far. Hes three points away from Jaromir Jagrs scoring record for an 18-year-old rookie in the postseason. MacKinnon played 21:56 in Game Five, second among Colorado forwards, behind only Paul Stastny (22:38); MacKinnon, who averaged 17:21 of ice time per game during the regular season, has played more than 21 minutes in four of five games in the series. The sequence leading up to the game-tying goal was contentious. There was an apparent holding penalty that was ignored in Colorados defensive zone, with the net empty, and then the Avalanche appeared to be offside seconds before PA Parenteau buried the tying goal. This is what is known as "getting the breaks." Avalanche LW Cody McLeod scored a goal, but was obliterated at even-strength, on for two shot attempts for and 17 against (10.5%). With McLeod off the ice, the Avalanche had 56.2% of the 5-on-5 shot attempts. For the Wild, LW Matt Moulson, C Mikko Koivu and D Marco Scandella were on the ice for at least two-thirds of the shot attempts at 5-on-5, though Scandella was unfortunate enough to get beaten by McKinnon on the winning goal. Its been a close series to this point, though the Wild have held the possession edge, and it shifts back to Minnesota for Game Six. Colorado may get a boost with C Matt Duchene nearing a return to the lineup. If Duchene is ready to go, that ought to help Colorados cause. KINGS KEEP ON KEEPING ON Facing elimination for the second time in the series, the Los Angeles Kings stormed into San Jose Saturday night and outplayed the Sharks from the get-go, building a 3-0 lead 22 seconds into the second period before chilling out and taking that lead to games end. The Kings had outshot the Sharks 19-6 by the time they had taken that 3-0 lead, chasing Sharks G Antti Niemi. Not only did the Sharks lose the game, but also lost D Marc-Edouard Vlasic was hurt (upper-body injury) on a run-in with Kings C Jarret Stoll. It cost Stoll two minutes for roughing, but Vlasic did not return. If hes not able to play in Game Six, thats a huge blow to the Sharks.. Kings RW Justin Williams had a game-high seven shots on goal (11 shot attempts), leading the Kings on ice for 73.1% of shot attempts. Kings G Jonathan Quick stopped all 30 shots sent in his direction, a decided improvement after some rough games earlier in the series. While the Kings are trying to make the unlikeliest of comebacks, from being down 3-0, if they manage to get good goaltending from Quick, maybe its possible. Scott Cullen can be reached at Scott.Cullen@bellmedia.ca and followed on Twitter at http://twitter.com/tsnscottcullen. For more, check out TSN Fantasy on Facebook. ' ' '
ns, and former world No. 1 Venus Williams posted o » « Playing in his fourth quarterfinal of the season,


Slim, sleek, and functional, the leather Hailu is the perfect wallet. It features 8 card slots + 2 slip pockets for cash, receipts, or a checkbook. The back exterior pocket contains a removable zippered coin pouch, and it is also the perfect size to store an iPhone.


text:
\section{Introduction}

	Baumslag-Solitar groups are two-generator one-relator groups given by the presentation
	\[ BS(n,m) =  \langle a,b \;\lvert \; ba^n =  a^m b\rangle,\]
	for $n,m\in\mathbb{Z}$. They were introduced in 1962 by Baumslag and Solitar\footnote{The group $BS(1,2)$ was already around at the time and appears in a 1951 paper by Higman \cite{higman}.} to describe the first examples of non-Hopfian finitely generated one-relator groups. A group is Hopfian whenever $G/N\cong G$ implies that $N=\{1\}$. The specific result \cite{BS} states that $BS(n,m)$ is Hopfian if and only if
	\begin{itemize}
		\item $n$ divides $m$ or vice versa; or
		\item $n$ and $m$ have the same prime divisors.
	\end{itemize}
	There were some errors in the original proofs. This issue was later resolved by Collins and Levin in \cite{collinslevin}. It is also possible to determine which of these groups are residually finite \cite{meskin}. This is the case when $\lvert m \rvert = \lvert n \rvert$, or $\lvert n\rvert =1$ or $\lvert m \rvert =1$. A last interesting remark is that, whenever $\lvert n\rvert, \lvert m\rvert \ne 1$, these groups are HNN extenstions of $\mathbb{Z}$ with respect to the automorphism $\mu: n\mathbb{Z}\rightarrow m\mathbb{Z}$, sending $n$ to $m$.\newline

	We only focus on the Baumslag-Solitar group $BS(2,3)= \langle a,b \;\lvert \; ba^2 =  a^3b\rangle$ and the following homomorphism which is surjective, but not injective:
	\[   \phi :  BS(2,3) \rightarrow BS(2,3) :
	\begin{cases}
	a \mapsto a^2\\
	b\mapsto b \\
	\end{cases}
	. \]
	The element $[a^b,a]$ is non-trivial, and inside the kernel of $\phi$. As mentioned before we notice that $BS(2,3)/ \Ker(\phi)$ is isomorphic to $BS(2,3)$. And thus their Cayley graphs are isomorphic up to change of generators. We will first describe this quotient from a graph-viewpoint. Our other aim is to show that $\Ker(\phi)$ is a free group of infinite rank. The proof is geometric. We will give an explicit set of generators and show that their action on the Bass-Serre tree associated to $BS(2,3)$ is free. Finally we present how $\phi$ appears as a morphism on fundamental groups, induced by a continuous map.\newline

	It is useful to remind ourselves of Britton's lemma. We do not give its general version, but directly apply it to $BS(2,3)$.

	\begin{lemma}[Britton's lemma]
	Let $\omega= a^{\alpha_0}\prod_{i=1}^{n} b^{\beta_i}a^{\alpha_i}$ in $BS(2,3)$ such that $\beta_i\ne 0$. If $\omega =1$, then either
	\begin{itemize}
	\item $n=0$ and $\alpha_0=0$;
	\item or $n>0$  and $\exists i\in \{1\dots n-1\} $ such that
	\begin{enumerate}
	\item $\beta_i>0,  \beta_{i+1}<0$ and $\alpha_{i}\in 2^{\min\{\beta_i,\lvert\beta_{i+1}\rvert\}} \mathbb{Z}$;
	\item $\beta_i<0,  \beta_{i+1}>0$ and $\alpha_{i}\in 3^{\min\{\lvert\beta_i\rvert,\beta_{i+1}\}} \mathbb{Z}$.
	\end{enumerate}
	\end{itemize}
	\end{lemma}

	\section*{Acknowledgements}
	The author would like to thank Gilbert Levitt for some very interesting comments. In particular the point of view presented in Section \ref{levittpres} is due to Levitt, as is Figure \ref{Levittpic}.

	\section{An interpretation of non-Hopfianness and a quotient of graphs}

	We first calculate the kernel of $\phi$, then remind the reader what the Cayley graph of $BS(2,3)$ looks like, in order to finally describe its quotient induced by the morphism $\phi$.

	\begin{definition}
	Given a word $\omega= a^{\alpha_0} \prod_{i=1}^{n} a^{\alpha_i}b^{\beta_i}$ in $\{a,b\}^*$, define $\rho(\omega)=\sum_{i=1}^{n} \beta_i$ and $\rho_{a}(\omega) = \sum_{i=1}^{n} \lvert\beta_i\rvert$.
	\end{definition}

	Note that, as a consequence of Britton's lemma, we see that if $\omega$ represents the trivial element, then $\rho(\omega)=0$.

	\begin{lemma}
		$\Ker(\phi)$ is normally generated by $\{ [a^b,a], [a^b, a^{-1}], [a^b,a^2], [a^b, a^{-2}]  \}$.
	\end{lemma}
	\begin{proof}

	Consider a word $\omega$ in the generators of $BS(2,3)$. If $\omega\in \Ker(\phi)$, then $
	\rho(\omega)=0$, since $\rho(\phi(\omega))=0$ and $\phi$ does not affect $b$. Given that a proof is a sequence of trivialities, we can
	check the following:

	\begin{itemize}
		\item If $\rho_a(\omega)=2$, then an easy calculation shows that if $\phi(\omega)=1$, then  $\omega$ represents the trivial element.
		\item If $\rho_a(\omega)=4$, one can show that $\omega$ represents one of the generators given above (up to
		cyclic permutation of the letters).
		\item Consider the group element $b^\beta a^\alpha b^{\beta'}$, where $\beta,\beta'>0$. By Bachet-Bezout, there exist $\lambda,\mu\in \mathbb{Z}$ such that $\lambda 2^\beta + \mu 3^{\beta'} =1$. Hence $b^\beta a^\alpha b^{\beta'} = b^\beta a^{\alpha\lambda 2^\beta + \alpha\mu 3^{\beta'}} b^{\beta'} $, which equals $a^{\alpha\lambda 3^{\beta} } b^{\beta+\beta'} a^{\alpha\mu 2^{\beta'} }$.
		\item By repeatedly applying the previous item, any word $\omega$ (up to cyclic permutation) can be replaced by a word
		$\omega'$ that represents the same element in $BS(2,3)$ and is of the form $a^{\alpha_1}b^{\beta_1}a^{\alpha_1}b^{\beta_2}\dots a^{\alpha_n}b^{\beta_n} $, where none of the exponents are zero and exponents of subsequent $b$'s have opposite sign.
		\item We specialise even more, in the sense that we minimise the exponents $\beta_i$. This means that
		 $ba^{2\alpha} b^{-1}$ can be replaced by $a^{3\alpha}$ and $b^{-1} a^{3\alpha}b$ by $a^{2\alpha}$.
		\item So suppose $\omega$ is of the prescribed minimal form, such that $\rho_a(\omega)>4$ and is in
		$\Ker(\phi)$, then the image is  $a^{2\alpha_1}b^{\beta_1}a^{2\alpha_2}b^{\beta_2}\dots a^{2\alpha_n}b^{\beta_n} $.  We apply Britton's lemma. Suppose there is some $i\in\{ 1\dots n\}$ such that $\beta_i>0,  \beta_{i+1}<0$ and $2\alpha_{i}\in 2^{\min\{\beta_i,\lvert\beta_{i+1}\rvert\}} \mathbb{Z}$. Without loss of generality we suppose $\beta_i$ is the minimum. Then $\alpha_i= 2^{\beta_i-1}\alpha_i'$, and thus $b^{\beta_i} a^{\alpha_i}b^{\beta_{i+1}}= b a^{3^{\beta_i-1}\alpha_i'} b^{\beta_{i+1}+\beta_i-1}$. Note that this contradicts minimality of the word $\omega$ if $\beta_i>1$. The reader can check that a similar contradiction is found for the second case of Britton's lemma.
		\item Because of the previous point, there is some $b$ whose exponent is either $1$ or $-1$. Up to taking inverses and cyclicly permuting the word, we may suppose it is $-1$ and it appears on the second position, i.e. $\omega$ can be chosen of the form $b^{\beta_1}a^{\alpha_2}b^{-1}a^{\alpha_3} b^{\beta_3}\dots$
			\item Adding minimality to the argument, this can be rewritten as
			\[b^{\beta_1 - 1} a^{3(\frac{\alpha_2-1}{2}) + \alpha_3 -t} b a b^{-1} a^{t}  b a^{-1}b^{-1 } a^{- t}  a^{t} b a b^{\beta_3 -1} \dots\]
			where $t$ is either $1$ or $2$, such that $\alpha_3-t\in 3\mathbb{Z}$. Note that the first part of our rewritten word is exactly  a generator described in the statement of the lemma. Suppose $\omega'$ is the remaining word, after deleting the generator. One notices that $\rho_a(\omega')=\rho_a(\omega)-2$. Hence an induction argument based on the value of $\rho_a$ finishes the proof.
		\end{itemize}

	\end{proof}

	\begin{corollary}
		$\Ker(\phi)$ is normally generated by $[a^b,a]$.
	\end{corollary}
	\begin{proof}
		Since all other generators can by normally generated by $[a^b,a]$:
		\begin{itemize}
			\item $[a^b,a^{-1}] = a^{-1}[a^b,a]^{-1}a$,
			\item $[a^b,a^{2}] = [a^b,a] a [a^b,a]a^{-1}$, and
			\item $[a^b,a^{-2}] = a^{-1}[a^b,a]^{-1} a^{-1} [a^b,a]^{-1} a^2$.
		\end{itemize}
	\end{proof}

	A description of $BS(1,2)$ and its Cayley graph is a pretty standard introduction to the world of Baumslag-Solitar groups (f.e. \cite{meier2008groups}). For self-containment purposes, we quickly describe the Cayley graph of $BS(2,3)$, which follows a slight variation of the classical example $BS(1,2)$. The standard building block is given in Figure \ref{buildingblock}, where red edges are labeled by $a$ and blue ones by $b$. This is the cycle induced by the single relator. These building blocks fit nicely together into an upper-half plane given in Figure \ref{plane}, based at a copy of $\mathbb{Z}$ generated by $a$. We may now construct the Cayley graph. Consider the base line generated by $a$, then we can insert three planes here such that upward arrows do not overlap. One can see this as attaching a plane based at $1$, $a$ and $a^2$. In a similar way we construct lower half planes (one can see this by interpreting Figure \ref{plane} with the baseline on top). Two of these can be attached at $\mathbb{Z}$ such that blue arrows do not overlap. Locally this looks like Figure \ref{side3dview}. In general consistenly applying this construction to each left coset of $\langle a \rangle$, which all appear as a Cayley graphs of $\mathbb{Z}$, gives us the Cayley graph of $BS(2,3)$. Now project such that each left coset of $\langle a \rangle$ becomes a point, we obtain a tree of degree five, as can be seen in Figure \ref{sideview}. This is the Bass-Serre tree associated to the HNN extension $BS(2,3) = \text{HNN}(\mathbb{Z},2\mathbb{Z}\sim 3\mathbb{Z})$. Choosing a specific vertex of this tree, one sees that three edges are directed upward and two downward (the halfplanes attached upward and downward respectively). This `side view' will be important for our visual interpretation later. We first define the following sets.

\begin{minipage}{\linewidth}
		\centering
		\begin{minipage}{0.3\linewidth}
			\begin{figure}[H]
				\centering
				\begin{tikzpicture}
				\filldraw (0,0) circle (1.5pt);
				\draw[->-, very thick,red] (0.1,0) -- (1.4,0);
				\filldraw (1.5,0) circle (1.5pt);
				\draw[->-, very thick,red] (1.6,0) -- (2.9,0);
				\filldraw (3,0) circle (1.5pt);
				\draw[->-, very thick,red] (3.1,0) -- (4.4,0);
				\filldraw (4.5,0) circle (1.5pt);

				\draw[->-, very thick,blue] (0,0.1) -- (0,1.4);
				\filldraw (0,1.5) circle (1.5pt);
				\draw[->-, very thick,blue] (4.5,0.1) -- (4.5,1.4);
				\filldraw (4.5,1.5) circle (1.5pt);

				\draw[->-, very thick,red] (0.1,1.5) -- (2.15,1.5);
				\filldraw (2.25,1.5) circle (1.5pt);
				\draw[->-, very thick,red] (2.35,1.5) -- (4.4,1.5);
				\end{tikzpicture}
				\caption{Building block of $BS(2,3)$.}
				\label{buildingblock}
			\end{figure}
		\end{minipage}
		\begin{minipage}{0.64\linewidth}
			\begin{figure}[H]
				\centering
				\begin{tikzpicture}[scale=0.6]
				\filldraw (0,0) circle (1.5pt)
				\draw[->-, very thick,red] (0.1,0) -- (1.4,0);
				\filldraw (1.5,0) circle (1.5pt);
				\draw[->-, very thick,red] (1.6,0) -- (2.9,0);
				\filldraw (3,0) circle (1.5pt);
				\draw[->-, very thick,red] (3.1,0) -- (4.4,0);
				\filldraw (4.5,0) circle (1.5pt);
				\draw[->-, very thick,red] (4.6,0) -- (5.9,0);
				\filldraw (6,0) circle (1.5pt);
				\draw[->-, very thick,red] (6.1,0) -- (7.4,0);
				\filldraw (7.5,0) circle (1.5pt);
				\draw[->-, very thick,red] (7.6,0) -- (8.9,0);
				\filldraw (9,0) circle (1.5pt);
				\draw[->-, very thick,red] (9.1,0) -- (10.4,0);
				\filldraw (10.5,0) circle (1.5pt);
				\draw[->-, very thick,red] (10.6,0) -- (11.9,0);
				\filldraw (12,0) circle (1.5pt);
				\draw[->-, very thick,red] (12.1,0) -- (13.4,0);
				\filldraw (13.5,0) circle (1.5pt);

				\draw[->-, very thick,blue] (0,0.1) -- (0,1.4)
				\filldraw (0,1.5) circle (1.5pt);
				\draw[->-, very thick,blue] (4.5,0.1) -- (4.5,1.4);
				\filldraw (4.5,1.5) circle (1.5pt);
				\draw[->-, very thick,blue] (9,0.1) -- (9,1.4);
				\filldraw (9,1.5) circle (1.5pt);
				\draw[->-, very thick,blue] (13.5,0.1) -- (13.5,1.4);
				\filldraw (13.5,1.5) circle (1.5pt);

				\draw[->-, very thick,red] (0.1,1.5) -- (2.15,1.5)
				\filldraw (2.25,1.5) circle (1.5pt);
				\draw[->-, very thick,red] (2.35,1.5) -- (4.4,1.5);

				\draw[->-, very thick,red] (4.6,1.5) -- (6.65,1.5);
				\filldraw (6.75,1.5) circle (1.5pt);
				\draw[->-, very thick,red] (6.85,1.5) -- (8.9,1.5);

				\draw[->-, very thick,red] (9.1,1.5) -- (11.15,1.5);
				\filldraw (11.25,1.5) circle (1.5pt);
				\draw[->-, very thick,red] (11.35,1.5) -- (13.4,1.5);

				\draw[->-, very thick,blue] (0,1.6) -- (0,2.9)
				\filldraw (0,3) circle (1.5pt);

				\draw[->-, very thick,blue] (6.75,1.6) -- (6.75,2.9);
				\filldraw (6.75,3) circle (1.5pt);

				\draw[->-, very thick,blue] (13.5,1.6) -- (13.5,2.9);
				\filldraw (13.5,3) circle (1.5pt);

				\draw[->-, very thick,red] (0.1,3) -- (3.275,3)
				\filldraw (3.375,3) circle (1.5pt);
				\draw[->-, very thick,red] (3.475,3) -- (6.65,3);

				\draw[->-, very thick,red] (6.85,3) -- (10.025,3);
				\filldraw (10.125,3) circle (1.5pt);
				\draw[->-, very thick,red] (10.225,3) -- (13.4,3);

				\draw[->-, very thick,blue] (0,3.1) -- (0,4.4)
				\filldraw (0,4.5) circle (1.5pt);

				\draw[->-, very thick,blue] (10.125,3.1) -- (10.125,4.4);
				\filldraw (10.125,4.5) circle (1.5pt);

				\draw[->-, very thick,red] (0.1,4.5) -- (4.9625,4.5)
				\filldraw (5.0625,4.5) circle (1.5pt);
				\draw[->-, very thick,red] (5.1625,4.5) -- (10.025,4.5);

				\draw[->-, very thick,red] (10.225,4.5) -- (13.5,4.5);
				\end{tikzpicture}
				\caption{Construction of a plane in $BS(2,3)$.}
				\label{plane}
			\end{figure}
		\end{minipage}
	\end{minipage}
	\newline

	\begin{minipage}{\linewidth}
			\centering
			\begin{minipage}{0.25\linewidth}
				\begin{figure}[H]
					\centering
					  \includegraphics[width=\linewidth]{side3dview.pdf}
					\caption{Side 3D view of $BS(2,3)$.}
					\label{side3dview}
				\end{figure}
			\end{minipage}
			\begin{minipage}{0.64\linewidth}
							\begin{figure}[H]
								\centering
								\includegraphics[width=\linewidth]{sideview.pdf}
								\caption{Projection of cosets of $\langle a\rangle$.}
								\label{sideview}
							\end{figure}
						\end{minipage}
		\end{minipage}
		\newline

	We define the height of an element to be its image by the map
	\[  \alpha: BS(2,3) \rightarrow \mathbb{Z}: \left\{  \begin{array}{lll}
	a &\mapsto& 0\\
	b &\mapsto& 1
	\end{array}\right.,  \]
	and we define what it means for two cosets to be neighbours.
	\begin{definition}
		Two left cosets $\lambda$ and $\mu$ of $\langle a \rangle$ are same height neighbours if for $l\in \lambda$ either $l a^b$, $l a a^b$ or $l a^2 a^b$ are in $\mu$. Notation: $\lambda \sim \mu$.
	\end{definition}
	If we consider the graph defined by those cosets as vertices and edges between two cosets that are same height neighbours, then we obtain a forest (read: disjoint union) of $3$-valent trees. Note that such a tree has a unique $2$-colouring. This colouring can be seen in Figure \ref{coloringH} and thus also the visual interpretation of the same height neighbour relation becomes clear. Presented is a limited sideview of $BS(2,3)$, where the cosets are coloured by the $2$-colouring. Two coloured points (representing left cosets of $\langle a \rangle$) that are two edges apart, are neighbours. From the visual it is clear that a point has exactly three neighbours. Now we can define the following sets:
	\begin{definition}
		Given $\lambda$ a left coset of $\langle a \rangle$, then
		\begin{itemize}
			\item $H_\lambda$ is the connected component of $\lambda$ of the forest induced by the same height neigbour relation;
		\end{itemize}
		Since $H_\lambda$ is a $3$-valent tree it has a unique $2$-colouring.
		\begin{itemize}
			\item  $H_\lambda^+$ are the cosets in $H_\lambda$, that have the same colour as $\lambda$;
			\item $H_\lambda^-:= H_\lambda\backslash H_\lambda^+$.
		\end{itemize}
	\end{definition}
		Note that if $\mu \in H_\lambda^-$, then clearly $H_\lambda^- =  H_\mu^+$.

\begin{figure}
  \includegraphics[width=\linewidth]{coloringH.pdf}
  \caption{Coloring of $H_\lambda$.}
  \label{coloringH}
\end{figure}

	We now have the tools to show what happens when we quotient by $\Ker(\phi)$. Write \linebreak $G=  BS(2,3)/ \Ker(\phi)$, which we will not consider as an abstract group, but rather as the specific manifestation of $BS(2,3)$ as a quotient of itself. Note that $\phi(a)=a^2$. This means that $\overline{a}$, in $G$, will behave like $a^2$ in $BS(2,3)$. In a sense we would like to take the root of $\overline{a}$ (note that the reason for this awkward phrasing comes from the visual interpretation we will see soon). This is why we look for a pre-image of $a$, one can take for example the commutator $[b,a]= bab^{-1}a^{-1}$. So the specific group element $\overline{[b,a]}$ in  $G$ will behave as the abstract group element $a$ in $BS(2,3)$. For brevity we write $\tilde{a}:= \overline{[b,a]}$.\newline

	\begin{figure}
		\centering
	\begin{tikzpicture}
	\foreach \x in {0,...,3
	{\draw[->-, very thick,blue] (4*\x,0) -- (4*\x,3);
		\draw[->-, very thick,blue] (4*\x,0) -- (4*\x +2,3);
		\draw[->-, very thick,blue] (4*\x,0) -- (4*\x -2,3);
		\filldraw[white] (4*\x,3) circle (2.5pt);
		\filldraw[white] (4*\x+2,3) circle (2.5pt);
		\filldraw[white] (4*\x-2,3) circle (2.5pt);
		\filldraw (4*\x,3) circle (1.5pt);
		\filldraw (4*\x+2,3) circle (1.5pt);
		\filldraw (4*\x-2,3) circle (1.5pt);
		\filldraw[white] (4*\x,0) circle (2.5pt);
		\filldraw (4*\x,0) circle (1.5pt);
	}

	\draw[->-,green,very thick] (0,0) to[bend right=20] (2,3);
	\draw[->-,green,very thick] (2,3) to[bend right=20] (4,0);
	\draw[->-,green,very thick] (4,0) to[bend right=20] (6,3);
	\draw[->-,green,very thick] (6,3) to[bend right=20] (8,0);
	\end{tikzpicture}
	\caption{The identification of the element $[a^b,a]$.}
	\label{treething}
	\end{figure}

	We will construct the quotient graph by identifying points in the original graph, following a principle we will call `horizontal identification'. Let us see which effect, identifying two points that are at a `distance' $\omega=[a^b,a^{-1}]$ has. We follow the path (in green) of $\omega$ in Figure \ref{treething}. Suppose we start on the line $\langle a \rangle$, then we go up a level to $b\langle a \rangle $. Next we move over an edge labelled $a$, which means that when we take the edge $b^{-1}$, we do not take the same edge back (i.e. we follow the second green arrow). After this, moving by $a^{-1}$, we go up one of the two other planes. Then we go down again. Note that since we identify two  points of $\langle a \rangle$  and $bab^{-1}a^{-1}ba^{-1}b^{-1} \langle a \rangle$, in fact those lines are completely identified. By symmetry all cosets of $H_{\langle a \rangle}^+$, will become equivalent to $\langle a \rangle$ in $G$. Note that by symmetry also all lines in $H_\lambda^-$ are identified with one another. Let us consider the behaviour of $\tilde{a}$. Note that it is exactly equal to the the first four letters of $\omega$. This means that we are basically following the first two green arrows of $\omega$ in Figure \ref{treething}. Next we apply $\tilde{a}$ a second time, we arrive on the line $\omega\langle a \rangle$, which has been identified to $\langle a \rangle$. Specifically the point we arrive in is identified with $a$, since $\tilde{a}=\overline{a}$. One notices that viewing $G$ wrt the new generators $\{\overline{b},\tilde{a} \}$ will give us an intertwining of the lines $\langle a \rangle$ and $bab^{-1} \langle a \rangle$. With respect to the sets $H_{\langle a \rangle}^+$ and $H_{\langle a \rangle}^-$, which are each just one line in $G$, this means that the generator $\tilde{a}$ will alternate points of $H_{\langle a \rangle}^+$ and $H_{\langle a \rangle}^-$. The result can be seen in Figure \ref{horizontalidentification}. Here points on the lower line are obtained by alternating points of $\langle a\rangle$ and $[b,a^{-1}]\langle a \rangle$, where we start by $1$ and $[b,a]$ respectively. Out of this new `baseline' once again three planes open up. We describe one, since the other two are analogous by symmetry. Consider the plane based at $\overline{1}$ in Figure \ref{horizontalidentification}. We take one step back and look at Figure \ref{loweridentification}, before identification. Then we see that the intertwining respects the upper brown points, in the sense that if the first green point from the left is $1$, then the first yellow point from the left is $[b,a]$. This means that the fourth point is exactly $[b,a]a$. Which is underneath the second brown point. This means that, after identification, a position opens up in between consecutive brown points. This makes sense, since by symmetry of the cayley graph, also the line $H_{b\langle a \rangle}^+$, must intertwine with $H_{b\langle a \rangle}^-$. This fills up the gap. Note that points in $H_{b\langle a \rangle}^-$ are exactly the orange points in Figure \ref{horizontalidentification}.

\begin{figure}
  \includegraphics[width=\linewidth]{loweridentification.pdf}
  \caption{Two lines are interleaved if at a `distance' $bab^{-1}a^{-1}$.}
  \label{loweridentification}
\end{figure}

\begin{figure}
  \includegraphics[width=\linewidth]{Horizontalidentification.pdf}
  \caption{A part of the Cayley graph of $G$, where $\tilde{a}$ is now a generator. Choices of colours are consistent with previous  drawings.}
  \label{horizontalidentification}
\end{figure}

Note that since $\Ker(\phi)$ is normally generated by $[a^b,a]$, this is the only `real' identification that takes place. Other identifications are due to taking conjugates of this element. I.e. if two cosets of $\langle a\rangle$ are identified, of course the tree structure above and below it must become identified too.

\section{A free group of infinite rank}

Let us now prove that the kernel of $\phi$ is a free group of infinite rank. In order to do this in an accessible way, we first introduce some terminology. For illustrations of the concepts, also see Figure \ref{paths}.

\begin{definition}
	Let $\mathcal{P}$ be the set of geodesic paths starting at $\langle a\rangle$ in the Bass-Serre tree in Figure \ref{sideview} (these are the left cosets of $\langle a \rangle$). Let $\{ a,b \}^*$ be all words in $a,b$ and their inverses, and $n,m\in\mathbb{Z}$. We define:
	\begin{enumerate}
		\item Let $\omega\in \{ a,b \}^*$, then the associated path in the Bass-Serre tree is the projection of the associated path in the Cayley graph of $BS(2,3)$. This path may contain backtracking.
		\item A good representative of a path $p\in\mathcal{P}$ is a word $\omega$ such that $p$ is its associated path in the Bass-Serre tree. Equivalently: such that $\rho_a(\omega)$ is minimal and $\omega$ represents the left-coset at the end of $p$.
		\item A path contains a tip if a good representative $\omega$ contains a subword of the form $ba^n b^{-1}$.
		\item A path contains a valley if a good representative $\omega$ contains a subword of the form $b^{-1}a^n b$.
		\item A path is end-essential if all good representatives do not end in a tip.
		\item A path $p\in\mathcal{P}$ is swiss if a good representative $\omega$ contains a subword of the form $b^{-1}a^n ba^mb^{-1}$ or $ba^nb^{-1}a^m b$. A path that is not swiss is called nepalese.
		\item Let $\omega$ be a good representative of $p\in \mathcal{P}$, an end-essential nepalese path with a tip. Let $\omega_1$ be the first part of $\omega$, let $B$ be the tip and $\omega_2$ the remainder. Then $p$ has two triplets at $B$, namely the paths associated to $\omega_1 a B a^{-1} \omega_2$ and $\omega_1 a^{-1} Ba \omega_2$.
		\item Let $\omega$ be a good representative of $p\in \mathcal{P}$, an end-essential nepalese path with a valley. Let $\omega_1$ be the first part of $\omega$, let $V$ be the valley and $\omega_2$ the remainder. Then $p$ has a twin at $V$, namely the path associated to $\omega_1 a^{-1} V a b^{-1} \omega_2$.
		\item Let $\sim$ be the sibling relation defined on end-essential nepalese paths, induced by setting $p\sim q$ if they belong to the same set of triplets or the same set of twins.
		\item We denote by $G_\sim$ the graph induced by the sibling relation on end-essential nepalese paths. Two paths in the same connected component are called relatives.
		\item For a path $p\in\mathcal{P}$ define the function $c:\mathcal{P}\rightarrow\mathbb{N}$ by setting $c(p)$ equal to the number of valleys and tips.
 	\end{enumerate}
\end{definition}

We note that the sibling relation is not an equivalence relation, since it is not transitive and not reflexive. But since it is symmetric $\sim$ clearly defines a graph on the set of end-essential nepalese paths. This graph is not connected. We show the following.

\begin{figure}
	\centering
	\subcaptionbox{Swiss.\label{swiss}}[0.45\linewidth][l]
	{\includegraphics[width=0.45\textwidth]{Swisspath.pdf}}
	\hspace{20pt}
	\subcaptionbox{Nepalese.\label{nepalese}}[0.45\linewidth][r]
	{\includegraphics[width=0.45\textwidth]{Nepalesepath.pdf}}
	\caption{An artistic interpretation of a swiss and nepalese path. The reader should think of these paths as paths in the Bass-Serre tree \ref{sideview}. Note that the swiss path has valleys and tips that share an edge, while for the nepalese path they are disjoint. Moreover, in this example, the nepalese path is end-essential while the swiss path is not.}\label{paths}
\end{figure}

\begin{theorem}
	$\Ker(\phi) \simeq \mathbb{F}_\infty$. In particular, let $\mathcal{O}$ be a set of words in $\{a,b\}^*$ such that
	\begin{itemize}
		\item Each word represents an end-essential nepalese path and each such path is represented by at most one word.
		\item For each connected component of $G_\sim$, there is exactly one path that is represented by a word in $\mathcal{O}$.
	\end{itemize}
	Then the symmetric set
	\[ \mathcal{S}= \{ \omega a^{i} [a^b,a]^{j} a^{-i}\omega^{-1} \;\vert\; j\in \{-1,1 \},i\in\{ 0,1\}, \omega\in\mathcal{O}\} \]
	freely generates  $\Ker(\phi)$. In particular this shows that the action of $\Ker(\phi)$ on the Bass-Serre tree is free.
\end{theorem}
\begin{proof}
	{\em Generation.}
	We first show that $\mathcal{S}$ is generating. Let $\tilde{\mathcal{O}}$ be a set containing for each geodesic path in the Bass-Serre tree exactly one word which is a good representative. Then it can easily be seen that $\Ker(\phi)$ is generated by
	\[\mathcal{S}= \{ \omega a^{i} [a^b,a]^{j} a^{-i}\omega^{-1} \;\vert\; j\in \{-1,1 \},i\in\{ 0,1 \}, \omega\in\tilde{\mathcal{O}}\}. \]
	Hence we can reduce the number of paths needed. Suppose $\omega$ represents a swiss path, then it is of the form $\omega_1 bab^{-1}ab\omega_2$ or $\omega_1 b^{-1}abab^{-1}\omega_2$ (up to possibly taking other exponents of the $a$'s). We treat these two cases for $i=0$ and $j=1$, the other cases are similar. Note that for the first calculation we consider the conjugated group element and in the second we only consider the group element representing the path. As we are conjugating the paths this leads to redundant notation. We write the first in full for the readers convenience.
	\begin{align*}
		\omega [a^b,a] \omega^{-1} 	&= 	\omega_1 bab^{-1}ab\omega_2 [a^b,a] \omega_2^{-1} b^{-1}a^{-1}ba^{-1}b^{-1}\omega_2\\
									&=	\omega_1 [a^b,a] \omega_1^{-1} \omega_1 aba \omega_2 [a^b,a] \omega_2^{-1} a^{-1}b^{-1} a^{-1} \omega_1^{-1} \omega_1 [a^b,a]^{-1} \omega_1^{-1}, \text{ or }\\[10pt]
		\omega 	&= 	\omega_1 b^{-1}abab^{-1}\omega_2 \\
									&=	\omega_1 a^{-1} b^{-1} [a^b,a] ba \omega_1^{-1} \omega_1 a^{-1}b^{-1}a^4 \omega_2 .
 	\end{align*}
 	Hence we have reduced the problem to the shorter paths $\omega_1$ and $\omega_1 aba \omega_2$, and $\omega_1 a^{-1} b^{-1}$ and \linebreak $\omega_1 a^{-1}b^{-1}a^4 \omega_2$ respectively. An induction hypothesis on the length of the paths finishes the job.\newline

 	Hence we may suppose our paths are nepalese. We suppose they are not end-essential. So they are represented by a word $\omega$ of the form $\omega_1 bab^{-1}$. Note that the element $bab^{-1}a^i[a^b,a]a^{-i}ba^{-1}b^{-1}$ is in the subgroup $\mathbb{F}_2$ generated by $\{ [a^b,a], a[a^b,a]a^{-1}\}$. Hence, because of its specific form, it can be written as one of the generators (or their inverses), or a product of at most four generators (or their inverses). Say $g_1$ or $g_1g_2g_3g_4$. In particular  we can rewrite $\omega a^i [a^b,a]a^{-i} \omega^{-1}$ as $\omega_1 g_1 \omega_1^{-1}$ or $\omega_1 g_1\omega_1^{-1} \omega_1 g_2 \omega_1^{-1}\omega_1 g_3 \omega_1^{-1}\omega_1 g_4 \omega_1^{-1}$, where $\omega_1$ is an end-essential path.\newline

 	\begin{figure}
 		\centering
 		\subcaptionbox{Twins.\label{twinpaths}}[0.45\linewidth][l]
 		{\includegraphics[width=0.45\textwidth]{Twinpaths.pdf}}
 		\hspace{20pt}
 		\subcaptionbox{A group element in $\Ker(\phi)$ represented as a path.\label{conjugate}}[0.45\linewidth][r]
 		{\includegraphics[width=0.45\textwidth]{conjugatepath.pdf}}
 		\caption{On the left: The black path (correspongding to $\omega$) has a valley $V$. Its twin at $V$ is given by the blue path. Note that the red paths corresponds to $\omega_1 a^{-1}b^{-1}$ and $ba\omega_1^{-1}$, and the orange path to the normal generator $[a^b,a]$. On the right: An element in $\Ker(\phi)$, red paths are the paths defined by $\omega$, the orange path corresponds to $[a^b,a]$, the second edge  $E$ of the commutator path seperates the Bass-Serre tree into two connected components.}\label{sexypaths}
 	\end{figure}

  	Finally consider $\omega$, a good representative for an end-essential nepalese path $p$. Let $\omega\notin \mathcal{O}$. We will show that $\omega$ can be written as a product of one (any) of its siblings and paths $q$ such that $c(q)< c(p)$. Consider two triplets $\omega_1 B \omega_2 $ and $\omega_1 a B a^{-1} \omega_2$, where $B$ is a tip. Then
	\begin{align*}
		\omega 			&= 	\omega_1 B\omega_2 \\
						&=	\omega_1 B a B^{-1} a^{-1} \omega_1^{-1} \omega_1 a B a^{-1} \omega_2 \\
						&= 	\omega_1 [a^b,a] \omega_1^{-1} \omega_1 (a B a^{-1} \omega_2) \text{, or}\\[10pt]
		\omega 			&= 	\omega_1 B\omega_2 \\
						&=	(\omega_1 a^{-1}) a B a^{-1} B^{-1} (a \omega_1^{-1}) \omega_1 a^{-1} B a \omega_2 \\
						&= 	(\omega_1 a^{-1}) [a^b,a]^{-1} (a \omega_1^{-1})( \omega_1 a^{-1} B a \omega_2).
	\end{align*}
 	Clearly $c(\omega_1)< c(\omega)$. On the other hand consider two twins $\omega_1  V  b^{-1} \omega_2$ and $\omega_1 a^{-1} V a b^{-1} \omega_2$. Note that the valley can either be of the form $b^{-1}ab$ or $bab^{-1}$. We suppose it is the former, the argument for the latter is similar. Note also Figure \ref{twinpaths} for a visual interpretation. One has that
 	\begin{align*}
 		\omega 			&= 	\omega_1 b^{-1} a b\omega_2 \\
 						&=	(\omega_1 a^{-1}b^{-1}) ba b^{-1}ab a^{-1} b^{-1}a^{-1} (ba\omega_1^{-1}) \omega_1 a^{-1}b^{-1} aba\omega_2\\
 						&=  (\omega_1 a^{-1}b^{-1}) [a^b,a] (ba\omega_1^{-1}) \omega_1 a^{-1}b^{-1} aba\omega_2.
 	\end{align*}
	The case $b^{-1}a^{-1}b$ is similar, obtaining a conjugate of $ [a^b,a]^{-1}$ in the first term. Since each connected component in $G_\sim$ is represented in $\mathcal{O}$ by some element, each end-essential nepalese path can be written as a product of the designated relative in $\mathcal{O}$ and elements with less tips or valleys. One ends the argument by induction on the number of tips and valleys. \newline

 	{\em Freeness.} We are now ready to prove the given basis is free. Let $\omega\in\mathcal{O}$, then $\omega a^i [a^b,a]a^{-i}\omega^{-1}$ does not necessarily represent a geodesic path (we call the path $p$), it may have backtracking when the last edge of $\omega$ and the first of $[a^b,a]$ overlap. Similarly it might also overlap on the last edge of $[a^b,a]$ and the first of $\omega^{-1}$. Let $E$ be the second edge of $[a^b,a]$ as a subpath of $p$ (see Figure \ref{conjugate}), then it seperates the tree in two conncected components (say $E^-$ and $E^+$). We claim that applying any other element $\tilde \omega a^s [a^b,a]a^{-s} \tilde\omega^{-1}$of $\mathcal{S}$, one will never traverse $E$ again. There are several possibilities.
 	\begin{itemize}
 		\item Firstly suppose $\tilde \omega = \omega$, then this reduces to the fact that $\langle [a^b,a], a[a^b,a]a^{-1} \rangle$ is free on two generators.
 		\item Consider the concatenated path $\omega a^i [a^b,a]a^{-i}\omega^{-1}\tilde \omega a^s [a^b,a]a^{-s} \tilde\omega^{-1} $. Since $\tilde \omega$ is end-essential and nepalese, this path remain in $E^+$, once having crossed $E$. Let $E_{\tilde \omega}$ be the second edge in the second commutator $[a^b,a]$. Suppose $E_{\tilde \omega}$  is not part of the path defined by $\omega^{-1}$. In particular, to cross $E$, we first need to cross $E_{\tilde \omega}$, which means we have merely displaced the problem.
 		\item So we may suppose $E_{\tilde \omega}$ is part of $\omega^{-1}$. Note that this is only possible when $\omega^{-1}$ contains a tip or a valley. In this case the composition $\omega^{-1} \tilde\omega [a^b,a]\tilde{\omega}^{-1}$ defines a sibling of $\omega^{-1}$. It is impossible, using a sequence of siblings, to cross $E$, since for every connected component of the sibling relation we only allowed one path to be contained in $\mathcal{O}$, which is $\omega$.
 	\end{itemize}
\end{proof}

\begin{remark}
	Note that if one is willing to give up the generating set, then there is a quick and clean argument to show that $\Ker(\phi)$ acts freely on the Bass-Serre tree. The stabiliser of each vertex is given by a conjugate of $\langle a \rangle$. So to show that $\Ker(\phi)$  acts freely, it is enough to show that the intersection with each of the stabilisers is trivial. Given that $\Ker(\phi)$ is normal, it suffices to show that its intersection with $\langle a \rangle$ is trivial. Since $a^n$ is mapped to $a^{2n}$ this is true.
\end{remark}

Furthermore we look at the limit case of applying the morphism $\phi$.
\begin{proposition}
$\Ker(\phi^n)$ is normally generated by $\{ [b^m a b^{-m}, a] \;\lvert\; 0<m\le n    \}$
\end{proposition}
\begin{proof}
By induction.
\end{proof}

\begin{corollary}\label{limit}
$BS(2,3) / \bigcup_n \Ker(\phi^n) = \langle a,b \;\lvert \; [b^m a b^{-m}, a] \text{ where } m\in \mathbb{N} \rangle.$
\end{corollary}

Call this group $L$. Note that $\forall n\in\mathbb{N}$ there is an element $\alpha\in L$, such that $\alpha^{2^n}= \overline{a}$. I.e $\overline{a}$ has infinitely many roots. More specifically the presentation induced by $\alpha$ and $\overline{b}$ is the same as the one given in Corollary \ref{limit}. One sees that the group generated by $\langle \overline{b^m a b^{-m}} \;\vert\; m\in\mathbb{N} \rangle $ is isomorphic to the dyadic rationals $\mathbb{Z}[1/2]$.

\section{Another interpretation of non-Hopfianness and a continuous map}\label{levittpres}

We present another interpretation on the non-Hopfianness of $BS(2,3)$, which was communicated to the author by Gilbert Levitt. In some sense, this is a more topological version of Remark 6.10 in \cite{levitt}. The paper treats generalised Baumslag-Solitar groups, which are fundamental groups of graphs of groups. This is also the setting of Remark 6.10. These groups can be  retrieved as the fundamental groups of certain 2-dimensional simplicial complexes (see \cite[Section 7]{delgado}). We will not introduce graphs of groups and associated concepts, but will rather immediately take the topological point of view adapted to $BS(2,3)$. As a last remark note that the forthcoming example is not particular to the case of $BS(2,3)$, but can easily be altered to apply to other Baumslag-Solitar groups. The same is true for Figure \ref{Levittpic}. \newline

First as small detour. We note the following fact: in $BS(2,3)$, the elements $a^2$ and $a^3$ are ``the same" since they are conjugates.  What we mean is that they behave in the same way, which is made precise in the following lemma.
\begin{lemma}
	The group $BS(2,3)$ has an abstract group presentation, where the generators can be interpreted as either $\{a^2,b\}$ or $\{a^3,b\}$. This presentation is given by $\langle \lambda, \mu \;\vert\; \lambda^2 = \mu \lambda \mu^{-1}\lambda^{-1}\mu\lambda \mu^{-1}\rangle$.
\end{lemma}
\begin{proof}
	Starting from the usual presentation $\langle a,b \;\vert \; ba^2b^{-1} =  a^3\rangle$, one can apply two sequences of Tietze transformations to obtain the presentation above. We present the calculations for $a^2$:
	\begin{align*}
	\langle a,b \;\vert \; ba^2b^{-1} =  a^3\rangle &= \langle a, \lambda, b \;\vert \; \lambda= a^2,  \;  ba^2b^{-1} =  a^3\rangle\\
	&= \langle a, \lambda, b \;\vert \; \lambda= a^2,  \;  ba^2b^{-1} =  a^3,  \;  a= b\lambda b^{-1} \lambda^{-1}\rangle\\
	&= \langle \lambda, b \;\vert \; \lambda= b\lambda b^{-1} \lambda^{-1}b\lambda b^{-1} \lambda^{-1},  \;  b\lambda b^{-1} =  b\lambda b^{-1} \lambda^{-1}\lambda\rangle\\
	&= \langle \lambda, b \;\vert \; \lambda^2= b\lambda b^{-1} \lambda^{-1}b\lambda b^{-1}\rangle\\
	&= \langle \lambda, \mu\;\vert \; \lambda^2= \mu\lambda \mu^{-1} \lambda^{-1}\mu\lambda \mu^{-1}\rangle.
	\end{align*}
	We can obtain the same presentation for $a^3$ by doing similar calculations.
\end{proof}
We define $\phi'$ to be the morphism
\[    BS(2,3) \rightarrow BS(2,3) :
\begin{cases}
a \mapsto a^3\\
b\mapsto b \\
\end{cases}
. \]
From a group presentation point of view the following is true: since $a^2$ and $a^3$ behave in a completely similar way the morphisms $\phi$ and $\phi'$ are in fact also the same morphism! As a consequence they have the same kernel.\newline

\begin{figure}[b]
	\centering
	\includegraphics[width=0.5\textwidth]{LevittpictureFinal.pdf}
	\caption{$BS(2,3)$ as a fundamental group and $\phi$ as a morphism induced by a continuous self-map (in red) of the space $B$.}\label{Levittpic}
\end{figure}

We start by describing a space $B$ such that $BS(2,3)$ is its fundamental group. The basic building blocks are two circles $S_0$ and $S_1$, and two annuli $A_0:= S_a^0 \times [0,1]$ and $A_1:= S_a^1 \times [1,2]$. Now we glue the blocks together. For $A_0$ we attach $S_a^0\times 0$ to $S_0$ along the identity map $e^{i\theta}\mapsto e^{i\theta}$. We do the same for $S_a^0\times 1$ to $S_1$. For $A_1$ we attach  $S_a^1\times 1$ to $S_0$ along the map $e^{i\theta} \rightarrow e^{3i\theta}$, and $S_a^1\times 2$ to $S_1$ along the map $e^{i\theta} \rightarrow e^{2i\theta}$. This space is depicted two times in Figure \ref{Levittpic}, being once given by the outer annuli and circles, and a second time by the inner annuli and circles. The reader can easily verify that the fundamental group of $B$ at $(1,\frac{1}{2})$ is generated by two loops
\begin{align*}
&\gamma_1: [0,1]\rightarrow B: \theta\mapsto \left(e^{2\pi i \theta}, \frac{1}{2}\right)\\
&\arraycolsep=1.4pt\def\arraystretch{1.5}
\gamma_2: [0,1]\rightarrow B: x \mapsto \left\{\begin{array}{ll} \left(1,\frac{1}{2}-2x\right) & \text{ if } x\in  \left[0,\frac{1}{4}\right] \\
\left(1,1+2x-\frac{1}{2}\right) & \text{ if } x\in \left[ \frac{1}{4},\frac{3}{4} \right] \\
\left(1,\frac{5}{2}-2x\right) & \text{ if } x\in \left[\frac{3}{4},1\right]
\end{array}\right. ,
\end{align*}

where $\pi_1(B, (1,\frac{1}{2}))$ is isomorphic to $BS(2,3)$ by mapping $a$ to $[\gamma_1]$ and $b$ to $[\gamma_2]$. \newline

Finally the map $\tilde \phi: B\rightarrow B$ is given in red in Figure \ref{Levittpic}. We define the map piecewise
\begin{itemize}
	\item $A_0\rightarrow A_1: (e^{i\theta},x)\mapsto (e^{i\theta},2-x)$
	\item $A_1\rightarrow A_0: (e^{i\theta},x)\mapsto (e^{6i\theta},2-x)$
	\item $S_0\rightarrow S_1: e^{i\theta} \mapsto e^{2i\theta}$
	\item $S_1\rightarrow S_0: e^{i\theta} \mapsto e^{3i\theta}$ .
\end{itemize}

It is easy to verify that this map respects the gluing of the building blocks, hence is continuous on $B$. In particular $\tilde \phi$ induces a map on the fundamental group of $B$, being $\tilde{\phi}_*: \pi_1(B, (1,\frac{1}{2}))\rightarrow \pi_1(B, (1,\frac{3}{2}))$. One could check that $\Ker(\tilde{\phi}_*)$ is exactly $\Ker(\phi)$ (and thus also $\Ker(\phi')$). So from a group presentation point of view, we can say that $\tilde \phi$ induces the group morphism $\phi$. However, suppose we really want to recover $\phi$, then we would need a map from $\pi_1(B, (1,\frac{1}{2}))$ to itself. Consider the path
\[\arraycolsep=1.4pt\def\arraystretch{1.5}
 p^1: [0,1]\rightarrow B : x \mapsto \left\{\begin{array}{ll}  \left(1, \frac{1}{2} + x\right)   & \text{ if } x\in 							\left[0,\frac{1}{2}\right] \\
\left(1, 2-x\right)   & \text{ if } x\in \left[\frac{1}{2},1\right]
\end{array}\right..  \]
This path imposes an isomorphism $p^1_*: \pi_1(B, (1,\frac{3}{2})) \rightarrow \pi_1(B, (1,\frac{1}{2})) $. Now composing, we obtain that $p^1_*\circ \tilde{\phi}_*$ is exactly $\phi$. If, on the other hand, we want to recover $\phi'$, we only need to compose  $\tilde{\phi}_*$ with $p^2_*$, where $p^2$ is the path given by
\[\arraycolsep=1.4pt\def\arraystretch{1.5}
 p^2: [0,1]\rightarrow B : x \mapsto \left\{\begin{array}{ll}  \left(1, \frac{1}{2} - x\right)   & \text{ if } x\in 							\left[0,\frac{1}{2}\right] \\
\left(1, \frac{1}{2} + x\right)   & \text{ if } x\in \left[\frac{1}{2},1\right]
\end{array}\right. .  \]

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


text:
Sudan Relief Operations Endangered
Rebel demands cause agencies to curtail efforts.
By Tony Carnes| April 1, 2000
Just as the international spotlight reached the persecuted Christians of southern Sudan, more than a dozen humanitarian groups—including World Vision, Oxfam, and CARE—withdrew their operations over a dispute with rebel leader John Garang.
The southern commander demanded that humanitarian groups sign a memorandum of understanding by March 1 recognizing his Sudanese People's Liberation Army (SPLA) as the governing authority on aid efforts in southern Sudan.
Some groups say the agreement would force them to take sides in the civil war.
"We feel our objectivity is at stake here. The agreement is primarily aimed at aligning non-government organizations or humanitarian groups with political factions," says Bruce Wilkinson, senior vice president of World Vision.
The aid groups also object to terms that would give the SPLA the right to restrict aid groups' public meetings, oversee their budgets, and access their trucks and relief supplies that then could be used in the war.
"Garang has chosen to shoot himself in the foot by making himself look more interested in the politics of the situation than the needs of his people," warns Robert Seiple, U.S. Ambassador for Religious Freedom.
The withdrawal of "blue chip" relief groups may mean that aid to southern Sudan will drop by a third, Seiple says. Hundreds of thousands of southern Sudanese may still be at risk of severe malnutrition, according to the U.N. World Food Program.
Most groups are staying, however—including World Relief and Franklin Graham's Samaritan's Purse—placing Christians agencies on both sides of the issue. U.N. relief efforts throughout the country will continue, as will World Vision's efforts in northern Sudan.
On the deadline day for signing with the SPLA, Sudanese government forces bombed a hospital run by Samaritan's Purse in southern Sudan. Since then it has been bombed three more times. A hospital sponsored by Voice of the Martyrs and Far Reaching Ministries was also bombed, killing one aid worker.
The dispute raises the troublesome issue of when a rebel movement becomes a legitimate state. A senior U.S. official warns that Africa "is a continent with lots of rebel leaders. If the aid groups sign this memo, they would be asked to do so elsewhere."
But, says Steve Wondu, SPLA representative to the U.S., "These are our families we are fighting for. We ought to be able to have a say where you are spending millions of dollars for our children."
"When we started working in Sudan, we decided not to let the Khartoum government dictate whom we could and couldn't help," says Franklin Graham. "Since 1998, our hospital has helped more than 100,000 Sudanese patients, and our doors will remain open."
Tony Carnes is Senior News Writer for Christianity Today.
Related Elsewhere
See our earlier coverage of this issue, "Bombs Continue to Fall on Ministry Hospitals in Sudan | Samaritan's Purse hit for fourth time, two killed in Voice of the Martyrs bombing" (Mar. 24, 2000)
See press releases from World Vision and CARE about why they left Sudan, as well as information from Samaritan's Purse and World Relief about the challenges of staying.
Copyright © 2000 Christianity Today. Click for reprint information.
International | Sudan
More From: By Tony Carnes
A Poet for 'Bruised Evangelicals'
Why Are There So Many Angry Theologians?
Naming Names in the Abortion Debate
Malcolm Guite has found himself a sort of tribal elder for younger generations of Christians.
Kara Bettis
The 50 Countries Where It's Hardest to Follow Jesus in 2023
Latest report on Christian persecution finds Nigeria and Sub-Saharan Africa the epicenter of jihadist violence, while China leads effort to redefine religious rights.
CT Editors
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View cart "Eagle, Colorado – A Walking Tour of a Historic Mountain Town" has been added to your cart.
This beautiful coffee-table style book features never-before published photos of the Gore Valley from 1908 – 1950s. Kirby, whose grandparents homesteaded a ranch in what is now West Vail, shares stories and photos of life before Vail.


Proverbs 2:2-- He that hath a Baleful Eye knows not the want that shall come upon him.
When the Guzzlewhomper abandons a host, tho' hangover free, they bear the telltale mark of possession, the Baleful Eye. This woefully blood shoot orb is the only evidence of the debauch the proceeding evening.
MacThoys adorn their encampments with the Baleful Eye in the hopes of warding off the Guzzlewhomper, by implying another 'whomper had been and gone and left no beer behind.
The Guzzlewhomper and, by default, the Baleful Eye, is the predecessor of the Evil Eye. Archaeological finds of the Baleful Eye have been dated by Thorgrim Laboratories back to the Paleolithic age. The design has been identified on a variety of objects, ranging from wine goblets to cod pieces.
The Baleful Eye also has a reputation also as a bad luck charm. It was thought to impede Mojo (better known as Monkey Luv)-- which is believed to give rise (no pun intended) to tales of the MacThoi Widows killing stare.
MacThois scholars believe that the digitus impudicus became a sign of defiance through its show of virility in the face of the Baleful Eye.
Then, with incantations, she invoked the Spirits of Death, the swift hounds of Hades who feed on souls and haunt the lower air to pounce on living men . . . .She steeled herself with their malignity and bewitched the eyes of Talos with the evil in her own. She flung at him the full force of her malevolence, and in an ecstasy of rage she plied him with images of death.--Apollonius of Rhodes writing on Medea.
Laughter is another best defense for the Baleful Eye. Mooning has also been known to break the spell. The Baleful Eye was placed on jewelry and homes in the belief the artificial would displace an attack from the real one, similar to a Star Wars defense.
Other defenses against the Baleful Eye seen adorning MacThoi environs include, the Clan Crest (the Digitus Impudicus), and a rendering of the Barley Stein.


text:
Link sAUsage
Link sAUsage: 19 September 2016
News and notes from the Auburn corners of the Internet
By Walt Austin@AU_GS_Walt Sep 19, 2016, 9:39am CDT
Share All sharing options for: Link sAUsage: 19 September 2016
John Reed-USA TODAY Sports
The women finished tied for 5th at the Mason Rudolph Championship in their second event of the season. The men, meanwhile, will be opening their season at the Inverness Intercollegiate today and tomorrow.
The Tigers dropped a tough SEC match to South Carolina last night, 0-1. Yes, Bruce Pearl was there serving BBQ to students before the match. You will not find another coach at Auburn who does more to promote and help out every other sport. He's an amazing ambassador for the school.
So, here's my immediate post-game recap and my later thoughts on the game from Saturday. I'm still mulling over some other thoughts I have on the game that I may write about later, today.
Blake has TAKES for you. They're brief and, no, the article isn't unfinished. I'm pretty much on board with exactly his feelings. Though there is a need for something to change, soon.
There were bright spots in the game. Tony Stevens is really starting to live up to his potential at wide receiver. Then there's Kamryn Pettway, who somehow managed over 100 yards for the second game in a row.
Kevin Steele spoke to the media last night and he wants the defense to create more turnovers. He said there were good and bad things for the Tigers in Saturday's game, and missed tackles were a big issue. Also finishing sacks.
So, I didn't write anything on Landon Rice. I probably should have, but it's a touchy topic. He's gone because he was accused of rape. He left on his own. The accusation came months after the event. He's not currently charged. It sounds a lot like it's a "he-said-she-said" situation and he's walked away from Auburn on his own. Honestly, and please don't take this the wrong way, I believe every charge should be investigated as if true, but in today's world some universities are quicker to punish even without the law and maybe that's why Rice decided to walk away from Auburn rather than deal with the long and lengthy battle he would likely face with the school, since the incident supposedly happened on school grounds. If it's true, then I hope the victim is able to receive justice. If it's not, then I hope Landon is able to move on with his life. With what little info I've seen so far, though, I don't know if either instance will be the case, and that's sad.


text:
Marcílio Marques Moreira (Río de Janeiro, 25 de noviembre de 1931) es un político y diplomático brasileño, ministro de Hacienda durante la presidencia de Fernando Collor de Mello.

Cursó la escuela en varios países centroeuropeos. Egresado en 1954 del Instituto Rio Branco y posteriormente estudió Derecho en la Universidad del Estado de Río de Janeiro, egresando en 1957. En 1963 aprobó una maestría en Ciencia Política en la Universidad de Georgetown.

Fue Ministro de Hacienda de Brasil durante el gobierno de Fernando Collor de Mello y presidente del Consejo de Ética Pública del gobierno de Lula.

Escritos
Autor
 Indicações para o Projeto Brasileiro (Río de Janeiro, Tempo Brasileiro Ed., 1971)
 Poder, Liberdade e Desenvolvimento: Indicações para o Debate Brasileiro (Río de Janeiro, Tempo Brasileiro Ed., 1980)
 De Maquiavel a San Tiago (Brasília, Universidade de Brasília Ed., 1981)
 Confissão Liberal (Río de Janeiro, P.E.N. Clube, 1984)
 The Brazilian Quandary (New York, Twentieth Century Fund, 1986)
 Diplomacia, Política e Finanças: Depoimento ao CPDOC-FGV (Río de Janeiro, Objetiva, 2001)

Organizador
 Perfil Parlamentar: San Tiago Dantas (Brasília, Câmara dos Deputados, 1983)
 Atualidade de San Tiago Dantas (São Paulo, Lettera.doc, 2005)
 Cultura das Transgressões no Brasil: Lições da História, org. em conjunto com Fernando Henrique Cardoso (Editora Saraiva, 2008)

Colaborador
 Aggressive Unilateralism, org. por Jagdish Bhagwati and Hugh T. Patrick (Ann Arbor: The University of Michigan Press, 1990)
 O Futuro do Brasil: A América Latina e o fim da Guerra Fria, org. por José Álvaro Moiséis (Río de Janeiro: Paz e Terra, 1992)
 In the Shadow of the Debt: Emerging Issues in Latin America (New York: The Twentieth Century Fund Press (1992)
 Desenvolvimento, Tecnologia e Governabilidade, org. por João Paulo dos Reis Veloso (São Paulo: Nobel, 1994)
 Brazil: At the Turning Point, editado por Colleen S. Morton (La Jolla, Institute of the Americas, 1995)
 Diplomacia para o Desenvolvimento, 2º vol. de Sessenta Anos de Política Externa: 1930-1990 org. por José Augusto Guílhon Albuquerque (São Paulo: USP, 1996)
 National Perspectives on the New Regionalism in the South, United Nations University, publicado por Palgrave-Macmillan Group, UK, 2000.
 Globalização, Desenvolvimento e Equidade, Fundação Calouste Gulbenkian (Lisboa: Publicações D. Quixote, 2001).
 Investimentos em Educação, Ciência e Tecnologia, organizado por Jorge Werthein e Celio da Cunha (Brasília, UNESCO, Ministério da Educação, Instituto Sangari, 2004).
 Por Que o Brasil Não É Um País de Alto Crescimento?, org. por João Paulo dos Reis Veloso (Jose Olympio, 2006)
 Chegou a Vez do Brasil?: Oportunidade para a Geração de Brasileiros que Nunca Viu o País Crescer, org. por João Paulo dos Reis Veloso (Jose Olympio, 2007)
 A Crise Global e o Novo Papel Mundial dos BRICS, org. por João Paulo dos Reis Veloso (Jose Olympio, 2009)

Prefaciador
 Brilho nos Olhos: Dez Anos de transformações na visão de grandes executivos (São Paulo: A.T.Kearney, 2003)
 O Encilhamento: Anatomia de uma Bolha Brasileira, de Ney Carvalho (Río de Janeiro e São Paulo: Comissão Nacional de Bolsa e BOVESPA, 2004).

Referencias

Enlaces externos

 Ministro Marcílio Marques Moreira

Nacidos en Río de Janeiro
Alumnado de la Universidad de Georgetown
Politólogos de Brasil
Diplomáticos de Brasil
Embajadores de Brasil en Estados Unidos
Ministros de Hacienda de Brasil

redpajama_set_name: RedPajamaWikipedia


Maidstone's first micro-pub, albeit a bit larger than most, housed in part of the old Style & Winch brewery cellars (now forming part of a small trading estate). Total of seven rustic high wooden benches under the low barrel-vaulted roof, with decor largely an ever-expanding selection of pump clips. However, there aren't actually any handpumps, with all beers dispensed by gravity from casks on stillages in the cool room to the rear of the small counter in one corner. On this first visit I found four cask beers on offer: Old Dairy Red Top, Animal Sea lion (saison), Redemption Big Chief and Brumaison London Porter (£3.60). With a friendly barman and generally good atmosphere too, this is well worth seeking out (with entrance via a rear gate out of normal business hours).


The current close value 103.0 is between the support and resistance channel of 100 and 104. The stock has a Sell Signal , 100 becomes a critical support to stop the fall and 104 is a immediate resistance on the upside.
NMDC Limited is an India-based exploration-stage company. The Company is engaged in the exploration and production of iron ore along with diamond, production and sale of sponge iron and generation and sale of wind power. The Company operates business in two segments: iron ore and other minerals & services. The Company operates in Chhattisgarh, Karnataka, Madhya Pradesh and Andhra Pradesh. The Company's subsidiaries include J&KMDC Limited, NMDC-CMDC Limited, NMDC SARL, NMDC Power Limited and Jharkhand National Mineral Development Corporation Limited.
NMDC Limited has Sell Signal. What is your opinion on it?


Weebly has a free EDU package.
My school uses Google Apps for Education (GAFE).
GAFE has a Weebly App for the Google Chrome browser.
The students sign in to Weebly through our GAFE account.
Weebly EDU allows me to create a class.
The class keeps my students' work private if I so choose.
I can monitor the progress of my students because they are not logging in with individual accounts.
Weebly is easy to use, yet offers some super Web design features.
The free education plan (education.weebly.com) limits student Web sites to five pages, but I can live with that. Teacher sites have no limit on the number of pages in a site, and they can create up to 10 separate sites. Students are allowed two Web sites. Also with the free EDU plan, teachers can add any number of classes, but are allowed only 40 total student accounts. For teachers with more than 40 students, additional accounts can be purchased, or earned by recruiting other teachers to sign up for Weebly Education. The teacher has total control of the student accounts including setting up the user name and password.
And I am guessing your biggest question is this: If you are a GAFE school, why aren't you using Google Sites for Web design? First of all, the Google Sites learning curve is (arguably) a good deal longer than Weebly's. I also don't have the teacher management controls in Google Sites. And, I'm sorry, but most Google Sites I've seen are just plain ugly. I hearken back to the days of Microsoft FrontPage templates - UGH!
Clippingmagic.com - remove image backgrounds easily to create a "cut-out" .png image with a transparent background! These are great when you want to put an image on a colored or pattern background without having the hideous white box (or any other background) around it.
Awesome Screenshot and Techsmith Snagit extensions for Chrome - both allow you to capture a still shot of your screen (or a portion of it) and then annotate it with shapes, arrows, text, etc. The image can then be saved and placed on a Web page or in a document.


Most of us assume that the laws about what is and what isn't legally obtained evidence to be black and white; however, there are actually quite a few gray areas when it comes to illegally obtained evidence in California.
One example of the murkiness surrounding how evidence is obtained is a police officer who happened to notice what looked an awful lot like a drug-related transaction. Unable to ignore the situation, the officer approached the car and after a brief skirmish, the officer found that the car was full of drugs, ammo, and an alarming number of weapons. The question is, since the officer didn't have a warrant, could the evidence be used as proof of guilt.
Whenever there's a case involving a car and evidence, things always get complicated. Technically, an officer can't go through your car looking for proof of illegal activities; however, if they pull you over for speeding and there's a bag full of drugs on your backseat, they can arrest you and use the baggie of drugs as evidence.
If you're involved in a case and don't think the evidence was obtained with a legal search warrant, it is in your best interest to hire a good defense lawyer who has a strong understanding of the Fourth Amendment.


text:
Sexy Time
Is it healthy to have sexual fantasies about someone other than your partner? We asked an expert
Gina Florio
We don't talk about it much, but sexual fantasies are a part of life that all of us have experienced before, whether we consciously invent them or they come to us in our dreams. Even though you can't help who you have a sexual fantasy about, it's easy to feel worried or guilty when you start daydreaming about someone who isn't your partner.
HelloGiggles spoke with Wendy Strgar, relationship expert, founder of Good Clean Love and author of forthcoming book SEX THAT WORKS: An Intimate Guide to Awakening Your Erotic Life. Although it may seem problematic to fantasize about someone other than your S.O., Strgar says that it may not be as bad as you think.
"We can use [fantasies] for fuel to awaken our libido," Strgar tells HG. "We have the ability to direct that sexual arousal back towards our intimate partner."
Anyone who has been in a longterm relationship knows that the kind of sex you have changes over time. In fact, you might find yourself in a sexual rut, which could be the very reason you have "fantasies of secret desire," as Strgar calls them.
However, just because you're having sexual fantasies about someone else doesn't mean you and your SO are doomed.
It doesn't mean you're about to cheat or that you don't love your partner anymore. It just means you've got some pent-up sexual energy that needs to be channeled somewhere. Therefore, in order for your fantasy not to turn into something dangerous you act on, Strgar strongly advises doing a little self-reflection about why you're imagining that hot scenario in the first place.
"We should do a lot of self inquiry about what we are wanting in our primary relationship and what it is that is pulling us towards sexual ideas about another," she tells HG.
"In and of itself it is very normal, but without reflection can also become a destructive force in the sexual relationship that you are cultivating with your partner," she continues.
Figure out what it is that's sparking these fantasies and see if there's something in particular you're lacking in your sex life. Also, don't feel like you have to confess to your partner about your sexual fantasies. Strgar says this is a conversation that can "become very hurtful very quickly."
Strgar suggests treading lightly and doing some "inner searching before acting on or sharing these kinds of fantasies with partners." Duly noted.
By Gina Florio
Popular in Love
How to introduce sex toys in the bedroom without it getting awkward
6 ways to know if you're in an unhealthy relationship
You're putting too much effort into blow jobs, according to men
We asked all the longtime-married couples we know to share their advice for a healthy, lasting relationship
The Right Swipe: Meet Nayda and Alex, a happily married couple that met on Tinder
How to talk to a new partner about your sexual assault
The Right Swipe: Meet Anna and Jon, a committed long-term couple that met on Bumble
Why heartbreak and music go hand in hand
How to deal if you'll be separated from your significant other over the holidays (and you don't want to be)
Include these phrases in your dating profile if you want to swipe right on a great booty call
A beginner's guide to role play: How to choose a sexual fantasy—and actually play it out
Under $10k Weddings: Kris and Iris's sweet San Francisco wedding
I tried using vegan condoms. Here's how they measured up
Everything you need to know about sleeping with another woman for the first time
What not to ask a plus-size bride, according to a plus-size bride
What I've learned about sex after 10 years of marriage
Is breakup sex ever a good idea? Here's what therapists really think
What two years of accidental celibacy taught me about independence
This is exactly what you should do to prepare for anal sex, according to experts
8 myths you've probably believed about STIs
All Topics in Love
Having Sex


Celebrate Your American Spirit in Style this 4th of July!
Red, white and blue are the colors of the day every July 4th, but this year the party can be different than last year's. July 4th party rentals can help liven up the celebration with new items designed to share the day's spirit of independence.
Enjoy the parades and fireworks with no worries when our event rental company supplies tables, chairs and more. Rent party equipment to save time and effort gearing up for the celebration. Table linens, china, glassware and other decor pieces will make the party space festive and honor the meaningful colors of American freedom.
With a wide array of chair rentals to choose from helps assist the party-giver in setting the mood and reminding guests of the reason for this annual celebration. July 4th party rentals also include the choice of table rentals and linens rentals to keep each guest comfortable and relaxed.
Red, White & Blue, of Course!
Hosts can rent party equipment of coordinated color blends that match the American holiday with reds, whites and blues. It is one less detail to take up time better spent socializing. Linen table cloths, cloth overlays and napkins are tasteful and fitting for fun and conversation. Glassware comes in several colors as does the china for refreshing variation.
Our event rental company adds convenience and ease to a potentially hectic time. The party can be ready before or after a holiday concert so guests feel welcome and everyone can relax, including the hosts. A red, white and blue dessert suggestion may also brighten the celebration.
Stylish flatware is yet another option for the well-dressed table at a festive party. Diners can eat with flair in either casual or dressy attire. There are many ways to salute the flag with these decor choices, so the hostess can focus on more immediate party needs.
Grilled foods and picnic dishes look even better on an attractive table, adding a nice touch to a day and evening of holiday celebration. All of these ideas contribute to a pleasant time of catching up with family and friends, so make it easier this year.
After the party, everything goes back and the clean-up is done. Find out more by contacting Gaston Rentals today and speak with an Event Planning specialist about your ideas.
Party hosts deserve to enjoy their own celebration, and July 4th party rentals make it possible. Planning ahead is a snap with Gaston Rentals. Call and ask about all the necessary details and get suggestions if needed. Get ahead of the crowds!


text:
Artist 101: Charlie Puth
Caroline November 30, 2017, 7:42 pm
Your go-to-guide for everything you need to know about an American singer, songwriter and record producer Charlie Puth.
When and where was Charlie born?
December 2, 1991,
Rumson, New Jersey, U.S.
What are his most popular songs?
We Don't Talk Anymore (feat. Selena Gomez)
5 facts about the artist
As a two-year-old, Puth's right eyebrow was permanently scarred in a nearly fatal dog-bite incident.
Puth was horrendously unpopular in school and was often more commonly seen at home than socializing
Tootsie Pops are supposedly his favorite treat
Charlie loves coding
He couldn't tell the time until he was 12 years old
What else do I need to know about Charlie Puth?
Puth is an American singer, songwriter and record producer. His debut song "See You Again", which he wrote, co-produced, and performed with Wiz Khalifa for the Furious 7 soundtrack as a tribute to Paul Walker, became number one for 12 non-consecutive weeks in the United States. His initial exposure came from the viral success of his song videos uploaded to YouTube. After the success of "See You Again", he subsequently gained wider recognition for multiple subsequent releases, including his debut single "Marvin Gaye", which featured vocals from Meghan Trainor. The song topped charts in New Zealand, Ireland, and the United Kingdom.
What are his social media links?
Twitter: @charlieputh
Instagram: @charlieputh
Facebook: @charlieputh
Snapchat: notcharlieputh
Written by Caroline
When nothing is going right, go left.
EMAIL: zos.caroline@gmail.com
INSTAGRAM: https://www.instagram.com/karolina9c/
Disney+ UK: The Complete List Of EVERYTHING You Can Watch On Disney Plus UK Upon Launch
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Toy Story 4 Trailer Spotlights New Character
Why Fans Love: Little Mix
"Incredibles 2" Passes $1B at Global Box Office
The Incredibles 2 Has an Incredible Weekend at the Box Office


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To benefit the AdvoCare Foundation, AdvoCare, a Plano-based health and wellness company, is hosting its second annual GenerationFit event on Saturday, February 23, at 6:15 a.m. The early morning workout and fundraiser, which will be held at the Ford Center of The Star in Frisco, is open to the public and all fitness levels.
Held on the Saturday of AdvoCare's annual Success School, GenerationFit will include VIP special guests: Andy Dalton, Cincinnati Quarterback and 3-time Pro-Bowl athlete, and Rich Froning, 4-time Reebok Crossfit Games Champion. The workout will be led by David Thompson, former NFL player and local Dallas fitness instructor/owner of CampDaveFit.
The AdvoCare Foundation will use event proceeds to award grants to nonprofit organizations that improve the physical health and wellness of children in areas where it's needed most.
Event registration is required, and there are various giving levels offering different AdvoCare Foundation workout gear and opportunities, including a VIP Experience offering breakfast with Rich Froning and Andy Dalton.


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March Issue '16
Cymbals showcases student artistic talents
Caroline Bernstein, Arts Editor
If you have not had the chance to look through an issue of Cymbals, Princeton Day School's literary magazine, then you are missing out on an impressive collection of work created by Princeton Day School students. Cymbals highlights student work in a variety of different mediums—writing, photography, studio art, and even architecture. When choosing pieces to put into the next volume, Cymbal's staff likes to consider which pieces most accurately refect the diverse and creative environment that PDS is. "I try to look around and see what makes PDS special, and so things that I don't think other schools might necessarily have, we need to shine a spotlight on," said Upper School English teacher and faculty advisor to Cymbals for the past fve years Jamie McCulloch. He cited initiatives such as our architecture program and our commitment to sustainability as things he would like to see highlighted in the magazine.
This year, Cymbals was named frst place among high school magazines by the American Scholastic Press Association. This is no small honor, nor is it one that is new to Cymbals. This is the second consecutive year that Cymbals has ranked frst and the year before that it had been ranked second.
"There's still another level that we're chasing, which is frst [place] with something like honorable distinction, something that's even more wonderful," said Mr. McCulloch. With an impressive collection of students working on and editing the magazine, this seems to be an attainable goal for Cymbals. However, being authentic and real is the staff's number one priority, not winning awards. "I think that we always have to weigh the sort of needs or desires of the staff who edits Cymbals against the contest and to see whether a contest, conforming to it, would cramp our vision," explained Mr. McCulloch, adding, "I don't think that it has, I think it actually has strengthened our vision to submit to that contest because it helps clarify who's doing what in the magazine and makes us think about how the artists and writers are featured in the magazine so that the spotlight shines upon them and what kind of work they're doing better."
For anyone with an interest in submitting to Cymbals, (and all students are highly encouraged!) it is a very simple process. Just go to cymbals.submittable. com, and follow the steps online for submitting your work. Submissions are then reviewed by Cymbals staff and ultimately voted on to see whether or not they will be put into the upcoming volume. "I think it is a good way for people to showcase what they haven't already in English. If someone is talented at something they don't necessarily have a place for at school, they can showcase it here," replied Cymbal's staff member junior Catie Higgins when asked why people should submit to the magazine.
Revived this year is an online edition of the magazine, which will allow students even more opportunities to see their work published. The staff is currently in the process of choosing which pieces to place in their online version, hoping to fnd the same success online as they have with the print.
What role should PDS play in patrolling student social media?
Should students be able to request teachers?
Gun ownership must be limited
Students grow more aware of large-scale issues
Are we ready to think outside the box?
Zen and the art of college acceptance
New York Fashion Week highlights new trends
Something Rotten! lights up Broadway
Student predicts next Oscar winners
Rapping from beyond the grave


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Texas Hockey
Texas Stars Wrap Up Weeks Nine and Ten: Condra Feel the Love Tonight
Erik Condra, good grief.
By Shepard Price@shepardprice Dec 16, 2018, 10:26am CST
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Erik Condra #22 of the Texas Stars
Texas Stars Photography
Erik Condra had an 11-game point streak, one that didn't end until Friday night. Over the past seven games alone, he's scored 12 points and 11 of them were primary. He is driving offense and doing a lot with (and without) his linemates — Travis Morin and especially Joel L'Esperance — right now, and that's paying dividends for the Texas Stars.
Derek Laxdal, the Texas Stars' head coach had this to say:
The line of Condra, Morin, and L'Esperance has been really good for us. It's a line that can generate a lot of o-zone play. Since we put them together they've had a really good run for our group.
L'Esperance has five goals and three assists playing with Condra over the past six games, and Morin has two goals and five assists. Nobody is generating shots like L'Esperance right now — 22 in the past seven games — but Morin comes close with 20. Third place is Ben Gleason with 19 shots, and while he doesn't have a goal in the past six games (he does have five assists), that's something bound to change.
Texas was on a four-game win streak, and is on a five-game point streak, and has five wins in the past seven games. That hot streak has brought the Stars back to fourth place in the Central Division and tied with the Chicago Wolves.
Lest I forget, the Dallas Stars' most recent call-up, Denis Gurianov, has one goal and seven assists (six primary) over his past six games. He's flourishing this season with 28 points in 23 games and is tied for fourth in the AHL (Condra is second). He's become a much more well-rounded player and looks competent in his own zone as well.
Landon Bow has also been spectacular over his past six games, coming away with a .910 save percentage and a 4-1-1 record.
"I really like the way Bow is playing right now. He's our number one goalie, and he understands that, and he's playing like it. That's really good for us and really bad for other teams," Laxdal says.
The Texas Stars have gotten a lot out of multiple players to have become this hot. This includes a five-goal, two-assist performance from Adam Mascherin (Dallas's fourth round overage pick from the 2018 draft) over the past seven games. In addition, Michael Mersch has returned with three points in his last six games and Roope Hintz has four points in three games since being sent back down to Cedar Park.
The power play is converting at a critical rate as well, converting on seven of 27 chances (25.9 percent) in this recent stretch of games. The penalty kill is doing — less well, with a 67.9 percent kill rate. That's not a primary concern for a team doing this well, but it is something to watch as the season progresses, especially for a penalty kill that is fifth-best in the division.
At the end of the day, however, Condra's incredible streak was the main talking point about the Texas Stars over this past two weeks, as it should have been.


Sheffield-born 37 year old Jonathan Eyre is well known as an inspiring and colourful liturgical organist, choral conductor and theatre organ performer.
He will perform this weekend at the final concert on Rye's famous Wurlitzer organ before major reburbishment begins in its home at the Milligan Theatre at Rye Academy.
Jonathan spent four years as Assistant Organist at Bradford Cathedral, where he recorded a series of CD's of music by J.S.Bach. He is currently sub-organist at the Priory Church of St Bartholomew the Great in the City of London.
accompany silent films on both piano and theatre organs.
Jonathan's concert on Sunday November 25 begins at 2.30pm with a refreshment interval included in the admission price of £9.50. Doors open at 1.30pm.
The Sunday concerts arranged by the Friends of Rye Wurlitzer will not resume until June 2019.


This stained glass window depicts St. Leonard holding a skull. This is one of a set of sixteen stained glass windows from the old Poor Claire's Monastery of Chicago built in about 1870 and demolished in the early 1970s. Windows are made in the Munich style.


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Are You Making These Wholesale Nfl Jerseys Errors?
shawnholiman24
Line Wile Ahmad Brooks will absent the game
San Francisco 49 people will rush to New York to the challenge of New York giants. But the line of Ahmad Brooks, but since his sister has decided not to organize.
With Jason's regression, the offensive group of cowboy has been further enhanced. From May 6 this year, Jason 37-year-old birthday was selected by Dallas Cowgirl at the 2003 draft meeting, after which he spent 15 seasons in the team, two times selected for the best lineup Entry, 11 times.
Ganobo and Breddy have worked together for 3 seasons, and they have gone in the 17 games, of which he pays the outside world in the 2016 season to pay attention to his strength. Although the New England Patriots Trainer Group has always focused on the details in the game, Garobo pointed out that Braddy told him the importance of these details.
However, it seems that Brradi still kills the quartet team to win the champion, and Garobo is injured and needs to prove that they can play all the potential. Original Garobo was considered to make breakthroughs last season, but the knee crucified torn tear made his 2018 season to reimburse.
After Jason announced, the location of the cowboy is once a unstable factor of the team attack, and the team has lost Jason and External Hands after DEZ BRYANT. No reliable alternative In the case of entering the training camp.
Hill did not clearly indicate when they want to get the opportunity to be a starter. So he may still stay in Brisbane Saints continue to serve on the bench. However, Hill has clearly expressed a desire to serve as the starting quarterback in the future. At present we do not know when he will be able to do so.
"This is not the coach tells you," he said. "It is the player to master these doorways in the game. This is some small things, all of which are some details, and how to integrate these details is the place that really stands out."
Brooks is a professional bowl of professional bowls in 2013. It has been playing at 49 people in 2009, and the basic full-service team has completed 287 cockroaches and 42 times four-point guard. But the current 4 games only completed a quarter-free killing.
Demales – Thomas feels yourself return youth
Beijing July 10th is 30 years old. It is a spree in the state of running. Although this is not so severe, it is like Demaryuis Thomas, which feels that he returns to youth. See.
US Time May 3, 2018, Jason announced his retirement and joined ESPN's "Monday Late Rugby" program as an explanation. US Time Thursday, he announced that he will re-enter the rivers and lakes, re-join Dallas cowboy and prepare for the 16th season of your career.
Defending champion Seattle Seahawks in the Super Bowl last week lost to rival Denver Broncos, they bounce back this week. Rely on a high level of team play, with 41:14 big win over the San Diego Chargers Seattle guest, ushered in the first win of the new season. Main quarterback Russell – Wilson feels good, his 13 pass 11, advancing 121 yards, he had two running touchdowns. Seahawks two more touchdowns, but also by the complete running back. Defending champion demonstrated his powerful ground attack capability.
"I certainly see themselves as the starting quarterback," Hill said in a recent interview. "I think when you consider the other question:?? Is the New Orleans Saints or the other team when you enter the free agent market, which is you begin to know how other people think of you when we become a free agent long enough it can not really let us know how other people look at me, we will face this problem on becoming a free agent. "
"I have been playing for a long time after 30 years old," Thomas said, "I feel that I can still continue to make a good performance for several years. I have said before, now I feel the best when I feel the whole career. The most restricted hip The injury feels no longer exists. "
In Drew headed star – the case of Brisbane (Drew Brees) absence, the Saints still rely on the team steady play won. After the renewal Jimmy – Graham once again demonstrated his ability, his five catches for success for the team to get two touchdowns, can be described as the first hero of the Saints win. In addition, the team's backup quarterback two are doing well. Griffin 19 pass 13, and sends 179 yards two touchdowns. McQuarrie outdone, cheap nfl Jerseys his 20 pass 12, fed 117 yards and two touchdowns. I believe under the presence of Brisbane, be sure to show both of them are satisfied.
Cowboys took off the outer handle Machael Gallup at last season, and traded Amari Cooper in the transaction deadline. However, in the near-terminal position, the team and a stable output point like Jason, Brake Jarwin, a total of 27th, a total of 27 codes, promoted 307 yards, gain 3 A ball reaches.
"I will continue to play rugby and competition in the heart," Jason said in the statement released through the cowboy official website. "We have a group of rising new stars, I want to help them compete for the championship. This is all my personal decisions, and I feel very good about this. I am very expected to return to the stadium. "According to Jason, Jason will receive $ 5 million from the cowboy.


All the team at UKMA would like to thank YOU the anglers who've support our websites. Over the past year the team has grown yet again, with over thirty team members now supplying fact based information, reviews and features on match, carp, specialist and predator angling, that hopefully assist in making your time spent on the bank as rewarding as possible. Our goal in 2014 is to continue adding as much information to you the angler as possible and keep to our policy of only publishing the facts based on individual and team findings. You are part of an ever growing worldwide angling audience, which has from its humble beginnings back in 2006, now grown into a free online media source for anglers, that now has the support over 40,000 visitors per month. Your continued support and input enables us to source the information you need, exactly when you need it!!
On behalf of all the team members at UKMA, we would like to say many thanks for your support over the past year, a Merry Christmas to all and a prosperous new year!!